Hales–Jewett theorem


In mathematics, the Hales–Jewett theorem is a fundamental combinatorial result of Ramsey theory named after Alfred W. Hales and Robert I. Jewett, concerning the degree to which high-dimensional objects must necessarily exhibit some combinatorial structure; it is impossible for such objects to be "completely random".
An informal geometric statement of the theorem is that for any positive integers n and c there is a number H such that if the cells of a H-dimensional n×n×n×...×n cube are colored with c colors, there must be one row, column, or certain diagonal of length n all of whose cells are the same color. In other words, the higher-dimensional, multi-player, n-in-a-row generalization of a game of tic-tac-toe cannot end in a draw, no matter how large n is, no matter how many people c are playing, and no matter which player plays each turn, provided only that it is played on a board of sufficiently high dimension H. By a standard strategy stealing argument, one can thus conclude that if two players alternate, then the first player has a winning strategy when H is sufficiently large, though no practical algorithm for obtaining this strategy is known.
More formally, let WnH be the set of words of length H over an alphabet with n letters; that is, the set of sequences of of length H. This set forms the hypercube that is the subject of the theorem.
A variable word w over WnH still has length H but includes the special element x in place of at least one of the letters. The words w, w,..., w obtained by replacing all instances of the special element x with 1, 2,..., n, form a combinatorial line in the space WnH; combinatorial lines correspond to rows, columns, and diagonals of the hypercube. The Hales–Jewett theorem then states that for given positive integers n and c, there exists a positive integer H, depending on n and c, such that for any partition of WnH into c parts, there is at least one part that contains an entire combinatorial line.
For example, take n = 3, H = 2, and c = 2. The hypercube WnH in this case
is just the standard tic-tac-toe board, with nine positions:
A typical combinatorial
line would be the word 2x, which corresponds to the line 21, 22, 23; another combinatorial line is xx, which is the line
11, 22, 33. In this particular case, the Hales–Jewett theorem does not apply; it is possible to divide
the tic-tac-toe board into two sets, e.g. and, neither of which contain
a combinatorial line. On the other hand, if we increase
H to, say, 8, and partition this board
into two sets, then one of the two sets must contain a combinatorial line. For a proof, see below.

Proof of Hales–Jewett theorem (in a special case)

We now prove the Hales–Jewett theorem in the special case n = 3, c = 2, H = 8 discussed above. The idea is to
reduce this task to that of proving simpler versions of the Hales–Jewett theorem. One can prove the general case of the Hales–Jewett theorem by similar methods, using mathematical induction.
Each element of the hypercube W38 is a string of eight numbers from 1 to 3, e.g. 13211321 is an element of the hypercube. We are assuming that this hypercube is completely filled with "noughts" and "crosses". We shall use a proof by contradiction and assume that neither the set of noughts nor the set of crosses contains a combinatorial line. If we fix the first six elements of such a string and let the last two vary, we obtain an ordinary tic-tac-toe board, for instance 132113?? gives such a board. For each such board abcdef??, we consider the positions
abcdef11, abcdef12, abcdef22. Each of these must be filled with either a nought or a cross, so by the pigeonhole principle two of them must be filled with the same symbol. Since any two of these positions are part of
a combinatorial line, the third element of that line must be occupied by the opposite symbol. In other words, for each choice of abcdef
, there are six possibilities:
  1. abcdef11 and abcdef12 are noughts; abcdef13 is a cross.
  2. abcdef11 and abcdef22 are noughts; abcdef33 is a cross.
  3. abcdef12 and abcdef22 are noughts; abcdef32 is a cross.
  4. abcdef11 and abcdef12 are crosses; abcdef13 is a nought.
  5. abcdef11 and abcdef22 are crosses; abcdef33 is a nought.
  6. abcdef12 and abcdef22 are crosses; abcdef32 is a nought.
Thus we can partition the six-dimensional hypercube W36 into six classes, corresponding to each of the above six possibilities. .
Now consider the seven elements 111111, 111112, 111122, 111222, 112222, 122222, 222222 in W36. By the pigeonhole principle, two of these elements must fall into the same class. Suppose for instance
111112 and 112222 fall into class, thus 11111211, 11111222, 11222211, 11222222 are crosses and 11111233, 11222233 are noughts. But now consider the position 11333233, which must be filled with either a cross or a nought. If it is filled with a cross, then the combinatorial line 11xxx2xx is filled entirely with crosses, contradicting our hypothesis. If instead it is filled with a nought, then the combinatorial line 11xxx233 is filled entirely with noughts, again contradicting our hypothesis. Similarly if any other two of the above seven elements of W36 fall into the same class. Since we have a contradiction in all cases, the original hypothesis must be false; thus there must exist at least one combinatorial line consisting entirely of noughts or entirely of crosses.
The above argument was somewhat wasteful; in fact the same theorem holds for H = 4.
If one extends the above argument to general values of n and c, then H will grow very fast; even when c = 2 the H given by the above argument grows as fast as the Ackermann function. The first primitive recursive bound is due to Saharon Shelah, and is still the best known bound in general for the Hales–Jewett number H = H.

Connections with other theorems

Observe that the above argument also gives the following corollary: if we let A be the set of all
eight-digit numbers whose digits are all either 1, 2, 3,
and we color A with two colors, then A contains at least one arithmetic progression of length three, all of whose elements are the same color. This is simply because all of the combinatorial lines appearing in the above proof of the Hales–Jewett theorem, also form arithmetic progressions in decimal notation. A more general formulation of this argument can be used to show that the Hales–Jewett theorem generalizes van der Waerden's theorem. Indeed the Hales–Jewett theorem is substantially a stronger theorem.
Just as van der Waerden's theorem has a stronger density version in Szemerédi's theorem, the Hales–Jewett theorem also has a density version. In this strengthened version of the Hales–Jewett theorem, instead of coloring the entire hypercube WnH into c colors, one is given an arbitrary subset A of the hypercube WnH with some given density 0 < δ < 1. The theorem states that if H is sufficiently large depending on n and δ, then the set A must necessarily contain an entire combinatorial line.
The density Hales–Jewett theorem was originally proved by Furstenberg and Katznelson using ergodic theory. In 2009, the Polymath Project developed a new proof of the density Hales–Jewett theorem based ideas on from the proof of the corners theorem. Dodos, Kanellopoulos, and Tyros gave a simplified version of the Polymath proof.