R.J.Millour
1st Prize

a)Where was Bc8 captured? / Где был взят Bc8 ?
b) In what order were the Knights captured? / В каком порядке были взяты кони?
c) Can 2 pieces have been captured on d2-square of the same board? / Возможно, что 2 фигуры были взяты на поле d2 одной и той же доски?
Monochrome chess Alice chess A(1+4) B(1+1)

In what follows, the pieces appearing in square brackets (“[ ]”] are original ones. So, [Sb1] refers to the WS that was on b1 at the start of the game, whereas Sb1 is the BS presently on b1. Boards A/B are referred to as A/B states. The squares where pieces are captured in B state are underlined. So, c2- c4:[Pd7]d5 means that WPc, in B state on c4 after a doublestep, captures BPd, itself in B state on d5 after a double-step. In Monochrome Alice, 0-0-0 is forbidden and a K appearing in A state on the a-c-e-g files, or in B state on the b-d-f-h files, cannot have played 0-0. On the other hand, a K appearing in B state on the a-c-e-g files, or in A state on the b-d-f-h files, must have castled. Here, White and Black played 0-0! 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., and promoting a P requires 4 or 6 captures depending on whether it began with a doublestep. A pawn can capture an opponent pawn only if they both are in the same state right before the capture; in view of that, a P beginning with a double-step may capture an opponent P itself using the double-step, but may neither capture e.p. nor capture at home an opponent P. Conversely, a P that does not use the double-step may not capture an opponent P using it, but may capture e.p. and also capture at home an opponent P. In Monochrome, the Knights never move; therefore, Sb1 is promoted. The fact that this S is in A state implies in Alice a 6- move excelsior, which in Monochrome must be 6 captures one of which may include an e.p. capture. This excelsior ends with b3:a/c2:b1=S.
- The piece captured in A state on a/c2 is possibly a WP having never moved, in actual fact with f7-f5:[Pe2]e4 and c2-c4:[Pd7]d5, [Pa2] will be taken at home; - the piece captured in B state on b1 cannot be [Sb1], inevitably taken in A state on account of having never moved; therefore, the 6 victims of the excelsior are [Qd1], [Rh1], [Bf1], [Pb2] e.p., [Pa2] and promotion of [Pg2]!
When the last move c7:b6+ is retracted, the Ps on a7 and c7 show that the promotion to Q or R of a WP was needed to capture [Sb8]. This new excelsior does not involve [Rh8], which is available for capture only after the black 0-0, which is possible only after capturing [Sg8] by promoting [Pg2], which promotion implies first the capture of [Ra8], and in actual fact [Ra8] is just freed by capturing [Sb8]! Thus, the 4 victims of this new excelsior will be [Bf8], [Qd8], [Pe7] and [Pg7]. One P will disappear in B state on 5th rank, but the other one must promote because, as a P, it can neither be taken at home, nor on 6th rank in B state as the opponent P is on 5th rank in A state! Thus, finally we have 4 excelsiors according to the following chronology:
1) Release of [Ra1] → d7-d5 releases [Bc8] and c2-c4 allows B:[Sb1]b1. Also e/g7-e/g5 allows B:[Pd2]d2. Here [Pd2] disappears at home in A state, at the end of the move the capturing B piece turns to B state, so that [Ke1] is not in check. [Bc1] is released and, with b1 and c1 unoccupied, [Ra1] escapes via c1.
2) 1st excelsior → Thanks to [Ra1], e/g7-e/g5:[Pf/h2]f/h4:[Ra1]g3:[Bc1]h2:[Sg1]g1=X is possible. Sooner or later, c4:[Pd7]d5 occurs, and also f7-f5:[Pe2]e4 freeing [Bf1]: with g1 and f1 unoccupied, White may castle and capturing [Rh1] becomes possible.
3) 2nd excelsior → Thanks to X, one can do h/f2-h/f4:[Pg/e7] g/e5:[Pe/g7=X]f6:[Bf8]e7:[Qd8]d/f8=Y and Y:[Sb8]b8, freeing [Ra8]. Also [Pd2] could achieve this excelsior giving Y, but we intentionally consider [Pd2] is captured on d2 [see “1)” above] in connection with the 3rd question asked in this problem.
4) 3rd excelsior → Thanks to [Ra8], g2-g4:[Ph7]h5:[Ra8]g6:[Bc8]h7:[Sg8]g8=Z is possible. With g8 and f8 unoccupied, Black may castle.
5) 4th excelsior → Thanks to Z, b7:[Qd1]a/c6:[Rh1]b5:[Bf1]a/c4:[Pb2e.p.]b3:[Pa2]a2: [Pg2=Z]b1=S can take place.
6) Last moves → The WK or Y captures [Rh8], then Y is itself captured by c7:[Ph/f2=Y]b6+. At point "2)" [Ra1], which in Monochrome cannot access the 2nd rank, is inevitably captured on the 3rd. Consequently, [Bc1] disappears on h2, not on f2 where the capturing P would give check the WK before it played 0-0, which could not yet have occurred since [Sg1] has not yet been captured ! For similar reasons, at point "4)" [Bc8] disappears on h7, not on f7. The chronology shows the Knights disappeared in the following order: 1) [Sb1] 2) [Sg1] 3) [Sb8] 4) [Sg8].
"1)" and "3)" show [Pd2] is possibly captured on d2 in A state. Having castled, and therefore in B state on e1, e3, c3 or c1, the WK is not able to take an A-state piece on d2, but "6)" shows Y may capture [Rh8], why not on d2 and why not in A state! So, at first sight, it seems possible that 2 pieces were captured on d2 in the same state. But mind, please! Now we know the states of all the 25 missing pieces at the moment they disappeared and therefore we can determine whether the number of displacements for a side was odd or even. For that, it is sufficient to examine to number of B pieces (present + captured) for the side: if this number is odd, the number of piece displacements is odd, if it is even, the number of piece displacements is even. Here, with his Ka5 White has 1 unit in B state and 6 units were captured in B state at f/h4, h2, e4, b5, b3 and b1, yielding an odd number of piece displacements. But having castled (= 2 displacements in only 1 move), White played an even number of moves. With his Pb6 Black has 1 unit in B state and 5 units were captured in B state at d5, g/e5, e7, h5 and h7, yielding an even number of piece displacements. But having castled, Black played an odd number of moves. This cannot be! Both sides have made the same number of moves; c7xb6+ shows it is White to play! That means that it is not possible for 2 pieces to have been captured on d2 in the same state. 2 pieces could have been captured on d2: [Pd2] in A state, but [Rh8] inevitably in B state!
Finally, the answers are / Итак, ответы: a) [Bc8] was captured on h7! /Вс8 был взят на h7 b) 1) [Sb1] 2) [Sg1] 3) [Sb8] 4) [Sg8] c) 2 pieces could not have been captured on d2 on the same board! / Нет, 2 фигуры не могли быть взяты на поле d2 одной и той же доски.
Possible proof game / Доказательная партия: 1.c4→B d5→B 2.Qa4→B Bf5→B 3.Qa6→A Bg6→A 4.e4→B B:[Sb1]b1→B 5.h4→B g5→B 6.h:g5→A Bh6→B 7.g4→B Be3→A 8.Be2→B B:[Pd2]d2→B 9.Bf4→B e5→B 10.Bg3→A f5→B 11.Bh2→B f:e4→A 12.f4→B h5→B 13.Rc1→B Bf5→A 14.Rc3→A Be6→B 15.Ra3→B e:f4→A 16.Rg3→A f:g3→B 17.g:h5→A g:h2→A 18.Bf3→A h:[Sg1]g1=X→B 19.Be2→B Xg7→A 20.0-0→B Bc3→A 21.Rf5→A Be5→B 22.Rb5→B Bf6→A 23.g:f6→B Xe7→B 24.f:e7→A Bh3→A 25.e:d8=Y→B Bg4→B 26.Yb6→A b:a6→B 27.Y:[Sb8]b8→B Rc8→B 28.c:d5→A Rc6→A 29.Bc4→A Re6→B 30.Kf2→A Rg6→A 31.Ke3→B Bf5→A 32.h:g6→B Bh7→B 33.g:[Bc8]h7→A a:b5→A 34.h:[Sg8]g8=Z→B b:c4→B 35.b4→B c:b3e.p.→A 36.Yh2→A b:a2→B 37.Zb3→A 0-0→B 38.Zb1→B a:b1=S→A 39.Kd4→A Rf2→A 40.Kc5→B Rd2→B 41.Yf2→B Kh7→A 42.Y:[Rh8]d2→A Kg6→B 43.Yb4→B Kf5→A 44.Yb6→A Kg4→B 45.Kb4→A Kh5→A 46.Ka5→B c:b6→B+.
Масштабный замысел! При очень ограниченном материале проходит четкая хронология 4-х эксцельсиоров и строгий порядок взятий коней.

B.Rothmann
2nd Prize

J.Golha
3rd Prize

7#                                   4+3

h4=                               3+3

Marscirce Poursuite

ParrainCirce EinsteinChess
b)Ke7→e6

G.Bakcsi & L.Zoltán
4th Prize

J.Golha
1st Honourable Mention

h15=                                   3+4

h4#                               2+4

MirrorCirce Black must check

ParrainCirce AntiEinsteinChess
b,c)Kc2→c6,d5

G.Bakcsi & L.Zoltán
2nd Honourable Mention

J.Štúň
3rd Honourable Mention

h8#                                   3+3

h3#                               3+3

Circe White must check
b)-Sg3

ChameleonCirce Madrasi
b)Bc5→g5

S.Shifrin
4th Honourable Mention

D.Müller
M.Pfannkuche D.Speer
Commendation

s5#                                   3+4

h3#                               3+3

Maximummer Circe

Circe Rochadeschach

I.Bryuhanov
Commendation

B.Rothmann
Commendation

s7#                                   3+3

14=                               3+4

UltraSchachZwang Circe

Marscirce Poursuite

S.Shifrin
Commendation

K.Mlynka
Commendation

s7#                                   2+4

sh6#                               1+5

Maximummer Circe

AntiAndernachChess
ReversalEinsteinChess

J.Lörinc
1st-2nd Prize

D.Novomeský
1st-2nd Prize

8#                                   2+3

h8#                               3+1

Koeko Maximummer

Koeko Maximummer
Royal Queen - c8,b4

K.Mlynka
3rd Prize

I.Bryuhanov
4th Prize

h3#                                   1+1

s9#                               2+2

BrunnerChess Sentinelles

Circe Poursuite
b)Kd4→e4

J.Lörinc
1st Honourable Mention

J.Štúň
2nd Honourable Mention

h10,5=                                   2+3

sh7#                               2+3

Koeko Maximummer

ChameleonChess ParrainCirce
b)Qd7→e1

E.Huber
3rd Honourable Mention

J.Golha
4th Honourable Mention

h4#                                   1+1+2

h4#                               1+2

ParrainCirce Glasgowchess

EinsteinChess Sentinelles
b,c)Kf1→e1,d1

G.Bakcsi & L.Zoltan
Commendation

N.A.Bakke
Commendation

h7#                                   2+3

sh15#                               4+1

MirrorCirce White must check

Circe Rochadeschach

K.Mlynka
Commendation

B.Horstmann
Commendation

h2#                                   2+3

5=                               2+2

a)
AntiAndernachChess
ReversalEinsteinChess
b)
ChameleonChess
AntiAndernachChess

Maximummer Circe
b)mirror a1↔h8