## Generated lace-towers for the .5.3.3.-family

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### Generated lace-towers for the .5.3.3.-family

Hello,
Recently I wrote a program that made some calculations on the .5.3.3.-symmetry, to estimate an upper bound for the 600-cell diminishings.
I just realized this could could well be used to generate lace-towers for the polytopes of this group.
Here is the c++ source code of the program:
Code: Select all
`#include <iostream>#include <fstream>using namespace std;class fint{//integers of the ring ax+bf, where x=1 and f=phi, the golden ratio.    public:    short int x;    short int f;    fint(){x=0;f=0;}    fint(int sx){x=sx;f=0;}    fint(int sx,int sf){x=sx;f=sf;}    fint operator*(fint& b){        fint r;        r.x=this->f*b.f+this->x*b.x;        r.f=this->f*(b.f+b.x)+this->x*b.f;        return r;    }    bool operator==(fint x){return(x.x==this->x&&x.f==this->f);}    bool operator!=(fint x){return(x.x!=this->x||x.f!=this->f);}    fint operator=(const int& x){fint r;r.x=x;return r;}    fint operator=(const fint& z){x=z.x;f=z.f;return z;}    fint operator+(fint b){        fint r;        r.x=this->x+b.x;        r.f=this->f+b.f;        return r;    }};const fint o(0,0);//0const fint x(1,0);//1const fint y(-1,0);//-1const fint f(0,1);//phi/*int orders[30];//the number of occurrence for every orbitint vertes[120];//a group-element for every vertexfint vertices[120][4];//how the bottom row of the matrix of a group element looks like, for every vertex.*/class fintmatrix{//the group elements: orthogonal matrices in R^4 (or actually not R, but Z+fZ)    public:    fint mat[4][4];    short vertex;    int det;    short order;    bool operator==(fintmatrix mat){for(int i=0;i<4;i++)for(int j=0;j<4;j++)if(mat.mat[i][j]!=this->mat[i][j])return false;return true;}    fintmatrix(){for(int i=0;i<4;i++)for(int j=0;j<4;j++)mat[i][j].x=(i==j);det=1;}    explicit fintmatrix(int sdet){for(int i=0;i<4;i++)for(int j=0;j<4;j++)mat[i][j]=(i==j);det=sdet;}    fintmatrix operator*(fintmatrix right){        fintmatrix r;for(int i=0;i<4;i++)for(int j=0;j<4;j++){            fint q;            for(int k=0;k<4;k++)q=q+this->mat[i][k]*right.mat[k][j];            r.mat[i][j]=q;            }r.det=this->det*right.det;        return r;    }};class thegroup{//the group itself.    public:    fintmatrix group[14400];    int last;    thegroup(){last=5;        group[1].det=-1;        group[1].mat[0][0]=y;        group[1].mat[0][1]=f;        group[2].det=-1;        group[2].mat[1][0]=f;        group[2].mat[1][1]=y;        group[2].mat[1][2]=x;        group[3].det=-1;        group[3].mat[2][1]=x;        group[3].mat[2][2]=y;        group[3].mat[2][3]=x;        group[4].det=-1;        group[4].mat[3][2]=x;        group[4].mat[3][3]=y;        for(int k=0;k<5;k++)            for(int i=0;i<4;i++){                for(int j=0;j<4;j++){                    cout<<group[k].mat[i][j].x<<',';                    }cout<<endl;            }cout<<endl;    }    int isthere(fint mat[][4]){        int i=last;        bool joepie;        do{joepie=true;            for(int k=0;k<4;k++)                for(int j=0;j<4;j++)                    if(group[i].mat[k][j]!=mat[k][j])                        joepie=false;            if(joepie)                return i;            i--;        }while(i>=0);return -1;}    void add(fintmatrix& mat){if(last>=14400)cout<<"help :'-(";group[last]=mat;last++;}};void ab(int a, int b, int q, thegroup& group){//doing group-operation a and b consecutively.    fintmatrix mat=group.group[q];    do{mat=group.group[a]*mat;        if(group.isthere(mat.mat)==-1){            group.add(mat);        }        mat=group.group[b]*mat;        if(group.isthere(mat.mat)==-1){            group.add(mat);        }    }while(!(mat==group.group[q]));}/*void giveorders(int q, thegroup& group){//calculate the orbit of group-operation q.    fintmatrix mat;    fintmatrix id;    int i=0;    do{mat=group.group[q]*mat;        i++;    }while(!(mat==id));    group.group[q].order=i;    cout<<i;orders[i]++;}*/void makegroup(thegroup& group){//make all elements of the group.    ofstream output("C:/Users/Wout/Documents/group.txt",ios::out);    for(int i=1;i<14400;i++){        ab(1,2,i,group);        ab(1,3,i,group);        ab(2,3,i,group);        ab(1,4,i,group);        ab(2,4,i,group);        ab(3,4,i,group);        if(!(i%100))            cout<<i<<','<<group.last<<":)"<<endl;    }    for(int i=0;i<14400;i++){        output<<endl;        for(int j=0;j<4;j++){            for(int k=0;k<4;k++)                output<<group.group[i].mat[j][k].x<<' '<<group.group[i].mat[j][k].f<<(k<3?";":"");            output<<endl;        }        output<<endl;    }    cout<<group.last<<endl;    cout<<"file weggeschreven, nu door naar berekenen.";}/*bool search(fint current[],int last,int& where){//search for 'current' in 'vertices'.    bool rtrnv;    for(where=0;where<last;where++){        rtrnv=true;        for(int j=0;j<4;j++)            if(vertices[where][j]!=current[j])                rtrnv=false;        if(rtrnv)            return true;    }    return false;}void make600(thegroup& group){//assign a vertex-number to every group-element.    int lastvertex=1;    fint current[4];    int where;    vertices[0][3]=x;    for(int i=0;i<14400;i++){        for(int j=0;j<4;j++)            current[j]=group.group[i].mat[3][j];        if(search(current,lastvertex,where)){            group.group[i].vertex=where;        }else{            lastvertex++;            group.group[i].vertex=where;            vertes[where]=i;            cout<<lastvertex<<'.'<<where<<"::"<<endl;            for(int j=0;j<4;j++){                cout<<current[j].x<<current[j].f<<' ';                vertices[where][j]=current[j];            }        }    }}void biggerorders(thegroup& group){//give the length of the orbits of all vertices of the 600-cell.    int orders[31][50];    int order[30];    bool found;    int k;    int last=0;    for(int i=0;i<31;i++){        for(int j=0;j<50;j++)        orders[i][j]=0;        order[i]=0;    }    for(int i=0;i<14400;i++){        for(int g=0;g<30;g++)            order[g]=0;        for(int j=0;j<120;j++){            k=0;            fintmatrix mat=group.group[vertes[j]];            for(int i=0;i<4;i++)                for(int j=0;j<3;j++)                    mat.mat[j][i]=0;            do{mat=mat*group.group[i];                found=true;                for(int i=0;i<4;i++)                    if(mat.mat[3][i]!=group.group[vertes[j]].mat[3][i])                        found=false;                k++;            }while(!found);            order[k-1]++;        }        k=0;        do{            found=true;            for(int j=0;j<30;j++)                if(orders[j][k]!=order[j])                    found=false;            k++;        }while(!found&&k<=last);        if(k>last){            for(int i=0;i<30;i++)                cout<<(orders[i][last]=order[i])<<',';            cout<<endl;//cin.get();            orders[30][last]++;            last++;        }else{             /*for(int i=0;i<30;i++)                cout<<order[i]<<',';cout<<endl;            orders[30][k-1]++;        }        if(!(i%100))            cout<<i;    }    for(int i=0;i<last;i++){        for(int j=0;j<30;j++)            if(orders[j][i])                cout<<j+1<<':'<<orders[j][i]<<' ';        cout<<orders[30][i]<<endl;    }cin.get();    int cf=0;    for(int i=0;i<last;i++)        cf+=orders[30][i];    cout<<cf;}void adjustedbiggerorders(thegroup& group){//gives other order information, quite usefull    int orders[33][40];    int order[30];    bool found;    int k;    int last=0;    int orbited;    bool inorbit;    for(int i=0;i<33;i++){        for(int j=0;j<40;j++)        orders[i][j]=0;        order[i]=0;    }cin.get();    for(int i=0;i<14400;i++){        orbited=3;        for(int g=0;g<30;g++)            order[g]=0;        for(int j=0;j<120;j++){            k=0;            fintmatrix mat=group.group[vertes[j]];            for(int i=0;i<4;i++)                for(int j=0;j<3;j++)                    mat.mat[j][i]=0;            do{mat=mat*group.group[i];                inorbit=true;                found=true;                for(int i=0;i<4;i++){                    if(mat.mat[3][i]!=group.group[vertes[j]].mat[3][i])                        found=false;                    if(j==0||j==1){                        if(!(mat.mat[3][i]==group.group[vertes[2]].mat[3][i]||(j==0&&mat.mat[3][i]==group.group[vertes[1]].mat[3][i])))                            inorbit=false;                    }else                        inorbit=false;                }                if(inorbit&&((orbited==3&&j==1)||j==0)){                    orbited--;                    if(j!=1)cout<<i<<','<<j<<' '<<orbited<<endl;                }                k++;            }while(!found);            order[k-1]++;        }        k=0;        do{            found=true;            for(int j=0;j<30;j++)                if(orders[j][k]!=order[j])                    found=false;            k++;        }while(!found&&k<=last);if(orbited!=2&&orbited!=3)cout<<orbited<<' ';        if(k>last){            for(int i=0;i<30;i++)                /*cout<</*(orders[i][last]=order[i]);/*<<',';            cout<<endl;//cin.get();*            orders[29+orbited][last]++;            last++;        }else{             /*for(int i=0;i<30;i++)                cout<<order[i]<<',';cout<<endl;*            orders[29+orbited][k-1]++;        }        if(!(i%100))            cout<<endl;//i;    }    for(int i=0;i<last;i++){        for(int j=0;j<30;j++)            if(orders[j][i])                cout<<j+1<<':'<<orders[j][i]<<' ';        cout<<'('<<orders[30][i]<<','<<orders[31][i]<<','<<orders[32][i]<<')'<<endl;    }cin.get();    int cf=0;    for(int i=0;i<last;i++)        cf+=orders[30][i]+orders[31][i]+orders[32][i];    cout<<cf;}*/void givecoordinates(int p, thegroup& group){    //ofstream output("C:/Users/Wout/Documents/533tower.txt"),ios::out);    fint vertices[4][3][1200];    fint dynkin[4];    fint currvertex[4];    int last=0;    bool hurray;    bool write;    for(int i=0;i<4;i++){        dynkin[i]=(p%2?x:o);        p/=2;        cout<<dynkin[i].x<<":)";    }cin.get();    for(int symmetry=0;symmetry<4;symmetry++){        last=0;        for(int i=0;i<14400;i++){            for(int j=0;j<4;j++)                currvertex[j]=o;            for(int j=0;j<4;j++){                for(int k=0;k<4;k++)                    currvertex[j]=currvertex[j]+dynkin[k]*group.group[i].mat[k][j];            }//cout<<"i:"<<i<<'\t';            hurray=true;            for(int j=0;j<4;j++)                if(j!=symmetry&&currvertex[j].x+currvertex[j].f*1.618033988749894848204586834365638117720309179805762862135<0){                    hurray=false;                    //if(j!=symmetry)cout<<currvertex[j].x<<','<<currvertex[j].f<<":)"<<endl;                }            if(hurray){                write=true;                for(int j=0;j<=last;j++)                    for(int k=0;k<3;k++){                        int l=k+(symmetry<=k);                        if(currvertex[l]!=vertices[symmetry][k][j])                            break;                        if(k==2)                            write=false;                    }                if(write){                    //cout<<":) ";for(int q=0;q<4;q++)cout<<currvertex[q].x<<','<<currvertex[q].f<<';';cout<<endl;                    last++;                    for(int j=0;j<3;j++)                        vertices[symmetry][j][last]=currvertex[j+(symmetry<=j)];                }            }//cout<<"i:"<<i<<"last:"<<last<<' ';            //cin.get();        }        cout<<p<<','<<symmetry<<','<<last<<endl;        for(int i=0;i<=last;i++){            for(int j=0;j<3;j++)                cout<<vertices[symmetry][j][i].x<<','<<vertices[symmetry][j][i].f<<';';            cout<<endl;        }    }}int main(){//main    cout<<":)"<<endl;    thegroup group;    makegroup(group);    for(int i=1;i<16;i++)        givecoordinates(i,group);    return 1;}`
However, my compiler gives me errors concerning the ofstream in the givecoordinates-subroutine (I am quite ignorant concerning stream handling ) I hope someone with more programming-experience could help me out .
The error I get says: invalid declaration of 'std::basic_ios<char>::out'

Now what the program does (But doesn't output to a file ):
It has the following classes:
fint: numbers that look like a+b*f, because all uniforms of .5.3.3. have towers that use just these numbers, and now we can do integer calculations.
fintmatrix: This is a matrix that changes the CD-diagram, according to the EKF-rules. Defaults to the identity matrix.
thegroup: this is a set of fintmatrices, that form a group. The first 5 matrices are hardcoded to be the identity, and changing of all 4 nodes.

First we make the complete .5.3.3.-group, that is, we determine all matrices you can get by multiplying the default matrices.
This gives us a set of 14400 matrices, that all correspond to a way of changing any CD-diagram. This can also be interpreted as a symmetry of R^4 according to a weird basis.
Now what's great, is that when we take such a matrix, and multiply it with some CD-dyagram, say x5o3o3o, we get a different cd-diagram (say (-x)5f3o3o. However, this diagram still is part of the x5o3o3o-polytope. When this diagram has only 1 negtaive node, the polytope the other nodes are giving is part of the lace tower. (So when we have (-x)5f3o3o, we know the tower according to .3.3.-symmetry has an f3o3o-part.)

This basically is what the program does: it takes a CD-diagram (any numer 1-16 is converted to binary, a 0 means o, 1 means x, so 4 gives 0010->o5o3x3o), then this is multiplied by all matrices of thegroup, and then tower-elements are written down in the vertices-array.

Now the negative node of such a dynkin-diagram tells you how far down the tower you should be. However, I couldn't implement this in the programm without getting multiple instances of the same tower-element.
When the program is run aas-is, it will give the tower-elements in an near-correct order, but whenever one tower-element should occur multiple times in the tower, it inconveiniently gets ommited.
I hope someone can help me out, then the towers of .5.3.3.-symmetry on Klitzings site can be determined once and for all.
student91
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Joined: Tue Dec 10, 2013 3:41 pm

### Re: Generated lace-towers for the .5.3.3.-family

student91 wrote:... then the towers of .5.3.3.-symmetry on Klitzings site can be determined once and for all.

Good news!
So we'd be a big step nearer to our aimed for publication. For, then we'll get finally on an authoritative ground.
And I think, this could make up a manageable set of EKFs to be detailed within that first publication.

So far we just had hand-sieved finds. Sure, the ones already found ought to be valid. But we were not too sure on having overlooked some candidates. Most "candidates" clearly already could be sieved out by purpose for several reasons of impossibility. But then some had to be evaluated manually ...

And then, the next idea would be to set up the same with "q" instead of "f" and then run through both .4.3.3. and .3.4.3., hehe
And according follow-up publications...

--- rk
Klitzing
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Location: Heidenheim, Germany

### Re: Generated lace-towers for the .5.3.3.-family

Klitzing wrote:
student91 wrote:[..]

Good news!
I've tweaked my program a little, and now I have results. However, the list is quite massive, so whenever I try to put it in-line (using [code]-tags of course) and press the 'preview'-button, my posts gets deleted and i get the error-message: "your post contains too few characters."
therefore, I put it in a drive-file, and here is the link.
The results are grouped in 2-tuples, the first number gives the x-value, and the second number the f-value. The last number, in parentheses, gives information about where a tower-part must go. If you have two tower-parts A and B, A being a_1 a_2 a_3 (a_4), and B being b_1 b_2 b_3 (b_4), then A should be above B if either (a_1 a_2 a_3) has a bigger radius than (b_1 b_2 b_3) and a_4<b_4, or if (a_1 a_2 a_3) has a smaller radius than (b_1 b_2 b_3) and a_4>b_4.
P.S. This ay we can only find towers of the symmetries .5o3o3o, o5.3o3o, o5o3.3o and o5o3o3. (the latter is written as o5o3o3o in my file for some reason)
student91
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Joined: Tue Dec 10, 2013 3:41 pm

### Re: Generated lace-towers for the .5.3.3.-family

I see. That provided info then is nothing else what already is provided for the regular and quasiregular ones in my website (then labeled "Vertex layers").
There I usually provide some arbitrary layer count. That one should translate somehow to your 4th number pair. I have to admit that I don't understand the content of those, so far.
As far as I've scrolled down, I've seen that you run through all Wythoffians of that hyic symmetry, each along the 4 main axes.

Not too bad as a first start. But nothing essentially new. Sure not all truncates, rhombates etc. already were elaborated in my files, but still the process is straight forward. But, at least, you now have a programmed way to derive these. Next we will have to elaborate an automated process to derive "possible" quirks in any combination, and to sieve out a priori impossibilities...

--- rk
Klitzing
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### Re: Generated lace-towers for the .5.3.3.-family

Klitzing wrote:I see. That provided info then is nothing else what already is provided for the regular and quasiregular ones in my website (then labeled "Vertex layers").
There I usually provide some arbitrary layer count. That one should translate somehow to your 4th number pair. I have to admit that I don't understand the content of those, so far.
Indeed I didn't just yet program the Holy Grail, that would take some more time .
Furthermore, I think the reason you didn't understand the "last" nodes yet is because I didn't explain them. This node gives the value that should be on the missing node of the CD-diagram in order to make the tower-part coincide with a vertex of the given polytope.
See, when you write something like (-x)5F3x3o, you are actually giving a point in a skew coordinate system, (-x)-distace away from 1 hyperplane, distance F to another hyperplane etc.
So when you see 1,1 1,0 0,0 (-1,0) in the tower of x5x3x3o, in .5o3o3o-symmetry, this means the first o has a value of (1x+1f)=F, the second o a value of (1x+0f)=x, and the third o has a value of o, therefore we have that F3x3o is part of the lace-tower of x5x3x3o. In fact, the last number tells us that (-1x+0f)=(-x) should be at the place of the ., thus we know (-x)5F3x3o is part of the vertex-set of x5x3x3o, and that is why we know F3x3o is part of the tower.
As far as I've scrolled down, I've seen that you run through all Wythoffians of that hyic symmetry, each along the 4 main axes.
Indeed that's just what I've done so far. But there's more! It has been very easy to change my program to 433 and 343-symmetries, so I have towers for those as well.
Code: Select all
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`x3o4o3o.3o4o3o-symmetry:0,0 0,0 0,0 (1,0)1,0 0,0 0,0 (-1,0)0,0 0,0 0,1 (-1,0)1,0 0,0 0,0 (-2,0)0,0 0,0 0,0 (-1,0)o3.4o3o-symmetry:1,0 0,0 0,0 (0,0)0,0 0,1 0,0 (-1,0)1,0 0,0 0,1 (-1,0)2,0 0,0 0,0 (-1,0)1,0 0,1 0,0 (-2,0)0,0 0,0 0,1 (-1,0)1,0 0,0 0,0 (-1,0)o3o4.3o-symmetry:1,0 0,0 0,0 (0,0)0,0 1,0 0,1 (0,-1)1,0 1,0 0,0 (0,-1)1,0 0,0 0,1 (0,-1)0,0 1,0 0,0 (0,-1)o3o4o3.-symmetry:1,0 0,0 0,0 (0,0)0,0 1,0 0,0 (0,-1)1,0 0,0 0,0 (0,-1)o3x4o3o.3o4o3o-symmetry:1,0 0,0 0,0 (0,0)0,0 0,1 0,0 (-1,0)1,0 0,0 0,1 (-2,0)2,0 0,0 0,0 (-3,0)1,0 0,0 0,1 (-3,0)0,0 0,1 0,0 (-3,0)1,0 0,0 0,0 (-3,0)o3.4o3o-symmetry:0,0 0,0 0,0 (1,0)1,0 0,1 0,0 (-1,0)2,0 0,0 0,1 (-1,0)1,0 0,1 0,1 (-2,0)3,0 0,0 0,0 (-1,0)0,0 0,0 0,2 (-1,0)1,0 0,2 0,0 (-3,0)3,0 0,0 0,1 (-2,0)2,0 0,1 0,1 (-3,0)1,0 0,0 0,2 (-2,0)3,0 0,1 0,0 (-3,0)0,0 0,2 0,0 (-3,0)1,0 0,1 0,1 (-3,0)3,0 0,0 0,0 (-2,0)2,0 0,1 0,0 (-3,0)1,0 0,0 0,1 (-2,0)0,0 0,0 0,0 (-1,0)o3o4.3o-symmetry:0,0 1,0 0,0 (0,0)1,0 1,0 0,1 (0,-1)1,0 0,0 0,2 (0,-1)2,0 1,0 0,0 (0,-1)1,0 2,0 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2,0 1,2 (-3,-2)2,1 1,0 1,1 (-3,-3)3,1 1,0 1,0 (-4,-3)1,1 1,0 1,2 (-3,-3)1,0 2,0 1,2 (-4,-3)1,1 1,0 1,2 (-4,-3)2,1 1,0 1,1 (-5,-3)3,1 1,0 1,0 (-5,-3)1,0 2,1 1,0 (-5,-3)2,0 2,0 1,1 (-5,-3)1,0 1,0 2,2 (-4,-3)2,0 1,0 2,1 (-5,-3)1,0 1,1 2,0 (-5,-3)1,0 1,1 1,0 (-5,-3)2,0 1,0 1,0 (-5,-3)1,0 1,0 1,0 (-4,-3)o3.4o3o-symmetry:1,0 1,0 1,0 (1,0)2,0 1,1 1,0 (-1,0)1,0 1,2 1,0 (-2,0)1,1 1,2 2,0 (-2,-1)2,1 1,1 2,0 (-1,-1)3,1 1,0 2,1 (-1,-1)2,2 2,1 1,0 (-1,-2)1,0 1,2 3,1 (-3,-1)1,1 1,2 2,1 (-3,-1)1,2 2,2 1,0 (-2,-2)3,1 1,0 2,2 (-2,-1)2,1 1,1 2,2 (-3,-1)1,2 2,2 1,1 (-3,-2)3,2 2,0 1,1 (-1,-2)3,3 1,0 1,1 (-1,-2)3,2 2,0 1,2 (-2,-2)2,2 2,1 1,2 (-3,-2)4,3 1,0 1,0 (-1,-2)1,0 1,0 3,3 (-2,-1)2,0 1,1 3,2 (-3,-1)1,0 2,2 3,1 (-3,-2)2,0 2,1 3,2 (-3,-2)1,2 3,2 1,1 (-3,-3)1,1 3,2 2,1 (-3,-3)1,2 3,3 1,0 (-4,-3)3,3 1,0 1,2 (-2,-2)2,1 3,1 2,2 (-3,-3)2,2 3,1 1,2 (-3,-3)4,3 1,0 1,2 (-3,-2)5,3 1,0 1,1 (-3,-2)5,3 1,0 1,0 (-2,-2)3,2 3,1 1,2 (-4,-3)3,3 2,1 1,2 (-4,-3)4,3 2,0 1,2 (-3,-3)1,0 2,0 3,3 (-2,-2)1,1 3,3 2,0 (-4,-3)1,0 4,3 1,0 (-4,-3)1,0 1,0 4,3 (-2,-2)2,0 1,0 4,3 (-3,-2)2,1 2,0 3,3 (-3,-3)3,1 2,1 3,2 (-4,-3)3,1 3,1 2,2 (-4,-3)3,2 3,2 1,1 (-5,-3)2,2 3,3 1,0 (-5,-3)3,1 3,2 2,1 (-5,-3)2,1 3,3 2,0 (-5,-3)4,3 1,0 2,2 (-3,-3)5,3 2,0 1,1 (-3,-3)5,3 2,1 1,0 (-4,-3)5,3 1,0 2,1 (-3,-3)3,3 2,2 1,1 (-5,-3)3,3 1,1 2,2 (-4,-3)3,2 1,1 3,2 (-4,-3)4,3 2,2 1,0 (-5,-3)3,3 1,2 2,1 (-5,-3)2,0 4,3 1,0 (-5,-3)1,0 4,3 1,0 (-5,-3)1,0 1,0 4,3 (-3,-2)1,1 2,0 3,3 (-3,-3)2,2 1,0 3,3 (-3,-3)1,1 2,1 3,2 (-4,-3)2,1 2,2 3,1 (-5,-3)3,1 2,2 3,1 (-5,-3)1,2 1,0 3,3 (-3,-3)3,2 1,2 3,1 (-5,-3)1,0 3,3 2,0 (-5,-3)2,0 3,2 2,1 (-5,-3)5,3 1,0 1,0 (-3,-3)5,3 1,1 1,0 (-4,-3)5,3 1,1 2,0 (-4,-3)4,3 1,2 2,0 (-5,-3)2,2 1,2 3,1 (-5,-3)1,2 1,1 3,2 (-4,-3)4,3 1,2 1,0 (-5,-3)3,3 1,2 1,0 (-5,-3)3,2 1,2 2,0 (-5,-3)2,2 1,2 2,1 (-5,-3)1,0 3,3 1,0 (-5,-3)1,0 3,1 2,2 (-4,-3)1,0 3,1 1,2 (-4,-3)1,1 2,1 1,2 (-4,-3)1,2 1,1 2,2 (-4,-3)2,0 3,2 1,1 (-5,-3)2,1 2,2 1,1 (-5,-3)1,2 1,0 2,2 (-3,-3)4,3 1,0 1,0 (-3,-3)3,3 1,1 1,0 (-4,-3)3,2 1,1 2,0 (-4,-3)3,1 2,2 1,0 (-5,-3)2,2 1,0 2,1 (-3,-3)3,1 2,1 1,0 (-4,-3)1,1 2,0 1,2 (-3,-3)1,0 1,0 1,2 (-3,-2)2,1 2,0 1,1 (-3,-3)2,0 1,0 1,1 (-3,-2)1,0 1,0 1,0 (-2,-2)o3o4.3o-symmetry:1,0 1,0 1,0 (1,0)1,0 1,1 2,0 (-1,0)2,0 1,1 2,1 (-1,-1)2,1 1,0 3,1 (-1,-1)1,0 2,1 2,2 (-1,-2)1,0 1,2 1,0 (-2,0)2,0 1,2 1,1 (-2,-1)1,0 2,2 1,2 (-2,-2)1,1 2,0 3,2 (-1,-2)2,2 1,0 3,1 (-2,-1)2,2 1,1 2,1 (-3,-1)2,1 1,2 1,1 (-3,-1)1,1 1,0 3,3 (-1,-2)1,2 2,0 3,2 (-2,-2)1,0 1,0 4,3 (-1,-2)1,2 2,1 2,2 (-3,-2)1,1 2,2 1,2 (-3,-2)3,1 1,2 1,0 (-3,-1)1,1 3,2 1,2 (-3,-3)1,0 3,3 1,2 (-4,-3)1,2 1,0 3,3 (-2,-2)3,3 1,0 1,0 (-2,-1)3,2 1,1 2,0 (-3,-1)1,2 3,1 2,2 (-3,-3)2,2 3,1 2,1 (-3,-3)3,2 2,1 2,0 (-3,-2)2,1 3,2 1,1 (-3,-3)3,1 2,2 1,0 (-3,-2)1,0 1,0 5,3 (-2,-2)1,1 1,0 5,3 (-3,-2)1,2 1,0 4,3 (-3,-2)1,2 2,0 4,3 (-3,-3)1,2 2,1 3,3 (-4,-3)1,2 3,1 3,2 (-4,-3)1,1 2,0 5,3 (-3,-3)1,0 2,1 5,3 (-4,-3)2,0 3,3 1,1 (-4,-3)1,0 3,3 2,2 (-5,-3)1,1 3,2 3,2 (-5,-3)1,0 4,3 1,0 (-4,-3)2,0 3,3 2,1 (-5,-3)4,3 1,0 1,0 (-2,-2)3,3 2,0 1,0 (-2,-2)2,2 3,1 3,1 (-4,-3)4,3 1,0 2,0 (-3,-2)3,3 2,0 2,1 (-3,-3)3,2 2,1 3,1 (-4,-3)2,1 3,2 3,1 (-5,-3)2,1 1,0 5,3 (-3,-3)2,2 1,0 4,3 (-3,-3)2,2 1,1 3,3 (-4,-3)1,1 2,2 3,3 (-5,-3)1,0 2,2 4,3 (-5,-3)3,2 1,1 3,2 (-4,-3)2,1 1,2 3,3 (-5,-3)2,0 1,1 5,3 (-4,-3)1,0 1,1 5,3 (-4,-3)2,0 1,2 4,3 (-5,-3)1,0 1,2 4,3 (-5,-3)1,0 4,3 2,0 (-5,-3)3,1 2,2 3,1 (-5,-3)3,1 1,2 3,2 (-5,-3)1,0 4,3 1,0 (-5,-3)2,0 3,3 1,0 (-5,-3)2,1 3,2 2,0 (-5,-3)3,1 2,2 2,1 (-5,-3)4,3 1,0 1,0 (-3,-2)3,3 1,0 2,2 (-3,-3)3,3 2,0 1,1 (-3,-3)3,3 1,0 1,2 (-3,-3)3,2 2,1 1,1 (-4,-3)1,0 1,0 5,3 (-3,-3)3,2 1,1 1,2 (-4,-3)3,1 1,2 2,2 (-5,-3)2,1 1,2 2,2 (-5,-3)2,0 1,2 3,2 (-5,-3)1,0 1,2 3,3 (-5,-3)1,0 1,0 4,3 (-3,-3)1,0 1,1 3,3 (-4,-3)1,0 2,2 3,1 (-5,-3)1,0 3,3 1,0 (-5,-3)1,1 3,2 2,0 (-5,-3)2,2 3,1 1,0 (-4,-3)1,1 2,2 2,1 (-5,-3)1,2 3,1 1,0 (-4,-3)2,2 1,0 1,2 (-3,-3)2,2 1,1 1,2 (-4,-3)1,2 2,1 1,1 (-4,-3)2,0 1,1 3,2 (-4,-3)2,1 1,0 2,2 (-3,-3)1,0 2,1 3,1 (-4,-3)1,2 2,0 1,1 (-3,-3)1,1 2,0 2,1 (-3,-3)1,1 1,0 2,0 (-3,-2)1,0 1,0 1,0 (-2,-2)1,2 1,0 1,0 (-3,-2)o3o4o3.-symmetry:1,0 1,0 1,0 (1,0)1,0 1,0 2,0 (-1,0)1,0 1,1 1,0 (-2,0)2,1 1,0 2,0 (-3,-1)2,0 1,1 1,0 (-2,-1)1,0 2,1 1,0 (-2,-2)1,1 1,0 2,1 (-3,-3)1,1 2,0 2,0 (-3,-2)2,2 1,0 1,0 (-3,-1)1,2 2,0 1,0 (-3,-2)1,0 1,0 3,1 (-4,-3)1,2 1,0 1,1 (-3,-3)1,0 1,0 3,1 (-5,-3)1,1 1,0 2,1 (-5,-3)1,2 1,0 1,1 (-4,-3)1,2 2,0 1,0 (-4,-3)1,1 2,0 2,0 (-5,-3)2,1 1,0 2,0 (-5,-3)2,2 1,0 1,0 (-4,-3)1,0 2,1 1,0 (-5,-3)2,0 1,1 1,0 (-5,-3)1,0 1,0 2,0 (-5,-3)1,0 1,1 1,0 (-5,-3)1,0 1,0 1,0 (-4,-3)`
Not too bad as a first start. But nothing essentially new. Sure not all truncates, rhombates etc. already were elaborated in my files, but still the process is straight forward. But, at least, you now have a programmed way to derive these. Next we will have to elaborate an automated process to derive "possible" quirks in any combination, and to sieve out a priori impossibilities...

--- rk
That's for later. You're welcome to suggest how you would program this. I will give it a try some time, but for now I have no idea how to make a program that could check a polytope to be CRF. (In fact, I don't even know how to calculate a circumradius of e.g. x5(-f)3q3F, that would help a lot for sure)
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### Re: Generated lace-towers for the .5.3.3.-family

The spreadsheet I wrote for Klitzing will handle negative numbers correctly.
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### Re: Generated lace-towers for the .5.3.3.-family

wendy wrote:The spreadsheet I wrote for Klitzing will handle negative numbers correctly.

I know, but I want to know how I can implement the spreadsheet into a program. You see, it is quite difficult to make a program that changes spreadsheet-imput to get to results, whereas when you just let the program do the same thing as the spreadsheet, without actually using the spreadsheet, your program works much better.
But frankly, I have no idea how and why the spreadsheet works, I only use it as a magical tool that answers my questions (esp. to understand how an inverted matrix is related to the radius of a polytope I am not well-educated enough. )
So if someone would explain to me what that magical inversemat-command does, and why it works with such an odd matrix, I would be very happy .
student91
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### Re: Generated lace-towers for the .5.3.3.-family

Wendy herself once described the coded content of her spreadsheet by:
Wendy wrote:The coordiates you enter for the polytopes are vectors in an oblique coordinate system. The matrix you enter from the dynkin symbol is the 'dynkin' matrix. It is used to calculate the stott matrix, (by simple reciprocal), the stott matrix is used to do a simple dot product between the two vectors.

--- rk
Klitzing
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### Re: Generated lace-towers for the .5.3.3.-family

student91 wrote:Furthermore, I think the reason you didn't understand the "last" nodes yet is because I didn't explain them. This node gives the value that should be on the missing node of the CD-diagram in order to make the tower-part coincide with a vertex of the given polytope.

Thus, ain't it better to write that node part in its correct place then?
Kind of:
Code: Select all
`x4o3o3o.4o3o3o-symmetry:(1,0) 0,0 0,0 0,0(-1,0) 0,1 0,0 0,0(-1,0) 0,0 0,1 0,0(-1,0) 0,0 0,0 0,1(-1,0) 0,0 0,0 0,0o4.3o3o-symmetry:1,0 (0,0) 0,0 0,01,0 (0,-1) 0,1 0,01,0 (0,-1) 0,0 0,11,0 (0,-1) 0,0 0,0etc.`

Thus it looks like that you calculate all possible Wythoff polytopes which use some given vertex set. And then you choose some subsymmetry (by omission of any individual of the nodes) and then select from that first list all those which will not be degenerate in the remaining nodes part (i.e. don't provide an according degenerate facet). Perhaps further stronger restrictions for sieving? (E.g. non-reverse nodes within the remainder?) - But then it looks more like an arbitrary compilation. It would not provide a sequence of ever deeper layers. - Or did you sort the outcome accordingly as well?

Furthermore all provided coordinates lines of each block thus are designed to provide all the same circumradius within the given dynkin base (e.g. by using that said spreadsheet). This clearly is just because of the first statement of the previous paragraph.

--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

Perhaps it would be nice to calculate (and then provide, for sure) the (signed) height of the respective facet above/below the polychoral center each in addition?
--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

The formula for a specific polytope-group, like {5,3,3}, can be done if you have access to the family of polytopes

x5o3o3o, o5x3o3o, o5o3x3o, and o5o3o3x, on the same coordinate system (ie the same symmetry).

Suppose you start with a tetrahedron of o5o3o3x. You will then get four points, say A, B, C, D. Then the coordinates of the remainig figures are

x3o3o5o gives A
o3x3o5o gives A+B
o3o3x5o gives A+B+C
o3o3o5x gives (A+B+C+D)/f

Given these coordinates, populate the matrix S_ij by taking the dot product of P_i P_j. This works if the edge is set to 2, which is the default form of the polytope. What this gives then is a matrix which can be used as a basis for a matrix-dot product. The length of any vector V_i is then S_ij V_i V_j = length² V, where V is the named vector. You can use the dot product to find the angle between two vectors by S_ij V_i W_j = length V length W cos A, and divide this product by the lengths of V_i and V_j

Hope that helps.

The Stott matrix for 3,3,5 is this matrix, multiplied by 2/(5-3f).

4-2f, 3-f, 2 , f
3-f , 6-2f , 4 , 2f
2 , 4 , 6 , 3f
f , 2f , 3f , 4
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### Re: Generated lace-towers for the .5.3.3.-family

Klitzing wrote:Or did you sort the outcome accordingly as well?

Definitely not!
--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

Dear Wout,

I'm still in the run of checking your programmed output wrt. my manual lists.

But now I found an interesting discrepancy, which as such is not a wrong claim on either side, just a different way of representation. But that then surely would provide different results, when being used for further researches!
So we'd have to spot on. Esp. whether that discrepancy occured just by hazard, or is for purpose, i.e. knowingly implemented in your program. - ?!?

The example was the equatorial section of rox = o5o3x3o wrt. o5o . o subsymmetry:
Code: Select all
`...o5o3x3o...o5o3.3o-symmetry:...0,1 0,1 0,2 (-1,-3)0,0 2,1 0,0 (-2,-2)1,0 1,0 2,2 (-2,-2)...2,1 0,0 0,0 (-1,-2)...`

whereas my list there does provide
Code: Select all
`...o . F5FV . f5fC . x5x...`

(where x=1,0 / f=0,1 / F=1,1 / V=0,2 / A=2,1 / C=2,2),
i.e. I provide the decagon F5F, while you provide its 2 inscribed pentagons A5o + o5A separately...

Such a decomposition, when allowed resp. aimed for, then should be possible in further occasions too !?

--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

Oops
that provided example turns ought to be none...
At the one hand F5F and A5o have different circumradii, at the other the hull of the compound of A5o + o5A has the alternate orientation, kind like x10o and o10x.
Looks like I just stumbled upon a further error in my manually assembled listings...
(Esp.: A5o provides the correct (equatorial) circumradius, while F5F does not...)
--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

Klitzing wrote:Dear Wout,student91

I'm still in the run of checking your programmed output wrt. my manual lists.
I hope you won't go through all unsorted lists, by the time I will have an implemented radius-calculator, I will be able to sort them the way we want, and thus I would wait with the big towers if I were you.
Besides, I still don't really understand how the circumradius-calculation works, though I can of course implement it without actually knowing how it works.
The part I'm having difficulty with is the interpretation of the Stott-vectors. I don't see what these are, I only see you get them by inverting a matrix that I do actually understand the meaning of (I guess )
Well, I guess I'll just implement it, and you should expect the sorted lists then before next weekend.
student91
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### Re: Generated lace-towers for the .5.3.3.-family

I will try to explain Stott vectors for student91.

If you take the point (1,2,3) it represents a point in the first octant. Physicists talk of 'position vector', which is a vector from (0,0,0) to (1,2,3). The vector sum of two position vectors is a position vector, simply by adding the coordinate of the points, so (1,2,3) + (4,5,6) = (5,7,9), &c.

The points (1,g,g) represent a plane that is 1 cm from the plane x=0. We intersect it with a second plane (g,2,g), which is a plane 2 cm distant from y=0, and this gives a line (1,2,g). And so forth. A point is an intersection of 'equal signs', ie planes.

Because x is already loaded with meanings, I am going to use 'g' to represent a random or variable value that usually one uses 'x' for.

We make a 'position polytope', by picking the points x=1, y=2, z=3, with all changes of sign. This would give a rectangular box whose sides are 2 cm, 4 cm, 6 cm, the centre of which is at the origion (0,0,0). The particular process is 'AC'. (all change of sign). We can make equally a snub figure by putting a coordinate marked EC (even change of sign) and OC odd change of sign.

In terms of the dynkin symbol, this is the polytope (1)2(2)2(3), of the group ".2.2." .

The same trick works if you slope the mirrors inwards. We still have a point giving a 'position vector' and a 'position polytope'. But we don't have right angles.

A polytope like (1)3(2)5(3) of .3.5. , still has edges that are 2cm, 4cm and 6cm long, but the shape is no longer a prism, but a generalised omnitruncate. The 2cm edge is the (1) reflected in a mirror, the edge runs between the hexagon and rectangle. The (2) edge is 4 cm, runs between the hexagon and decagon, and the (3) edge runs between the rectangle and decagon. The 2cm comes from reflecting (1) to (-1) &c.

The vector goes from the centre of the polytope to a vertex. Mrs Stott showed that if you increase the (1) edge to say (1.5), the edge rises to 3 cm, but the decagon is unchanged, in essence 'pushed away' from the centre without change of size. You can contract the edge to zero length, so if (3) went to (0), the ends of the line would meet, and the vertex would fall on that mirror. Knowing that the points (1,g,g) is still a plane 1 cm distant (ie a 1 cm thick slab fits between), from the (0,h,h) plane, [h is a different g], and so forth, then a change in the second or third coordinates does not affect the length edge perpendicular to the first, just the location.

The main difference here is, that when we do 'AC', instead of doing just the eight octants of .2.2., we are going into all of the 120 symmetries of .3.5., and 'EC' and 'OC' go into the black and white sectors of .3.5. when we use alternating colours, so the vertex of the snub dodecahedron is some s3s5s = g3g5g (EC), for three different numerals.

MATRIX-DOT

The axies of our new coordinate goes from (1,0,0) = x3o5o, (0,1,0) = o3x5o, (0,0,1) = o3o5x. The verticies of the remaining four figures, plus the centre, form a parallelogram in each of the 120 symmetries, so (1,0,1) = (1,0,0) + (0,0,1), since the first has a vertex, which allows a 1 cm slab between it and the first and third mirrors, and the 1 cm slab represents a 1 coordinate.

The axis coordinate is oblique (slanted, or at angles to each other), which means the usual matrix properties of vectors do not work. Specifically, we have to do something to make the dot product give the length * length * cos(angle). Normally, if x,y,z is one point, and p,q,r is another, then px + qy + rz = (XYZ)(PQR) cos [xyz - 000 - pqr], where () represents the root-sum-square, ie rss(x,y,z)=(XYZ), and in the angle, xyz is the point, 000 is the centre, and pqr is the other point.

The fix for this is to either convert both to rectangular coordinates, and then do the ordinary dot product there. Using the dot matrix costs a grid per product, the rectangular coordinates a grid per point. If you are doing a full simplex in 4d or 5d, the rectangular coordinates is cheaper, but for random coordinates, the grid is cheaper.

THE DYNKIN MATRIX

This is frightfully clever stuff that I missed it until i calculated the individual arrows for a problem. But the process is quite intuitive.

Let's suppose you have the plane as described above. You put on a plane, a line at (0,0,0) pointing perpendicular to (1,g,g). In general, this will not be zeros. This is the 'normal' or perpendicular to the first plane. You do this to the second and third plane too. So you have three rays pointing from (0,0,0). Now, if you make the angle between the planes smaller, the normals will rotate with the plane, so if the x-plane is held constant, and the y-plane rotates into the first quadrant, the x-normal is still pointing to (0,1), but the y-normal moves towards (0,-1). The angle between these normals is the supplement (180-a) of the angle between the two mirrors, and the dot product is then the -cos(a).

Since the dynkin graph consists of mirrors (ie nodes or dots), and the angles between the mirrors (ie the branches), the matrix representing the verticals to the planes is then 1 when i=j (i and j th coordinates), and -cos(pi/a) for the branches. It is in effect, a way of writing the Dynkin symbol as a matrix.

Now, if we solve what vectors are normal to the mirrors in stott-coordinates, then we expect that the dot of the first normal with any other normal is zero, and with the first normal gives one. A full matrix if the dot products of the i th normal and the j-th normal gives a unit matrix (all zero, except 1 when i=j). This means that

stott matrix * dynkin matrix = unit matrix.

Since the stott matricies are hard to work out by hand (it can be done: i did most of them this way), and hard to describe by values, it is easier to use the general dynkin matrix, and use an inversion technique to get the reciprocal. This is what the spreadsheet does.

For the limited number of symmetries like 3,3,p, there is a generic form to fill in, this extends infinitely for the 3,..3,P family.

Code: Select all
`         [ 4-2b   3-b    2     a  ]                2     [ 3-b   6-2b    4    2a  ]       2    [  3-b   2     a  ]    2   [ 2   a ]  ---    [  2      4     6    3a  ]     -----  [   2    4    2a  ]  ----  [ a   2 ]  5-3b   [  a     2a    3a     4  ]      4-2b  [   a    2a    3  ]   3-b                  4d                                 3d                      2d                         STOTT MATRICIES for (p), (3,p) and (3,3,p).                    p     2     3    4    5    6         -----------------------------         a     0     1    q    f    h    short chord          b    -1     0    1    f    2   third chord = a²-1`
The dream you dream alone is only a dream
the dream we dream together is reality.

wendy
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### Re: Generated lace-towers for the .5.3.3.-family

Thank you wendy, I think I do understand now why the stott-matrix works, and how it works etc.

Meanwhile, I have been trying to sort the results according to their height. This proved not to be easy, as I calculate the square height by [Height^2=(radius of the polytope)^2-(radius of the element)^2], and thus I can't get the coordinates signed.
So I tried to sort them according to the distance of them to the 1st vertex, and this worked out quite well. However, some of the layers in big polytopes like x5x3x3x are still not sorted according to their height yest, i.e. you have things sorted like 1-1-1.2-1.2-1.3-1.2-1.3-1.4 etc.
I guess next thing I will do is try to sort the elements according to their distance to a point whose vector is normal to the symmetry-plane, and hope this gives better results.
These vectors of course look like e.g. A5o3o3o when in the .5o3o3o-symmetry. Do you have any idea on how I could initialize A such that I won't get overflow-errors (I work with shorts ) but the vector is still outside the polytope?
Any help would be appreciated.
However, this list is much better than my previous list, so I will upload it again.
It is the same list as before, but now a bit better sorted, and with the squared height in the second column.
student91
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### Re: Generated lace-towers for the .5.3.3.-family

student91 wrote:These vectors of course look like e.g. A5o3o3o when in the .5o3o3o-symmetry. Do you have any idea on how I could initialize A such that I won't get overflow-errors (I work with shorts ) but the vector is still outside the polytope?
Any help would be appreciated.

Well, if you just consider x-edged Wythoffians, then the largest polychoron would be x3x3x5x. Thus take the circumradius of that. That distance then would be farer away from the polychoral center than the hyperplane of every non-degenerate cell or facet of any such Wythoffian.
--- rk
Klitzing
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### Re: Generated lace-towers for the .5.3.3.-family

Dear student91

When you come to calculate height, the trick is not to work with the indicated radii, but imagine that the edge slopes from a position to a position. So the drift is vertex1 minus vertex2 in the same symmetry, you will get some negatives, but the matrix will handle these perfectly correctly.

So the edge for xo3oo5ox&#x, in the base plane, moves from (1,0,0) to (0,0,1), so it is (-1,0,1). This is the vector you feed into the matrix. It does not matter which way you do it, since (-1,0,1) and (1,0,-1) give the same result. This is the 'drift' or projection of the edge on the base plane. The height then comes from drift² + height² = edge² (along with some cursed powers of two, which i always stuff up.). But the section I give is the second section of the {3,3,5}, which can also be found elsewhere.
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### Re: Generated lace-towers for the .5.3.3.-family

Sorry Wendy, but student91 was after the signed heights. His program provides all facets orthogonal to a given direction (the normal) in the form of some cell of some faceting polytope. Thus it is not only interesting to get the actual unsigned distance of the supporting hyperplane from the polytopal center (origin), but the additional Information, whether it is situated above or below, then would be appreciated as well.

OTOH, Wendy is surely right. As you (i.e. student91) could refer the current hyperplane/facet wrt. the uppermost hyperplane, i.e. the cell of the convex one. Again the spreadsheet calculates either the height as a function of the lacing edge length, or conversely the lacing edge lengths as a function of a given height. As you already should know supporting vertices of the respective planes (by way of construction of the respective faceting: the respective seed-point of kaleidoscopical construction), you thus ought know therefrom the lacing edge size between those stacked bases (the 2 cells of consideration). Thus the spreadsheet (or a corresponding programmed logic) could calculate the height of that lace prism, i.e. the distance between the 2 considered cells. And as the uppermost cell is a constant of that research, you could calculate its (absolut, i.e. unsigned) height. But as that one corresponds to a CRF, it clearly is positive. So the signed height then could be deduced as "absolut height of uppermost cell - unsigned height of lace prism", which then becomes a signed number.

--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

Klitzing wrote:Wendy is surely right. As you (i.e. student91) could refer the current hyperplane/facet wrt. the uppermost hyperplane, i.e. the cell of the convex one. Again the spreadsheet calculates either the height as a function of the lacing edge length, or conversely the lacing edge lengths as a function of a given height. As you already should know supporting vertices of the respective planes (by way of construction of the respective faceting: the respective seed-point of kaleidoscopical construction), you thus ought know therefrom the lacing edge size between those stacked bases (the 2 cells of consideration). Thus the spreadsheet (or a corresponding programmed logic) could calculate the height of that lace prism, i.e. the distance between the 2 considered cells. And as the uppermost cell is a constant of that research, you could calculate its (absolut, i.e. unsigned) height. But as that one corresponds to a CRF, it clearly is positive. So the signed height then could be deduced as "absolut height of uppermost cell - unsigned height of lace prism", which then becomes a signed number.

--- rk

Well, if you just consider x-edged Wythoffians, then the largest polychoron would be x3x3x5x. Thus take the circumradius of that. That distance then would be farer away from the polychoral center than the hyperplane of every non-degenerate cell or facet of any such Wythoffian.
--- rk

Oh well, I guess I could do it that way... But wouldn't I then have to subtract two square-roots? That would mean that I would get out of the Q+fQ-arithmetric that I've been using, and thus trying to program it that way will decrease the accuracy of the calculations.

But I've done what you suggested, and now I've sorted the list correctly. I have even been able to sign the numbers according to their distance to the far-awaypoint, and I've sorted the list accordig to the (squared)height. Note that the square of the height is given as an exact number in the squared brackets, and the value next to that has a minus-sign if the square-root of that should be negative. The actual squared height of course isn't negative .

Now if I'd have matrices
student91
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### Re: Generated lace-towers for the .5.3.3.-family

Hello again!

I have re-written the program, it can now generate any Coxeter group (as long as it is 4-dimensional and the branches are labeled either 2,3,4 or 5). Furthermore, it can look for group-homeomorphisms from a given group G to a (4-dimensional) Coxeter subgroup H, and then use this to represent a polytope of group G in the coordinate-system of group H.
This sounds real usefull, and it is! After recalling that [2,2,2]-symmetry gives standard cartesian coordinates, one sees that the thus-described homeomorphism will be able to represent any polytope in cartesian coordinates, given that [2,2,2]-symmetry is a subgroup. This means I will be able to auto-generate coordinates from 4-dimensional CD-diagrams, for every symmetry except [3,3,3]. But there's more! As I have a homeomorphism from [5,3,3] to [3,3,3], i can invert this to get a function from [3,3,3] to [5,3,3], which then can be mapped to [2,2,2], so I will soon be able to auto-generate coordinates from CD-diagrams for every 4-dimensional group. For this, I only need to invert the matrix used to put [5,3,3] polytopes in [3,3,3]-symmetry. For now, I think I will make everyone happy by giving the polytopes of the [5,3,3]-family in [3,3,3], [2,2,2], [5,2,5], [3,2,3], [31,1,1] and [5,3,2]-symmetry.

Here you have them embedded in: [2,2,2], [3,2,3], [5,2,5], [5,3,2], [3,3,3], [31,1,1]. The format is the same as in previous files, i.e. (a,b) means ax+bf, and the matrices at the top represent the CD-diagrams, out of which the order of the coordinates can be understood.

Besides, this way I got x3o4o3o embedded in a quite weird way in [2,2,2]-symmetry (i.e. acs).
My program gives the coordinates
acs<(x,o,x,q),(o,q,o,q),(o,o,u,o),(x,q,x,p),(u,o,o,o)>.
It's quite asymmetrical. Quickfur, could you do a quick render on this to see if it really is a 24-cell, or does anyone already recognise this representation of ico?
student91
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### Re: Generated lace-towers for the .5.3.3.-family

student91 wrote:... Besides, this way I got x3o4o3o embedded in a quite weird way in [2,2,2]-symmetry (i.e. acs).
My program gives the coordinates
acs<(x,o,x,q),(o,q,o,q),(o,o,u,o),(x,q,x,p),(u,o,o,o)>.
It's quite asymmetrical. Quickfur, could you do a quick render on this to see if it really is a 24-cell, or does anyone already recognise this representation of ico?

Well, I did already know of ico representations according to
oxo4ooq oxo4qoo&#zx (i.e. a tegum sum of 2 perpendicular q-{4}'s and a tes) and as
uooox ouoox oouox oooux&#zx (tegum sum of 4 pairwise perpendicular u-lines and a tes),
just scroll down to the bottom of the target page and cf. the last 2 given matrices.

Yours seem to separate these symbols into halves each: s.th. like the first half of the first and the second half of the second then should describe yours in its 2nd and 4th resp. 1st and 3rd coordinates.

--- rk
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### Re: Generated lace-towers for the .5.3.3.-family

The spreadsheet that I provided to richard is different to the one that he is using. He has made a lot of imporvements and changes, which is in keeping with the style of conventional mathematics, but different to what I wrote. The main distinction is that klitzing counts circumradii where i calculate circumdiameter.

Richard is using a german version of excel, so the pages come out with the generic german headings (even in english). I use an english table.

So if you get a known case, and poke the values in, and you are out by a factor of 2, ie you get 2x where you're expecting x, then this is the most likely cause.

(I have no problem with people hacking proggies to meet their own needs. I've even hacked the input tables of my own proggies to match a different section's input forms).

If you are looking at hints on how to do arithmetic on strange things like matricies, there is the REXX AAT, which has directly cuttable rexx code that does most of this matrix stuff, square roots, &c. You can get it in "english" and in "czechy" as a suite of web pages that you can localise.
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