Join Bridge Winners
Computer Bridge: Realizing that a Layout Is Impossible
(Page of 19)

I. Introduction

Computer-bridgeplayers seem to have trouble discerning when an inference about a concealed hand's holding has been rendered impossible by the cards played. This could be addressed in part by setting upper and lower bounds for each concealed hand's "lawful" suit lengths (where "lawful" means, "assuming that no concealed hand has revoked") and adjusting those bounds as the play unfolds.

To simplify the discussion, the computer-bridgeplayer in question is assumed to be declaring.

Added after publication. I envision a scheme that would enable a computer-bridgeplayer to assemble enough of the single-dummy information available to a given concealed hand, to permit the computer-bridgeplayer to then draw inferences as to why that concealed hand made or did not make a particular call or play.

This article presents one small element of this scheme. When I feel that I have figured out how to articulate my mental picture of the scheme, I will write an article that attempts to do that and explains how the element presented in the present article fits in. (Shouldn't I have written the next article before this one? Quite possibly.)

II. A Five-card Ending

Consider a deal where the defenders were dealt 16 major-suit cards, these have been played to the first eight tricks, the defenders were also dealt and still hold five diamonds and five clubs, and it is Declarer's lead at Trick 9.

Being that neither defender has yet played a minor-suit card, we have no information that would constrain the minors' lawful layouts. This means that either defender could hold from zero to five diamonds and from zero to five clubs. This can be expressed as

LHO

  • : ?????
  • : ?????

RHO

  • : ?????
  • : ?????,

where the question marks indicate potential length in the suit.

The number of diamonds dealt to LHO will be denoted by number_of_diamonds_dealt_LHO, and the number dealt to RHO will be denoted by number_of_diamonds_dealt_RHO. The club suit will have an analogous pair: number_of_clubs_dealt_LHO and number_of_clubs_dealt_RHO. (The majors will also have upper and lower bounds. On this deal, each defender's upper and lower bounds for a given major will be equal because the play would have revealed the distribution of both majors.)

The upper and lower bounds for LHO's diamonds can be expressed as

  • number_of_diamonds_dealt_LHO <= 5
  • number_of_diamonds_dealt_LHO >= 0

or the more compact

  • 0 <= number_of_diamonds_dealt_LHO <= 5.

The analogous upper and lower bounds for the other combinations of suit and defender are

  • 0 <= number_of_diamonds_dealt_RHO <= 5
  • 0 <= number_of_clubs_dealt_LHO <= 5
  • 0 <= number_of_clubs_dealt_RHO <= 5.

III. A Five-card Ending after the First Defender's Turn

Suppose that Declarer leads a club and LHO follows. This brings to one, the number of clubs played by LHO. Because this exceeds the present lower bound of the corresponding variable, the latter needs to be adjusted:

LHO

  • : ????? -> X????,

where "->" indicates that the character string at its left is changed to the one at its right, "X" is a placeholder for length that the defender must hold, and the bold font indicates what has changed.

This can also be expressed as

  • 1 <= number_of_clubs_dealt_LHO <= 5.

(Whenever iconic and arithmetic representations appear together, they are equivalent.)

This leaves four clubs outstanding. Because this is lower than the present upper bound of the number of clubs dealt to RHO, the latter needs to be adjusted:

RHO

  • : ????? -> ????[?],

where the square brackets indicate that one of the question marks dropped out for lack of a chair.

This can also be expressed as

  • 0 <= number_of_clubs_dealt_RHO <= 4.

LHO's holding at least one club leaves one fewer slot for LHO to hold a diamond:

LHO

  • : ????? -> ????[?]
  • 0 <= number_of_diamonds_dealt_LHO <= 4.

RHO's holding at most four clubs means that at least one slot in RHO's hand must be occupied by a non-club. Being that RHO's lengths in the majors were already precisely determined, that non-club must be a diamond:

RHO

  • : ????? -> X????
  • 1 <= number_of_diamonds_dealt_RHO <= 5.

IV. The Five-card Ending after the Second Defender's Turn

From the previous page, the first defender's turn at Trick 9 yielded

LHO

  • : ????
  • : X????

RHO

  • : X????
  • : ????
  • 0 <= number_of_diamonds_dealt_LHO <= 4
  • 1 <= number_of_diamonds_dealt_RHO <= 5
  • 1 <= number_of_clubs_dealt_LHO <= 5
  • 0 <= number_of_clubs_dealt_RHO <= 4.

Suppose that after Dummy plays, the second defender also follows by playing a club. This brings to one, the number of clubs played by RHO. Because this exceeds the present lower bound of the corresponding variable, the latter needs to be adjusted:

RHO

  • : ???? -> X???
  • 1 <= number_of_clubs_dealt_RHO <= 4.

This leaves three clubs outstanding. If all of them were held by LHO, this would mean that LHO started with four clubs. However, because this is lower than the present upper bound of the number of clubs dealt to LHO, the latter needs to be adjusted:

LHO

  • : X???? -> X???[?]
  • 1 <= number_of_clubs_dealt_LHO <= 4.

RHO's holding at least one club leaves one fewer slot for RHO to hold a diamond:

RHO

  • : X???? -> X???[?]
  • 1 <= number_of_diamonds_dealt_RHO <= 4.

LHO's holding at most four clubs means that at least one slot in LHO's hand must be occupied by a non-club, and that non-club must be a diamond:

LHO

  • : ???? -> X???
  • 1 <= number_of_diamonds_dealt_LHO <= 4.

V. The Four-card Ending after the First Defender's Turn

From the previous page, the play at Trick 9 yielded

LHO

  • : X???
  • : X???

RHO

  • : X???
  • : X???
  • 1 <= number_of_diamonds_dealt_LHO <= 4
  • 1 <= number_of_diamonds_dealt_RHO <= 4
  • 1 <= number_of_clubs_dealt_LHO <= 4
  • 1 <= number_of_clubs_dealt_RHO <= 4.

Suppose that Declarer won Trick 9 and leads a diamond at Trick 10 and that LHO follows. This brings to one, the number of diamonds played by LHO. Because this does not exceed the present lower bound of the corresponding variable, the latter is left intact:

LHO

  • : X??? -> X???
  • 1 <= number_of_diamonds_dealt_LHO <= 4.

This leaves four diamonds outstanding. Because this is not lower than the present upper bound of the number of diamonds dealt to RHO, the latter is left intact:

RHO

  • : X??? -> X???
  • 1 <= number_of_diamonds_dealt_RHO <= 4.

Because the lower and upper bounds on LHO's slots for diamonds haven't changed, those for LHO's clubs don't need to change either:

LHO

  • : X??? -> X???
  • 1 <= number_of_clubs_dealt_LHO <= 4.

Similarly, the lower and upper bounds on RHO's slots for clubs can remain intact:

RHO

  • : X??? -> X???
  • 1 <= number_of_clubs_dealt_RHO <= 4.

VI. The Four-card Ending after the Second Defender's Turn

From the previous page, the first defender's turn at Trick 10 yielded

LHO

  • : X???
  • : X???

RHO

  • : X???
  • : X???
  • 1 <= number_of_diamonds_dealt_LHO <= 4
  • 1 <= number_of_diamonds_dealt_RHO <= 4
  • 1 <= number_of_clubs_dealt_LHO <= 4
  • 1 <= number_of_clubs_dealt_RHO <= 4.

Suppose that after Dummy plays, the second defender also follows by playing a diamond. This brings to one, the number of diamonds played by RHO. Because this does not exceed the present lower bound of the corresponding variable, the latter is left intact:

RHO

  • : X??? -> X???
  • 1 <= number_of_diamonds_dealt_RHO <= 4.

This leaves three diamonds outstanding. If West held all three, this would imply that West was dealt four diamonds. Because this is not less than the present upper bound of the corresponding variable, the latter is left intact:

LHO

  • : X??? -> X???
  • 1 <= number_of_diamonds_dealt_LHO <= 4.

Because the lower and upper bounds on RHO's slots for diamonds haven't changed, those for RHO's clubs don't need to change either:

RHO

  • : X??? -> X???
  • 1 <= number_of_clubs_dealt_RHO <= 4.

Because the lower and upper bounds on LHO's slots for diamonds haven't changed, those for LHO's clubs don't need to change either:

LHO

  • : X??? -> X???
  • 1 <= number_of_clubs_dealt_LHO <= 4.

VII. The Three-card Ending after the First Defender's Turn

From the previous page, the play at Trick 10 yielded

LHO

  • : X???
  • : X???

RHO

  • : X???
  • : X???
  • 1 <= number_of_diamonds_dealt_LHO <= 4
  • 1 <= number_of_diamonds_dealt_RHO <= 4
  • 1 <= number_of_clubs_dealt_LHO <= 4
  • 1 <= number_of_clubs_dealt_RHO <= 4.

At Trick 11, Declarer (who let's say won the previous trick) leads a club and LHO shows out, discarding a diamond. This means that the number of clubs dealt to LHO is determined to be the present lower bound of the corresponding variable. Therefore, the upper bound must be decreased to equal the present lower bound:

LHO

  • : X??? -> X[?][?][?]
  • 1 <= number_of_clubs_dealt_LHO <= 1.

This leaves three clubs outstanding, which must be held by RHO. To reflect this, the lower bound on the number of clubs held by RHO must be increased to equal the present upper bound:

RHO

  • : X??? -> XXXX
  • 4 <= number_of_clubs_dealt_RHO <= 4.

Because LHO remains with only non-clubs and the only unplayed non-clubs are diamonds, the lower bound on the number of diamonds held by LHO must be increased to equal the present upper bound:

LHO

  • : X??? -> XXXX
  • 4 <= number_of_diamonds_dealt_LHO <= 4.

To balance this, the upper bound on the number of diamonds held by RHO must be decreased to equal the present lower bound.

RHO

  • : X??? -> X[?][?][?]
  • 1 <= number_of_diamonds_dealt_RHO <= 1.

Because each defender's suit lengths are now known exactly, there can be no further adjustment to the lower and upper bounds of the corresponding variables.

VIII. Constraining the Defenders' Suit Lengths Based on the Declaring Side's Holdings

The constraints on the defenders' suit lengths would be first set when our computer-bridgeplayer views its hand. Suppose it holds

A A432 A432 A432.

Its holding a singleton spade means that each defender's spade length would range from 0 to 12. For each other suit, each defender's length would range from 0 to 9. This translates to

  • 0 <= number_of_spades_dealt_LHO <= 12
  • 0 <= number_of_spades_dealt_RHO <= 12
  • 0 <= number_of_hearts_dealt_LHO <= 9
  • 0 <= number_of_hearts_dealt_RHO <= 9
  • 0 <= number_of_diamonds_dealt_LHO <= 9
  • 0 <= number_of_diamonds_dealt_RHO <= 9
  • 0 <= number_of_clubs_dealt_LHO <= 9
  • 0 <= number_of_clubs_dealt_RHO <= 9.

(Analogous length constraints would be set also for Partner.)

The next adjustment of the constraints on the defenders' suit lengths would occur when Dummy was tabled. (We're still assuming that our computer-bridgeplayer becomes Declarer.) If Dummy were

2 KJT KJT9 KJT98,

then combining its suit lengths with those of the closed hand

A A432 A432 A432

would constrain the defenders' suit lengths as follows:

  • 0 <= number_of_spades_dealt_LHO <= 11
  • 0 <= number_of_spades_dealt_RHO <= 11
  • 0 <= number_of_hearts_dealt_LHO <= 6
  • 0 <= number_of_hearts_dealt_RHO <= 6
  • 0 <= number_of_diamonds_dealt_LHO <= 5
  • 0 <= number_of_diamonds_dealt_RHO <= 5
  • 0 <= number_of_clubs_dealt_LHO <= 4
  • 0 <= number_of_clubs_dealt_RHO <= 4.

(Adjusting suit-length constraints based on Dummy before adjusting suit-length constraints based on the opening lead seems more straightforward than making these adjustments in chronological sequence.)

IX. Using the Opening Lead to Constrain the Defenders' Suit Lengths

From the previous page, the declaring side's holdings yielded the suit-length constraints

LHO

  • : ???????????
  • : ??????
  • : ?????
  • : ????

RHO

  • : ???????????
  • : ??????
  • : ?????
  • : ????
  • 0 <= number_of_spades_dealt_LHO <= 11
  • 0 <= number_of_spades_dealt_RHO <= 11
  • 0 <= number_of_hearts_dealt_LHO <= 6
  • 0 <= number_of_hearts_dealt_RHO <= 6
  • 0 <= number_of_diamonds_dealt_LHO <= 5
  • 0 <= number_of_diamonds_dealt_RHO <= 5
  • 0 <= number_of_clubs_dealt_LHO <= 4
  • 0 <= number_of_clubs_dealt_RHO <= 4.

Suppose the opening lead is the K. This brings to one the number of spades played by LHO. Because this exceeds the present lower bound of the corresponding variable, the latter needs to be adjusted:

LHO

  • : ??????????? -> X??????????
  • 1 <= number_of_spades_dealt_LHO <= 11,

This leaves 10 spades outstanding. If all of them were held by RHO, this would mean that RHO started with 10 spades. However, because this is lower than the present upper bound of the number of spades dealt to RHO, the latter needs to be adjusted:

RHO

  • : ??????????? -> ??????????[?]
  • 0 <= number_of_spades_dealt_RHO <= 10.

The constraints on the defenders' dealt lengths in the other three suits are unaffected. A rigorous way to determine this is provided on the next page.

X. Using RHO's Card at Trick 1 to Constrain the Defenders' Suit Lengths

From the previous page, the opening lead yielded

LHO

  • : X??????????
  • : ??????
  • : ?????
  • : ????

RHO

  • : ??????????
  • : ??????
  • : ?????
  • : ????
  • 1 <= number_of_spades_dealt_LHO <= 11
  • 0 <= number_of_spades_dealt_RHO <= 10
  • 0 <= number_of_hearts_dealt_LHO <= 6
  • 0 <= number_of_hearts_dealt_RHO <= 6
  • 0 <= number_of_diamonds_dealt_LHO <= 5
  • 0 <= number_of_diamonds_dealt_RHO <= 5
  • 0 <= number_of_clubs_dealt_LHO <= 4
  • 0 <= number_of_clubs_dealt_RHO <= 4.

Suppose that RHO shows out on the spade lead at Trick 1, playing the 5. This means that the number of spades dealt to RHO is known to equal the present lower bound of the corresponding variable. Therefore, its upper bound is set to its present lower bound.

RHO

  • : ?????????? -> [?][?][?][?][?][?][?][?][?][?]
  • 0 <= number_of_spades_dealt_RHO <= 0.

This leaves 10 spades outstanding, which must be held by LHO. To reflect this, the lower bound on the number of spades held by LHO must be raised to equal its present upper bound:

LHO

  • : X?????????? -> XXXXXXXXXXX
  • 11 <= number_of_spades_dealt_LHO <= 11.

Because the upper bound on RHO's spade length has decreased, the lower bounds for RHO's remaining non-spades could be affected. This is determined by considering for each non-spade suit, whether the other suits could combine to give RHO 13 cards.

If RHO held no hearts, the other suits could provide at most zero spades, five diamonds, and four clubs, which sums to only nine. Therefore, RHO must hold at least four hearts:

RHO

  • : ?????? -> XXXX??
  • 4 <= number_of_hearts_dealt_RHO <= 6.

If RHO instead held no diamonds, the other suits could provide at most zero spades, six hearts, and four clubs, which sums to only 10. Therefore, RHO must hold at least three diamonds:

RHO

  • : ????? -> XXX??
  • 3 <= number_of_diamonds_dealt_RHO <= 5.

If RHO instead held no clubs, the other suits could provide at most zero spades, six hearts, and five diamonds, which sums to only 11. Therefore, RHO must hold at least two clubs:

RHO

  • : ???? ->XX??
  • 2 <= number_of_clubs_dealt_RHO <= 4.

The above has produced the following length constraints for RHO:

  • :
  • : XXXX??
  • : XXX??
  • : XX??
  • 0 <= number_of_spades_dealt_RHO <= 0
  • 4 <= number_of_hearts_dealt_RHO <= 6
  • 3 <= number_of_diamonds_dealt_RHO <= 5
  • 2 <= number_of_clubs_dealt_RHO <= 4.

Now the upper bounds on LHO's non-spade lengths need to be decreased to be consistent with the new lower bounds on RHO's lengths in those suits (that is, in a given non-spade suit, LHO's maximum length is the number of cards available in that suit after deducting RHO's minimum length from the defenders' supply in that suit):

  • : XXXXXXXXXXX
  • : ?????? -> ??[?][?][?][?]
  • : ????? -> ??[?][?][?]
  • : ???? -> ??[?][?]
  • 11 <= number_of_spades_dealt_LHO <= 11
  • 0 <= number_of_hearts_dealt_LHO <= 2
  • 0 <= number_of_diamonds_dealt_LHO <= 2
  • 0 <= number_of_clubs_dealt_LHO <= 2.

XI. Suit-length Inferences from the Concealed Hands' Choices

Suppose that our computer-bridgeplayer is dealt 2=3=3=5, thereby constraining the concealed hands' suit lengths as follows:

LHO

  • : ???????????
  • : ??????????
  • : ??????????
  • : ????????
  • 0 <= number_of_spades_dealt_LHO <= 11
  • 0 <= number_of_hearts_dealt_LHO <= 10
  • 0 <= number_of_diamonds_dealt_LHO <= 10
  • 0 <= number_of_clubs_dealt_LHO <= 8,

and likewise for the other two concealed hands.

Now suppose that LHO dealt and begins the auction with a call that shows at least five hearts and says nothing about length in the other suits. The constraints inferred from the concealed hands' choices (such as this one) could be expressed similarly to the lawful constraints. To avoid confusion between these two types of constraints, the lawful constraints will be designated as "Lawful" and the inferences from the concealed hands' choices will be designated as "Agreement-consistent", "Agreement-inconsistent", or "Agreement-unspecified" (which are illustrated below).

Once an inference from the concealed hands' choices has been definitively determined as true or false by the cards played, it will be also designated as "Confirmed" or "Impossible," respectively.

The concealed hands' length constraints that are based solely on the computer-bridgeplayer's hand will be qualified accordingly as "Lawful":

  • 0 <= number_of_spades_dealt_LHO <= 11, Lawful
  • 0 <= number_of_hearts_dealt_LHO <= 10, Lawful
  • 0 <= number_of_diamonds_dealt_LHO <= 10, Lawful
  • 0 <= number_of_clubs_dealt_LHO <= 8, Lawful,

and likewise for the other two concealed hands.

Where inferences from the concealed hands' choices don't conflict with any other inferences or lawful constraints, then the minimum-length constraint on LHO's hearts inferred from LHO's call is

  • 5 <= number_of_hearts_dealt_LHO, Agreement-consistent.

This would leave at most five hearts for RHO:

  • number_of_hearts_dealt_RHO <= 5, Agreement-consistent,

and at most five hearts for Partner.

If LHO holds at least five hearts, then LHO's hand could accommodate at most eight non-hearts:

  • 0 <= number_of_spades_dealt_LHO <= 8, Agreement-consistent
  • 0 <= number_of_diamonds_dealt_LHO <= 8, Agreement-consistent
  • 0 <= number_of_clubs_dealt_LHO <= 8, Agreement-consistent.

The agreement-consistent non-heart length constraints for the other two concealed hands would be unaffected. This would be determined by a procedure analogous to that of the previous section.

XII. Testing Suit-length Inferences during the Play

Each inference about a concealed hand's suit length will be reexamined whenever the suit's lawful constraints have been tightened or an additional inference is taken from a concealed hand's action.

In the previous example, the computer-bridgeplayer held three hearts, yielding the following lawful length constraint on LHO:

  • 0 <= number_of_hearts_dealt_LHO <= 10, Lawful,

and LHO had shown at least five hearts, yielding the following length-constraint on LHO:

  • 5 <= number_of_hearts_dealt_LHO, Agreement-consistent.

Suppose that Dummy (Partner) tables six hearts, which would affect the upper bound on the number of hearts LHO could hold:

  • 0 <= number_of_hearts_dealt_LHO <= 4, Lawful.

Among the inferences made about the concealed hands is that LHO holds at least five hearts, but that's now disproved:

  • 5 <= number_of_hearts_dealt_LHO, Agreement-consistent, Impossible.

Rather than abandon this inference entirely, we could replace it with the inference that LHO holds at least four hearts, which is only one fewer than the number shown by LHO's initial call during the bidding.

  • [5 <= number_of_hearts_dealt_LHO, Agreement-consistent, Impossible.]
  • 4 <= number_of_hearts_dealt_LHO, Agreement-consistent.

This possibility could have been anticipated when formalizing the inference about LHO's heart length based on LHO's initial call.

  • 5 <= number_of_hearts_dealt_LHO, Agreement-consistent
  • [4 <= number_of_hearts_dealt_LHO, Agreement-consistent],

where the brackets designate the inference as inactive unless the inference immediately above it were to be disproved. (A symbol is needed to show these two inferences as connected, I just haven't decided on one.)

That way, if the cards played were to show that LHO could not have been dealt five or more hearts but could have been dealt precisely four, then the first of the above expressions would be rejected but the second would become active and be valid.

  • [5 <= number_of_hearts_dealt_LHO, Agreement-consistent, Impossible]
  • 4 <= number_of_hearts_dealt_LHO, Agreement-consistent.

For selected minimum-length constraints derived from concealed hands' actions, our computer-bridgeplayer could set also a constraint for that length minus one, to allow for the often rational choice to treat a suit as one card longer. This might be done also for selected maximum-length constraints, to allow for a concealed hand's having treated a long suit as having one fewer card.

XIII. Inferences on the Lengths of More than One Suit

Returning to LHO's starting the auction by showing a heart suit of at least five cards, inferences can be expressed about LHO's other suit lengths relative to that of hearts, assuming that the relevant information on the opponents' system is available.

  • number_of_hearts_dealt_LHO > number_of_spades_dealt_LHO, Agreement-consistent
  • number_of_hearts_dealt_LHO >= number_of_diamonds_dealt_LHO, Agreement-consistent
  • number_of_hearts_dealt_LHO >= number_of_clubs_dealt_LHO, Agreement-consistent.

The first of these would be retested whenever a defender's played card resulted in adjustment of any of the constraints on the defenders' lengths in hearts or spades. The other two would be treated analogously for their suits of interest.

XIV. Inferences on Suit-length Relative to a Card Played

Suppose that the opening lead is the 5 and the defenders' spades are the K, J, 9, 7, 5, 4, 3, and 2. The lead could lawfully be any of the following (if not a singleton):

  • Highest of two, three, or four cards
  • Second-highest of two, three, four, or five cards
  • Third-highest of three, four, five, or six cards
  • Fourth-highest of four, five, six, or seven cards
  • Fifth-highest of five, six, seven, or eight cards.

Against notrump contracts, these opponents have agreed to lead fourth best, top of three spots, top of a doubleton, and top of a sequence of three or more honors. In order to more conveniently represent any instances where not all leads within a given bullet item would warrant the same "Agreement-" qualifier, the above list will be recast so that each bullet depicts only one possibility:

  • Higher of two cards
  • Highest of three cards
  • Highest of four cards
  • Lower of two cards
  • Second-highest of three cards
  • Second-highest of four cards
  • Second-highest of five cards
  • Third-highest of three cards
  • Third-highest of four cards
  • Third-highest of five cards
  • Third-highest of six cards
  • Fourth-highest of four cards
  • Fourth-highest of five cards
  • Fourth-highest of six cards
  • Fourth-highest of seven cards
  • Fifth-highest of five cards
  • Fifth-highest of six cards
  • Fifth-highest of seven cards
  • Fifth-highest of eight cards.

Next, because the defenders' agreement covers three spots but not three cards headed by an honor, these need to be distinguished:

  • Higher of two cards
  • Highest of three spots
  • Highest of three cards headed by an honor
  • Highest of four cards
  • Lower of two cards
  • Second-highest of three spots
  • Second-highest of three cards headed by an honor
  • Second-highest of four cards
  • Second-highest of five cards
  • Third-highest of three spots
  • Third-highest of three cards headed by an honor
  • Third-highest of four cards
  • Third-highest of five cards
  • Third-highest of six cards
  • Fourth-highest of four cards
  • Fourth-highest of five cards
  • Fourth-highest of six cards
  • Fourth-highest of seven cards
  • Fifth-highest of five cards
  • Fifth-highest of six cards
  • Fifth-highest of seven cards
  • Fifth-highest of eight cards.

These can now be placed under the three "Agreement" groupings.

Agreement-consistent:

  • Higher of two cards
  • Highest of three spots
  • Fourth-highest of four cards
  • Fourth-highest of five cards
  • Fourth-highest of six cards
  • Fourth-highest of seven cards.

Agreement-unspecified:

  • Highest of three cards headed by an honor
  • Second-highest of three cards headed by an honor
  • Third-highest of three cards headed by an honor.

Agreement-inconsistent:

  • Highest of four cards
  • Lower of two cards
  • Second-highest of three spots
  • Second-highest of four cards
  • Second-highest of five cards
  • Third-highest of three spots
  • Third-highest of four cards
  • Third-highest of five cards
  • Third-highest of six cards
  • Fifth-highest of five cards
  • Fifth-highest of six cards
  • Fifth-highest of seven cards
  • Fifth-highest of eight cards.

The agreement-consistent and agreement-unspecified lead+holding combinations will now be formed into an expression using the exclusive-or (XOR) operator (which by using in this fashion, I might be taking a liberty):

  • Higher of two cards, Agreement-consistent XOR Highest of three spots, Agreement-consistent XOR Highest of three cards headed by an honor, Agreement-unspecified XOR Second-highest of three cards headed by an honor, Agreement-unspecified XOR Third-highest of three cards headed by an honor, Agreement-unspecified XOR Fourth-highest of four cards, Agreement-consistent XOR Fourth-highest of five cards, Agreement-consistent XOR Fourth-highest of six cards, Agreement-consistent XOR Fourth-highest of seven cards, Agreement-consistent.

XV. Updating Inferences on Suit-length Relative to a Card Played

From the previous page, the spades dealt to the opponents were the K, J, 9, 7, 5, 4, 3, and 2, the 5 was led, and the lead interpretations not inconsistent with the defenders' agreement were expressed as

  • Higher of two cards, Agreement-consistent XOR Highest of three spots, Agreement-consistent XOR Highest of three cards headed by an honor, Agreement-unspecified XOR Second-highest of three cards headed by an honor, Agreement-unspecified XOR Third-highest of three cards headed by an honor, Agreement-unspecified XOR Fourth-highest of four cards, Agreement-consistent XOR Fourth-highest of five cards, Agreement-consistent XOR Fourth-highest of six cards, Agreement-consistent XOR Fourth-highest of seven cards, Agreement-consistent.

Supposing that RHO follows with the J at Trick 1, this would leave available the K, 9, 7, 4, 3, and 2 for LHO to possibly hold and would not invalidate any of the above lead interpretations.

On the next trick in which at least one defender plays a spade, RHO plays the 3, which leaves available the K, 9, 7, 4, and 2 for LHO to possibly hold and therefore makes impossible the lead interpretation

  • Fourth-highest of seven cards, Agreement-consistent, Impossible.

The next spade played by the defenders is the K by LHO, which invalidates the lead interpretations

  • Higher of two cards, Agreement-consistent, Impossible
  • Highest of three spots, Agreement-consistent, Impossible
  • Highest of three cards headed by an honor, Agreement-unspecified, Impossible.

The interpretations of the opening lead would be reexamined whenever further information became available on the defenders' lawful spade holdings, until either an interpretation was found Confirmed or every interpretation was found Impossible (the latter of which would indicate that the lead was Agreement-inconsistent).

XVI. Constraints on Number of HCP Dealt to Each Defender

If our computer-bridgeplayer were dealt 14 HCP, the HCP constraints on the concealed hands could be depicted as follows:

  • 0 <= number_of_HCP_dealt_LHO <= 26,

and likewise for the other two concealed hands.

These lawful HCP constraints would be reexamined when Dummy comes down and whenever a concealed hand plays a court card (A, K, Q, or J). Adjusting any lawful HCP constraint would trigger the reexamination of any action-based inferences about the concealed hands' HCP constraints. The process is similar to that for suit lengths and is not presented here.

XVII. Action-based Inferences about Specific Cards

Suppose that during the play, the club and heart aces are outstanding and the defenders' actions suggest that each concealed hand holds one of them, but there has been no indication as to which defender is more likely to hold a given ace. The assertion that LHO holds the club ace could be depicted as

  • A_dealt_LHO,

and likewise for the other cards and the other concealed hand. Unlike the terms for denoting suit length and HCP, which translated to numbers, the terms for whether a given player holds a particular card translate to "true" or "false".

To depict that two or more inferences would be jointly true, the keyword "AND" will be used, as in

  • (A_dealt_LHO) AND (A_dealt_RHO).

The inference that one ace was dealt to one defender and the other ace to the other defender could then be expressed (assuming no information that suggests that one allocation of aces is the more likely) as

  • ((A_dealt_LHO) AND (A_dealt_RHO)) XOR ((A_dealt_RHO) AND (A_dealt_LHO)).

XVIII. When Action-based Inferences Conflict

If some conflicting inferences were based on Partner's actions and some were based on actions by one or both opponents, the computer-bridgeplayer would regard as valid only those based on Partner's actions.

If the conflicting inferences were all by the opponents, then they would be treated as mutually exclusive by being joined in an expression using XOR.

XIX. Some of the Areas Needing Further Definition

Areas needing further definition include the following:

  • How to handle conflicting inferences from Partner's actions
  • Constructing inferences based on formulas, such as an opening bid criterion based on the sum of HCP and the lengths of the two longest suits
  • How to assign weight to inferences from the concealed hands' actions
  • How to apply inferences that are not yet Confirmed or Impossible.
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