Johnsonian Polytopes

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

Re: Johnsonian Polytopes

Postby Marek14 » Wed Jan 04, 2012 8:22 pm

wintersolstice wrote:
Marek14 wrote:
Second of all, not all of the diminishings can be considered augmentations of this shape. Think of bidiminishings: when you diminish one vertex from a certain 600-cell, the second vertex can be from another one. Of course, if you will continue, you will encounter the final (undiminishable) result sooner than after 120 diminishings.



But that's only if the 720 vertices are grouped into "fixed sets" of 120 vertices so that a vertex is only part of one 600-cell but surely you can regroup them so that the second vertex can be taken from the same 600-cell rather than a different set?

If it were possible to choose less than or equal than 120 vertices that can't be contained inside a single 600-cell (given that no two are adjacent)regardless of how the vertices are grouped, then it would be true. I've no idea?

btw I didn't discover spidrox


That such diminishings exist seems trivial to me. Let's say you diminish two adjacent vertices, A and B (of course, you can't do that). Each of them must be on a different 600-cell even if there are different mappings (which I'm not sure is proven). So, what if you go one 600-cell vertex away from A to C and diminish C and B? Can you put a 600-cell through A and C?

Related question? How many different distances between two vertices exists? If there's more than vertex-vertex distances on a 600-cell (and with five times that many vertices, yet the same symmetry, there SHOULD be), then some vertex pairs must lie on different 600-cells no matter what.
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Re: Johnsonian Polytopes

Postby wintersolstice » Wed Jan 04, 2012 8:52 pm

Marek14 wrote:Related question? How many different distances between two vertices exists? If there's more than vertex-vertex distances on a 600-cell (and with five times that many vertices, yet the same symmetry, there SHOULD be), then some vertex pairs must lie on different 600-cells no matter what.


Might be worth looking into that :D however it would be useful to remember that the edge-length for each 600-cell won't be the same as that for the rectified 600-cell (which means they'll be at different scales) because the the distance between two ajdacent vertices is an edge length of the rectified 600-cell but when you group them into 600-cell no two are adjacent.

Like I said in my original post I've no idea :D
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Re: Johnsonian Polytopes

Postby Marek14 » Wed Jan 04, 2012 9:23 pm

wintersolstice wrote:
Marek14 wrote:Related question? How many different distances between two vertices exists? If there's more than vertex-vertex distances on a 600-cell (and with five times that many vertices, yet the same symmetry, there SHOULD be), then some vertex pairs must lie on different 600-cells no matter what.


Might be worth looking into that :D however it would be useful to remember that the edge-length for each 600-cell won't be the same as that for the rectified 600-cell (which means they'll be at different scales) because the the distance between two ajdacent vertices is an edge length of the rectified 600-cell but when you group them into 600-cell no two are adjacent.

Like I said in my original post I've no idea :D


It doesn't matter whether the edge-length is the same or not. What matters is how many distinct classes of vertices there are if you mark one vertex as "special".

In case of 600-cell, we find:

12 closest vertices
20 vertices in next layer
Then 12
Then 30
Then 12 again
20 again
12 again
and finally 1 opposite vertex.

So that's 8 possible relationships between two vertices of 600-cell.

Now, I tried it for rectified 600-cell and there are MUCH more possible relations between two vertices than 8. This means that it's easy to slice off two vertices which don't belong to the same 600-cell.
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Re: Johnsonian Polytopes

Postby quickfur » Wed Jan 04, 2012 11:32 pm

wintersolstice wrote:I noticed it said in the wiki than a 5,n-duoprism can be augented with pentagonal prism pyramid for any number up to 38.
I don't think it's right.

I did a calculation for find the angle between the pentagonal pryamid cell and the pentagonal prism base and the angle is 18 degrees.

Actually, to be more accurate, the angle is A=arcsin(sqrt((5-2*sqrt(5))/5)), which comes out to about 18.96°, that is, approximately 19°. This is important, because it is the deciding factor for whether the 5,10-duoprism can be fully augmented.

There is no reason not to allow adjacent augmentations, so the only requirement being that 2*A + B ≤ 180°, where B is the internal angle of an n-gon. Since the internal angle of an n-gon is (n-2)*180°/n, we have:

Code: Select all
2*A + (n-2)*180°/n ≤ 180°
(n-2)*180°/n ≤ 180° - 2*A
(n-2)/n ≤ 1 - 2*A/180°
... (some basic algebra later) ...
n ≤ 180°/A

Plugging in the value of A, we get the RHS roughly equal to 9.49, which means that we can have adjacent augmentations of n,5-duoprisms up to n=9. When n=10, adjacent augments become non-convex.

Now, if augmentations are spaced at least one prism apart, then we can go up slightly higher (note that it's OK for the pentagonal prism pyramid to become coplanar with an adjacent pentagonal prism, since they will just form cells in the shape of elongated pentagonal pyramids -- the pentagonal prisms in the duoprism are joined by their pentagonal faces -- although this case doesn't actually occur):

Code: Select all
A + (n-2)*180°/n ≤ 180°
(n-2)*180°/n ≤ 180° - A
(n-2)/n ≤ 1 - A/180°
...
n ≤ 2*180°/A = 360°/A

Plugging in the value of A, we get the RHS approximately equal to 18.98. This means that for non-adjacent augments, we can go up to 5,18-duoprisms.
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Re: Johnsonian Polytopes

Postby wintersolstice » Thu Jan 05, 2012 12:52 pm

quickfur wrote:
wintersolstice wrote:I noticed it said in the wiki than a 5,n-duoprism can be augented with pentagonal prism pyramid for any number up to 38.
I don't think it's right.

I did a calculation for find the angle between the pentagonal pryamid cell and the pentagonal prism base and the angle is 18 degrees.

Actually, to be more accurate, the angle is A=arcsin(sqrt((5-2*sqrt(5))/5)), which comes out to about 18.96°, that is, approximately 19°. This is important, because it is the deciding factor for whether the 5,10-duoprism can be fully augmented.

There is no reason not to allow adjacent augmentations, so the only requirement being that 2*A + B ≤ 180°, where B is the internal angle of an n-gon. Since the internal angle of an n-gon is (n-2)*180°/n, we have:

Code: Select all
2*A + (n-2)*180°/n ≤ 180°
(n-2)*180°/n ≤ 180° - 2*A
(n-2)/n ≤ 1 - 2*A/180°
... (some basic algebra later) ...
n ≤ 180°/A

Plugging in the value of A, we get the RHS roughly equal to 9.49, which means that we can have adjacent augmentations of n,5-duoprisms up to n=9. When n=10, adjacent augments become non-convex.

Now, if augmentations are spaced at least one prism apart, then we can go up slightly higher (note that it's OK for the pentagonal prism pyramid to become coplanar with an adjacent pentagonal prism, since they will just form cells in the shape of elongated pentagonal pyramids -- the pentagonal prisms in the duoprism are joined by their pentagonal faces -- although this case doesn't actually occur):

Code: Select all
A + (n-2)*180°/n ≤ 180°
(n-2)*180°/n ≤ 180° - A
(n-2)/n ≤ 1 - A/180°
...
n ≤ 2*180°/A = 360°/A

Plugging in the value of A, we get the RHS approximately equal to 18.98. This means that for non-adjacent augments, we can go up to 5,18-duoprisms.


that's strange because when I did it it turned out to be 18 degrees exactly! should I post my workings out? how did you work it out? You see my formula was different it involved arccosine not arsine.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 05, 2012 4:33 pm

wintersolstice wrote:[...]that's strange because when I did it it turned out to be 18 degrees exactly! should I post my workings out? how did you work it out? You see my formula was different it involved arccosine not arsine.

Yes, that would help. I may have made a mistake, though I don't think so.
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Re: Johnsonian Polytopes

Postby wintersolstice » Thu Jan 05, 2012 6:00 pm

Image

Image

Image

Hopefully you can read my writing (I had to correct it a bit as you can probaly tell!) the edge length (or the same) is assumed to be one (which shouldn't matter :D )

right at the end you have the cosine of the angle, everytime I do that on a calculator it comes up with "18" no decimals (and it's in normal mode)

quickfur wrote: I may have made a mistake, though I don't think so.


I can say that aswell LOL
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 05, 2012 7:32 pm

OK, you are right, I made a mistake; I wrongly assumed that the hypotenuse of the apex-prism triangle was equal to the edge length, which it is not, because it actually traces out a line from the apex of the pentagonal pyramid to its base, not its vertices. So the angle is indeed 18°.

My bad.

So given an 18° angle, let's follow through with the previous derivation: for adjacent augments, we have the requirement 2*A + B ≤ 180°, where A=18° and B is the internal angle of the n-gon, which is (n-2)*180°/n. So we have: (this time i'm showing the full derivation in case i goofed up somewhere else too)
Code: Select all
2*A + (n-2)*180°/n ≤ 180°
(n-2)*180°/n ≤ 180° - 2*A
(n-2)/n ≤ 1 - 2*A/180°
1-2/n ≤ 1 - A/90°
-2/n ≤ -A/90°
2/n ≥ A/90°
n/2 ≤ 90°/A
n ≤ 180°/A
n ≤ 10

So this means that we can fully augment 5,n-duoprisms up to (and including) n=10. So we actually have more possible augmentations than I previously concluded, because my wrong calculation excluded the 5,10-duoprism.

For non-adjacent augmentations, we have the requirement A+B≤180°, so that lets us go a bit higher:
Code: Select all
A+(n-2)*180°/n ≤ 180°
(n-2)/n ≤ 1 - A/180°
1-2/n ≤ 1 - A/180°
-2/n ≤ -A/180°
n/2 ≤ 180°/A
n ≤ 360°/A
n ≤ 20

So we can actually go all the way up to n,20-duoprisms if we avoid adjacent augmentations. That's a lot of possible combinations of augmentations.

I was going to do a count of the total number of possibilities, but I suck at combinatorics, so maybe somebody else who has a better grasp of Polya's enumeration theorem can help us out here. :)
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Jan 05, 2012 8:11 pm

How does the augmentation look for the special case of (5,5) duoprism where you can augment prisms in both rings?
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 05, 2012 8:34 pm

Marek14 wrote:How does the augmentation look for the special case of (5,5) duoprism where you can augment prisms in both rings?

I have yet to confirm this, but I believe the 5,5-duoprism should allow all combinations of augments in both rings. A fully augmented 5,5-duoprism should have 20 pentagonal pyramids and 50 square pyramids as cells.

Currently I'm working on enumerating all duoprism augments: for the m<5, n<5 cases the combinations are quite restricted and should be within reach of manual enumeration. :P
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Re: Johnsonian Polytopes

Postby Marek14 » Thu Jan 05, 2012 10:24 pm

quickfur wrote:
Marek14 wrote:How does the augmentation look for the special case of (5,5) duoprism where you can augment prisms in both rings?

I have yet to confirm this, but I believe the 5,5-duoprism should allow all combinations of augments in both rings. A fully augmented 5,5-duoprism should have 20 pentagonal pyramids and 50 square pyramids as cells.

Currently I'm working on enumerating all duoprism augments: for the m<5, n<5 cases the combinations are quite restricted and should be within reach of manual enumeration. :P


I put together a quick Python script for the enumeration (basically, it uses Erastothenes sieve mechanism to eliminate identical combinations).

If you want to use it, change the "duoprism_size" and "path" lines on top to whatever you want :)

Based on results, there should be 14 augments for (5,6), 18 for (5,7), 30 for (5,8), 46 for (5,9) and 78 for (5,10) -- one of them, though, is always the base, unaugmented duoprism.

For (5,5), there is 8 basic augments, which means that if both sets of rings can be augmented independently, there's 8*7/2 = 28 possible augments, one of which is unaugmented (5,5) duoprism.

Code: Select all
// write your own duoprism size and path for this to work
duoprism_size = 7
path = "c:\\programs\\Python32\\"

def fill(listing):
    for i in range(2**duoprism_size):
        listing.append(True)

def reverse(x):
    k = 0
    y = x
    for i in range(duoprism_size):
        if y >= 2**(duoprism_size - 1 - i):
            k = k + 2**(i)
            y = y - 2**(duoprism_size - 1 - i)
    return k

def sieve(listing, x):
    y = x
    z = reverse(x)
    for i in range(duoprism_size):
        if y != x:
            listing[y] = False
        if z != x:
            listing[z] = False
        if y >= 2**(duoprism_size - 1):
            y = (y - 2**(duoprism_size - 1)) * 2 + 1
        else:
            y = y * 2
        if z >= 2**(duoprism_size - 1):
            z = (z - 2**(duoprism_size - 1)) * 2 + 1
        else:
            z = z * 2

def writeout(x):
    y = x
    t = ""
    for i in range(duoprism_size):
        if y >= 2**(duoprism_size - 1 - i):
            t = t + "1"
            y = y - 2**(duoprism_size - 1 - i)
        else:
            t = t + "0"
    return t

outfile = open(path + "augment" + str(duoprism_size) + ".txt", "w")
listing = []
fill(listing)
for run in range(2**duoprism_size):
    if listing[run]:
        sieve(listing, run)
count = 0
for run in range(2**duoprism_size):
    if listing[run]:
        count += 1
        w = writeout(run)
        print(w)
        outfile.write(w + "\n")
outfile.write("\n" + str(count) + "\n")
outfile.close()


EDIT: corrected the algorithm

EDIT2: As the algorithm for enumerating non-adjacent augmentations is very simple, I ran it as well. Here is the algorithm:

Code: Select all
duoprism_size = 20
path = "c:\\programs\\Python32\\"

def fill(listing):
    for i in range(2**duoprism_size):
        listing.append(True)

def reverse(x):
    k = 0
    y = x
    for i in range(duoprism_size):
        if y >= 2**(duoprism_size - 1 - i):
            k = k + 2**(i)
            y = y - 2**(duoprism_size - 1 - i)
    return k

def sieve(listing, x):
    y = x
    z = reverse(x)
    adjacent = False
    for i in range(duoprism_size):
        if y != x:
            listing[y] = False
        if z != x:
            listing[z] = False
        if y >= 2**(duoprism_size - 1):
            y = (y - 2**(duoprism_size - 1)) * 2 + 1
        else:
            y = y * 2
        if z >= 2**(duoprism_size - 1):
            z = (z - 2**(duoprism_size - 1)) * 2 + 1
        else:
            z = z * 2
        if (y + 1) / 4 == (y + 1) // 4:
            adjacent = True
    if adjacent:
        listing[x] = False

def writeout(x):
    y = x
    t = ""
    for i in range(duoprism_size):
        if y >= 2**(duoprism_size - 1 - i):
            t = t + "1"
            y = y - 2**(duoprism_size - 1 - i)
        else:
            t = t + "0"
    return t

outfile = open(path + "augment_na_" + str(duoprism_size) + ".txt", "w")
listing = []
fill(listing)
for run in range(2**duoprism_size):
    if listing[run]:
        sieve(listing, run)
count = 0
for run in range(2**duoprism_size):
    if listing[run]:
        count += 1
        w = writeout(run)
        print(w)
        outfile.write(w + "\n")
outfile.write("\n" + str(count) + "\n")
outfile.close()


And here are the results:

(5,11): 16 augments
(5,12): 26 augments
(5,13): 31 augments
(5,14): 49 augments
(5,15): 64 augments
(5,16): 99 augments
(5,17): 133 augments
(5,18): 209 augments
(5,19): 291 augments
(5,20): 455 augments

Remember that this includes the base duoprism, so the true number of augmented duoprisms is 1 less in each case.
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Re: Johnsonian Polytopes

Postby quickfur » Thu Jan 05, 2012 11:04 pm

OK, I have enumerated all possible augmentations of 3,n-duoprisms.

Some things to note:
  • n-gonal prism pyramids have two different dihedral (dichoral) angles that concern us: (1) the angle of the n-gonal pyramid with the prism, which lies in the plane of the ring of n-gonal prisms in an m,n-duoprism, which I call the "intra-ring angle"; and (2) the angle of the square pyramid with the prism, which interfaces with the other ring, which I call the "cross-ring angle". This distinction is important, since some pyramids are steep in the intra-ring angle but relatively shallow in the cross-ring angle, which makes them behave differently within the ring vs. across the rings.
  • In order for two augments in different rings to be convex, the sum of their cross-ring angles must be ≤90° (since the two rings of a duoprism are at 90° relative to each other, so if the angle sum is > 90° then the augments would be non-convex).
  • It turns out that the cross-ring angle is always ≤45° for our purposes: the triangular prism pyramid (36.7°), cubical pyramid (45°), and pentagonal prism pyramid (13.28°); so we can always augment both rings simultaneously as long as the augments within each ring are valid. (Hexagonal prisms and above don't concern us, since their pyramids cannot be CRF.)
  • For same-ring adjacent augmentations to be convex, they must satisfy the constraint 2*A+B≤180°, where A is the intra-ring angle and B is the internal angle of the n-gon, where n is the number of members in the ring. Solving for n, this works out to be equivalent to: n≤180°/A. For the case of 3,n-duoprisms, the RHS works out to be slightly more than 3 but less than 4, so the 3,3-duoprism is the only case where adjacent augmentations are allowed.
  • For same-ring non-adjacent augmentations to be convex, they must satisfy the constraint A+B≤180°, which works out to be n≤360°/A. For the case of 3,n-duoprisms, the RHS works out to be between 6 and 7, so we can have non-adjacent augments up to the 3,6-duoprism.
  • Due to the symmetry of the duoprism, combinations of augmentations in either ring are completely independent of each other (every cell in one ring is adjacent to all cells in the second ring, so it doesn't matter what permutation of augmentations are on the second ring, all cells in the first ring "see" exactly the same thing). This cuts down on the number of combinations significantly.

So here's the list of 3,n-duoprism augmentations:

1) 3,3-duoprism:
- 1 augment
- 2 augments, same ring
- 1+1 augments (1 on each ring)
- 3 augments, same ring
- 2+1 augments (2 on one ring, 1 on the other)
- 3+1 augments
- 2+2 augments
- 3+2 augments
- 3+3 augments (fully augmented 3,3-duoprism)
Total: 9 augmentations

From here on, I'll adopt the convention of writing the augments as x+y, where x is the number of augments on the first ring (3 n-prisms) and y is the number of augments on the second ring (n triangular prisms). (This wasn't necessary for the 3,3-duoprism since the two rings were identical.)

2) 3,4-duoprism:
From here on, the triangular prisms can no longer have adjacent augments. So the only possible augmentations on the second 4-membered ring is no augment, 1 augment, or two antipodal augments. The 3 cubes in the first ring can have any combination of augments. So:
- 0+1 augments
- 1+0 augments
- 0+2 augments
- 1+1 augments
- 2+0 augments
- 1+2 augments
- 2+1 augments
- 3+0 augments (note: 0+3 not possible)
- 3+1 augments
- 2+2 augments
- 3+2 augments
Total: 11 augmentations

3,5-duoprism:
Again, only non-adjacent augments allowed for the second ring. A pentagon can only accomodate up to 2 non-adjacent augments, so that's as high as it can go here. The first ring can have any combination of augments.
- So this means it's exactly the same list of augmentations as the 3,4-duoprism, so I'm omitting it to prevent boring y'all with redundant information.
Total: 11 augmentations

3,6-duoprism:
Here, no augmentations are possible in the first ring because they are hexagonal prisms. Augmentations in the second ring must be non-adjacent, and furthermore augmentations in a 6-membered ring can be spaced 1 cell apart (meta) or 2 cells apart (para). So:
- 0+1 augment
- 0+2 augments, meta
- 0+2 augments, para
- 0+3 augments, meta (para not possible with ≥ 3 augments)
Total: 4 augmentations

NOTE: 3,7-duoprisms and above cannot be augmented, because the intra-ring angle of the pyramids is too large to remain convex in rings of 7 members or greater.

So the grand total of 3,n-duoprism augmentations is 35 (39 if you count the non-augmented duoprisms themselves).

Next up: 4,n-duoprisms.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 06, 2012 12:19 am

Alright, here are the 4,n-duoprism augments:

4,4-duoprism (aka tesseract)
The tesseract can be decomposed into two identical rings of cubes, so the number of combinations is actually less than it might seem at first glance (the same configuration of augments on one ring maps to the other ring via a symmetry operation). In each ring, you can have either adjacent (a) or antipodal (p) augments. Both rings can be augmented independently. In addition, the extra symmetry of the tesseract collapses quite a few cases into identical augmentations, further reducing the number of combinations.
- 1 augment
- (1+1 augments -- IDENTICAL TO 2a due to extra symmetry of tesseract)
- 2a augments (2a = 2 adjacent augments on the same ring)
- 2p augments (2p = 2 antipodal augments on the same ring)
- 3 augments (same ring)
- 1+2a augments
- (1+2p augments -- IDENTICAL TO 3 augments due to extra symmetry of tesseract)
- 1+3 augments
- (2a+2a augments -- IDENTICAL TO 1+3)
- 2a+2p augments
- (2p+2p augments -- IDENTICAL TO 4)
- 4 augments
- 4+1 augments
- 2a+3 augments
- (2p+3 augments -- IDENTICAL TO 4+1)
- (2a+4 augments -- IDENTICAL TO 3+3)
- 2p+4 augments
- 3+3 augments
- 3+4 augments
- 4+4 augments == 24-cell
Total: 13 (+1 if you include the 24-cell).

4,5-duoprism:
No adjacent augments allowed in the second ring; a 5-membered ring allows only 0, 1, or 2 augments.
All augmentations allowed in first ring. A 4-membered has either adjacent (a) or antipodal (p) augments.
So:
- 0+1 augments
- 1+0 augments
- 0+2 augments
- 1+1 augments
- 2a+0 augments
- 2p+0 augments
- 1+2 augments
- 2a+1 augments
- 2p+1 augments
- 3+0 augments
- 2a+2 augments
- 2p+2 augments
- 3+1 augments
- 4+0 augments
- 3+2 augments
- 4+1 augments
- 5+0 augments
- 5+1 augments
- 5+2 augments
Total: 20 augmentations

4,6-duoprism:
Only the second ring can take augmentations (first ring has hexagonal prisms; no CRF pyramid possible). All augments must be non-adjacent. A 6-membered ring can have meta (m) or para (p) augments (cf. benzene substitution positions).
- 1 augment
- 2m augments
- 2p augments
- 3 augments
Total: 4 augmentations

4,7-duoprism:
Only the 2nd ring can be augmented; all augments must be non-adjacent. The 2-augment case has meta/para "isomers". The 3-augment case is always meta-meta-para, so it only produces 1 combination:
- 1 augment
- 2m augments
- 2p augments
- 3 augments
Total: 4 augmentations

4,8-duoprism:
An 8-membered ring can have meta (m: 1 cell between), para (p: 2 cells between), or antipodal (x: 3 cells between) augments. The 3-augment case has two distinct isomers: meta-meta (mm: 3 of the 4 vertices of an inscribed square) or meta-para (mp: 2 augments meta to each other, the 3rd para to both).
So:
- 1 augment
- 2m augments
- 2p augments
- 2x augments
- 3mm augments
- 3mp augments
- 4 augments
Total: 7 augmentations.

NOTE: 4,9-duoprisms and above cannot be augmented, because the cubical pyramids will be nonconvex.

Grand total: 13 + 20 + 4 + 4 + 7 = 48 augmentations (49 if you include the 24-cell).
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 06, 2012 12:33 am

Marek14 wrote:[...]
For (5,5), there is 8 basic augments, which means that if both sets of rings can be augmented independently, there's 8*7/2 = 28 possible augments, one of which is unaugmented (5,5) duoprism.
[...]

Oh? I enumerated 35 for the 5,5-duoprism. Did I make a mistake? Here's my list: (2a = 2 adjacent augments in 1 ring; 2m = 2 non-adjacent augments; 3aa = 3 augments in a row; 3am = 2 augments in a row plus 1 one space away; x+y means x augmentations in one ring, y augmentations in other ring):
Code: Select all
   - 1 augment
   - 2a augments
   - 2m augments
   - 1+1 augments
   - 3aa augments
   - 3am augments
   - 2a+1 augments
   - 2m+1 augments
   - 4 augments
   - 3aa+1 augments
   - 3am+1 augments
   - 2a+2a augments
   - 2a+2m augments
   - 2m+2m augments
   - 5 augments
   - 4+1 augments
   - 3aa+2a augments
   - 3aa+2m augments
   - 3am+2a augments
   - 3am+2m augments
   - 5+1 augments
   - 4+2a augments
   - 4+2m augments
   - 3aa+3aa augments
   - 3aa+3am augments
   - 3am+3am augments
   - 5+2a augments
   - 5+2m augments
   - 4+3aa augments
   - 4+3am augments
   - 5+3aa augments
   - 5+3am augments
   - 4+4 augments
   - 5+4 augments
   - 5+5 augments
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Re: Johnsonian Polytopes

Postby Marek14 » Fri Jan 06, 2012 6:44 am

quickfur wrote:
Marek14 wrote:[...]
For (5,5), there is 8 basic augments, which means that if both sets of rings can be augmented independently, there's 8*7/2 = 28 possible augments, one of which is unaugmented (5,5) duoprism.
[...]

Oh? I enumerated 35 for the 5,5-duoprism. Did I make a mistake? Here's my list: (2a = 2 adjacent augments in 1 ring; 2m = 2 non-adjacent augments; 3aa = 3 augments in a row; 3am = 2 augments in a row plus 1 one space away; x+y means x augmentations in one ring, y augmentations in other ring):
Code: Select all
   - 1 augment
   - 2a augments
   - 2m augments
   - 1+1 augments
   - 3aa augments
   - 3am augments
   - 2a+1 augments
   - 2m+1 augments
   - 4 augments
   - 3aa+1 augments
   - 3am+1 augments
   - 2a+2a augments
   - 2a+2m augments
   - 2m+2m augments
   - 5 augments
   - 4+1 augments
   - 3aa+2a augments
   - 3aa+2m augments
   - 3am+2a augments
   - 3am+2m augments
   - 5+1 augments
   - 4+2a augments
   - 4+2m augments
   - 3aa+3aa augments
   - 3aa+3am augments
   - 3am+3am augments
   - 5+2a augments
   - 5+2m augments
   - 4+3aa augments
   - 4+3am augments
   - 5+3aa augments
   - 5+3am augments
   - 4+4 augments
   - 5+4 augments
   - 5+5 augments


Nah, your list is correct. I used wrong triangle number so I didn't count the situations with same augments on both rings.
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Re: Johnsonian Polytopes

Postby wintersolstice » Fri Jan 06, 2012 1:21 pm

I did a caculation on the 3,(3/4) - duoprisms and realised you can't augment both girdles at the same time! the angle in the triangular prism pyramid (for the square pyramids) is actually about 66 degrees not 36.7! it makes sense because as the number of sides increase the angle decreases

You can however augment both girdles of the 3,5 - duoprism

btw that's just my working out (I could have gone wrong :D )

the workings out I used on the pentagonal case should indicate how I did it :D (it was more or less the same)

also the elongated triangular and pentagonal (bi)prism pyramid account for 4 of the aumented duoprisms since the 4,n - duoprisms are prism prisms :D

second you didn't need to do the tesseract I had already done it (in a earlier post) complete with names for them all. :D

I could if anyone os interest give names to all these augementations:D Do you think it could be a good idea? (complete with my own notation for each girdle. I have more prefixes than just "para" and "meta"

btw I say when I say "girdle" not "ring" in case your wondering what I've said :D
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 06, 2012 3:57 pm

wintersolstice wrote:I did a caculation on the 3,(3/4) - duoprisms and realised you can't augment both girdles at the same time! the angle in the triangular prism pyramid (for the square pyramids) is actually about 66 degrees not 36.7! it makes sense because as the number of sides increase the angle decreases

I must be getting old. :oops: I don't know how I got the 36.7 figure... I must've been half-asleep or something when I did that. I just checked again, you're right that the actual cross-ring angle is about 65.91°, that is, arctan(sqrt(5)). Oops. So now I have to go back and revise my enumerations.

Actually, I really should start double-checking all my calculations from now on... i seem to be making little slip ups all the time.

Or perhaps I should be actually rendering these things... nothing dispels mistaken assumptions more than the convex hull algo producing the wrong shape when you give it wrong coordinates. Laziness is my downfall. :mrgreen:
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 06, 2012 4:33 pm

OK, the only needed revisions were for the 3,3-duoprism and the 3,4-duoprism, which cannot take augments in both rings at the same time. So that reduces their augmentation counts to 3 and 5, respectively, which makes the correct total number of augmentations of 3,n-duoprisms 23, not 35.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 06, 2012 8:37 pm

Marek14 wrote:[...] Based on results, there should be 14 augments for (5,6), 18 for (5,7), 30 for (5,8), 46 for (5,9) and 78 for (5,10) -- one of them, though, is always the base, unaugmented duoprism.

OK, just to make sure that we have accurate counts here, I wrote my own program to compute the counts, based on a slightly different method (explained below). I can vouch for all your counts except for (5,6), for which I got 13. Was that a typo, or perhaps my program has a bug? :)

[...]
(5,11): 16 augments
(5,12): 26 augments
(5,13): 31 augments
(5,14): 49 augments
(5,15): 64 augments
(5,16): 99 augments
(5,17): 133 augments
(5,18): 209 augments
(5,19): 291 augments
(5,20): 455 augments

Remember that this includes the base duoprism, so the true number of augmented duoprisms is 1 less in each case.

All these counts are confirmed by my program.

Here's the algorithm I used: given an n-membered ring, I represent a particular augmentation by an n-digit binary number where 1=augmented, 0=not. To enumerate all augmentations, simply loop from 0 to 2n-1. This, of course, produces many augmentations that are equivalent under rotation and reflection. To count only unique configurations, I noted that we can divide the numbers from 0 to 2n-1 into equivalence classes based on whether they can be mapped to each other by some sequence of rotations/reflections. For each equivalence class, I choose the lowest member as its "canonical" member. So given a particular number representing an augmentation, I perform n rotations on it, and if at any time I get an augmentation numerically less than the original, then it cannot be canonical, so I discard it. After testing all n rotated copies, I mirror-image it and repeat the test. In this way, I can short-circuit the test for many of the numbers. If no smaller number is found via rotation/reflection, then it's proven that it's the smallest member of its equivalence class, so we count it.

Using this scheme, each unique augmentation pattern can be assigned a unique odd number (an even number can always be rotated right by 1 bit to make a smaller number, so all even numbers are non-canonical), that describes the exact pattern of augmentations on the n-ring. This makes it possible for us assign all duoprism augmentations a unique, canonical identifier in a consistent way, that also allows mechanical retrieval of structure. For example, we can name an m,n-duoprism augmented by pattern x and pattern y on each respective ring, as "x,y-augmented m,n-duoprism", where x and y are the canonical numbers for the respective augmentation pattern. The only problem is that for large 5,n-duoprisms these numbers get quite large and ugly, so we probably still want to have alternative, human-pronunciable names for them. :P
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Re: Johnsonian Polytopes

Postby Marek14 » Fri Jan 06, 2012 9:15 pm

quickfur wrote:
Marek14 wrote:[...] Based on results, there should be 14 augments for (5,6), 18 for (5,7), 30 for (5,8), 46 for (5,9) and 78 for (5,10) -- one of them, though, is always the base, unaugmented duoprism.

OK, just to make sure that we have accurate counts here, I wrote my own program to compute the counts, based on a slightly different method (explained below). I can vouch for all your counts except for (5,6), for which I got 13. Was that a typo, or perhaps my program has a bug? :)

[...]
(5,11): 16 augments
(5,12): 26 augments
(5,13): 31 augments
(5,14): 49 augments
(5,15): 64 augments
(5,16): 99 augments
(5,17): 133 augments
(5,18): 209 augments
(5,19): 291 augments
(5,20): 455 augments

Remember that this includes the base duoprism, so the true number of augmented duoprisms is 1 less in each case.

All these counts are confirmed by my program.

Here's the algorithm I used: given an n-membered ring, I represent a particular augmentation by an n-digit binary number where 1=augmented, 0=not. To enumerate all augmentations, simply loop from 0 to 2n-1. This, of course, produces many augmentations that are equivalent under rotation and reflection. To count only unique configurations, I noted that we can divide the numbers from 0 to 2n-1 into equivalence classes based on whether they can be mapped to each other by some sequence of rotations/reflections. For each equivalence class, I choose the lowest member as its "canonical" member. So given a particular number representing an augmentation, I perform n rotations on it, and if at any time I get an augmentation numerically less than the original, then it cannot be canonical, so I discard it. After testing all n rotated copies, I mirror-image it and repeat the test. In this way, I can short-circuit the test for many of the numbers. If no smaller number is found via rotation/reflection, then it's proven that it's the smallest member of its equivalence class, so we count it.

Using this scheme, each unique augmentation pattern can be assigned a unique odd number (an even number can always be rotated right by 1 bit to make a smaller number, so all even numbers are non-canonical), that describes the exact pattern of augmentations on the n-ring. This makes it possible for us assign all duoprism augmentations a unique, canonical identifier in a consistent way, that also allows mechanical retrieval of structure. For example, we can name an m,n-duoprism augmented by pattern x and pattern y on each respective ring, as "x,y-augmented m,n-duoprism", where x and y are the canonical numbers for the respective augmentation pattern. The only problem is that for large 5,n-duoprisms these numbers get quite large and ugly, so we probably still want to have alternative, human-pronunciable names for them. :P


Actually, from the description, you used the same algorithm as me, except that I make reflection right away, and check the rotations of the original and the reflection simultaneously.

Checking the (5,6)-duoprism, I found there was a bug from old version of the program which occasionally didn't sieve out the reflection. The correct number is indeed 13: 1 without augmentation, 1 monoaugmented, 3 biaugmented, 3 triaugmented, 3 tetraaugmented, 1 pentaaugmented and 1 hexaaugmented.

The use of programs reminds me of the time when I was studying uniform hyperbolic tilings :) I made programs at that time to assemble them.
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Re: Johnsonian Polytopes

Postby quickfur » Fri Jan 06, 2012 11:05 pm

quickfur wrote:[...]
4,5-duoprism:
No adjacent augments allowed in the second ring; a 5-membered ring allows only 0, 1, or 2 augments.
All augmentations allowed in first ring. A 4-membered has either adjacent (a) or antipodal (p) augments.
So:
- 0+1 augments
- 1+0 augments
- 0+2 augments
- 1+1 augments
- 2a+0 augments
- 2p+0 augments
- 1+2 augments
- 2a+1 augments
- 2p+1 augments
- 3+0 augments
- 2a+2 augments
- 2p+2 augments
- 3+1 augments
- 4+0 augments
- 3+2 augments
- 4+1 augments
- 5+0 augments
- 5+1 augments
- 5+2 augments
Total: 20 augmentations
[...]

Caught another error: the first ring has only 4 members! So the last 3 augmentations are invalid; the correct count is 17.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 07, 2012 1:35 am

While we're on the topic of augmentations, I wondered if the 24-cell can be augmented. So I calculated the dihedral angle of the 24-cell to be approx 97.18°, and the tetrahedron-octahedron dihedral angle of the octahedral pyramid to be approx 54.74°. Since their sum is < 90° (provided I didn't make another stupid mistake like I seem to be consistently doing recently), this means a 24-cell augmented by octahedral pyramids will be convex.

However, adjacent augments are not allowed, because 2*54.74° + 97.18° > 180°, so the result would be non-convex.

Anyone care to enumerate all possible 24-cell augmentations? ;)

Also on the topic of augmentation, not all non-convex augmentations are non-CRF; consider the grand antiprism, for example. Erecting pyramids on its pentagonal antiprism cells actually makes it non-convex, but it just so happens that two adjacent pyramids have their apices exactly one edge-length apart, so we insert an edge between the apices and fill in the non-convex gap between the pyramids with a ring of 5 tetrahedra, thus forming a part of the 600-cell's surface.

So we shouldn't throw out non-convex augmentations on sight, since some of them might actually lead to valid CRFs!

For duoprisms, however, this case is not possible because the width of the prism is equal to the edge length, so it's impossible for any pyramidal augments to be exactly one edge-length apart except in the case of the infinite duoprism, which is not valid in the first place. It works for the grand antiprism because the height of a pentagonal antiprism is quite short compared to its edge length, so it's possible to join two augments by an edge. It may be possible to do this with rings of other antiprisms too, but then we'll have to worry about how the margins of the ring of antiprisms can be filled up in a CRF-compatible way. It may not be possible. But it might be worth a look.
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Jan 07, 2012 11:16 am

Hmm, if I remember correctly, the vertices of a 24-cell can be expressed as (2,0,0,0) and all permutations and sign changes of it (8 vertices) plus (1,1,1,1) and all sign changes (16 vertices) with edge length 2, edges being of type (2,0,0,0)-(1,1,1,1) or (1,1,1,1)-(-1,1,1,1).

Since 24-cell is self-dual, we can use the same structure for cells/faces.

We should also notice that there are three tesseractic-symmetry groups of cells embedded in the structure. One of them are orthogonal faces ((2,0,0,0) and its ilk), second is (+-1,+-1,+-1,+-1) with even sign parity, third is the same, but with odd sign parity.

Let's start by augmenting the "top" face, (2,0,0,0). This eliminates all faces (+1,+-1,+-1,+-1), so next face to augment must be (0,<+-2,0,0>) [<+-2,0,0> indicates permutation], (-1,+-1,+-1,+-1) or (0,0,0,-2).

Let's make a handy table:
Fixed: (2,0,0,0)
Possible 1: (0,2,0,0),(0,-2,0,0),(0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (first "ring", the same group as first cell)
Possible 2: (-1,1,1,1),(-1,-1,1,1),(-1,1,-1,1),(-1,-1,-1,1),(-1,1,1,-1),(-1,-1,1,-1),(-1,1,-1,-1),(-1,-1,-1,-1) (second "ring", different groups than first cell)
Possible 3: (-2,0,0,0) (antipode of first cell)

This leads to three biaugmented 24-cells:

1: orthobiaugmented 24-cell
Fixed: (2,0,0,0),(0,2,0,0)
Possible 1: (0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to both fixed cells)
Possible 2: (-1,-1,1,1),(-1,-1,-1,1),(-1,-1,1,-1),(-1,-1,-1,-1) (all remaining cells in other groups, in "meta" position to both fixed cells)
Possible 3: (-2,0,0,0),(0,-2,0,0) (antipodes to two fixed cells, "ortho" position to one and "para" to other)

2: metabiaugmented 24-cell:
Fixed: (2,0,0,0),(-1,1,1,1)
Possible 1: (0,-2,0,0),(0,0,-2,0),(0,0,0,-2),(-1,1,-1,-1),(-1,-1,1,-1),(-1,-1,-1,1) (cells in "ortho" position to one fixed cell and "meta" to the other)
Possible 2: (-1,-1,-1,-1) (sort of "antipode" to the particular "meta" combination, it has "meta" position to both fixed cell)

3: parabiaugmented 24-cell:
Fixed: (2,0,0,0),(-2,0,0,0)
Possible: (0,2,0,0),(0,-2,0,0),(0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (the remaining group of the fixed cells, all in "ortho" position to both)

This leads to 4 possible triaugments:

1: tesedge-triaugmented 24-cell (ooo)
Name explanation: it augments three cells in one tesseractic group around the "edge" of this fictious tesseract.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to all three fixed cells)
Possible 2: (-1,-1,-1,1),(-1,-1,-1,-1) (all remaining cells in other groups, in "meta" position to all three fixed cells)
Possible 3: (-2,0,0,0),(0,-2,0,0),(0,0,-2,0) (antipodes to the three fixed cells, "ortho" position to two of them and "para" to the third)

2: assymetrical triaugmented 24-cell (omm)
Name explanation: it's the least symmetrical of the four.
Fixed: (2,0,0,0),(0,2,0,0),(-1,-1,1,1)
Possible 1: (0,0,-2,0),(0,0,0,-2) ("ortho" position to two fixed cells, "meta" to one)
Possible 2: (-1,-1,-1,-1) ("ortho" position to one cell, "meta" to two)

3: arc-triaugmented 24-cell (oop)
Name explanation: The three augmented cells lie on the same tesseract and form an arc of two antipodes and one lateral cell
Fixed: (2,0,0,0),(0,2,0,0),(-2,0,0,0)
Possible 1: (0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to all three fixed cells)
Possible 3: (0,-2,0,0) (antipode to the unpaired cell, "ortho" to two fixed cell and "para" to the third)

4: symmetrical triaugmented 24-cell: (mmm)
Name explanation: the most symmetrical triaugmentation using one augment from each tesseractic group
Fixed: (2,0,0,0),(-1,1,1,1),(-1,-1,-1,-1)
This augment cannot be augmented any further.

Now for tetraaugments. There is 5 of those:

1: tesvertex-tetraaugmented 24-cell
Name explanation: it augments four cells in one tesseractic group around the "vertex" of this fictious tesseract.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2)
Possible 1:(-2,0,0,0),(0,-2,0,0),(0,0,-2,0),(0,0,0,-2) (antipodes to the four fixed cells, "ortho" position to three of them and "para" to the fourth)
Possible 2:(-1,-1,-1,-1) (the only cell in other group, in "meta" position to all four fixed cells)

2: edge+1-tetraaugmented 24-cell
Name explanation: it augments three cells in one tesseractic group around the "edge" of this fictious tesseract plus one cell from another group.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-1,-1,-1,1)
Possible:(0,0,0,-2) (the only possible augmentation cell, "ortho" position to three fixed cells and "meta" to the fourth)

3: tesassymetrical-tetraaugmented 24-cell
Name explanation: it augments four cells in one tesseractic group in "assymetrical" configuration as opposed to "vertex" or "ring" configuration
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-2,0,0,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (two cells in "ortho" position to all four fixed cells)
Possible 2: (0,-2,0,0),(0,0,-2,0) (antipodes to two of the fixed cells, "ortho" position to three fixed cells and "para" to the fourth)

4: skewed tetraaugmented 24-cell
Name explanation: augments two and two cells in two different tesseractic groups
Fixed: (2,0,0,0),(0,2,0,0),(-1,-1,1,1),(-1,-1,-1,-1)
Cannot be augmented any further.

5: ring-triaugmented 24-cell
Name explanation: The four augmented cells lie on the same tesseract and form a ring aroung it.
Fixed: (2,0,0,0),(0,2,0,0),(-2,0,0,0),(0,-2,0,0)
Possible: (0,0,2,0),(0,0,-2,0),(0,0,0,2),(0,0,0,-2) (four cells in "ortho" position to all four fixed cells)

Now for pentaaugmentations. There are 3 of them.

1: tesantiedge-pentaaugmented 24-cell
Name explanation: it augments five cells in one tesseractic group such that three cells around one "edge" remains unaugmented.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0)
Possible:(0,-2,0,0),(0,0,-2,0),(0,0,0,-2) (antipodes to three unpaired fixed cells, "ortho" position to four fixed cells and "para" to the fifth)

2: vertex+1-pentaaugmented 24-cell
Name explanation: it augments four cells in one tesseractic group around the "vertex" of this fictious tesseract plus one cell in another group.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-1,-1,-1,-1)
Cannot be augmented any further.

3: tesantiarc-pentaaugmented 24-cell
Name explanation: it augments five cells in one tesseractic group such that the unaugmented cells form an "arc".
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-2,0,0,0),(0,-2,0,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (two cells in "ortho" position to all five fixed cells)
Possible 2: (0,0,-2,0) (antipode to the only unpaired cell, "ortho" position to four fixed cells and "para" to the fifth)

Now for hexaaaugmentations. There are 2 of them.

1: tesantiface-hexaaugmented 24-cell
Name explanation: it augments six cells in one tesseractic group such that two cells around one "face" remains unaugmented.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0),(0,-2,0,0)
Possible: (0,0,-2,0),(0,0,0,-2) (antipodes to two unpaired fixed cells, "ortho" position to five fixed cells and "para" to the sixth)

2: tesantipode-hexaaugmented 24-cell
Name explanation: it augments six cells in one tesseractic group such that the unaugmented cells are antipodal.
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(-2,0,0,0),(0,-2,0,0),(0,0,-2,0)
Possible 1:(0,0,0,2),(0,0,0,-2) (two cells in "ortho" position to all six fixed cells)

Now for the only possible heptaaugmentation:

1: heptaaugmented 24-cell
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0),(0,-2,0,0),(0,0,-2,0)
Possible: (0,0,0,-2) (antipodes to the only unpaired fixed cell, "ortho" position to six fixed cells and "para" to the seventh)

And the only octaaugmentation:
1: octaaugmented 24-cell
Fixed: (2,0,0,0),(0,2,0,0),(0,0,2,0),(0,0,0,2),(-2,0,0,0),(0,-2,0,0),(0,0,-2,0),(0,0,0,-2)
Cannot be augmented any further.

So, for number of augments:
1 - 1
2 - 3
3 - 4
4 - 5
5 - 3
6 - 2
7 - 1
8 - 1

total: 20
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 07, 2012 4:01 pm

Marek14 wrote:Hmm, if I remember correctly, the vertices of a 24-cell can be expressed as (2,0,0,0) and all permutations and sign changes of it (8 vertices) plus (1,1,1,1) and all sign changes (16 vertices) with edge length 2, edges being of type (2,0,0,0)-(1,1,1,1) or (1,1,1,1)-(-1,1,1,1).
[...]
So, for number of augments:
1 - 1
2 - 3
3 - 4
4 - 5
5 - 3
6 - 2
7 - 1
8 - 1

total: 20

Cool, thanks! I've added this info to the CRF wiki page.

On another note, I've just realized that even though we can't augment 6,n-duoprisms with pyramids because hexagonal prisms can't have CRF pyramids, we can augment them with one of the Klietzing segmentotopes, namely triangle||hexagonal_prism, which consists of a hexagonal prism base, two triangular cupolae that meet at their triangular faces, and a ring of square pyramids and triangular prisms. Similarly, 8,n-duoprisms should be augmentable by square||octagonal_prism, and 10,n-duoprisms by pentagon||decagonal prism.

So there are yet more augmented duoprisms that we haven't considered!!
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Jan 07, 2012 4:53 pm

quickfur wrote:So there are yet more augmented duoprisms that we haven't considered!!


So the next step would be to check out the dichoral angles and then the same program I used for (5,n) duoprisms could be also used for these. And we need to keep eyes for possible double-ring augments on (5,6), (5,8), (5,10), (6,6), (6,8), (6,10), (8,8), (8,10) and (10,10)!
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 07, 2012 5:18 pm

Marek14 wrote:
quickfur wrote:So there are yet more augmented duoprisms that we haven't considered!!


So the next step would be to check out the dichoral angles and then the same program I used for (5,n) duoprisms could be also used for these. And we need to keep eyes for possible double-ring augments on (5,6), (5,8), (5,10), (6,6), (6,8), (6,10), (8,8), (8,10) and (10,10)!

Keep in mind that augmenting with n-gon||2n-prisms have orientation to account for; for example, when augmenting an 8,n-duoprism with two square||octagonal prisms, the augments can either have the square in the same orientation, or in dual orientations. But if there's only a single augment, then the orientation doesn't matter. So you may not be able to reuse your code that easily (same goes for my version of the program). Also, this means that having an augment in one ring imposes an orientation on the other ring too, so that two augments on different rings can either be lined up by the square edges, or have square edge matching square vertex.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 07, 2012 5:39 pm

Actually, it's even more subtle than that. When you have two augments on different rings, call them A1 and A2, then A1 could be positioned relative to A2's square's vertex, or A2's square's edge, and for each of these positions, it could be oriented so that its square has vertex lined with with A2 or edge lined up with A2. So you have 3 distinct configurations: vertex-vertex, vertex-edge, or edge-edge.

So more interesting combinations are possible than with simple pyramidal augments. Whew! Duoprisms could keep us occupied for a while yet. ;)
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Re: Johnsonian Polytopes

Postby Marek14 » Sat Jan 07, 2012 5:53 pm

quickfur wrote:Actually, it's even more subtle than that. When you have two augments on different rings, call them A1 and A2, then A1 could be positioned relative to A2's square's vertex, or A2's square's edge, and for each of these positions, it could be oriented so that its square has vertex lined with with A2 or edge lined up with A2. So you have 3 distinct configurations: vertex-vertex, vertex-edge, or edge-edge.

So more interesting combinations are possible than with simple pyramidal augments. Whew! Duoprisms could keep us occupied for a while yet. ;)


Well, it could be resolved simply by using ternary system (0 - no augmentation, 1 and 2 - various augmentation orientations0 instead of binary.
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 07, 2012 6:35 pm

Alright, here's my first stab at this.

First, triangle||hexagonal_prism is essentially an expanded 3-prism pyramid (expand the triangles of the 3-prism outwards, turning it into a 6-prism, and the apex turns into a triangle), so the intra-ring dichoral angle is the same as the latter: atan(5/3) ~= 52.23°. (This can be verified by manual calculation of coordinates for triangle||hexagonal_prism.)

It also adds another complication: it has two distinct cross-ring angles, one for its square pyramid cells (for vertex-aligned cross-ring augs), which is about 57.69°, and one for its triangular prism cells (for edge-aligned cross-ring augs), which is about 65.91° -- same as cross-ring angle for the 3-prism pyramid.

Because of its large intra-ring dichoral angle, adjacent augments are only allowed for the 3,6-duoprism, and non-adjacent augments are allowed up to the 6,6-duoprism. Sadly, the 6,6-duoprism doesn't allow cross-ring augments because both the vertex-aligned and edge-aligned angle sums are > 90°.

However, the 5,6-duoprism does allow simultaneous augmentations with 5-prism pyramids and triangle||6-prism: vertex-aligned augments have cross-ring angle sum 70.97° and edge-aligned augments have cross-ring sum 79.19°, so both are allowed. A lot of combinations are possible: for example, if we have one triangle||6-prism augment, then we can add a 5-prism pyramid in the other ring either vertex-aligned, or edge-aligned. So this increases the number of distinct combinations of 5-prism pyramid augmentations in the other ring.

Furthermore, for two triangle||6-prism augments on the first ring, we have two possible orientations, either ortho (o) or gyro (g).

Sadly, 6,n-duoprisms larger than n=6 cannot be augmented, so potentially very numerous combinations like the 6,10-duoprism are excluded. :( And the only 6,n-duoprism that can take cross-ring augments is the 6,5-duoprism; all others have their cross-ring angle sums > 90°. So this is the only really interesting case of the 6,n-duoprisms.

I haven't looked at 8,n-duoprisms and 10,n-duoprisms yet, but those do look more promising. :)
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Re: Johnsonian Polytopes

Postby quickfur » Sat Jan 07, 2012 6:37 pm

Marek14 wrote:
quickfur wrote:Actually, it's even more subtle than that. When you have two augments on different rings, call them A1 and A2, then A1 could be positioned relative to A2's square's vertex, or A2's square's edge, and for each of these positions, it could be oriented so that its square has vertex lined with with A2 or edge lined up with A2. So you have 3 distinct configurations: vertex-vertex, vertex-edge, or edge-edge.

So more interesting combinations are possible than with simple pyramidal augments. Whew! Duoprisms could keep us occupied for a while yet. ;)


Well, it could be resolved simply by using ternary system (0 - no augmentation, 1 and 2 - various augmentation orientations0 instead of binary.

True. Although things get quite hairy when you include pyramid augmentations. (See my previous post.. can't believe i spend almost an hour writing that.)
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