Main     Puzzle 8M61 by Richard Pavlicek    

Ruff-day the 13th

What is the most ruffs you have ever seen on a bridge deal? Probably in the 9 to 11 range, but it’s easy to construct a deal with 13 ruffs, the obvious maximum. Suppose you bid like a bat out of Hell to reach this grand slam off three aces.

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

The play is as easy as skinning a black cat: seven ruffs in hand and six in dummy, the ultimate crossruff. Note also that each ruff is necessary, i.e., there is no other way to make the contract unless West gifts a trick by leading a club.

Curiously, a club lead by West would prevent 13 ruffs, even if declarer refused the gift. With North obliged to ruff first, the 13th trick would consist of a trump led by South, hence not a ruff. Ghoulish undoings, to be sure, but I want to clarify in the puzzle below that “best defense” means trying to win the most tricks as usual, not trying to stop 13 ruffs.

Construct a deal where South can make 4 S (no more) against best defense with 13 necessary ruffs.

Obviously, only 10 ruffs need be obtained by North-South, but each ruff must be necessary for the side obtaining it to achieve its goal.

Many solutions exist. A further goal is for the sum of all North-South card ranks to be as high as possible, and secondarily for the North and South totals to be as close as possible.

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Tim Broeken Wins Again!

In October 2015 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Perhaps it was too bizarre, as only 26 persons gave it a try. But hey, that’s twice as many participants as necessary ruffs; or as I’ve sometimes been accused of before, playing with half a deck. Only the 10 listed below produced valid solutions.

No surprises at the top. Tim Broeken continued his winning ways as the first of five solvers to produce the optimal construction (highest N-S card rank total). Nine wins out of 26 puzzle contests is a formidable record, but have no doubts, it will be ‘broeken’ often. Next most wins by one person is three for Dan Gheorghiu (aka Dan Dang) whose solutions have been optimal for six months in a row. Past double winners are Leigh Matheson, Dean Pokorny and Manuel Paulo.

RankNameLocationN-S TotalDisparity
1Tim BroekenNetherlands2691
2Tom SlaterEngland2691
3Jurijs BalasovsLatvia2691
4Dan GheorghiuBritish Columbia2691
5Tina DenleeQuebec2691
6Grant PeacockMaryland2680
7Jon GreimanIllinois2620
8Leigh MathesonAustralia2564
9Nicholas GreerEngland2520
10Richard SteinCalifornia2411

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Solution

Before unveiling the great pumpkin, let’s look at some other noble attempts — or maybe if spelt ‘Nobel’ I could lure some Swedish prizes into these contests. Our correspondent from Down Under provided this entry, with a sage observation as well.

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

Leigh Matheson: To prevent an overtrick East-West must crossruff the first three tricks. Then on a club exit, declarer must crossruff the rest. If the trump position were more fluid, there would be a seesaw squeeze alternative.

Leigh alludes to the 4-card ending reached above, in which declarer has no option but to continue his merry crossruff.

But suppose the S A and S K were switched. The ending would then be as shown at right.

Declarer can now win all the tricks by leading the S K, because West is seesaw-squeezed. If West pitches a club, one ruff establishes the C K. If he pitches a heart, declarer overtakes, and one ruff establishes the H J.

Hence, 13 ruffs would not be necessary per the puzzle conditions.
South
leads
S A Q
H J 10
D
C
S
H Q 7
D
C A Q
Table S
H
D 6 5 4 3
C
S K 8
H
D
C K J

All valid entries (10 listed) were carefully checked for alternatives, such as the seesaw squeeze, and everyone passed with flying pumpkins. A wise group we have here.

Ornithological studies

I appreciate my fine feathered friends, so with Gareth taking the month off, I was glad to have Grant come trick-or-treating with a worthy solution. (These are the times when I really miss Barry Crane.) While perfect in disparity as North-South each add to 134, the 268 total fell short by just a single point from being optimal.

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

The play is essentially the same as in Leigh’s example, ruffing until the cows come home; thrice on defense and 10 times as declarer, with each ruff being necessary as required.

Several respondents offered similar layouts but missed out by making South’s diamonds too strong. Note that D J-9-8-7 is tops, else declarer would have a ruffing finesse available as an alternative to crossruffing.

Curiously, declarer can reach the ending at right, which is the basic matrix of a seesaw squeeze. Leading the S 10 forces West to shorten a red suit, then declarer can duck or overtake accordingly. Why won’t it work?

Because the only sequence of play to reach the ending has North on lead.
Seesaw
squeeze
S J 9
H K Q
D
C
S
H A 5
D Q 10
C
Table S
H
D
C K 8 7 6
S 10 8
H
D J 9
C

Grant Peacock: Wait! Would overruffs count as necessary ruffs?

Definitely, though no one pursued this approach. An overruff is a ruff and it’s usually necessary. The problem is for the ruff that is overruffed to be necessary, which seems plausible only if trump tight. Below is a valid solution, albeit not a tie-breaking contender.

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

Best defense is to cash three clubs, else 4 S is easily made with a loser-on-loser play. West exits with a diamond, and five tricks are routinely crossruffed (East following suit) to reach the ending below:

North leads a heart, which is unnecessary for East to ruff, so he discards — might as well pitch the D A — as South ruffs. Even though the D K is high, declarer must ruff it, as East pitches his club.

The last three tricks consist of two overruffs and an underruff, with all ruffs by both sides being necessary.

Yes, underruffs count too.
North
leads
S A K
H K Q J
D
C
S
H A 8 7 6
D 5
C
Table S J 9 7
H
D A
C 4
S Q 10 8
H
D K Q
C

Prime time construction

The optimal solution gives North-South a card rank total of 269, an odd number, so the best disparity is 1 (135-134). In fact 269 is prime, which could also describe the five solvers who found it, three of whom are regulars. From the two new kids on the block, I’ll favor the entry from the distaff side, since over 95 percent of my participants are male.*

*Bridge puzzles obviously appeal more to men. In the current series, out of 144 names on my successful solver lists, only five are women based on first-name recognition. These are Audrey Kueh (England) twice, Christina Syrakopoulou (Greece) twice, and this month Tina Denlee (Quebec). Please correct me if my gender assumptions are wrong or if I missed anyone.

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

Tina Denlee: West must give East a heart ruff, else declarer can [win 12 tricks]. East must return a diamond ruff, else declarer can [win 11 tricks]. Then West, out of trumps, cannot return a club without giving declarer 1 club + 1 spade + 1 heart + 8 ruffs = 11 tricks, so must lead a heart for East to ruff… Finally, East returns a club, and N-S must crossruff the rest, because all ruffing finesses lose, and all E-W honors are guarded enough not to be ruffed out.

Hard to improve on that explanation. Other perfect solvers produced the same layout, varying only in side-suit identity and alternate divisions of the N-S trump ranks.

Tom Slater: The initial crossruff defense is needed [for E-W to win three tricks], leaving declarer no choice but to crossruff the last 10.

Dan Gheorghiu: Any other play by E-W to the first three tricks would result in declarer’s overtricks. With hands for slam [and a likely grand] North-South have to settle for 4 S. Ghoulish undoings indeed, professor! … I was struggling with East’s dilemma at Trick 4: A club lead looks safe trying to win a trick, but so does a diamond (gift) trying to prevent 13 ruffs, which [I see] we were not supposed to do. …

Rough time for ruffs

Jon Greiman: I can’t decide if this is the roughest contest, or the ruffiest.

Certainly, it was during my roughest time. Thanks to those who have expressed condolences at my time of great sorrow. The contests will continue, as Mabel would have wanted them to. She was my inspiration and always will be.

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© 2015 Richard Pavlicek