# coloring a $4\times 4$ square with $4$ colors

Today one of my friends asked me the following problem:

Suppose we have a $4\times 4$ table and we want to color it with four colors such that all four colors are used and also no two neighbouring squares (sharing an edge) have the same color. How many colorings are possible? (if we reach a coloring from rotating or reflecting another one, then those two are different)

I guess there is a solution using brute force and a lot of case checking, but is there a clean solution to this problem?

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One way is to calculate the chromatic polynomial of $P_4 \Box P_4$ and substitute $t=4$, which gives the answer 6000732. – jp26 Feb 23 '12 at 20:27
The polynomial can be calculated using Maple or it is tabulated here: staff.hs-mittweida.de/~peter/research/grids.html For exactly 4 colours to be used we need to subtract the 7812 colourings using 3 or fewer colours, giving 5992920. – jp26 Feb 23 '12 at 20:38
@jp26, may I suggest you combine your comments and post them as an answer? – Gerry Myerson Feb 23 '12 at 23:02

As requested, using the polynomial $$f(t):= t ( t-1 ) ( {t}^{14}-23\,{t}^{13}+253\,{t}^{12}-1762 \,{t}^{11}+8675\,{t}^{10}-31939\,{t}^{9}+90723\,{t}^{8}-202160\,{t}^{7 }+355622\,{t}^{6}-492434\,{t}^{5}+529770\,{t}^{4}-430857\,{t}^{3}+ 251492\,{t}^{2}-94782\,t+17493 )$$ and calculating $f(4)-4f(3)+6f(2)$, the answer is 5969496 by Jonas' observation. The 4 is the four different sets of three of the colours and the 6 is the pairs of two colours.

The calculations for calculating the chromatic polynomial of the $4\times 4$ grid are too involved to post here, though. Hopefully I can show the ideas behind a $3\times 3$ and you can imagine the work required to do the work for the question by hand.

My chromatic polynomial reduction rules are given on pages 24 and 25 of my course notes

Using deletion-contraction on a central edge (1:2- 2:2) you get that $g(t) = g_1(t) - g_2(t)$ where $G_1$ and $G_2$ are the following graphs:

We can then use deletion-contraction again on the opposite edge in both $G_1$ and $G_2$ and get $f(t) = h_1(t) - 2 h_2(t) + h_3(t)$ where the graphs are as follows:

From here we can use complete intersection and maximum valency for $H_2$ and $H_3$, respectively so that $h_2(t) = (t-2)^2 \times p(C_6,t) = t \left( t-1 \right) \left( {t}^{4}-5\,{t}^{3}+10\,{t}^{2}-10\,t+5 \right) \left( t-2 \right) ^{2}$ and $h_3(t) = t \times p(P_3 \cup P_3,t-1) = t \left( t-1 \right) ^{2} \left( t-2 \right) ^{4}$, using the standard results for polynomials of cycles, paths and unions.

One more deletion-contraction gives $h_1 = t \left( t-1 \right) \left( {t}^{7}-9\,{t}^{6}+36\,{t}^{5}-84\,{t}^{ 4}+126\,{t}^{3}-124\,{t}^{2}+76\,t-23 \right)$ and putting them together gives us $$g(t) = t \left( t-1 \right) \left( {t}^{7}-11\,{t}^{6}+55\,{t}^{5}-161\,{t} ^{4}+298\,{t}^{3}-350\,{t}^{2}+244\,t-79 \right)$$

Note that $g(2)=2$ since the graph is bipartite, $g(3) =246$ and $g(4) =9612$, so that there are 8640 colourings using exactly 4 colours.

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This is really neat! I look forward to thinking through the details. One quick question: shouldn't the final answer be $f(4)-4f(3)+6f(2)$? (There are 4 different 3-colorings we want to exclude, so we subtract $4f(3)$. But then we've subtracted each 2-coloring twice, so we add back $6f(2)$.) – Jonas Kibelbek Feb 24 '12 at 1:45
Indeed, Jonas, thanks for that observation! – jp26 Feb 24 '12 at 2:30
really nice! thanks a lot. the intresting point is, this question was posed for a mathematical olympiad!!! – Goodarz Mehr Feb 25 '12 at 13:05