Ending Puzzle Dots and Boxes
8 replies. Last post: 2005-12-11Reply to this topic Return to forum
Knox (Computer) at 2005-12-05
The following position was reached in Knox vs. ypercube game of the 6'th championship.
6 + + + +--+ + | | |5 +--+--+--+ +--+ |4 +--+ + + + + |Bl| |3 +--+--+--+--+--+ | | |2 + + + + + + | |1 + +--+--+ +--+ A B C D E FBlue to play and win.
Notation: A dot is indicated by a column letter followed by a row number. For example, A1 is the dot in the lower-left corner. A move is indicated by specifying the two ending dots separated by a dash. For example, A1-A2 is the move that connects the lower-left dot to the dot just above it.
Knox (Computer) at 2005-12-08
Has anybody worked on this yet?
Don't be fooled by the simple resolved-looking nature of the position; it's surprisingly tricky. In the actual game, the blue player (rated over 1900) did not win!
ypercube ★ at 2005-12-08
The poor Blue player did not find the move.
If he did, it would be the first time that either Knox or its cousin KnoxB would be defeated by a simple mirror playing strategy.
michael at 2005-12-08
Well, with the 1 box advantage blue already has, a third quad is most likely to secure victory :-)
ypercube ★ at 2005-12-09
A very good point which gives us a clue why the Blue player has 1900 rating while michael a 2200.
Knox (Computer) at 2005-12-10
Since there is only one long chain, the blue player has control. Generally speaking, the controlling player wants the long chains as long as possible and especially wants to avoid quads. The blue player can do both by sacrificing a box to break up the potential quad in the middle. So the obvious play is d4-e4 (or d3-d4, c4-d4, or d4-d5). But after the response d3-d4, e4-e5 we have the following.
6 + + + +--+ + | | |5 +--+--+--+ +--+ | |4 +--+ + +--+ + |Bl| |R |3 +--+--+--+--+--+ | | |2 + + + + + + | |1 + +--+--+ +--+ A B C D E F
The ending of 2 quads and one long chain of length 8 will split 8-8 with best play. There is a 1-1 split of already captured boxes. So the winner will be the winner of the exchange of short chains and this player cannot be the controlling player, hence Red wins (The final score will be 13-12.).
The only winning move for Blue is 1.e4-e5! Red cannot afford to sac a box to destroy the quad. The ending would look the same as in the previous try but the sacrificed box would convert the 13-12 win to a 12-13 loss. So Red has nothing better than a free move. Blue can safely respond with a free move and again Red has nothing better than to take the last free move resulting in the following position.
6 + + + +--+ + | | |5 +--+--+--+ +--+ | |4 +--+ + + + + |Bl| |3 +--+--+--+--+--+ | | |2 + + + + + + | | |1 +--+--+--+--+--+ A B C D E FBlue to Play
At first glance, it appears that Blue has simply delayed the inevitable as apparently Blue must sac a box creating the same losing ending. Instead, Blue wins with the counterintuitive d5-d6! (this could have been played one move earlier with the same effect). Blue has control and instead of sacrificing one box to get one long chain of length 8, blue sacrifices more boxes to strong>shorten the chain length and to create another dreaded quad.
After taking the two sacrificed boxes. Red will be ahead by one box. Red will gain a second box during the exchange of short chains. But the ending with three quads and a 3-chain attached to one of the quads favors the controlling player, Blue, by three boxes! Hence, Blue wins by a single box with best play (Red should open up the 3-chain before opening up a quad, otherwise Blue wins the ending by four boxes instead of three).
In the original position, Blue's immediate d5-d6 loses! Red can defeat this play by a counter-sacrifice that destroys the quad creating an 8-chain. As mentioned previously, Red can't afford this sacrifice after e4-e5 but after d5-d6 he can! The reason is because d4-d5 gives Red two more boxes.
Knox (Computer) at 2005-12-11
During the game, I made the original comment that Knox(B) hadn't been defeated by a mirror strategy. Ypercube was simply repeating my comment under the natural assumption that I would know. I tweaked KnoxB's opening play after the referenced game was played and I simply forgot about games against the earlier version of KnoxB. So I should have said, since the
opening tweak …blah blah blah.
KnoxB's opening play was so rigid (it's still very rigid) that it was possible to defeat KnoxB every time going second by playing the same exact opening over and over. The referenced game shows one such method. This problem was revealed even more clearly in two games versus
oord2000 where he found a way to defeat KnoxB going second by playing essentially the identical game over and over. The first 10 moves of this method would always result in either the central row or central column being one long chain with no other moves being played. The second player could then win the same game repeatedly simply by following the mirror strategy until all the free moves were gone.