Main     Puzzle 8M13 by Richard Pavlicek    

Diamonds in Distress

The diamond suit gets more than its fair share of accidental contracts, because its artificial use in many contexts causes misunderstandings. Case in point: The multi 2 D opening to show a weak two-bid in either major, while firmly ingrained in Europe, is new to the American scene. As the following deal illustrates, our mastery of this convention is a long way off, though results are promising — alas, for puzzle material but not for bridge.

South opened a multi 2 D and North responded 2 NT, an artificial inquiry to clarify South’s hand; 3 C then showed a maximum with hearts (3 D would show spades, and three of a major would be natural and minimum). Oops! North forgot they were playing multi and interpreted 3 C as a feature with a natural weak 2 D bid; hence key-card Blackwood to the hopeless slam in diamonds.

6 D South S A K 6 3
H K
D A Q 9
C A Q 4 3 2
Both Vul

West

Pass
Pass
Pass


North

2 NT
4 NT
6 D


East

Pass
Pass
All Pass


South
2 D
3 C
5 D
S Q J 10
H 10 6 4
D K J 10 8
C J 8 5
Table S 9 7 5
H J 7 5
D 7 6 5 4
C 10 9 7
Lead: S Q S 8 4 2
H A Q 9 8 3 2
D 3 2
C K 6

Well, almost hopeless. Declarer won his eight side tricks, finessing diamonds en route (West perforce splitting) to reach the ending at right. A spade was ruffed in hand as both opponents helplessly followed, then a heart achieved a trump promotion: West chose to ruff high, so declarer pitched the club and won the last two tricks with a finesse. Believe it or not!

(Smart money is on not.)
North
leads
S 6
H
D A 9
C 4
S J
H
D K J 8
C
Table S 9
H
D 7 6 5
C
S
H 9 8 3
D 3
C

The above deal shows the weakest trump suit that can make a slam against any defense. As verification, note that if North-South held only four trumps, one opponent must have at least five, so at most eight side tricks could be won, which would require four trump tricks. The only way to accomplish this is to have at least A-K-J-7* in one hand, which adds to 45. The above case (A-Q-9-3-2) adds to 40. Trying to reduce the strength of five trumps is also futile, though an alternate arrangement of A-Q-9-3 opposite 2 is possible.

*Thanks to Martin Sinot, Netherlands, for this improvement. I had original written A-K-J-9, but A-K-J-7 is sufficient if West has 6-5-4-3-2 and East has Q-10-9-8. Declarer wins eight side tricks, ruffs with the seven as West perforce underruffs, then exits to score A-K-J.

Perhaps more widely known is the weakest trump suit that can make a grand slam (say 7 D), which is D A-J-9 opposite K-2, adding to 49. Each opponent has 3=3=4=3 shape with the D Q-10 onside; so regardless of the lead declarer can cash nine side winners, finesse the D 9, and crossruff the last three tricks. Alternately, trumps could be A-K-J-9 opposite 2, allowing a finesse and a coup.

Now, for the current puzzle:

What is the weakest diamond suit that can make 5 D against any defense?

A further goal is for the sum of all North-South cards to be as low as possible, and secondarily for the North and South sums to be equal or as close as possible. All cards are summed by rank: Ace = 14, king = 13, …, two = 2.

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Tim Broeken Wins!

In June 2015 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Interest took an upturn this month with 44 participants. Go figure; last month a normal 3 NT draws all of 19, and now a ridiculousD more than doubles the turnout. Could this reflect the status of bridge? Only 26 of the 44 entries were valid (5 D makable against any defense) of which only the 16 listed below achieved a trump-suit total of 35 or less.

Congratulations to Tim Broeken, Netherlands, who was the first of four to submit the maximal, or more appropriately, minimal construction for both tie-breakers. He’s back in the saddle again… This is Tim’s seventh puzzle-contest win, far and away my leading solver.

RankNameLocationDiamondsNorth-SouthDisparity
1Tim BroekenNetherlands321780
2Tom SlaterEngland321780
3Dan GheorghiuBritish Columbia321780
4Grant PeacockMaryland321780
5Adam DickinsonScotland321820
6Nicholas GreerEngland321820
7Jamie PearsonOntario321820
8Dan BakerTexas321840
9Xavier DantanFrance321848
10Charles BlairIllinois3218420
11Dean PokornyCroatia321860
12Leigh MathesonAustralia331757
13David KenwardEngland331791
14Wayne SomervilleNorthern Ireland331920
15Julien ReichertFrance341826
16John PortwoodEngland351960

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Solution

There are at least three ways to define the weakest trump holding to make 5 D: (1) fewest combined trumps, (2) fewest trumps in one hand, or (3) lowest sum of rank counts. The conditions of this contest clearly dictated Option 3, but for curiosity sake let’s first look at Option 1, the solution to which is a subset of the picture atop this page. John Portwood provided the following deal:

5 D S 5 4 3 2
H A Q 10
D
C A Q 7 6 5 4
Leader
1. W
2. S
3. N
4. S
5. N
6. S
7. N
8. N
Lead
S 8
H 2
S 3
H 3
S 4
C 2
C A
H A
2nd
2
9
J
J
K
J
10
8
3rd
9
10
Q
Q
A
Q
3
4
4th
10
6
6
7
7
9
K
K
S 8 7 6
H K J 9
D 8 7 6 5 4
C K J
Table S K J 9
H 8 7 6
D K J 10 3 2
C 10 9
S A Q 10
H 5 4 3 2
D A Q 9
C 8 3 2

After any plain-suit lead, declarer can reach the ending at right. North leads a spade, and East must ruff with the 10 or jack (else South scores the nine and exits to win two more); South discards. The diamond return is won as cheaply as possible, then South exits with his last plain-suit card to win the last two.

After a diamond opening lead (North pitches a club) South captures an honor from East; then the trump promotion develops similarly but only needing to endplay East once.
North
leads
S 5
H
D
C 7 6 5 4
S
H
D 8 7 6 5 4
C
Table S
H
D K J 10 3 2
C
S
H 5
D A Q 9
C 8

If the criterion were Option 2 (fewest trumps in one hand) it is possible to succeed with two trumps opposite two, or four total, even with the obvious trump lead to curb ruffing ability. The minimal construction is left to the (crazed) reader, but this solver was right on target:

Tom Slater: The weakest four-card trump suit I can manage is A-J opposite 8-2.

Lowest rank sum

All that mattered for this contest was the sum of the card ranks, no matter how many trumps were held or how they were divided. The above solution (A-Q-9) adds to 35, which is well off the mark. Various improvements are possible as shown in the winner list, and the lowest is 32, or specifically K-10-4-3-2. Nicholas Greer, England, demonstrates this with a North-South total of 182, perfectly split at 91 each.

5 D S A Q 10 5
H 5 4 3
D K 10 4
C 5 4 2
Leader
1. W
2. S
3. N
4. S
5. N
6. S
7. N
8. S
9. N
10. S
Lead
H 8
D 2
C 2
S 2
C 4
S 3
H 4
S 4
H 5
C A
2nd
3
J
9
6
J
J
9
K
J
8
3rd
K
K
10
10
Q
Q
10
A
Q
5
4th
A
5
6
7
7
8
6
9
7
K
S K J 6
H 8 7 6
D A Q J 8
C 8 7 6
Table S 9 8 7
H K J 9
D 9 7 6 5
C K J 9
S 4 3 2
H A Q 10 2
D 3 2
C A Q 10 3

Nicholas Greer: This trump suit can be held to two losers by leading a plain suit after cashing all the side tricks, having used three spade entries and the D K to take all four rounded-suit finesses.

Diamonds must be led once, forcing West to split his honors to force the king, though the exact order of play can be varied to reach the ending at right. A heart or club lead then promotes the D 10 into an 11th trick.

If West chose to win the D A early and return the queen, the D 10 is still promoted but in a two-card ending.
South
leads
S 5
H
D 10 4
C
S
H
D A Q 8
C
Table S
H
D 9 7 6
C
S
H 2
D 3
C 3

The ultimate solution

Not surprisingly, our leading solver and three others found the way to reduce the North-South total. Rather than require the 10 in each side suit for finessing, a ruff enables a third trick in one. This requires the N-S trumps to be divided 4-1, else two rounds of trumps would be fatal. The following deal submitted by Tim Broeken has the lowest possible N-S total of 178, split 89-89.

5 D S 5 4 3 2
H A Q 10
D K 10 4 3
C 5 4
Leader
1. W
2. S
3. N
4. S
5. N
6. S
7. N
8. S
9. N
10. N
Lead
C J
D 2
C 5
C 2
S 2
H 2
S 3
H 3
H A
S 4
2nd
4
Q
8
10
9
9
J
J
8
K
3rd
7
K
A
D 3
10
10
Q
Q
4
A
4th
Q
5
9
K
6
6
7
7
K
8
S 8 7 6
H K J 9
D A Q J 9
C J 10 9
Table S K J 9
H 8 7 6
D 8 7 6 5
C K 8 7
S A Q 10
H 5 4 3 2
D 2
C A Q 6 3 2

Tom Slater: …Ruffing in the hand with longer trumps removes the need for the C 10.

Dan Gheorghiu: There are eight side tricks by finessing everything, the D K by leading a diamond from South, one club ruff in North, and the D 10 is promoted at Trick 11 or 12.

Grant Peacock: South recently talked his partner into playing Roman 2 D and is eager to try it out.

Mathematical proof

Charles Blair: I wonder whether the problem answer would be different if you used a sum from ace = 13 to two = 1, which, I think, is a more natural summation by rank.

Only difference I can see is a reduction from 32 to 27, however (as certified by Professor Irwin Corey) the answer “three cubed” would be a blunder of exponential proportion. Clearly, the weakest ruffing* suit to lose only two tricks in five diamonds must be two to the fifth power. Case closed.

*Due to derogatory remarks against the Mexican people, the use of ‘trump’ suit is henceforth abolished.

Leigh Matheson: It’s a diamond heist and they can’t stop it!

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© 2015 Richard Pavlicek