Puzzle 8K10 Main |
| by Richard Pavlicek |
Rudolph with your nose so bright, did you set this slam tonight?
Six spades they tried, the reindeer replied, I led a heart and declarer was fried.
I see, said the master of Yuletide arts, But what if the slam is played in hearts?
No! cried Rudolph, about to stampede, Thats even set if a trump is the lead.
On Christmas morning the fog lifted, but only the North-South cards remained:
6 South Both Vul | Q 9 8 7 6 5 K 10 9 8 7 9 2 | Rudolph West Pass Pass Pass Pass | Donner North 2 3 6 | Ronald East Pass Pass Pass | Blitzen SOUTH 2 2 NT 4 Pass | ||
? ? ? ? | ? ? ? ? | ||||||
A K J Q J 6 5 A K J 10 5 A |
Help! Nobody can remember the other hands, so its up to you:
Construct a West hand for Rudolph that fits the dialogue.
Multiple solutions exist. Further goals (tiebreakers for the December contest) are for the HCP total and spot-card sum (in that order of priority) for West and East to be as near to equal as possible.
In December 2010 this puzzle was presented as a contest, with 76 participants from 42 locations. Thanks to everyone who entered, and congratulations to the 25 successful solvers. Ranking is by the most even division of the E-W HCP, secondarily by the most even spot-card sums, and lastly by date-time of entry.
Kudos to Eddy Choi, Hong Kong, who was the first of 11 solvers to give East-West identical HCP totals and spot card sums.
Rank | Name | Location | HCP | Spots |
---|---|---|---|---|
1 | Eddy Choi | Hong Kong | 6-6 | 55-55 |
2 | Charles Blair | Illinois | 6-6 | 55-55 |
3 | Jonathan Mestel | England | 6-6 | 55-55 |
4 | Gareth Birdsall | England | 6-6 | 55-55 |
5 | Manuel Paulo | Portugal | 6-6 | 55-55 |
6 | John Auld | England | 6-6 | 55-55 |
7 | Jim Munday | California | 6-6 | 55-55 |
8 | Pavel Striz | Czech Republic | 6-6 | 55-55 |
9 | Leif-Erik Stabell | Zimbabwe | 6-6 | 55-55 |
10 | Julian Wightwick | England | 6-6 | 55-55 |
11 | Howard Liu | Illinois | 6-6 | 55-55 |
12 | Mike Chanter | England | 6-6 | 53-57 |
13 | James Lawrence | England | 6-6 | 48-62 |
14 | Ufuk Cotuk | England | 5-7 | 58-52 |
15 | Sid Ismail | South Africa | 2-10 | 55-55 |
16 | Barry Rigal | New York | 2-10 | 55-55 |
17 | Paul Gilbert | England | 10-2 | 59-51 |
18 | Jan Kuipers | Netherlands | 10-2 | 51-59 |
19 | Richard Stein | California | 2-10 | 62-48 |
20 | Barry Goodman | England | 2-10 | 62-48 |
21 | Wayne Somerville | Northern Ireland | 2-10 | 63-47 |
22 | Ryan Leung | Singapore | 10-2 | 47-63 |
23 | Sebastien Louveaux | Belgium | 2-10 | 63-47 |
24 | Harold Goodman | Georgia (US) | 2-10 | 63-47 |
25 | Rik ter Veen | Netherlands | 2-10 | 63-47 |
Puzzle 8K10 Main | Top One Foggy Christmas Eve |
Per the dialogue, the object is to construct a West hand, and thereby a complete deal, in which a heart lead will defeat both 6 and 6 . It is easy to see how 6 can be defeated by a heart ruff; but if such a layout exists, how could a heart lead defeat 6 ? Indeed, it could not, so a more devious approach is needed.
Ten solvers (places 16-25) went for the brazen solution of a 10-0 club division, allowing the defense to get a club ruff against either 6 or 6 . While undeniably valid, this construction requires the A to be in the same hand as the long clubs for a heart lead to beat both slams, which means a lopsided HCP division and therefore no contention for the top spot. (A 10-card club suit also makes the auction implausible, but this has no bearing on the puzzle solution.)
A more subtle construction allows an equal arrangement of 6 HCP for West and East, as well as an equal spot-card sum of 55. The first solver to submit an optimal solution was Eddy Choi (Hong Kong):
6 South | Q 9 8 7 6 5 K 10 9 8 7 9 2 | Trick 1 W 2 E | Lead 4 3 | 2nd 7 | 3rd A | 4th 5 | W-L L1 | ||
4 2 Q 9 8 7 6 4 3 2 K J 10 | 10 4 3 2 A 3 Q 8 7 6 5 4 3 | ||||||||
Lead: 4 | A K J Q J 6 5 A K J 10 5 A | Lose 1 trump trick |
Six spades is defeated by the defense shown, since declarer cannot reach dummy to draw the last trump; East must score the 10. Six hearts is defeated in more straightforward fashion by the imminent spade ruff after any lead (including a trump).
Eddys solution was duplicated by Jonathan Mestel, John Auld (can I add Lang Syne? as I write this just an hour into 2011), Julian Wightwick and Howard Liu. Alas, this was not my intended solution translation: I overlooked it but of course equally correct.
Howard Liu: Neat puzzle! Nice usage of the 9 to enforce the tiebreaker, in the spirit of dividing the twelve points of Christmas.
Not sure how to interpret this, but next time Im makin a list and checkin it twice!
The solution I intended was submitted by the six of the top 11, including the winner of my last contest, Charles Blair (Illinois):
6 South | Q 9 8 7 6 5 K 10 9 8 7 9 2 | Trick 1 W 2 E | Lead 2 3 | 2nd 7 | 3rd A | 4th 5 | W-L L1 | ||
4 3 2 Q 9 8 7 6 4 2 K J 10 | 10 4 3 2 A 3 Q 8 7 6 5 4 3 | ||||||||
Lead: 2 | A K J Q J 6 5 A K J 10 5 A | Lose 1 trump trick |
Six spades is defeated by the defense shown, similarly garnering a trump trick for East. It is also curious, though irrelevant to the puzzle, that a heart lead is required to beat 6 , whereas any lead suffices on the previous deal. Six hearts is defeated with any lead.
And finally, to avoid possible ambiguity in the instructions, Leif-Erik Stabell (Zimbabwe) left no antlers unturned as he rafted the mighty Zambezi River, producing this layout:
6 South | Q 9 8 7 6 5 K 10 9 8 7 9 2 | Trick 1 W 2 E | Lead 2 9 | 2nd 7 | 3rd A | 4th 5 | W-L L1 | ||
4 3 2 Q 8 7 6 4 3 2 K J 5 | 10 4 3 2 A 9 Q 10 8 7 6 4 3 | ||||||||
Lead: 2 | A K J Q J 6 5 A K J 10 5 A | Lose 1 trump trick |
Leif-Erik Stabell: This assumes high cards are included in counting spots as ace = 1, king = 13, queen = 12, jack = 11, etc. If high cards are included but ace = 14, swap the 9 and 3, in which case East-West cannot have the same spot count. If high cards are not included in the spot count, swap the 10 and 5 in addition.
I think I see it now: Three cases covered, two subtle swaps and a partridge in a pear tree.
Charles Blair: Using the Four Aces Point Count (ace = 3, king = 2, queen = 1, jack = 0.5) your solution fails the equal HCP goal.
Wonderful. Instead of a Yuletide with three wise men, I get one wise guy.
Pavel Striz: Merry Christmas, reindeer!
Then all the reindeer loved him, as they shouted out with glee,
Rudolph the red-nosed reindeer, youll go down in his-tor-y!
Puzzle 8K10 Main | Top One Foggy Christmas Eve |
Acknowledgments to song lyricist, Johnny Marks
© 2010 Richard Pavlicek