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New: 5 x 3 bg puzzle(SPOILERS)
Posted By: Stein Kulseth In Response To: New: 5 x 3 backgammon puzzle (Stein Kulseth)
Date: Tuesday, 29 June 2010, at 2:47 p.m.
Again a very rich puzzle from Nack!
The nice thing about it is that it is possible for any one side to move forwards to a proper end position in 5 moves. Heres alternative (a) starting with a non-double:
65: 24/13
11: 6/5(4)
11: 8/7(3) 6/5
66: 24/12 13/7(2) 11: 13/12(4)This solution is pretty unique apart from some leeway in the exact order you play the rolls in, and which 6 pips are played as 5+1 instead of using a single 6, but the fact that it moves from mid to mid with 11, makes it impossible to use this as a symmetrical solution for 10 rolls.
However we can also move forwards in 5 rolls as second player, allowing 5 doubles, as in this alternative (b):
66: 24/18(2) 13/7(2)
44: 6/2(4)
44: 18/10 18/14 6/2
11: 14/13 8/7(3)
33: 13/10(4)Again pretty unique, apart from the order of the rolls and which 4 pips are played as 3+1 instead of as 4. But on inspection (a) requires being able to play 18/12 with at least one of the 6's, while (b) requires playing 13/7 with at least one of the 6's. Thus we have an unavoidable collision, and (a) and (b) cannot be combined, and the best we can do going forwards is 11 rolls.
Either (a) or (b) can be used as player 2 with player 1 rolling:
62: 8/2 6/4
44: 13/9(4)
44: 13/9 8/4 6/2(2)
55: 24/14(2)
55: 14/2(2)
22: 8/4 6/4(2)
There is however the possibility of having one player go backwards. This would require having 3 checkers blotted, hit and entered. 2 single dice are needed to blot the 8 point checkers, 3 to enter them, 2 to move the 24 point, and 10 to move the mid and 6 points to new locations. In all 17 moves are needed, which can fit into 5 moves!
Furthermore, if the entering number is double N, then by rolling double N twice, we may enter all three blots and move one of the stacks N pips, and by rolling double (N-1) twice we may move the 24-pt onto the entered checkers, and move the other stack N-1 pips (... and have a spare N-1 that may be a problem, but maybe we can stick that somewhere ...)
Tragically, (a) above is particularily unsuited to hitting the blots. (b) on the other hand are able to hit them all, but as all hit blots need to enter on two rolls as outlined above, it will be necessary on some roll to hit two on the same move, and (b) cannot do that - on the points where these blots end up.
So it seems, that whichever way we look at it 11 rolls is the best we can do, yet there is a minor twist left, yet, allowing us to produce this game which I suggest as the proper solution:
14: 8/7 8/4 33: 24/21* 13/10(3) 44: bar/21 13/9(3) 44: 6/2(4) 33: 24/21(2) 8/5 6/3 11: 21/20* 8/7(3) 22: - 66: 24/18* 20/14 13/7(2) 44: bar/21(2) 13/9(2) 44: 18/10 14/10 6/2 33: 6/3(4) While still 11 rolls, at least one of the rolls is a dancing roll and is not actually used apart from providing the necessary tempo.
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