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Eight-point prime Puzzle -- Solution

Posted By: Nack Ballard
Date: Monday, 10 May 2010, at 2:11 a.m.

Object Position

The OBJECTIVE of this puzzle is to build an eight-point prime from the 9pt to the 2pt (as shown above) in FIVE rolls. It was originally posted here.

The puzzle comes in six parts.

PART 1: Blue calls his rolls and plays five times in a row. In the position below, can he build the prime? If so, how?

Part 1: Unblocked position
Make eight-prime, 9pt to 2pt, five rolls

Part 1, Logic: 100 pips are required to move the sixteen checkers to the object position (eight-point prime). Also, a minimum of twenty "prossovers" (my term for a distance traveled that brings a checker one fewer subplay closer to a prime point) are required: four from the 24pt, three from the 23pt, three from the 22pt, six from the midpoint, one from the 8pt and three from the 6pt. If a solution exists, then it must involve five doublets (composed of four subplays each) totalling 100 pips.

The three 6pt spares cannot all end up on the same point. One spare must move no more than 4 pips and another no more than 3 pips. Hence, one doublet can be no larger than 44 and another can be no larger than 33. OTOH, those two rolls cannot be smaller than 44 and 33, because as it is they sum to 28 pips, and even when supplemented by three maximal rolls (66 66 66) the total is 100 pips. Therefore, the group of five rolls must exactly be 66 66 66 44 33.

Part 1, Answer: Yes, Blue can make an eight-point prime from 9pt to 2pt with those five rolls in many orders and in many ways.

Part 1, Solution: For example,

66: 24/12 13/7(2)
66: 23/5 13/7
66: 22/4 13/7
44: 13/9 7/3 6/2(2)
33: 12/9 8/5 7/4 6/3

PART 2: Same as part 1, except White occupies her starting points, as shown in the diagram BELOW. Blue still gets five rolls and makes all the plays. Now can he build the prime? If so, how?

Part 2: Blocked position
Make eight-prime, 9pt to 2pt, five rolls

Part 2, Logic: Joe (who provided the proper group of rolls for Part 1) observed, "It would take 21 parts of a move to complete the task and you only have 20 parts of a move in 5 rolls."

That's right. The 24pt checker is not any worse off than it was; 24/21 or 18/15 followed by 6s still reaches a prime point in the same number of prossovers. The culprit is the 23pt checker: it squanders a prossover hurdling White's 8pt, having to first play 23/20. If White did not own her devilish 8pt, Blue could accomplish his objective with the vital 23/5.

Part 2, Answer: No. With White in the way it is impossible for Blue to build an eight-point prime from his 9pt to his 2pt in five rolls.

PART 3: Again, five rolls by Blue. In the LEFT-hand position BELOW, can he build the prime? If so, how?

Part 3, Logic: 96 pips are required to get all checkers to the object position (eight-point prime). This time (with the 24pt checker moved up to the 20pt), only nineteen prossovers are required, but (given that four doublets plus one non-doublet permit only eighteen subplays) five doublets are still needed. Also, due to the 6pt spares (as explained in Part 1), one doublet cannot exceed 44 and another cannot exceed 33.

With those restrictions, the group of rolls must be either 66 66 55 44 33 or 66 66 66 44 22. However, it cannot be the latter, because ten (of the sixteen) checkers start on light-colored points and only eight can finish on light points. With only even numbers on the dice, it is impossible to alter the parity of the checkers' point colors -- although it's amusing to try to make the prime anyway :)

I'll be giving the solution to parts 3 and 4 together.

Part 3: Unblocked position Make eight-prime, 9pt to 2pt, five rolls

Part 4: Blocked position Make eight-prime, 9pt to 2pt, five rolls

PART 4: Same as part 3, except White occupies her starting points, as shown in the RIGHT-hand diagram. Now can Blue build his prime? If so, how?

Part 4, Logic: The same logic applies here as in part 3. White's points prevent some solutions, but with the diversity of four different doublets, many specific solutions are still possible.

Parts 3 and 4, Answer: Yes, Blue can.

Parts 3 and 4, Solution: Joe was the only solver, so here is the roll sequence he provided (I'm filling in the plays):

55: 23/8 13/8
66: 22/4 20/14
66: 14/2 13/7(2)
33: 8/5(2) 6/3(2)
44: 13/9(2) 8/4 6/2

PART 5: In the position shown BELOW, Blue and White alternate rolls and play five rolls each. Can they both build their eight-point primes from 9pt to 2pt? If so, how?

Part 5, Answer: Yes, they can.

Part 5, Solution: Joe gave the (Blue/White) sequence 55/55, 66/66, 44/44, 66/66, 33/33. Actually, though, the roll sequence he listed for parts 3 and 4 (similarly doubled up and if played as I indicated) works for part 5, too.

Parts 5 and 6: Interactive
Both sides make eight-prime, five rolls each

PART 6: Same as part 5, except that neither player is allowed to roll the same number twice in a row, and neither player is allowed to match the number the opponent just rolled. Can both players build their eight-point primes? If so, how?

This is by far the most difficult part, and evidently nobody was able to solve it (even though the conditions as originally worded, allowing a consecutive 44 or 33, made it a bit easier).

Part 6, Logic: First of all, neither side can be hit. Both sides need a 44 and a 33 (to unload the 6pt), and if the remaining rolls are 66 66 66, a hit could only occur on the 4pt (setting the opponent back exactly 4 pips), but no back checker is on (nor can it reach) that point. Therefore, no hit can occur, and the remaining rolls are 66 66 55.

Blue and White together have ten rolls, and arbitrarily we'll pick Blue to go first. Any matching pair of rolls must be at least three apart in the sequence (else a player would have matched the opponent's or his own previous roll, violating a stated condition). The four 66s, therefore, must be maximally spread: 1st, 4th, 7th and 10th in the sequence. Employing the format "Blue/White, Blue/White...," our skeletal sequence looks like this:

66/xx, xx/66, xx/xx, 66/xx, xx/66.

Including the extra "prossover" because her 23/5 is blocked, White needs all twenty of the subplays afforded by five rolls. There is no additional prossover that can be wasted on 22/18; it must escape with a 6. However, at least one Blue 4 must be played 13/9 (no other way to make the 9pt) and one Blue 3 must be played 6/3 (no other way to use three 6pt spares); therefore, to avoid her 22pt checker from hitting (on the 16pt) or being hit (on the 22pt), White must roll 66 (which first occurs on her second roll) before Blue rolls 44 or 33.

We already know Blue's second roll is not 66, and we've just proven it is neither 44 nor 33; thus it must be 55. That takes care of another piece of the puzzle, giving us

66/xx, 55/66, xx/xx, 66/xx, xx/66.

With his initial 66, Blue cannot escape the checker on White's 2pt, and White must play at least one of her 6pt spares to her 2pt without hitting, so her first roll cannot be 44. But it also cannot be 66 (same as her following roll) nor 55 (that would make two 55s in a row). By process of elimination, White's first roll must therefore be 33, producing...

66/33, 55/66, xx/xx, 66/xx, xx/66.

There are two ways to fill in the rest of the sequence so that the 44s are not adjacent:

(a) 66/33, 55/66, 44/55, 66/44, 33/66
(b) 66/33, 55/66, 33/44, 66/55, 44/66

Part 6, Answer: Yes, they can!

Part 6, Solution: Both (a) and (b) have specific solutions. The trick is that, because White's first roll is 33, she must play 23/20 with one of her 3s and later two big numbers (instead of using 23/18 to escape) to bring the 23pt checker to her outer board. Only in this way can she avoid being hit (by Blue's 44) or blocked (by his second 66).

I'll walk you through a sequence of plays for the (a) roll sequence:

Blue will play 66: 22/4 20/14

White will play 33: 23/20 13/10 8/5 6/3

Blue will play 55: 23/8 13/8

White will play 66: 22/4 20/14

Blue will play 44: 13/9(2) 8/4 6/2

White will play 55: 20/15 14/9 13/3

Blue will play 66: 14/2 13/7(2)

White will play 44: 15/11 13/9 6/2(2)

Blue will play 33: 8/5(2) 6/3(2)

White will play 66: 13/7(2) 11/5 10/4

Final position after playing five rolls apiece

Please forgive the clumsy formatting, but here is the play sequence, summarized:

66: 22/4 20/14... 33: 23/20 13/10 8/5 6/3
55: 23/8 13/8... 66: 22/4 20/14
44: 13/9(2) 8/4 6/2... 55: 20/15 14/9 13/3
66: 14/2 13/7(2)... 44: 15/11 13/9 6/2(2)
33: 8/5(2) 6/3(2)... 66: 13/7(2) 11/5 10/4

Nack