This puzzle was originally posted here. Feel free to follow the thread there, but it's pushed far enough down now that I decided to re-post here.
From the position below, left (opening position plus an ace-split), Blue is allowed to call his rolls and play four times in a row.
OBJECTIVE: Make a seven-point prime from the 8pt to the 2pt.
Extra credit: If White occupies her usual starting points (as shown below, right), can Blue still accomplish his objective?
Extra extra credit: Identify all possible ways (groups and orders of four rolls) with which Blue can make this seven-point prime in each of the two positions.
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As Daniel Murphy pointed out, Black must advance 72 pips to close the seven-point prime from 8pt to 2pt. Actually, at least 72 pips are needed. This can be accomplished by excluding the 24pt checker and using the other fourteen checkers to build the prime.
Four checkers (two on each of the 6pt and 8pt) are already in place. Let's determine the minimum number of (1- to 6-pip) checker subplays necessary for the other ten checkers to reach the remaining destinations. The 23pt checker needs three, the midpointers eight, the 8pt spare one, and the 6pt spares three. That's a total of fifteen subplays. This tells us that four doublets are required. (Three doublets plus one non-doublet would generate only 4 + 4 + 4 + 2 = 14 subplays.)
As Joe Russell observed, it is necessary that two of these doublets be 4s or smaller, in order to place all three 6pt spares on the inside points. Precisely speaking, one doublet cannot be larger than 44 and another cannot be larger than 33.
Having determined these restrictions (at least 72 pips from four doublets without exceeding 44 and 33 for two of them), only four roll combinations fill the bill. Of these, 66 66 33 33 can be eliminated because either the 6pt or 3pt would be stuck with a third checker.
That leaves three legitimate roll groups to try:
(a) 66 55 44 33
(b) 66 66 44 22
(c) 66 66 44 33
Let's start with (a). To utilize the 6pt spares, at least one 4 must be 6/2 and one 3 must be 6/3. And two 6s must be played 13/7(2) -- only way to make the 7pt. And clearly at least one 5 must come down 13/8. Also, a 4 must be played 8/4 and a 3 must be played 8/5 -- otherwise Blue can't (while retaining the 8pt and 7pt) make both the 5pt and 4pt with what's left. These prerequisite subplays give us the partially formed prime below:
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| Forced subplays made | |
Let's reorganize all the subplays by roll:
66: 13/7(4)
55: 23/8 13/8
44: 8/4 7/3 6/2(2)
33: 8/5(2) 7/4 6/3
Matt and Joe both gave the solution 66: 13/7(4), 55: 23/3, 44: 13/5 6/2(2), 33: 8/5 7/4(2) 6/3. Careful scrutiny reveals a difference of my 13/8 8/5 (5 and 3) transposing with their 13/5 (4 and 4), and my 8/4 7/3 (4 and 4) transposing with their 8/3 7/4 (5 and 3), and the rest is the same. Both are valid solutions.
Can the rolls be played in any order? Yes! Granted, I cannot (for example) play a 3 from 7/4 until a 6 has been played 13/7, nor the second 3 from 8/5 without the support of a 5 played 13/8, but if I roll 33 first from the initial position, I play 13/10(3) 6/3, and then play two 5s to the 5pt and one 6 to the 4pt as the rolls occur. What matters in this case is that two 3s and two 5s are combined to travel from 13pt to 5pt, and that one 3 and one 6 are combined to travel from 13pt to 4pt; in that way, any order can be handled.
Finally for (a), do all roll orders work in the blocked position? (White owns her starting points, repeated below in the right-hand diagram.) Daniel found solutions for at least half of the twenty-four 66 55 44 33 orders, but in any case the answer is: Yes, all roll orders work.
For example, White's points don't interfere with any roll order using the subplays made in my or Matt/Joe's solutions. Granted, if you were only able to find a solution such as 66: 23/11 13/7(2), 55: 13/8(3) 11/6, 44: 8/4(2) 6/2(2), 33: 8/5(2) 6/3(2), you might conclude that 66 cannot be rolled first, but that's why I walked you through the original reasoning process; finding the forced subplays makes it easier to organize an implementation that does work, even when there are White points to circumvent.
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66: 13/7(4)
66: 23/5 13/7
44: 7/3(2) 6/2(2)
22: 7/5 8/4 6/4
... and the order of rolls doesn't matter; just find the proper transposition themes for the subplays.
What about the blocked position? In this case, the 23pt checker can escape only with a 2 played 23/21 and a 6 played 21/15. So, it is impossible to make the prime if both sets of double 6s precede the double 2s. Precisely, only 66 66 44 22 and 66 66 22 44 do not work; the other ten permutations do.
Consider, for example, 66 44 22 66. Instead of mimicking the solution listed above (in jumbled order), you can play 13/7(4), 8/4 7/3 6/2(2), 23/21 13/11 7/5 6/4, and 21/3 11/5. The key, again, is to plan ahead that the 2 will be played up and prepare the slotting of the points accordingly.
Finally, there is (c) 66 66 44 33, which it seems Joe alone found. Any roll order works in the unblocked position. Here is perhaps the most straightforward example:
66: 13/7(4)
66: 23/5 8/2
33: 7/4(2) 6/3(2)
44: 13/9 6/2 24/20
Because the (c) rolls sum to 4 pips more than those of (a) or (b), the extra slack is taken up by 24/20. Without the 24pt checker on the board, there would be no solution for (c)!
For (c)'s blocked position, there is no solution, regardless of the order of rolls. With eight 6s crashing the position, three of them must be 23/5 or Blue can't make his 5pt; but White's 8pt blocks 23/5.
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In the unblocked position: All permutations of rolls work for (a), (b) or (c).
In the blocked position:
(a) 66 55 44 33: All roll permutations work.
(b) 66 44 22 66: Only 66 66 44 22 and 66 66 22 44 fail.
(c) 66 66 44 33: No roll permutation works.
Nack
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