
BGonline.org Forums
Another threebyfive problem
Posted By: Stein Kulseth
Date: Sunday, 22 June 2014, at 12:01 p.m.
In Response To: Another threebyfive problem (Nack Ballard)
Yes, joy and frustration go hand in hand on this one.
And I have reached a dead end, having by now deduced that the problem cannot be solved, which more often than not means that there is yet a thing to be discovered about the puzzle.
For what it is worth, here are my deductions in approximately logical order, and I would appreciate it if as a hint you could give the number of the first one I have got wrong:
 White moves first
 Blues three rolls are all doubles
 Blue's three rolls must be 33, 22 and 11
 ... in that sequence
 One of the checkers originally on the 6pt have been hit
 Two of the checkers on the 5pt have indeed come from the 8pt
 Two checkers are hit on each of White's 2nd and 3rd roll  and entered on Blue's roll 2 and 3
 The two checkers peeled off the mid point must have been one each in roll 1 and 2
 Blue's first move is either:
 33: 13/7 8/5(2), Leaving blots on the 7pt and 8t, both of which must be hit
 33: 13/10 8/5(2) 6/3, blotting on 3pt, 8pt and 10pt, two of which must be hit
 After the first of these, Blue's second move must be 22: b/23(2) 13/11, 6/4 blotting on the 4pt and 11pt, both must be hit
After the second move, Blue's move is b/23(2) 13/11, and then the last deuce is either 11/9, or played with a remaining blot 10/8 or maybe even 3/1*. Anyway, must leave two blots, both must be hit. White's roll 2, 3 and 4 must all be doubles
 Of the 14 moves available for White, only 3 are available for "extra moves", meaning moves that do not hit a checker, or put a checker in its final position
 Then, Blue's first move can not be 33: 13/7 8/5(2)
 And Blue's first move can not be 33: 13/10 8/5(2) 6/3 either!
Of course, If I were right in all of these, 13 and 14 would contradict 9 and prove the puzzle insolvable
FWIW, here's my reasoning about the puzzle (LONG, and perhaps interesting to NOBODY):
 If Blue moves first, White must move 62 pips in three rolls. 62 > 24+24+11, so all three White rolls must then be doubles. As 62 is not a multiple of 4, White must either pick up another 2,6, or 10 pips to move by being hit, or miss out on 6 pips by being blocked.
Last one first  with the two point made 6s are blocked, and the White position can be reached by 66 twice and 55 once. However this allows for only one Blue checker to be hit which is not even enough to close the Blue 23 point needed to block the last 6.
On being hit  the solution to position 1 allowed for a hit on Blues 6 point to allow the White position in three moves. However this solution also allows for a max of three Blue checkers to be hit (one on the bar pt, and twice on the Blue 6pt), so cannot be used. Being hit on Blue's 2pt requires White to roll a 1, which would then not allow him to move the 64 pips needed. Likewise a hit on Blue's 10pt would require White to move 72 pips, eg 66 thrice, which would not allow White to get to the 10pt to be hit. Some Blue moves are indicated, Blue must have moved 4 checkers off the bar, 2 checkers off the mid pt, 2 checkers off the 8pt (the last may have been hit there), and 3 checkers off the 6pt. In total at least 11 checkers moved, so all three rolls must have been doubles.
 As we have counted 11 moves from bar or start position, at most 1 Blue checker can have reached final position from something other than bar or its starting position. Thus the entered checkers already indicate that two of Blue's rolls must be 11 and 22. For the 3rd roll, first assume that it is 11. Then only the 22 roll would allow Blue to blot two checkers off the mid point by playing it b/23 13/11 13/9. Then we can count the remaining pips needed to produce the Blue position, and Blue would need 11 (eleven) pips which is not doable with two rolls of 11.
Thus, at most two of the Blue checkers must have come from the 6pt (two aces needed to enter), so at least one of the checkers on the 5 must have come directly from the 8pt with a roll of 3. So Blue's rolls are 11, 22 and 33 Obviously the 33 roll must be the first, otherwise we would see checkers on the 22pt. Also, the 11 must be last, as it cannot be used to blot the needed checkers. Two aces are needed for entering, and at least one for 6/5. The last could possibly be used 8/7 to blot two checkers, but then we would not be able to fill the 5pt with 3 more checkers out of which two would need to come from the 8pt. So the sequence must be 33, 22, 11.
 Now, with only one roll of 11 available, out of which 2 are needed to enter, only 2 remains for 6/5(2). The third checker must have moved using a 2 or a 3, and have been hit off the 4pt or 3pt.
 By now this seems obvious, but we must double check the obvious as well. We have found that at most 2 checkers on the 5pt can come from the 6pt, whereas at least one comes directly from the 8pt. If the other does not come from playing 8/5 we have a problem getting our remaining threes to play. Then we must have three 3s play off the 6 and mid points. But, note that in proving statement 3 we deduced that at most 1 checker can reach its destination in two moves, and this must now be the 4th checker on the 5pt. So 3 3s off 6pt and midpt would mean that either the 10pt or the 3pt will be made , and the checkers can not be hit and recirculated.
 Again fairly obvious. Both the checkers entered on 24pt must obviously have been hit the roll before. We have deduced before that at least one of the checkers on the 23pt must be entered with the 22 roll. If the second isn't then it must be entered on the last 11 roll, played b/24(2) b/23. But this does not allow any aces to be played 6/5 as is absolutely needed (in addition to other problems ...)
 These checkers have both been hit, and as we have deduced that 2 2s are played b/23(2) so we just cannot play two checkers off mid with 22 as that would make the 11pt point and not allow the checkers to be hit. We could possibly do it with the initial 33 roll, as that could allow us to split them later with a 2 in order to get them hit. However, that would make the opening play 8/5(2) 13/10(2) leaving only one blot for White to hit in roll 2, which is in conflict with statement 7.
 Follows from the above.
 As does this.
 Counting checker moves White must have made, 5 moves to clear the mid pt, 4 moves to hit Blue checkers, and as only one Blue checker can have been hit on the 14pt, White will need at least 3 more moves to get these to 7pt or 8pt.
Thus there is some possibility that White could make do with only two doubles. Then all the first moves with the back checkers must hit something, we would need White's opening roll to be played from mid leaving the back checkers in place, the most promising would be 65: 13/8 13/7 which also puts two checkers in right places. Then Blue's first move would have to be 33: 13/7 8/5(2) in order to let White hit one of the outfield blots from the 24pt. Both blots would have to be hit so: 61 b/18*/17*, followed by Blue 22: b/23(2) 13/10 6/4. Again both blots must be hit, so White would have to roll 33, hitting both blots, eg. 24/21*/15 17/14*. Then for the last roll, the best White can do is 66: 13/7(3) 14/8, filling almost the desired position, but leaving the last checker stranded on 15. In any case, accounting for other variants, White would be at least 8 pips in the red. Follows from the above
 In order to hit both blots on Blues 7pt and 8pt with a double on roll two, White must split the back checkers. Most promising is 51: 13/8 24/23, which later allows 66: 24/18* 23/17* 13/7(2), then 22: 18/16/14* 17/15/13 picking up the blot Blue will leave on 14, and then finally 66: 14/8 13/7(3) leaving the desired White position, but leaving a Blue blot unpicked up on the 4pt ...
For White to pick up the blot Blue will leave on the 4pt after his second move, White will need to be hit back off the 14pt, and he cannot get there by his 66 move. Neither can he do so if he opens 12: 24/23 24/22, and hits blots with a subsequent 55.
He can be hit back though, after 23: 24/22 24/23, and then 44: 22/18*/14 23/17*/13, but now he has already used 4 extra moves, in conflict with Statement 12 3 of the doubles, 22, 44, and 55 cannot even hit two Blue checkers as required. 66 may hit, after an openening 13: 24/23 24/21, and then 66: 23/17* 21/15* 13/7(2), but then there is no way to get back to the blot on Blue's 3pt.
33 may hit after an opening 62: 24/22 24/18, after which Blue hits on his 3pt, and two of White's subsequent 3s play b/22* and 18/15*. However, the two remaining threes will count as "extra moves", and as White already has used up 2 of his extras on the opening roll, this conflicts statement 12.
11 also may hit twice as long as 24/18 is part of the opening roll. However, again this uses 1 "extra move", and in playing 11 where only 2 aces are used to hit, the remaining 2 extra moves are used. Then next roll, in order to pick up the remaining two checkers, another small roll is needed (There are some variants here, but the maximum distance between the blots are 3, and the maximum distance between White's checker in Blues outfield and a blot will be 2, so in no case can any roll larger than 33 be used  infact I think the only roll to pick up 2 is another 11). Thus more "extra moves" will be needed, and White hasn't got any.

BGonline.org Forums is maintained by Stick with WebBBS 5.12.