1)

Raymond Smullyan, Chess problems with Sherlock Holmes, 1979

White : Kc8; Bg1; Ph2 : 3

Black : Ka8 : 1

Last move ? White to play.

Ka7xNa8

2)

Jean-Claude Dumont & Jean-Claude Gandy, Europe Echecs, 1977

White : Ka4; Rb4; Bg1, g6; Pc4, c6, h2 : 7

Black : Ka6; Pa7, b7 : 3

Mate in 2 moves. Last moves ?

White plays forward 1.Bg6-f5 and mates by 2.Bf5-c8# or 2.c4xb5#

depending on black move.

In the position of the diagram, black played the last move. It can
only

have been Kb6(x)a6, eventually capturing on a6. The double check by
the

wRb4 and wBg1 on b6 is explained by the en passant capture b5xc6 e.p.

The previous black move is therefore c7-c5. Then the check by wBg1
can

only be explained by Nc5(x)a6. So we see that the bK captured a wNa6.
To

avoid retro-pat by the black, this wN must have captured a bNa6 to
allow

bNb8(x)a6. Here ends the sequence of moves that we can deduce. So the

last moves were :

White : Ka4; Rb4; Bg1, g6; Nc5; Pb5, c4, h2 : 8 (and eventually

something on a6)

Black : Kb6; Nb8; Pa7, b7, c7 : 5

1. ...... Nb8(x)a6

2. Nc5xa6+ c7-c5

3.b5xc6 e.p.++ Kb6xa6 diagram.

3)

Gideon Husserl, Israel Ring Tourney, 1966 - 1971

White : Kc8, d6; Qc6; Rd8, f6; Ba8; Ne8; Pa7, b7 : 9

Black : 0

Color the pieces. Last move ?

The Qc6 and the Rd8 give to check to both Kings. This can only be

explained by these men being white and the last move being c7xd8=R++

with bKc8 and wKd6. Pb7, Ne8 and Rf6 must not give check. So we have

bPb7, wNe8 and wRf6. Because of bPb7, the position of the Ba8 can only

be explained by and under-promotion : wBa8 and wPa7.

The black piece captured at d8 cannot be a Rook or a Queen because
they

make an unexplained check to the wK. If it is a bB, black doesn't have
a

previous move. So it must be a bN which played bNe6(x)d8. The last
move

is completely determined : c7xNd8=R++.

4)

Tibor Orban, Die Schwalbe, 1976

White : Ke1; Qd1; Ra1, h1; Bc1; Nb1, g1; Pa2, b2, c2, d2, e4, f2, g2,
h2

: 15

Black : Ke8; Qd8; Ra8, h8; Bc8, f8, Nb8, g8; Pa7, b7, c6, e6, f7, g7,
h7

: 15

Proof game in exactly 4 moves.

There are easy solutions in 3 moves, e.g. : 1.e4 c6 2.Bc4 e6 3.Bxe6
dxe6

diagram.

But in 4 moves the solution is unique : 1.e4 e6 2.Bb5 Ke7! 3.Bxd7 c6

4.Be8 Kxe8 diagram.

5)

The intention is (backward) :

1.Bf7-e6+ g4xPf3

2.Bh6-g5 g5-g4

3.Qh5-g6 g6-g5

4.Rg5-f5 f5-f4

5.Rd6-f6 f6-f5

6.Fe6-f7 f7-f6

7.Nf6-g8 Kg8-f8

8.Ng4-f6+ etc...

The 3rd Knight comes from the promotion on a8 after 2 captures axNb
and

bxRa7-a8=N.

The wPg has been captured on his column.

The 15 last single moves are completely determined.

6)

Black captured Ph7xg6xf5xe4xd3xc2. After that white could play d2-d3
and

then the wBc1 could get captured by the bPa on b6 or b4. After that
the

wPa promoted on a8 to replace the wBe2 or the wRh1 already captured
by

the bP.

We will suppose that white may castle. So the wRh1 never moved. So the

wPa promoted to the wB currently at e2. I will now show that this

implies that the bK must have moved.

When black played h7xg6, the bPf7 must already have been on f6 so that

the move Px? isn't ambiguous. When black played g6xf5, the bPe7 is
still

at e7 for the same reason.

The U-Chess rules make that the first move of each side can only be

played by the d or e pawn. The wPd played after the bPh reached the
c2

square. The bPe7 moved after bPh7 reached f5. So we see that this game

begun by 1.e3 d6.

What is important here is the first black move. We deduce that when

black played e7-e6 to free the bBf8, the bPf was at f6 and the bPd
at

d6. This Bishop passed by the squares e7, d8 and over b6 to reach the

his destination of the diagram c5.

At the moment of the first move of the promoted white bishop Ba8-b7,
the

bPb6 is already there. This means that the bBc5 is also at his final

position and so we know that black already played c7-c6. Because the

wPc4 and wPd3 must have reached there final square to allow the wBc1
to

be captured by the bPa, the only way for the for the promoted wBa8
to

reach e2 is through b7, c8, d7, e8 at least, showing that the bK must

have evacuated his square.

So we showed that if white may castle, black cannot. If we suppose that

black can castle, this prouve us that the promoted piece on a8 is the

wRh1 and so white cannot castle.

This is an interesting problem like they were composed in those years.

7)

Dr. Luigi Ceriani, problem, 1952

White : Kg2; Qg1; Rh2; Bc1, f8, h1; Nd1; Pb2, c2, d2, e3, f2, g3, h3
:

14

Black : Ke2; Pb7, d5, e6, e7, g7, h5 : 7

What was the first move of the black king ?

The wPa2 captured a2xb3xc4xd5xe6xf7-f8=B. The bBf8 was captured at home.

The bPd7 captured on e6 after the wPa2 arrived at f7. The bPd5 comes

from c7 capturing the second missing white piece.

To open the South cage, the only way to put a piece on f1 and to retract

Ke1-e2. This piece can only have been a bR captured on e1 by a wN.
This

wN was captured by the bPc7.

The presence of the bR behind the wall of the wPawns is explained by
the

cross-captures h2xg3 and g2xh3. We now know where all the pieces have

been captured.

The last moves are : 1. ... c6xNd5 2.Nb4-d5 c7-c6 3.Nd3-b4 h6-h5

4.Ne1-d3 h7-h6 5.Nd3xRe1 Rf1-e1 etc...

We can now deduce that bPf7 has been captured at f7 without moving.
So

the wPa2 didn't capture the bRh8 (currently at e1), the bBc8 and the

bQd8, because they couldn't leave the 8th row before bPd7xe6.

At the moment of the move f7-f8=B the pieces must have been at : bBc8,

bQd8, bRe8 and bKh8.

The place of the BRe8 and bKh8 imposes as there first move : O-O !

Harry Nelson mentions that Black's a-pawn must promote in this problem.
His shortest game for this position is 43 moves. Is that minimal?

8)

Gianni Donati, Die Schwalbe, 1998

Dedicated to Danna

White : Ke1; Qd1; Ra1, h1; Bc1, f1; Ng1; Pa2, b2, c2, d2, f2, g2, h2
:

14

Black : Kg8; Qa4; Ra8, h8; Be2, g3; Nb8, g6; Pa7, b4, c7, d7, f7, g7,
h7

: 15

Shortest Proof Game in 12 moves.

1.Nc3 b5 2.Nd5 Ba6 3.Nxe7 Nxe7 4.e4! Ng6 5.e5 Bd6 6.e6 Kf8 7.e7+ Kg8

8.e8=N! Bg3 9.Nd6 Qh4 10.Nf5 Qa4 11.Nd4 b4 12.Nde2 Bxe2 diagram.

The fact that after his third move white must play an odd number of

moves imposes him to promote his e pawn to a Knight and this N must
come

back to its original square to be captured ! A very beautifull problem.

Jim Boyce's solutions to 5 and 6:

Hi,

The last several moves leading to the Scherer position were

ng4-f6+ kg8-f8

nf6-g8 pf7-f6

be6-f7 pf6-f5

rd6-f6 pf5-f4

rg4-f4 pg6-g5

qh5-g6 pg5-g4

bh6-g5 pg4xpf3

bf7-e6+

Reasoning:

1) Black still has all his pawns and six of eight pieces. He is

missing a knight and a rook. The rook spen all it time on a8,
a7, or

b8.

2) White has promoted a pawn to a knight. White is missing one
unit,

which was captured on the f-file by the g-pawn.

3) Where are White's a- and b-pawns? One is on b6. The other
was

either captured on the f-file or is the extra knight. In order
to get

to the f-file, it must have been promoted. So one of the a- and

b-pawns has been promoted. It took a path through b6, a7 (capturing

the black rook), and a8. The other capture moved the a-pawn to
the

b-file.

4) So the White f-pawn stayed on the f-file until it was captured by

the Black g-pawn.

5) Now we can start retracting moves. The final move was a discovered

check. The moving piece was the bishop, and not a pawn being
promoted

to a knight.

6) The only move black could have made was to capture on f3, and to

captured unit was a pawn.

7) White's next several retractions must leave Black with a legal

retraction. White retracts moves leaveing squares for the f-and

g-pawns to have moved from.

8) Finally, White retracts his two knight moves, the earliest of which

was a check.

-----

Now for the U-chess problem.

1) White has lost 6 units; Black has made six captures with pawns.

Five of those captures were made on white squares (g6, f5, e4, d3,
and

c2). White's dark-square bishop was captured by the a-pawn.

2) White's d-pawn did not move until the five white square captures

were done. Then the d-pawn moved to d3. By this time, the
c-pawn has

gone to c4, and freed the dark-square bishop to go die on the b-file.

3) White has not captured anything. So White's a-pawn was promoted

only after Black cleared the a-file. So the promoted a-pawn is
still

on the board.

4) Now for some U-chess stuff. The Black f-pawn was already on
f6

when the g6 capture happened. The Black e-pawn was still on e7
when

the f5 capture happened.

5) Black's first move was p-q3.

6) Since the pawns were already on f6 and d6 when the pawn moved to

e6, the path for the Black dark-square bishop to get to c5 goes

through e7 and d8 and over c7 and b6. So the c-pawn moved to
c6

before the b-pawns moved to b6.

7) Which is White's promoted piece?

7a) If it is the rook, then White cannot castle.

7b) If it is the bishop, then Black cannot castle, because the path

from a8 to e2 must pass through e8. Several squares were already

obstructed before the b-pawn moved to b6: c2, c4, c6, and e6.

8) It seems clear that a games can be constructed to show that White

can castle, or that Black can castle. Which one depends on what
the

a-pawn was promoted to.

Additional problem

Pascal Wassong

Europe Echecs, 1994

version

Dedicated to Babette

White : Kc5; Qa2; Rb1, b2; Be1, h8; Na6, d2;
Pa4, b3, b4, c2, e3, f2 :

14

Black : Kc3; Qc1; Rd1, e2; Ba7; Na1; Pc6, c7,
d7, e6, f3, g7 : 12

What where the last 56 single moves ?

I don't give all the explanations. This is a complicated
problem, were

some tries are not so easy to eliminate. Here
is the untangled position

:

White : Kc5; Qa4; Ra1, c1; Be1; Nb8, d2; Pa2,
b3, b4, c2, e3, f2, h2 :

14

Black : Kb2; Qa5; Rd1, e2; Ba7, b5; Nb6; Pc6,
c7, d7, e6, f7, g7 : 14

And now forward :

1. ... Kb2-c3

2.Qa4-a3 Qa5-a4

3.Rc1-b1 a6-a5

4.Qa3-c1 Qa4-a3

5.Nb8-a6 Qa3-b2

6.Na6-b8 a5-a4

7.Nb8-a6 a4-a3

8.Na6-b8 Bb5-c4

9.Nb8-a6 Ba7-b8

10.h2-h3 Nb6-a4+

11.b3xa4 Bc4-b3

and from now on the moves are unique. Before
we could have added extra

moves by the wNa6 and the bBc4, so that the previous
move could have

been Bd5-b3.

12.a2xb3 a3-a2

13.h3-h4 Qb2-a3

14.Rb1-b2 f7-f6

15.Ra1-b1 a2-a1=N

16.Rb2-a2 Qa3-b2

17.Ra2-a3 Qb2-a2

18.Rb1-b2 Qa2-b1

19.Rb2-a2 Qb1-b2

20.Qc1-b1 Qb2-c1

21.Ra2-b2 f6-f5

22.Qb1-a2 Qc1-b1

23.h4-h5 Qb1-c1

24.Rb2-b1 Qc1-b2

25.Rb1-c1 Qb2-b1

26.h5-h6 Qb1-b2

27.Qa2-b1 Qb2-a2

28.h6-h7 Qa2-b2

29.Ra3-a2 Qb2-a3

30.Ra2-b2 Qa3-a2

31.h7-h8=B Qa2-a3

32.Qb1-a2 f5-f4

33.Rb2-b1 Qa3-b2

34.Qa2-a3 Qb2-a2

35.Rb1-b2 Qa2-b1

36.Rb2-a2 Qb1-b2

37.Rc1-b1 Qb2-c1

38.Ra2-b2 f4-f3

39.Qa3-a2 Bb8-a7+ diagram