I'm trying to figure out if there is a way to determine how many unique patterns exist for a given $N\times N$ grid if you choose N points on the grid. For example, for a $2\times 2$ grid we can get two unique patterns from the six possible combinations. The rest are just rotations and mirrors of the two unique patterns below

[x] [x]

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[x] [ ]

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Is there a mathematical way of determining a unique number of patterns for a NxN grid where N=3,4,5,6,7,8?

I figured for a 3x3, there are 14 unique patterns for picking 3 random points on the grid, but it gets tedious after that.

N:   N^2 : N^2 Choose N Unique pattern
2    4     6            2             
3    9     84           14            
4    16    1820         ????          
5    25    53130        ????          
6    36    1947792      ????          
7    49    85900584     ????          

  • $\begingroup$ one must choose $N$ points on a binary grid? $\endgroup$ – nbubis Nov 17 '13 at 4:46
  • $\begingroup$ yes, forgot to add that in $\endgroup$ – user109380 Nov 17 '13 at 4:52
  • 2
    $\begingroup$ Are you familiar with group theory? Polya enumeration could be the way to go here. $\endgroup$ – Gerry Myerson Nov 17 '13 at 5:26

We can actually do a bit more and compute the cycle index $Z(G_N)$ for general $N$, where we have to distinguish between $N$ even and $N$ odd. This will permit lookup in the OEIS, which in turn leads to more material about this interesting problem.

We proceed to enumerate the permutations of $G_N$ by their cycle structure. For $N$ even, we get the identity, which contributes $$a_1^{N^2}.$$ There is a vertical reflection, which contributes $$a_2^{N^2/2},$$ the same for a horizontal reflection, i.e. $$a_2^{N^2/2}.$$ The reflection in the rising diagonal contributes $$a_1^N a_2^{(N^2-N)/2},$$ the same for the other diagonal, i.e. $$a_1^N a_2^{(N^2-N)/2}.$$

What remains are the rotations. Two of these contribute (recall that $N$ is even) $$2\times a_4^{N^2/4}$$ and one of them, $$a_2^{N^2/2}.$$ This gives for even $N$ the cycle index $$Z(G_N) = \frac{1}{8} \left( a_1^{N^2} + 3 a_2^{N^2/2} + 2 a_1^N a_2^{(N^2-N)/2} + 2 a_4^{N^2/4}\right).$$ For $N$ odd, we get the identity, which is $$a_1^{N^2}.$$ The two reflections now contribute $$2\times a_1^N a_2^{(N^2-N)/2}.$$ The reflection in the two diagonals are unchanged and contribute $$2\times a_1^N a_2^{(N^2-N)/2}$$ What remains is the rotations, two of which have cycle structure $$2\times a_1 a_4^{(N^2-1)/4}$$ and the last one has cycle structure $$a_1 a_2^{(N^2-1)/2}.$$

This gives for odd $N$ the cycle index $$Z(H_N) = \frac{1}{8} \left( a_1^{N^2} + 4 a_1^N a_2^{(N^2-N)/2} + 2 a_1 a_4^{(N^2-1)/4}+a_1 a_2^{(N^2-1)/2}\right).$$ Evidently for $N$ marks being placed on the grid we seek to compute $$[z^N] Z(G_N)(1+z) \quad\text{and}\quad [z^N] Z(H_N)(1+z),$$ alternating between the two for $N$ even and $N$ odd.

This produces the sequence $$1, 2, 16, 252, 6814, 244344, 10746377, 553319048, 32611596056, 2163792255680,\ldots$$ which is A019318 from the OEIS.

Here is the Maple program that was used to compute these cycle indices.


pet_varinto_cind :=
proc(poly, ind)
           local subs1, subs2, polyvars, indvars, v, pot, res;

           res := ind;

           polyvars := indets(poly);
           indvars := indets(ind);

           for v in indvars do
               pot := op(1, v);

               subs1 :=

               subs2 := [v=subs(subs1, poly)];

               res := subs(subs2, res);


G :=
        if type(N,odd) then return FAIL; fi;

        2*a[1]^N*a[2]^((N^2-N)/2) + 2*a[4]^(N^2/4));

H :=
        if type(N,even) then return FAIL; fi;

        a[1]*a[2]^((N^2-1)/2) + 2*a[1]*a[4]^((N^2-1)/4));

v :=
        option remember;
        local p, k, gf;

        if type(N, even) then
            gf := expand(pet_varinto_cind(1+z, G(N)));
            gf := expand(pet_varinto_cind(1+z, H(N)));

        coeff(gf, z, N);

Here is another interesting MSE cycle index computation I. This MSE cycle index computation II is relevant also.

  • $\begingroup$ I want to ask about the notation $Z(G_N)(1+z).$ I assume this means that $a_1$ gets replaced by $1+z,$ $a_2$ gets replaced by $1+z^2,$ and $a_4$ gets replaced by $1+z^4,$ but I wanted to double check. $\endgroup$ – Will Orrick Jan 1 '14 at 15:27
  • $\begingroup$ That is correct. Wikipedia has useful data on this: Cycle Index and Polya Enumeration Theorem. $\endgroup$ – Marko Riedel Jan 1 '14 at 15:46

Here's an answer that leads to explicit formulas whose most complicated term is a sum over a product of binomial coefficients. I haven't yet made any attempt to connect it to the excellent solutions in the answers given earlier.

First some notation. The dihedral group of the square, $D_4,$ has elements $\{e,\rho,\rho^2,\rho^3,h,v,d_1,d_2\},$ where $e$ is the identity, $\rho$ is the $90^\circ$ rotation, $h$ is the reflection about the horizontal axis, $v$ is the reflection about the vertical axis, $d_1$ is the reflection about the forward diagonal, and $d_2$ is the reflection about the backward diagonal.

(Added: Having never learned Pólya enumeration properly, I made my answer overly complicated. I leave my previous answer, but add a more direct route to equation (**) which appears below. In this problem we have $D_4$ acting on the set $X,$ consisting of the $\binom{n^2}{n}$ ways of coloring $n$ small squares in the $n\times n$ grid black, leaving the remaining squares white. The orbit counting theorem leads immediately to (**) since $$ K(n)=\lvert\mathcal{O}\rvert=\frac{1}{8}\sum_{g\in D_4}C(n,n,\langle g\rangle) $$ is identical to (**) when one realizes that $\langle\rho\rangle=\langle\rho^3\rangle=C_4.$ The orbit counting theorem is described in many places. For convenience I've appended some background information about it to the end of this post.)

Since the order of $D_4$ is $8,$ all subgroups are of order $1,$ $2,$ $4,$ or $8.$ Either of the two elements of order $4,$ $\rho$ and $\rho_3,$ generates $C_4,$ the rotation group of the square. Each of the five elements of order $2$, $\rho^2,$ $h,$ $v,$ $d_1,$ and $d_2,$ generates a subgroup of order $2.$ There are two additional subgroups of order $4,$ isomorphic to $C_2\times C_2.$ They are $\{e,\rho^2,h,v\}$ and $\{e,\rho^2,d_1,d_2\}.$ Any other group generated by two elements of order $2$ is all of $D_4.$

There are therefore $10$ subgroups with inclusions given by the following Hasse diagram.

enter image description here

Define $C(n,b,G)$ to be the set of configurations of the $n\times n$ grid with $b$ marked squares that are invariant under the group $G.$ We are ultimately interested in $b=n,$ but will start out with slightly greater generality. Define $\overline{C}(n,b,G)$ to be the subset of $C(n,b,G)$ consisting of elements that are not invariant under any group that properly contains $G.$ The number we wish to compute, which we denote $K(n),$ is $$ \begin{aligned} (*)\quad K(n)=&\frac{\lvert\overline{C}(n,n,D_4)\rvert}{1}+\frac{\lvert\overline{C}(n,n,\{e,\rho^2,h,v\})\rvert}{2}+\frac{\lvert\overline{C}(n,n,C_4)\rvert}{2}+\frac{\lvert\overline{C}(n,n,\{e,\rho^2,d_1,d_2\})\rvert}{2}\\ &+\frac{\lvert\overline{C}(n,n,\{e,h\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e,v\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e,\rho^2\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e,d_1\})\rvert}{4}\\ &+\frac{\lvert\overline{C}(n,n,\{e,d_2\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e\})\rvert}{8}. \end{aligned} $$

(Added: The set $C(n,n,G)$ is the set of colorings whose stabilizer contains $G,$ while $\overline{C}(n,n,G)$ is the set of colorings whose stabilizer equals $G.$ The formula (*) for $K(n),$ the number of orbits, comes from counting each coloring with a weight equal to the reciprocal of the size of its orbit, the weight having been computed using the orbit-stabilizer theorem. (See the end of the post.) For example, if $x$ has stabilizer $D_4,$ then the weight is $1/1=8/8;$ if $x$ has stabilizer $\{e,\rho^2,h,v\},$ then the weight is $1/2=4/8.$ Of course, as noted above, the use of (*) and the inclusion-exclusion argument below is rendered superfluous by the use of the orbit counting theorem. It is interesting that although (*) involves all ten subgroups of $D_4,$ the final expression (**) involves only the seven cyclic subgroups. My naive derivation does not explain this, but the orbit counting theorem makes it manifest that that must be the case.)

From the diagram, we obtain expressions for the $\overline{C}(n,b,G)$ in terms of the $C(n,b,G):$ $$ \begin{aligned} \lvert\overline{C}(n,b,D_4)\rvert&=\lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e,\rho^2,h,v\})\rvert&=\lvert C(n,b,\{e,\rho^2,h,v\})\rvert- \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,C_4)\rvert&=\lvert C(n,b,C_4)\rvert- \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e,\rho^2,d_1,d_2\})\rvert&=\lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert- \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e,h\})\rvert&=\lvert C(n,b,\{e,h\})\rvert- \lvert C(n,b,\{e,\rho^2,h,v\})\rvert\\ \vert\overline{C}(n,b,\{e,v\})\rvert&=\lvert C(n,b,\{e,v\})\rvert- \lvert C(n,b,\{e,\rho^2,h,v\})\rvert\\ \vert\overline{C}(n,b,\{e,d_1\})\rvert&=\lvert C(n,b,\{e,d_1\})\rvert- \lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert\\ \vert\overline{C}(n,b,\{e,d_2\})\rvert&=\lvert C(n,b,\{e,d_2\})\rvert- \lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert\\ \vert\overline{C}(n,b,\{e,\rho^2\})\rvert&=\lvert C(n,b,\{e,\rho^2\})\rvert- \lvert \overline{C}(n,b,\{e,\rho^2,h,v\})\rvert- \lvert \overline{C}(n,b,C_4)\rvert\\ &\quad- \lvert \overline{C}(n,b,\{e,\rho^2,d_1,d_2\})\rvert- \lvert \overline{C}(n,b,D_4)\rvert\\ &=\lvert C(n,b,\{e,\rho^2\})\rvert- \lvert C(n,b,\{e,\rho^2,h,v\})\rvert- \lvert C(n,b,C_4)\rvert\\ &\quad- \lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert+2 \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e\})\rvert&=\lvert C(n,b,\{e\})\rvert- \lvert \overline{C}(n,b,\{e,h\})\rvert- \lvert \overline{C}(n,b,\{e,v\})\rvert- \lvert \overline{C}(n,b,\{e,\rho^2\})\rvert\\ &\quad- \lvert \overline{C}(n,b,\{e,d_1\})\rvert- \lvert \overline{C}(n,b,\{e,d_2\})\rvert- \lvert \overline{C}(n,b,\{e,\rho^2,h,v\})\rvert\\ &\quad- \lvert \overline{C}(n,b,C_4)\rvert- \lvert \overline{C}(n,b,\{e,\rho^2,d_1,d_2\})\rvert- \lvert \overline{C}(n,b,D_4)\rvert\\ &=\lvert C(n,b,\{e\})\rvert- \lvert C(n,b,\{e,h\})\rvert- \lvert C(n,b,\{e,v\})\rvert- \lvert C(n,b,\{e,\rho^2\})\rvert\\ &\quad- \lvert C(n,b,\{e,d_1\})\rvert- \lvert C(n,b,\{e,d_2\})\rvert+2\lvert C(n,b,\{e,\rho^2,h,v\})\rvert\\ &\quad+2\lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert \end{aligned} $$ These imply that $$ \begin{aligned} (**)\quad K(n)=&\frac{\lvert C(n,n,C_4)\rvert}{4}+\frac{\lvert C(n,n,\{e,h\})\rvert}{8}+\frac{\lvert C(n,n,\{e,v\})\rvert}{8}+\frac{\lvert C(n,n,\{e,\rho^2\})\rvert}{8}\\ &+\frac{\lvert C(n,n,\{e,d_1\})\rvert}{8}+\frac{\lvert C(n,n,\{e,d_2\})\rvert}{8}+\frac{\lvert C(n,n,\{e\})\rvert}{8}. \end{aligned} $$

It remains to compute the quantities $C(n,b,G)$ for the seven groups that appear in the expression above. (Added: It is usual in Pólya enumeration at this point to use the cycle index and generating function techniques. My more bare-handed approach yields the final expressions in a fairly direct fashion.) We have $$ C(n,b,\{e\})=\binom{n^2}{b}. $$ In general, the counting depends on whether $n$ is even or odd. If $n$ is odd, then the action of $C_4$ fixes the center square; every other square has an orbit of size $4.$ Therefore, a configuration that is invariant under the action of $C_4$ is specified by choosing the markings of the center square and of one square from each of the $\frac{n^2-1}{4}$ orbits of size $4.$ (The other three squares in an orbit of size $4$ must be marked the same way as the first square.) Since the total number of marked squares must be $b,$ we have $$ C(n,b,C_4)=\begin{cases}\binom{(n^2-1)/4}{b/4} & \text{if $b\equiv0\pmod{4}$}\\ \\ \binom{(n^2-1)/4}{(b-1)/4} & \text{if $b\equiv1\pmod{4}$}\\ \\ 0 & \text{otherwise.}\end{cases} $$ The center square is unmarked when $b\equiv0\pmod{4}$ and marked when $b\equiv1\pmod{4}.$ If $n$ is even then, under the action of $C_4,$ every square has an orbit of size $4.$ Therefore $$ C(n,b,C_4)=\begin{cases}\binom{n^2/4}{b/4} & \text{if $b\equiv0\pmod{4}$}\\ \\ \binom{n^2/4}{(b-1)/4} & \text{if $b\equiv1\pmod{4}$}\\ \\ 0 & \text{otherwise.}\end{cases} $$ Specializing to $b=n,$ we get $$ C(n,n,C_4)=\begin{cases}\binom{n^2/4}{n/4} & \text{if $n\equiv0\pmod{4}$}\\ \\ \binom{(n^2-1)/4}{(n-1)/4} & \text{if $n\equiv1\pmod{4}$}\\ \\ 0 & \text{otherwise.}\end{cases} $$

The analysis for $G=\{e,\rho^2\}$ is similar. The orbit sizes are now $2$ instead of $4,$ with the exception of the center square in the $n$ odd case, which has orbit size $1.$ The result, for $b=n,$ is $$ C(n,n,\{e,\rho^2\})=\begin{cases}\binom{n^2/2}{n/2} & \text{if $n$ even}\\ \\ \binom{(n^2-1)/2}{(n-1)/2} & \text{if $n$ odd.}\end{cases} $$

When $G=\{e,h\}$ and $n$ is odd, the action of $G$ fixes the $n$ squares on the horizontal centerline. All other squares have orbit size $2.$ A configuration invariant under the action of $\{e,h\}$ is specified by marking $j$ of these orbits and $b-2j$ squares on the centerline. When $n$ is even, all squares have orbit size $2,$ so we need only mark $b/2$ of these orbits. The analysis when $G=\{e,v\}$ is similar. In the $b=n$ case, this gives $$ C(n,n,\{e,h\})=C(n,n,\{e,v\})=\begin{cases}\binom{n^2/2}{n/2} & \text{if $n$ even}\\ \\ \sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j} & \text{if $n$ odd.}\end{cases} $$

When $G=\{e,d_1\}$ or $\{e,d_2\},$ it is a diagonal that is fixed by the action of $G.$ So the analysis is the same as for $G=\{e,h\}$ with $n$ odd. We have $$ C(n,n,\{e,d_1\})=C(n,n,\{e,d_2\})= \sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j}. $$ Mathematica recognizes this sum, which we denote by $S,$ as a generalized hypergeometric function with argument $-1:$ $$ S=\sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j} = \ _3F_2\left[\begin{matrix}1/2-n/2 & -n/2 & (n-n^2)/2\\1/2 & 1 & \end{matrix};-1\right]. $$ Indeed, $$ \begin{aligned} S=&\sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j} =\sum_j\binom{(n^2-n)/2}{j}\binom{n}{2j}\\ =&\sum_j\frac{\frac{n^2-n}{2}(\frac{n^2-n}{2}-1)\ldots(\frac{n^2-n}{2}-j+1)}{j!}\frac{n(n-1)\ldots(n-2j+1)}{2j(2j-1)\ldots1}\\ =&\sum_j\frac{n-n^2}{2}\left(\frac{n-n^2}{2}+1\right)\ldots\left(\frac{n-n^2}{2}+j-1\right)\frac{\frac{-n}{2}\frac{-n+1}{2}\ldots\frac{-n+2j-1}{2}}{j(j-\frac{1}{2})\ldots\frac{2}{2}\frac{1}{2}}\frac{(-1)^j}{j!}\\ =&\sum_j\frac{n-n^2}{2}\left(\frac{n-n^2}{2}+1\right)\ldots\left(\frac{n-n^2}{2}+j-1\right)\\ &\times\frac{\frac{-n}{2}\left(\frac{-n}{2}+1\right)\ldots\left(\frac{-n}{2}+j-1\right)\frac{1-n}{2}\left(\frac{1-n}{2}+1\right)\ldots\left(\frac{1-n}{2}+j-1\right)}{\frac{1}{2}\frac{3}{2}\ldots\left(\frac{1}{2}+j-1\right)\cdot1\cdot2\ldots\cdot j}\frac{(-1)^j}{j!}, \end{aligned} $$ which, by definition, is the stated generalized hypergeometric function.

Combining these results, we have $$ K(n)=\begin{cases} \frac{1}{8}\binom{n^2}{n} + \frac{1}{4}\binom{n^2/4}{n/4} + \frac{3}{8}\binom{n^2/2}{n/2} + 
\frac{1}{4}S & \text{for $n\equiv0\pmod{4}$}\\ \frac{1}{8}\binom{n^2}{n} + \frac{1}{4}\binom{(n^2-1)/4}{(n-1)/4} + \frac{1}{8}\binom{(n^2-1)/2}{(n-1)/2} + 
\frac{1}{2}S & \text{for $n\equiv1\pmod{4}$}\\ \frac{1}{8}\binom{n^2}{n} + \frac{3}{8}\binom{n^2/2}{n/2} + 
\frac{1}{4}S & \text{for $n\equiv2\pmod{4}$}\\ \frac{1}{8}\binom{n^2}{n} + \frac{1}{8}\binom{(n^2-1)/2}{(n-1)/2} + 
\frac{1}{2}S & \text{for $n\equiv3\pmod{4}.$} \end{cases} $$ This answer agrees with that of Marko Riedel and the OEIS sequence he cites. (Added: These formulas can be obtained without much difficulty by extracting the relevant coefficient from the final generating function in Marko Riedel's answer.)

Added: (Some background about the orbit-stabilizer theorem and the orbit counting theorem.) Let $G$ be a group acting on a set $X.$. The stabilizer of $x\in X$ is the set of elements of $G$ that fix $x,$ that is, the set of $g\in G$ such that $gx=x.$ The orbit of $x\in X$ is the set of elements of $X$ to which $x$ is sent by some element of $G.$

The orbit-stabilizer theorem states that, for any $x\in X,$ $$ \lvert G\rvert=\lvert\text{orbit}(x)\rvert\cdot\lvert\text{stabilizier}(x)\rvert. $$ It is a consequence of the fact that $\text{stabilizer}(x)$ is a subgroup of $G$ and the fact that the elements of $\text{orbit}(x)$ are in one-to-one correspondence with the left cosets of $\text{stabilizer}(x).$

The action of $G$ on $X$ partitions $X$ into disjoint orbits. Let $\mathcal{O}$ be the set of orbits. The fixed set of $g\in G$ is the set of elements of $X$ fixed by $g,$ that is, the set of $x\in X$ such that $gx=x.$ The orbit counting theorem, which is often called Burnside's lemma, states that the number of orbits equals the average (over all $g\in G$) of $\lvert\text{fixed}(g)\rvert.$ It is proved by considering the set $P$ of pairs $(g,x)$ satisfying $gx=x.$ We may count the elements of $P$ either by running over $G$ or by running over $X:$ $$ \begin{aligned} \lvert P\rvert=\sum_{g\in G}\lvert\text{fixed}(g)\rvert&=\sum_{x\in X}\lvert\text{stabilizer}(x)\rvert\\ &=\sum_{x\in X}\frac{\lvert G\rvert}{\lvert\text{orbit}(x)\rvert}\\ &=\lvert G\rvert\sum_{O\in\mathcal{O}}\sum_{x\in O}\frac{1}{\lvert\text{orbit}(x)\rvert}\\ &=\lvert G\rvert\sum_{O\in\mathcal{O}}1\\ &=\lvert G\rvert\cdot\lvert\mathcal{O}\rvert, \end{aligned} $$ where the orbit-stabilizer theorem was used to obtain the second line. This implies that $$ \lvert\mathcal{O}\rvert=\frac{1}{\lvert G\rvert}\sum_{g\in G}\lvert\text{fixed}(g)\rvert, $$ which is the desired result.


You're looking for the number of orbits of the dihedral group $D_N$ acting on the $N\times N$ grid with $N$ chosen cells. (Caution: This group is often expressed as $D_{2N}$ as it has $2N$ elements.)

You want to use the Cauchy-Frobenius-Burnside Lemma (aka Burnside Lemma, aka "Not Burnside Lemma") to compute this. That is, you have a group action of $D_N$ on the set $\Omega$ of all such patterns ($N\times N$ grids with $N$ marked cells), and the number of orbits in this action (which is what you want) is then given by $$\frac{1}{2N}\sum_{g\in G} |\Omega^g|$$ where $|\Omega^g|$ is the number of elements in $\Omega$ fixed by $g$.

  • $\begingroup$ and I agree with Gerry (above) that Polya enumeration is an efficient way of doing this. (See also, cycle index.) $\endgroup$ – Doc Nov 17 '13 at 6:03
  • $\begingroup$ For the uninitiated (and probably the OP as well), would you like to present the final formula? $\endgroup$ – nbubis Nov 17 '13 at 6:34
  • $\begingroup$ @nbubis, by "formula" I'm interpreting that you mean a closed form for the answer. I doubt this is even possible. However, your comment inspired me to cook up an example (below) that hopefully clarifies most of the procedure. $\endgroup$ – Doc Nov 17 '13 at 16:50

Consider the action of the group $G$ on a set $\Omega$ of cardinality $n$. The cycle index of $(G,\Omega)$ is the polynomial defined by $$Z_G(X_1,X_2,\dots, X_n)=\frac{1}{|G|} \sum_{g\in G} X_1^{c_1(g)}X_2^{c_2(g)}\cdots X_n^{c_n(g)}$$ where $c_i(g)$ denotes the number of $i$-cycles in the cycle decomposition of the element $g$ acting on $\Omega$.

For example, if we consider the action of $D_4$ on the square, we would first label the vertices of the square by $\{1,2,3,4\}$ in some order. Let's do this in cyclic clockwise order, starting from the upper left corner. Then we get the following table. \begin{array}{c|c|c} g & \mbox{permutation} & \mbox{cycle index term}\\ \hline \mbox{identity} & (1)(2)(3)(4) & X_1^4 \\ 90^o \mbox{rotation} & (1,2,3,4) & X_4^1 \\ 180^o \mbox{rotation} & (1,3)(2,4) & X_2^2 \\ 270^o \mbox{rotation} & (1,4,3,2) & X_4^1 \\ \mbox{horizontal reflection} & (1,4)(2,3) & X_2^2\\ \mbox{vertical reflection} & (1,2)(3,4) & X_2^2\\ \mbox{diagonal reflection} & (1,3)(2)(4) & X_1^2X_2\\ \mbox{opposite diag reflection} & (2,4)(1)(3)& X_1^2X_2 \end{array}

Thus we obtain as cycle index $$Z_G(X_1,X_2,\dots, X_n)=\frac{1}{8}(X_1^4 + 2X_1^2X_2+3X_2^2+2X_4). $$

Now consider $$Z_G(b+1,b^2+1,b^3+1,b^4+1).$$ The coefficient of $b^i$ will give the number (up to isomorphism!) of positions in which there are precisely $i$ marked squares. Returning to the grid problem, we are only interested in the coefficient of the $b^2$ term. This turns out to be $\frac{1}{8}(16) = 2$.



Above is a link to a puzzlemaker referred to as Math Square.

It allows you to create matrice puzzle.

For example, you can create a 6x6 grid and the rows and columns are maths equation having multiplication, division, addition and subtraction. The answers to the equation are available on the extreme right and the extreme bottom of the grid. However, no number in the grid is provided. Can such a puzzle be logically or mathematically solved using any of the above methods mentioned in this thread.


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