Bridge Sudoku

The starting point is to notice that:

• The cages record a full game, therefore, we can find all 52 cards in the cages (tricks).

• There are totally 4 "2"s and 4 "J"s in all cages.

• We can only have a "2" or a "J" in each Sudoku box.

• All 13 cages are in the 8 Sudoku boxes other than box 5.

• Therefore, for all boxes other than box 5, "2" or "J" must be in the cages.

• The same argument holds for ("3", "Q"), ("4", "K") and ("5", "A").

• We can try to mark all cells that could be 2345 (or JQKA) on the grid.

• On the other hand, except for box 5, all cells that are not in the cage can only be 1,6,7,8,9.

The result should be something like the picture, where red cells are cells with 2345 (or JQKA), and blue cells are cells with 6789 and 1 (or 10).

The next thing to notice is, for 1st ~ 12th tricks, we know who won each trick, because we know the leader of 2nd ~ 13th tricks.

• Let's mark these cells with dotted squares.

Since we know West + East won 11 out of 13 tricks in total, the 13th trick is either won by West or East.

Since there is no trump suits, the cells with dotted squares must be the same suit as the cells with dotted circles in each cage.

• Also, the cards in dotted squares must have a higher rank than the cards in dotted circles.

When we consider the ranking between dotted squares and dotted circles, we can mark some more squares as red or blue.

♦️10: it's the only card in 6789,10 that can beat 9.

9 in box 8: the 9 must be in cage 7, and if we put the 9 in r7c5, there is no card can be played by East.

We can use regular Sudoku tricks to put a 1 in box 1, and pencil mark some cells.

That there are totally 4 "10"s (or "1"s) in the cage. All digits (1~9) in box 4 are in cages. We can count the number of "10"s that are already in a cage, and determine whether others can be in cages as well in other boxes.

We already found "10"s in cages of box 3, box 4, 7 and 8. Therefore, "1"s in box 2 and 6 must be outside of cages.

The same logic works for "7"s, we have found them in cages of box 2, 4, 8 and 9. So all other "7"s must be outside of cages.

Combined with Sudoku rules, we can have something like the photo.

10 in box 4: We can't place 10 in r5c2, otherwise we will have 2 "♦️10" in the grid (the other one is in box 3).

By the rule of contract bridge, if a player has at least one card of the same suit as the leader, a card of that suit must be played.

Therefore, when we see North played a ♦️ in the 8th trick, we know the north player must have played a ♦️ in 2nd and 4th trick too.

In 2nd, 4th and 8th tricks there are already 7 cards of ♦️. And we know 4 of them are "blue", and exactly one of r4c2, r5c2 is blue as well.

• This is 5 blue cells in total, and all of them are ♦️.

A "blue cell that is ♦️" must be one of 6, 7, 8, 9, 10. There's exactly 5 choices. Therefore, we have found all "blue cell that is ♦️".

One of them must be ♦️7, and the only choice is r5c2.

The other two blue cells with ♦️ must be 6 or 8. (And therefore r1c7, r4c3 form a 68 pair)

Now we can mark all red and blue cells.

Also notice that we have found 7 out of 8 "red cell with ♠️" now.

• And r8c4 can only be ♠️J, because it cannot be ♠️Q nor ♠️A ("3"s and "5"s in 1st and 3rd tricks must be ♠️), and it cannot be ♠️K (by normal Sudoku).

• I use "20", "30", "40", "50" to represent "2 or J", "3 or Q", "4 or K", and "5 or A".

"9" in box 3 can only be in row 2.

Now we can place all 6,7,8,9,10 (or 1) in the grid.

• There is a "68 pair" across box 3 and box 4, they must have different digits.

• The 68 pairs in box 2 and box 7 are resolved by counting the number of "6"s and "8"s that are already in cages.

Since East played a ♠️ in 12th trick, East also had to play a ♠️ in 1st trick.

In cage 1, the only card that can beat ♠️K is ♠️A. This gives us some more digits.

The only cell that can be "♠️9" is r6c3, because,

• r5c7 is not a ♠️, otherwise there will be two ♠️J.

• r7c4 is not a ♠️, otherwise there will be two ♠️10. (North played a ♠️ later in the game, so North must have played a ♠️ if the 9 is ♠️9.)

On the other hand, the leading card of 8th trick cannot be ♠️8, nor ♦️8. Because North played a ♠️ later, and ♦️8 is already in the grid.

So we know that the leading card must be ♥️8 or ♣️8. And North and South didn't have any card in that suit since 8th trick.

South played a ♥️7 in 12th trick, therefore, the leading card of 8th trick must be ♣️8. (Otherwise South should have played a ♥️ in 8th trick)

So we know that North and South didn't have any card of ♣️ since 8th trick.

Furthermore, the ♥️2 in 2nd trick shows that the South didn't have a ♦️ since 2nd trick. Therefore, the South didn't have a ♦️ since 2nd trick, nor a ♣️ since 8th trick.

The 4 in r2c7 must be ♠️4, because South played ♠️9 in 8th trick. And we have found all red cells of ♠️.

Therefore, all red cells in 9th~13th trick that are played by South must all be ♥️!

The leading card of 6th trick is an 8, which can't be ♦️8 nor ♣️8.

It can't be a ♠️8 either, otherwise there will be two ♠️Q in the grid. (There is one in either 1st trick or 3rd trick).

Therefore, it must be ♥️8, and the Q must be ♥️Q.

The ♥️Q makes r8c7 a "♥️3". And r7c1 now must be a ♥️4. (It can neither be ♥️3 nor ♥️Q)

r9c6 can only be ♥️5 or ♥️A now. ♥️3, ♥️4, ♥️Q, ♥️K are already played.

We can also determine the suit of all "7"s and "8"s in the grid.

r7c5 is a ♣️6.

• It can't be ♦️6, ♥️6, because they are already in the grid.

• It can't be ♠️6, otherwise North had to play a ♠️10, which is also on the grid.

7th trick must be ♣️3, ♥️10, ♣️9, ♣️6.

• West played a ♣️K in 8th trick, so West had to play ♣️, and they didn't win the trick, so it is ♣️3 instead of ♣️Q.

• North didn't win the trick either, so they must have played ♥️10, instead of ♣️10.

r6c7 now must be ♠️6.

r6c2 now must be ♣️10.

r9c4 must be ♦️.

• It can't be ♠️3 nor ♠️Q. Because both of them are already placed in the grid.

• It can't be ♥️3 nor ♥️Q for the same reason.

• It can't be ♣️3 nor ♣️Q, because North player didn't have any ♣️ since 7th trick.

At least one of r9c3 and r9c5 is ♦️ to win 13th trick.

That is, there is either a ♦️J or ♦️K in 13th trick.

Now, the 25 pairs in 5th trick cannot be ♦️ anymore.

• If they are ♦️, r5c8 must be ♦️K. (because North played a ♦️ later in the game).

• We already have a ♦️2 and ♦️4 in 4th trick.

• We cannot place a ♦️J nor ♦️K in 13th trick.

Therefore, the 25 pairs in 5th trick must be ♣️.

• It can't be ♥️, otherwise South would have played ♣️7 in 5th trick, but a ♥️Q in 6th trick.

Since North played a ♦️ in 13th trick, the first card of 10th trick cannot be a ♦️.

• North played a ♠️8 in 10th trick.

• Both ♠️Q and ♥️Q are in the grid.

Therefore, 10th trick must be ♣️Q, ♠️8, ♣️5, ♥️K.

The 9 in 9th trick can only be ♥️9 now. And East played a ♥️J.

Therefore, East played a ♥️5 in 6th trick.

Now we have placed ♣️5 in the grid, one of r4c8 and r6c8 must be ♣️A. And the winner of the trick must have played ♣️A.

One of r4c1, r6c1 is a "3" or "Q", which can't be ♠️, ♥️, nor ♣️. So it must be ♦️3 or ♦️Q.

Therefore we know the leading card of 11th trick must be a ♦️. And North must follow.

r8c3 must be ♦️A, because all other "A"s are placed in the grid.

Therefore, in 11th trick, the only choice are: ♦️J, {♦️3 or ♦️Q}, ♦️5 for 2, 3, 5 pencil marks.

♦️5 can never win the trick, so it must be placed in r5c1. And the 23 pairs in column 1 is resolved by normal sudoku. So they must be ♦️J and ♦️Q for East to win the trick.

We have put all digits in the grid, and we can use contract bridge rules to determine the rest of cards.

r2c5, r2c6 must be ♠️3 and ♠️2 so West can win the trick.

r9c3 must be ♣️J, it is the only J not on the board.

r9c4 must be ♦️3, because ♦️Q had been played.

r9c5 must be ♦️K, because ♦️4 had been played. And East must have played a ♦️ here to so West and East can win 13th trick.

r5c8 is ♣️4 (the only 4 or K not in the grid).

r6c8 is ♣️2 (♣️J is in r9c3).