Puzzle Corner

Introduction

It has been a year since I reviewed the criteria used to select solutions for publication. Let me do so now.

As responses to problems arrive, they are simply put together in neat piles, with no regard to their date of arrival or postmark. When it is time for me to write the column in which solutions are to appear, I first weed out erroneous and illegible responses. For difficult problems, this may be enough; the most publishable solution becomes obvious. Usually, however, many responses still remain. I next try to select a solution that supplies an appropriate amount of detail and that includes a minimal number of characters that are hard to set in type. A particularly elegant solution is, of course, preferred, as are contributions from correspondents whose solutions have not previously appeared. I also favor solutions that are neatly written, typed, or sent via e-mail, since these produce fewer typesetting errors.

Problems

J/A 1. Larry Kells reports on an “oddball-sounding remark” he overheard at his bridge club. “That was an awfully risky grand slam you bid! You sure were lucky to get the six-zero trump break you needed to make it.” Naturally he was very curious but never did find out what had occasioned that remark. We are asking for help in solving this mystery.

J/A 2. Ken Rosatto has 12 identical coins and another that looks the same but weighs either more or less. He wants to find the oddball in three weighings using only a balance.

J/A 3. Rocco Giovanniello has a pyramidal variant of the tetrahedral “wink jumping” puzzle we published last year. The pyramid has four square layers. The top layer has one space, the second has four, the third nine, and the fourth 16. Each of the spaces except the top is occupied with a wink; the top is empty. The goal is to eliminate all but one wink by a sequence of 3-D checkerlike jumps.

Speed Department

David E. Brahm reports that a popular game in Naples, FL (at least, between his parents), is a dice game called farkle. A player gets points on his first throw if his dice either a) contain a one or a five, b) form three or more of a kind, or c) form three pairs. Otherwise the roll is a “farkle,” and the turn is over. What is the probability of rolling a farkle if six dice are used?

Solutions

Mar 1. I wonder whether Larry Kells was a Course I major: he is such an expert on bridge(s). Today he reports on an argument that arose at his bridge club; neither side was vulnerable. After South’s weak 2 opening, East-West bid up to 7 and made it for +1,510. Afterward, one of the kibitzers expressed an opinion that a 7 sacrifice would have been worthwhile for North-South. The four players all objected, saying that as long as West led his trump, 7 was obviously doomed to go down seven for 1,700, more than the value of the successful grand slam. Who was right? Here’s the deal:

David Cipolla enjoyed this problem and writes, “The kibitzer is right. With ideal execution by North and South, even after a West lead of the four of trumps, the sacrifice could have gone down only four and not seven as the players argued. The requirement is for North and South to give up a trick that they could win, thereby allowing a cross-ruff strategy to become effective. At trick one, North and South must play the three and two, respectively, losing to West’s four. With West still on lead, he must play and win the next three heart tricks while North (or South) drops three diamonds and South (or North) drops three clubs. Note that East cannot win the first round of hearts even if he did not previously discard the two and so is unable to gain the lead. When West leads his fourth heart, South ruffs and North drops his last diamond. South then plays any diamond with North ruffing. North then plays a club with South ruffing. This continues until North and South are out of clubs or diamonds. They each have one spade at the end of the cross ruff, and either North or South wins that trick. So East and West win only four tricks: one trump trick and three heart tricks. North and South win nine tricks: one heart ruff, four diamond ruffs, three club ruffs, and a trump trick at the end. Down four.

Mar 2. Andrew Russell offers an interesting variant of an old problem. You have nine coins all of equal weight. Some material has been removed from one coin and added to another so that the total weight is unchanged. You have four weighings on a balance scale. Can you find four weighings that permit you to determine the lighter and heavier coin without previous weighings?

Tom Terwilliger sent us the following detailed solution.

Weighing #1  1, 2, 3 vs. 4, 5, 6    Note that 9 is never weighed!
Weighing #2  1, 4, 7 vs. 2, 5, 8
Weighing #3  1, 5 vs. 2, 4
Weighing #4  3, 6 vs. 7, 8

You have 72 possibilities of light and heavy coins (9 x 8). In four weighings you have 81 possible results (34). The trick is to place the proper number of coins on each pan of the scale. Each weighing must divide the above 72 possibilities nearly in thirds, or else there won’t be a unique solution. I started by listing the results of using various numbers of coins in each pan as follows:

One coin in each pan: There are 15 ways each pan can be heavy and 42 ways the scale can balance. If you weigh 1 vs. 2, then either 1 can be heavy (eight possibilities) or 2 can be light (seven possibilities, not counting the case of 1 heavy and 2 light twice). Since seven coins are not weighed, the scale will balance any time both odd coins are in the group of seven (7 x 6). Clearly this is no good, as after this weighing, there are 42 possibilities left if the scale balances, and there are only 27 results possible on the remaining three weighings.

Two coins in each pan: Weigh 1, 2 vs. 3, 4. If 1, 2 is heavy, then the following are possible:

1 heavy  3, 4, 5, 6, 7, 8, 9 light  7 possibilities
(If 2 is light the scale balances)
2 heavy  3, 4, 5, 6, 7, 8, 9 light 7 possibilities
3  light  5, 6, 7, 8, 9 heavy 5 possibilities
(Don’t count 1 heavy and 3 light twice!)
4  light  5, 6, 7, 8, 9 heavy  5 possibilities
Total:  24 possibilities

There are also 24 ways pan 3, 4 can be heavy, and by subtraction 24 ways the scale can balance.

5, 6, 7, 8, 9 containing both odd coins  5 x 4 = 20 possibilities
1 heavy 2 light or the reverse 2 possibilities
3 heavy 4 light or the reverse 2 possibilities

Here, each weighing divides the possibilities exactly in thirds. I then tried to solve this with four weighings, each containing two coins per pan. I wrote a computer program to try all combinations and found that there are none, so I tried three coins in each pan.

Weigh 1, 2, 3 vs. 4, 5, 6. Pan 1, 2, 3 will be heavy if

1, 2, 3 heavy  4, 5, 6, 7, 8, 9 light  6 x 3 = 18 possibilities
7, 8, 9 heavy  4, 5, 6 light   3 x 3 =  9 possibilities
Total: 27 possibilities

The scale will balance any time the two odd coins are in the same group of three, which will happen 6 x 3 = 18 times.
This is not as good as 24/24/24, but since four weighings of two coins each didn’t work, the solution must involve combining two- and three-coin weighings. Now there is an easy way to do two (symmetric) weighings of three coins in each pan, namely the rows and columns of a square:

#1  1, 2, 3 vs. 4, 5, 6  #2   1, 4, 7 vs. 2, 5, 8

There are no more than nine possibilities for each result. In fact, the scale can never balance both times, and there are exactly nine ways each of the other eight results can be obtained:

1, 2, 3 and 1, 4, 7 pans heavy:
1  heavy  5, 6, 8, 9 light   4 possibilities
3  heavy  5, 8 light   2 possibilities
7  heavy  5, 6 light   2 possibilities
9  heavy  5 light   1 possibility

1, 2, 3 heavy and 1, 4, 7 balanced
1  heavy  4, 7 light    2 possibilities
2  heavy  5, 8 light    2 possibilities
3  heavy  6, 9 light    2 possibilities
7  heavy  4 light    1 possibility
8  heavy  5 light    1 possibility
9  heavy  6 light    1 possibility

The only remaining step is to show that there is no solution if you weigh three coins in each pan once and then two coins in each pan three times. This was done easily by modifying the above-mentioned computer program.

Better Late Than Never

Y2002. Avi Ornstein noticed that we mistakenly omitted the parentheses from 10 = (20+0)/2.

2002 Oct 3. Tom Harriman and Gunnar Bergman provided very different proofs from the one given. Note that the trigonometry was omitted from the published solution due to space considerations; it was present in Hess’s solution. Also I just noticed a typo: the correct formula for q is 2 sin-1 (9/16) + 3 sin-1 (1/8).

Other Respondents

P. W. Abrahams, J. Astolfi, D. Bator, H. M. Blume Jr., C. K. Brown, C. Cheng, D. Cipolla, G. Coram, J. Dieffenbach, I. Gershkoff, D. Gross, J. Grossman, J. E. Hardis, T. J. Harriman, R. Hess, H. Hochheiser, J. Hoebel, S. Hsu, S. Kanter, D. King, S. Klein, P. Latham, R. Lax, B. Layton, R. Marks, R. D. Marshall, L. J. Nissim, L. Peters, C. Polansky, J. Rudy, L. Sartori, I. Shalom, G. Steele, C. Swift, L. Villalobos, and G. Waugh.

Proposer’s Solution to Speed Problem

The only way to roll a six-dice farkle is to form two pairs and two singletons from the numbers 2, 3, 4, and 6. There are (  ) = 6 ways to distribute 2/3/4/6 among pairs and singletons and 6 x 5 x (  ) = 180 ways to order them. There are 6 x 180 = 1,018 farkles out of 66 = 46,656 possible rolls, giving a farkle probability of 1,080 / 46,656 2.3%.

Send problems, solutions, and comments to Allan Gottlieb, New York University, 715 Broadway, 7th Floor, New York NY 10003, or to gottlieb@nyu.edu.