Tuesday, April 29, 2008

Roman (1)

We present now problems with three related themes: Theme Roman, Theme Hamburg, Theme Dresden. In these themes there exist the following common elements:

(a) The problems have at least one try.
(b) There is a General plan (Hauptplan), which becomes obvious with the moves after the try (the virtual play), but is not successful.
(c) A Pre-plan (Vorplan) is applied, which starts with the key and continues with modification of the defensive moves. Then the general plan reappears and is finally applied.
(d) The themes require at least 3 moves.

Theme Roman : A black piece, which can defend adequately, is relocated to a square, from where it can defend again, but not sufficiently.


The Roman theme has got its name from a problem, which was dedicated by the German composers Johannes Kohtz and Carl Kockelkorn to a composer of Rome in 1905.



(Problem 135)
Johannes Kohtz and Carl Kockelkorn,
Dedicated to A. Guglielmetti of Rome,
”Deutsches Wochenschach”, 1905
White plays and mates in 4 moves
#4 (6+2)
[8/1S2b3/8/1B6/K2P4/2k1P3/5Q2/8]

Tries: {1.Sc5? Bxc5!}, {1.Bd3? Kxd3!}, {1.Be8? Bh4!}, {1.Bd7? Bh4!}, {1.Bc6? Bh4!}, {1.Qe1+? Kc2!}, {1.Qf1? Ba3!}, {1.Qb2+? Kxb2!}, {1.Qc2+? Kxc2!}, {1.Qd2+? Kxd2!}, and
Try: {1.Qe2? [2.Bd3 [3.Qc2#]]
1...Bg5! (aiming to capture the pawn 2...Bxe3}

The general plan of white, revealed in the last try, is not successful. A pre-plan of ‘relocating the Bishop to another diagonal’ will be applied, to weaken its defense.
Key: 1.Sd6! [2.Se4#]
1...Bxd6

Now the general plan can be applied.
2.Qe2 Bf4 (aiming to capture the pawn, Bxe3 as after the try, but unfortunately has come too close and allows a new continuation for the white)
3.exf4 Kxd4
4.Qe5#


After extensive research in older problems, it was found that the mechanism of the Roman theme first appeared in a problem of the British composer Kidson in 1858.


(Problem 136)
H. E. Kidson,
”Cassell’s Family Paper”, 1858
White plays and mates in 3 moves
#3 (7+9)
[1r6/1p2p1r1/1S1kp3/S1R3bQ/1B6/P2b4/4s3/K7]

Tries: {1.Qxe2? Bxe2!}, {1.Qf3? / Qe8? Bf6+!}, {1.Qxg5? Rxg5!}, {1.Rc1+? / Rc2+? / Rc3+? / Rc4+? Ke5!}, {1.Rb5+? Kc7!}, {1.Rc8+? / Rc6+? Ke5!}, {1.Rxg5+? / Rf5+? / Rd5+? Kc7!}, {1.Sc4+? Bxc4!}, {1.Sxb7+? Rxb7!} and
Try: {1.Qh8? [2.Qxb8#]. This is the general plan. Rb8 can not move because of Sxb7#)
1...Bf6+!}

Key: 1.Qh2+! Bf4 (pre-plan for the Bishop-relocation)
2.Qh8 Be5 (the general plan is applied, but now the defensive move of the black Bishop blocks a flight of the King)
3.Rc3#


[This post in Greek language].

Bi-Valve

Theme Bivalve : The black, opening a line of one black piece, closes another line of another black piece.



(Problem 73)
B. G. Mansfield,
First Prize, ”Olympic Tourney”, 1936
White plays and mates in 2 moves
#2 (6+11)
[8/KS5b/pR6/8/k1p1Q2r/4s3/PPp5/3r1sbq]

Tries: {1.Sc5+? Ka5!}, {1.Rb4+? Kxb4!}, {1.Qc6+? Qxc6!}, {1.Qe8+? Rd7!}, {1.b3+? Ka3!}.
Key: 1 a3! [2.Rb4#]
The threat is neutralized by lifting Se3 on air, because Bg1 pins Rb6, but on which square will the black Knight come down?
1...Sg2 2.Qc6# (one line of Qh1 has been closed)
1...Sg4 2.Qxc4# (one line of Rh4 has been closed)
1...Sf5 2.Qxc2# (one line of Bh7 has been closed)
1...Sd5 2.Qe8# (one line of Rd1 has been closed and one line of Qh1 has been closed)


In the prized problem-74, product of the cooperation of the great composers Loshinsky and Umnov, we see that Black tries (playing differently) to correct his defense (=black correction), and then tries to correct again (=secondary correction), and then tries to correct again (=tertiary correction), but eventually he becomes checkmated.


(Problem 74)
L. I. Loshinsky – E. Umnov,
First Prize, ”Western Morning News”, 1930
White plays and mates in 2 moves
#2 (9+8)
[B4K2/4pQ2/6p1/2p1kS2/6r1/2SRsbP1/1B3q2/4R3]

Tries: {1.Kxe7? / Bd5? Qxg3!}, {1.Qd5+? Bxd5!}, {1.Rd5+? Bxd5!}, {1.Qe6+? Kxe6!}, {1.Qxe7+? Kxf5!}, {1.Sxe3? / Rdxe3+? / Rexe3+? Qxe3!}, {1.Se2+? Rd4!}, {1.Se4+? Qxb2!}.
Key: 1 Sh6! [2.Qxe7#].
When Bf3 is lifted on air the threat is neutralized, because a line of Qf2 opens and the Qf7 is pinned, but on which square will the black Bishop come down?
1...Bh1 2.Sxg4# (Rg4 has been left unprotected)
1...Bd1 (black correction, protects Rg4, but Se3 has been left pinned) 2.Rd5#
1...Be4 (secondary black correction, does not protect directly Rg4, but unpins Se3 which guards g4. The bishop however closes the line of Rg4 (bi-valve) towards d4) 2.Se2#
1...Be2 (tertiary black correction, supports Rg4 and unpins Se3, closes however the line of Qf2 (valve) towards Bb2) 2.Se4#


[This post in Greek language].

Monday, April 28, 2008

Valve

Theme Valve : The black opens a line for one of his pieces, closing another line of the same piece.



(Problem 71)
C. S. Kipping,
First Prize, ”S. Glasnik”, 1947
White plays and mates in 3 moves
#3 (8+11)
[rb4br/2p2p2/1p4p1/2p1P3/1p2BQ2/4PS2/4SKP1/3k4]


In problem-71, by Kipping, we see two valves.
Tries: {1.Qh2? Rxh2!}, {1.Qf5? gxf5!}, {1.Qg4? f5!}, {1.Bc2+? Kxc2!}.
Key: 1.Qf6! [2.Qd8#]
1...Bh7 2.Qh4 [3.Qh1#] Bg8 3.Qd8#
1...Ba7 2.e6 [3.Qa1#] Bb8 3.Qd8#


In the next Problem-72, by Matthews, the Bishop opens a Rook line, but closes another Rook line, (with the Rook staying on its square, or moving to a better defensive position).


(Problem 72)
R. C. O. Matthews,
”The Problemist”, 1950
White plays and mates in 3 moves
#3 (10+11)
[r4b2/6p1/1p6/3B1pP1/4pP1B/2P1pS1p/2PSK1kp/4R2b]

Tries: {1.Bf2? exf2!}, {1.Kxe3? Bc5+!}, {1.Sf1? / Sc4? Bc5!}, {1.Rxh1? Kxh1!}, {1.Rf1? Ra1!}.
Key: 1.Sxe4! [2.Bf2 [3.Sh4#]].
The pair of the Knights has formed a half-battery with the Bishop so, if one of the Knights is lifted and then the other is lifted, a check from Bd5 is discovered.
It is not good to play 1...fxe4 2.Bxe4 Ra4 / Re8 3.Sd4# / Se5#
If the Bishop Bf8 is lifted, then the Rook Ra8 will be able to run to h8 to void the threat of the key. But the black Bishop can not stay on the air, but must come down on some square... Let us see five variations.

(a) 1...Be7 2.Sd4
(The plan is to lift also Se4 and Bd5 to check Kg2. The Rook Ra8 can not interfere in the diagonal d5 – g2 reaching e4, because the row is closed by the Sd4 and the file is closed by Be7. So the Rook will attack Bd5).
2...Ra5 / Rd8 3.Sc5# / Sd6#

(b) 1...Bd6 2.Sc5
(The plan is to lift also Sf3 and Bd5 to check Kg2. The Rook Ra8 can not attack Bd5, because the row is closed by Sc5 and the file is closed by Bd6. So the Rook will interfere in the diagonal d5 – g2 reaching e4).
2...Ra4 / Re8 3.Sd4# / Se5#

(c) 1...Bc5 2.Sd6
(The continuation is exactly the same as in variation-b).

(d) 1...Bb4 2.Se5
(The continuation is exactly the same as in variation-a).

(e) 1...Ba3 2.Rf1 and 3.Se1#
(Because Ra8 can not reach a1).



[This post in Greek language].

Plachutta intersection

How is related the Nowotny intersection with the Grimshaw intersection? There is a Grimshaw intersection, then we put on it a white piece, to force mutual black interferences, and we name this square Nowotny intersection.

Well, the Plachutta intersection is similarly related with the Wurzburg – Plachutta intersection.

Theme Plachutta intersection : The white moves a piece onto the intersection of two black linear pieces (with similar way of movement), that is on a Wurzburg – Plachutta intersection, to force mutual black interferences.


The theme Plachutta creates two simultaneous threats and needs at least three moves.

Let us see a simple presentation of this theme in problem-69, by Manolas: The Rook Re3 stops the checkmate Sa3#. The Queen Qd1 stops the checkmate Bd7#. The lines of action of the two black pieces intersect on square d3.

(Problem 69)
Manolas Emmanuel,
published in Award Manolas-60 JT, 12-07-2010
White plays and mates in 3 moves
#3 (7+8)
[8/Kp4br/1P6/pk2S3/2R3BR/4r3/2p5/1S1q4]

Tries: {1.Bd7+? Qxd7!}, {1.Sc3+? Rxc3!}, {1.Sa3+? Rxa3!}.
Key: 1.Sd3! [2.Sa3# / Bd7#]
The key is a Plachutta move. It gives a flight to the black king, sacrificing a rook.
1...Qxd3 2.Sa3+ Qxa3 3.Bd7#
1...Rxd3 2.Bd7+ Rxd7 3.Sa3#

The Bg7 may try to defend leaving square g7, then the Rook Rh7 guards d7, thus the Bishop must move to a square from which it can guard a3.
1...Bb2 2.Bd7+ Rxd7 3.Rc5#

The black Bishop corrects the arrival square, (this is called black correction), and guards a3 from another diagonal.
1...Bf8 2.Bd7+ Rxd7 3.Sc3#

The black King in rage starts capturing white pieces:
1...Kxc4 2.Be6+ and now
___2...Kb5 3.Bd7# (capturing the Rook, the King becomes target of the indirect battery Bishop-Rook, which was aiming at a square next to the King, not directly at the King).
___2...Kxd3 (another black correction, the hyper-active King has captured two pieces, but...) 3.Bc4#


The outstanding German composer Rehm, who is a specialist on more-movers, shows in his Problem-70 the theme Plachutta, twice!

(Problem 70)
Hans-Peter Rehm,
Fifth Prize, ”Deutsche Schachzeitung“, 1963
White plays and mates in 5 moves
#5 (8+9)
[s2B4/6p1/4R1B1/3p3s/1K1k4/1P1Pp1S1/p6r/S3r3]

Tries: {1.Bb6+? Sxb6!}, {1.Bc7? Sxc7!}, {1.Be7? Rc1!}, {1.Sf5+? Kxd3!}, {1.Sc2+? Rxc2!}.
Key: 1.Bg5! (with threat a Plachutta move 2.Se2 and double threat 3.Bxe3# / Sc2#, resulting in
2...Ree2 3.Sc2+ and 4.Bxe3#
or 2...Rhe2 3.Bxe3 and 4.Sc2#)

1...Rb2
The black passes Rh2 over the critical square e2 and takes care not to play 1...Rd2?? (bad move with continuation 2.Bxe3 Rxe3 3.Sf5 and 4.Rxe3#).
2.Be7 Rc1
The black defends as he did in one of the tries. But now the stage is ready for the second Plachutta on square c2.
3.Sc2+
and 3...Rcxc2 4.Se2+ Rxe2 5.Bc5#
or 3...Rbxc2 4.Bc5+ Rxc5 5.Se2#


[This post in Greek language].

Saturday, April 26, 2008

Wurzburg – Plachutta intersection

Theme Wurzburg – Plachutta intersection : two black pieces with similar linear movement mutually interfere (mutual Holzhauzen interference) on the same square.
If the pieces are white, we have white Wurzburg – Plachutta intersection.


The theme first appeared in 1909, and it needs at least 3 moves.
In one variation, with the first move we force a black piece-A to interfere to a similar black piece-B. On the second move we lure piece-A to play the role of piece-B. On the third move we achieve our goal.
In another variation, piece-B interferes to piece-A, piece-B is lured, and then we give mate.
The piece that arrives on the square of interference assumes the duty of guarding two squares and we say that it is overloaded.



(Problem 67)
Jan Hartong,
First Prize, ”Tijdschrift v. d. Nederlandse Schaakbond“, 1943
White plays and mates in 3 moves
#3 (8+11)
[b6s/2SpQ3/r7/p1pr1p2/S1k1PP2/1pBsK3/8/1B6]

Tries: {1.Qxc5+? Sxc5!}, {1.Qxd7? Rxd7!}, {1.exd5? Re6+!}, {1.Bxd3+? Rxd3+!}.
Key: 1.Bd2! [2.Qxc5 Rxc5 / Sxc5 3.Bxd3# / Sb2#] (The Queen Qe7 must not reach c5).

There are two thematic defenses:
(a) 1...Rad6 2.Qxd7 [3.Qb5#], (and now...)
2...Rc6 / Bc6 3.Qxd5# / Sb6# (because Ra6, which must guard b6, has passed over the critical square c6 and this became a black Grimshaw intersection)
or 2...R6xd7 (there is not R5xd7) 3.Sb6#

(b) 1...Rdd6 2.Qf6 ([3.Qc3#], (and now...)
2...Rd4 / Se5 2.Qxa6# / Sb2#
or 2...Rdxf6 (there is not Raxf6) 3.Bxd3#

Let us see two non thematic defenses:
(c) 1...d5 2.Qe5 (the rook is pinned and there is a threat Bxd3. If Sd3 plays, allows 2.Sb2, and the continuation is ...) 2...b2 3.Ba2#

(d) 1...fxe4 2.Qxe4+ Rd4 3.Bxd3# (without interesting play).


In the more recent problem-68, by Chepizhni, we see Wurzburg – Plachutta interferences with changed play in three phases.


(Problem 68)
Viktor Ivanovich Chepizhni,
First Prize e.a., ”40th Anniversary Tourney, Shakhmatny v. SSSR”, 1966
(There is set play). White plays and mates in 3 moves
* #3 (9+11)
[Q5b1/P1Kp4/p1S1S3/1p1p1p2/2k2p2/PR1Rs3/2P5/b3q3]

Phase of set play: (*)
1...Bc3 2.Rb4+ Bxb4 3.Se5# (mate A)
1...Qc3 2.Rd4+ Qxd4 3.Sa5# (mate B)

Phase of virtual play: Try: {1.Kb6? [2.Qc8 [3.Sc~]] Sxc2!
1...Bc3 2.Rb4+ Bxb4 3.Se5# (mate A)
1...Qc3 2.Se5+ Qxe5 3.Rb4# (mate C) }

Some more tries: {1.Qf8? d6!}, {1.Qc8? Sxc2!}, {1.Sa5+? Qxa5+!}, {1.Se5+? Bxe5+!}, {1.Rd4+? Bxd4!}, {1.Rb4+? Qxb4!}.

Phase of actual play: Key: 1.Kd6! (same threat 2.Qc8, forms battery with Sc6)
1...Bc3 2.Sa5+ Bxa5 3.Rd4# (mate C)
1...Qc3 2.Rd4+ Qxd4 3.Sa5# (mate B)
1...Sc2 2.Sa5+ Qa5 3.Qd5#

We have noted, inside parentheses, with (A, B, C) the mates that follow the moves (Bc3, Qc3), in order to show their combinatorial change : (AB, AC, CB).

Now, about the solving method, an experienced solver, observing the initial setting of the pieces, he/she could think in the following way:
The position is almost symmetrical and the composer wants Ba1 (which stops Se5) and Qe1 (which stops Sa5) to mutually interfere on c3, which becomes Wurzburg – Plachutta intersection.
Since Se6 (which holds c5) can be easily captured to create a flight for Kc4, probably Kc7 comes closer. From which direction does the King approach?
If we plan to play Kb6 as key, the Queen Qe1 can not play Qe3 to check. The composer, who put Bishop on a1 and Queen on e1 (Ba1 and Qe1), could equally have put Queen on a1 and Bishop at e1 (Qa1 and Be1), but why did he choose the former?
A possible explanation : If the correct key-move is Kd6, then a Queen on a1 could capture pawn a3, give check and ruin the solution.
From that point on, the solver discriminates which is the try and which is the key and solves the problem.


[This post in Greek language].

Wednesday, April 23, 2008

Holzhausen intersection

The theme Grimshaw, which we have already seen, was related with interferences between pieces with dissimilar linear movement.
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].

Task (3), with Babson promotions

As we have said, task is the composition that achieves the maximum number of variations with a specific characteristic.

Theme: Babson task : White plays first. Black defends with four promotions (Q, R, B, S) (AUW allumwandlung). White answers with four corresponding promotions (Q, R, B, S).


The American composer A. J. Babson proposed this theme in 1925, for corresponding promotion of black and white pawns. The years passed, and the composers managed to present Heterodox problems with the Babson task but failed to construct Orthodox (direct-mates) problems.
In 1934 André Chéron, a specialist in end-games and in problems with promotions declared that an Orthodox Babson will never be composed.
In 1964 the composer Pierre Drumare put as object of his life the Search for the Impossible, that is to compose an Orthodox Babson. What he presented after years of work was a monstrous construct – the pieces of your chess-board are not enough to set the position, because it needs nine Rooks and nine Bishops. In 1982 he quitted, declaring that an Orthodox Babson will never be composed.
In 1983 the unknown until then composer Leonid Vladimirivitch Yarosh published a problem (his third published problem!) and it was a completely correct and legal Orthodox Babson! Indeed Yarosh had a few more Babsons unpublished!



(Problem 122)
L. V. Yarosh,
1983,
Published in magazine ‘Shakhmatny v. SSSR’
White plays and mates in 4 moves
#4 (14+8)
[1q5R/P2S4/5p2/2p2p2/2Pk1b1s/6P1/PSpP1P1K/BQ3R1B]

Key: 1.Rxh4! (ok, this is not a very nice key, but in the next problem Yarosh took care for this also...)
1...cxb1=Q 2.axb8=Q [3.Qd6] Qxb2 3.Qb3 Qc3 4.Qxc3#
1...cxb1=R 2.axb8=R [3.Rxf4] Rxb2 3.Rb3 Kxc4 4.Rxf4#
1...cxb1=B 2.axb8=B [3.Ba7] Be4 3.Bxf4 Bxh1 4.Be3#
1...cxb1=S 2.axb8=S [3.Sc6] Sxd2 3.Sc6+ Kc3 4.Rc1#



(Problem 123)
L. V. Yarosh,
1983,
First Prize, ‘Shakhmatny v. SSSR’
White plays and mates in 4 moves
#4 (16+8)
[Bq1B1K2/3PpS2/P3Pp2/P1p2P2/2Pk1b1R/1p7/pS1P1P2/QR6]

Key: 1.a7!
1...axb1=Q 2.axb8=Q Qxb2 3.Qxb3 Qc3 4.Qaxc3#
1...axb1=R 2.axb8=R Rxb2 3.Rxb3 Kxc4 4.Qa4#
1...axb1=B 2.axb8=B Be4 3.Bxf4 Bxa8 4.Be3#
1...axb1=S 2.axb8=S Sxd2 3.Qc1 Se4 4.Sc6#

(Since 1983 more than a dozen Babsons have been composed by various composers.
The content of this post comes from texts of Tim Krabbé, whose interesting web page you may find at this address).


[This post in Greek language].

Task (2), with Full Knight Wheel

As we have said, task is the composition that achieves the maximum number of variations with a specific characteristic.

Theme: Task with Full Knight Wheel is the composition which has 8 thematic variations with a Knight reaching (with one step) eight different squares.


In the next problems we see one full Knight wheel in Iatridis’s problem-63 and two full Knight wheels in Petrovic’s problem-64.


(Problem 63)
Iatridis Stavros,
”To Skaki”, 1944
White plays and mates in 3 moves
#2 (8+6)
[K5B1/2p5/2SP1k2/5P2/3Sp3/6s1/1Q1B2b1/s7]

The Olympic winner Stavros Iatridis presents a task with a full white-knight wheel, where the battery Qb2 with Sd4 is ready to fire:

Key: 1.Se7! (Nice key! It gives one more flight to the black King, and also unmasks the battery Pa4-Bg2 so, if pawn e4 moves, Bg2 checks the white King).
1...e3+ (but this move stops the guarding of g5 by the Bishop Bd2) 2.Sf3#
1...Se2 [2...Sc3] 2.Sxe2#
1...Sc2 [2...Sd4] 2.Sxc2#
1...Sb3 [2...Sd4] 2.Sxb3#
1...Ke5 2.Sb5#
1...cxd6 2.Sc6#
1...Kg7 2.Se6#
1...Sxf5 [2...Sd4] 2.Sxf5#


Two years after the publication of Iatridis’s problem, the excellent composer Nenad Petrovic presented a problem with two full Knight wheels, one full white-knight wheel and one full black-knight wheel! He was compelled to use all sixteen white pieces to achieve the double task, with an admirable result.


(Problem 64)
Nenad Petrovic,
”The Chess”, 1946
White plays and mates in 2 moves
#2 (16+6)
[3B2r1/2P3P1/KR3P1R/2PskS2/2P3p1/2P2PPB/1Q4S1/5s1r]

Try: {1.c8=Q? Re8!}
Key: 1.Rh5! (The Rook with the Knight form a battery, which gives mate if the Knight moves, for example 2.Sh6#)

black S wheelwhite S wheel
1...Sxc7+ 2.Bxc7#1...Rxg7 2.Sxg7#
1...Sxb6 2.Sd4#1...Rh8 2.Sh6#
1...Sb4+ 2.cxb4#1...Rxh3 2.Sh4#
1...Sxc3 2.Qxc3#1...Sxg3 2.Sxg3#
1...Sde3 2.Sfxe3#1...Sfe3 2.Sfxe3#
1...Sf4 2.gxf4#1...Sxb6 2.Sd4#
1...Sxf6 2.Bxf6#1...gxf3 2.Sd6#
1...Se7 2.Sxe7#1...Se7 2.Sxe7#


[This post in Greek language].

Tuesday, April 22, 2008

Iatridis Stavros

Iatridis Stavros, (1887 – 1976), was golden Olympic winner in composition of chess problems. Apart from his objectively successful occupation with chess composition, which brought for him a lot of first prizes and many distinctions, he was a very fast solver and he was called “The Dragon of Problems”.
Iatridis was strong player of over-the-board game, having the title of Master with excellent ability in game analysis. In the first unofficial Greek championship in 1935 he won every game he played, but professional obligations kept him away from the last two rounds.
On November 25, 1964, in the 16th Chess Olympiad in Tel Aviv, the colonel of the Greek Army Iatridis Stavros was nominated golden winner in composition, in the category direct-mate three-mover. This was the fourth golden Olympic medal in the modern history of Greece, (after Tsiklitiras, Louis, and ex-king Constantine).

In 1967 chess was recognized by special legislation as intellectual sport in Greece.
Iatridis served as president of the Greek Chess Federation (G.C.F., Elliniki Skakistiki Omospondia) for almost one year.
He invited the Grand Master Dr. Petar Trifunovic as trainer for the national team.
He organized tourneys with sponsors commercial companies (Phillips, Shell).
He organized the Acropolis international tourney.
He invited Bobby Fischer in Greece for a handicap simultaneous demonstration.
Having connections with the army, he found easily office-rooms for G.C.F.
He wrote the column “Kallitexniko Skaki” (=artistic chess) in the monthly magazine “O Skakistis” (=the chess player), published by Harvatis Costas.
The enterprising and effective president Iatridis resigned because of ill health.

In October 1969 he was appointed chairman of the Preliminary Games for the World championship (Zonal 3) which were held in Athens.
In 1971, the retired colonel Iatridis, honorable president of G.C.F., co-founded the Chess Club of Ambelokipi (Skakistikos Omilos Ambelokipon).

In 1976, the likable uncle-Stavros was killed in a car accident. He was 89 years old.

We translate here an excerpt from the preface of the booklet “Kanonismos Zatrikiou” (=Chess Regulation) by Stavros D. Iatridis, publishing house Astir Papadimitriou, 1940:
We believe that we offer service to the players of Chess, especially to young persons, with the publication of the present International Regulation translated into Greek, because now contestations and gripes will be avoided, in the quietest, most intellectual and most enjoyable game of world, a game fairly considered as the King of the games.
The playing of Chess is allowed in all the clubs, in the houses of poor and in the palaces of rich, because it is a polite and very attractive amusement. It sharps the brain and teaches what can someone achieve with patience, combination and forecast, and usually protects youth from acquiring other, bad habits. For this reason, a lot of States support its distribution with suitable efforts and subsidies...


Those were the writings of Iatridis, back in 1940. Twenty four years later he became Olympic winner and three years after that the Greek state recognized chess as intellectual sport! For the subsidies he mentioned, the Greek problemists still wait, sixty eight years later.


The golden three-mover


(Problem 119)
Iatridis Stavros,
Gold metal, “16th Olympiad”, Tel Aviv, 1964
White plays and mates in 3 moves
#3 (7+8)
[8/1Ks1p3/1p5S/4p1q1/4s1S1/3R2p1/Qs2B3/7R]

The problem-119 was awarded with a gold metal, because it is a rare beauty. Watch the solution:
Tries: {1.Sf2+? gxf2!}, {1.Qd5+? Sxd5!}, {1.Qc4+? Sxc4!}, {1.Qa4+? Sxa4!}.

The problem starts as a threat-problem:
Key: 1.Rh4! (The knight Sg4 and the rook Rh4 form a battery which will fire when Sg4 moves, for example 2.Sf6#).
if 1...Kf4 2.Rf3+ Ke4 3.Sf2#

The black Queen Qg5 can capture the rook, or can step between King and Knight:
1...Qxh4 2.Qxb2 (and mate follows in the next move, with Qb4# ή Qxe5#).
1...Qf4 2.Kxb6!!
This move brings black in a zugzwang situation, and the problem is transformed to a waiter-problem! With this last move the white King is exposed to eight checks!

Four checks are given by the black Knights and are answered by the white Queen:
2...Sa4+ 3.Qxa4#
2...Sc4+ 3.Qxc4#
2...Sd5+ 3.Qxd5#
2...Sa8+ 3.Qxa8#

Four checks are given by the black Queen and are answered by the white Knight:
2...Qe3+ 3.Sxe3#
2...Qf2+ 3.Sxf2#
2...Qf6+ 3.Sxf6#
2...Qxh6+ 3.Sxh6#

Two more variations with pinning of the black Queen:
2...e6 3.Sf6#
2...g2 3.Sf2#


Now, for the completeness of this presentation of Stavros Iatridis, we post an educational game of his, that shows how Iatridis took advantage from the mistakes of his opponent during the opening and reached an end of game that he won easily. Note the comment at the tenth move, that reveals the analytic mind of Iatridis:

White: P. P., Black: Stavros Iatridis
Athens, 1937
Queen’s gambit
1.d4 d5 2.Sf3 Sf6 3.c4 e6 4.Sbd2 Be7 5.e3 0-0
6.Bd3 b6 7.0-0 Bb7 8.b3 Sbd7 9.Bb2 Se4 10.Se5

At this point Master Iatridis announces to his opponent that after the captures of the pieces they will reach an endgame with same-coloured bishops and he will win with his passed pawn.
10...Sxe5 11.dxe5
If 11.Bxe4 dxe4 12.dxe5 Qd3 13.Sb1 Rfd8 and black wins,
or 13.Re1 Rfd8 14.Bc1 Bb4 15.Re2 Bc3 16.Qxb1 and black wins.
11...Sxd2 12.Qd2 dxc4 13.bxc4 Be4 14.Rfd1 Qxd3 15.Qxd3 Bxd3
16.Rxd3 Rfd8 17.Rad1 Rxd1+ 18.Rxd1 Rd8 19.Rxd8+ Bxd8 20.Ba3 c5
21.Kf1 Kf8 22.Ke2 Ke7 23.Kd3 Kd7 24.Kc2 Kc6 25.Bb2 a6
26.a4 b5 27.cxb5 axb5 28.axb5+ Kxb5 29.Kd3 c4+ 30.Kc2 Ba5
31.Bc3 Bxc3 32.Kxc3 Kc5 33.e4 Kb5 34.f4 Kc5 35.g4 Kb5
36.h4 Kc5 37.f5 Kb5 38.g5 Kc5 39.h5 g6

Here white resigns.
If 40.fxg6 fxg6
41.hxg6 hxg6 42.Kc2 Kd4 43.Kd2 Kxe4 44.Kc3 Kxe5 and black wins.


[This post in Greek language].

Task (1), with Organ Pipes

Task is the composition that shows the maximum number of different variations having a specific characteristic.


Theme: Organ Pipes task. It is a composition with Organ Pipes having 8 thematic variations.

In the theme Organ Pipes, which contains the arrangement of pieces B-R-R-B, there are four Grimshaw intersections. Depending on which of the 2 pieces (R or B) reaches each of the 4 intersections, we should have 8 different thematic variations. This is a task still not achieved, as far as I know. (But someday someone will make it a reality).

In the following problems we see the first presentation of this theme by the amazing problemist Sam Loyd (with 4 thematic variations) and the record by Jan Hartong (with 6 thematic variations).

(Problem 61)
Sam Loyd,
”Boston Globe”, 1859
White plays and mates in 2 moves
#2 (6+10)
[2brrb2/8/p7/7Q/1p1kpPp1/1P1pS1P1/3K4/8]

Tries: {1.Sc2+? dxc2!}, {1.Qe5+? Rxe5!}, {1.Qd5+? Rxd5!}, {1.Qb5? [2.Qc4#] axb5!}.
Key: 1.Qa5! (zugzwang)
1...Bd7 / Bd6 2.Qd5#
1...Rd7 / Re6 2.Sf5#
1...Be7 / Be6 2.Qe5#
1...Re7 / Rd6 2.Qb4#
Non thematic variations are those which do not relate with the theme:
1...Bc5 2.Qa1#
1...Bb7 2.Sf5# etc.


Please study in the next Problem-62 how has Hartong used some mechanisms (pins, line closures) to improve the appearance of this theme and to achieve 6 thematic variations.

(Problem 62)
Jan Hartong,
First Prize, ”Problem”, 1951
White plays and mates in 2 moves
#2 (10+11)
[1BbrrbQ1/KS5p/5p1p/4P2R/1p1pk1B1/1P1S2R1/8/3s4]

Tries: {1.Sbc5+? / Sdc5+? Bxc5+!}, {1.Bf5+? Bxf5!}, {1.Sf2+? Sxf2!}.
Key: 1.exf6! (zugzwang)
1...Bd7 / Bd6 2.Qd5#
1...Rd7 2.Bf5#
1...Rd6 2.Sbc5#
1...Be7 / Be6 2.Re5#
1...Re7 2.Sdc5#
1...Re6 2.Qh7#


[This post in Greek language].

Monday, April 21, 2008

Exercise No.5 for Solvers

Dear readers,
in this exercise No.5 twelve direct-mate problems are included, and you are kindly requested to solve them. (Now, if you solve some but not all, it is quite all right. Send less than twelve solutions. Write only the key for two-movers, write first and second move for three-movers, write the full variation for the more-mover. Avoid anonymous comments. Last date for comments: 30/04/2008).

The first six problems have Pickaninny pawn on squares b7 thru g7 correspondingly.
In the problem-55 there is set-play and tries relevant with Grimshaw and Nowotny intersections.
In the problem-56 there is initially a disturbing pawn f7, but how can we deal with the four moves of Pickaninny e7?
In the problem-57 we see the revealer rook Ra1 and we understand which is the key.
In the problem-58 we spot instantly the Nowotny key-move.
In the problem-59 we do not need extra strong forces to achieve mate.
In the problem-60 there is a double Grimshaw (one Bishop and two Rooks).

The solutions will be posted next month.


(Problem 49)
Comins Mansfield,
1916
White plays and mates in 2 moves
#2 (6+4)
[KR2B3/Bp6/Q7/8/8/8/5S1p/6kr]
 
(Problem 50)
Marble & Bettmann,
White plays and mates in 2 moves
#2 (5+2)
[7Q/2p5/1P1S3R/3k4/8/2K5/8/8]
 

(Problem 51)
Bettmann,
1915
White plays and mates in 2 moves
#2 (13+2)
[4B3/3p4/2P1PQ2/P1k2P1R/4P1S1/1K4R1/5S1B/8]
 
(Problem 52)
Κ. Comins Mansfield,
3rd Prize, Memorial L. Segal, Themes-64, 1962
White plays and mates in 2 moves
#2 (7+3)
[8/4p2K/3P4/5kP1/2R1Sp2/1B3Q2/8/8]
 
(Problem 53)
C. Morse,
”The Problemist”, 1962
White plays and mates in 2 moves
#2 (9+3)
[5Q2/4Bp2/4P1R1/4R3/4p1BP/4P1k1/8/6K1]
 
(Problem 54)
Unknown composer
Lexorama #10
White plays and mates in 2 moves
#2 (10+8)
[6Q1/1S4p1/2R2S1P/5k2/3P1s1p/5RbP/5p1p/2B2K1s]
 
(Problem 55)
W. Tura,
First Prize, ”Europe Echecs”, 1962
(Set play). White plays and mates in 2 moves
* #2 (10+11)
[2B1S1q1/2S1Pb2/1pP3rs/b3k1p1/r2p2R1/BR1p2K1/5Q2/8]
 
(Problem 56)
M. Niemeijer,
First Prize, ”Tijdschrift v. d. N. S. B.”, 1928
White plays and mates in 3 moves
#3 (9+8)
[2b3K1/1pPkpp2/1P1P1RP1/3p1p1S/3S1p2/5P2/8/8]
 

(Problem 57)
Ren A.
White plays and mates in 3 moves
#3 (6+5)
[8/8/6s1/8/8/2s3R1/4P2p/RbSQK2k]
 

(Problem 58)
Mertes
White plays and mates in 2 moves
#2 (4+3)
[8/6r1/5Q1b/7k/8/5S1K/4S3/8]
 

(Problem 59)
Allan Werle,
”Tidskrift för Schack“, 1945
White plays and mates in 4 moves
#4 (2+2)
[8/4P3/8/8/8/8/3p1K2/7k]
 
(Problem 60)
J. A. Schiffmann,
”T N S B Chess Assn. Tourney”, 1929
White plays and mates in 2 moves
#2 (12+8)
[8/1s1Pp3/b3PkSP/R3S1p1/r2Q2P1/p5K1/B4B2/3r1R2]
 


[This post in Greek language].

Pawns Albinos and Pickaninnies (2)

In the next problem-47 we see a double Pickaninny.


(Problem 47)
N. G. G. van Dijk,
”American Chess Bulletin”, 1958
White plays and mates in 2 moves
#2 (14+7)
[SKb5/Bp1pp3/P1PP1P2/q2k1B1S/R1R1pQ2/2P5/1P6/8]

We observe that there are many tries: {1.Sb6+? Qxb6!}, {1.Sc7+? Qxc7+!}, {1.Bd4? / Bb6? / fxe7? / Sg7? Qb4!}, {1.Bxe4+? Ke6!}, {1.Qd2+? / Qe5+? K(x)e5!}, {1.Qxe4+? Kxd6!}, {1.Rc5+? Qxc5!}, {1.Rd4+? Kxc6!}, {1.Rxa5+? Kxc4!}.
Key: 1.b3!
The meaning of the key-move is to support Rc4 so, if black plays Qa4 or Qb4, the mate Sc7# is valid. The key brings black in a zugzwang situation. Any moves of the Pickaninny b7 or of the Pickaninny e7 or of any black piece are answered with mate.


In the next problem-48 we watch an Albino-Pickaninny duel.


(Problem 48)
G. H. Drese,
Second Prize, ”Tijdschrift v. d. K. N. S. B.”, 1935
White plays and mates in 3 moves
#3 (8+6)
[3S4/4pp2/2RS1B2/KB1k1P2/3p4/4s1s1/5P2/8]

There are a few tries: {1.Rc5+? Kxc5!}, {1.Bc4+? Sxc4+!}, {1.fxe3? exd6!}.
Key: 1.Bd3!
The key brings black in a zugzwang situation. If Se3 moves, there is Bc4#. If Sg3 moves, there is Be4#. The Pickaninny e7 tries to defend, but its defenses are voided by the Albino f2:
1...e6 2.fxe3 (not 2.fxg3? exf5! and there is no mate).
1...e5 2.fxg3 (not 2.fxe3? dxe3! and Bf6 does not see the flight d4).
1...exd6 2.f3 (protects again the square e4).
1...exf6 2.f4 (protects again the square e5).


[This post in Greek language].

Friday, April 18, 2008

Pure mates, Ideal mates, etc.

Some composers pay special attention to the form of mate. There is a school of such composers, named Bohemian School. (The composers do not come all from Bohemia! They just have similar criteria for compositions). The composers give to their problems comfortable positions with a sense of freedom and in at least two variations the mates are ideal.

Pure mate: The square on which the mated king stands and every square attached to it, is under threat of one hostile force or it is blocked by a friendly piece, (not threat and block simultaneously).
Economical mate: All the white pieces (with possible exclusion of King and Pawns) are taking part in the mating net.
Model mate: A mate pure and economical.
Ideal mate: It is a model mate where all pieces, black and white, take part. It is also called Perfect mate.
Mirror mate: It is a model mate where all the squares around the king are empty.


(Problem 45)
Manolas Emmanuel,
Newspaper “Allagi” of Patras, 14/06/1983
White plays and mates in 2 moves
#2 (5+2)
[8/1Q6/2p1S1K1/3k4/8/8/2PS4/8]

In the problem-45, by Manolas, we see two echo-mates and a Perfect mirror mate.

Tries: {1.Qxc6+? Kxc6!}, {1.Qd7+? Ke5!}, {1.Kf6? / Kf5? Kd6!}.

Key: 1.Sc4!
(while one of the knights is in threat, with the sacrificial key we put the second knight in danger, too)
1...Kxc4 / Kxe6 2.Qb3# / Qf7# (These are the echo-mates).
1...Ke4 2.Qxc6# (This is the perfect mirror mate).

Update 07-ii-2017 : I have learned today, after 34 years, that this problem of mine is anticipated. Prior compositions are (1) Durebert, 1925, bulletin de la FRE, [8/2Q5/3p1S1K/4k3/8/8/3PS3/8] and (2) Fredriksson E. G., 1929, Schackvorlden, [8/5Q2/K1S1p3/3k4/8/8/3SP3/8].
I still remain very satisfied that I have composed this problem. Manolas Emmanuel.


(Problem 46)
Godfrey Heathcote,
”The Observer”, 1927
White plays and mates in 3 moves
#3 (10+9)
[8/KpQ5/2P5/1Pk5/s1P5/Pp1qp1rr/3P1Pb1/2RR4]

In Heathcote’s problem-46 we have four model mates with pinning of the black Queen, combined with an Albino pawn at d2.

Key: 1.f3! [2.Qe5+ / Qe7+ / cxb7+]
1...Qc2 2.Qe5+ Kxc4 3.d3# (the queen Qc2 is pinned by Rc1)
1...Bxf3 2.Qe7+ Qd6 3.d4# (the queen Qd6 is pinned by Qe7)
1...Sc3 2.cxb7+ Kd4 3.dxc3# (the queen Qd3 is pinned by Rd1)
1...Bxf3 2.Qe7+ Kd4 3.dxe3# (the queen Qd3 is pinned by Rd1)


(Problem 134)
V. Pachman,
First prize, ”Schach-Echo”, 1959
White plays and mates in 5 moves
#5 (3+7)
[8/2p5/8/5p2/8/1p3p1K/kp6/qS3Q2]

In Pachman’s more-mover problem-134 we see model mates, characteristics of Bohemian school. The open position creates possibilities for tries, but the solution has a strategic content:
Tries: {1.Kh4? / Kg3? / Kh2? / Qd3? / Qg1? f2!}
Tries: {1.Qa6+? / Qb5? / Qc4? / Qxf3? / Qf2? Kxb1!}
Try: {1.Sc3+? Ka3!}
Key: 1.Qe1! [2.Sc3+ Ka3 3.Qe7+ c5 4.Qxc5#]
1...c5 2.Qf1 [3.Sc3+ Ka3 4.Qa6+ Kb4 5.Sd5#]
2...c4 3.Qe1 [4.Sc3+ Ka3 5.Qe7#], (a threat which is actually realized).

[This post in Greek language].

Pawns Albinos and Pickaninnies (1)

Theme Albino : A white pawn makes from its starting square four different moves (one step forward, two steps forward, capture left, capture right).


Theme Pickaninny : A black pawn makes from its starting square four different moves (one step forward, two steps forward, capture left, capture right).



(Problem 41)
A. C. Reeves,
”Probleemblad”, 1965
White plays and mates in 2 moves
#2 (9+10)
[1s1b1B2/1Q2p3/1p1S1Rp1/S1k5/p3R3/Kp1p4/1sP5/5B2]

In problem-41 there are several tries: {1.Qd5+? Kxd5!}, {1.Qc6+? Sxc6!}, {1.Rf5+? gxf5!}, {1.Sxb3+? axb3!}, {1.Rxa4? exd6!}, {1.Rc4+? Sxc4+!}, {1.Re5+? Kd4!}, but we specifically note that four tries Albino are answered with four defenses Pickaninny:
Try: {1.c3? [2.Re5#] exf6!}
Try: {1.c4? [2.Qd5#] e6!}
Try: {1.cxb3? [2.b4#] exd6!}
Try: {1.cxd3? [2.d4#] e5!}
Key: 1.Rb4! [2.Se4#]
1...Sc4 2.Rxc4#
1...exd6 2.Bxd6#


In problem-42 black tries to defend with Pickaninny, but white has four continuations with Nowotny intersections.


(Problem 42)
N. Littlewood,
First Prize, ”B.C.P.S. Ring Tourney”, 1966
White plays and mates in 3 moves
#3 (12+11)
[bb2RSs1/r4p2/qr4PP/p1P1B1k1/3S4/2sP1RpQ/5PK1/8]

There are several tries: {1.Sh7+? Kxg6!}, {1.Bxg3? Bxg3!}, {1.Bf4+? Bxf4!}, {1.Bxb8? Sxh6!}, {1.Qf5+? Kxh6!}, {1.Qg4+? Kxg4!}, {1.Qxg3+? Kh5!}, {1.Qh5+? Kxh5!}, {1.Qh4+? Kxh4!}, {1.fxg3? Rb2+!}.
Key: 1.Re6! ( > 2 Sh7#)
1...f6 2.Rd6 (Nowotny on d6) Rxd6 / Bxd6 3.Bf4# / Sde6#
1...f5 2.Rc6 (Nowotny on c6) Rxc6 / Bxc6 3.Rxg3# / Sde6#
1...fxe6 2.c6 (Nowotny on c6) Rxc6 / Bxc6 3.Rxg3# / Sde6#
1...fxg6 2.Bc7 (Nowotny on c7) Rxc7 / Bxc7 3.Re5# / Sh7#


In problem-43 we see that the mates for Pickaninny are changed in two phases.


(Problem 43)
A. C. Reeves,
”Die Schwalbe”, 1965
(Set play). White plays and mates in 2 moves
* #2 (13+6)
[R2Sr3/RB1pk1P1/PpP1P1P1/1K2p1P1/3Q4/2Sr4/8/8]

First Phase: (*) (Set play)
1...d6 / d5 / dxc6+ / dxe6
2.Sd5# / Qb4# / Bxc6# / Bc8#

If we start with checks, we do not achieve our goal:
Tries: {1.Qb4+? d6!}, {1.Qxd7+? Rxd7!}, {1.Sd5+? Kd6!}

Last Phase: Key: 1.Qxe5!
1...d6 / d5 / dxc6+ / dxe6
2.Qf6# / Qc7# / Sxc6# / Qxe6# (four mates are changed).


In the problem-44 the King Kf6 with the Bishop Bg7 form a royal battery which fires with a move of Kf6.


(Problem 44)
Henry D’ Oyly Bernard,
Second Prize ex aequo, ”Good Companions”, 1917
(Set play). White plays and mates in 2 moves
* #2 (14+10)
[2s3b1/6B1/p3PKP1/p1pp1R2/QbPk1pr1/4sP2/1PPS1P2/3R4]

First Phase: (*) (set play)
1...Bb~ (makes a random move) / Bxd2 2.Sb3# / Rxd2#

There are many tries : {1.Bh8? Rh4!}, {1.Rxf4+? Rxf4+!}, {1.Rxd5+? Sxd5+!}, {1.Qb3? dxc4!}, {1.Qb5? axb5!}, {1.Qa3? Bxa3!}, {1.Qxb4? cxb4!}, {1.fxg4? Sxg4+!}, {1.fxe3+? Kxe3!}, {1.Sb1+? Sxd1!}, {1.Se4+? Sxd1!}, {1.Sb3+? Kxc4!}, {1.c3+? Kd3!}.

Last Phase: Key: 1.Qa1!
The key brings black in zugzwang situation, despite the fact that black has many available moves. The four moves of Bb4 are handled by the Albino b2.
1...Ba3 / Bxd2 / Bc3 / Bxa5
2.bxa3# / b3# / bxc3# / b4#


[This post in Greek language].

Thursday, April 17, 2008

Nowotny intersection (and the first Nowotny)

The first Nowotny


(Problem 125)
A. Nowotny,
1854
White plays and mates in 4 moves
#4 (6+4)
[1s5r/8/6b1/2R5/1B1k4/8/1R1K1S2/5S2]


Key: 1.Sg3! [2.Se2#] Re8
2.Rbc2 [3.R2c4#] Bxc2
3.Sfe4 [4.Bc3#]. (The Springer (=Knight) move aims to cut the influence lines of the black Rook and Bishop and, if Sf4 is captured, to threaten mate with Sg3 from two places).
3...Bxe4 / Rxe4
4.Se2# / Sf5#


Theme Nowotny intersection : it is a square, which is Grimshaw intersection for black pieces with dissimilar linear movement, and on which a white piece is placed in order to cut the black lines and to force black pieces (when they capture the white piece) to mutually interfere.
White Nowotny : it is an intersection of white pieces with dissimilar linear movement, and on this intersection a black piece is placed, in order to force white pieces to mutually interfere.



(Problem 40)
M. Lipton,
Second Prize, BCPS Ring Tourney, 1966
White plays and mates in 2 moves
#2 (6+5)
[2K1k3/1R4BS/3R4/5Ss1/8/2b5/4r3/3r4]

With a Nowotny intersection, the white has simultaneously two threats.

In Lipton’s problem we must locate which square is the Nowotny intersection and also discover which piece must be sacrificed there!

(Should become Nowotny the Grimshaw intersection e5?)
Try: 1.Be5? [2.Sg7# / Re7#] Sxh7!

(Or should become Nowotny the Grimshaw intersection d4?)
Try: 1.Sd4? [2.Sf6# / Rd8#] Re7!
Try: 1.Rd4? [2.Sf6# / Sd6#] Re6!

Key: 1.Bd4! [2.Sg7# / Rd8#] Se6 2.Re7#


[This post in Greek language].

Wednesday, April 16, 2008

Grimshaw in 2 and in 3 phases


(Problem 38)
Janos Kiss,
Second Prize, ”Probleemblad”, 1953
White plays and mates in 2 moves. (Set play).
* #2 (12+11)
[8/bS1B1s2/r1p5/1pp1BPpK/1P2k1S1/3R3P/2Pp1QR1/3br3]

The asterisk [*] in the problem stipulation shows that there exists set play, thus there are at least two phases:

First phase: Looking at the position on the chessboard, we observe that, in some black moves the white has ready answers with mates. This is called set play.
* 1...Rb6 / Bb6 / Re2 / Be2
2.Sxc5# / Bxc6# / Qf3# / Qe3#

Second phase: We perform the key – move. If the black defending play the same moves, as in the first phase, and the white change the answers that give mate, this is a very good element of the problem.
Key: 1.Qxc5! (Black is in zugzwang situation).
1...Rb6 / Bb6 / Re2 / Be2
2.Qd4# / Qxc6# / Sf6# / Re3#
We see four changed mates on a double Grimshaw, which is a very important achievement of the composer.


Theme Mutate (Changed Mates) : In at least two phases of play, for the same defensive black moves, white changes the moves that give mate

Comment : The problem has some weak points:
(a) The key captures pawn c5, trying to achieve zugzwang situation.
(b) There is an unprovided check (1...Bxg4+), for which there is no set mate before the key. Thus the key (obviously) must protect Be5, so the move (2.Rxg4) can give mate.


Let us see now Grimshaw in three phases of the solution:


(Problem 39)
Mircea Manolescu,
First Prize, Meredith section ”Revista de Sah”, 1956
White plays and mates in 2 moves. (Set play).
* #2 (7+5)
[8/8/K4P2/1R1P2Br/p1k3S1/r7/3Q4/b7]

The solution of this problem has three phases. There is set play before the key, there is trial play after a try, and there is actual play after the key.

First Phase: (*) 1...Bc3 / Rc3 2.Se3# / Se5# (Grimshaw intersection at c3)

Second Phase: Try: 1.Sf2? [2.Qb4#] Rb3!
1...Bc3 / Rc3 2.Qd3# / Qf4# (the mates on Grimshaw intersection are changed)

Third Phase: Key: 1.Be3! [2.Rb4#]
1...Bc3 / Rc3 2.Qe2# / Qd4# (the mates on Grimshaw intersection are changed again)
1...Rb3 / Rd3 / Rxe3 2.Rc5# / Qb4# / Sxe3#

We saw here a frame of presentation of themes, which is called Zagoruyko, and rises considerably the value of the problem.


Frame Zagoruyko : In at least three phases and in at least two (always the same) answers of black, the continuations of white are changed.


In Problem-39 the composer Manolescu presented a three-phased Zagoruyko with Grimshaw.


[This post in Greek language].

The First Grimshaw

Walter Grimshaw was a composer of chess problems who has lived in the nineteenth century. In 1854 he was the winner of the first solving contest for chess problems in London, England.
The problemists are familiar with his name, from the theme Grimshaw intersection, which first appeared in the following problem of his, an orthodox more-mover (mate in five).



(Problem 37)
Walter Grimshaw,
”Illustrated London News”, 1850
White plays and mates in 5 moves
#5 (7+8)
[4r1r1/8/B2S4/1P1k2s1/1K2pPb1/2Q2p1s/2P5/8]

Tries: {1.Sf5? / Sf7? / Sxe8? / Qd2+? / Qc6+? / Qc5+? Ke6!}, {1.Qb3+? / Qc4+? Kxd6!}, {1.Qf6? Re6!}, {1.Qe5+? Rxe5!}.
Key: 1.Bc8! [2.Qc5# / Qd2#]
1...Bxc8
White bishop is sacrificed.
2.Qf6 [3.c4#].
Now appears the Grimshaw intersection : black can cut the line of the queen to d6, playing 2...Be6 (or even 2...Se6), but, doing so, black interferes with the rook guarding e5, allowing 3.Qe5#. Trying to stop this, black plays...
2...Re6
but this interferes with the bishop guarding f5, and the continuation is...
3.Qd4+ Kxd4
and so the white queen is lost...
4.Sf5+ Kd5
5.c4#
mate.


[This post in Greek language].

Friday, April 11, 2008

Multiple Grimshaw Intersections

The great composer Loshinsky created the following problems, each containing three Grimshaw intersections:



(Problem 35)
Lev Ilych Loshinsky,
”L' Italia Scacchistica”, 1930
White plays and mates in 2 moves
#2 (10+8)
[5S2/2K2B2/P1p3r1/1pk5/1S5R/2pP3b/r1Q4B/b4R2]

Problems having many tries make the life of solvers difficult, but here the situation is rather simple:
Tries: {1.Sd7+? Bxd7!}, {1.Se6+? Rxe6!}, {1.Sxg6? Bg4!}, {1.Bxg6? Bg4!}, {1.Rc4+? bxc4!}, {1.d4+? Kxb4!}, {1.Bg1+? Rxg1!}, {1.Bd6+? Rxd6!}, {1.Qxc3+? Bxc3!}, {1.Qf2+? Rxf2!}, {1.Rf5+? Bxf5!}.

Key: 1.Rb1! [2.d4#]
Now every defense creates a Grimshaw interference, and the black can not take the white piece that gives mate. The black defenses and the white continuations are:

Grimshaw intersection at e6
1...Re6 2.Sd7# (2...Bxd7 is not possible).
1...Be6 2.Bd6# (2...Rxd6 is not possible. Another black defense, 1...Rd6, results in a similar mate picture after 2.Bxd6#).

Grimshaw intersection at g4
1...Rg4 2.Se6# (2...Bxe6 is not possible).
1...Bg4 2.Bg1# (2...Rxg1 is not possible).

Grimshaw intersection at b2
1...Rb2 2.Qxc3# (2...Bxc3 is not possible).
1...Bb2 2.Qf2# (2...Rxf2 is not possible).


The second orthodox two-mover by Loshinsky presented here, is one of the famous chess problems because it is a complete block.

Theme Complete block : In the initial position, any black move can be answered with mate by the white.



(Problem 36)
Lev Ilych Loshinsky,
”Tijdschrift v. d. Nederlandse Schaakbond”, 1930
White plays and mates in 2 moves
#2 (6+7)
[b3K2b/r1R2p1r/4k3/1S1s4/2BP1Q2/8/8/8]

There are several tries: {1.Kd8? Bf6+!}, {1.Rc6+? Bxc6+!}, {1.Rxa7? Bc6+!}, {1.Rb7? Rxb7!}, {1.Re7+? Rxe7+!}, {1.Qd6+? Kf5!}, {1.Qe5+? Bxe5!}, {1.Qe4+? Kf6!}, {1.Qxf7+? Rxf7!}, {1.Qf5+? Kxf5!}, {1.Ba2? Rxa2!}.
Key: 1.Bb3!
The key preserves the blocked position, does not threatens anything, but black finds himself in a zugzwang situation. Black has six thematic defenses on three intersections and we see two simple Grimshaws and a pawn Grimshaw:

Grimshaw intersection at b7
1...Rb7 2.Rc6# (2...Bxc6 is not possible).
1...Bb7 2.Re7# (2...Rxe7 is not possible).

Grimshaw intersection at g7
1...Rg7 2.Qe5# (2...Bxe5 is not possible).
1...Bg7 2.Qxf7# (2...Rxf7 is not possible).

Pawn Grimshaw intersection at f6
1...Bf6 2.Qg4# (2...f5 is not possible).
1...f6 2.Qe4# (2...Be5 is not possible).

There are some more, not thematic, defenses of black :
1...f5 2.Qd6#, 1...Rxc7 2.Sxc7#, 1...Bxd4 2.Sxd4#.


[This post in Greek language].

Wednesday, April 09, 2008

Organ Pipes

Theme Organ Pipes: In the problem there is an arrangement of pieces (Bishop – Rook – Rook – Bishop), which is known as Organ Pipes. This arrangement forms four intersections and, theoretically, can have eight discrete thematic variations (with Grimshaw intersections).



(Problem 34)
Taverner T.,
First Prize, ”L’ arti”, 1881
White plays and mate in 2 moves
#2 (10+7)
[3brrb1/2S4B/8/2p4Q/2p2k2/5P2/4P1KR/2S2RB1]

In Taverner’s orthodox two-mover we see the arrangement B-R-R-B. We present with a condensed notation the various tries we make to solve this problem in two moves:
Tries: {1.Sd5+? Bxd5!}, {1.Se6+? Rxe6!}, {1.Sxe8? Bg5!}, {1.Qg4+? Ke5!}, {1.Qf5+? Rxf5!}, {1.e4? Bg5!}, {1.e3+? Rxe3!}, {1.Bd4? cxd4!}, {1.Sd3+? cxd3!}, {1.Rh4+? Bxh4!}, {1.Rh3? Bg5!}.

Key: 1.Rh1!
This move is not easy to find because it is not a threat, but brings black in a zugzwang situation, a situation where whatever black plays is bad for him, nevertheless black must play something.
For example, if black defend with 1...Bxh7, the square d5 stays unguarded, and white mates with 2 Sd5#. Or, if black plays 1...Re5, blocks this flight of the king allowing 2 Qg4#. If black could forfeit moving, white would not have a mating move.
Everyone of the nineteen legal answers of the black if followed instantly with mate.
The variations are : 1...c3 2.Sd3#, 1...Bxc7 2.Rh4#, 1...Bh4 2.Rxh4#, 1...Bg5 2.Qh2#, 1...Bf6 2.Qf5#, 1...Be7 2.e3#, 1...Rxe2+ 2.Sxe2#, 1...Re3 2.Bh2#, 1...Re4 2.fxe4#, 1...Re5 2.Qg4#, 1...Re6 2.Sd5#, 1...Re7 2.Rh4#, 1...Rf5 2.Qxf5#, 1...Rf6 2.Rh4#, 1...Rf7 2.Sd5#, 1...Bd5 2.Sxd5#, 1...Be6 2.e3#, 1...Bf7 2.Qf5#, 1...Bxh7 2.Sd5# .

The thematic approach in finding the solution is to observe that in the initial position black is almost in zugzwang.
If black is forced to play first, only two moves exist (Re3 and Bg5) without a mate. But, each of these moves blocks a critical flight of the black king. The move Re3 blocks the square e3 that is guarded by Bg1, and the move Bg5 blocks the square g5 that is guarded by Qh5.
After removing properly the rook from h2, white will be able to bring Bg1 or Qh5 on this square to give mate: 1...Re3 2.Bh2# and 1...Bg5 2.Qh2# .

The arrangement Organ Pipes (B-R-R-B) is well known to problemists, who understand that the black pieces will have mutual interferences during the solution. For example, see what happens when black play 1...Bf7. White now mates with 2.Qf5#, a move made possible only because the black bishop cut the action of black Rf8, which were guarding f5. We call self-interference what black B makes to black R.
Similarly, if black answers 1...Rf7, this creates interference on the guarding of d5 by Bg8. The result is that white gives mate with 2.Sd5#. Mutual interferences like this, between two black pieces (of dissimilar linear movement) on an empty square, characterize the Grimshaw intersections.
There are four Grimshaw intersections in this problem, (but the solution does not achieve the task of eight different mate pictures, it shows only four: Qf5#, Sd5#, e3#, Rh4#).

Grimshaw intersection on f7 : 1...Bf7 / Rf7 2.Qf5# / Sd5#
Grimshaw intersection on e6 : 1...Be6 / Re6 2.e3# / Sd5#
Grimshaw intersection on e7 : 1...Be7 / Re7 2.e3# / Rh4#
Grimshaw intersection on f6 : 1...Bf6 / Rf6 2.Qf5# / Rh4#

[This post in Greek language].

Monday, April 07, 2008

Grimshaw intersection

The classical chessmen (with the exception of knight and king) when examined in pairs, perform
Α. dissimilar linear movement, as in these pairs:
(1st) rook and bishop,
(2nd) rook and queen moving like a bishop,
(3rd) bishop and queen moving like a rook,
(4th) bishop and pawn making its first move with two steps
Β. similar linear move, as in these pairs:
(1st) rook and rook,
(2nd) rook and queen moving like a rook,
(3rd) bishop and queen moving like a bishop.


If the lines, on which move the two pieces that perform linear movement, have a common square, this square is called intersection.
If the first piece moves onto the free intersection, then it creates interference because it hinders the full action of the second piece. If the second piece, in another variation, can also go to the free intersection, then we observe mutual interference of the pieces.
We have various types of intersections. The Grimshaw intersection was first appeared in a chess problem of the composer Walter Grimshaw, and bears his name.


Theme: Grimshaw intersection : is the free square on which two black pieces with dissimilar linear movement are mutually interfered. Particularly the intersection with bishop and pawn is called pawn Grimshaw.
White Grimshaw is the free square on which two white pieces with dissimilar linear movement are mutually interfered.


If we put a white piece on a Grimshaw intersection to force the black pieces (by taking the white) to mutually interfere, then the intersection is called Nowotny intersection.
For interferences of black pieces with similar linear movement, we will see Holzhauzen intersection (simple interference), and Wűrzburg-Plachutta intersection (mutual interference), and Plachutta intersection (placing of a white piece on the intersection to force mutual interference of the black pieces).



(Problem 32)
A. G. Corrias,
”Good Companion”, 1917
White plays and mates in 2 moves
#2 (4+7)
[8/B2K3Q/5p2/3k4/2p2P2/p6p/r7/b7]

To make clear the theme of the Grimshaw intersection, we will study the solution of problem-32, by Corrias, with intersection of rook and bishop lines:

Key: 1.Qb1! [2.Qb7#] (Black has three ways of defense).
1...c3 (gives to the black king a new flight c4, but leaves d3 unguarded, allowing to white to give mate) 2.Qd3#
The other two defenses create thematic variations which present the theme Grimshaw intersection:
1...Bb2 (and the black bishop stops the white queen from reaching b7. But the bishop blocks one line of Ra2, which now can not reach g2 to stop the white from giving mate) 2.Qh1#
1...Rb2 (and the black rook stops the white queen from reaching b7. But the rook blocks one line of Ba1, which now can not reach e5 to stop the white from giving mate) 2.Qf5#

This mutual interference between two black pieces of dissimilar linear movement (here bishop and rook) on an intersection (the intersection here is the square b2) is the theme Grimshaw, which appears often enough in orthodox problems.



(Problem 33)
Frank Janet,
”St. Louis Globe Democrat”, 1916
White plays and mates in 2 moves
#2 (6+8)
[6b1/2s1p2p/8/8/r4kSK/4Sp2/2B2P2/3Q1s2]

In problem-33, by Janet, we see a Grimshaw intersection between a pawn, which makes its two-step initial move, and a bishop:
Key: 1.Qd7! [2.Qf5#] (The black may answer by stopping the queen from reaching its destination square f5, but two of the black defenses have the fatal defect of the mutual interference of two black pieces).
1...Be6 (interferes with pawn e7, which can not move to e5, as defense for ...) 2.Qxc7#
1...e6 (interferes with Bg8, which can not move to c4, as defense for ...) 2.Qxa4#
This mutual interference between bishop and pawn on an intersection (here the intersection is the square e6) is the pawn Grimshaw.

The other (not thematic) variations of the problem are : 1...Se6 2.Sd5#, 1...Ra5 2.Qd4#, 1...Sxe3 2.fxe3#, 1...Sg3 2.fxg3# .

[This post in Greek language].