Main Puzzle 8N43 by Richard Pavlicek
[Friday, March 31] PavCo attorneys today filed a class-action suit against Walt Disney Studios for its unlawful appropriation of a bridge term for an animated film. Financial recovery is expected to exceed $500 million, and qualified claimants must register by April 30, midnight GMT. To be eligible you must be a member in good standing of a national bridge organization, or a homeless derelict; we dont really care. The key is to be alive, which puts us one up on Disney. Register now, and well turn this ice-capade into our own Scrooge McDuck.
Card players have been aware of frozen suits before Mickey Mouse hit the drawing board, and the term was in use when I began to play in the 60s. A suit is considered frozen if, and only if, both sides can gain a trick if the suit is led first by either opponent instead of themselves. Its like the zugzwang in chess; whoever moves first loses. Consider the major suits on the following deal:
|3 NT South|| J 10 6 4|
Q 9 3
A K J 8
| K 7 3|
10 7 5
Q 10 9 5
6 5 2
| Q 9 5|
K 8 4 2
K J 8 7
| A 8 2|
A J 6
A 4 3
Q 10 9 7
The spade suit is frozen. Declarer can gain a trick if either defender leads a spade, and the defense can gain a trick if declarer leads a spade from either hand. More specifically, the side that first leads a spade wins only one trick, and the other side wins two. (I am ignoring Norths long spade, which will always be a winner if established, regardless of what happens on previous rounds.)
The heart suit might also look frozen, but not quite. Call it semi-frozen, because it is frozen only to one side, in this case the defenders, who cannot lead a heart without losing a trick. Note that declarer is able to lead hearts once (low to the jack) without disturbing its frozen state to the defense, after which the suit becomes truly frozen.
The difference between frozen and semi-frozen is often slight. For instance, swapping the 7 and 9 makes the spade suit semi-frozen, as declarer can now run the jack (losing to the king) and the suit remains frozen to the defense, and is then truly frozen.
The play in 3 NT is irrelevant, but declarer can succeed, even after the best lead of a diamond, by availing his frozen assets.
Now its your turn! Using different high-card arrangements than my examples:
|Construct a frozen spade suit and a semi-frozen heart suit.|
The semi-frozen hearts must be frozen to the defense but not declarer (as my example). As a further goal (tiebreakers for the April 2017 contest) try to keep the rank sum of each suit holding as close to average (26) as possible, and the North-South HCP as low as possible.
In April 2017 this puzzle was presented as a challenge with no help provided inviting anyone who wished to submit a solution. Participation was down, expected perhaps for a frozen theme in April (Im weather challenged), or maybe my title was too short, or its a subtle message for me to take a hike. Only 39 brave souls gave it a try, of which only the nine listed were correct on both constructions.
Congratulations to Duncan Bell, England, who is on the hottest RPbridge streak of all-time, winning three months in a row, and four of the last five. Being fast and clever, the only way to beat this guy might be to avail our frozen assets and freeze his hard drive. Also achieving a perfect score were Tina Denlee, Foster Tom and Lief-Erik Stabell.
Countrywise, the win goes to Canada with three placings, versus only two each for the U.K. and U.S. Considering the puzzle theme, I guess this was predictable, as I envision their entries being delivered by dogsled along the Iditerod Trail.
|3||Foster Tom||British Columbia||0||5|
|7||Dan Gheorghiu||British Columbia||0||9|
Before showing solutions, lets theorize a bit. There are 67,108,864 (413) possible distributions of a single suit among four hands, all of which must have a constant rank sum. Counting each card at face value (extending to jack 11, queen 12, king 13, ace 14) this sum must be 104, which is easily verified by simple addition, or the formula to sum consecutive integers: number * (lowest + highest) / 2, or specifically 13*(2+14)/2. Therefore, the average rank sum of the suit holding in each hand is 104/4, or 26.
Winning this puzzle required each hand to be exactly average, which is possible in a variety of ways but rare my count shows only 1728 of the 67+ million layouts to be equivalued. Only one doubleton (A-Q) adds to 26, but this cannot exist in a frozen or semi-frozen layout, so each hand must have 3-4 cards, hence a 4-3-3-3 distribution, which reduces the number to 840; but this includes 24 permutations of each combination, so there are effectively only 35 distinct combinations of four suit holdings that each add to 26. (I would list them here, but Im afraid I might be committed.)
An interpretation problem arose with this puzzle. Consider the following suit layout, which Tom Slater submitted as frozen (spades) and Charles Blair submitted as semi-frozen (hearts). Which is it? Or should it be classified as something else?
|Notrump||K 9 4|
|Q 8 6||A 10 2|
|N-S lead||J 7 5 3|
The suit might be called super-frozen, because if N-S lead it entirely, they win nothing (ignoring Souths long card); but if E-W lead it twice, N-S win two tricks a two-trick difference. Sounds pretty frozen! But one could also argue that N-S are entitled to zero tricks, so they can lead the suit once safely (cant do worse than zero) and still gain a trick if E-W lead the next round; so in that sense it could be deemed semi-frozen.
Tom Slater: This suit is doubly frozen. Declarer has nothing by force, and the defense has just the ace.
Charles Blair: The suit will be frozen after South leads and plays any card from dummy.
PavCo attorneys assured me that whichever way I ruled, they could manufacture the legalese to prevail in court. After all, they put Disney on ice*, so what chance could Slater and Blair possibly have? Oh, well; enough litigation. Considering the desolate leaderboard this month, I decided to accept both.
*which brings to mind the old legend that Walt Disney was cryonically preserved after death, which is not true; but if it were I would be keeping my house temperature very low just in case.
About a dozen respondents lost out because they overlooked a finessing option by declarer, typically a backward finesse. Most of these layouts were perfectly balanced (zero variance) alas, but not frozen. For instance:
|Notrump||J 9 6|
|A 8 4||K 10 3|
|N-S lead||Q 7 5 2|
North-South can establish a trick by force if North leads the nine, which East covers: 10, queen, ace. On the second round South leads and finesses against the eight, so E-W can win only their two tops. Even if you swap the six and four to foil the backward finesse, the suit wouldnt be frozen but only semi-frozen, since low to the jack (or simply leading the jack) keeps the suit frozen to the defense.
All it takes is a slight change in the previous layout to effect a perfect solution. The following spade suit was submitted by Tina Denlee and Foster Tom, both of whom live in the great white north (albeit 5000 kilometers apart) so they should know all about freezing.
|Notrump||Q 8 6|
|K 9 4||A 10 2|
|N-S lead||J 7 5 3|
Each hand totals 26, and the suit is completely frozen. If North or South leads any card from either hand, they lose three fast tricks; but if East or West starts the suit, N-S must score the queen or jack.
Foster Tom: East-West have two tricks and N-S only have their 13th card, unless someone breaks the ice. In the play, one need only remember second hand low and cover an honor. It is barely possible to give N-S only 2 HCP [with 26 sums all around], but I could only semi-freeze that.
The above layout contains the fewest possible N-S HCP for a frozen suit with four 26-sum holdings, but an equally good alternative is J-8-7 opposite Q-6-5-3, as discovered by Duncan Bell, Lief-Erik Stabell and Tim Broeken. In either case, the N-S and/or E-W hands can be swapped.
Duncan Bell: If N-S lead, E-W cover honors to set up a finessing position at trick 2. If E-W lead, N-S play low as second hand to guarantee a trick.
For Part 2, the object was to create a layout frozen only to East-West, i.e., semi-frozen such that North or South can lead the suit once without disturbing its frozen state to the defense. Again the goal was to equalize the rank totals of each hand at 26, which Dan Gheorghiu, British Columbia, achieved with this curious layout:
|Notrump||K J 2|
|Q 10 4||A 7 5|
|N-S lead||9 8 6 3|
North-South are entitled to one trick (ignoring Souths long card) if only they lead; but if East or West starts the suit with any card, N-S can develop two tricks with direct play, e.g., if East leads the five, South covers with the six. What deems the suit semi-frozen is that South can lead it once without loss, specifically the nine (or eight) covered by West and North, then the suit remains frozen whether East takes the ace or not.
Dan Gheorghiu: If either defender leads, the Q-10 can be double-finessed for two tricks. If South leads (i.e., nine, 10, jack) the finesse assures only one trick, but the remaining two tricks are frozen.
Only one semi-frozen layout fits the optimal conditions (balanced pips, fewest N-S HCP) except for the obvious 180-degree rotation that applies to any layout in this puzzle. This was discovered by the top four solvers, and a few others who missed out on Part 1. Our winner, Duncan Bell, England, gave the 2 HCP to South:
|Notrump||10 9 7|
|A 8 4||K J 2|
|N-S lead||Q 6 5 3|
North-South are entitled to zero tricks (ignoring Souths long card) if only they lead. If East or West leads first (any card), N-S can establish a trick by force, so the suit is clearly frozen to E-W. It is not frozen to N-S, however, because North can lead the 10 or nine (East covers with the king) after which the suit remains frozen to E-W. Hence, semi-frozen.
Duncan Bell: North can lead the 10 or nine, which must be taken with the king to set up a frozen position. (If East plays the J or 2, N-S can guarantee one of the first three tricks.)
Leif-Erik Stabell: North can lead the 10 to the king, then the suit is frozen.
Foster Tom: North can lead the ten or nine, which West must carefully cover with the king to keep the position intact. Giving N-S only 1 HCP would leave the A-K-Q for E-W, forcing A-Q doubleton for a 26 rank sum, which Im sure admits no frozen position.
Tina Denlee: I imagine a deal where N-S cash seven minor winners while E-W follow suit, then exit in a major to force E-W to break the ice, but E-W exit in the same major to zugzwang declarer in the other. Down one!
And the rest of us imagine Tina in a cage or at least a wall around Canada.
Dan Gheorghiu: Since I can still fog a mirror, I signed up for the PavCo class-action suit, and I eagerly await my cut from the Donald family.
The duck stops here.
Apologies to Donald and nephews Huey, Dewey and Louie
© 2017 Richard Pavlicek