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In the following discussion, pre-colored hexagons are referred to asbound-ary. Uncolored hexagons are calledinterior. Without loss of generality, wemay assume that the edges of the board are made up of pre-colored hexagons(see figure). Thus, the interior hexagons are surrounded by hexagons on allsides.Theorem 10.2.6.For a completed standard Hex board with non-empty in-terior and with the boundary divided into two disjoint yellow and two disjointblue segments, there is always at least one crossing between a pair of seg-ments of like color.Proof.Along every edge separating a blue hexagon and a yellow one, insertan arrow so that the blue hexagon is to the arrow’s left and the yellow oneto its right. There will be four paths of such arrows, two directed towardthe interior of the board (call these entry arrows) and two directed awayfrom the interior (call these exit arrows), see Fig. 10.12.Fig. 10.12. On an empty board the entry and exit arrows are marked. Ona completed board, a blue chain lies on the left side of the directed path.Now, suppose the board has been arbitrarily filled with blue and yellowhexagons.Starting with one of the entry arrows, we will show that it ispossible to construct a continuous path by adding arrows tail-to-head alwayskeeping a blue hexagon on the left and a yellow on the right.In the interior of the board, when two hexagons share an edge with anarrow, there is always a third hexagon which meets them at the vertextoward which the arrow is pointing. If that third hexagon is blue, the nextarrow will turn to the right. If the third hexagon is yellow, the arrow willturn to the left. See (a,b) of Fig. 10.13.Loops are not possible, as you can see from (c) of Fig. 10.13. A loop cir-cling to the left, for instance, would circle an isolated group of blue hexagons
198Combinatorial gamesbacFig. 10.13. In (a) the third hexagon is blue and the next arrow turns tothe right; in (b) — next arrow turns to the left; in (c) we see that in orderto close the loop an arrow would have to pass between two hexagons of thesame color.surrounded by yellow ones. Because we started our path at the boundary,where yellow and blue meet, our path will never contain a loop.Becausethere are finitely many available edges on the board and our path has noloops, it eventually must exit the board using via of the exit arrows.All the hexagons on the left of such a path are blue, while those on theright are yellow. If the exit arrow touches the same yellow segment of theboundary as the entry arrow, there is a blue crossing (see Fig. 10.12). If ittouches the same blue segment, there is a yellow crossing.10.2.3 Hex and YThat there cannot be more than one crossing in the game of Hex seemsobvious until you actually try to prove it carefully. To do this directly, wewould need a discrete analog of the Jordan curve theorem, which says thata continuous closed curve in the plane divides the plane into two connectedcomponents. The discrete version of the theorem is slightly easier than thecontinuous one, but it is still quite challenging to prove.