Main     Puzzle 8M43 by Richard Pavlicek    

The Nonagon

Over 40 years ago, David Sachs of Baltimore, Maryland, composed a curious 9-card problem called “The Nonagon.” It is presented below, though I have used aces for top cards to improve readability. In notrump, declarer must win eight tricks with South to lead.

NT win 8/9 S 9 8
H A K 9
D A Q 9
C 2
Leader
1. S
2. N
3. N
4. S
5. S
Lead
D 2!
H K!
C 2
S A
D 3
2nd
10
10
9
J
J
3rd
Q
2
Q
8
A
4th
C 8
S 10
4
Q
C 10
S J 10
H
D K J 10
C J 6 5 4
Table S K Q
H Q J 10
D
C K 10 9 8
South leads S A
H 3 2
D 3 2
C A Q 7 3

Declarer begins with a diamond finesse, which triple-squeezes East; he must keep a heart stopper, and a spade discard would succumb West to a spade-diamond squeeze (club finesse, three rounds of hearts) so East lets go his fourth-round club stopper. At Trick 2, a top heart is led to triple-squeeze West; he must keep a diamond stopper, and a club discard would allow declarer to establish his fourth club, so West’s best defense is to abandon his spade stopper.

Next comes the club finesse, S A, and a diamond to North, as East easily parts with another club, leaving the position at right.

Finally, the losing D 9 triple-squeezes East again. Parting with either major stopper is a certain loss, so East lets go the C K. West is then endplayed in clubs.
North
leads
S 9
H A 9
D 9
C
S
H
D K
C J 6 5
Table S K
H Q J
D
C K
S
H 3
D
C A 7 3

The Nonagon consists of three triple squeezes (hence the name) followed by an endplay that is virtually invisible from the start. Beautiful, except for one serious flaw: It could never occur in bridge. The 9-card ending is unreachable from a full deal without a revoke.

Illegal Ending Proof: All four hands have a spade, so the six missing spades allow only one trick with a spade lead. Three hands have a heart, so the five missing hearts allow only one trick with a heart lead. Same with diamonds. Hence there are only three possible suit leads for the four preceding tricks, which proves the ending is impossible with legal plays.

Your mission is to resurrect the ending for bridge:

Construct a legal Nonagon to win eight out of nine tricks.

Note that finesses like the D A-Q and C A-Q above are not required. All that’s necessary is to retain the essence of the original problem, i.e., three triple squeezes and an endplay, assuming East-West put up the strongest defense.

A further goal (tie-breaker for the November 2015 contest) is for the North-South hands to be as weak as possible, judged by the sum of all card ranks: Ace = 14, King = 13, Queen = 12, Jack = 11, etc.

Dan Gheorghiu Wins

In November 2015 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Participation dropped to a new low, aside from the extremely difficult first puzzle in this series (Seesaw Recall with just four attempts). Only 17 persons gave it a try, but that doesn’t bother me. Quality beats quantity. (That’s my story and I’m stickin’ to it, despite rumors I’m fading fast.) Only the four listed below produced valid layouts.

Dan Gheorghiu is on an amazing streak, submitting optimal puzzle solutions for seven months in a row, and this time being the only person to find the lowest possible North-South total. This is Dan’s fourth puzzle-contest win.

RankNameLocationN-S Total
1Dan GheorghiuBritish Columbia99
2Tom SlaterEngland102
3Nicholas GreerEngland102
4Tim BroekenNetherlands107

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Solution

One workaround to create a legal ending is to eliminate spades from one hand, effectively allowing two previous spade tricks instead of one. Our all-time leading solver did just that with the following valid construction.

NT win 8/9 S 7 2
H 9 8 2
D 10 5 4
C 2
Leader
1. S
2. N
3. N
4. S
5. S
6. N
Lead
H 3!
S 7
C 2
D 9
D 2
D 5!
2nd
S 3
5
8
6
7
C Q
3rd
9
C 3
K
4
10
3
4th
5
4
4
C 9
C 10
8
S 4 3
H
D 8 7 6
C J 6 5 4
Table S 6 5
H 7 6 5
D
C Q 10 9 8
South leads S
H 4 3
D 9 3 2
C A K 7 3

Tim Broeken: After a heart start, a diamond discard by West gives directly the eighth trick; and a club discard allows East to be triple-squeezed later by giving away a club at Trick 2. Only a spade does not directly cost a trick. Now after a high spade (club discard South) and a high club, East has to discard clubs on the next three diamonds (high-high-low); otherwise either North’s S 2 or H 2 becomes high. West has no choice but to win the third diamond and [lead from C J-6-5 into South’s A-7].

Note the importance of cashing one top club before the throw-in, else West would escape the endplay, and East could not be squeezed discarding after North. The order of Tricks 2-5 could be altered in many ways, but that’s fine; my puzzle conditions did not require a unique path to success. West was triple-squeezed, then East, then East again as West was endplayed.

A novel approach

Another workaround to obtain legality is to reduce the number of cards in hearts or diamonds. (The original Nonagon held eight of each, which, split between three hands, allowed only one previous trick each.) Tom Slater took that approach to produce this neat arrangement, featuring a different kind of endplay:

NT win 8/9 S 5 3
H 10 9 2
D 7 6 2
C 2
Leader
1. S
2. N
3. N
4. N
5. S
6. S
7. N
Lead
H 3!
D 7
D 6
C 2
C A
H 5
D 2!
2nd
C 5
S 6
S 7
9
7
C 8
3rd
9
H 4
C 3
Q
H 2
10
4th
6
3
4
6
10
7
S 8 4
H
D 5 4 3
C 8 7 6 5
Table S 7 6
H 8 7 6
D
C K J 10 9
South leads S 9 2
H 5 4 3
D
C A Q 4 3

Tom Slater: A heart lead forces West to discard a club to prevent a major-suit simple squeeze. Two rounds of diamonds force East to discard his spade stopper and endplay guard, else clubs can be established. Now two rounds of clubs, then a heart triple-squeezes West out of his exit card. Finally, throw West in with a diamond [to lead from his S 8-4].

Tom’s clever construction even exceeds the requirements, as East was arguably triple-squeezed twice, first to release his spade stopper, then to release his guard. West was also triple-squeezed twice, first to release his club stopper, then to release his exit card — and all this followed by an endplay. A dodecagon, perhaps?

Nona plays the 99er game

Only one respondent managed to find the lowest possible North-South rank sum of 99 — who else but our man on a 7-month perfecta streak. Call it the mini Nonagon:

NT win 8/9 S Q 4 3
H 9 4 3
D 8 6 3
C
Leader
1. S
2. N
3. S
4. N
5. E
6. S
Lead
D 2!
H 3
H 2
H 4
S 9
C J!
2nd
4
5
C 5
7
J
3rd
6
8
9
C 2
6
4th
S 5
C 4
6
C 10
3
S 10 8 6
H
D 7 5 4
C 10 5 4
Table S 9 7 5
H 7 6 5
D
C 9 8 7
South leads S J 2
H 8 2
D 2
C J 6 3 2

Dan Gheorghiu: The Nonagon is a saturated squeeze (menaces in all four suits) but the count must be rectified. This layout might be a double Nonagon, functioning on which discard East chooses at Trick 1. East cannot discard a heart (lone stopper), and a club leads to the original Nonagon (no need to detail further); but what about a spade? [Three rounds of hearts] then force West to pitch all his clubs, then South’s last black jack squeezes West for an eighth trick. Certainly there is a nagging question whether West’s club discard at Trick 2 is a legitimate triple squeeze or just an idle card. [Likewise for Tricks 3-4].

I would say an idle card. Once East chooses the inferior spade discard at Trick 1, West becomes the sole guardian of spades and diamonds. Hence the effective play elements are one triple squeeze (Trick 1), an endplay (Trick 4) to rectify the count, and a simple squeeze (Trick 6). So it wouldn’t be a “double” Nonagon, but still a true Nonagon, since East is presumed to take the strongest path (club discard).

Nonagon in real life

In order to prove the Nonagon a reality, below is a complete deal to show how it could arise in actual play. Suppose South opens 2 NT and later winds up in 6 NT after North uses Stayman to discover no major fit and shows a diamond suit. Rather than risk leading from an honor, West chooses a passive club lead.

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

Declarer pitches a spade from dummy and wins the C K, then the D K reveals the Hawaiian split. The D J is led next, covered by West, to the ace in dummy; then declarer returns to hand with a heart. This routine play leaves the 9-card ending below — yes, the Nonagon.

Declarer finesses the D 8 to triple-squeeze East, who does best to pitch a club. Next comes a heart to the queen to triple-squeeze West, who does best to pitch a spade.

Declarer cashes the S K, crosses to the S A and cashes the D 10, as East easily parts with another club. Finally, the losing D 4 is led to triple-squeeze East again. His only hope is to surrender his last club, but this leaves West endplayed in clubs.
South
leads
S A 5 4
H K 5 4
D 10 8 4
C
S Q 9 8
H
D 9 7 6
C 8 5 4
Table S J 10 7
H 10 9 8
D
C Q J 10
S K 2
H Q 2
D 2
C A 7 3 2

Note that the above ending is generically identical to Dan’s construction but with realistic card ranks.

Nona gone, what’s next?

Coming soon will be the Triskaidecagon, featuring three triple and two double squeezes. Yeah, right. More likely it’s time to return to Earth. But wait! If there were five suits, I could throw in a quadruple squeeze or two — as the men in white coats circle my house.

Dan Gheorghiu: Going from a full deal to a legal 9-card ending was another diabolical assignment — for triple-squeezing one’s brain.

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© 2015 Richard Pavlicek