Solution to Table for Four

by Chris Luhrs and Noah Snyder

This puzzle consists of seven scrambled bughouse problems. You and your partner are playing the bottom boards and are up on time. You must mate your opponent (on one of the boards) using no more than the specified number of moves on each of the boards. The trick is that you don't know which boards pair up. The correct pairing and mates is below.

The correct pairing is A-3, B-5, C-1, D-2, E-7, G-6. Taking the squares a3, b5, etc. on the letter board at the top, and reading left to right gives BALFOUR, the answer to this puzzle.

Here are the required mates:
----A and 3:

1.(dropped) Ng7 Rg8xg7
2.(dropped) Rf8++

1. Nf5-e3

if ... Na3 moves or f2-f4: 2. Ne3-c2 wins a knight on move 3
else: Bc8xNa3.

----B and 5

1. Ra1xa5+ anything
2. Bc5xd6

1. (dropped) f2+ Kxf2
2. Nf6-g4+ Ke1
3. (dropped) f2++

----C and 1

if fxg6: (dropped) Qf7++
else: (dropped) Qf8++

1. ... (wait until black moves on the other board then play) Qf1+
if ... 2. Kd2 Qxd1
2. Kxf1 Ne3+
3. fxd3 Nxe3+
4. anything Nxd1

----D and 2
1. Ne4xf6+
if ... Nxf6: 2. (dropped) Qe7++
if ... Qxf6: 2. (dropped) Qf8++
if ... Kd8: 2. (dropped) Qe8++

if... Kxf2: 2. (dropped) Ne4+ wins queen
if... Kd1: 2. (dropped) Re1+ wins queen

----E and 7

1. Qxd5

1.(dropped)a2 anything
2. a2-a1++

----F and 4
if ... Rxf7 2.(dropped)R/Qe8++
else ... (dropped)R/Qh8++

1. (dropped)e2
if ... Qxe2 2. Nxc2 (Qa6 loses queen, else loses rook, white can't win a piece on move 3)
else ... 2. e2xR/Q

----G and 6
1. Nxf7
if ... Kxf7: 2. Bdg6+ anything 3. Qxd8
else: NxR/Q

1. (dropped) Ne2+ Kh1
2. (dropped) Q/Rg2+ Rxg2
3. Nxf2++

Here is a chain of logic explaining why this is the only pairing:

Board E is mate in 1. The only things that white can do in that move is take a bishop or take a pawn. He cannot mate no matter what piece is handed to him. The only board that can mate with just a bishop or a pawn is board 7. Thus the match must be E-7.

Board 3 cannot mate or force a capture of the queen. Thus all it can do is force a capture of two pawns or of a knight. The only board that is useful on is the knight forcing mate on board A. Thus the match must be A-3.

At this point on no board can white mate without using a major piece.

Board 6 black cannot capture a major piece. Thus black must force a mate. This can only be done with a rook or queen. Said rook or queen can only be captured using board G. Thus the match must be G-6.

At this point white cannot capture any major pieces. Thus 5 is the only remaining board where black can mate. Boards 1, 2, and 4 must lead to partner mating. But the remaining boards for them to partner with are B, C, D, F, and white cannot mate on B. Thus we must have the pair B-5.

All that remains is to find out which of boards 1, 2, and 4 can get pieces allowing boards C, D, and F to mate. F requires a rook or queen without losing a piece. Thus F-4 is the only option. C and D both require a queen, but differ in that D passes partner a knight before it requires the queen, thus only D can pair with 2, yielding D-2 and C-1.

2006 MIT Mystery Hunt