Puzzle 8S71   Main

Trump Moves to Lilliput

by Richard Pavlicek

Leaving the White House was never foreseen by our 45th President. Obviously this was only possible after a fake election, not only with almost every Democrat voting twice but with ballots right out of Sixth Sense: “I see dead people.” Notwithstanding, the Trump coup failed, and a forced eviction ensured the Trump ending. Now the problem was where to put him. Alas, weeks went by, and no country stepped forward. Where would someone of such small stature be welcome? Of course! The island of Lilliput.

We’ll stop by there shortly, but first let’s look at a few other trump endings. The first of these is common:

Trump elopement	K Q — K
A Q K — —		— — A K A
win 2 South leads	— A Q Q

With spades trump, South to lead needs two of the three remaining tricks. If South leads the Q, West pitches his heart, and declarer can win only one trick. If South leads the Q, West can either ruff or pitch for the same result. Instead, South must cash the A then lead a diamond.* North elopes with the K, either by overruffing the queen, or pitching if West ruffs high.

*The actual trump elopement requires only a two-card ending, but the need to cash a winner first is often crucial.

The next trump ending is rare, because its preconditions are restrictive:

Smother play	A Q J — —
K 10 — — A		— A A K —
win 2 South leads	Q J K — —

South again is on lead and needs two tricks with spades trump. If a trump is led, declarer wins only the A; but if he exits with his losing heart, the defense is helpless. West’s only hope is to pitch his club as East wins. On the forced diamond return, West’s K is smothered, as dummy can overruff or discard depending on West’s play.

Each of the above examples used extremely high card ranks. In actual play, these endings would typically contain middle ranks — but in light of our country’s own Trump ending, now is the time to think small. Your task is to effect “trump moves to Lilliput” with two endings.

Create a three-card ending for (1) a trump elopement and (2) a smother play, using low card ranks.

South leads with spades trump and must win two tricks against any defense. Further, there must be only one card (of three) which South can lead to succeed. Correct solutions will be ranked by the sum of all card ranks* used in both endings, and ties will be broken by the sum of only South’s card ranks. Lower is better.

*Ace = 14, King = 13, …, Three = 3, Two = 2.

Andrew Spooner Wins

This puzzle contest ran from November 23 (2020) to New Year’s Day 2021. There were 33 entries from 19 persons (multiple entries were allowed from the same person, but only the latest one counted). Not exactly a robust turnout but an improvement over the recent Jolly Old Saint Nicholas.

Fifteen of the 19 persons submitted a valid solution for at least one ending, but only five were successful on both. Hats off to Andrew Spooner, who was the only solver to submit optimal solutions (lowest rank sum) for both. Andrew is the winner of Diamond Stack (a puzzle still open but allowing only one entry per person) and has other recent high finishes. The five successful solvers are listed below by total rank sum, with ties broken by South rank sum and date-time of entry.

Winner List

Rank	Name	Location	Total	South
1	Andrew Spooner	Australia	75	16
2	Grant Peacock	Maryland	77	15
3	Charles Blair	Illinois	77	15
4	Joseph DiMuro	California	77	15
5	Jean-Christophe Clement	France	77	17

Solution

This puzzle was a takeoff on Lilliputian Squeezes (2015) where the object was to create the smallest triple squeeze. Today Americans are more obsessed with Trump endings — albeit four years overdue — so that was the obvious follow-up.

Trump elopement

The first of my chosen trump endings was the trump elopement, for which most solvers found the lowest possible rank sum (36) using only twos, threes and fours in each suit. Alas, several submitted bogus layouts like the following:

Sum 36	3 — 2 3
— 3 3 4		4 4 4 —
win 2 South leads	2 2 — 2

South ruffs a heart in dummy, then ruffs a diamond with his last trump. Two tricks, and only one path to success; but is this a trump elopement? Not by my definition, which requires leading a plain suit in which next hand is void and has a high trump. This is simply a crossruff.

Four of the five winners managed to give South three deuces (optimal solution):

Sum 36	3 3 — 2
— — 3 4 3		4 4 4 —
win 2 South leads	2 2 2 —

Andrew Spooner: Ruff a diamond in the North hand, then lead a club to elope with South’s trump.

Joseph DiMuro: South must lead a diamond, ruffed in dummy, then return a club; South’s trump wins a trick by elopement.

My favorite of the optimal solutions contained a squeeze element:

Sum 36	3 — 3 3
4 — 4 4		— 4 3 — 2
win 2 South leads	2 2 2 —

South leads a heart to elope with the 3, while West is squeezed. West cannot afford to ruff (North discards a diamond and declarer crossruffs the rest), and whichever four he pitches establishes North’s three.

Charles Blair: Geza Ottlik described a similar elopement as West being squeezed in the four of clubs.*

*In Ottlik’s scenario West could safely pitch a diamond (East had a winner) so the “club-four squeeze” was appropriate; pitching it would establish the three, and keeping it would allow the three to be ruffed.

Smother play

The second trump ending, the smother play, proved to be a stumbling block for almost half the solvers. In their attempts to reduce the rank sum they overlooked alternate solutions. To wit:

Sum 39	6 2 3 —
5 2 3 — —		— — 4 3 2
win 2 South leads	4 3 — 2 —

The smother play works perfectly: South leads the losing diamond, West’s only hope is to pitch a heart, and the forced club return smothers West’s 5. Alas, but so does the simple play of crossing in trumps and ruffing the heart. Since my conditions required South’s first lead to be unique, this was clearly invalid.

One way to render a valid solution was to reduce my example to its lowest terms:

Sum 41	6 4 3 — —
5 2 — — 2		— 5 3 2 —
win 2 South leads	4 3 2 — —

Joseph DiMuro: This is just the diagram you provided, with the ranks lowered as far as they will go; so I know I don’t have the optimal solution… I may try to improve it later.

Joe “knows” right but apparently got caught up in the holidays — or kidnapped by aliens — as he didn’t follow up.

Considering that five trumps are required for a smother play, the lowest theoretical sum is 39 ( 6-5-4-3-2 + 3-2 + 3-2 + 3-2 + any 4) but achieving it is tricky — in fact impossible when South has two trumps without leaving an alternate solution. The secret was to invert the layout, no doubt an easy task for our winner; after all, living “Down Under” what else? The optimal solution:

Sum 39	4 3 — — 3
— 3 2 3 —		5 2 — — 4
win 2 South leads	6 — 2 2

Andrew Spooner: South’s only winning play is to lead a diamond and discard a club from North. East can either ruff, allowing declarer to win two trumps easily, or discard a club for the smother ending.

Andrew not only achieved the optimal total for both endings (36 + 39 = 75) but also for the South tiebreaker (6 + 10 = 16). Well done, and the only solver to prevent my second “puzzle victory.” In all my Bridge Puzzles, only once was the best solution not found by at least one participant. That was World Series of Bridge in 2016.

Final Notes

The Donald: This is your last fake puzzle, Mister RP-loser! As soon as I overturn this fake election, I will order the FCC to block this f***ing site forever.

Terminology 101

Trump control	Democrats’ futile effort
Trump promotion	Twitter’s main function
Trump suit	Litigation by Rudy Giuliani
Trump Echo	Nevada casino soon to open
Trump a loser	Is there a comma missing?

Enough about losers. My next puzzle will honor greatness: I Have a Dream.. ends on MLK Day (January 20). Hey! That’s also the official Trump ending, so it eliminates a loser as well. All aboard for Lilliput!

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© 2020 Richard Pavlicek