[Retros] Orbit 2004 Solutions

[Ivan Denkovski]

Here are the solutions of retros published in Orbit during 2004.

1232. Gligor Denkovski
1.Nc3 d5 2.Nxd5 Nd7 3.Nxe7 Ndf6 4.Nxg8 Nxg8 5.g4 Qh4 6.g5 Ke7 7.g6 Kf6
8.gxh7 Kg6 9.hxg8=N Kh7

1233. Rustam Ubaidullaev
1.g3 g5 2.Bh3 Bg7 3.Bxd7+ Kxd7 4.Kf1 Ke6 5.Kg2 Kf5 6.Kf3 Be6 7.Ke3 Kg4 8.Nf3
Kh3 9.Ng1+ Kg2 10.Qf1+ Kxf1 11.Kf3 Ke1 12.Kg2 Kd1 13.Kf1 Bd4 14.Nf3 Bb6
15.Ne1 c5 16.Kg2 Nc6 17.Kf3 Qd5+ 18.Ke3 Rd8 19.Rg1 Qh1 20.Rg2 Bc8 21.Nf3 Qe1
22.Rg1 Rd7 23.Rh1 Nd8 24.Ng1 c4+ 25.Kf3 g4+ 26.Kg2

1293. Marco Bonavoglia
1.Nf3 Nc6 2.Ne5 Nd4 3.Nc6 dxc6 4.g4 Bf5 5.gxf5 e6 6.f6 Ne7 7.fxe7 Kd7 8.e8=N
Be7 9.Nd6 Bg5 10.Nf5 Qf6 11.Nh4 Rad8 12.Nf3 Kc8 13.Ng1

1294. Mario Parrinello
1.f4 Nh6 2.f5 Rg8 3.f6 Nf5 4.fxg7 f6 5.gxf8=N Rg3 6.c4 Rc3 7.c5 Rxc1 8.c6
Rc5 9.cxb7 Nc6 10.bxc8=N Rb8 11.Nxa7 Rb6 12.Nc8 Qxc8 13.Nxh7 Qa6 14.Nf8 Kxf8

1295. Andrey Frolkin & Kostas Prentos
1.f4 Nc6 2.f5 Ne5 3.f6 Ng6 4.fxe7 f6 5.a4 Kf7 6.e8=R Bb4 7.Re6 c5 8.Rc6 dxc6
9.Ra3 Qd5 10.Re3 Bd7 11.Re8 Rxe8 12.Kf2 Rxe2+ 13.Kg3 Rxd2 14.Be2 N8e7 15.Bh5
Rc8 16.Nf3 Rc7 17.Re1 Nc8 18.Re8 Bxe8

1296. Thomas Volet
13 White Units, so all Black captures by Ps, on b-file, c-file, and at g6.
13 Black Units, so all White captures by Ps, on c-file, at f3, and on
g-file.  All 8 pawns of each color are in the diagram, so there are no
promotions.
In order for a King to retract from its square in the diagram position, the
position will have to unlock by Pe7-e6.  (Rb7-b6 does not help, as the P at
b4 cannot retract.)
Before Pe7-e6, however, we need to return the BKB to f8 and the BKR to g8
(the R cannot return through the h-file because the WN can retract only
after Ph7xg6, which itself blocks h7).  The BKB does not appear in the
diagram position, but can be uncaptured at g3 after Pg3-g4.
However the uncaptured BKB cannot reach f8 without retracting to e7 (it
cannot occupy d6 with the WK at c5), and a Black B cannot retract to e7 with
the WK at c5 unless there is some unit providing a retroscreen at d6 for the
WK.  That screening unit could only be (i) a White unit uncaptured at g6
(either the WQ or the missing WR), or (ii) the WN at h8 (after Ph7xg6).
Furthermore, the BKR must retract to g8 or h8 before the BKB is locked into
f8 by Pe7-e6.  On the corridor of squares d6-d7-e7-e8-f8-g8-h8, a Rook
cannot pass another Rook, so if it is a WR that screens on d6, the BKR must
be waiting in the Northeast corner during the screening maneuver at d6 and
the retraction of the BKB to e7 and f8.
The only way to restore the BKB to the board is by an uncapture at g3.
Therefore the BKR was not captured at g3, and it must be the R at e2 in the
diagram position.  But the only way to release that R from e2 is if
immediately after its retraction e2 is occupied by a White unit.  But the
only units available for that (the P at f3 cannot retract to e2 with the BK
at d3) are the unit that can be uncaptured at g6, and the N at h8.
Let us first consider the unpin at e2.  An uncaptured WQ could not retract
to e2 with the BK at d3, which leaves the WN or an uncaptured WR. But a WR
can only retract to e2 from the northerly side of the e-file, which would
require the BR could to retract to e1.  However, once having retracted to
e1, that R can never escape the Southwest corner to reach g8 or h8.
Therefore, the unit that retracts to e2 to free the BKR must be the WN at
h8.
The WN at h8 could not have provided both the unpin at e2 and the screen at
d6  The unpin freezes the unpinning unit at e2 for a while, so if the WN did
both maneuvers, it must have first (in the retroplay) have provided the
screen.  But the screen must be in place as the BKB retracts to e7 (to which
it must retract before retracting to f8) and the screening unit can only
retract off d6 if another unit takes its place as a screen.  That could be
the BKB, retracting to e7, but in that case the BKB and BR would have to
have reached f8 and h8 (or g8), respectively, which cannot happen until the
unpin by the WN frees the BR at e2.
If we suppose that a WR uncaptured at g6 provides the screen at d6, then the
BKR must be waiting in the Northeast corner, which means that the WN is
locked in at e2.  But for the WR to retract to d6 it must first retract to
d5.  For a WR to retract to d5, there must be some unit to provide a screen
at d4 for the BK.  With the WN stuck at e2 and the BR in the Northeast
corner, there is no other available unit to screen at d4.  Thus the WR
cannot retract to d5 in a position that can continue the retroplay to a
release.
Therefore the WQ must be uncaptured at g6 and provide the retroscreen at d6.
In order for the WQ to retract to d6 (to which it can retract from several
squares including g3, f4, e5 and e7), there must be a unit providing a
screen at d4 or d5 for the BK.  Fortunately the BQB, at h3 in the diagram,
is available to retract to d5 at the appropriate time.
The WQ must remain on d6 for two Black retromoves (arrival of BKB at e7, and
Bf8-e7).  The WR at b6 in the diagram supplies the needed White tempo.
No Black unit can have made the last move, unless it was Ph7xg6.  (Not
Pa5xb4, blocking the retraction of the WPa7, nor Pe7-e6, locking out the
BKB.)   But we have concluded that the unit captured at g6 was the WQ.  If
the very last move to reach the diagram was Ph7xQg6, White must have
delivered that Queen check on the very last White move.  (In retroplay, 1.
Ph7xQg6  2. Qg5-g6 (removing the check).) But that hypothetical sequence
would leave no possible preceding Black move.
Therefore White moved last, and the only possible last move for White that
allows for immediately preceding play for Black is Pg3-g4, freeing the BQB.
Thus we have:  1.Pg3-g4 Ph7:Qg6  2.Qf6-g6 Bf5-h3  3.Ng6-h8 Bg4-f5 4.Ne7-g6
Bh5-g4 5.Nf5-e7 Bg4-h5 6.Nd4-f5 Re4-e2 7.Ne2-d4 Rd4-e4  8.Ph2:Bg3 Rd7-d4
9.Qe5-f6 Re7-d7  10.Qf6-e5 Re8-e7  11.Qh6-f6 Rh8-e8  12.Qh5-h6 Bf5-g4
13.Qh6-h5 Be4-f5  14.Qh5-h6 Bd5-e4  15.Qe5-h5 Bh4-g3  16.Qd6-e5 Be7-h4
17.Rb7-b6 Bf8-e7  18.Qg3-d6 Pe7-e6  19.Rb6-b7 Be6-d5  20.Kd5-c5 Bc8-e6,
etc...

1357. Dmitrij Baibikov
Draw by 50-move rule. Critical position: White: Ke1, Qd4, Rc4, f4, Bc1, e4,
Pb4, c5, d5, e5, f2, f5, g4 (13), Black: Kg1, Rc3, d2, Ba1, Pb3, b7, c2, d3,
e3, f3, g3, h2 (12). Genesis: 0.h1Q! 1.tempo Qh8! 2.-16.tempos K->a4 17.Kf1
Re2 18. Bd2 Ka3 19.Be1 Kb2 20.Kg1 Kc1 21.Bd2+ Kd1 22.Bc1 Bb2 23.-24.tempos
Q->a1 25.Kf1 Ba3 26.Bb2 Qc1 27.Kg1 Qd2 28.Ba1 Kc1 29.Bb2+ Kb1 30.Bc1 Bb2
31.-33.K->f1 K->a4 34.Kg1 Ba1 35.Kf1 Kb5 36.-49.tempos K->h1 50.Bb2 Qc1+

1358. Gligor Denkovski
1.Nc3 Nf6 2.Ne4 Nxe4 3.g3 Nxd2 4.g4 Nxf1 5.Bh6 Nxh2 6.Qd2 Nxg4 7.OOO Nxh6
8.Qe1 Ng8

1359. Marco Bonavoglia
1.a4 h5 2.Ra3 Rh6 3.Rf3 Rb6 4.Rf6 exf6 5.a5 Bd6 6.axb6 Bg3 7.hxg3 axb6 8.Rh4
Ra5 9.Ra4 h4 10.Ra1 Rh5 11.Na3 Rh8

1435. Enzo Minerva
1.d3 g5 2.Bxg5 Bh6 3.Bxe7 Bc1 4.Bf8 Qe7 5.Qd2 Kd8 6.Kd1 Qe8 7.Qe1

1436. Enzo Minerva
1.e3 h5 2.Qxh5 d5 3.Kd1 Bg4+ 4.Qxg4 Qd7 5.Qe2 Kd8 6.Qe1 Qe8

1437. Diyan Kostadinov
1.Nf3 g6 2.Ne5 Bh6 3.Nxd7 Bxd2+ 4.Bxd2 h6 5.Nb6 Qd7 6.Nxa8 Kd8 7.Nxc7 Qe8
8.Ba5+ Nd7 9.Ne6#

1438. Gligor Denkovski
1.g4 a5 2.Bg2 a4 3.Be4 a3 4.Nf3 axb2 5.OO bxa1=N 6.Bb2 Nxc2 7.Bxg7 Ne3
8.dxe3 Bxg7 9.Qxd7+ Kf8 10.Nbd2 Qe8 11.Ra1 Bxa1 12.Qa4 Kg7 13.Qd1 Kf6 14.Bc2

1439. Christoph Fieberg
1.e3 Nc6 2.Qf3 Nd4 3.exd4 c5 4.Qc6 c4 5.f3 c3 6.Kf2 cxd2 7.Kg3 d1=N 8.Bh6
gxh6 9.Bb5 h5 10.c4 h4+ 11.Kg4 Nh6+ 12.Kh5 Nf5 13.g4 Ng3+ 14.hxg3 h3

1440. Paul  Raican
1.h4 a5 2.h5 a4 3.h6 a3 4.Rh5 axb2 5.a4 Ra6 6.a5 Rf6 7.a6 b5 8.a7 b4 9.Rb5
d5 10.a8=B d4 11.Bd5 d3 12.Ra8 dxc2 13.d4 b3 14.Nd2 b1=B 15.Ba3 b2 16.Nb3
c1=B 17.Qd3 Bg5 18.Qh3 Bg6 19.f4 b1=B 20.fxg5 Bbf5 21.e4 e5 22.exf5 e4
23.fxg6 e3 24.gxh7 e2 25.Kd2 e1=B+ 26.Kd1 Bh4 27.g3 Bd6 28.gxh4