Post-Heptagons from the 5th Dimension
(Page of 3)

This is the final part of a 3-part series begun a while back (part 1, part 2).

Part 2 ended like this:

So, in 3-suited bridge, double squeezes (as squeezes are traditionally presented), are somewhat more complicated than in 4-suited bridge.  And, some of them could really handle 5 outstanding stoppers instead of the usual 4 present in a double squeeze, except that just can't arise in that game.  Viewing these endings as post-compounds from the 4th dimension is actually simplifying, and highlights a useful technique of preserving multiple superimposed squeezes as long as possible.

Obviously where I'm going next is to explore what 5-suited bridge can tell us about our game.  As I'm sure you've guessed, the next article is about post-heptagons from the 5th dimension.

So what can we learn from the 5th dimension?  Consider this 5-suited ending:

Position 1, with 5th suit missing

West
A
QJ10
Q
North
AK2
3
East
K
976
9
QJ
South
Q
8
A9
D

The diagram doesn't support a 5th suit.  Call it Bolts, lower ranking than clubs, and true all the hands up to 7 cards with BAKQ in the north hand and small B's in the West and South hands (East already has 7 cards).

East stops 4 suits, West stops 3, and none of the doubly-stopped menaces have entries.  And yet, the power of having 4 menaces is immense.  Run the unshown 5th suit to get everyone down to 4 cards (this reaches Position 3 a bit below).  Most likely East will shake 3 hearts (not that it matters), then eventually West must unguard spades or diamonds.  Now A, K will effect a simple squeeze against East.

In 4 dimensions, heptagon squeezes don't exist.  But, it's not because we can't overcome 3 doubly-stopped suits and a basic menace, but merely because we can't fit 7 stoppers into the opponents' hands.  We do have saturated squeezes (to my knowledge, first described by Don Kersey in the August 2003 Bridge World.  He also describes them on his terrific website here), where one opponent stops 3 suits, and the other stops the 4th and one or two others.  These have some fairly complex requirements about extensive recessing and menaces accompanied by multiple winners.  For example, a 4-suited saturated squeeze that reaches the same 4 card ending as position 1 might look like:

Position 2

West
A
QJ10
QJ10
North
AK2
AK
32
East
K
976
9
QJ10
South
Q
832
AK9
D

You have menaces in 4 suits, with each opponent stopping 3, and no simpler squeeze available.

Cash the K and the A, K and you come down to the same 4 card ending:

Position 3

West
A
QJ10
Q
North
AK2
3
East
K
976
9
QJ
South
Q
8
A9
D

East & West are shown with lots of extra cards so that they still have 7 stoppers.  However they choose to discard down to 4 (including allowing revokes!), though, declarer can get home -- if no suit is good, the A, K will simple squeeze East.

Saturated squeezes seem like rare, fragile beasts, with dozens of variants.  While that is true, if we view them as post-heptagons from the 5th dimension, they become substantially simpler.  Focusing on the requirements for a saturated squeeze to be present misses the boat.  Instead, our focus should be maximizing the chance of getting that extra trick.   If you're lucky enough to have menaces in every suit and only one loser, don't look for a saturated squeeze, look for a post-heptagon from the 5th dimension.  Even if in practice it might just amount to a simple squeeze or even a 3-3 split.

A key advantage of this approach is that there are many fewer variants.  In position 2, a "solitary" saturated squeeze, there are a number of variants on the minor suit winners.  And possibly some variants that aren't true saturated squeezes because some simpler squeeze turns out to be present.  But, in all these cases the right play is to reach position 3 and then read the position.  As far as I can tell, all of the saturated squeezes boil down to 4 post-squeeze positions.  Position 3 is one, the others are on the next page...

Position 4

North
A2
A2
3
South
3
A2
A2

[South to lead A, or, symmetrically, North the A]

This is an amazing position.  4 different 7-suit squeezes are possible (well, much squeezing has already occurred).  All you need is one suit that is over its sole stopper (a basic menace), and any other (starting) arrangement of stoppers is survivable.

At the point shown, anyone with an original 3 (or even 4 in 5-suit bridge) stoppers has been squeezed down to 2.  If all 4 suits are still stopped, this means that each opponent has sole responsibility for 2 suits, and some simple squeeze is possible.  It's not possible for East to have the majors and West the minors since we stipulated there was some basic menace.  That leaves five possible cases.  For example, if West covers the blacks, you don't have the entries to squeeze him, but East will have the reds and you're all set there:  A A A.  Other layouts are similar.

According to Kersey's classification, there are 3 kinds of saturated squeezes with 2 menaces in each hand:  "Symmetric", "Asymmetric", and "Segregated", each with various permutations of entries and recessing.  All of them have the potential to reach an ending like this one (though Asymmetric more commonly reaches Position 5, and Segregated can reach Position 6).

If you were in the North hand, playing the A is entirely symmetrical.

Perhaps there was a simple squeeze, or even just a 3-3 split all along.  Certainly there were never really 7 stoppers.  This does not detract from the utility of the technique.

A 3rd useful post-heptagon:

Position 5

North
2
AK2
3
South
3
A2
A2

This position results from an asymmetric saturated squeeze where hearts is the basic menace.  Play the A and, unless you think it is good after seeing West's discard, discard the 2.   That leaves a 4 card ending where West might stop a black suit but can't stop both and can't stop diamonds.  Now A, K will squeeze East in diamonds and some black suit.  Even if you've misread the position and it turns out East stops clubs and hearts, the same line of play will effect a spade-diamond squeeze on West.

If we know West has the sole spade stopper but can't stop both minors, then if he follows to the A, now change gears:  pitch a diamond from dummy then play A.  One way or another, you'll get home.

Most saturated squeezes can be operated by coming down to one of these 3 endings, and in 5-suited bridge these all correspond to multiple post-septagon endings.  With reciprocal compounds (previous article) many were post-compound squeezes (or even 2 or 3 of those), but some were simply double squeezes entirely contained in 3 suits.  Similarly, there are some saturated squeezes contained in 4 suits that cannot survive a 7th stopper (not demonstrated here).

Finally, back to the 2-way sure-tricks problem linked to at the top...

If I've succeeded in my mission, this should be a bit easier now:

Position 7

North
K
A432
A432
J654
South
AQ43
KQ
KQ5
AK32

You declare 7N on a diamond lead (say you win Q).  When you play K, the stiff Q drops (I promise no Grosvenor).  You can now claim (CORRECTION: almost -- you can always make, but one variant requires reading the position).

Say RHO drops the Q.   First, let's focus on East:

West
North
K
A432
A43
J65
East
J1098
J1098
J109
South
AQ43
KQ
K5
A32
D

With a holding like this, East is already rectified for a squeeze.  With sole responsibility for both red suits, East must release a spade on the A.

Meanwhile, cashing K and red winners ending with the K will squeeze West in the blacks:

West
765
109
North
4
43
J5
East
South
AQ4
K
3
D

With sole responsibility for clubs, West also must unguard spades.  If both opponents release spades, the 4 is your 13th trick.

The main way that the layout might be different is that West had more red cards.  In that case, a club-red squeeze was set up from the beginning.  If it's club-diamond, you will have proved it and can play double dummy.  If East stops diamonds, but the heart guard is ambiguous, the ending might look like this:

West
7x
x
109
North
4
4
J5
East
J10x
x
J
South
AQ4
3
D

(One opponent has a small spade and the other a small heart, so strike a x from one and a x from the other)

Cash A pitching a worthless diamond, and then on Q either West follows and the long spade is good, or West is squeezed in clubs and hearts.

That was the relatively traditional way of analyzing this ending.  However, aficionados of 5-suit bridge will of course recognize this as an opportunity to "play it as a heptagon squeeze".

Cash a bunch of winners to reach this ending with South on lead:

North
4
2
J6
South
AQ4
3

This is equivalent to position 3 from before.  Since you'll know who stops diamonds, this is a fairly easy sure-tricks problem.  Even if there were ambiguity, the position would always be guessable -- West can't stop spades (because he has 2 clubs), and if East does, East can only stop one red suit -- play to squeeze West in clubs and the red suit that East doesn't stop by cashing AQ.

What happened here is you had several possible simpler squeezes that were all kept in play in this 4 card ending:  they are either still available (if needed), or already happened:

1. West in clubs & spades:  this already happened on the last red winner.

2. West in clubs & hearts:  this is set up to happen on the AQ

3. West in clubs & diamonds:  similar to #2

4. East in spades, hearts, and diamonds:  this already happened on the 2nd round of clubs.

5. If no one of those is sufficient, East relinquishing a suit in the triple squeeze means that one of the simples against West will suffice.

That is the technique I advocate for all variety of complex squeezes:  look for multiple simpler squeezes that you can simultaneously keep in play while tightening the position.  This is just the generalization of playing a simple as a double, or even just combining your chances.  Play a triple-single as a septagon.  Easy!

What about the other variant, where on K, LHO drops the Q.  Now cash all your non-Aces ending in dummy to reach:

North
A4
A2
6
South
A4
5
A3

This is equivalent to the amazing position 4.  Even though you haven't proved any suit except clubs, this position is also sure-tricks.  Assume RHO still has clubs stopped (or the play is easy).  Then, play A:

a)  RHO shows out:  pitch spade, A, A for a type-R double squeeze (or maybe 3-3 diamonds all along, or maybe 4 is good).  It's not really a double squeeze at this point as the opponents have already relinquished too many stoppers.  But playing it that way copes with the remaining ambiguity.

b)  RHO follows:  You either have a spade-diamond squeeze (pitch a club, then A effects it), or a minor suit squeeze (pitch a spade, A, A).  Unfortunately, you still need to read the position.   [This variant was CORRECTED; previously asserted it was sure tricks]

Similar to the previous position, you've kept several simpler squeezes in play.

As shown, this puzzle is a saturated triple-simple squeeze (2 different kinds, depending on who stops clubs).  But, notice that if the defenders had some extra red cards, they still would be squeezed out of them, and playing it as a triple-triple copes with the Mystery Menace in hearts.  The limitations of having only 13 cards prevented a defender from stopping 4 suits, or in this case even having 2 defenders stop the red suits, but there was no need to discover or guess who had the stoppers since with overwhelming pressure you could survive anything.