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Steklov Institute of Mathematics at Saint-Petersburg
27 Fontanka, Saint-Petersburg, 191011, Russia
E-mail: yumat@pdmi.ras.ru
URL: http://logic.pdmi.ras.ru/~yumat

This note presents some nonevident ways for determining the colorability of a graph and the number of its colorings via algebraic manipulations of polynomials in many variables.

Suppose we have a graph G and are interested in learning whether one can properly color its vertices in k colors. In other words, we want to know whether one can assign a color c[i] to the i-th vertex in such a way that the ends of each edge of G would get different colors, c[i] belonging to some k-element set Colors[k]. Could we find the answer with general computer algebra systems?

On one hand, such systems are universal in the sense that they can compute everything that can be computed in principle, so the answer is definitely positive.

On the other hand, computer algebra systems were initially intended for dealing with mathematical formulas, so the required program might be rather cumbersome. To make our problem more challenging, let us state it in the following way:

Write the shortest possible program ColorableQ which would determine, for a given graph G and a natural number k, whether graph G is vertex colorable in k colors.

Pay attention that we do not ask for the most efficient program; in fact, the intrinsic computational complexity of graph coloring is not yet known.

The solution of our problem depends, of course, on the particular system exploited. A specialized system can contain ColorableQ as a standard function, making the solution of our problem trivial. Using the Mathematica standard add-on package Combinatorica, one can easily define ColorableQ as follows.

Thus, in order to keep our problem challenging, we need to restrict the admissible tools to basic algebraic manipulations with formulas.

Another point to be taken into account is the representation of graphs. Combinatorica uses an adjacency matrix in its canonical representation for graphs. However, it also allows a more compact representation via an unordered set of adjacent vertices. We will be looking for a solution to our problem with this alternative representation.

Here is one possible program. Is it the shortest one based on "first principles" only?

Let us see why it works. To this end we first rewrite the program in a lengthy form with notation for the intermediate objects.

The Map (applied at level 2) transforms G, a list of numbers of adjacent vertices, into a list of pairs of colors which should be different.

The first Apply produces a list of differences of colors which should all be different.

As soon as we replace all c[i] by some values (from our k-element set Colors[k]) that do not produce a coloring of the graph G, the list P2 will contain a zero.

The second Apply transforms P2 into a polynomial.

This polynomial is equal to zero for any choice of values of c[i] from Colors[k] that is not a coloring of the graph.

P4 is another representation for the polynomial P3, so it is equal to zero for each choice of values of c[i] from the set Colors[k] if and only if the graph G has no colorings in k colors. The size of P4 is exponentially greater than the size of P3, and this is the point that makes our program computationally nonefficient.

The substitution DegreeReduction is the heart of the program. In graph theory, colors are not specified. We are free to select for Colors[k] any set with k elements, and our choice is the set of all k-th roots of unity.

Under this interpretation of colors, the left-hand side of DegreeReduction is equal to its right-hand side, so the values of the polynomials P4 and P5 coincide as long as the values of c[i] are taken from the set Colors[k].

We saw that if the graph G has no colorings in k colors, then both polynomials P4 and P5 are equal to zero for each choice of values of c[i] from the set Colors[k]. But the polynomial P5 has degree at most k-1 in each variable, and by the interpolation theorem there is only one such polynomial. In other words, the graph G has no colorings in k colors if and only if the polynomial P5 is identical to the zero polynomial, and that is what is checked in the last line of the program.

Thus, there is no magic behind ColorableQ. But what is required if we ask about the number of colorings rather than about their mere existence?

Using only basic tools for formula manipulation, write the shortest possible program NumberOfColorings which would determine, for a given graph G and a natural number k, the number of vertex colorings of the graph G in k colors.

It turns out that a slight modification of ColorableQ enables us to solve this more difficult problem. First of all, we need to replace the standard Subtract by a special function.

Similar to Subtract, the value of ColorSubtract[c1, c2] is equal to 0 as soon as c1 and c2 are equal elements from Colors[k]. However, if c1 and c2 are unequal elements of this set, then the value of ColorSubtract[c1, c2] is equal to 1, because in this case the sum is equal to .

Modifying the final part of the former program, we get the desired new one.

P6 is just the free term of P5. It turns out to be equal to the number of colorings divided by , where V[G] is the number of vertices of graph G. Why?

Due to the fact that the value of ColorSubtract is either 1 or 0, the value of P4 and P5 is equal to 1 or 0 depending on whether the values of c[i] produce a coloring or not. Having a small degree in each variable, the polynomial P5 can again be restored from these values by the interpolation theorem. The Lagrange form of the interpolating polynomial would consist of exactly NumberOfColorings[G, k] summands, and it is not difficult to check that the free term of each summand is equal to .

Removing redundant notation, we get the following short form of NumberOfColorings.

The latter output can be compared with

In the author's HTML paper, http://logic.pdmi.ras.ru/~yumat/Journal/Triangular/triang.htm, one can find more nonevident algebraic ways for calculating the number of colorings of plane triangulations, giving new restatements of the Four-Color Theorem.

About the Author

Yuri Matiyasevich is the head of the Laboratory of Mathematical Logic at Steklov Institute of Mathematics in St. Petersburg, Russia. His most known achievement was the negative solution of Hilbert's tenth problem (http://logic.pdmi.ras.ru/Hilbert10/stat/stat-eng.html) about which he wrote a book (http://logic.pdmi.ras.ru/~yumat/H10Pbook/index.html, English translation published by the MIT Press in 1993). Besides logic, Yuri Matiyasevich works in number theory and discrete mathematics.


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