## Construction of BT-polytopes via partial Stott-expansion

Discussion of known convex regular-faced polytopes, including the Johnson solids in 3D, and higher dimensions; and the discovery of new ones.

### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:Thus so far, the ones
Code: Select all
`BDAABDDAABCDABCDDAABCDD`
haven't been investigated. The ones that are CRF are:
Code: Select all
`ABABDDBCBCD`
and of course B, which is not new. Lacing edges are provided in this post, thus so far the following have been proven not to be CRF
Code: Select all
`AABD(ABC)`

Some further investigations gave: (BD) is CRF, same lacings as ABDD, (AABDD) is not, as 4 and 5 are completely isolated. (AABCD) has the same problem. (ABCDD) on the other hand has a connection between 4 and 5, but these are then not connected to the others. (AABCDD) then has 4 and 5 completely isolated again.

Indeed, for the B-ed ones we have:
Code: Select all
`(B)      = BAFox2oxofo3oooox5ooxoo&#zx = oxoofooxo3oooxoxooo5ooxoooxoo&#xt (CRF, done)(BD)     = BAFox2xoxFx3oxoox5ooxoo&#zx = xoxxFxxox3oxoxoxoxo5ooxoooxoo&#xt (CRF)(AAB)    -> non-unit edges within the "layers"(AABD)   = BAFox2xoxFo3oxooo5ooxof&#zx -> non-unit edges between the "layers"(AB)     = BAFox2oxofx3xxxxo5ooxof&#zx = oxoxfxoxo3xxxoxoxxx5ooxfofxoo&#xt (CRF)(BDD)    -> non-unit edges within the "layers"(ABDD)   = BAFox2ooofx3xoxxo5ofxof&#zx = oooxfxooo3xoxoxoxox5ofxfofxfo&#xt (CRF)(BC)     = BAFox2oxofo3oofox5xxoxx&#zx = oxoofooxo3oofxoxfoo5xxoxxxoxx&#xt (CRF)(ABD)    -> non-unit edges within the "layers"(AABDD)  = BAFox2xxxFo3xoxxx5ofxof&#zx -> non-unit edges between the "layers"(BCD)    = BAFox2xoxFx3oxfox5xxoxx&#zx = xoxxFxxox3oxfxoxfxo5xxoxxxoxx&#xt (CRF)(AABC)   -> non-unit edges within the "layers"(AABCD)  = BAFox2xoxFo3oxfoo5xxoxF&#zx -> non-unit edges between the "layers"(ABC)    = BAFox2oxofx3xxFxo5xxoxF&#zx -> non-unit edges between the "layers"(BCDD)   -> non-unit edges within the "layers"(ABCDD)  = BAFox2ooofx3xoFxo5xFoxF&#zx -> non-unit edges between the "layers"(ABCD)   -> non-unit edges within the "layers"(AABCDD) = BAFox2xxxFo3xoFxx5xFoxF&#zx -> non-unit edges between the "layers"`

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:
Code: Select all
`(B)      =  BAFox 2 oxofo 3 oooox 5 ooxoo &#zx  =  oxoofooxo 3 oooxoxooo 5 ooxoooxoo &#xt  (CRF, done)(BD)     =  BAFox 2 xoxFx 3 oxoox 5 ooxoo &#zx  =  xoxxFxxox 3 oxoxoxoxo 5 ooxoooxoo &#xt  (CRF)(AB)     =  BAFox 2 oxofx 3 xxxxo 5 ooxof &#zx  =  oxoxfxoxo 3 xxxoxoxxx 5 ooxfofxoo &#xt  (CRF)(ABDD)   =  BAFox 2 ooofx 3 xoxxo 5 ofxof &#zx  =  oooxfxooo 3 xoxoxoxox 5 ofxfofxfo &#xt  (CRF)(BC)     =  BAFox 2 oxofo 3 oofox 5 xxoxx &#zx  =  oxoofooxo 3 oofxoxfoo 5 xxoxxxoxx &#xt  (CRF)(BCD)    =  BAFox 2 xoxFx 3 oxfox 5 xxoxx &#zx  =  xoxxFxxox 3 oxfxoxfxo 5 xxoxxxoxx &#xt  (CRF)`

Here now comes the first interesting one of that subsymmetry list:
Code: Select all
`BAFox2xoxFx3oxoox5ooxoo&#zx (BD)1-2 = x1-3 = f1-4 = F1-5 = fq2-3 = x2-4 = fq2-5 = f3-4 = f3-5 = x4-5 = x=> just the linear "layer" sequence 123545321 remains as tower:xoxxFxxox3oxoxoxoxo5ooxoooxoo&#xt1-2 = ike || id, 2-3 = id || srid, 3-4 = srid || ti4-5-4 = 12-augm. tipeo........3o........5o........     & | 24  *   *   *  * |  5   5   0   0   0   0   0  0   0   0  0 |  5  5   5  0   0   0   0  0  0  0   0   0   0  0   0  0  0  0 | 1  5  1  0  0  0  0  0  0  0 verf = pip.o.......3.o.......5.o.......     & |  * 60   *   *  * |  0   2   4   4   0   0   0  0   0   0  0 |  0  1   4  2   2   4   2  0  0  0   0   0   0  0   0  0  0  0 | 0  2  2  2  1  0  0  0  0  0 verf = pip..o......3..o......5..o......     & |  *  * 120   *  * |  0   0   0   2   2   2   2  0   0   0  0 |  0  0   0  0   2   1   2  1  2  1   2   1   2  0   0  0  0  0 | 0  0  1  1  2  1  2  1  0  0...o.....3...o.....5...o.....     & |  *  *   * 120  * |  0   0   0   0   0   0   2  1   2   1  1 |  0  0   0  0   0   0   0  0  0  0   2   2   1  2   2  1  2  1 | 0  0  0  0  0  2  1  1  2  2....o....3....o....5....o....       |  *  *   *   * 12 |  0   0   0   0   0   0   0  0   0  10  0 |  0  0   0  0   0   0   0  0  0  0   0   0   0  0  10  0  0  5 | 0  0  0  0  0  0  0  2  0  5 verf = pip------------------------------------+------------------+------------------------------------------+---------------------------------------------------------------+-----------------------------x........ ......... .........     & |  2  0   0   0  0 | 60   *   *   *   *   *   *  *   *   *  * |  2  1   0  0   0   0   0  0  0  0   0   0   0  0   0  0  0  0 | 1  2  0  0  0  0  0  0  0  0oo.......3oo.......5oo.......&#x  & |  1  1   0   0  0 |  * 120   *   *   *   *   *  *   *   *  * |  0  1   2  0   0   0   0  0  0  0   0   0   0  0   0  0  0  0 | 0  2  1  0  0  0  0  0  0  0......... .x....... .........     & |  0  2   0   0  0 |  *   * 120   *   *   *   *  *   *   *  * |  0  0   1  1   0   1   0  0  0  0   0   0   0  0   0  0  0  0 | 0  1  1  1  0  0  0  0  0  0.oo......3.oo......5.oo......&#x  & |  0  1   1   0  0 |  *   *   * 240   *   *   *  *   *   *  * |  0  0   0  0   1   1   1  0  0  0   0   0   0  0   0  0  0  0 | 0  0  1  1  1  0  0  0  0  0..x...... ......... .........     & |  0  0   2   0  0 |  *   *   *   * 120   *   *  *   *   *  * |  0  0   0  0   1   0   0  1  1  0   1   0   0  0   0  0  0  0 | 0  0  0  1  1  1  1  0  0  0......... ......... ..x......     & |  0  0   2   0  0 |  *   *   *   *   * 120   *  *   *   *  * |  0  0   0  0   0   0   1  0  1  1   0   0   1  0   0  0  0  0 | 0  0  1  0  1  0  1  1  0  0..oo.....3..oo.....5..oo.....&#x  & |  0  0   1   1  0 |  *   *   *   *   *   * 240  *   *   *  * |  0  0   0  0   0   0   0  0  0  0   1   1   1  0   0  0  0  0 | 0  0  0  0  0  1  1  1  0  0...x..... ......... .........     & |  0  0   0   2  0 |  *   *   *   *   *   *   * 60   *   *  * |  0  0   0  0   0   0   0  0  0  0   2   0   0  2   0  1  0  0 | 0  0  0  0  0  2  1  0  2  0......... ...x..... .........     & |  0  0   0   2  0 |  *   *   *   *   *   *   *  * 120   *  * |  0  0   0  0   0   0   0  0  0  0   0   1   0  1   1  0  1  0 | 0  0  0  0  0  1  0  1  1  1...oo....3...oo....5...oo....&#x  & |  0  0   0   1  1 |  *   *   *   *   *   *   *  *   * 120  * |  0  0   0  0   0   0   0  0  0  0   0   0   0  0   2  0  0  1 | 0  0  0  0  0  0  0  1  0  2...o.o...3...o.o...5...o.o...&#x    |  0  0   0   2  0 |  *   *   *   *   *   *   *  *   *   * 60 |  0  0   0  0   0   0   0  0  0  0   0   0   0  0   0  1  2  1 | 0  0  0  0  0  0  0  0  2  2------------------------------------+------------------+------------------------------------------+---------------------------------------------------------------+-----------------------------x........3o........ .........     & |  3  0   0   0  0 |  3   0   0   0   0   0   0  0   0   0  0 | 40  *   *  *   *   *   *  *  *  *   *   *   *  *   *  *  *  * | 1  1  0  0  0  0  0  0  0  0xo....... ......... .........&#x  & |  2  1   0   0  0 |  1   2   0   0   0   0   0  0   0   0  0 |  * 60   *  *   *   *   *  *  *  *   *   *   *  *   *  *  *  * | 0  2  0  0  0  0  0  0  0  0......... ox....... .........&#x  & |  1  2   0   0  0 |  0   2   1   0   0   0   0  0   0   0  0 |  *  * 120  *   *   *   *  *  *  *   *   *   *  *   *  *  *  * | 0  1  1  0  0  0  0  0  0  0.o.......3.x....... .........     & |  0  3   0   0  0 |  0   0   3   0   0   0   0  0   0   0  0 |  *  *   * 40   *   *   *  *  *  *   *   *   *  *   *  *  *  * | 0  1  0  1  0  0  0  0  0  0.ox...... ......... .........&#x  & |  0  1   2   0  0 |  0   0   0   2   1   0   0  0   0   0  0 |  *  *   *  * 120   *   *  *  *  *   *   *   *  *   *  *  *  * | 0  0  0  1  1  0  0  0  0  0......... .xo...... .........&#x  & |  0  2   1   0  0 |  0   0   1   2   0   0   0  0   0   0  0 |  *  *   *  *   * 120   *  *  *  *   *   *   *  *   *  *  *  * | 0  0  1  1  0  0  0  0  0  0......... ......... .ox......&#x  & |  0  1   2   0  0 |  0   0   0   2   0   1   0  0   0   0  0 |  *  *   *  *   *   * 120  *  *  *   *   *   *  *   *  *  *  * | 0  0  1  0  1  0  0  0  0  0..x......3..o...... .........     & |  0  0   3   0  0 |  0   0   0   0   3   0   0  0   0   0  0 |  *  *   *  *   *   *   * 40  *  *   *   *   *  *   *  *  *  * | 0  0  0  1  0  1  0  0  0  0..x...... ......... ..x......     & |  0  0   4   0  0 |  0   0   0   0   2   2   0  0   0   0  0 |  *  *   *  *   *   *   *  * 60  *   *   *   *  *   *  *  *  * | 0  0  0  0  1  0  1  0  0  0......... ..o......5..x......     & |  0  0   5   0  0 |  0   0   0   0   0   5   0  0   0   0  0 |  *  *   *  *   *   *   *  *  * 24   *   *   *  *   *  *  *  * | 0  0  1  0  0  0  0  1  0  0..xx..... ......... .........&#x  & |  0  0   2   2  0 |  0   0   0   0   1   0   2  1   0   0  0 |  *  *   *  *   *   *   *  *  *  * 120   *   *  *   *  *  *  * | 0  0  0  0  0  1  1  0  0  0......... ..ox..... .........&#x  & |  0  0   1   2  0 |  0   0   0   0   0   0   2  0   1   0  0 |  *  *   *  *   *   *   *  *  *  *   * 120   *  *   *  *  *  * | 0  0  0  0  0  1  0  1  0  0......... ......... ..xo.....&#x  & |  0  0   2   1  0 |  0   0   0   0   0   1   2  0   0   0  0 |  *  *   *  *   *   *   *  *  *  *   *   * 120  *   *  *  *  * | 0  0  0  0  0  0  1  1  0  0...x.....3...x..... .........     & |  0  0   0   6  0 |  0   0   0   0   0   0   0  3   3   0  0 |  *  *   *  *   *   *   *  *  *  *   *   *   * 40   *  *  *  * | 0  0  0  0  0  1  0  0  1  0......... ...xo.... .........&#x  & |  0  0   0   2  1 |  0   0   0   0   0   0   0  0   1   2  0 |  *  *   *  *   *   *   *  *  *  *   *   *   *  * 120  *  *  * | 0  0  0  0  0  0  0  1  0  1...x.x... ......... .........&#x    |  0  0   0   4  0 |  0   0   0   0   0   0   0  2   0   0  2 |  *  *   *  *   *   *   *  *  *  *   *   *   *  *   * 30  *  * | 0  0  0  0  0  0  0  0  2  0......... ...x.x... .........&#x    |  0  0   0   4  0 |  0   0   0   0   0   0   0  0   2   0  2 |  *  *   *  *   *   *   *  *  *  *   *   *   *  *   *  * 60  * | 0  0  0  0  0  0  0  0  1  1...ooo...3...ooo...5...ooo...&#x    |  0  0   0   2  1 |  0   0   0   0   0   0   0  0   0   2  1 |  *  *   *  *   *   *   *  *  *  *   *   *   *  *   *  *  * 60 | 0  0  0  0  0  0  0  0  0  2------------------------------------+------------------+------------------------------------------+---------------------------------------------------------------+-----------------------------x........3o........5o........     & | 12  0   0   0  0 | 30   0   0   0   0   0   0  0   0   0  0 | 20  0   0  0   0   0   0  0  0  0   0   0   0  0   0  0  0  0 | 2  *  *  *  *  *  *  *  *  * ikexo.......3ox....... .........&#x  & |  3  3   0   0  0 |  3   6   3   0   0   0   0  0   0   0  0 |  1  3   3  1   0   0   0  0  0  0   0   0   0  0   0  0  0  0 | * 40  *  *  *  *  *  *  *  * oct......... oxo......5oox......&#x  & |  1  5   5   0  0 |  0   5   5  10   0   5   0  0   0   0  0 |  0  0   5  0   0   5   5  0  0  1   0   0   0  0   0  0  0  0 | *  * 24  *  *  *  *  *  *  * gyepip.ox......3.xo...... .........&#x  & |  0  3   3   0  0 |  0   0   3   6   3   0   0  0   0   0  0 |  0  0   0  1   3   3   0  1  0  0   0   0   0  0   0  0  0  0 | *  *  * 40  *  *  *  *  *  * oct.ox...... ......... .ox......&#x  & |  0  1   4   0  0 |  0   0   0   4   2   2   0  0   0   0  0 |  0  0   0  0   2   0   2  0  1  0   0   0   0  0   0  0  0  0 | *  *  *  * 60  *  *  *  *  * squippy..xx.....3..ox..... .........&#x  & |  0  0   3   6  0 |  0   0   0   0   3   0   6  3   3   0  0 |  0  0   0  0   0   0   0  1  0  0   3   3   0  1   0  0  0  0 | *  *  *  *  * 40  *  *  *  * tricu..xx..... ......... ..xo.....&#x  & |  0  0   4   2  0 |  0   0   0   0   2   2   4  1   0   0  0 |  0  0   0  0   0   0   0  0  1  0   2   0   2  0   0  0  0  0 | *  *  *  *  *  * 60  *  *  * trip......... ..oxo....5..xoo....&#x  & |  0  0   5   5  1 |  0   0   0   0   0   5  10  0   5   5  0 |  0  0   0  0   0   0   0  0  0  1   0   5   5  0   5  0  0  0 | *  *  *  *  *  *  * 24  *  * gyepip...x.x...3...x.x... .........&#x    |  0  0   0  12  0 |  0   0   0   0   0   0   0  6   6   0  6 |  0  0   0  0   0   0   0  0  0  0   0   0   0  2   0  3  3  0 | *  *  *  *  *  *  *  * 20  * hip......... ...xox... .........&#x    |  0  0   0   4  1 |  0   0   0   0   0   0   0  0   2   4  2 |  0  0   0  0   0   0   0  0  0  0   0   0   0  0   2  0  1  2 | *  *  *  *  *  *  *  *  * 60 squippy`

That is, the total cell list here is

24+24 = 48 gyepips (gyroelongated pentagonal pyramids, J11),
20 hips,
2 ikes,
40+40 = 80 octs,
60+60 = 120 squippies,
40 tricues,
60 trips.

This CRF polychoron is one of the simpler ones. Not only that it neither contains bilbiros, thawros nor pocuros. Instead it has 3-stratic polar caps which are known segmentochora stacks: first ike || id, then id || srid, and finally srid || ti. And the remaining equatorial bistratic layer then is nothing but a tipe (= prism of truncated ike), which is augmented at each of its pips by pip-pyramids.

Note furthermore that the peppies and the paps of the 2 outermost layers each here combine into one set of gyepips (J11), and the same subcells of the 3rd and 4th layer each into a second set of gyepips.

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:
Klitzing wrote:
Code: Select all
`(B)      =  BAFox 2 oxofo 3 oooox 5 ooxoo &#zx  =  oxoofooxo 3 oooxoxooo 5 ooxoooxoo &#xt  (CRF, done)(BD)     =  BAFox 2 xoxFx 3 oxoox 5 ooxoo &#zx  =  xoxxFxxox 3 oxoxoxoxo 5 ooxoooxoo &#xt  (CRF)(AB)     =  BAFox 2 oxofx 3 xxxxo 5 ooxof &#zx  =  oxoxfxoxo 3 xxxoxoxxx 5 ooxfofxoo &#xt  (CRF)(ABDD)   =  BAFox 2 ooofx 3 xoxxo 5 ofxof &#zx  =  oooxfxooo 3 xoxoxoxox 5 ofxfofxfo &#xt  (CRF)(BC)     =  BAFox 2 oxofo 3 oofox 5 xxoxx &#zx  =  oxoofooxo 3 oofxoxfoo 5 xxoxxxoxx &#xt  (CRF)(BCD)    =  BAFox 2 xoxFx 3 oxfox 5 xxoxx &#zx  =  xoxxFxxox 3 oxfxoxfxo 5 xxoxxxoxx &#xt  (CRF)`

Here now comes the first interesting one of that subsymmetry list:

Let's continue with the next, i.e. (AB):
Code: Select all
`BAFox2oxofx3xxxxo5ooxof&#zx (AB)1-2 = x1-3 = f1-4 = F1-5 = fq2-3 = x2-4 = sqrt(A)=1.9021132-5 = f3-4 = f3-5 = x4-5 = x=> = tower 123545321:oxoxfxoxo3xxxoxoxxx5ooxfofxoo&#xt1-2 = id || ti,  2-3 = ti || tido........3o........5o........     & | 60   *   *   *  * |   4   2  0   0   0   0  0   0   0   0  0  0 |  2  2  1   4  0  0   0   0   0  0   0   0   0  0   0  0   0  0 | 1  2  2  0  0  0  0  0  0  0.o.......3.o.......5.o.......     & |  * 120   *   *  * |   0   1  1   2   2   0  0   0   0   0  0  0 |  0  0  1   2  2  1   2   2   1  0   0   0   0  0   0  0   0  0 | 0  2  1  2  1  1  0  0  0  0..o......3..o......5..o......     & |  *   * 120   *  * |   0   0  0   0   2   2  1   2   0   0  0  0 |  0  0  0   0  0  0   1   2   2  1   1   2   2  0   0  0   0  0 | 0  0  0  1  1  2  1  1  0  0...o.....3...o.....5...o.....     & |  *   *   * 120  * |   0   0  0   0   0   0  0   2   2   2  1  0 |  0  0  0   0  0  0   0   0   0  0   2   1   2  1   1  2   2  0 | 0  0  0  0  0  1  1  2  1  1....o....3....o....5....o....       |  *   *   *   * 60 |   0   0  0   0   0   0  0   0   0   4  0  2 |  0  0  0   0  0  0   0   0   0  0   0   0   2  0   4  0   2  1 | 0  0  0  0  0  2  0  1  0  2------------------------------------+-------------------+---------------------------------------------+----------------------------------------------------------------+-----------------------------......... x........ .........     & |  2   0   0   0  0 | 120   *  *   *   *   *  *   *   *   *  *  * |  1  1  0   1  0  0   0   0   0  0   0   0   0  0   0  0   0  0 | 1  1  1  0  0  0  0  0  0  0oo.......3oo.......5oo.......&#x  & |  1   1   0   0  0 |   * 120  *   *   *   *  *   *   *   *  *  * |  0  0  1   2  0  0   0   0   0  0   0   0   0  0   0  0   0  0 | 0  2  1  0  0  0  0  0  0  0.x....... ......... .........     & |  0   2   0   0  0 |   *   * 60   *   *   *  *   *   *   *  *  * |  0  0  1   0  2  0   2   0   0  0   0   0   0  0   0  0   0  0 | 0  2  0  2  1  0  0  0  0  0......... .x....... .........     & |  0   2   0   0  0 |   *   *  * 120   *   *  *   *   *   *  *  * |  0  0  0   1  1  1   0   1   0  0   0   0   0  0   0  0   0  0 | 0  1  1  1  0  1  0  0  0  0.oo......3.oo......5.oo......&#x  & |  0   1   1   0  0 |   *   *  *   * 240   *  *   *   *   *  *  * |  0  0  0   0  0  0   1   1   1  0   0   0   0  0   0  0   0  0 | 0  0  0  1  1  1  0  0  0  0......... ..x...... .........     & |  0   0   2   0  0 |   *   *  *   *   * 120  *   *   *   *  *  * |  0  0  0   0  0  0   0   1   0  1   0   1   0  0   0  0   0  0 | 0  0  0  1  0  1  1  0  0  0......... ......... ..x......     & |  0   0   2   0  0 |   *   *  *   *   *   * 60   *   *   *  *  * |  0  0  0   0  0  0   0   0   2  0   0   0   2  0   0  0   0  0 | 0  0  0  0  1  2  0  1  0  0..oo.....3..oo.....5..oo.....&#x  & |  0   0   1   1  0 |   *   *  *   *   *   *  * 240   *   *  *  * |  0  0  0   0  0  0   0   0   0  0   1   1   1  0   0  0   0  0 | 0  0  0  0  0  1  1  1  0  0...x..... ......... .........     & |  0   0   0   2  0 |   *   *  *   *   *   *  *   * 120   *  *  * |  0  0  0   0  0  0   0   0   0  0   1   0   0  1   0  1   0  0 | 0  0  0  0  0  0  1  1  1  0...oo....3...oo....5...oo....&#x  & |  0   0   0   1  1 |   *   *  *   *   *   *  *   *   * 240  *  * |  0  0  0   0  0  0   0   0   0  0   0   0   1  0   1  0   1  0 | 0  0  0  0  0  1  0  1  0  1...o.o...3...o.o...5...o.o...&#x    |  0   0   0   2  0 |   *   *  *   *   *   *  *   *   *   * 60  * |  0  0  0   0  0  0   0   0   0  0   0   0   0  0   0  2   2  0 | 0  0  0  0  0  0  0  2  1  1......... ....x.... .........       |  0   0   0   0  2 |   *   *  *   *   *   *  *   *   *   *  * 60 |  0  0  0   0  0  0   0   0   0  0   0   0   0  0   2  0   0  1 | 0  0  0  0  0  2  0  0  0  1------------------------------------+-------------------+---------------------------------------------+----------------------------------------------------------------+-----------------------------o........3x........ .........     & |  3   0   0   0  0 |   3   0  0   0   0   0  0   0   0   0  0  0 | 40  *  *   *  *  *   *   *   *  *   *   *   *  *   *  *   *  * | 1  1  0  0  0  0  0  0  0  0......... x........5o........     & |  5   0   0   0  0 |   5   0  0   0   0   0  0   0   0   0  0  0 |  * 24  *   *  *  *   *   *   *  *   *   *   *  *   *  *   *  * | 1  0  1  0  0  0  0  0  0  0ox....... ......... .........&#x  & |  1   2   0   0  0 |   0   2  1   0   0   0  0   0   0   0  0  0 |  *  * 60   *  *  *   *   *   *  *   *   *   *  *   *  *   *  * | 0  2  0  0  0  0  0  0  0  0......... xx....... .........&#x  & |  2   2   0   0  0 |   1   2  0   1   0   0  0   0   0   0  0  0 |  *  *  * 120  *  *   *   *   *  *   *   *   *  *   *  *   *  * | 0  1  1  0  0  0  0  0  0  0.x.......3.x....... .........     & |  0   6   0   0  0 |   0   0  3   3   0   0  0   0   0   0  0  0 |  *  *  *   * 40  *   *   *   *  *   *   *   *  *   *  *   *  * | 0  1  0  1  0  0  0  0  0  0......... .x.......5.o.......     & |  0   5   0   0  0 |   0   0  0   5   0   0  0   0   0   0  0  0 |  *  *  *   *  * 24   *   *   *  *   *   *   *  *   *  *   *  * | 0  0  1  0  0  1  0  0  0  0.xo...... ......... .........&#x  & |  0   2   1   0  0 |   0   0  1   0   2   0  0   0   0   0  0  0 |  *  *  *   *  *  * 120   *   *  *   *   *   *  *   *  *   *  * | 0  0  0  1  1  0  0  0  0  0......... .xx...... .........&#x  & |  0   2   2   0  0 |   0   0  0   1   2   1  0   0   0   0  0  0 |  *  *  *   *  *  *   * 120   *  *   *   *   *  *   *  *   *  * | 0  0  0  1  0  1  0  0  0  0......... ......... .ox......&#x  & |  0   1   2   0  0 |   0   0  0   0   2   0  1   0   0   0  0  0 |  *  *  *   *  *  *   *   * 120  *   *   *   *  *   *  *   *  * | 0  0  0  0  1  1  0  0  0  0..o......3..x...... .........     & |  0   0   3   0  0 |   0   0  0   0   0   3  0   0   0   0  0  0 |  *  *  *   *  *  *   *   *   * 40   *   *   *  *   *  *   *  * | 0  0  0  1  0  0  1  0  0  0..ox..... ......... .........&#x  & |  0   0   1   2  0 |   0   0  0   0   0   0  0   2   1   0  0  0 |  *  *  *   *  *  *   *   *   *  * 120   *   *  *   *  *   *  * | 0  0  0  0  0  0  1  1  0  0......... ..xo..... .........&#x  & |  0   0   2   1  0 |   0   0  0   0   0   1  0   2   0   0  0  0 |  *  *  *   *  *  *   *   *   *  *   * 120   *  *   *  *   *  * | 0  0  0  0  0  1  1  0  0  0......... ......... ..xfo....&#xt & |  0   0   2   2  1 |   0   0  0   0   0   0  1   2   0   2  0  0 |  *  *  *   *  *  *   *   *   *  *   *   * 120  *   *  *   *  * | 0  0  0  0  0  1  0  1  0  0...x.....3...o..... .........     & |  0   0   0   3  0 |   0   0  0   0   0   0  0   0   3   0  0  0 |  *  *  *   *  *  *   *   *   *  *   *   *   * 40   *  *   *  * | 0  0  0  0  0  0  1  0  1  0......... ...ox.... .........&#x  & |  0   0   0   1  2 |   0   0  0   0   0   0  0   0   0   2  0  1 |  *  *  *   *  *  *   *   *   *  *   *   *   *  * 120  *   *  * | 0  0  0  0  0  1  0  0  0  1...x.x... ......... .........&#x    |  0   0   0   4  0 |   0   0  0   0   0   0  0   0   2   0  2  0 |  *  *  *   *  *  *   *   *   *  *   *   *   *  *   * 60   *  * | 0  0  0  0  0  0  0  1  1  0...ooo...3...ooo...5...ooo...&#x    |  0   0   0   2  1 |   0   0  0   0   0   0  0   0   0   2  1  0 |  *  *  *   *  *  *   *   *   *  *   *   *   *  *   *  * 120  * | 0  0  0  0  0  0  0  1  0  1......... ....x....5....o....       |  0   0   0   0  5 |   0   0  0   0   0   0  0   0   0   0  0  5 |  *  *  *   *  *  *   *   *   *  *   *   *   *  *   *  *   * 12 | 0  0  0  0  0  2  0  0  0  0------------------------------------+-------------------+---------------------------------------------+----------------------------------------------------------------+-----------------------------o........3x........5o........     & | 30   0   0   0  0 |  60   0  0   0   0   0  0   0   0   0  0  0 | 20 12  0   0  0  0   0   0   0  0   0   0   0  0   0  0   0  0 | 2  *  *  *  *  *  *  *  *  * idox.......3xx....... .........&#x  & |  3   6   0   0  0 |   3   6  3   3   0   0  0   0   0   0  0  0 |  1  0  3   3  1  0   0   0   0  0   0   0   0  0   0  0   0  0 | * 40  *  *  *  *  *  *  *  * tricu......... xx.......5oo.......&#x  & |  5   5   0   0  0 |   5   5  0   5   0   0  0   0   0   0  0  0 |  0  1  0   5  0  1   0   0   0  0   0   0   0  0   0  0   0  0 | *  * 24  *  *  *  *  *  *  * pip.xo......3.xx...... .........&#x  & |  0   6   3   0  0 |   0   0  3   3   6   3  0   0   0   0  0  0 |  0  0  0   0  1  0   3   3   0  1   0   0   0  0   0  0   0  0 | *  *  * 40  *  *  *  *  *  * tricu.xo...... ......... .ox......&#x  & |  0   2   2   0  0 |   0   0  1   0   4   0  1   0   0   0  0  0 |  0  0  0   0  0  0   2   0   2  0   0   0   0  0   0  0   0  0 | *  *  *  * 60  *  *  *  *  * tet......... .xxox....5.oxfo....&#xt & |  0   5  10   5  5 |   0   0  0   5  10   5  5  10   0  10  0  5 |  0  0  0   0  0  1   0   5   5  0   0   5   5  0   5  0   0  1 | *  *  *  *  * 24  *  *  *  * pocuro..ox.....3..xo..... .........&#x  & |  0   0   3   3  0 |   0   0  0   0   0   3  0   6   3   0  0  0 |  0  0  0   0  0  0   0   0   0  1   3   3   0  1   0  0   0  0 | *  *  *  *  *  * 40  *  *  * oct..oxfxo.. ......... ..xfofx..&#xt   |  0   0   4   8  2 |   0   0  0   0   0   0  2   8   4   8  4  0 |  0  0  0   0  0  0   0   0   0  0   4   0   4  0   0  2   4  0 | *  *  *  *  *  *  * 30  *  * bilbiro...x.x...3...o.o... .........&#x    |  0   0   0   6  0 |   0   0  0   0   0   0  0   0   6   0  3  0 |  0  0  0   0  0  0   0   0   0  0   0   0   0  2   0  3   0  0 | *  *  *  *  *  *  *  * 20  * trip......... ...oxo... .........&#x    |  0   0   0   2  2 |   0   0  0   0   0   0  0   0   0   4  1  1 |  0  0  0   0  0  0   0   0   0  0   0   0   0  0   2  0   2  0 | *  *  *  *  *  *  *  *  * 60 tet`

That is, the total cell count here is:

30 bilbiros (J91),
2 ids,
40 octs,
24 pips,
24 pocuros (J32),
60+60 = 120 tets,
40+40 = 80 tricues, and
20 trips.

This one now is the next interesting case. We already had only pocuros (ADD), as well as pocuros with thawros (CD).
(AB) now combines pocuros with bilbiros.

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:
Klitzing wrote:
Klitzing wrote:
Code: Select all
`(B)      =  BAFox 2 oxofo 3 oooox 5 ooxoo &#zx  =  oxoofooxo 3 oooxoxooo 5 ooxoooxoo &#xt  (CRF, done)(BD)     =  BAFox 2 xoxFx 3 oxoox 5 ooxoo &#zx  =  xoxxFxxox 3 oxoxoxoxo 5 ooxoooxoo &#xt  (CRF)(AB)     =  BAFox 2 oxofx 3 xxxxo 5 ooxof &#zx  =  oxoxfxoxo 3 xxxoxoxxx 5 ooxfofxoo &#xt  (CRF)(ABDD)   =  BAFox 2 ooofx 3 xoxxo 5 ofxof &#zx  =  oooxfxooo 3 xoxoxoxox 5 ofxfofxfo &#xt  (CRF)(BC)     =  BAFox 2 oxofo 3 oofox 5 xxoxx &#zx  =  oxoofooxo 3 oofxoxfoo 5 xxoxxxoxx &#xt  (CRF)(BCD)    =  BAFox 2 xoxFx 3 oxfox 5 xxoxx &#zx  =  xoxxFxxox 3 oxfxoxfxo 5 xxoxxxoxx &#xt  (CRF)`

Here now comes the first interesting one of that subsymmetry list: ...

Let's continue with the next, i.e. (AB): ...

You'r still tuned for the other ones? - Okay here now comes (ABDD):
Code: Select all
`BAFox2ooofx3xoxxo5ofxof&#zx (ABDD)1-2 = x1-3 = f1-4 = F1-5 = fq2-3 = x2-4 = fq2-5 = f3-4 = f3-5 = x4-5 = x=> = tower 123545321:oooxfxooo3xoxoxoxox5ofxfofxfo&#xt1-3 = bistratic id-cap of rahi, 3-5 = layers 6,9b,11 of roxo........3o........5o........     & | 60  *   *   *  * |   4   2   0   0  0   0   0   0  0  0 |  2  2   4  1   0  0  0   0   0   0  0   0  0   0  0 | 1  2  2  0  0  0  0  0  0.o.......3.o.......5.o.......     & |  * 40   *   *  * |   0   3   3   0  0   0   0   0  0  0 |  0  0   3  3   3  0  0   0   0   0  0   0  0   0  0 | 0  1  3  1  0  0  0  0  0..o......3..o......5..o......     & |  *  * 120   *  * |   0   0   1   2  1   2   0   0  0  0 |  0  0   0  1   2  1  2   1   2   2  0   0  0   0  0 | 0  0  2  1  1  1  2  0  0...o.....3...o.....5...o.....     & |  *  *   * 120  * |   0   0   0   0  0   2   2   2  1  0 |  0  0   0  0   0  0  0   2   1   2  1   1  2   2  0 | 0  0  0  0  1  2  1  1  1....o....3....o....5....o....       |  *  *   *   * 60 |   0   0   0   0  0   0   0   4  0  2 |  0  0   0  0   0  0  0   0   0   2  0   4  0   2  1 | 0  0  0  0  0  1  2  0  2------------------------------------+------------------+--------------------------------------+-----------------------------------------------------+--------------------------......... x........ .........     & |  2  0   0   0  0 | 120   *   *   *  *   *   *   *  *  * |  1  1   1  0   0  0  0   0   0   0  0   0  0   0  0 | 1  1  1  0  0  0  0  0  0oo.......3oo.......5oo.......&#x  & |  1  1   0   0  0 |   * 120   *   *  *   *   *   *  *  * |  0  0   2  1   0  0  0   0   0   0  0   0  0   0  0 | 0  1  2  0  0  0  0  0  0.oo......3.oo......5.oo......&#x  & |  0  1   1   0  0 |   *   * 120   *  *   *   *   *  *  * |  0  0   0  1   2  0  0   0   0   0  0   0  0   0  0 | 0  0  2  1  0  0  0  0  0......... ..x...... .........     & |  0  0   2   0  0 |   *   *   * 120  *   *   *   *  *  * |  0  0   0  0   1  1  1   0   1   0  0   0  0   0  0 | 0  0  1  1  1  0  1  0  0......... ......... ..x......     & |  0  0   2   0  0 |   *   *   *   * 60   *   *   *  *  * |  0  0   0  1   0  0  2   0   0   2  0   0  0   0  0 | 0  0  2  0  0  1  2  0  0..oo.....3..oo.....5..oo.....&#x  & |  0  0   1   1  0 |   *   *   *   *  * 240   *   *  *  * |  0  0   0  0   0  0  0   1   1   1  0   0  0   0  0 | 0  0  0  0  1  1  1  0  0...x..... ......... .........     & |  0  0   0   2  0 |   *   *   *   *  *   * 120   *  *  * |  0  0   0  0   0  0  0   1   0   0  1   0  1   0  0 | 0  0  0  0  1  1  0  1  0...oo....3...oo....5...oo....&#x  & |  0  0   0   1  1 |   *   *   *   *  *   *   * 240  *  * |  0  0   0  0   0  0  0   0   0   1  0   1  0   1  0 | 0  0  0  0  0  1  1  0  1...o.o...3...o.o...5...o.o...&#x    |  0  0   0   2  0 |   *   *   *   *  *   *   *   * 60  * |  0  0   0  0   0  0  0   0   0   0  0   0  2   2  0 | 0  0  0  0  0  2  0  1  1......... ....x.... .........       |  0  0   0   0  2 |   *   *   *   *  *   *   *   *  * 60 |  0  0   0  0   0  0  0   0   0   0  0   2  0   0  1 | 0  0  0  0  0  0  2  0  1------------------------------------+------------------+--------------------------------------+-----------------------------------------------------+--------------------------o........3x........ .........     & |  3  0   0   0  0 |   3   0   0   0  0   0   0   0  0  0 | 40  *   *  *   *  *  *   *   *   *  *   *  *   *  * | 1  1  0  0  0  0  0  0  0......... x........5o........     & |  5  0   0   0  0 |   5   0   0   0  0   0   0   0  0  0 |  * 24   *  *   *  *  *   *   *   *  *   *  *   *  * | 1  0  1  0  0  0  0  0  0......... xo....... .........&#x  & |  2  1   0   0  0 |   1   2   0   0  0   0   0   0  0  0 |  *  * 120  *   *  *  *   *   *   *  *   *  *   *  * | 0  1  1  0  0  0  0  0  0......... ......... ofx......&#xt & |  1  2   2   0  0 |   0   2   2   0  1   0   0   0  0  0 |  *  *   * 60   *  *  *   *   *   *  *   *  *   *  * | 0  0  2  0  0  0  0  0  0......... .ox...... .........&#x  & |  0  1   2   0  0 |   0   0   2   1  0   0   0   0  0  0 |  *  *   *  * 120  *  *   *   *   *  *   *  *   *  * | 0  0  1  1  0  0  0  0  0..o......3..x...... .........     & |  0  0   3   0  0 |   0   0   0   3  0   0   0   0  0  0 |  *  *   *  *   * 40  *   *   *   *  *   *  *   *  * | 0  0  0  1  1  0  0  0  0......... ..x......5..x......     & |  0  0  10   0  0 |   0   0   0   5  5   0   0   0  0  0 |  *  *   *  *   *  * 24   *   *   *  *   *  *   *  * | 0  0  1  0  0  0  1  0  0..ox..... ......... .........&#x  & |  0  0   1   2  0 |   0   0   0   0  0   2   1   0  0  0 |  *  *   *  *   *  *  * 120   *   *  *   *  *   *  * | 0  0  0  0  1  1  0  0  0......... ..xo..... .........&#x  & |  0  0   2   1  0 |   0   0   0   1  0   2   0   0  0  0 |  *  *   *  *   *  *  *   * 120   *  *   *  *   *  * | 0  0  0  0  1  0  1  0  0......... ......... ..xfo....&#xt & |  0  0   2   2  1 |   0   0   0   0  1   2   0   2  0  0 |  *  *   *  *   *  *  *   *   * 120  *   *  *   *  * | 0  0  0  0  0  1  1  0  0...x.....3...o..... .........     & |  0  0   0   3  0 |   0   0   0   0  0   0   3   0  0  0 |  *  *   *  *   *  *  *   *   *   * 40   *  *   *  * | 0  0  0  0  1  0  0  1  0......... ...ox.... .........&#x  & |  0  0   0   1  2 |   0   0   0   0  0   0   0   2  0  1 |  *  *   *  *   *  *  *   *   *   *  * 120  *   *  * | 0  0  0  0  0  0  1  0  1...x.x... ......... .........&#x    |  0  0   0   4  0 |   0   0   0   0  0   0   2   0  2  0 |  *  *   *  *   *  *  *   *   *   *  *   * 60   *  * | 0  0  0  0  0  1  0  1  0...ooo...3...ooo...5...ooo...&#x    |  0  0   0   2  1 |   0   0   0   0  0   0   0   2  1  0 |  *  *   *  *   *  *  *   *   *   *  *   *  * 120  * | 0  0  0  0  0  1  0  0  1......... ....x....5....o....       |  0  0   0   0  5 |   0   0   0   0  0   0   0   0  0  5 |  *  *   *  *   *  *  *   *   *   *  *   *  *   * 12 | 0  0  0  0  0  0  2  0  0------------------------------------+------------------+--------------------------------------+-----------------------------------------------------+--------------------------o........3x........5o........     & | 30  0   0   0  0 |  60   0   0   0  0   0   0   0  0  0 | 20 12   0  0   0  0  0   0   0   0  0   0  0   0  0 | 2  *  *  *  *  *  *  *  * idoo.......3xo....... .........&#x  & |  3  1   0   0  0 |   3   3   0   0  0   0   0   0  0  0 |  1  0   3  0   0  0  0   0   0   0  0   0  0   0  0 | * 40  *  *  *  *  *  *  * tet......... xox......5ofx......&#xt & |  5  5  10   0  0 |   5  10  10   5  5   0   0   0  0  0 |  0  1   5  5   5  0  1   0   0   0  0   0  0   0  0 | *  * 24  *  *  *  *  *  * pero.oo......3.ox...... .........&#x  & |  0  1   3   0  0 |   0   0   3   3  0   0   0   0  0  0 |  0  0   0  0   3  1  0   0   0   0  0   0  0   0  0 | *  *  * 40  *  *  *  *  * tet..ox.....3..xo..... .........&#x  & |  0  0   3   3  0 |   0   0   0   3  0   6   3   0  0  0 |  0  0   0  0   0  1  0   3   3   0  1   0  0   0  0 | *  *  *  * 40  *  *  *  * oct..oxfxo.. ......... ..xfofx..&#xt   |  0  0   4   8  2 |   0   0   0   0  2   8   4   8  4  0 |  0  0   0  0   0  0  0   4   0   4  0   0  2   4  0 | *  *  *  *  * 30  *  *  * bilbiro......... ..xox....5..xfo....&#xt & |  0  0  10   5  5 |   0   0   0   5  5  10   0  10  0  5 |  0  0   0  0   0  0  1   0   5   5  0   5  0   0  1 | *  *  *  *  *  * 24  *  * pero...x.x...3...o.o... .........&#x    |  0  0   0   6  0 |   0   0   0   0  0   0   6   0  3  0 |  0  0   0  0   0  0  0   0   0   0  2   0  3   0  0 | *  *  *  *  *  *  * 20  * trip......... ...oxo... .........&#x    |  0  0   0   2  2 |   0   0   0   0  0   0   0   4  1  1 |  0  0   0  0   0  0  0   0   0   0  0   2  0   2  0 | *  *  *  *  *  *  *  * 60 tet`

That is, the total cell count here is:

30 bilbiros,
2 ids,
40 octs,
24+24 = 48 peros,
40+40+60 = 140 tets,
20 trips.

--- rk[/quote]
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:
Klitzing wrote:
Klitzing wrote:
Klitzing wrote:
Code: Select all
`(B)      =  BAFox 2 oxofo 3 oooox 5 ooxoo &#zx  =  oxoofooxo 3 oooxoxooo 5 ooxoooxoo &#xt  (CRF, done)(BD)     =  BAFox 2 xoxFx 3 oxoox 5 ooxoo &#zx  =  xoxxFxxox 3 oxoxoxoxo 5 ooxoooxoo &#xt  (CRF)(AB)     =  BAFox 2 oxofx 3 xxxxo 5 ooxof &#zx  =  oxoxfxoxo 3 xxxoxoxxx 5 ooxfofxoo &#xt  (CRF)(ABDD)   =  BAFox 2 ooofx 3 xoxxo 5 ofxof &#zx  =  oooxfxooo 3 xoxoxoxox 5 ofxfofxfo &#xt  (CRF)(BC)     =  BAFox 2 oxofo 3 oofox 5 xxoxx &#zx  =  oxoofooxo 3 oofxoxfoo 5 xxoxxxoxx &#xt  (CRF)(BCD)    =  BAFox 2 xoxFx 3 oxfox 5 xxoxx &#zx  =  xoxxFxxox 3 oxfxoxfxo 5 xxoxxxoxx &#xt  (CRF)`

Here now comes the first interesting one of that subsymmetry list: ...

Let's continue with the next, i.e. (AB): ...

You'r still tuned for the other ones? - Okay here now comes (ABDD): ...

Next then is (BC):
Code: Select all
`BAFox2oxofo3oofox5xxoxx&#zx (BC)1-2 = x1-3 = f1-4 = F1-5 = fq2-3 = x2-4 = sqrt(A)2-5 = f3-4 = f3-5 = x4-5 = x=> = tower 123545321:oxoofooxo3oofxoxfoo5xxoxxxoxx&#xt1-2 = doe || srid, 4-5-4 = 12-augm. tiddipo........3o........5o........     & | 40   *  *   *  * |  3   3   0   0   0   0   0  0   0  0  0 |  3   3   6  0  0  0   0   0   0  0  0   0   0  0  0   0  0 | 1  1  3  3  0  0  0  0  0  0.o.......3.o.......5.o.......     & |  * 120  *   *  * |  0   1   2   2   2   0   0  0   0  0  0 |  0   2   2  1  2  1   2   1   2  0  0   0   0  0  0   0  0 | 0  1  2  1  1  2  1  0  0  0..o......3..o......5..o......     & |  *   * 60   *  * |  0   0   0   0   4   2   0  0   0  0  0 |  0   0   0  0  0  0   2   4   2  1  0   0   0  0  0   0  0 | 0  0  0  0  2  1  2  0  0  0...o.....3...o.....5...o.....     & |  *   *  * 120  * |  0   0   0   0   0   1   2  1   2  1  0 |  0   0   0  0  0  0   0   2   0  1  1   2   2  2  1   2  0 | 0  0  0  0  1  0  2  1  2  2....o....3....o....5....o....       |  *   *  *   * 60 |  0   0   0   0   0   0   0  0   4  0  2 |  0   0   0  0  0  0   0   0   0  0  0   2   4  0  0   2  1 | 0  0  0  0  0  0  2  0  1  2------------------------------------+------------------+-----------------------------------------+------------------------------------------------------------+-----------------------------......... ......... ........x     & |  2   0  0   0  0 | 60   *   *   *   *   *   *  *   *  *  * |  2   0   2  0  0  0   0   0   0  0  0   0   0  0  0   0  0 | 1  0  1  2  0  0  0  0  0  0oo.......3oo.......5oo.......&#x  & |  1   1  0   0  0 |  * 120   *   *   *   *   *  *   *  *  * |  0   2   2  0  0  0   0   0   0  0  0   0   0  0  0   0  0 | 0  1  2  1  0  0  0  0  0  0.x....... ......... .........     & |  0   2  0   0  0 |  *   * 120   *   *   *   *  *   *  *  * |  0   1   0  1  1  0   1   0   0  0  0   0   0  0  0   0  0 | 0  1  1  0  1  1  0  0  0  0......... ......... .x.......     & |  0   2  0   0  0 |  *   *   * 120   *   *   *  *   *  *  * |  0   0   1  0  1  1   0   0   1  0  0   0   0  0  0   0  0 | 0  0  1  1  0  1  1  0  0  0.oo......3.oo......5.oo......&#x  & |  0   1  1   0  0 |  *   *   *   * 240   *   *  *   *  *  * |  0   0   0  0  0  0   1   1   1  0  0   0   0  0  0   0  0 | 0  0  0  0  1  1  1  0  0  0..oo.....3..oo.....5..oo.....&#x  & |  0   0  1   1  0 |  *   *   *   *   * 120   *  *   *  *  * |  0   0   0  0  0  0   0   2   0  1  0   0   0  0  0   0  0 | 0  0  0  0  1  0  2  0  0  0......... ...x..... .........     & |  0   0  0   2  0 |  *   *   *   *   *   * 120  *   *  *  * |  0   0   0  0  0  0   0   1   0  0  1   1   0  1  0   0  0 | 0  0  0  0  1  0  1  1  1  0......... ......... ...x.....     & |  0   0  0   2  0 |  *   *   *   *   *   *   * 60   *  *  * |  0   0   0  0  0  0   0   0   0  1  0   0   2  0  1   0  0 | 0  0  0  0  0  0  2  0  0  2...oo....3...oo....5...oo....&#x  & |  0   0  0   1  1 |  *   *   *   *   *   *   *  * 240  *  * |  0   0   0  0  0  0   0   0   0  0  0   1   1  0  0   1  0 | 0  0  0  0  0  0  1  0  1  1...o.o...3...o.o...5...o.o...&#x    |  0   0  0   2  0 |  *   *   *   *   *   *   *  *   * 60  * |  0   0   0  0  0  0   0   0   0  0  0   0   0  2  1   2  0 | 0  0  0  0  0  0  0  1  2  2......... ......... ....x....       |  0   0  0   0  2 |  *   *   *   *   *   *   *  *   *  * 60 |  0   0   0  0  0  0   0   0   0  0  0   0   2  0  0   0  1 | 0  0  0  0  0  0  2  0  0  1------------------------------------+------------------+-----------------------------------------+------------------------------------------------------------+-----------------------------......... o........5x........     & |  5   0  0   0  0 |  5   0   0   0   0   0   0  0   0  0  0 | 24   *   *  *  *  *   *   *   *  *  *   *   *  *  *   *  * | 1  0  0  1  0  0  0  0  0  0ox....... ......... .........&#x  & |  1   2  0   0  0 |  0   2   1   0   0   0   0  0   0  0  0 |  * 120   *  *  *  *   *   *   *  *  *   *   *  *  *   *  * | 0  1  1  0  0  0  0  0  0  0......... ......... xx.......&#x  & |  2   2  0   0  0 |  1   2   0   1   0   0   0  0   0  0  0 |  *   * 120  *  *  *   *   *   *  *  *   *   *  *  *   *  * | 0  0  1  1  0  0  0  0  0  0.x.......3.o....... .........     & |  0   3  0   0  0 |  0   0   3   0   0   0   0  0   0  0  0 |  *   *   * 40  *  *   *   *   *  *  *   *   *  *  *   *  * | 0  1  0  0  1  0  0  0  0  0.x....... ......... .x.......     & |  0   4  0   0  0 |  0   0   2   2   0   0   0  0   0  0  0 |  *   *   *  * 60  *   *   *   *  *  *   *   *  *  *   *  * | 0  0  1  0  0  1  0  0  0  0......... .o.......5.x.......     & |  0   5  0   0  0 |  0   0   0   5   0   0   0  0   0  0  0 |  *   *   *  *  * 24   *   *   *  *  *   *   *  *  *   *  * | 0  0  0  1  0  0  1  0  0  0.xo...... ......... .........&#x  & |  0   2  1   0  0 |  0   0   1   0   2   0   0  0   0  0  0 |  *   *   *  *  *  * 120   *   *  *  *   *   *  *  *   *  * | 0  0  0  0  1  1  0  0  0  0......... .ofx..... .........&#x  & |  0   1  2   2  0 |  0   0   0   0   2   2   1  0   0  0  0 |  *   *   *  *  *  *   * 120   *  *  *   *   *  *  *   *  * | 0  0  0  0  1  0  1  0  0  0......... ......... .xo......&#x  & |  0   2  1   0  0 |  0   0   0   1   2   0   0  0   0  0  0 |  *   *   *  *  *  *   *   * 120  *  *   *   *  *  *   *  * | 0  0  0  0  0  1  1  0  0  0......... ......... ..ox.....&#x  & |  0   0  1   2  0 |  0   0   0   0   0   2   0  1   0  0  0 |  *   *   *  *  *  *   *   *   * 60  *   *   *  *  *   *  * | 0  0  0  0  0  0  2  0  0  0...o.....3...x..... .........&#x  & |  0   0  0   3  0 |  0   0   0   0   0   0   3  0   0  0  0 |  *   *   *  *  *  *   *   *   *  * 40   *   *  *  *   *  * | 0  0  0  0  1  0  0  1  0  0......... ...xo.... .........&#x  & |  0   0  0   2  1 |  0   0   0   0   0   0   1  0   2  0  0 |  *   *   *  *  *  *   *   *   *  *  * 120   *  *  *   *  * | 0  0  0  0  0  0  1  0  1  0......... ......... ...xx....&#x  & |  0   0  0   2  2 |  0   0   0   0   0   0   0  1   2  0  1 |  *   *   *  *  *  *   *   *   *  *  *   * 120  *  *   *  * | 0  0  0  0  0  0  1  0  0  1......... ...x.x... .........&#x    |  0   0  0   4  0 |  0   0   0   0   0   0   2  0   0  2  0 |  *   *   *  *  *  *   *   *   *  *  *   *   * 60  *   *  * | 0  0  0  0  0  0  0  1  1  0......... ......... ...x.x...&#x    |  0   0  0   4  0 |  0   0   0   0   0   0   0  2   0  2  0 |  *   *   *  *  *  *   *   *   *  *  *   *   *  * 30   *  * | 0  0  0  0  0  0  0  0  0  2...ooo...3...ooo...5...ooo...&#x    |  0   0  0   2  1 |  0   0   0   0   0   0   0  0   2  1  0 |  *   *   *  *  *  *   *   *   *  *  *   *   *  *  * 120  * | 0  0  0  0  0  0  0  0  1  1......... ....o....5....x....       |  0   0  0   0  5 |  0   0   0   0   0   0   0  0   0  0  5 |  *   *   *  *  *  *   *   *   *  *  *   *   *  *  *   * 12 | 0  0  0  0  0  0  2  0  0  0------------------------------------+------------------+-----------------------------------------+------------------------------------------------------------+-----------------------------o........3o........5x........     & | 20   0  0   0  0 | 30   0   0   0   0   0   0  0   0  0  0 | 12   0   0  0  0  0   0   0   0  0  0   0   0  0  0   0  0 | 2  *  *  *  *  *  *  *  *  * doeox.......3oo....... .........&#x  & |  1   3  0   0  0 |  0   3   3   0   0   0   0  0   0  0  0 |  0   3   0  1  0  0   0   0   0  0  0   0   0  0  0   0  0 | * 40  *  *  *  *  *  *  *  * tetox....... ......... xx.......&#x  & |  2   4  0   0  0 |  1   4   2   2   0   0   0  0   0  0  0 |  0   2   2  0  1  0   0   0   0  0  0   0   0  0  0   0  0 | *  * 60  *  *  *  *  *  *  * trip......... oo.......5xx.......&#x  & |  5   5  0   0  0 |  5   5   0   5   0   0   0  0   0  0  0 |  1   0   5  0  0  1   0   0   0  0  0   0   0  0  0   0  0 | *  *  * 24  *  *  *  *  *  * pip.xoo.....3.ofx..... .........&#xt & |  0   3  3   3  0 |  0   0   3   0   6   3   3  0   0  0  0 |  0   0   0  1  0  0   3   3   0  0  1   0   0  0  0   0  0 | *  *  *  * 40  *  *  *  *  * teddi.xo...... ......... .xo......&#x  & |  0   4  1   0  0 |  0   0   2   2   4   0   0  0   0  0  0 |  0   0   0  0  1  0   2   0   2  0  0   0   0  0  0   0  0 | *  *  *  *  * 60  *  *  *  * squippy......... .ofxo....5.xoxx....&#xt & |  0   5  5  10  5 |  0   0   0   5  10  10   5  5  10  0  5 |  0   0   0  0  0  1   0   5   5  5  0   5   5  0  0   0  1 | *  *  *  *  *  * 24  *  *  * pocuro...o.o...3...x.x... .........&#x    |  0   0  0   6  0 |  0   0   0   0   0   0   6  0   0  3  0 |  0   0   0  0  0  0   0   0   0  0  2   0   0  3  0   0  0 | *  *  *  *  *  *  * 20  *  * trip......... ...xox... .........&#x    |  0   0  0   4  1 |  0   0   0   0   0   0   2  0   4  2  0 |  0   0   0  0  0  0   0   0   0  0  0   2   0  1  0   2  0 | *  *  *  *  *  *  *  * 60  * squippy......... ......... ...xxx...&#x    |  0   0  0   4  2 |  0   0   0   0   0   0   0  2   4  2  1 |  0   0   0  0  0  0   0   0   0  0  0   0   2  0  1   2  0 | *  *  *  *  *  *  *  *  * 60 trip`

That is, the total cell count here is:

2 does,
24 pips,
24 pocuros,
60+60 = 120 squippies,
40 teddies,
40 tets,
60+20+60 = 140 trips.

--- rk
Klitzing
Pentonian

Posts: 1377
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Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:You'r still tuned for the other ones?
I am always, you are doing a great job figuring out the structure of all these polytopes , and I wish I could have helped, but I'm leaving on a vacation for three weeks now, and thus won't be able to actively participate to this forum (though I guess I will be able to read a post sometime).
The reason I haven't helped that much is because the deadline for my school-project is getting really close, and I have spent quite some time of the last three weeks on that. Furthermore any incmat I tried to make had at least four errors where ai,j*ai,i =/= aj,i*aj,j, which made me annoyed. Anyway, I tried to make the incmat for the double axial expansion, but so far I got stuck at calculating the lacing edges. I have calculated them, but things 7 and 8 gave me loads of weird numbers and some ?'s. I haven't been able to spot the error in these layers (there has to be one), so I will just post what I had so far:
Code: Select all
`x2o2A2F | o2x2F2A | V2o2x2x | o2V2x2x | F2o2F2o | o2F2o2F | o2f2o2A | f2o2A2o | F2x2x2F | x2F2F2x | x2f2x2A | f2x2A2x | o2o2B2x | o2o2x2B | F2f2o2x | f2F2x2o | f2f2F2F   1    |    2    |    3    |    4    |    5    |    6    |    7    |    8    |    9    |    10   |    11   |    12   |    13   |    14   |    15   |    16   |    17  A=F+x; B=V+x=2f+x; C=sqrt(f+u)=longest diagonal ike; D=q*f; n=noncomformal number; ?=number<1; E=fsqrt(3); G=fsqrt(3/2)  | 1 |2 | x | 2 |3 | C | D | 3 |4 | D | C | D | 4 |5 | f | D | x | D | 5 |6 | D | f | D | x | F | 6 |7 | n | n | D | f | n | ? | 7 |8 | n | n | f | D | ? | n | E | 8 |9 | f | f | x | C | f | f | n | G | 9 |10| f | f | C | x | f | f | G | n | f | 10|11| f | x | C | f | D | x | ? | n | x | f | 11|12| x | f | f | C | x | D | n | ? | f | x | C | 12|13| x | f | D | D | f | F | F | x | D | f | D | x | 13|14| f | x | D | D | F | f | x | F | f | D | x | D | D | 14|15| D | D | x | f | f | f | n | n | x | f | f | C | F | D | 15|16| D | D | f | x | f | f | n | n | f | x | C | f | D | F | x | 16|17| x | x | f | f | f | f | f | f | x | x | x | x | f | f | f | f | 17|`

As you can see, there are quite some x-edges, so this is probably a CRF, and thus we can list it's cell count. (personally I dont think you need the whole incmat for this, but just the cells and the faces. this would provide enough info and is particulary usefull for things like this which would otherwise get a really enormous incmats) I would like to do this when I'm back from my vacation. I am really sorry I won't be able to help you the next three weeks.

student91
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian

Posts: 317
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### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:
Klitzing wrote:
Klitzing wrote:
Klitzing wrote:
Klitzing wrote:
Code: Select all
`(B)      =  BAFox 2 oxofo 3 oooox 5 ooxoo &#zx  =  oxoofooxo 3 oooxoxooo 5 ooxoooxoo &#xt  (CRF, done)(BD)     =  BAFox 2 xoxFx 3 oxoox 5 ooxoo &#zx  =  xoxxFxxox 3 oxoxoxoxo 5 ooxoooxoo &#xt  (CRF)(AB)     =  BAFox 2 oxofx 3 xxxxo 5 ooxof &#zx  =  oxoxfxoxo 3 xxxoxoxxx 5 ooxfofxoo &#xt  (CRF)(ABDD)   =  BAFox 2 ooofx 3 xoxxo 5 ofxof &#zx  =  oooxfxooo 3 xoxoxoxox 5 ofxfofxfo &#xt  (CRF)(BC)     =  BAFox 2 oxofo 3 oofox 5 xxoxx &#zx  =  oxoofooxo 3 oofxoxfoo 5 xxoxxxoxx &#xt  (CRF)(BCD)    =  BAFox 2 xoxFx 3 oxfox 5 xxoxx &#zx  =  xoxxFxxox 3 oxfxoxfxo 5 xxoxxxoxx &#xt  (CRF)`

Here now comes the first interesting one of that subsymmetry list: ...

Let's continue with the next, i.e. (AB): ...

You'r still tuned for the other ones? - Okay here now comes (ABDD): ...

Next then is (BC): ...

Now to the final one of those "extended kaleido-facetings" according to the [2,3,5] subsymmetry of ex.

We have been going a two step research. The first step has been to investigate for these "quirks" (kaleido-facetings), introducing some edge inversions (x -> (-x)), resulting in no longer convex, subsymmetrically faceted figures. That step then has to be checked: whether (already in this un-expanded state) the required inter-layer vertex distances still show up unit distances, so that each layer can be connected to at least one of the others, and that this set of layers does not fall appart into subsets of unit connectability. - The final case (BCD) clearly passed that check.

Next we were applying the expansion. This then was designed such, that our figure of consideration becomes convex again. But here as well we will have to do a check. This time the inner-layer edges have to be investigated. Any non-unit edge of any layer shall not lie at the boundary. But it clearly is allowed as a mere vertex distance of a chord, i.e. when burried within the body of the figure. - So far we were lucky to spot such distances being u and immediately could drop that candidate. (BCD) did manage to pass that u-length check. But finally it turned out, that an f-length inner-layer edge here still would remain at its boundary. This hypersolid asks for several x-x-x-f trapeziums. In fact, you could spot 2 types of pocuro-like cell types on either hemiglome, one running down from the top polar base, the other running up from the equator. Both do connect the said elements by its cupolaic part. But the distance would be too short for twice a pocuro height. Thus those do intersect in their rotundaic part. The lateral pentagons thereby too get diminished. This is how those said trapeziums come in. - Therefore (BCD) now gets known to fail to be a CRF, after all.

None the less, we have been lucky to find within this [2,3,5] subsymmetry in total 11 true CRFs, all obtained as expanded kaleido-facetings of ex!

--- rk
Klitzing
Pentonian

Posts: 1377
Joined: Sun Aug 19, 2012 11:16 am
Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:Just for compilation:

So far we have the following systematical EKFs (extended kaleido-facetings), which evaluate to be CRFs:

A - of the icosahedron (ike, x3o5o)
[...]

B - of the hexacosachoron (ex, 600-cell, x3o3o5o)
[...]
where generally I used F=ff=f+x, V=F+v=2f, A=F+x=f+2x, B=V+x=2f+x
(Other combinatorical cases each either provide non-unit inter-layer or surviving non-unit inner-layer edges.)
For all those above ones incidence matrices could also be provided, if needed. (But I'm in the run to provide these too in my next homepage update, for sure. There even will be an extra subpage for those. - Stay tuned.)

B.5 - wrt. o3o2o3o
B.6 - wrt. o2o2o2o
B.7 - wrt. o2o3o3o
remain to be evaluated systematically - as far as I know.

Other individuals (or even systematicals) known so far?

--- rk
You have missed o5o2o5o-symmetry. I guess we then have all fourdimensional (reflectional) subsymmetries of ex. beside these we could also expand according to lower dimensional symmetries. I have already thought about these a bit. conclusion is that you can also expand along tet-symmetry, trip-symmetry, pip-symmetry.
Because ex has only odd numbers in its Schläfli-symbol (contrary to e.g. ico), and it doesn't have separated parts (like cartesian products have) it has only one kind of mirror. (a general cube has two kinds of mirrors, one kind placed along axial symmetry, and one kind along demicubic. ex on the other hand has only one kind of mirrors.) This means there is only one kind of every subdimensional sbsymmetry of ex (so there is only one .3.-subsymmetry, one .5.-subsymmetry and one .2.-subsymmetry, but also only one .2.3.-subsymmetry, one .2.5.-sybsymmetry and so on.) This is not true for things that have even numbers in the schlafli-symbol. (ico for example has two kinds of .3.-symmetry). This is because odd numbers change the mirrors. if you have e.g. the 3 mirrors of .3.-symmetry, a mirror that looks like the [.]3.-part on one side, is behaving as a .3[.]-mirror on the other side. This is really important to be able to know we have investigated all subdimensional symmetries. Furthermore this is important because it ables you to shortcut around quite some subdimensional smmetries: When you have a symmetry, e.g. o3o2o3o, and you expand it as x3o2o3o, this is the same as expansion along o3o-symmetry giving x3o. because we are already going to investigate o3o2o3o, o5o2o5o and o2o2o2o-symmetry, we already expand along all 2d-subsymmetries, and thus we only have to look for 3d-subsymmetries to get all subdimensional subsymmetries. This gives tet, trip and pip (and axial) as only conclusions

Now the expansion along these subsymmetries, they seem to be a little bit tricky. This post displays a mistake regarding this subject:
Klitzing wrote:... and for sure:

C - of the icosahedral pyramid (ikepy, ox3oo5oo&#x)
C.1 - wrt. o2o2o2o
C.1.1 - line || bilbiro (with 1 bilbiro, 4 peppies, 4 squippies, 4 tets, 2 trips)
C.2 - wrt. o2o2o3o
C.2.1 - {3} || thawro (with 1 oct, 3 peppies, 3 squippies, 9 tets, 1 thawro, 1 tricu, 3 trips)
C.3 - wrt. o2o2o5o
C.3.1 - {5} || pocuro (with 5 peppies, 2 pips, 1 pocuro, 10 squippies, 5 tets, 5 trips)

Btw., the latter 2 symmetries likewise will be open at case B.

--- rk
esp.
Btw., the latter 2 symmetries likewise will be open at case B.
Ex (and ikepy) doesn't have o2o2o3o-symmetry, nor o2o2o5o-symmetry (nor o2o3o3o). it rather has just o2o3o and o2o5o-subsymmetries (and ikepy only has o5o and o3o-symmetries) . (just as ike doesn't have o2o3o-symmetry, but only o3o-symmetry). Using this symmetry, you have to use the lace-tower tool &#xt or ||. (the symmetry of ex is some kind of antiprismatic, and of ikepy just pyramidial. you tried to describe them using prismatic symmetry, which they don't have this way). So in summary we have the followinng possible subsymmetries of ex:

4D:
o3o3o5o *
o2o3o5o *
o3o3o*b3o *
o3o3o3o *
o3o2o3o
o5o2o5o
o2o2o2o

3D:
o5o3o *+
o3o3o
o2o3o
o2o5o
o2o2o +

2D:
o5o +
o3o +
o2o +

1D:
o *+

+: will be contained in a higher dimensional investigation
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:4D:
[...]

3D:
[...]

2D:
[...]

1D:
[...]

+: will be contained in a higher dimensional investigation

This gives o3o2o3o, o5o2o5o, o2o2o2o, o3o3o, o2o3o and o2o5o to be investigated. Of these, only the 3-dimensionals don't have investigated possible expansions. These will be provided in this post.

o2o5o and o2o3o will be investigated simoultaneously, as they have a similar structure. they have the following possible expansions:
(+x)2o3/5o
o2(+x)3/5o
(+x)2(+x)3/5o
o2o3/5(+x)
(+x)2o3/5(+x)
o2(+x)3/5(+x)
(+x)2(+x)3/5(+x)
of these, (+x)2o3/5o, o2(+x)3/5o, o2o3/5(+x) and o2(+x)3/5(+x) will be contained in 4d-investigations. Furthermore the expansions (+x)2o3/5(+x) and (+x)2(+x)3/5o are similar because of the antiprismatic nature of these symmetries in ex. This thus gives us the expansions:
(+x)2(+x)3/5o
(+x)2(+x)3/5(+x)
which is conveniently few.

ex' representation in the given symmetries:
o2o5o: x2o5o||o2o5x||f2x5o||F2o5o||o2f5o||x2o5f||F2o5x||f2f5o||(o2x5x+V2o5o)||f2o5f||F2x5o||x2f5o||o2o5f||F2o5o||f2o5x||o2x5o||x2o5o

o2o3o: o2o3x||f2o3o||x2f3o||f2o3f||o2f3x||F2x3o||o2F3o||f2x3f||x2o3F||F2o3f||(o2f3f+V2o3o)||F2f3o||x2F3o||f2f3x||o2o3F||F2o3x||o2f3x||f2f3o||x2o3f||f2o3o||o2x3o

the first node must be quirked, as it must always be expanded. then o2o5o-symmetry only allows single quirks using only (-x)'s, because it doesn't have a 3. Therefore o2o5o has only two possible expansions. o2o3o has some more:
possible quirks:
A: o2(-x)3x (A1: o2x3(-x) )
AA: o2(-x)3o (AA1: o2o3(-x) )
B: o2(-x)3F (B1: o2F3(-x) )
C: F2(-x)3x (C1: F2x3(-x) )
CC: F2(-x)3o (CC1: F2o3(-x) )
D: f2(-x)3F (D1: f2F3(-x) )
Default: B,D and all things that start with x will be quirked in every expansion.
This thus gives us the following expansions for o2o3o:

(+x)2(+x)3o:
(A C + Default) : x2x3x || F2x3o || o2F3o || F2x3f || x2F3x || A2o3x || x2A3o || F2o3F || o2x3F || A2x3f || (x2F3f+B2x3o) || A2F3o || o2A3o || F2F3o || x2x3F || A2x3x || x2F3x || F2F3o || o2x3f || F2x3o || x2o3x
(A AA C + Default) : x2o3o || F2x3o || o2F3o || F2x3f || x2F3x || A2o3x || x2A3o || F2o3F || o2x3F || A2x3f || (x2F3f+B2x3o) || A2F3o || o2A3o || F2F3o || x2x3F || A2x3x || x2F3x || F2F3o || o2x3f || F2x3o || x2o3x
(A C CC + Default): x2x3x || F2x3o || o2F3o || F2x3f || x2F3x || A2o3x || x2A3o || F2o3F || o2x3F || A2x3f || (x2F3f+B2x3o) || A2F3o || o2A3o || F2F3o || x2x3F || A2o3o || x2F3x || F2F3o || o2x3f || F2x3o || x2o3x
(A AA C CC + Default): x2o3o || F2x3o || o2F3o || F2x3f || x2F3x || A2o3x || x2A3o || F2o3F || o2x3F || A2x3f || (x2F3f+B2x3o) || A2F3o || o2A3o || F2F3o || x2x3F || A2o3o || x2F3x || F2F3o || o2x3f || F2x3o || x2o3x

(+x)2(+x)3(+x):
(AA AA1 CC CC1 + Default): x2o3x || F2x3x || o2F3x || F2x3F || x2A3o || A2x3o || x2A3x || F2o3A || o2x3A || A2x3F || (x2F3F+B2x3x) || A2F3x || o2A3x || F2F3x || x2x3A || A2o3x || x2A3o || F2F3x || o2x3F || F2x3x || x2x3o
Others here give u.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: Construction of BT-polytopes via partial Stott-expansion

o3o3o-symmetry:
Code: Select all
`x3o3o || o3o3f || o3f3o || f3o3x || o3x3f || x3f3o || F3o3o || f3o3f || o3o3F || o3f3x || f3x3o || x3o3f || o3f3o || f3o3o || o3o3x  A   ||       ||       ||   B   ||   C   ||   D   ||       ||       ||       ||   D1  ||   C1  ||   B1  ||       ||       ||   A1`

quirks:
A: (-x)3x3o (A1: o3x3(-x) )
AA: o3(-x)3x (AA1: x3(-x)3o )
AAA: o3o3(-x) (AAA1: (-x)3o3o )
B: f3x3(-x) (B1: (-x)3x3f )
BB: F3(-x)3o (BB1: o3(-x)3F )
C: x3(-x)3F (C1: F3(-x)3x )
CC: (-x)3o3F (CC1: F3o3(-x) )
D: (-x)3F3o (D1: o3F3(-x) )

possible expansions:
(+x)3o3o
o3(+x)3o
(+x)3(+x)3o
o3o3(+x)
(+x)3o3(+x)
o3(+x)3(+x)
(+x)3(+x)3(+x)
of these, (+x)3o3o and o3o3(+x) resp. (+x)3(+x)3o and o3(+x)3(+x) are the same. Finally we then get:

(+x)3o3o:
(A B1 D)
(A AAA1 B1 D)
(A CC B1 D)
(A AAA1 CC B1 D)

o3(+x)3o:
(C C1)
(AA C C1) = (AA1 C C1)
(BB C C1) = (BB1 C C1)
(AA BB C C1) = (AA1 BB1 C C1)
(AA BB1 C C1) = (AA1 BB C C1)
(AA AA1 C C1)
(BB BB1 C C1)
(AA AA1 BB C C1) = (AA AA1 BB1 C C1)
(AA BB BB1 C C1) = (AA1 BB BB1 C C1)
(AA AA1 BB BB1 C C1)

(+x)3(+x)3o:
(AA BB1 CC C1 D)
(AA AA1 BB1 CC C1 D)
(AA AAA1 BB1 CC C1 D)
(AA BB BB1 CC C1 D)
(AA AA1 BB BB1 CC C1 D)
(AA AAA1 BB BB1 CC C1 D)

(+x)3o3(+x):
(A A1 B B1 D D1)
(AAA A1 B B1 D D1) = (A AAA1 B B1 D D1)
(AAA AAA1 B B1 D D1)
(A A1 B B1 CC D D1) = (A A1 B B1 CC1 D D1)
(AAA A1 B B1 CC D D1) = (A AAA1 B B1 CC1 D D1)
(A AAA1 B B1 CC D D1) = (AAA1 A B B1 CC1 D D1)
(AAA AAA1 B B1 CC D D1) = (AAA AAA1 B B1 CC1 D D1)
(A A1 B B1 CC CC1 D D1)
(AAA A1 B B1 CC CC1 D D1) = (A AAA1 B B1 CC CC1 D D1)
(AAA AAA1 B B1 CC CC1 D D1)

(+x)3(+x)3(+x):
(AAA AAA1 BB BB1 CC CC1 D D1)
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:
Klitzing wrote:Just for compilation:

So far we have the following systematical EKFs (extended kaleido-facetings), which evaluate to be CRFs:

A - of the icosahedron (ike, x3o5o)
[...]

B - of the hexacosachoron (ex, 600-cell, x3o3o5o)
[...]
where generally I used F=ff=f+x, V=F+v=2f, A=F+x=f+2x, B=V+x=2f+x
(Other combinatorical cases each either provide non-unit inter-layer or surviving non-unit inner-layer edges.)
For all those above ones incidence matrices could also be provided, if needed. (But I'm in the run to provide these too in my next homepage update, for sure. There even will be an extra subpage for those. - Stay tuned.)

B.5 - wrt. o3o2o3o
B.6 - wrt. o2o2o2o
B.7 - wrt. o2o3o3o
remain to be evaluated systematically - as far as I know.

Other individuals (or even systematicals) known so far?

--- rk
You have missed o5o2o5o-symmetry. ...

Not at all, cf.:
Klitzing wrote:B.4 - wrt. o5o2o5o => ex = xfooxo 5 xofxoo 2 oxofox 5 ooxofx &#zx
B.4.1 - oFxxox 5 Fofxfo 2 oxofox 5 ooxofx &#zx (with 10 gyepips, 25 ikes, 10 pips, 150 squippies, 75 tets, 50 trips)
B.4.2 - oFxFox 5 AxFoFx 2 oxofox 5 ooxofx &#zx (with 50 bilbiroes, 10 dips, 10 gyepips, 5 pips, 75 squippies)

student91 wrote:... I guess we then have all fourdimensional (reflectional) subsymmetries of ex. beside these we could also expand according to lower dimensional symmetries. I have already thought about these a bit. conclusion is that you can also expand along tet-symmetry, trip-symmetry, pip-symmetry. ...

tetrahedral I (intentionally) mentioned as B.7 (TBD), and onto the others I added a comment in my subsequent post, i.e. C.2 and C.3:
Klitzing wrote:Btw., the latter 2 symmetries likewise will be open at case B.

You're right with your comment on the wrong leading o2- prefixes for sure.
--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:o3o3o-symmetry: ...

(+x)3o3o:
(A B1 D)
(A AAA1 B1 D)
(A CC B1 D)
(A AAA1 CC B1 D)

Either I got lost in 0's and 1's, i.e. quite very, very huge incidence matrices, or its true that none of these will be CRF. As far as I got, it seems that already the partial stacks (A D ...) and (A CC D ...) look like that. - Please cross-check!

o3(+x)3o:
(C C1)
...

That one then not only is top-down anti-symmetrical, it even is CRF!
The total cell count of   xoofxxFfooFxofo 3 xxFxoFxxxFoxFxx 3 ofoxFoofFxxfoox &#xt   comes out to be:

16  octs
24  ikes
72  squippies
198 tets
24  tricues
48  trips
2   tuts

One striking configuration in here is that of the 6 equatorial tets (featuring digonal antiprisms), which are completely surrounded by ikes each. - Btw., could you spot those ikes in the above lace tower? - Probably not, as those only occur off-symmetrically. In fact, they appear as   .oofxx.foo..... 3 ............... 3 ............... &#xt .

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

I'll do that anytime soon when I have time. I guess though that it will be after the 12th of september, but it may be sooner.

o3(+x)3o:
(C C1)
...

That one then not only is top-down anti-symmetrical, it even is CRF!
The total cell count of   xoofxxFfooFxofo 3 xxFxoFxxxFoxFxx 3 ofoxFoofFxxfoox &#xt   comes out to be:

16  octs
24  ikes
72  squippies
198 tets
24  tricues
48  trips
2   tuts

One striking configuration in here is that of the 6 equatorial tets (featuring digonal antiprisms), which are completely surrounded by ikes each. - Btw., could you spot those ikes in the above lace tower? - Probably not, as those only occur off-symmetrically. In fact, they appear as   .oofxx.foo..... 3 ............... 3 ............... &#xt .

--- rk

Rather interesting indeed!! I would really like to see all these things rendered

Anyway, in the meantime I thought looking at the .2.5.-expansions. The result whas quite inconvenient as well
First I made the distance-matrix for ex in this symmetry. the distance-matrix gives the lacing distance between any two layers.
Code: Select all
`x2o5o||o2o5x||f2x5o||F2o5o||o2f5o||x2o5f||F2o5x||f2f5o||(o2x5x+V2o5o)||f2o5f||F2x5o||x2f5o||o2o5f||F2o5o||f2o5x||o2x5o||x2o5o  1  ||  2  ||  3  ||  4  ||  5  ||  6  ||  7  ||  8  ||  9  ||  10  ||  11 ||  12 ||  13 ||  14 ||  15 ||  16 ||  17 ||  18Values:x=1; f=phi; C=sqrt(A)=sqrt(f+u); qf=phi*sqrt(2); F=f+x=phi²; hf=phi*sqrt(3); fC=f*sqrt(A); V=2fA=F+x=f+2x; B=V+x=2f+xHeights:1-2=3-4=5-6=7-8=0.162459848 =height of peppypy2-3=6-7=8-(9;10)=0.262865556 =f*height of peppypy4-5=0.100405708 =v*height of peppypy1 | 1 2 | x | 23 | x | x | 34 | x | f | x | 45 | f | x | x | C | 56 | f | f | x | C | x | 67 | f | f | x | x | f | x | 78 | C | f | x | f | x | x | x | 89 |qf | f | f |qf | x | x | f | x | 9*10| C |qf | f | x |qf | C | x | f |qf | 10*11|qf | C | f | C | f | x | x | x | x | f |1112|qf |qf | f | f | C | f | x | f | f | x | x |1213| F |qf | C |qf | f | f | f | x | x | C | x | x |1314|hf |qf |qf | F | C | f | C | f | x |qf | x | f | x |1415| F |hf |qf | C | F |qf | f | C |qf | x | f | x | C | C |1516|hf | F |qf |qf |qf | C | f | f | f | f | x | x | x | x | x |1617|fC |hf | F |hf |qf |qf |qf | C | f |qf | f | f | f | x | f | x |1718|fC |fC |hf | F |hf | F |qf |qf |qf | C | C | f | f | f | x | x | x |181 | 1 2 | x | 23 | x | x | 34 | x | f | x | 45 | f | x | x |   | 56 | f | f | x |   | x | 67 | f | f | x | x | f | x | 78 |   | f | x | f | x | x | x | 89 |   | f | f |   | x | x | f | x | 9*10|   |   | f | x |   |   | x | f |   | 10*11|   |   | f |   | f | x | x | x | x | f |1112|   |   | f | f |   | f | x | f | f | x | x |1213|   |   |   |   | f | f | f | x | x |   | x | x |1314|   |   |   |   |   | f |   | f | x |   | x | f | x |1415|   |   |   |   |   |   | f |   |   | x | f | x |   |   |1516|   |   |   |   |   |   | f | f | f | f | x | x | x | x | x |1617|   |   |   |   |   |   |   |   | f |   | f | f | f | x | f | x |1718|   |   |   |   |   |   |   |   |   |   |   | f | f | f | x | x | x |181 | 1 2 | x | 23 | x | x | 34 | x |   | x | 45 |   | x | x |   | 56 |   |   | x |   | x | 67 |   |   | x | x |   | x | 78 |   |   | x |   | x | x | x | 89 |   |   |   |   | x | x |   | x | 9*10|   |   |   | x |   |   | x |   |   | 10*11|   |   |   |   |   | x | x | x | x |   |1112|   |   |   |   |   |   | x |   |   | x | x |1213|   |   |   |   |   |   |   | x | x |   | x | x |1314|   |   |   |   |   |   |   |   | x |   | x |   | x |1415|   |   |   |   |   |   |   |   |   | x |   | x |   |   |1516|   |   |   |   |   |   |   |   |   |   | x | x | x | x | x |1617|   |   |   |   |   |   |   |   |   |   |   |   |   | x |   | x |1718|   |   |   |   |   |   |   |   |   |   |   |   |   |   | x | x | x |18`
This distance-matrix (esp. the latter two) could be used to derive incidence matrices. However, in these really big polytopes, I prefer just this matrix and then some graph-drawings and stuff etc. (you can even deduce numbers this way, by dividing the order of the full symmetry (20) by the order of the subsymmetry used (so o2.5x||x2.5o gives 20/4=5 tets) guess you already knew this, but anyway)
Now I derived the distance-matrix for the first expansion:
Code: Select all
`x2o5o||o2o5x||f2x5o||F2o5o||o2f5o||x2o5f||F2o5x||f2f5o||(o2x5x+V2o5o)||f2o5f||F2x5o||x2f5o||o2o5f||F2o5o||f2o5x||o2x5o||x2o5o(-x)2o5o||o2f5(-x)||f2(-x)5f||F2o5o||o2f5o||(-x)2o5f||F2o5x||f2f5o||(o2(-x)5F+V2o5o)||f2o5o||F2(-x)5f||(-x)2f5o||o2o5f||F2o5o||f2o5x||o2(-x)5f||(-x)2o5o(+x)2(+x)5oo2x5o||x2x5x||F2o5f||A2x5o||x2F5o||o2x5f||A2x5x||F2F5o||(x2o5F+B2x5o)||F2x5f||A2o5f||o2F5o||x2x5f||A2x5o||F2x5x||x2o5f||o2x5o  1  ||  2  ||  3  ||  4  ||  5  ||  6  ||  7  ||  8  ||  9  ||  10  ||  11 ||  12 ||  13 ||  14 ||  15 ||  16 ||  17 ||  18Heights:1-2=3-4=5-6=7-8=0.162459848 =height of peppypy2-3=6-7=8-(9;10)=0.262865556 =f*height of peppypy4-5=0.100405708 =v*height of peppypy1 | 1 2 | x | 23 | f | x | 34 | C | f | x | 45 | f | x | f | C | 56 | f | f | f |qf | x | 67 |qf | f | x | x | f | C | 78 |qf | f | f | f | x | f | x | 89 |qf | f | f |qf | f | x | f | f | 9*10| F |qf | f | x |qf | F | x | f |qf | 10*11| F | C | f | C | f | f | x | x | x | f |1112|hf |qf | f | f |qf |qf | x | f | f | x | x |1213| F |qf | F |hf | f | f |qf | f | f | F | f |qf |1314|hf |qf |qf | F | C | f | C | f | x |qf | x | f | x |1415|fC |hf |qf | C | F |hf | f | C |qf | x | f | x |qf | C |1516|fC | F |qf |qf |qf |qf | f | f | f | f | x | x | f | x | x |1617|fC | F | F |hf | F |qf |qf |qf | f |qf | f | f | f | x | f | x |1718|fC |fC |fC |fC |hf | F |hf | F |qf | F |qf |qf | f | f | C | f | x |181 | 1 2 | x | 23 | f | x | 34 |   | f | x | 45 | f | x | f |   | 56 | f | f | f |   | x | 67 |   | f | x | x | f |   | 78 |   | f | f | f | x | f | x | 89 |   | f | f |   | f | x | f | f | 9*10|   |   | f | x |   |   | x | f |   | 10*11|   |   | f |   | f | f | x | x | x | f |1112|   |   | f | f |   |   | x | f | f | x | x |1213|   |   |   |   | f | f |   | f | f |   | f |   |1314|   |   |   |   |   | f |   | f | x |   | x | f | x |1415|   |   |   |   |   |   | f |   |   | x | f | x |   |   |1516|   |   |   |   |   |   | f | f | f | f | x | x | f | x | x |1617|   |   |   |   |   |   |   |   | f |   | f | f | f | x | f | x |1718|   |   |   |   |   |   |   |   |   |   |   |   | f | f |   | f | x |181 | 1 2 | x | 23 |   | x | 34 |   |   | x | 45 |   | x |   |   | 56 |   |   |   |   | x | 67 |   |   | x | x |   |   | 78 |   |   |   |   | x |   | x | 89 |   |   |   |   |   | x |   |   | 9*10|   |   |   | x |   |   | x |   |   | 10*11|   |   |   |   |   |   | x | x | x |   |1112|   |   |   |   |   |   | x |   |   | x | x |1213|   |   |   |   |   |   |   |   |   |   |   |   |1314|   |   |   |   |   |   |   |   | x |   | x |   | x |1415|   |   |   |   |   |   |   |   |   | x |   | x |   |   |1516|   |   |   |   |   |   |   |   |   |   | x | x |   | x | x |1617|   |   |   |   |   |   |   |   |   |   |   |   |   | x |   | x |1718|   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   | x |18`
The last part (layers 13-18) show a configuration that has already been thought about before: two pocuro's attached at their pentagons with bilbiro's straddled around. just as then, it doesn't close up, as "13" only has a connection with 14 and thus yields non-CRF-ity . Therefore, this expansion isn't CRF. This also means that the (+x)2(+x)5(+x)-thing has the same problem as well (lacing distances don't decrease by expansion).
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:o3o3o-symmetry: ...

o3(+x)3o:
(C C1)
(AA C C1) = (AA1 C C1)
...
(AA AA1 C C1)
...

Next being (AA AA1 C C1) or   ooofxxFfooFxofx 3 oxFxoFxxxFoxFxo 3 xfoxFoofFxxfooo &#xt , again being anti-symmetrical. It also comes out to be CRF. Note that its middle part will be the same as before in (C C1). So it too has these 6 equatorial tets, which are completely surrounded by ikes. But the poles would differ here (additional AA + AA1 changes).

Its cell statistic evaluates now to:

24  ikes
16  octs
72  squippies
112 tets
8   thawroes
8   tricues
24  trips

Having said this, it becomes obvious that we can derive (AA C C1) or   ooofxxFfooFxofo 3 oxFxoFxxxFoxFxx 3 xfoxFoofFxxfoox &#xt   directly therefrom: applying the changes AA as in (AA AA1 C C1) only to one of the poles of (C C1).

Accordingly that one would have as cell count:

24 ikes
16 octs
72  squippies
155 tets
4   thawroes
16  tricues
36  trips
1   tut

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:o3o3o-symmetry: ...

o3(+x)3o:
...
(BB BB1 C C1)
...

That   xooFxxFfooFoofo 3 xxFooFxxxFooFxx 3 ofooFoofFxxFoox &#xt   one again has anti-symmetry and is CRF.
Its cell count is:

24 bilbiroes
24 mibdies
16 octs
24 squippies
46 tets
8  tricues
2  tuts

That one is really hard to conquer. The off-symmetrical ikes of the (C C1) here get diminished into mibdies. They then occur as   .oo.xx.foo..... ............... ...............&#xt   in here. Furthermore here bilbiroes come in (re-using the pentagons of the mibdies at 2 faces each, while the other 2 pentagonal faces will interconnect these bilbiroes). Those too are off-symmetrically alligned. One might spot them only as   ............... xxFooFxx....... ...............&#xt .

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:o3o3o-symmetry: ...

o3(+x)3o:
(C C1)
(AA C C1) = (AA1 C C1)
(BB C C1) = (BB1 C C1)
(AA BB C C1) = (AA1 BB1 C C1)
(AA BB1 C C1) = (AA1 BB C C1)
(AA AA1 C C1)
(BB BB1 C C1)
(AA AA1 BB C C1) = (AA AA1 BB1 C C1)
(AA BB BB1 C C1) = (AA1 BB BB1 C C1)
(AA AA1 BB BB1 C C1)

Having now done (C C1), (AA AA1 C C1), as well as (BB BB1 C C1), and knowing that either of these structures only applies to an hemiglome each, i.e. kind of (C| = |C1), (AA C| = |C1 AA1), resp. (BB C| = |C1 BB1) only, we clearly can mix these freely.

(AA C|C1) already was dealt with.

(BB C|C1) = xooFxxFfooFxofo 3 xxFooFxxxFoxFxx 3 ofooFoofFxxfoox &#xt   thus ought to have for cells:
12  bilbiroes
12  ikes
12  mibdies
16  octs
48  squippies
122 tets
16  tricues
24  trips
2   tuts
The equatorial 6 tets (featuring as digonal antiprisms) here will have 2 adjoined mibdies towards the upper hemiglome and 2 adjoined ikes towards the lower hemiglome, for sure.

(AA C|C1 BB1) = ooofxxFfooFoofo 3 oxFxoFxxxFooFxx 3 xfoxFoofFxxFoox &#xt   thus ought to have for cells:
12 bilbiroes
12 ikes
12 mibdies
16 octs
48 squippies
79 tets
4  thawroes
8  tricues
12 trips
1  tut
Same here for those equatorial tets. (Just that I used here BB1 rather than BB, so the references to the upper and lower hemiglomes clearly get reversed.)

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:o3o3o-symmetry: ...

o3(+x)3o:
(C C1) *
(AA C C1) = (AA1 C C1) *
(BB C C1) = (BB1 C C1) *
(AA BB C C1) = (AA1 BB1 C C1)
(AA BB1 C C1) = (AA1 BB C C1) *
(AA AA1 C C1) *
(BB BB1 C C1) *
(AA AA1 BB C C1) = (AA AA1 BB1 C C1)
(AA BB BB1 C C1) = (AA1 BB BB1 C C1)
(AA AA1 BB BB1 C C1)

Concluding that part:
all those marked by * are valide CRFs, already being elaborated and commented on.

But the remaining ones all look impossible again. - Why?

Already within (BB BB1 C C1) the vertices of layer 4 where connectable by unit size edges exclusively to ones of layer 1 or of layer 7. Within (AA AA1 BB BB1 C C1) now the shape of layer 1 (and its mirror image at the opposite pole) changes, thereby keeping its distance, and all the remainder is kept the same. This would ask for different inter-layer lacings connecting to vertices of layer 1. Esp. the hypothetical edges, connecting 1 and 4, now would need a size of f. Therefore these do not remain allowed edges in a CRF. But then the vertices of layer 4 can solely be connected to ones of layer 7 any longer. Thus they will form dead ends! (An inner-layer unit edge in that layer is also not possible.)

As this argument affects only one hemiglome, in fact either one of (AA BB C| = |C1 BB1 AA1), the same argument will hold for any of the other cases too.

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:o3o3o-symmetry:
Code: Select all
`x3o3o || o3o3f || o3f3o || f3o3x || o3x3f || x3f3o || F3o3o || f3o3f || o3o3F || o3f3x || f3x3o || x3o3f || o3f3o || f3o3o || o3o3x  A   ||       ||       ||   B   ||   C   ||   D   ||       ||       ||       ||   D1  ||   C1  ||   B1  ||       ||       ||   A1`

...

(+x)3o3(+x):
(A A1 B B1 D D1)
...

Moving on to that one, i.e. to   oxxFxoAFxxFoxFx 3 xofxxFoooFxxfox 3 xFxoFxxFAoxFxxo &#xt.
It's also CRF. Total cell count here will be:

24 bilbiroes
16 octs
48 squippies
24 teddies
30 tets
8  thawroes
16 tricues
48 trips
2  tuts

Both, teddies and bilbiroes occur in here in off-symmetrical orientation only:
............... .ofxx.oo....... ...............   resp.   ............... ............... ..xoFxxF.ox....
These require likewise off-symmetrical pentagons
..ooo.oo.......3..ooo.oo.......3..ooo.oo....... : cycle = (CDGHE), and
....oo.o.oo....3....oo.o.oo....3....oo.o.oo.... : cycle = (EFJKH) - where layers are counted alphabetically.

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

Meanwhile I elaborated a further convex regular-faced polychoron (CRF). Its total cell consist is:
Code: Select all
`24 bilunabirotundae (bilbiroes, J91) 1 cuboctahedron (co) 8 octahedra (octs)30 pentagonal prisms (pips)18 square pyramids (squippies, J1)24 tridiminished icosahedra (teddies, J63)12 tetrahedra (tets) 4 triangular hebesphenorotundae (thawroes, J92) 1 truncated octahedron (toe)28 triangular cupolae (tricues, J3)24 triangular prisms (trips)`

I.e. not only a convex polychoron with 11 different types of cells, but moreover 5 of those types are Johnson solids! - If I remember correctly, then either one of these counts beats anything known so far.

This polychoron has an overall axially tetrahedral symmetry. In fact, it can be described as the multilayered resp. multistratic lace tower  xxxFooAFxxAxxFx 3 oxFxxAxxxFooFxx 3 xfoxFoofFxxFoox &#xt,  or, provided layerwise:  x3o3x || x3x3f || x3F3o || F3x3x || o3x3F || o3A3o || A3x3o || F3x3f || x3x3F || x3F3x || A3o3x || x3o3F || x3F3o || F3x3o || x3x3x.

(Here x describes a unit edge, f a vertex distance of size tau = (1+sqrt(5))/2 = 1.618..., F = x+f, A = x+F. - Note that all these larger "edges" do not contradict to CRF, esp. not to an unit-edged polychoron: These just occur as vertex distances of the respective cross-section polyhedra.)

This polychoron not only has the same number of vertex layers as the hexacosachoron (ex, x3o3o5o) when oriented in corresponding axially tetrahedral orientation, and these parallel layers moreover not only are at the same pairwise distances, it rather was derived therefrom as an expanded kaleido-faceting (EKF). That is, some of the vertex layers of the hexacosachoron were no longer described convexially, but as some kaleido-faceting thereof (by inverting some of their edges). Then the whole stack of layers further got Stott expanded simultanuously (within that perp space direction only).

Details thereof can be read from student91's (mere combinatorical) listing. (Sure the existance of each individual then had to be proved or disproved. - This then was the true reason of my elaboration.) - The layerwise description which he provides in the first run is that of the starting with hexacosachoron in the desired orientation. By "quirks" he refers to the possible kaleido-facetings of the respective section polyhedron. And finally by "(+x)3(+x)3o" he provides the type of Stott expansion to be applied (uniformly).

student91 wrote:o3o3o-symmetry:
Code: Select all
`x3o3o || o3o3f || o3f3o || f3o3x || o3x3f || x3f3o || F3o3o || f3o3f || o3o3F || o3f3x || f3x3o || x3o3f || o3f3o || f3o3o || o3o3x  A   ||       ||       ||   B   ||   C   ||   D   ||       ||       ||       ||   D1  ||   C1  ||   B1  ||       ||       ||   A1`

quirks:
A: (-x)3x3o (A1: o3x3(-x) )
AA: o3(-x)3x (AA1: x3(-x)3o )
AAA: o3o3(-x) (AAA1: (-x)3o3o )
B: f3x3(-x) (B1: (-x)3x3f )
BB: F3(-x)3o (BB1: o3(-x)3F )
C: x3(-x)3F (C1: F3(-x)3x )
CC: (-x)3o3F (CC1: F3o3(-x) )
D: (-x)3F3o (D1: o3F3(-x) )

...

(+x)3(+x)3o:
(AA BB1 CC C1 D)
...

The above being mentioned respective cells can be seen within that lace tower description, i.e. within  xxxFooAFxxAxxFx 3 oxFxxAxxxFooFxx 3 xfoxFoofFxxFoox &#xt  in the following way:
Code: Select all
`x..............3o..............3x..............     |  1 coxx.............3ox............. ...............&#x  |  4 tricuesxxx............ ............... xfo............&#xt |  6 pips............... oxFx...........3xfox...........&#xt |  4 thawroes.x..o..........3.x..x.......... ...............&#x  |  4 tricues.xxFoo.Fxx..... ............... ...............&#zx | 12 bilbiroes............... .x.xx..xx...... ...............&#xr | 12 pips............... ............... ..ox.oof.x.....&#zx | 12 teddies............... ...x..x........3...x..o........&#x  |  4 tricues............... ...x..xx....... ...............&#x  | 12 trips....o...x......3....x...x...... ...............&#x  |  4 tricues.....o...x..... ............... .....o...x.....&#x  |  6 squippies............... ......x...o....3......o...x....&#x  |  4 octs............... ......xx..o.... ...............&#x  | 12 squippies............... ............... ......of.xx.oo.&#zx | 12 teddies............... .......xxFooFxx ...............&#zx | 12 bilbiroes........x..x...3........x..o... ...............&#x  |  4 tricues........xx.xx.x ............... ...............&#xr | 12 pips.........x..x.. ............... .........x..o..&#x  |  6 trips............... ..........o..x.3..........x..o.&#x  |  4 octs...........x..x3...........o..x ...............&#x  |  4 tricues............x.x ............... ............o.x&#x  |  6 trips............... ............... ............oox&#x  | 12 tets............... .............xx3.............ox&#x  |  4 tricues..............x3..............x3..............x     |  1 toe`

But, for sure, not only this essence was considered for the proof of existance. Rather the full, very huge incidence matrix was derived to that end. And lots of individual vertex distances have been calculated to support this evaluation. - E.g. it even was checked that the first 2 types of tricues do not combine to cuboctahedra, even so they do connect at the hexagon in the corresponding way! They are not co-realmic.

--- rk
Last edited by Klitzing on Sun Sep 21, 2014 10:12 pm, edited 1 time in total.
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

Interesting -- do you have coordinate files?
Marek14
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### Re: Construction of BT-polytopes via partial Stott-expansion

Marek14 wrote:Interesting -- do you have coordinate files?

No, not so far. But the sectional polyhedra should be straight forward. And the 4th coordinate, the height, then should be obtainable from the corresponding sections of the hexacosachoron.

In contrast to student91 I personally lable these layers as A-B-C-D-E-F-G-H-g-f-e-d-c-b-a. (Note that the layer distances are centrally symmetric!) Then we get for the primary heights (and some of the deduced ones - based on unit edges for sure):
Code: Select all
`AB =      CD =      EF =      GH = sqrt[3-sqrt(5)]/4 =   0.218508 =  L     BC =                FG      = sqrt[7-3*sqrt(5)]/4 = 0.135045 =     S               DE                = 1/sqrt(8) =           0.353553 =  L+ S=>AC = BD =           EG = FH =                            0.353553 =  L+ S                              Gg =                       0.437016 = 2LAD =      CE = DF = EH = Fg =                            0.572061 = 2L+ S     BE =      DG =      Ff =                            0.707107 = 2L+2S          CF =      Eg =                                 0.790569 = 3L+ SAE = BF = CG = DH = Ef =                                 0.925615 = 3L+2S`

--- rk
Klitzing
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### Re: Construction of BT-polytopes via partial Stott-expansion

Had a first look into EKFs of rox (expanded kaleido-facetings of the rectified hexacosachoron).

There  rox = o3x3o5o  can be given as  DCBAVFfxoo 2 xoxFofofVx 3 oxoofoxxoo 5 ooxooxxoof &#zx, where
F=f+x=ff,
V=2f,
A=f+2x=F+x,
B=2f+x=V+x=fff,
C=2f+2x=2F,
D=3f+x=F+V.

Considered first the individual layers and their quirks (kaleido facetings) each:
Code: Select all
`A: D2x3o5o -> A2: D2(-x)3x5o -> A23: D2o3(-x)5fB: C2o3x5o -> B3: C2x3(-x)5f -> B32: C2(-x)3o5fC: B2x3o5x -> C2: B2(-x)3x5x -> C23: B2o3(-x)5F                             -> C24: B2(-x)3F5(-x)           -> C4: B2x3f5(-x) -> (C42 = C24)D: A2F3o5oE: V2o3f5oF: F2f3o5x -> F4: F2f3f5(-x)G: f2o3x5x -> G3: f2x3(-x)5F -> G32: f2(-x)3o5F           -> G4: f2o3F5(-x)H: x2f3x5o -> H1: (-x)2f3x5o -> H13: (-x)2F3(-x)5f           -> H3: x2F3(-x)5f -> (H31 = H13)I: o2V3o5oi: o2x3o5f -> i2: o2(-x)3x5f -> i23: o2o3(-x)5V`

I used 'I' and 'i' (instead of 'J') here, as both these layers fall into the same plane o2... .

So far business as usual. But then the set of combinations possible from all these layers and quirks multiplies to a count of 4320 cases, theoretically to be investigated - for this single subsymmetry only! - Halas, by sieving out the cases where a (-x) and an x would occure at the same node position (resulting in a u-edge after according expansion) would drop this huge count back to manageable 70.

From these 70 still it seems that lots are impossible fellows, i.e. not resulting in a CRF, because the cells of rox (being octs and ikes) are considerably larger than those of ex (being tets). Here individual quirks cannot be placed independantly at every layer, they would have to fit together somehow - at least within the reach of those starting cells. So it took a while to find a new CRF in here.

But finally a got one. Here it is:

Quirks B3G3H3i23 result in:  DCBAVFfxoo 2 xxxFofxFVo 3 o(-x)oofo(-x)(-x)o(-x) 5 ofxooxFfoV &#zx .
Expansion wrt. node position 3 then provides:  DCBAVFfxoo 2 xxxFofxFVo 3 xoxxFxooxo 5 ofxooxFfoV &#zx .
And that 1st found fellow (when evaluated) results in a total cell count of
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`30  ids120 octs60  pips48  pocuroes (J32)60  squippies (J1)220 tets40  thawroes (J92)2   ties80  tricues (J3)120 trips`

--- rk
Last edited by Klitzing on Sun Apr 26, 2015 2:30 pm, edited 1 time in total.
Klitzing
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Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Now also have looked thru all the remaining cases here:
we already went down from 4320 to 70;
now we result in a further decrease down to 2:

1. the rox itself
2. the one find, described recently

all other cases seem to be not a CRF for the one or other reason ...
Thus subsymmetry o2o3o5o has a very transparent number of cases only.

Even if we would look for . o3o5o subsymmetry, then we would have to consider the further case consisting from half rox and half of that other fellow in addition only. But that one then would ask for u-edges at least, so again impossible. Thus even within that much broader setup we still get only 2 possible ones.

--- rk
Klitzing
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Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Hello again, I just had some time left, and elaborated the double axial expansion of ex. As always, I did it a bit different than you would do, but it almost is the same as making an incidence matrix, just without the incidence matrix. First I calculated the edge-distances in ex in axial symmetry:
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`f2f2f2f + o2o2o2V + o2o2V2o + o2V2o2o + V2o2o2o + o2x2f2F + x2o2F2f + f2F2o2x + F2f2x2o + o2f2F2x + f2o2x2F + F2x2o2f + x2F2f2o + x2f2o2F + f2x2F2o + o2F2x2f + F2o2f2x   1    |    2    |    3    |    4    |    5    |    6    |    7    |    8    |    9    |   10    |   11    |   12    |   13    |   14    |   15    |   16    |   17   |1 | 12 | f | 23 | f |qf | 34 | f |qf |qf | 45 | f |qf |qf |qf | 56 | x | x | f | C |qf | 67 | x | f | x |qf | C | x | 78 | x | C |qf | x | f | f | C | 89 | x |qf | C | f | x | C | f | x | 910| x | C | x | f |qf | x | x | f | f |1011| x | x | C |qf | f | x | x | f | f | f |1112| x | f |qf | C | x | f | f | x | x | C | x |1213| x |qf | f | x | C | f | f | x | f | x | C | f |1314| x | x |qf | f | C | x | f | x | f | f | x | x | f |1415| x |qf | x | C | f | f | x | f | x | x | f | f | f | C |1516| x | f | C | x |qf | x | f | x | f | f | f | f | x | x | f |1617| x | C | f |qf | x | f | x | f | x | f | x | x | f | f | x | C |171 | 12 | f | 23 | f |   | 34 | f |   |   | 45 | f |   |   |   | 56 | x | x | f |   |   | 67 | x | f | x |   |   | x | 78 | x |   |   | x | f | f |   | 89 | x |   |   | f | x |   | f | x | 910| x |   | x | f |   | x | x | f | f |1011| x | x |   |   | f | x | x | f | f | f |1112| x | f |   |   | x | f | f | x | x |   | x |1213| x |   | f | x |   | f | f | x | f | x |   | f |1314| x | x |   | f |   | x | f | x | f | f | x | x | f |1415| x |   | x |   | f | f | x | f | x | x | f | f | f |   |1516| x | f |   | x |   | x | f | x | f | f | f | f | x | x | f |1617| x |   | f |   | x | f | x | f | x | f | x | x | f | f | x |   |171 | 12 | f | 23 | f |   | 34 | f |   |   | 45 | f |   |   |   | 56 |   |   | f |   |   | 67 |   | f |   |   |   |   | 78 |   |   |   |   | f | f |   | 89 |   |   |   | f |   |   | f |   | 910|   |   |   | f |   |   |   | f | f |1011|   |   |   |   | f |   |   | f | f | f |1112|   | f |   |   |   | f | f |   |   |   |   |1213|   |   | f |   |   | f | f |   | f |   |   | f |1314|   |   |   | f |   |   | f |   | f | f |   |   | f |1415|   |   |   |   | f | f |   | f |   |   | f | f | f |   |1516|   | f |   |   |   |   | f |   | f | f | f | f |   |   | f |1617|   |   | f |   |   | f |   | f |   | f |   |   | f | f |   |   |171 | 12 |   | 23 |   |   | 34 |   |   |   | 45 |   |   |   |   | 56 | x | x |   |   |   | 67 | x |   | x |   |   | x | 78 | x |   |   | x |   |   |   | 89 | x |   |   |   | x |   |   | x | 910| x |   | x |   |   | x | x |   |   |1011| x | x |   |   |   | x | x |   |   |   |1112| x |   |   |   | x |   |   | x | x |   | x |1213| x |   |   | x |   |   |   | x |   | x |   |   |1314| x | x |   |   |   | x |   | x |   |   | x | x |   |1415| x |   | x |   |   |   | x |   | x | x |   |   |   |   |1516| x |   |   | x |   | x |   | x |   |   |   |   | x | x |   |1617| x |   |   |   | x |   | x |   | x |   | x | x |   |   | x |   |17`
Now as we do the quirks, the new distances will also be the distances in the expanded form. This means that we can copy all distances where no quirks are made, or where two quirks are made on the same node, or where one thing has a quirk and the other one has an o (which means it could be quirked, you wouldn't notice!) Here are all distances in the double expansion (comparable to giving the edges in an incidence matrix)
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`F2F2f2f + x2x2o2V + x2x2V2o + x2B2o2o + B2x2o2o + x2o2f2F + o2x2F2f + F2A2o2x + A2F2x2o + x2F2F2x + F2x2x2F + A2o2o2f + o2A2f2o + o2F2o2F + F2o2F2o + x2A2x2f + A2x2f2x   1    |    2    |    3    |    4    |    5    |2   6    |1   7    |    8    |    9    |   10    |   11    |2  12    |1  13    |1  14    |2  15    |   16    |   17   |1:7;13;14}{2;3;4;6;10;162:6;12;15}{2;3;5;7;11;17symmetry transition: (1) (2,3) (4,5) (6,7) (8,9) (10,11) (12,13) (14,15) (16,17)1 | 12 | f | 23 | f |qf | 34 | f |qf |qf | 45 | f |qf |qf |qf | 56 |   | x | f |   |qf | 67 |   | f | x |qf |   | x | 78 | x | C |qf | x | f |   |   | 89 | x |qf | C | f | x |   |   | x | 910| x | C | x | f |qf |   | x | f | f |1011| x | x | C |qf | f | x |   | f | f | f |1112|   | f |qf |   | x | f |   |   |   |   | x |1213|   |qf | f | x |   |   | f |   |   | x |   |   |1314|   | x |qf | f |   |   | f |   |   | f |   |   | f |1415|   |qf | x |   | f | f |   |   |   |   | f | f |   |   |1516| x | f | C | x |qf |   | f | x | f | f | f |   | x | x |   |1617| x | C | f |qf | x | f |   | f | x | f | x | x |   |   | x | C |171 | 12 | f | 23 | f |qf | 34 | f |qf |qf | 45 | f |qf |qf |qf | 56 | f | x | f | V |qf | 67 | f | f | x |qf | V | x | 78 | x | C |qf | x | f |qf |qf | 89 | x |qf | C | f | x |qf |qf | x | 910| x | C | x | f |qf | f | x | f | f |1011| x | x | C |qf | f | x | f | f | f | f |1112| f | f |qf | V | x | f |qf | C | f |qf | x |1213| f |qf | f | x | V |qf | f | f | C | x |qf |qf |1314| f | x |qf | f | V | f | f | f |qf | f | f |qf | f |1415| f |qf | x | V | f | f | f |qf | f | f | f | f |qf | F |1516| x | f | C | x |qf | C | f | x | f | x | f |qf | x | x |qf |1617| x | C | f |qf | x | f | C | f | x | f | x | x |qf |qf | x | C |171 | 12 | f | 23 | f |   | 34 | f |   |   | 45 | f |   |   |   | 56 | f | x | f |   |   | 67 | f | f | x |   |   | x | 78 | x |   |   | x | f |   |   | 89 | x |   |   | f | x |   |   | x | 910| x |   | x | f |   | f | x | f | f |1011| x | x |   |   | f | x | f | f | f | f |1112| f | f |   |   | x | f |   |   | f |   | x |1213| f |   | f | x |   |   | f | f |   | x |   |   |1314| f | x |   | f |   | f | f | f |   | f | f |   | f |1415| f |   | x |   | f | f | f |   | f | f | f | f |   |   |1516| x | f |   | x |   |   | f | x | f | x | f |   | x | x |   |1617| x |   | f |   | x | f |   | f | x | f | x | x |   |   | x |   |171 | 12 |   | 23 |   |   | 34 |   |   |   | 45 |   |   |   |   | 56 |   | x |   |   |   | 67 |   |   | x |   |   | x | 78 | x |   |   | x |   |   |   | 89 | x |   |   |   | x |   |   | x | 910| x |   | x |   |   |   | x |   |   |1011| x | x |   |   |   | x |   |   |   |   |1112|   |   |   |   | x |   |   |   |   |   | x |1213|   |   |   | x |   |   |   |   |   | x |   |   |1314|   | x |   |   |   |   |   |   |   |   |   |   |   |1415|   |   | x |   |   |   |   |   |   |   |   |   |   |   |1516| x |   |   | x |   |   |   | x |   | x |   |   | x | x |   |1617| x |   |   |   | x |   |   |   | x |   | x | x |   |   | x |   |171 | 12 | f | 23 | f |   | 34 | f |   |   | 45 | f |   |   |   | 56 | f |   | f |   |   | 67 | f | f |   |   |   |   | 78 |   |   |   |   | f |   |   | 89 |   |   |   | f |   |   |   |   | 910|   |   |   | f |   | f |   | f | f |1011|   |   |   |   | f |   | f | f | f | f |1112| f | f |   |   |   | f |   |   | f |   |   |1213| f |   | f |   |   |   | f | f |   |   |   |   |1314| f |   |   | f |   | f | f | f |   | f | f |   | f |1415| f |   |   |   | f | f | f |   | f | f | f | f |   |   |1516|   | f |   |   |   |   | f |   | f | f | f |   |   |   |   |1617|   |   | f |   |   | f |   | f |   | f |   |   |   |   |   |   |17`
Next we determine the faces. A + means it is used in two different surtopes, and a ++ means it is used twice by a surtope of one kind. This is used to see if all surtopes are found in the next step.
The notation I use shoud be read as: ...a (1;3;9;10;15;17) => 4th node of parts 1,3,9,10,15&17 => .................2.................2.................2f.o.....oxox&#zx => teddi
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`faces:.... (1;10;3;15;17)+.... (1;11;2;14;16)+.... (1;10;7;6;11) +.... (1;11;17) +.... (1;17;9) +.... (1;8;9) +.... (1;8;16) +.... (1;10;16) +.... (10;13;16) +.... (4;13;16) +.... (4;8;16) +.... (5;9;17) +.... (5;12;17) +.... (11;12;17) +.... (2;6;11) +.... (3;7;10) +.a.. (11;12) +a... (10;13) +a... (2;6) + .a.. (3;7) +a... (2;14) ++.a.. (3;15) ++a... (3;10) +.a.. (2;11) +a... (3;7) +.a.. (2;6) +a... (4;13) ++.a.. (5;12) ++a... (4;16) +.a.. (5;17) +a... (6;7) +.a.. (6;7) +a... (7;10) +.a.. (6;11) +a... (10;16) +.a.. (11;17) +a... (10;13) +.a.. (11;12) +a... (13;16) +.a.. (12;17) +a... (14;16) +.a.. (15;17) +..a. (2;11) +...a (3;10) +..a. (4;16) +...a (5;17) +..a. (5;9) ++...a (4;8) ++..a. (8;9) +...a (8;9) +..a. (8;16) +...a (9;17) +..a. (11;12)+...a (10;13)+..a. (14;16) +...a (15;17) +aa.. (2) ++aa.. (3) ++a.a. (16) +.a.a (17) +a..a (10) +.aa. (11) +`
Then we can search the surtopes. As I've said, this is really similar to making an incidence matrix, but just without the incidence matrix.
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`found surtopes:a... (1;11;10;16;7;6;2;14) bilbiro 8×2..a. (1;5;8;9;11;12;17) gyepip 8×2..a. (8;4;16) tet 8×2aa.. (6;7) tet 4..aa (8;9) tet 4.aa. (11;12) squippy 4×2a... (10;13;16) squippy 8×2a... (10;3;7) squippy 8×2a... (4;13;16) squippy 8×2a.a. (16;14) squippy 4×2aa.. (2;6) trip 4×2a.a. (4;16) trip 4×2a..a (10;3) trip 4×2..a. (1;2;8;11;14;16) teddi 8×2derivations by symmetry transition:.a.. (1;10;11;17;6;7;3;15)...a (1;4;9;8;10;13;16)...a (9;5;17)a..a (10;13).a.. (11;12;17).a.. (11;2;6).a.. (5;12;17).a.a (17;15)aa.. (3;7).a.a (5;17).aa. (11;2)...a (1;3;9;10;15;17)`
This ables us to calculate the amount of surtopes we have. I do this by taking the order of the symmetry (16×2) and dividing this by the symmetry that is used to construct the surtope.
Code: Select all
`total:bilbiro: 16gyepip: 16tet: 24squippy: 64trip: 24teddi: 16`

Hurray hurray, I have finally given one new polytope (although I have talked about it before, and Klitzing has found far more of them. Good work Klitzing! )
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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### Re: Construction of BT-polytopes via partial Stott-expansion

And I can give you another one, this time of .3.2.-symmetry.
This polytope is the only expansion in .3.2.-symmetry that is CRF and doesn't coincide with an expansion of .3.-symmetry. Here it is:
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`x2o3x||F2x3o||o2x3f||F2F3o||x2F3x||A2x3x||x2x3F||F2o3F||o2A3o||A2F3o||(x2F3f+B2x3o)||A2x3f||o2x3F||F2F3x||x2A3o||A2o3x||x2o3F||F2x3f||o2F3o||F2x3o||x2x3x`

Here is the corresponding distance-matrix, cell list etc.:
Code: Select all
`C=sqrt(f+u)?=hf/(6+12f)=sqrt((7-3sqrt(5))/24)n=23-n? =(1,2)(4,5)(7,8)(9,10)(13,14)(15,16)(18,19)(21,22)f?=(2,3)(3,4)(5,6)(6,7)(8,9)(10,11/12)(11/12,13)(14,15)(16,17)(17,18)(19,20)(20,21)?=0.110....epansion on 1 and 2:A2+C1+H2+I1+i1+f2+e2+c1:              x2o3x||F2x3o||o2x3f||F2F3o||x2F3x||A2x3x||x2x3F||F2o3F||o2A3o||A2F3o||(x2F3f+B2x3o)||A2x3f||o2x3F||F2F3x||x2A3o||A2o3x||x2o3F||F2x3f||o2F3o||F2x3o||x2x3x  |  ?   |  f?  |  f?  |  ?   |  f?  |  f?  |  ?   |  f?  |  ?   |  f?             f?  |  ?   |  f?  |  ?   |  f?  |  f?  |  ?   |  f?  |  f?  |  ?   |  1  ||  2  ||  3  ||  4  ||  5  ||  6  ||  7  ||  8  ||  9  || 10  ||  11  || 12  || 13  || 14  || 15  || 16  || 17  || 18  || 19  || 20  || 21  || 22  A  ||  B  ||  C  ||  D  ||  E  ||  F  ||  G  ||  H  ||  I  ||  J  ||   K  ||  L  ||  j  ||  i  ||  h  ||  g  ||  f  ||  e  ||  d  ||  c  ||  b  ||  a         quirked:1:3,9,14,20}{1,5,7,11,16,18,222:1,8,17,18}{2,3,6,7,12,13,14,19,21,221 | 12 | x | 23 | x | f | 3 4 | f | x | f | 45 | f | f | x | x | 5  6 | f | x | C | x | f | 67 | f | C | x | f | x | f | 78 | f | f | f | f | f | x | x | 89 |qf |qf | f | f | x |qf | f |qf | 910|qf | x |qf | x | f | x | C | f | C |1011|qf |qf | f | f | x | C | x | f | x | f |11*12|qf | f | F | f |qf | x |qf | f | F | x |qf |12*13|qf | C |qf | f | C | x | f | x |qf | x | f | x |1314|qf | F | f |qf | f |qf | x | f | f |qf | x | F | C |1415| F |qf |qf | f | f | f | f | f | f | x | x | f | x | f |1516|hf | F |qf | C | f |qf | C |qf | x | f | x |qf | C | f | x |1617| F |qf |hf |qf | F | f |qf | f |hf | f |qf | x | x |qf | f |qf |1718| F |hf |qf | F |qf |qf | f | f |qf |qf | f |qf | f | x | f | C | f |1819|hf | F | F |qf |qf | C | C | f |qf | f | f | f | x | f | x | f | x | x |1920| C |Cf | F | F |qf |hf |qf | F | f |qf | f | F |qf | f | f | x |qf | f | f |2021|Cf |hf |Cf | F |hf |qf | F |qf | F | f |qf | f | x |qf | f | C | x | f | x | f |2122|Cf |Cf |hf |hf | F | F |qf |qf |qf |qf | C |qf | C | f | f | f | f | x | x | x | x |221 | 12 | x | 23 | x | f | 3 4 | f | x | f | 45 | f | f | x | x | 5  6 | f | x |   | x | f | 67 | f |   | x | f | x | f | 78 | f | f | f | f | f | x | x | 89 |   |   | f | f | x |   | f |   | 910|   | x |   | x | f | x |   | f |   |1011|   |   | f | f | x |   | x | f | x | f |11*12|   | f |   | f |   | x |   | f |   | x |   |12*13|   |   |   | f |   | x | f | x |   | x | f | x |1314|   |   | f |   | f |   | x | f | f |   | x |   |   |1415|   |   |   | f | f | f | f | f | f | x | x | f | x | f |1516|   |   |   |   | f |   |   |   | x | f | x |   |   | f | x |1617|   |   |   |   |   | f |   | f |   | f |   | x | x |   | f |   |1718|   |   |   |   |   |   | f | f |   |   | f |   | f | x | f |   | f |1819|   |   |   |   |   |   |   | f |   | f | f | f | x | f | x | f | x | x |1920|   |   |   |   |   |   |   |   | f |   | f |   |   | f | f | x |   | f | f |2021|   |   |   |   |   |   |   |   |   | f |   | f | x |   | f |   | x | f | x | f |2122|   |   |   |   |   |   |   |   |   |   |   |   |   | f | f | f | f | x | x | x | x |221 | 12 |   | 23 |   | f | 3 4 | f |   | f | 45 | f | f |   |   | 5  6 | f |   |   |   | f | 67 | f |   |   | f |   | f | 78 | f | f | f | f | f |   |   | 89 |   |   | f | f |   |   | f |   | 910|   |   |   |   | f |   |   | f |   |1011|   |   | f | f |   |   |   | f |   | f |11*12|   | f |   | f |   |   |   | f |   |   |   |12*13|   |   |   | f |   |   | f |   |   |   | f |   |1314|   |   | f |   | f |   |   | f | f |   |   |   |   |1415|   |   |   | f | f | f | f | f | f |   |   | f |   | f |1516|   |   |   |   | f |   |   |   |   | f |   |   |   | f |   |1617|   |   |   |   |   | f |   | f |   | f |   |   |   |   | f |   |1718|   |   |   |   |   |   | f | f |   |   |   |   | f |   | f |   | f |1819|   |   |   |   |   |   |   | f |   | f | f | f |   | f |   | f |   |   |1920|   |   |   |   |   |   |   |   | f |   | f |   |   | f | f |   |   | f | f |2021|   |   |   |   |   |   |   |   |   | f |   | f |   |   | f |   |   | f |   | f |2122|   |   |   |   |   |   |   |   |   |   |   |   |   | f | f | f | f |   |   |   |   |221 | 12 | x | 23 | x |   | 3 4 |   | x |   | 45 |   |   | x | x | 5  6 |   | x |   | x |   | 67 |   |   | x |   | x |   | 78 |   |   |   |   |   | x | x | 89 |   |   |   |   | x |   |   |   | 910|   | x |   | x |   | x |   |   |   |1011|   |   |   |   | x |   | x |   | x |   |11*12|   |   |   |   |   | x |   |   |   | x |   |12*13|   |   |   |   |   | x |   | x |   | x |   | x |1314|   |   |   |   |   |   | x |   |   |   | x |   |   |1415|   |   |   |   |   |   |   |   |   | x | x |   | x |   |1516|   |   |   |   |   |   |   |   | x |   | x |   |   |   | x |1617|   |   |   |   |   |   |   |   |   |   |   | x | x |   |   |   |1718|   |   |   |   |   |   |   |   |   |   |   |   |   | x |   |   |   |1819|   |   |   |   |   |   |   |   |   |   |   |   | x |   | x |   | x | x |1920|   |   |   |   |   |   |   |   |   |   |   |   |   |   |   | x |   |   | f |2021|   |   |   |   |   |   |   |   |   |   |   |   | x |   |   |   | x |   | x |   |2122|   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   |   | x | x | x | x |22x2o3x||F2x3o||o2x3f||F2F3o||x2F3x||A2x3x||x2x3F||F2o3F||o2A3o||A2F3o||(x2F3f+B2x3o)||A2x3f||o2x3F||F2F3x||x2A3o||A2o3x||x2o3F||F2x3f||o2F3o||F2x3o||x2x3x  |  ?   |  f?  |  f?  |  ?   |  f?  |  f?  |  ?   |  f?  |  ?   |  f?             f?  |  ?   |  f?  |  ?   |  f?  |  f?  |  ?   |  f?  |  f?  |  ?   |  1  ||  2  ||  3  ||  4  ||  5  ||  6  ||  7  ||  8  ||  9  || 10  ||  11  || 12  || 13  || 14  || 15  || 16  || 17  || 18  || 19  || 20  || 21  || 22  A  ||  B  ||  C  ||  D  ||  E  ||  F  ||  G  ||  H  ||  I  ||  J  ||   K  ||  L  ||  j  ||  i  ||  h  ||  g  ||  f  ||  e  ||  d  ||  c  ||  b  ||  afaces:... (1,2,3,4,5)+... (4,5,6,7,8)+... (4,5,10,11,15)+... (7,8,11,13,15)+... (11,14,15,18,19)+... (15,16,19,20,22)+... (2,4,6) +... (2,4,10) ++... (2,6,10) +... (3,5,7) +... (4,6,10) +... (5,7,11)+... (5,9,11)+... (6,10,12)+... (6,12,13)+... (6,8,13)+... (6,10,13)+... (7,11,14) +... (9,11,16) +... (10,12,13)+... (10,13,15)+... (11,15,16)+... (12,13,17)+... (13,17,19)+... (13,17,21)+... (13,19,21)+... (17,19,21)+... (18,19,22)+... (19,21,22)+a.. (1,3)+a.. (3,5)+a.. (3,7)+a.. (5,7)+a.. (5,9)+a.. (5,11)+a.. (7,11)+a.. (7,14)+a.. (9,11)+a.. (9,16)++a.. (11,14)+a.. (11,16)+a.. (16,20)++a.. (18,22)+a.. (20,22)+.a. (1,2)+.a. (1,3)+.a. (2,6)+.a. (3,7)+.a. (6,8)+.a. (6,12)+.a. (6,13)+.a. (7,8)+.a. (7,14)+.a. (8,13)+.a. (12,13)+.a. (12,17)+.a. (13,17)+.a. (13,19)+.a. (13,21)+.a. (14,18)+.a. (17,19)+.a. (17,21)+.a. (18,19)+.a. (18,22)+.a. (19,21)+.a. (19,22)+.a. (21,22)+..a (1,2)+..a (2,6)+..a (4,5)+..a (4,6)+..a (5,9)+..a (6,10)+..a (6,12)+..a (10,15)+..a (12,17)+..a (15,16)+..a (15,19,21)!+..a (17,21)+..a (20,22)+..a (21,22)+a.a (1)+.aa (1)+.aa (2)+a.a (5)+.aa (6)+aa. (7)+.aa (12)+.aa (17)+.aa (21)+a.a (22)+aa. (22)+.aa (22)+surtopes:aaa (1) trip 1aaa (22) hexp 1aa. (1,3) tet 3aa. (3,7) trip 3aa. (7,14) trip 3aa. (14,18) tet 3aa. (18,22) trip 3a.a (1,2,3,4,5) bilbiro 3a.a (5,9) squippy 3      a.a (5,11,16) -pip- 9-pya.a (20,22) squippy 3.aa (1,2) oct 2.aa (2,6) tricu 2.aa (6,12) tricu 2      .aa (6,10,13,17) -thawro- 12-py.aa (12,17) oct 2.aa (17,21) oct 2.aa (21,22) tricu 2a.. (3,5,7) squippy 6a.. (5,7,11) trip 6a.. (5,9,11) squippy 6a.. (7,11,14) squippy 6a.. (9,11,16) squippy 6a.. (11,14,15,16,18,19,20,22) bilbiro 6.a. (1,2,3,4,5,6,7,8) bilbiro 6.a. (7,8,11,13,14,15,18,19) bilbiro 6.a. (6,12,13) trip 6.a. (6,8,13) squippy 6.a. (12,13,17) squippy 6.a. (13,17,19) squippy 6.a. (13,17,21) squippy 6.a. (13,19,21) trip 6.a. (17,19,21) squippy 6.a. (18,19,22) squippy 6.a. (19,21,22) trip 6..a (2,4,6) tet 6..a (2,6,10) tet 6..a (4,6,10) tet 6..a (6,10,12) tet 6..a (4,5,9,10,11,15,16) mbidi 6..a (10,12,13,15,17,19,21) gyepip 6..a (15,16,19,20,21,22) teddi 6... (4,5,6,7,8,10,11,13,15) teddi 12... (6,10,12,13) tet 12... (13,17,19,21) tet 12... (2,4,6,10) tet 12total:trip: 1+3+3+3+6+6+6=28hip:1tet:3+3+6+6+6+6+12+12+12=66bilbiro:3+6+6+6=21squippy:3+3+6+6+6+6+6+6+6+6+6+6=66oct:2+2+2=6tricu:2+2+2=6mbidi:6gyepip:6teddi:6+12=18`
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`total cells:trip: 28hip: 1tet: 66bilbiro: 21squippy: 66oct: 6tricu: 6mbidi: 6gyepip: 6teddi: 18`
When the 6th layer is double-quirked, the teddi won't survive, and when the last layer is double-quirked, the other teddi won't. Therefore this is the only possible expansion in .3.2.-symmetry. In .2.2.2.-symmetry (my previous post), all expansions have also been found, as a third expansion will break a bilbiro into pieces.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian

Posts: 317
Joined: Tue Dec 10, 2013 3:41 pm

### Re: Construction of BT-polytopes via partial Stott-expansion

Hello there

We have now investigated all possible EKFs of ex (thanks to Klitzing). The only other uniform 4-polytope that will give BPT-EFKs is rox (Klitzing has already given an example). I have now determined rox' representation in all unknown subsymmetries. Other subsymmetries are on Klitzings site or (for the 532) a few posts back. Here they come:

[5,3,3]-symmetry:
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`o5o3x3o`
[31,1,1]-symmetry:
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`permutation:(A3X3B*b3C) (B3X3C*b3A) (C3x3A*b3B)parts:V3x3o*b3o  F3o3f*b3x  o3f3x*b3ftotal:VooFxfofx 3 xxxooofff 3 oVofFxxof *b3 ooVxfFfxo&#zx`
[3,3,3]-symmetry:
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`permutation:(A3B3C3D) (D3C3B3A)parts:o3x3o3B  o3x3V3o  V3x3o3f  x3f3f3x  o3f3x3F  f3F3o3x  F3o3f3ftotal:ooVofFxfxFfoB 3 xxxfFoffoxoVo 3 oVoxoffoFfxxx 3 BofFxfxFfoVoo&#zx`
[5,2,5]-symmetry:
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`permutation:(A5B2C5D) (B5A2D5C) (C5D2B5A) (D5C2A5B)parts:o5o2F5F  x5o2F5x  f5o2o5V  x5x2f5f  x5f2F5ototal:oxofoxxfFxFVofoF 5 ooxofxfxFFxoVfFo 2 FFxoVfFooxofoxxf 5 FxFVofoFooxofxfx&#zx`
[3,2,3]-symmetry:
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`permutation:(A3B2C3D) (B3A2D3C) (C3D2B3A) (D3C2A3B)parts:x3o2o3C  x3o2f3B  o3f2V3F  x3f2B3x  f3A2F3o  o3V2B3o  x3F2B3o  f3f2F3F  x3V2f3Ftotal:xoxoofxffAoVxFfxVCoBfFVxBAfoBoBFFf 3 oxoxfofxAfVoFxfVxoCfBVFBxfABoBoFfF 2 oCfBVFBxFFoBoBoFfFxoxoofxffAoVxFfxV 3 CoBfFVxBAfoBoBFFfoxoxfofxAfVoFxfVx&#zx`
[2,2,2]-symmetry:
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`permutation:(A2B2C2D) (A2C2D2B) (A2D2B2C) (B2A2D2C) (B2C2A2D) (B2D2C2A) (C2A2B2D) (C2B2D2A) (C2D2A2B) (D2A2C2B) (D2B2A2C) (D2C2B2A)parts:o2o2V2C  x2x2B2B  o2x2f2D  o2F2B2A  x2f2C2F  f2F2V2Btotal:oxooxfoxooxfoxooxfoxxFfFoxxFfFoxxFfFVBfBCVVBfBCVVBfBCVCBDAFBCBDAFBCBDAFB 2 oxxFfFVBfBCVCBDAFBoxooxfVBfBCVCBDAFBoxooxfoxxFfFCBDAFBoxooxfoxxFfFVBfBCV 2 VBfBCVCBDAFBoxxFfFCBDAFBoxooxfVBfBCVoxxFfFCBDAFBoxooxfVBfBCVoxooxfoxxFfF 2 CBDAFBoxxFfFVBfBCVVBfBCVCBDAFBoxooxfCBDAFBoxooxfoxxFfFoxxFfFVBfBCVoxooxf&#zx`

EDIT: And here now are the 3-dimensional subsymmetries (from klitzings site):
[2,5]-symmetry:
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`  1   |   2   |   3   |   4a  |   5a  |   5b  |   6a  |   6b  |   7a  |   7b  |   8a  |   9a  |  10a  |  10b  |  11a  |  11b  |  11c  |  12a  |  12b  |  13a  |  13b  |  14a  |  14b  |  15a  |  15b  |  16a  |  16b  |   17  |  18a  |  19a  |  19b  |  19c  |  20a  |  21a  |  22a  |  22b  |  23a  |  23b  |  24a  |  24b  |  25a  |  25b  |  26a  |  26b  |  27a  |  27b  |  27c  |  28a  |  28b  |  29a  |  30a  |  31a  |  31b  |  32a  |  32b  |  33a  |  33b  |  34a  |   35  |   36  |  37ao2o5o | x2x5o | o2o5f | F2o5x | f2x5x | A2o5o | o2x5f | V2f5o | x2F5o | B2o5x | f2o5F | F2f5x | x2f5f | B2x5x | o2V5o | V2o5F | C2x5o | f2o5V | A2F5o | F2f5f | D2o5o | o2x5F | C2o5f | x2F5x | B2x5f | F2V5o | D2x5o | f2x5F | B2F5o | o2F5F | V2f5f | C2x5x | B2o5F | f2F5x | F2o5V | D2o5x | x2x5F | B2f5x | o2F5x | C2f5o | F2f5f | D2o5o | f2V5o | A2o5F | o2o5V | V2F5o | C2o5x | x2f5f | B2x5x | F2x5f | f2F5o | x2o5F | B2x5o | o2f5x | V2o5f | f2x5x | A2o5o | F2x5o | o2f5o | x2o5x | o2o5o  1   |   2   |   3   |   4   |   5   |   6   |   7   |   8   |   9   |   10  |   11  |   12  |   13  |   14  |   15  |   16  |   17  |   18  |   19  |   20  |   21  |   22  |   23  |   24  |   25  |   26  |   27  |   28  |   29  |   30  |   31  |   32  |   33  |   34  |   35  |   36  |   37  |   38  |   39  |   40  |   41  |   42  |   43  |   44  |   45  |   46  |   47  |   48  |   49  |   50  |   51  |   52  |   53  |   54  |   55  |   56  |   57  |   58  |   59  |   60  |   61`
[3,3]-symmetry:
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`  1   |   2   |   3   |   4a  |   4b  |   5   |   6   |   7a  |   7b  |   8   |   9a  |   9b  |   10  |  11a  |  11b  |  12a  |  12b  |   13  |  14a  |  14b  |  15a  |  15b  |  16a  |  16b  |   17  |  18a  |  18b  |  19a  |  19b  |  20a  |  20b  |   21  |  22a  |  22b  |  23a  |  23b  |   24  |  25a  |  25b  |   26  |  27a  |  27b  |   28  |   29  |  30a  |  30b  |   31  |   32  |   33o3x3o | x3o3f | x3f3o | F3o3x | o3f3f | f3o3F | f3f3x | o3x3V | V3x3o | o3F3f | f3x3F | x3V3o | F3f3x | x3o3B | B3o3x | V3o3F | x3F3f | o3B3o | f3x3V | A3f3o | o3x3B | F3f3f | o3F3F | B3o3f | x3V3x | F3F3o | f3o3B | B3x3o | f3f3F | V3x3f | o3f3A | o3B3o | F3o3V | f3F3x | B3o3x | x3o3B | x3f3F | F3x3f | o3V3x | f3F3o | V3x3o | o3x3V | x3f3f | F3o3f | x3o3F | f3f3o | o3f3x | f3o3x | o3x3o  1   |   2   |   3   |   4   |   5   |   6   |   7   |   8   |   9   |   10  |   11  |   12  |   13  |   14  |   15  |   16  |   17  |   18  |   19  |   20  |   21  |   22  |   23  |   24  |   25  |   26  |   27  |   28  |   29  |   30  |   31  |   32  |   33  |   34  |   35  |   36  |   37  |   38  |   39  |   40  |   41  |   42  |   43  |   44  |   45  |   46  |   47  |   48  |   49`
[3,2]-symmetry:
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`  1   |   2   |   3   |   4   |   5a  |   6a  |   6b  |   6c  |   7a  |   8a  |   8b  |   9a  |   9b  |  10a  |  11a  |  11b  |  12a  |  13a  |  13b  |  13c  |  14a  |  15a  |  16a  |  16b  |  16c  |   17  |  18a  |  18b  |  18c  |  19a  |  19b  |  20a  |  21a  |  21b  |  21c  |  22a  |  22b  |  23a  |  23b  |  23c  |  24a  |  24b  |  25a  |  25b  |  26a  |  26b  |  27a  |  27b  |  27c  |  28a  |  28b  |  29a  |  29b  |  29c  |  30a  |  30b  |  31a  |  31b  |  32a  |  32b  |  32c  |  33a  |  34a  |  34b  |  34c  |   35  |   36  |  37a  |  37b  |  37c  |   38  |  39a  |  39b  |   40  |  41a  |  41b  |  42a  |  42b  |   43  |  44a  |  44b  |  44c  |   45  |   46  |   47  |   48  |   49o3x2o | x3o2f | x3f2x | o3f2F | F3o2f | o3V2o | F3x2o | f3o2V | f3f2F | f3F2x | o3x2B | V3x2f | o3F2A | B3o2o | o3B2x | f3x2B | x3V2F | f3A2o | F3f2V | x3o2C | F3F2f | B3o2F | B3x2x | V3o2B | x3F2B | x3B2f | V3F2o | o3B2V | f3x2C | A3f2F | o3x2D | F3f2B | f3B2o | C3o2o | o3F2C | F3V2f | B3o2A | B3f2x | o3C2x | x3V2B | x3B2F | f3o2D | F3F2V | f3f2C | B3x2F | o3f2D | f3B2x | C3o2x | V3x2B | V3F2f | o3B2A | B3f2o | o3C2o | F3o2C | f3F2B | f3f3o | f3A2F | x3o2D | F3V2o | B3o2V | x3f2C | B3x2f | x3B2x | o3V2B | F3x2B | o3B2F | F3F2f | A3f2o | f3F2V | o3x2C | V3x2F | B3o2x | x3f2B | o3B2o | x3V2f | F3o2A | F3f2x | x3o2B | f3f2F | V3o2o | x3F2o | o3f2V | o3F2f | f3o2F | f3x2x | o3x2f | x3o2o  1   |   2   |   3   |   4   |   5   |   6   |   7   |   8   |   9   |   10  |   11  |   12  |   13  |   14  |   15  |   16  |   17  |   18  |   19  |   20  |   21  |   22  |   23  |   24  |   25  |   26  |   27  |   28  |   29  |   30  |   31  |   32  |   33  |   34  |   35  |   36  |   37  |   38  |   39  |   40  |   41  |   42  |   43  |   44  |   45  |   46  |   47  |   48  |   49  |   50  |   51  |   52  |   53  |   54  |   55  |   56  |   57  |   58  |   59  |   60  |   61  |   62  |   63  |   64  |   65  |   66  |   67  |   68  |   69  |   70  |   71  |   72  |   73  |   74  |   75  |   76  |   77  |   78  |   79  |   80  |   81  |   82  |   83  |   84  |   85  |   86  |   87`
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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Posts: 317
Joined: Tue Dec 10, 2013 3:41 pm

### Re: Construction of BT-polytopes via partial Stott-expansion

Hi again, I have found a new expansion of rox, it is in demitesseractic symmetry, on the central node. here it is:

rox in demitesseractic symmetry:
Code: Select all
`permutation:A3o3B*b3C: (ABC)|(CAB)|(BCA)             1  |  2  |  3parts:V3x3o*b3o | F3o3f*b3x | o3f3x*b3f    A     |     B     |     Ctotal:1:V3x3o*b3o | F3o3f*b3x | o3f3x*b3f    A     |     B     |     C2:o3x3V*b3o | x3o3F*b3f | f3f3o*b3x    A     |     B     |     C3:o3x3o*b3V | f3o3x*b3F | x3f3f*b3o    A     |     B     |     C`
initial distance-matrix:
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`distances:1-1=2-2=3-3:A| AB| x | BC| C | q | C1-2:A|qf | AB| x | C | BC| x | x | x | C1-3:A| AB| f | BC| q | x | C2-3:A| AB| C | BC| x | x | C`
considered expansion:
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`(o)3(+x)3(o)*b3(o):A1A2A3:A1A2A3B1(Co):etc.A1A2A3:symmetry preservedA becomes a "loose end"invariances:Bquirks:Athings:1:B3o3x*b3x | F3x3f*b3x | o3F3x*b3f    A     |     B     |     C    2:                               x3o3B*b3x | x3x3F*b3f | f3F3o*b3x    A     |     B     |     C    3:                               x3o3x*b3B | f3x3x*b3F | x3F3f*b3o    A     |     B     |     C    distances:A| AB| x | BC|qf | q | C1-2:A|qf | AB| x | C | BC| x | f | x | C1-3:A| AB| f | BC| C | x | C2-3:A| AB| C | BC| f | x | Cfaces:aa..:(A2).aa.(A3).a.a(A1)+aa..:(A3).aa.(A1).a.a(A2)+aa..:(B2).aa.(B3).a.a(B1)+a.a.:(A3)a..a(A2)..aa(A1)+a...:(A2B2)..a.(A3B3)...a(A1B1)+a...:(A3B3C1)..a.(A1B1C2)...a(A2B2C3)+a...:(B2C3)..a.(B3C1)...a(B1C2)a...:(C1C3)..a.(C1C2)...a(C2C3)+....:(B1B2B3)+....:(C1C2C3)+....:(B1B2C3)....:(B2B3C1)....:(B1B3C2)+....:(B1C2C3)....:(B2C1C3)....:(B3C1C2)+cells:....:(B1C1B2C2B3C3)                             oct     192             192   oct-derivativea.aa:(C1C2C3)                                   ike     192/8           24   ike-derivativeaaa.:(A3)aa.a:(A2).aaa:(A1)                     co      192/24*3        24   oct-derivativeaa..:(C1B2A3B3).aa.:(A1B1C2B3).a.a:(B1A2B2C3)   thawro  192/6*3         96   ike-derivativeaa..:(A2B2).aa.:(A3B3).a.a:(A1B1)               tricu   192/6*3         96   oct-derivativea.a.:(C1A3B3)a..a:(A2B2C3)..aa:(A1B1C2)         pip     192/4*3         144a...:(B2C1C3)..a.(B3C1C2)...a:(B1C2C3)          squippy 192/2*3         288   oct-derivative.a..:(B1B2B3)                                   trip    192/2           96                                                total                   960     120×ike-derivative                                                                                600×oct-derivative                                                                                240×new element`
Thus finally the cells are:
Code: Select all
`192 octs24 ikes24 coes96 thawroes96 tricues144 pips288 squippies96 trips`
Furthermore the expansions of rox according to [3,3,3]-symmetry have also given nothing.
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
Tetronian

Posts: 317
Joined: Tue Dec 10, 2013 3:41 pm

### Re: Construction of BT-polytopes via partial Stott-expansion

student91 wrote:Hi again, I have found a new expansion of rox, it is in demitesseractic symmetry, on the central node. here it is:
[...]
Furthermore the expansions of rox according to [3,3,3]-symmetry have also given nothing.

Hew, you're quite fast. It took me up till today (even so not really working every day on it), to reproduce that one (already within days) and to deal all other combinations of that subsymmetry.

In fact, for the layers A-C there are 6 possibilities each, for D-F there are 3, and for G-I there are 2. Thus a total of 46,656 combinations. Therefrom all those, which would produce u-scaled lacing edges, can be omitted a priori, and the intrinsic symmetry of the used subsymmetry Dynkin diagram can be used to seave out further dublicates. And finally rox itself is known at least.

But then this still comes down to 174 to be evaluated remaining cases. One of which is yours. And indeed, all other ones finally would die either because of requiring partial lacing polygons, i.e. requiring for in-layer edges which are some polygonal chords greater than unity, or because of layers which can no longer be connected to any other one by use of unit edges only.

--- rk
Klitzing
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Location: Heidenheim, Germany

### Re: Construction of BT-polytopes via partial Stott-expansion

Klitzing wrote:
student91 wrote:[...]

Hew, you're quite fast. It took me up till today (even so not really working every day on it), to reproduce that one (already within days) and to deal all other combinations of that subsymmetry.

[...]
--- rk

Hello. It's great that you have done a thorough investigation of these cases, as my investigation had a very tiny bit of heuristics in them, (I was discarting EKF's that might exist if new lacings of length 1 would arise because of the expansion. This is very unlikely, but not entirely impossible. Your investigation proved this unlikely event didn't happen).
I frankly have done next to nothing on this field, I have been busy completing my scool project, and when that was finished I really wanted to take a break.
As I've had a long enough break now, I think we can continue our effort on trying to publish our result. I think a google drive-document will suit well for us if we want to collectively write an article. I will PM everyone with a link so they can contribute to the paper. (Although I think I will first do a little more investigations on possible constructions on EKF's, so that I can elaborate the most elegant construction in the paper. I will discuss that in a different threat I guess).
How easily one gives his confidence to persons who know how to give themselves the appearance of more knowledge, when this knowledge has been drawn from a foreign source.
-Stern/Multatuli/Eduard Douwes Dekker
student91
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Posts: 317
Joined: Tue Dec 10, 2013 3:41 pm

### Re: Construction of BT-polytopes via partial Stott-expansion

Is this what i am supposed to condense for the paper. I've read it somewhat.
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
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