PROBLEM #1

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 207 score: 0 Blue is Player 1
XGID=-a----D-B---eD---c-dbAB--B:0:0:1:00:0:0:3:0:10
Each side played twice

PROBLEM #2

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 209 score: 0 Blue is Player 1
XGID=-a----D-B---eD---c-db-BA-B:0:0:1:00:0:0:3:0:10
Each side played twice

PROBLEM #3

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 197 score: 0 Blue is Player 1
XGID=-a----E-C---eB---c-db-BA-B:0:0:1:00:0:0:3:0:10
Each side played twice

For each of the above three problems:

From the standard opening position, each side (legally) moved twice. How is this possible?

SOLUTION

Congratulations to Kevin McDonough, Roberto Litzenberger and Brian Lonergan, for solving problem #1; to Bob Koca, Roberto Litzenberger and Brian Lonergan, for solving problem #2; and to Bob Koca, Brian Lonergan and Roberto Litzenberger, for solving problem #3! (Kudos as well to Bob Koca and Brian Lonergan for solving problem #4, which will be briefly addressed.)

White's checker position is identical in all three target diagrams (see example below). Her pipcount is 147, which has progressed 20 pips from the starting position. Specifically, she has moved 6/5 = 1 pip, and 24/5 = 19 pips, which means one of the rolls contains an ace, and the other is a large doublet.

The obvious roll combination that comes to mind is that the small roll is 11 or 31 (4 pips) and the large roll is 44 (16 pips). To translate these rolls to 24/5 and 6/5, the 24pt checker must travel to the 5pt while being supplemented by three aces (or a 3) along the way. Taking into account that White's 19pt, 17pt and 12pt (Blue's 6pt, 8pt and midpoint, respectively) are blocked, there are exactly two ways to navigate home: 24/22 6/5(2) + 22/6, and 24/16 13/5 + 16/13 6/5.

There exists a subtle alternative: one of the 24pt checkers moves and gets hit -- sent back to the bar (25pt), and enters with 55, neatly landing on the 5pt. In this way, it travels a net of 20 – 1 = 19 pips, and 6/5 is the other pip of the 20-pip progress. Thus, the first roll is x1, splitting and slotting the 5pt, the split back checker is hit, and on White's second turn it covers the 5pt with bar/5. As Blue does not end up with a blot (or new point) on the near side, and White cannot reach her 10pt or 15pt in half a roll, a hit exchange must occur on her 20pt. In short, White plays 41W (24/20 6/5), Blue hits, and White hits back on her 20pt on the way to bar/5.

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 204 score: 0 Blue is Player 1
XGID=-a----D-C---eC---c-dbAA-AB:0:0:1:00:0:0:3:0:10
White's position (with a flexible Blue distribution)

The previous two paragraphs define three different approaches (or categories) for reaching White's final position, which I'll summarize and reorder as follows:

....(1) White plays 41E (24/20 6/5), is hit on her 20pt, and with 55 covers bar/5.
....(2) White plays 11E (24/22 6/5(2)), and with 44 plays 22/6.
....(3) White plays 44F (24/16 13/5), and with 11 or 31 plays 16/13 6/5.

With this White (147-pip) setup and certain conditions for Blue (two on roof, three other back checkers, no external blots, and only other points being 13pt, 8pt and 6pt), I identified 14 (four-roll) positions that can be arrived at uniquely, and 5 additional positions that can be arrived at in more than one way. Put another way, I found 27 qualifying positions, 8 of which turned out to be duplicates. (My hasty recap here was inaccurate.)

The above position, where Blue's 13pt-8pt-6pt distribution is balanced (3-3-4), can be arrived at with any of White's three approaches (Blue plays the opening roll):

.....(1) 51$-41E-33F-55P............................................[33F = bar/22 bar/21 13/10 8/5*]
.....(2) 63N-11E-22s-446 ..or.. 32n-11E-33B-446.......[33B = bar/22 24/21 13/7]
.....(3) 31$-44F-22s-11E.............................................[22s = bar/21 24/22 13/11]

I don't expect you to be fluent in Nactation, but White's part in these sequences match the set of three parallel lines above it, and with the visual aid of XG (or physical board) you can easily work it out, especially by consulting the additional diagrams that follow in this post.

In order to make each of White's three approaches unique for purposes of a problem set/trio and to thematically unite the positions by giving Blue a 22pt anchor, I had to make Blue's 3-3-4 distribution less balanced (4-2-4 in two of them, and 2-3-5 in the third).

Here is the sequence for Problem #1:

		White is Player 2 score: 0 pip: 162 Unlimited Game Jacoby Beaver pip: 184 score: 0 Blue is Player 1
XGID=-a---aD-C---eE---c-da-A-AA:0:0:1:33:0:0:3:0:10
Problem #1: Blue played 21W, White plays 41E

Blue opened with 21W (Wild, 24/22 6/5), and White replied with 41E (Each, 24/20* 6/5).

[Regarding Problem 4: If (instead of 21W) Blue opens with 41W (24/20 6/5), his 22pt blot is instead a second checker on the bar, and that is why his other back checker will stay on his 24pt instead of 21pt; that is, getting hit denies him a spare 3 to play 24/21 with his upcoming double 3s.]

With his 33, Blue plays as shown below:

		White is Player 2 score: 0 pip: 167 Unlimited Game Jacoby Beaver pip: 172 score: 0 Blue is Player 1
XGID=aa---AD-B-A-eD---c-daAB---:0:0:-1:55:0:0:3:0:10
Problem #1: Blue played 33F

With his 33, Blue played F (1:2:1 Field, bar/22 24/21 13/10 8/5*).

[By contrast, in the Problem 4 sequence, Blue played bar/22(2) 13/10 8/5*, and has no fifth 3 to play 24/21, and that is why his back blot ends up on the 24pt instead of the 21pt. This is a clarification of the bracketed note above the diagram.]

White now rolls a 55 joker; she hits twice with bar/20*/15*/10/5, thereby reaching the final position diagrammed below.

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 207 score: 0 Blue is Player 1
XGID=-a----D-B---eD---c-dbAB--B:0:0:1:00:0:0:3:0:10
Problem #1: Final position

Here is the sequence for Problem #2:

		White is Player 2 score: 0 pip: 163 Unlimited Game Jacoby Beaver pip: 185 score: 0 Blue is Player 1
XGID=-a-a--DAB---eE---c-cb---BA:0:0:1:22:0:0:3:0:10
Problem #2: Blue played 31n, White played 11E

With an opening 31, Blue played n (near, 8/7 6/3), and with her roll of 11 White replied with E (Each, bar/22* 6/5(2)).

Blue now rolls 22 and plays as shown below.

		White is Player 2 score: 0 pip: 163 Unlimited Game Jacoby Beaver pip: 177 score: 0 Blue is Player 1
XGID=-a-a--DAB--AeD---c-cb-BA--:0:0:-1:44:0:0:3:0:10
Problem #2: Blue played 22S

Blue played 22S (3:1 split, bar/23 24/22(2) 13/11), cooperatively adding a blot onto White's path.

From here, White rolls 44 and hits twice with 446 (6pt, 22/18*/14*/6), thereby reaching the final position for Problem #2 shown below.

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 209 score: 0 Blue is Player 1
XGID=-a----D-B---eD---c-db-BA-B:0:0:1:00:0:0:3:0:10
Problem #2: Final position

Here is the sequence for Problem #3:

		White is Player 2 score: 0 pip: 151 Unlimited Game Jacoby Beaver pip: 176 score: 0 Blue is Player 1
XGID=-a----E-CaA-dC---c-ea---BA:0:0:1:22:0:0:3:0:10
Problem #3: Blue played 43D, White played 44F

With his opening 43, Blue played D (Down, 13/10 13/9). With her 44, White replied with F (Field, 24/16* 13/5).

[Note that if Blue opens with 31$ (Slot, 13/10 6/5) instead of 43D, he reaches the same position with one more checker on the midpoint and one fewer checker on the 6pt, for a final 13pt-8pt-6pt distribution of 3-3-4 instead of 2-3-5. I would have preferred this more balanced distribution but the identical position could have been reached using a problem-2 approach with 63N-11E-22S-446.]

Blue now rolls 22 and plays as shown below:

		White is Player 2 score: 0 pip: 151 Unlimited Game Jacoby Beaver pip: 168 score: 0 Blue is Player 1
XGID=-a----E-CaAAdB---c-ea-BA--:0:0:-1:11:0:0:3:0:10
Problem #3: Blue played 22S

Blue played 22S (3:1 Split, bar/23 24/22(2) 13/11).

White rolls 11 and plays E (Each, 16/15*/14*/13 6/5), thereby reaching the final position below.

		White is Player 2 score: 0 pip: 147 Unlimited Game Jacoby Beaver pip: 197 score: 0 Blue is Player 1
XGID=-a----E-C---eB---c-db-BA-B:0:0:1:00:0:0:3:0:10
Problem #3: Final position

Here is a recap of the sequences for problems #1, 2 and 3 (with Blue playing first):

.....#1...21W-41E-33F-55P
.....#2...31n-11E-22S-446
.....#3...43D-44F-22S-11E

Nack

Messages In This Thread

• Twenty-two-point Mystery (finale)
  Nack Ballard -- Saturday, 30 March 2019, at 2:00 p.m.
  • A related problem 4?
    Bob Koca -- Sunday, 31 March 2019, at 4:58 a.m.
    • Problem #4 (diagrammed)
      Nack Ballard -- Sunday, 31 March 2019, at 9:55 a.m.
  • Twenty-two-point Mystery (finale) -- SOLUTIONS
    Nack Ballard -- Wednesday, 3 April 2019, at 9:48 p.m.

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