1. What is the specific shape of the winning South hand?
3-4-3-3 4-4-2-3 4-2-2-5 2-5-2-4 1-5-3-4
2. How many HCP in the winning South hand?
17 18 19 20 21 22 23
3. How many solvers found the optimal solution?
1 2 3 4 5 6 7
|  | Quit Top |
Dan Gheorghiu Wins
This puzzle contest, designated “April 2018” for reference, was open for over a year. Participants were limited to one try, unlike my usual contests that allowed entries to be revised with only the latest one counting. Participation was lean, no doubt a testament to the difficulty of the puzzle. There were six correct solutions, of which two were optimal (least North-South freakness and the most South HCP).
Congratulations to Dan Gheorghiu, who was first to submit the optimal solution. Dan is a brilliant solver with many past wins including The Case of the Four Aces, Third Best Blues, The Nonagon and Reese’s Pieces. Way back in 2011 he also won The Law of Total Trash.
| Rank | Name | Location | Freakness | South HCP |
|---|---|---|---|---|
| 1 | Dan Gheorghiu | British Columbia | 4 | 21 |
| 2 | Konrad Majewski | Poland | 4 | 21 |
| 3 | Jean-Christophe Clement | France | 4 | 20 |
| 4 | Duncan Bell | England | 4 | 20 |
| 5 | Tim Broeken | Netherlands | 6 | 20 |
| 6 | Bob Ondo | Michigan | 10 | 21 |
Solution
Creating a layout where 3 NT and 4 
both fail yet 4 
makes is quite a task, with few solutions — but that’s what puzzles are made of. Below is one valid solution:
| 4 South | A Q 4 3 2 | Trick | Lead | 2nd | 3rd | 4th | |
| makes 4 | J 2 | 1. W |  6 | 2 | 7 | Q | |
| J 3 2 | 2. S | J | K | A | 5 | ||
 4 3 2 | 3. N | J | K | A | 6 | ||
| K 10 9 8 |  | 7 6 5 | 4. S | Q | 7 | 2 | 8 |
| 7 6 | K 10 9 8 | 5. S | A | 5 | 3 | 8 | |
| 6 5 | K 10 9 8 7 | 6. S | 3 | 7 | 2 | 9 | |
| K 10 9 8 7 | 6 | 7. N | Q | 6 | 4 | 8 | |
| J | 8. N | 3 | 7 | 4 | 9 | ||
| A Q 5 4 3 | 9. W | 10 | 4 | 9 | 5
| ||
| South 20 HCP | A Q 4 | 10. W | 10 | 2 | 6 | J | |
| Freakness 6 | A Q J 5 | 11. S | 5 | 8 | 3 | 10
| |
| West is endplayed | |||||||
Four spades is made as shown. If West instead chooses to ruff at Trick 6, a diamond is pitched, and all variations lead to 10 tricks.
Three notrump is defeated with a diamond lead. If declarer next leads a heart to the jack, East wins and shifts to a spade. Declarer now cannot develop a ninth trick without the defense winning five. The same diamond lead also defeats 4 .
The optimal solution that follows was found only by Dan Gheorghiu and Konrad Majewski:
| 4 South | A 4 3 2 | Trick | Lead | 2nd | 3rd | 4th | |
| makes 4 | 3 2 | 1. W | K | A | 5 | J | |
| Q J 3 2 | 2. N | Q | K | A | 5 | ||
| 4 3 2 | 3. S | 4 | 6 | J | 7 | ||
| K 10 9 8 | | 7 6 5 | 4. N | 2 | 8 | J | 6 |
| 7 6 | K 10 9 8 | 5. S | A | 7 | 3 | 9 | |
| 6 5 | K 10 9 8 7 | 6. S | Q | 8 | 2 | 6 | |
| K 10 9 8 7 | 6 | 7. S | 4 | 7 | 3 | 10 | |
| Q J | 8. N | 4 | 7 | 5 | 9 | ||
| A Q J 5 4 | 9. W | 10 | 2 | 8 | Q
| ||
| South 21 HCP | A 4 | 10. W | 10 | 2 | 6 | J | |
| Freakness 4 | A Q J 5 | 11. S | 5 | 8 | 3 | 9
| |
| West is endplayed | |||||||
West would hardly lead the K in practice, but at double-dummy it’s the strongest start — and in fact necessary to beat 4 , although any spade is sufficient to beat 3 NT. Meanwhile, 4 cannot be stopped with any defense.
Triskaidekaphobes
Dan Gheorghiu: Insanely hard puzzle!
The Donald: “Only six correct solutions” my ass! Even Hillary sent more than that, and I submitted 13,
making four spades on every one of them. Simple play… trump a loser.
I would agree with an added comma. “Trump, a loser” is the correct form.
Professor Irwin Corey: The feasibility of this construction is an analytical catalyst which
exacerbates protocol over procedure to remain indigenous to its inception.