Puzzle 8M61   Main

# St. Valentine’s Hand

by Richard Pavlicek

Archaeologists made a startling discovery last month beneath the banks of the Tiber River near Via del Mare in southern Rome. A marble slab, carbon-dated as circa A.D. 250, contained 26 neatly arranged etchings, followed by three lines of text and the signature Valentinus Episcopus. Immediate research confirmed this to be the martyred bishop canonized centuries later as St. Valentine.

The text began with the sentence: Sex corda meridiem. Oh dear! Was Valentinus some kind of a sex pervert? No, wait, that’s Latin for “six,” and the other words indicate “hearts” and “south.”

The etchings were in Roman numerals ranging from I to XIII, separated into two groups of 13, one directly above the other, and each group had four partitions vertically aligned. Could these represent card ranks in four suits? No one could be sure, but a conversion to playing cards revealed no inconsistencies and produced the following diagram:

 Sex cordaMeridiem K 2 K Q J 10 K J 6 A K 4 3 A J 7 A 9 8 5 3 4 2 10 8 2

Wow! Could this be the origin of bridge? If so, it might explain our tendency to overbid. Sex corda is a poor contract, probably needing West to have A-Q, although just one high diamond onside might suffice with a lucky club lie.

The text continued: Fui occidentem et inveni genus solus ut evinco, which appears to translate “I was West and led the only suit to defeat.” So Valentinus found the killing lead. Very interesting, but what was his hand? Further excavation at the site found nothing.

Bridge history beckons! Can you help determine St. Valentine’s hand?

Construct a West hand to fit the story.

A further goal (tie-breaker for the February 2016 contest) is for the West hand to be as balanced as possible (lower freakness) and secondarily to be as weak as possible, judged by the sum of all card ranks: Ace = 14, King = 13, Queen = 12, Jack = 11, etc.

## Grant Peacock Wins

In February 2016 this puzzle was presented as a challenge, inviting anyone who wished to submit a solution. Participation picked up with 51 persons giving it a try. Go figure: Shorter month; more entries. Maybe there’s some mystique in that extra day we were all given.

Hold the aviary! Congratulations to Grant Peacock, Maryland, who was the first of seven to submit the optimal solution (lowest freakness and rank sum). Grant is a longtime, leading participant, winning my Ever More puzzle exactly one year ago, and he Kept the Ship Afloat way back in 2002 — but then he was from California, which bolsters my ornithological theory that eastern migrations exceed western due to tailwinds. Who knows… his next win may come from Iceland.

Winner List
RankNameLocationFreaknessRank Sum
1Grant PeacockMaryland292
2Leif-Erik StabellZimbabwe292
3Tim BroekenNetherlands292
4Tom SlaterEngland292
5Nicholas GreerEngland292
6Dan GheorghiuBritish Columbia292
7Jamie PearsonOntario292
8Jurijs BalasovsLatvia293
9Gareth BirdsallEngland376
10Tina DenleeQuebec376
11Jon GreimanIllinois376
12David BrooksAustralia680
13Jacco HopNetherlands682
14Dean PokornyCroatia684
15Leigh MathesonAustralia684
16Christina SyrakopoulouGreece684
17Charles BlairIllinois684
18Jim MundayMississippi791
19Jonathan MestelEngland799

 Puzzle 8M61   Main Top   St. Valentine’s Hand

## Solution

Many successful solutions hinge on the need of a diamond ruff when West has either ace-seventh or a blank queen. In the latter case West could have as little as 6-5-4-3 7-6-4-2 Q 9-7-6-5, which is indeed the lowest possible rank sum. Alas, the freakness was a notch too high in the priority tie-breaker, so these solutions had to settle for places 9-11.

Whether St. Valentine’s hand was a trivial one to score a diamond ruff, or something remarkable, we may never know, as his life remains a mystery. Indeed, even his actual existence is questioned, so the sainthood stuff may all be a myth — unlike my Uncle Cedric, of course, who was canonized in 1983 in front of many witnesses. Poor guy… fell into a laser printer; blinded for life.

What appeal to me most are the subtle constructions, even though not in contention for the top spot, like the following layout submitted by Dean Pokorny, Leigh Matheson and Charles Blair.

Charles Blair: I would not have found this deal if the contest had been held in February 2013, 2014 or 2015.

Yes, I’m generous like that, adding an occasional extra day for deadbeats.

 6 South K 2 K Q J 10 K J 6 A K 4 3 Trick1 W2 N3 N4 N5 S6 S7 N Lead 3? K Q 2 A 2 6 2ndK675 57Q 3rd435J 3K4 4th724 36108 W-LW1W2W3W4W5W6L1 3 4 2 A 9 8 7 5 3 J 7 6 5 Q 10 9 8 6 5 4 7 6 Q 10 Q 9 Lead: 3 A J 7 A 9 8 5 3 4 2 10 8 2

The need for a specific suit lead is hardly obvious, so to see the light, suppose West starts with his singleton spade. Declarer wins the K (finesses can wait at double-dummy), draws trumps, clears spades, and leads a diamond (West must duck) to the king. A low diamond then gives East the lead in the following ending:

 win allSuccess — J 10 J A K 4 Trick8 E9 N10 N11 S Lead Q J 10 9 2nd2 8 9? 3rd58A 4thA 9 6 W-LW1W2W3 — — A 9 J 7 6 5 Q 10 9 8 — — Q 9 East leads — A 9 8 — 10 8 2

A spade would yield the slam outright with a ruff-sluff, so East must lead a club. Either the Q or 9 allows declarer to finesse his way home, but no finesse is necessary because finishing trumps squeezes West in the minors.

The same ending is also easily reached after a trump or low diamond lead.

Leigh Matheson: An opening club lead is required to avert the looming endplay on East.

Spot on, as it not only allows East to exit safely in clubs, but it kills the entry for the squeeze. Also note that if declarer attempts to strip clubs before throwing East in with the second diamond, West can overtake with the A and cash the setting trick.

### Mate in Six

Who else but a chess grandmaster would spot a queen sacrifice, or imagine that a spade lead could be necessary in the following layout?

Jonathan Mestel: Here’s an early contribution to your pons asinorum (bridge of fools). I’m sure it can be improved upon, but I like the way West has to hold the Q and lead from it in order to break up the squeeze.

Yeah, considering it placed last on the leaderboard, my guess is it could be improved upon too. But whatever you lost in ranking, you gained in style points.

Suppose West makes the obvious open-cards lead of a low diamond. Declarer wins the king and leads four rounds of trumps (no spade ruff) as West comfortably pitches three diamonds and a club. On the last trump, West is triple-squeezed; a club surrenders two tricks immediately, a diamond allows a diamond to be established to squeeze West again in the black suits, so the best hope is a spade. Declarer now pitches a club from dummy and ducks a diamond, then the last spade squeezes West in the minors. Whew!

 6 South K 2 K Q J 10 K J 6 A K 4 3 Trick1 W2 N3 N4 N5 N Lead 3! K Q J 10 2ndK2467 3rd5358A 4th7 3 5 7 5 W-LW1W2W3W4W5 Q 4 3 — A 9 8 7 5 3 Q J 6 5 10 9 8 6 5 7 6 4 2 Q 10 9 7 Lead: 3 A J 7 A 9 8 5 3 4 2 10 8 2

Declarer wins the K in dummy, smartly refusing the finesse to keep communication fluid. (If declarer wins the J or ruffs it out early, the defense is easier, as East can retain a good spade with his late Q entry.) Then four rounds of trumps leaves the following position, where South leads the 9 to triple-squeeze West.

 win 7Failure 2 — K J 6 A K 4 3 Trick6 S7 S8 E Lead 9 2 10! 2nd A!8A 3rd 364 4th 6102 W-LW1L1W2 Q 4 — A 9 8 Q J 6 10 9 8 6 — Q 10 9 7 South leads A J 9 4 2 10 8 2

A club pitch is clearly fatal, so suppose West lets go a spade; declarer pitches a club from dummy, leads to the K and another to East (West must duck both) then the last spade winner squeezes West in the minors.

If West pitches a diamond, declarer leads to the K and another to set up the J, which squeezes West in the black suits.

But wait! West can pitch the diamond ace! Now East gains the lead with the Q for a spade through to kill the squeeze. Down one!

### Balanced Hand Solution

No solution is possible with West’s shape 4-3-3-3 (freakness 0) or 4-4-3-2 (freakness 1) but the next step up is achievable in a subtle manner, hinging on West holding all the intermediate diamond spots. This was found by the top eight solvers, and seven of them produced identical, optimal constructions (lowest rank sum 92). One carelessly included a higher spade spot for 93.

Before showing how to defeat it, let’s see how declarer can succeed with a vice squeeze. Suppose West chooses an innocuous trump lead. (Oh! Forget everything you learned in January about the merits of a trump lead, which only works in odd-numbered months.)

 6 South K 2 K Q J 10 K J 6 A K 4 3 Trick1 W2 N3 N4 N5 S6 S7 N8 N9 N10 N11 S Lead 2? Q J 10 9 2 A K K 2 A 2ndK7 6 6 399 8910? 3rd658A 3J287J 4th34 7 8 735J45 W-LW1W2W3W4W5W6W7W8W9W10 5 4 3 4 2 Q 10 9 8 7 Q J 5 Q 10 9 8 6 7 6 A 5 3 9 7 6 Lead: 2 A J 7 A 9 8 5 3 4 2 10 8 2

Nicholas Greer: On a heart lead declarer runs five hearts discarding a club from dummy, then plays a diamond to the jack. East’s best defense is to throw a spade and two clubs, then duck the diamond [else West is easily squeezed in the minors]. Now top clubs force East down to three spades and two diamonds [else declarer could duck a diamond to his blank ace]. Then spades are cashed [with a finesse] and West must either discard a winning club or blank the Q [allowing a diamond trick to be established]. On a club or diamond lead, the play is similar. An initial spade lead, however, tangles the entries…

Dan Gheorghiu: On any lead but a spade, declarer leads all the trumps (club pitch from dummy) and finesses the J. If East wins the A, three rounds of spades squeeze West in the minors. If East ducks, declarer cashes A-K to strip East of his fourth spade; then three rounds of spades force West to give up a diamond, so North’s K-6 wins a trick against East’s A-5. An original spade lead cuts the communication, and any squeeze fails.

Tom Slater: Ordinary play will see West squeezed in the minors after East takes the A. Better is to duck, but on most leads West is still caught in a vice squeeze, with East unable to retain a winning spade without baring the A. Only a spade lead defeats VI

Indeed it does, so let’s have a closer look:

 6 South K 2 K Q J 10 K J 6 A K 4 3 Trick1 W2 N3 N4 N5 N6 S7 S8 N Lead 3! K Q J 10 9 2 A 2ndK67 8 6 499 3rd6358A 3J2 4th724 7 8 735 W-LW1W2W3W4W5W6W7W8 5 4 3 4 2 Q 10 9 8 7 Q J 5 Q 10 9 8 6 7 6 A 5 3 9 7 6 Lead: 3 A J 7 A 9 8 5 3 4 2 10 8 2

Leif-Erik Stabell: Somewhat mysteriously, a spade lead messes up declarer’s communication and defeats the slam when East ducks the first round of diamonds.

This leaves the following ending:

 win 4Failure 2 — K 6 K 4 Trick9 N Lead K 2nd 5! 3rd8 4thJ W-LW1 5 — Q 10 Q J Q 10 9 — A 5 — North leads A J — 4 10 8

When the K is led, East can blank the A since dummy has no further entry. Running spades then squeezes West out of a diamond, alas, to no avail since East has a good spade to cash.

Or as Valentinus might describe it: Descendit unum est pons bonum.

Jamie Pearson: An original spade lead scuttles the throw-in, allowing East a safe exit or a second trick when he wins the A. How appropriate for February that hearts are used only to squeeze, and St. Valentine found the way to avoid it.

## Lost Marbles?

Dan Gheorghiu: Breaking news from Rome! Further probing of the marble slab with a quantum scanner revealed additional etchings under Valentinus’s signature: Iis qui inviso hoc perplexor erit deditus pons ad vitam, or “Those who look at this puzzle will be bridge addicts for life.” Prophecy or curse?

Tom Slater: All good archaeologists must consider the evidence. It’s a well-known fact that squeeze play wasn’t invented until well after the fall of the Roman Empire, so the solution must be very simple. What’s not so well-known is that the convention of treating the ace as the top card post-dates this marble slab, so the only conclusion is that the diagram was wrongly converted…

Tina Denlee: A marble slab with the West hand was finally discovered by students from the Bridge University of Tribliana on February 30, 3210, solving a millenary puzzle…

 Puzzle 8M61   Main Top   St. Valentine’s Hand