# A curious conjecture about self-reciprocal polynomials

Let $p$ be a polynomial
$p(z) = a_0 + a_1z+\dots + a_nz^n\ \ \ \ \ \ a_i\in {\bf C},$
and let $p^*$ denote the reciprocal polynomial
$p^*(z) = a_n + a_{n-1}z+\dots + a_0z^n=z^np(1/z).$
We say $p$ is self-reciprocal if $p=p^*$. For example,
$1+2z+3z^2+2z^3+z^4$
and
$1+2z+3z^2+3z^3+2z^4+z^5$
are self-reciprocal. Suppose
$p(z) = a_0 + a_1z+\dots + a_dz^d+a_{d+1}z^{d+1}+a_dz^{d}+\dots +a_{1}z^{2d+1}+a_0z^{2d+2},\ \ \ \ \ \ a_i\in {\bf C},$
or
$p(z) = a_0 + a_1z+\dots + a_dz^d+a_{d+1}z^{d+1}+a_{d+1}z^{d+2}+a_dz^{d+3}+\dots +a_{1}z^{2d+2}+a_0z^{2d+3},\ \ \ \ \ \ a_i\in {\bf C},$

The question is this: for which increasing sequences $a_0 do the polynomial roots $p(z)=0$ lie on the unit circle $|z|=1$?

Some examples:

This represents, when $d=1$ and $a_0=1$ and $a_1=1.1$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.2$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.3$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.4$ the largest $a_{d+1}$ which has this roots property is as in the plot.

This represents, when $d=1$ and $a_0=1$ and $a_1=1.5$ the largest $a_{d+1}$ which has this roots property is as in the plot.

Conjecture 1: Let $s:{\bf Z}_{>0}\to {\bf R}_{>0}$ be a ”slowly increasing” function.

1. Odd degree case.
If $g(z)= a_0 + a_1z + \dots + a_dz^d$, where $a_i=s(i)$, then the roots of $p(z)=g(z)+z^{d+1}g^*(z)$ all lie on the unit circle.
2. Even degree case.
The roots of
$p(z)=a_0 + a_1z + \dots + a_{d-1}z^{d-1} + a_{d}z^{d} + a_{d-1}z^{d+1} + \dots + a_{1}z^{2d-1}+a_0z^{2d}$
all lie on the unit circle.

The next conjecture gives an idea as to how fast a “slowly increasing” function can grow.

Conjecture 2:* Consider the even degree case only. The polynomial
$p(z)=a_0 + a_1z + \dots + a_{d-1}z^{d-1} + a_{d}z^{d} + a_{d-1}z^{d+1} + \dots + a_{1}z^{2d-1}+a_0z^{2d}$,
with $a_0=1$ and all $a_i>0$, is symmetric increasing if and only if it can be written as a product of quadratics of the form $x^2-2\cos(\theta)x+1$, where all but one of the factors satisfy $2\pi/4\leq \theta\leq 4\pi/3$. One of the factors can be of the form $x^2+\alpha x+1$, for some $\alpha \geq 0$.

* It was pointed out to me by Els Withers that this conjecture is false.

# Lester Hill’s “The checking of the accuracy …”, part 11

The field $F_{101}$

All essential points connected with the checking of telegraphic sequences by the methods proposed in this paper may be fully illustrated in one finite field. For our purposes, perhaps the most useful field is $F_{101}$, to which we shall confine our attention in the following sections. The elements of the field $F_{101}$ are the one hundred and one marks\footnote{Hill actually uses the symbol $X$ in place of $100$.} $0$, $1$, $2$, $\dots$, $100$. The operations of addition and multiplication are effected as explained in a previous example; and are abbreviated as suggested. To determine sums and products, we regard the marks of the field momentarily as integers of elementary arithmetic. Thus we have

$\sum_1^n f_i = f_h,\ \ \ \ \ \ (\prod_1^n f_i = f_k),$

the $f_i$ being $n$ marks of $F_{101}$, distinct or not, if, when the $f_i$ are momentarily regarded as integers of elementary arithmetic, the congruence

$\sum_1^n f_i \equiv f_h \pmod{101},\ \ \ \ \ \ (\prod_1^n f_i \equiv f_k \pmod{101}),$

holds. It will not be possible to provide a full multiplication table for the field $F_{101}$. But the following special table will be found convenient.

$\begin{array}{r|rrr} x & x^2 & 1/x & -x\\ 1 & 1 & 1 & 100 \\ 2 & 4 & 51 & 99 \\ 3 & 9 & 34 & 98 \\ 4 & 16 & 76 & 97 \\ 5 & 25 & 81 & 96 \\ 6 & 36 & 17 & 95 \\ 7 & 49 & 29 & 94 \\ 8 & 64 & 38 & 93 \\ 9 & 81 & 45 & 92 \\ 10 & 100 & 91 & 91 \\ 11 & 20 & 46 & 90 \\ 12 & 43 & 59 & 89 \\ 13 & 68 & 70 & 88 \\ 14 & 95 & 65 & 87 \\ 15 & 23 & 27 & 86 \\ 16 & 54 & 19 & 85 \\ 17 & 87 & 6 & 84 \\ 18 & 21 & 73 & 83 \\ 19 & 58 & 16 & 82 \\ 20 & 97 & 96 & 81 \\ 21 & 37 & 77 & 80 \\ 22 & 80 & 23 & 79 \\ 23 & 24 & 22 & 78 \\ 24 & 71 & 80 & 77 \\ 25 & 19 & 97 & 76 \\ 26 & 70 & 35 & 75 \\ 27 & 22 & 15 & 74 \\ 28 & 77 & 83 & 73 \\ 29 & 33 & 7 & 72 \\ 30 & 92 & 64 & 71 \\ 31 & 52 & 88 & 70 \\ 32 & 14 & 60 & 69 \\ 33 & 79 & 49 & 68 \\ \end{array}$

$\begin{array}{r|rrr} x & x^2 & 1/x & -x\\ 34 & 45 & 3 & 67 \\ 35 & 13 & 26 & 66 \\ 36 & 84 & 87 & 65 \\ 37 & 56 & 71 & 64 \\ 38 & 30 & 8 & 63 \\ 39 & 6 & 57 & 62 \\ 40 & 85 & 48 & 61 \\ 41 & 65 & 69 & 60 \\ 42 & 47 & 89 & 59 \\ 43 & 31 & 47 & 58 \\ 44 & 17 & 62 & 57 \\ 45 & 5 & 9 & 56 \\ 46 & 96 & 11 & 55 \\ 47 & 88 & 43 & 54 \\ 48 & 82 & 40 & 53 \\ 49 & 78 & 33 & 52 \\ 50 & 76 & 99 & 51 \\ 51 & 76 & 2 & 50 \\ 52 & 78 & 68 & 49 \\ 53 & 82 & 61 & 48 \\ 54 & 88 & 58 & 47 \\ 55 & 96 & 90 & 46 \\ 56 & 5 & 92 & 45 \\ 57 & 17 & 39 & 44 \\ 58 & 31 & 54 & 43 \\ 59 & 47 & 12 & 42 \\ 60 & 65 & 32 & 41 \\ 61 & 85 & 53 & 40 \\ 62 & 6 & 44 & 39 \\ 63 & 30 & 93 & 38 \\ 64 & 56 & 30 & 37 \\ 65 & 84 & 14 & 36 \\ 66 & 13 & 75 & 35 \\ \end{array}$

$\begin{array}{r|rrr} x & x^2 & 1/x & -x\\ 67 & 45 & 98 & 34 \\ 68 & 79 & 52 & 33 \\ 69 & 14 & 41 & 32 \\ 70 & 52 & 13 & 31 \\ 71 & 92 & 37 & 30 \\ 72 & 33 & 94 & 29 \\ 73 & 77 & 18 & 28 \\ 74 & 22 & 86 & 27 \\ 75 & 70 & 66 & 26 \\ 76 & 19 & 4 & 25 \\ 77 & 71 & 21 & 24 \\ 78 & 24 & 79 & 23 \\ 79 & 80 & 78 & 22 \\ 80 & 37 & 24 & 21 \\ 81 & 97 & 5 & 20 \\ 82 & 58 & 85 & 19 \\ 83 & 21 & 28 & 18 \\ 84 & 87 & 95 & 17 \\ 85 & 54 & 82 & 16 \\ 86 & 23 & 74 & 15 \\ 87 & 95 & 36 & 14 \\ 88 & 68 & 31 & 13 \\ 89 & 43 & 42 & 12 \\ 90 & 20 & 55 & 11 \\ 91 & 100 & 10 & 10 \\ 92 & 81 & 56 & 9 \\ 93 & 64 & 63 & 8 \\ 94 & 49 & 72 & 7 \\ 95 & 36 & 84 & 6 \\ 96 & 25 & 20 & 5 \\ 97 & 16 & 25 & 4 \\ 98 & 9 & 67 & 3 \\ 99 & 4 & 50 & 2 \\ 100 & 1 & 100 & 1 \end{array}$

Squares, reciprocals, negatives

Using the scheme of reciprocals shown in this Table, we may easily perform an rational operations in $F_{101}$.

Example:

Suppose, for example, that we wish to solve the system of equations:

$36x-79y=52,\ \ \ 90x+85y = 98.$

They may be written

$x-79y/36 = 13/9,\ \ \ \ x+17y/18 = 49/45$

and the fractions are quickly evaluated. Thus: $-79/36 = 96$. Determining the fractions in this manner, we write the two equations in the form:

$x+96y=80,\ \ \ \ x+29y=37,$

whence $y=73$ and $x=41$

The modulus $101$ is very convenient to work with. The residue, modulo $101$, of any integer is immediately obvious, at sight of the integer, and is therefore obtained without computation.

# The trouble with college math classes …

Ran across this terrific observation in Wallace’s book on set theory and mathematics, and thought I’d share:

The trouble with college math classes – which classes consist almost entirely in the rhythmic ingestion and regurgitation of abstract information, and are paced in such a way as to maximize this reciprocal data-flow – is that their shear surface-level difficulty can fool us into thinking we really know something when all we really “know” is abstract formulas and rules for their deployment. Rarely do math classes ever tell us if a formula is truely significant, or why, or where it came from, or what was at stake.

David Foster Wallace at the Hammer Museum in Los Angeles, January 2006. 2006 CC BY-SA 3.0

And, of course, rarely do students think to ask – the formulas alone take so much work to “understand” (i.e., be able to solve problems correctly with), we often aren;t aware that we don’t understand them at all. That we end up not even knowing that we don’t know if the most incidious part of most math classes.

- David Foster Wallace, in Everything and More (section 2a)

# More odd examples of p-ary bent functions

In an earlier post I discussed bent functions $f:GF(3)^2\to GF(3)$. In this post, I’d like to give some more examples, based on a recent paper with Caroline Melles, Charles Celerier, David Phillips, and Steven Walsh, based on computations using Sage/pythpn programs I wrote with Charles Celerier.

We start with any function $f:GF(p)^n\to GF(p)$. The Cayley graph of $f$ is defined to be the edge-weighted digraph

$\Gamma_f = (GF(p)^n, E_f ),$
whose vertex set is $V=V(\Gamma_f)=GF(p)^n$ and the set of edges is defined by

$E_f =\{(u,v) \in GF(p)^n\ |\ f(u-v)\not= 0\},$
where the edge $(u,v)\in E_f$ has weight $f(u-v)$. However, if $f$ is even then we can (and do) regard $\Gamma_f$ as an edge-weighted (undirected) graph.

We assume, unless stated otherwise, that $f$ is even.

For each $u\in V$, define

• $N(u)=N_{\Gamma_f}(u)$ to be the set of all neighbors of $u$ in $\Gamma_f$,
• $N(u,a)=N_{\Gamma_f}(u,a)$ to be the set of all neighbors $v$ of $u$ in $\Gamma_f$ for which the edge $(u,v)\in E_f$ has weight $a$ (for each $a\in GF(p)^\times = GF(p)-\{0\}$),
• $N(u,0)=N_{\Gamma_f}(u,0)$ to be the set of all non-neighbors $v$ of $u$ in $\Gamma_f$ (i.e., we have $(u,v)\notin E_f$),
• the support of $f$ is
${\rm supp}(f)=\{v\in V\ |\ f(v)\not=0\}$

Let $\Gamma$ be a connected edge-weighted graph which is regular as a simple (unweighted) graph. The graph $\Gamma$ is called strongly regular with parameters $v$, $k=(k_a)_{a\in W}$, $\lambda=(\lambda_a)_{a\in W^3}$, $\mu=(\mu_a)_{a\in W^2}$, denoted $SRG_{W}(v,k,\lambda,\mu)$, if it consists of $v$ vertices such that, for each $a=(a_1,a_2)\in W^2$

$|N(u_1,a_1) \cap N(u_2,a_2)| = \left\{ \begin{array}{ll} k_{a}, & u_1=u_2,\\ \lambda_{a_1,a_2,a_3}, & u_1\in N(u_2,a_3),\ u_1\not= u_2,\\ \mu_{a}, &u_1\notin N(u_2),\ u_1\not= u_2,\\ \end{array} \right.$
where $k$, $\lambda$, $\mu$ are as above. Here, $W= GF(p)$ is the set of weights, including $0$ (recall an “edge” has weight $0$ if the vertices are not neighbors).

This example is intended to illustrate the bent function $b_8$ listed in the table below
$\begin{array}{c|ccccccccc} GF(3)^2 & (0, 0) & (1, 0) & (2, 0) & (0, 1) & (1, 1) & (2, 1) & (0, 2) & (1, 2) & (2, 2) \\ \hline b_8 & 0 & 2 & 2 & 0 & 0 & 1 & 0 & 1 & 0 \\ \end{array}$

Consider the finite field
$GF(9) = GF(3)[x]/(x^2+1) = \{0,1,2,x,x+1,x+2,2x,2x+1,2x+2\}.$
The set of non-zero quadratic residues is given by
$D = \{1,2,x,2x\}.$
Let $\Gamma$ be the graph whose vertices are $GF(9)$ and whose edges $e=(a,b)$ are those pairs for which $a-b\in D$.

The graph looks like the Cayley graph for $b_8$ in the Figure below

Bent function b_8 on GF(3)^2

except

$8\to 0, 0\to 2x+2, 1\to 2x+1, 2\to 2x,$
$3\to x+2, 4\to x+1, 5\to x, 6\to 2, 7\to 1, 8\to 0.$
This is a strongly regular graph with parameters $(9,4,1,2)$.

$\begin{array}{c|ccccccccc} v & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ \hline N(v,0) & 3,4,6,8 & 4,5,6,7 & 3,5,7,8 & 0,2,6,7 & 0,1,7,8 & 1,2,6,8 & 0,1,3,5 & 1,2,3,4 & 0,2,4,5 \\ N(v,1) & 5,7 & 3,8 & 4,6 & 1,8 & 2,6 & 0,7 & 2,4 & 0,5 & 1,3 \\ N(v,2) & 1,2 & 0,2 & 0,1 & 4,5 & 3,5 & 3,4 & 7,8 & 6,8 & 6,7 \\ \end{array}$

The axioms of an edge-weighted strongly regular graph can be directly verified using this table.

Let $S$ be a finite set and let $R_0, R_1, \dots, R_s$ denote binary relations on $S$ (subsets of $S\times S$). The dual of a relation $R$ is the set

$R^* = \{(x,y)\in S\times S\ |\ (y,x)\in R\}.$
Assume $R_0=\Delta_S= \{ (x,x)\in S\times S\ |\ x\in S\}$. We say $(S,R_0,R_1,\dots,R_s)$ is an $s$-class association scheme on $S$ if the following properties hold.

• We have a disjoint union

$S\times S = R_0\cup R_1\cup \dots \cup R_s,$
with $R_i\cap R_j=\emptyset$ for all $i\not= j$.

• For each $i$ there is a $j$ such that $R_i^*=R_j$ (and if $R_i^*=R_i$ for all $i$ then we say the association scheme is symmetric).
• For all $i,j$ and all $(x,y)\in S\times S$, define

$p_{ij}(x,y) = |\{z\in S\ |\ (x,z)\in R_i, (z,y)\in R_j\}|.$
For each $k$, and for all $x,y\in R_k$, the integer $p_{ij}(x,y)$ is a constant, denoted $p_{ij}^k$.

These constants $p_{ij}^k$ are called the intersection numbers of the association scheme.

For this example of $b_8$, we compute the adjacency matrix associated to the members $R_1$ and $R_2$ of the association scheme $(G,R_0,R_1,R_2,R_3)$, where $G = GF(3)^2$,

$R_i = \{(g,h)\in G\times G\ |\ gh^{-1} \in D_i\},\ \ \ \ \ i=1,2,$
and $D_i = f^{-1}(i)$.

Consider the following Sage computation:

sage: attach "/home/wdj/sagefiles/hadamard_transform2b.sage"
sage: FF = GF(3)
sage: V = FF^2
sage: Vlist = V.list()
sage: flist = [0,2,2,0,0,1,0,1,0]
sage: f = lambda x: GF(3)(flist[Vlist.index(x)])
sage: F = matrix(ZZ, [[f(x-y) for x in V] for y in V])
sage: F  ## weighted adjacency matrix
[0 2 2 0 0 1 0 1 0]
[2 0 2 1 0 0 0 0 1]
[2 2 0 0 1 0 1 0 0]
[0 1 0 0 2 2 0 0 1]
[0 0 1 2 0 2 1 0 0]
[1 0 0 2 2 0 0 1 0]
[0 0 1 0 1 0 0 2 2]
[1 0 0 0 0 1 2 0 2]
[0 1 0 1 0 0 2 2 0]
sage: eval1 = lambda x: int((x==1))
sage: eval2 = lambda x: int((x==2))
sage: F1 = matrix(ZZ, [[eval1(f(x-y)) for x in V] for y in V])
sage: F1
[0 0 0 0 0 1 0 1 0]
[0 0 0 1 0 0 0 0 1]
[0 0 0 0 1 0 1 0 0]
[0 1 0 0 0 0 0 0 1]
[0 0 1 0 0 0 1 0 0]
[1 0 0 0 0 0 0 1 0]
[0 0 1 0 1 0 0 0 0]
[1 0 0 0 0 1 0 0 0]
[0 1 0 1 0 0 0 0 0]
sage: F2 = matrix(ZZ, [[eval2(f(x-y)) for x in V] for y in V])
sage: F2
[0 1 1 0 0 0 0 0 0]
[1 0 1 0 0 0 0 0 0]
[1 1 0 0 0 0 0 0 0]
[0 0 0 0 1 1 0 0 0]
[0 0 0 1 0 1 0 0 0]
[0 0 0 1 1 0 0 0 0]
[0 0 0 0 0 0 0 1 1]
[0 0 0 0 0 0 1 0 1]
[0 0 0 0 0 0 1 1 0]
sage: F1*F2-F2*F1 == 0
True
sage: delta = lambda x: int((x[0]==x[1]))
sage: F3 = matrix(ZZ, [[(eval0(f(x-y))+delta([x,y]))%2 for x in V] for y in V])
sage: F3
[0 0 0 1 1 0 1 0 1]
[0 0 0 0 1 1 1 1 0]
[0 0 0 1 0 1 0 1 1]
[1 0 1 0 0 0 1 1 0]
[1 1 0 0 0 0 0 1 1]
[0 1 1 0 0 0 1 0 1]
[1 1 0 1 0 1 0 0 0]
[0 1 1 1 1 0 0 0 0]
[1 0 1 0 1 1 0 0 0]
sage: F3*F2-F2*F3==0
True
sage: F3*F1-F1*F3==0
True
sage: F0 = matrix(ZZ, [[delta([x,y]) for x in V] for y in V])
sage: F0
[1 0 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0]
[0 0 0 1 0 0 0 0 0]
[0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 1 0 0]
[0 0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 1]
sage: F1*F3 == 2*F2 + F3
True


The Sage computation above tells us that the adjacency matrix of $R_1$ is

$A_1 = \left(\begin{array}{rrrrrrrrr} 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \end{array}\right),$
the adjacency matrix of $R_2$ is

$A_2 = \left(\begin{array}{rrrrrrrrr} 0 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 0 \end{array}\right),$
and the adjacency matrix of $R_3$ is

$A_3 = \left(\begin{array}{rrrrrrrrr} 0 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 0 & 1 & 1 \\ 1 & 0 & 1 & 0 & 0 & 0 & 1 & 1 & 0 \\ 1 & 1 & 0 & 0 & 0 & 0 & 0 & 1 & 1 \\ 0 & 1 & 1 & 0 & 0 & 0 & 1 & 0 & 1 \\ 1 & 1 & 0 & 1 & 0 & 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 0 \end{array}\right)$
Of course, the adjacency matrix of $R_0$ is the identity matrix. In the above computation, Sage has also verified that they commute and satisfy

$A_1A_3 = 2A_2+A_3$
in the adjacency ring of the association scheme.

Conjecture:
Let $f:GF(p)^n\to GF(p)$ be a bent function, with $p>2$. If the level curves of $f$ give rise to a weighted partial difference set then $f$ is weakly regular, and the corresponding (unweighted) partial difference set is of (positive or negative) Latin square type.

For more details, see the paper [CJMPW] with Caroline Melles, Charles Celerier, David Phillips, and Steven Walsh.

# Complements in the symmetric group

Almost 20 years ago I was asked a question by Herbert Kociemba, a computer scientist who has one of the best Rubik’s cube solving programs known. Efficient methods of storing permutations in $S_E$ and $S_V$ (the groups of all permutations of the edges $E$ and vertices $V$, respectively, of the Rubik’s cube) are needed, hence leading naturally to the concept of the complement of $S_m$ in $S_n$. Specifically, he asked if $S_8$ has a complement in $S_{12}$ (this terminology is defined below). The answer is, as we shall see, ”no.” Nonetheless, it turns out to be possible to introduce a slightly more general notion of a ”$k$-tuple of complementary subgroups” for which the answer to the analogous question is ”yes.”

This post is a very short summary of part of a paper I wrote (still unpublished) which can be downloaded here. This post explains the ”no” part of the answer. For the more technical ”yes” part of the answer, see the more complete pdf version.

Notation: If $X$ is any finite set then

• $|X|$ denotes the number of elements in $X$.
• $S_X$ denotes the symmetric group on $X$.
• $S_n$ denotes the symmetric group on $\{1,2,...,n\}$.
• $A_n$ denotes its alternating subgroup of even permutations.
• $C_n$ denotes the cyclic subgroup of $S_n$ generated by the $n$-cycle $(1,2,...,n)$.
• $M_{10}$ denotes ”the Mathieu group of degree $10$” and order $720=10!/7!$, which we define as the subgroup of $S_{10}$ generated by $(2,6,10,7)(3,9,4,5)$ and $(1,5)(3,8)(4,10)(7,9)$.
• $M_{11}$ denotes the Mathieu group of degree $11$ and order $7920=11!/7!$ generated by $(1,2,3,4,5,6,7,8,9,10,11)$ and $(3,7,11,8)(4,10,5,6)$.
• $M_{12}$ denotes the Mathieu group of degree $12$ and order $95040=12!/7!$ generated by $(2,3,5,9,8,10,6,11,4,7,12)$ and $(1,12)(2,11)(3,10)(4,9)(5,8)(6,7)$.
• For any prime power $q$, ${\Bbb F}_q$ denotes the finite field with $q$ elements.
• $AGL_n({{\Bbb F}}_q)$ denotes the affine group of transformations on ${\Bbb F}_q^n$ of the form $\vec{v}\longmapsto A\vec{v}+\vec{a}$, where $A\in GL_n({{\Bbb F}}_q)$ and $\vec{a} \in {\Bbb F}_q^n$.

If $G$ is a finite group and $G_1,G_2$ are subgroups then we say $G_2$ is the complement of $G_1$ when

• $G_1\cap G_2=\{1\}$, the identity of $G$,
and
• $G=G_1\cdot G_2=\{g_1g_2\ |\ g_1\in G_1,\ g_2\in G_2\}$.

Let $X$ denote a finite set. If $G$ is a subgroup of $S_X$ and $x\in X$ then we let $G_x$ denote the stabilizer of $x$ in $G$:
$G_x=\{ g\in G\ |\ g(x)=x\}.$

Let $G$ be a permutation group acting on a finite set $X$ (so $G$ is a subgroup of the symmetric
group of $X$, $S_X$). Let $k\geq 1$ be an integer and let
$X^{(k)}=\{{\rm distinct\ }k{\rm -tuples\ in\ }X\} =\{(x_1,x_2,...,x_k)\ |\ x_i\not= x_j,\ 1\leq i \textless j\leq k\}.$
We say $G$ acts $k$-transitively on $X$ if $G$ acts transitively on $X^{(k)}$ via the ”diagonal” action
$g:(x_1,x_2,...,x_k)\longmapsto (g(x_1),g(x_2),...,g(x_k)).$

If $G$ acts transitively on $X$ and $G_x=1$ for some (hence all) $x\in X$ then we say $G$ acts regularly on $X$. If $G$ acts $k$-transitively on $X$ and acts regularly on $X^{(k)}$ then we say $G$ acts sharply $k$-transitively on $X$.

The classification of $k$-transitive groups, for $k\geq 4$, is to Jordan: A sharply $k$-transitive group, $k\geq 4$, must be one of the following.

• $k \geq 6$: $S_k$ , $S_{k+1}$ and $A_{k+2}$ only.
• $k = 5$: $S_5$ , $S_6$ , $A_7$ and the Mathieu group $M_{12}$.
• $k = 4$: $S_4$ , $S_5$ , $A_6$ and the Mathieu group $M_{11}$.

We give a table which indicates, for small values of $n$, which $S_m$ have a complement in $S_n$.

 $n$ $m$ complement of $S_m$ in $S_n$ $4$ $2$ $A_4$ $4$ $3$ $C_4$ $5$ $2$ $A_5$ $5$ $3$ $\langle (1,2,3,4,5),(2,3,5,4)\rangle \cong AGL_1({{\Bbb F}}_5)$ size $20$ $5$ $4$ $C_5$ $6$ $2$ $A_6$ $6$ $3$ $\langle (2,3,4,5,6),(3,4,6,5)\rangle \cong PGL_2({{\Bbb F}}_5)$ size $120$ $6$ $5$ $C_6$ $7$ $2$ $A_7$ $7$ $5$ $\langle (1,2,6,5,3,7),(1,4,3,2,5,6)\rangle \cong AGL_1({{\Bbb F}}_7)$ size $42$ $7$ $6$ $C_7$ $8$ $2$ $A_8$ $8$ $5$ $\langle (1,5,8,6,7,4),(1,5,8,7)(2,3,4,6)\rangle \cong PGL_2({{\Bbb F}}_7)$ size $336$ $8$ $6$ $AGL_1({{\Bbb F}}_8)$ size $56$ $8$ $7$ $C_8$ $9$ $2$ $A_9$ $9$ $6$ $PGL_2({{\Bbb F}}_8)$ size $504$ $9$ $7$ $AGL_1({{\Bbb F}}_9)$ size $72$ $9$ $8$ $C_9$ $10$ $2$ $A_{10}$ $10$ $7$ $M_{10}$ size $720$ ($PGL_2({{\Bbb F}}_9)$ is another) $10$ $9$ $C_{10}$ $11$ $2$ $A_{11}$ $11$ $7$ $M_{11}$ size $7920$ $11$ $9$ $AGL_1({{\Bbb F}}_{11})$ size $110$ $11$ $10$ $C_{11}$ $12$ $2$ $A_{12}$ $12$ $7$ $M_{12}$ size $95040$ $12$ $9$ $PGL_2({{\Bbb F}}_{11})$ $=\langle (1,2,3,4,5,6,7,8,9,10,11), (1,2,4,8,5,10,9,7,3,6),$ $(1,10)(2,5)(3,7)(4,8)(6,9)(11,12)\rangle$ size $1320$ $12$ $11$ $C_{12}$

Proposition: $S_m$ has a complement in $S_n$ if and only if there is an subgroup $H$ of $S_n$ such that

1. $H$ acts $(n-m)$-transitively on $\{1,2,…,n\}$,
2. $|H|=n(n-1)...(m+1)=n!/m!$.

Example: $S_{10}$ has not one but two non-isomorphic subgroups, $M_{10}$ and $PGL_2({{\Bbb F}}_9)$, of order $720=10!/7!$, each of which acts $3$-transitively on $\{1,2,...,10\}$. Thus $S_7$ has two non-isomorphic complements in $S_{10}$.

The statement below is the main result.

Theorem: The following statements hold.

1. If $n>2$ is not a prime power or a prime power plus $1$ then the only $1 for which $S_m$ has a complement in $S_n$ are $m=2$ and $m=n-1$.
2. If $n>12$ is a prime power and not a prime power plus $1$ then the only $1 for which $S_m$ has a complement in $S_n$ are $m=2$, $m=n-2$ and $m=n-1$.
3. If $n>12$ is a prime power plus $1$ but not a prime power then the only $1 for which $S_m$ has a complement in $S_n$ are $m=2$, $m=n-3$ and $m=n-1$.
4. If $n>12$ is both a prime power plus $1$ and a prime power then the only $1 for which $S_m$ has a complement in $S_n$ are $m=2$, $m=n-3$, $m=n-2$ and $m=n-1$.
5. If $n\leq 12$, see the above table.

# Grobner bases and permutation puzzles, according to Martin Kreuzer

This post is about some research Prof. Martin Kreuzer, at the Univ. of Passau, does which combines Grobner bases and solving permutation puzzles, such as the Rubik’s cube.

Prof. Martin Kreuzer giving a talk in Annapolis, 2013-12-04

Non-commutative Grobner Bases and Twisty Puzzles was the title of his talk (the link takes you to his slides, which he kindly allowed me to post).

Grobner bases were introduced in the mid-1960s (by Bruno Buchberger and, independently, Heisuke Hironaka), but in hindsight it is amazing they weren’t invented earlier. Indeed, Grobner bases arise when one tried to generalize long division of polynomials in one variable to polynomials of several variables. Non-commutative Grobner bases arise when one does not allow the variables to commute with each other. The main setting is where
$G = \langle x_1,...,x_n\ |\ l_1=r_1,...,l_s=r_s\rangle$ is a finitely presented monoid,
$K \langle X \rangle= \oplus_{g\in G} Kg = K/I$ is a monoid ring, and where
$I = \langle l_1-r_1,...,l_s-r_s\rangle$ is a two-sided ideal generated by binomials.

First, Prof. Kreuzer introduces the non-commutative analogs of term ordering, leading term, and Grobner basis with respect to that ordering.

The non-commutative analog of the Buchberger procedure.

After developing enough theory, it turns out that the word problem in group theory can be re-expressed in terms of the membership problem for $K \langle X \rangle$.

Prof Kreuzer discussing aspects of computational group theory.

Martin Kreuzer defines a twisty puzzle as a special form of a permutation puzzle. Its pieces (usually called cubies) can be permuted by twisting certain faces of the puzzle. The cubies carry stickers defining their location in the solved puzzle. For example, with the Rubik’s Cube, the 54 stickers of a 3×3×3 cube can be permuted by 90 degree turns of the faces. The possible permutations form a group called the Rubik’s cube group. Then he gives the following definition and remarkable Theorem.

Definition: Let G be a twisty puzzle group given via a set X of generating twists.
(a) The graph whose vertices are the elements of G and whose edges are pairs of positions which differ by just one twist is called the Cayley graph of G.
(b) The diameter δ(G) of the Cayley graph of G is called God’s number for the twisty puzzle.
(c) An algorithm which produces for every scrambled position p an element of X∗ whose length is the distance (in the Cayley graph) from p to the solved position is called a God’s algorithm for the twisty puzzle.

Theorem (God’s Algorithm via Grobner Bases)
Let X be a set of twists (and their inverses) which generate the group of a twisty puzzle. Given a scrambled position p of the puzzle, perform the following steps:
(1) Compute a Grobner basis G of the defining ideal of the twisty puzzle group ring w.r.t. a length compatible word ordering.
(2) Find any word w ∈ X∗ representing a sequence of twists that result in the position p.
(3) Compute NRG(w) ∈ X∗ and return this word.

This is an algorithm which computes a word representing a shortest possible sequence of twists that produces p from the solved position.

This has even been implemented on a computer in Cocoa, for a permutation puzzle with a very small group.

Prof Kreuzer at his laptop running a “toy” implementation of God’s algorithm.

# Another example of a Prohibition era cipher message

The following message, with Elizebeth Friedman‘s decryption, is a message from “Consolidated Exporters” (a Vancouver Canada company illegally importing liquor into the US) to one of its fleet of about 50 “black ships”, the SS Noble. (They are called “black ships” because they “rum run” without lights at night to avoid detection.)

Many thanks to the George C. Marshall Foundation, Lexington, Virginia, for providing this reproduction!