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Spot Card Jungle

Last night at the Spotsylvania Bridge Club, I spotted a peculiar ending. South was declarer in 3 NT, a good spot despite some spotty play early on, and needed to win the last six tricks with South on lead. Amazingly, all the honor cards had been played. Declarer managed his spot cards with spotless technique to succeed against any defense.

Guess what my name is!
NT win 6  ?
 ?
 ?
 ?
S 7 6 3
H 9
D 8 4
C
Table S
H 4
D 7 3
C 7 6 5
South leads  ?
 ?
 ?
 ?

Now the spotlight is on you. Try to avoid a blind spot as you supply the missing spots to win the rest.

I’ll bet you can do it spot on. Run, Spot, run!

Complete the ending by assigning North and South each six spot cards (2-9 only, no honors) based on the story. Multiple solutions exist, so a further objective is to make the N-S hands as weak as possible, yet still be able to win six tricks against any defense. Weakness is judged as the sum of all N-S cards.

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Jonathan Mestel Wins!

In February-March 2011 this puzzle was presented as a contest, with 117 participants from 33 locations. Thanks to those who entered, and congratulations to the 84 who supplied N-S cards able to win six tricks. Listed below are the solvers who achieved a card sum of 65 or less. Ties are broken by the evenness of the N-S split, and lastly by date and time of entry.

No tiebreaker was needed for the top spot. Jonathan Mestel was the only one to find the optimal solution, which he submitted just a day before the contest ended, spoiling my shutout — grrr (LOL) — though it was hardly a surprise. Jonathan, a chess grandmaster since 1982, was the 1997 World Chess Solving Champion. When bridge caught his interest, the result was inevitable. Five years ago he won my Six Against The Rock contest (1056 entrants) and topped the final Leaderboard of the series with a 59.25 average (60 max).

The United Kingdom rocks — gold, silver and bronze!

RankNameLocationSumN-S
1Jonathan MestelEngland6131-30
2James LawrenceEngland6231-31
3Wayne SomervilleNorthern Ireland6231-31
4Dan BakerTexas6231-31
5Aurelien BoutinFrance6231-31
6Benjamin YipHong Kong6231-31
7Manuel PauloPortugal6231-31
8Licai YeoSingapore6231-31
9Zla KhadgarOhio6231-31
10Darek KardasPoland6231-31
11Gonzalo GodedSpain6231-31
12Jim MundayCalifornia6231-31
13Peter DuttonEngland6231-31
14Hendrik NigulEstonia6231-31
15Colin SchlossPennsylvania6231-31
16Julian WightwickEngland6231-31
17Reint OstendorfNetherlands6230-32
18John Lindsey IIMassachusetts6230-32
19Nan WangNew Jersey6230-32
20Tim BroekenNetherlands6235-27
21Charles BlairIllinois6224-38
22Jonathan WeinsteinIllinois6331-32
23John ReardonEngland6331-32
24David HodgeEngland6331-32
25Andrew BraginCalifornia6331-32
26Paul NelsonCalifornia6331-32
27Ufuk CotukEngland6331-32
28Jonathan FergusonTexas6331-32
29Joshua ParksVirginia6331-32
30Simon KearonEngland6332-31
31John KropinakOhio6333-30
32Horatio ChinHong Kong6329-34
33Alan ArdronEngland6329-34
34Dan IsraeliIsrael6329-34
35Jacqueline RappIllinois6334-29
36Sam NeedhamCalifornia6329-34
37Dan DangBritish Columbia6329-34
38Jeffery JettisonSingapore6328-35
39Lee WeinerGeorgia (US)6328-35
40Thomas HauklandNorway6327-36
41Gene OwensMichigan6327-36
42Tim DeLaneyIndiana6326-37
43Barry PaulNew York6429-35
44Jean-Christophe ClementFrance6532-33
45Ralph FisherNova Scotia6531-34
46Eric SumnerMinnesota6535-30
47Charles GrayArizona6530-35
48Doug SnowKentucky6535-30
49Pavel StrizCzech Republic6535-30
50Baxter CliffordFlorida6529-36
51Richard SteinCalifornia6525-40
52Donna WegnerSouth Dakota6541-24

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Solution

Because honor cards (A-K-Q-J-10) are not permitted, only five winning cards — S 9-8 D 9 C 9-8 — are available for North-South. Therefore, the sixth trick must be derived from a double finesse in spades or a squeeze. Most of the top solvers took advantage of both, finessing spades to gain one trick and squeezing out another, thus requiring only four of the five winners, as the second-place entry illustrates:

NT win 6 S 9 8 5
H 5
D 2
C 2
Trick
1. S
2. N
3. S
4. N
Lead
S 4
C 2
S 2
S 5
2nd
6
6
7
?
3rd
8
8
9
4th
C 5
S 3
H 4
S 7 6 3
H 9
D 8 4
C
Table S
H 4
D 7 3
C 7 6 5
James Lawrence
England

S 4 2
H
D 9 5
C 8 3

After a spade finesse (West splitting) declarer returns to hand with a club, on which West pitches a spade (he will be finessed anyway), then a second spade is won in dummy. Meanwhile, East safely pitches a club and a heart. The S 5 then squeezes both defenders: East must pitch a diamond to guard clubs, South the C 3, and West must pitch a diamond to guard hearts. A double squeeze, or as this solver quipped:

James Lawrence: Perhaps this should be called a “spot squeeze,” because it’s hard to spot.

Everyone who submitted a 62 sum equally split (places 2-16) duplicated the above construction, excepting the optional swap of the S 5/4 and C 3/2 to retain the 31-31 balance.

A completely different 62-sum layout was submitted by the winner of my recent Yarborough Fair contest. While lacking in evenness (35-27), the incorporation of a double guard squeeze is noteworthy:

Guess what my name is!
NT win 6 S 9 8
H 2
D 9 5
C 2
Trick
1. S
2. N
Lead
S 2
S 8
2nd
3
?
3rd
9
4th
D 3
S 7 6 3
H 9
D 8 4
C
Table S
H 4
D 7 3
C 7 6 5
Tim Broeken
Netherlands

S 2
H 3
D 2
C 9 8 3

On the first spade, suppose East pitches a diamond as shown. The next spade squeezes him in three suits: A club pitch is an instant loss, a heart subjects West to a red-suit squeeze, and a diamond subjects West to a finesse. East cannot alter his fate by pitching a heart on the first spade, as he must pitch a diamond next, allowing West to be squeezed. A similar guard squeeze was constructed by:

Charles Blair: I wonder if Spot would make a good guard dog.

Construction of bridge endings, as for this puzzle, will sometimes produce an illegal position, reachable only by a revoke. All entries in this contest were verified as legal by my software prior to submission. The test is surprisingly simple.

Legal Ending Test using the above ending to illustrate: Three hands have a spade, so 7 missing spades allow at most 2 spade tricks; all hands have a heart, so 9 missing hearts allow at most 2 heart tricks; all hands have a diamond, so 6 missing diamonds allow at most 1 diamond trick; three hands have a club, so 6 missing clubs allow at most 2 club tricks. These add to 7, which satisfies (barely) the number of played tricks, so the ending is legal.

In the jungle, the mighty jungle

The lion sleeps tonight. OK, OK, it’s a leopard, and its sleep was cut short. One solver saw through my red herrings of finessable spades and curious heart spots to discover the optimal solution (sum 61):

NT win 6 S 9 8 2
H 2
D
C 8 2
Trick
1. S
2. S
3. N
4. N
Lead
D 9
C 3
S 9
S 8
2nd
4
D 8
H 4
?
3rd
H 2
8
H 3
4th
3
5
3
S 7 6 3
H 9
D 8 4
C
Table S
H 4
D 7 3
C 7 6 5
Jonathan Mestel
England

S
H 3
D 9 2
C 9 4 3

The D 9 must be cashed first, pitching a heart. When a club is next led to dummy, West must unguard a red suit to protect spades; and whichever suit he gives up isolates South’s threat against East, who is squeezed on the second spade winner. This rare form of compound squeeze does not require the usual common-suit entry because of its threats in all four suits, a condition known as saturation.

The above solution is unique, assuming evenness counts. Limiting North to a sum of 31 requires the lowest spot options (H 2 C 8-2) so any swap with South would increase North’s sum. Further, it cannot be mirrored (e.g., swapping hands or black suits) for positional reasons and the need to cash the D 9 first.

Jonathan Mestel: I’ve always had a soft spot for PavCo.

Good thing, as my company could take you out in a heartbeat with a cell phone plant.

Reint Ostendorf: I will let my bidding be inspired by your Yarborough Fair and this contest. Interesting for you: “spotten” is a verb in Dutch meaning “to mock.”

Too late. I needed that word nine years ago when Rosalind Hengeveld tried to convince me that our “baseball” is your “honkball.”

Spot: Arf! Woof-ersehen!

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© 2011 Richard Pavlicek