Pieces with similar linear movement are Rook with Rook, Rook with Queen moving like a Rook, Bishop with Queen moving like a Bishop.
Let us suppose that a Rook Re1 guards square e8 and another Rook Ra3 guards h3. When Ra3 moves to e3, then Re3 guards both e8 and h3 (the rook Re3 is now overloaded). If we force Re3 to play the role of Re1 (that is, to move towards e8), then Re3 abandons its initial role (that is, to guard h3).
| Theme: Holzhausen intersection : A ‘linear’ piece interferes with a ‘linear’ piece with similar way of movement and it is forced to become its substitute. |
In the next problem-65, by A. P. Grin (pseudonym of Alexander Pavlovic Guliaev), we have 4 Holzhausen intersections corresponding with the arrival square of the Queen.
 | (Problem 65) A. P. Grin, Honourable Mention, ”British Chess Magazine”, 1967 White plays and mates in 3 moves #3 (8+15) |
| [4S3/1spp1pp1/SpqPkB2/r2p3Q/r5p1/8/5RB1/bbs4K] | |
Tries: {1.Sxg7+? Kxd6!}, {1.Bd4? f6!}, {1.Sxc7+? Qxc7!}, {1.Qxg4+? Rxg4!}, {1.Qxf7+? Kxf7!}, {1.Qxd5+? Rxd5!}, {1.Qe5+? Bxe5!}, {1.Qf5+? Bxf5!}, {1.Re2+? Sxe2!}.
Key: 1.Qg5! [2.Sg7+ Kd6 3.Be7]
(The black Queen leaves from c6 to create a flight for her King, keeping the guard on c7).
1...Qc5 2.Qxd5+ (d5 was guarded by Ra5) Qxd5 3.Saxc7#
1...Qc4 2.Qg4+ (g4 was guarded by Ra4) Qxg4 3.Saxc7#
1...Qc3 2.Qe5+ (e5 was guarded by Ba1) Qxe5 3.Saxc7#
1...Qc2 2.Qf5+ (f5 was guarded by Bb1) Qxf5 3.Saxc7#
In the next problem-66 the black avoids four Holzhausen intersections and falls in the trap of an unexpected Nowotny intersection.
 | (Problem 66) Jan Hannelius, Second Prize, ”Probleemblad”, 1967 White plays and mates in 3 moves #3 (13+13) |
| [3rrb2/BBS4b/8/1p5q/p1ks1R1p/2Psp1pP/PPS1Q1P1/3R3K] | |
Tries: {1.Be4? Qxe2!}, {1.Rxd4+? Rxd4!}, {1.Re4? Qxe2!}, {1.Qxd3+? Bxd3!}, {1.Sxe3+? Rxe3!}, {1.Sxd4? Be4!}, {1.Sa3+? Bxa3!}, {1.Rxd3? Bxd3!}.
Key: 1.Bc6! [2.Re4] (which is a Nowotny move having two threats 3.Qxd3# / Sxe3#, and if 2...Qxe2 3.Bxb5#).
1...Qc5 (if 2.Re4? Qxa7!) 2.Sa3+ (a3 was guarded by Bf8) Qxa3 3.Bxb5#
1...Qd5 (if 2.Re4? Qxe4!) 2.Rxd4+ (d4 was guarded by Rd8) Qxd4 3.Bxb5#
1...Qe5 (if 2.Re4? Qxe4!) 2.Sxe3 (e3 was guarded by Re8) Qxe3 3.Bxb5#
1...Qf5 (if 2.Re4? Qxe4!) 2.Qxd3 (d3 was guarded by Bh7) Qxd3 3.Bxb5#
1...Qg5 (if 2.Re4? Bxe4!) 2.Be4 (Nowotny) [3.Qxd3# / Sxe3#]
It is impressive, in this problem by Hannelius, that Bb7 goes to e4 with two steps, in order to give time to the opponent to destroy his defensive arrangement.
[This post in Greek language].
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