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StrateGems

No. 13, Jan-Mar 2001

R0052 - Pascal Wassong

StrateGems 13, Jan.Mar 2001

White captured axb and gxf. Black captured hxg and the cross captures bxc and cxb. White promoted the h-pawn at h8 and black promoted the a-pawn at a1. The first unpromotion cannot be at h8 since the wSg7 is trapped at the north of the board and cannot intercept at f3. The attempt to retract wS to g3 fails because the inevitable bS/wS collision forces white to retract e2-e3 before bS can reach f3. Therefore a1 is the first unpromotion.

Retract -1. Sh5-g7 Sc7-e8 -2. Sg7-h5 Sa6-c7- 3. Se8-g7 Sb8-a6 -4. Sc7-e8 Sa6-b8 -5. Sa8-c7 Sc7-a6 -6. e2-e3! Se8-c7 -7-19. Sg7-h5-g3-f1-h2-g4-h6-f7-d8-b7-a5-b3-a1 -20. Sa8-c7 a2-a1=S -21. Sc7-a8 a3-a2 -22. Se8-c7 a4-a3 -23. Sg7-e8 a5-a4 -24. a4xb5


R0053 - Michel Caillaud

StrateGems 13, Jan.Mar 2001

Retract: -1. Rg3-h3 Bh2-g1 -2. Rf3-g3 Bg1-h2 -3. Rf4-f3 Bh2-g1 -4. Re4-f4 Bg1-h2 -5. Re5-e4 Bh2-g1 -6. Re6-e5 Bg1-h2 -7. Rd6-e6 Bh2-g1 -8. Rc6-d6 Bg1-h2 -9. Rc7-c6 Bh2-g1 -10. Rc5-c7 Bg1-h2 -11. Ra5-c5 b3-b4 -12. Rc5xPa5 b2-b3 -13. Rc7-c5 Bh2-g1 -14. Rc6-c7 Bg1-h2 -15. Rd6-c6 Bh2-g1 -16. Re6-d6 Bg1-h2 -17. Re5-e6 Bh2-g1 -18. Re4-e5 Bg1-h2 -19. Rf4-e4 Bh2-g1 -20. Rf3-f4 Bg1-h2 -21. Rg3-f3 Bh2-g1 -22. Rh3-g3 Bg1-h2 -23. Rh1-h3 a4-a5 -24. h2-h1=R a3-a4 -25. h3-h2 a2-a3 -26. h4-h3 h3xRg4

Ten retro-screens by the black rook on the h2-b8 diagonal.


R0054 - René J. Millour

StrateGems 13, Jan.Mar 2001

In Monochrome Alice, OOO is forbidden and a king, appearing in A state on a-c-e-g files, or in B state on b-d-f-h files, cannot have played OO. On the other hand, a king appearing in B state on a-c-e-g files, or in A state on b-d-f-h files, must have castled. Here, white played OO, and black did not.

Since it could not have come from a1 in Monochrome, the wRd4 resulted from promotion. Only Rd4 and Kf4 are retro-mobile because of the Alice condition. Thus: (a) the retraction of a move, such as Bb7-a8 is a retro check, (b) the retraction of Kg2-h1 is illegal (in the forward play the king would have placed itself in check in A state before turning to B state). To avoid black retrostalemate, white must immediately uncapture.

Consider pawns. Getting a bP to 5th rank in A state excludes the possibility of a double-step, and implies 2 captures, in B state on 4th rank 3 captures, etc. Promoting a pawn requires 4 or 6 captures, depending on whether it began with a double-step.

Thus, we find that the last move cannot have been b7xa8B. This bishop in A state implies (in Alice) a 6-move excelsior, which in Monochrome must be 6 captures, which may include an en passant capture. In the case of a2xBb3xRc4xPd5xPc6(ep)xPb7xXa8=B, the bP captured in A state on d5 would itself have captured d/f7xBc/e6xRd5, and the piece X, captured at a8, would be the promoted pawn resulting from f/d7-f/d5xPe4xQd3x??c2xNb1=X.

Now, this last sequence is impossible: a capture is lacking for d3 or c2, the unmatched states of the Pawns making impossible, in a 5-move excelsior, both an e.p. capture (here of the wPd) and the capture of a pawn at home (here of the wPc).

This, the Ba8-Kh1 paralysis can be broken only by an interposition between a8 and h1. Now, there is no other white-squared piece on the board and the black-squared pieces can only uncapture black-square pieces! An exception saves the day: the wP promoted to rook captured e.p. the bPb on white square! The paralysis is thus broken by restoring Pb7, which is legal, because, in Alice forward play, the bishop (which did not result from a7-a8=B, illegal in Monochrome) could have crossed over, in B-state, the bPb7 before turning to A state on a8.

A wP capturing e.p. must be in B state on the 5th rank! An Alice Monochrome excelsior implies 6 moves and 6 captures. Note that the pawn, therefore, is in B state before promoting and a knight (here the Nb8) in A state because the knight never moves in monochrome and cannot be captured by the pawn! A simple excelsior: b2xQc3xBb4xPc5xPb6(ep)xPc7xRd8=R.

In addition to the bPb ep capture, the excelsior captures 5 black-square pieces: Q, R, B on squares inaccessible to the pawns, and 2 pawns on squares requiring the fewest captures on their part, thus one of the two pawns on its home square, and the other on the 5th rank, which itself captured 2 times because it is required to be in A state in order to be captured by a wP, itself in A state on the 4th rank.

One more pawn, also a black-squared one, must be uncaptured on the first retro move to avoid black retro-stalemate. The wK would produce a bPf4 in B state, which would have made 3 captures, which is in fact impossible (see below the number of captures by black on black squares).

The wR restores a bPd4 in A state, which came from the 7th rank by a double step and a single capture. Black therefore has only two retro moves and, without delay, white must uncapture a piece by unpromotion.

The last black move? Was it c5xd4 or e5xd4? If c5xd4, the pawn came from c7, and in that case, before it promotes to b8, the wP must have captured bPe on c5 (after e7xd6xc5), the bPb ep and the bPa on a7. Now the retroplay Rd8xPd4 c5xYd4 (the white piece Y being undetermined for the time being) Rb8-d8 c7-c5 a7xb8=R does not work because the rook is on b8 in B state, signifying an excelsior in 5 moves, which is inconsistent with ep capture needed on b6. To have the rook in A state on b8 requires an extra white retro move, which is impossible due to black retrostalemate! However, with promotion on d8, the retroplay Rd8xPd4 e5xYd4 Y-d4 e7-e5 c7xd8=R, or simply Rd8xPd4 e5xYd4 c7xd8=R is legal.

The last move Rd8xPd4 is thus known with precision. The white unit Y, involved in the move e5xYd4, will be determined later. Note also that on e5 the pawn retro-paralyses the wK without the retro check). If c7x[Q/R/B]d8=R replaces a unit to give black an infinity of retro moves, sooner or later whie must restore the bPc to c7. Then comes the uncapture by ep and the return to b7 of bPb, making possible the retraction Kg2-h1 and the retroplay of the Ba8. Thus there are restored a bP on c/a5 and 2 other units on the 4th and 3rd rank.

The bP, restored on c/a5 in A state, is neither the bPe captured by the wR on d4, nor the bPc captured by the wP on c7. Is it the bPa after a7xb6xc/a5? The 2 victims would have been pieces, not pawns, because pawns would have required too many black-square captures by white (6 black-square captures were already made by white: 5 in the excelsior, 1 in the last move). Thus the bPa would have captured: 1. wBc on b6, because the wRa cannot reach b6 in monochrome; 2. the wRa on c/a5. Thus, the capture of the bishop on b6 won't work because, in this case, the Nb8 would still have to be present. In fact, it could not be captured by the wP upon promoting, as shown above, nor by any of the following pieces:

- The wK because of their different states (capturing the knight would imply a king in A state on a7 or c7, which is not possible given that it has castled.

- Ra1, which can't reach the 8th rank in monochrome.

- Rd4, which, having only been recently promoted, played only the move Rd8xd4 (see above).

- Bc1, which, before dying by a7xBb6, was blocked from b8 by pawns at a7 and c7 (capturing the knight requires a bishop in A state, which could not cross over a pc7 in A state!).

One comes to the paradox that the bPa could not be captured as a pawn, not as a promoted piece, by the white excelsior, which required and extra bP. In fact, the Bc1 captured Nb8 via a7, before being captured by the bPg, which itself was then captured by the white excelsior, as we shall see.

g7xBf6xRe5xPd4 allows, theoretically, the promoting wP to capture Pg7 on d4, but the other wP, here captured in A state on d4, must itself capture 2 times. This would make too many captures. On the other hand, the excelsior of the bPg, in 5 moves and 4 captures (ep not possible) does not involve any capture by white: g7-g5xPf/h4xRg3xBh2xNg1X. The unmoved knight is captured on g1. A wP (f or h), uncapturable at home, is taken in B state on the 4th rank. The Ra1 ends up on the 3rd rank because the 2nd rank is unavailable to it in monochrome. Finally, the Bc1 cannot be captured at f2, where the capturing pawn would give check to wK before it played OO, which could not yet have occurred since the Ng1 has not yet been captured! So the Pg7 captured Bc1 on h2.

[Now that the promotion of the bPg has been shown, one discovers that the piece captured on c/a5 is not necessarily the promoted bPg: the capture squares of the bPg promoted piece and the 3 other original pieces (Q R B) can be interchanged!]

From the cpatured white pieces, one finds that the piece Y, which was captured on d4, was necessarily the wPd, which came straight from d2 without capturing. The last black move was therefore e5xPd4. So the answers are:


P0076 - C.C. Frankiss

StrateGems 13, Jan.Mar 2001

a) 1. h4 g5 2. Rh3 Bg7 3. Ra3 Be5 4. Rxa7 Nf6 5. Rxb7 Ra3 6. Rb3 Ba6 7. Rh3 Rg3 8. Rh2 Bd3 9. Rh1 Na6

b) 1. h4 g5 2. Rh3 Bg7 3. Ra3 Be5 4. Rxa7 Nf6 5. Rxb7 Ra5 6. Rb6 Ba6 7. Rxf6 Bd3 8. Rf3 Na6 9. Rh3 Bg3 10. Rh1 Re5

c) 1. h4 g5 2. Rh3 Bg7 3. Rb3 Be5 4. Rxb7 Nf6 5. Rxa7 Ba6 6. Rxa8 Bd3 7. Ra4 Na6 8. Rg4 Qa8 9. Rg3 Qf3 10. Rh3 Qg3 11. Rh1


P0077 - Michel Caillaud

StrateGems 13, Jan.Mar 2001

1. c4 a5 2. c5 a4 3. c6 a3 4. cxb7 axb2 5. a4 d5 6. a5 d4 7. a6 Qd5 8. a7 Nd7 9. b8=Q h5 10. Qb7 Rb8 11. a8=Q h4 12. Ra7 h3 13. Na3 b1=Q 14. Qa4 Qbe4 15. d3 hxg2 16. Be3 dxe3 17. h4 exf2+ 18. Kd2 gxf1=Q 19. h5 Qfg2 20. h6 f1=Q 21. hxg7 Qff3 22. Rh7 Nh6 23. g8=Q Qh1 24. Qg2 Rg8 25. Qac6


P0078 - Michel Caillaud

StrateGems 13, Jan.Mar 2001

1. c4 h5 2. c5 Rh6 3. c6 Rxc6 4. a4 Rxc1 5. Ra3 c5 6. Rc3 Qa5 7. Na3 Ra1 8. Rc1 Qc3 9. f4 a5 10. f5 Ra6 11. f6 Rxf6 12. Nh3 b6 13. Nf4 Ba6 14. Kf2 Bd3 15. Ke3 Ra2 16. Nd5 Rxf1 17. h4 f5 18. Rh3 Rh1 19. Rf3 Rh2 20. Rf1 f4+ 21. Kf2 f3 22. Ra1 Bb1 23. Rh1

Two square circuits by the white rooks with echoed motivations.