Study 75

Árpád Rusz
Magyar Sakkvilág
2013
- version -

White wins

1.axb4+ Kxa4 1...cxb4 2.Kb3 bxa6 3.Rbxd6! exd6 4.Rh5+ d5 5.Rxd5# 2.bxc5 Rxa6 We have reached the starting position of the original 2013 version. 3.cxd6! 3.Rxa6+? bxa6 4.cxd6 exd6=; 3.Rxb7? Nxb7= 3...Rxb6 3...exd6 4.Rxb7+- 4.dxe7! 4.Rh4+? Kb5 5.dxe7 Ra6+ 6.Kb3 Ra8 7.Rh8 Ng6! 8.Rxa8 Nxe7= 4...Rxh6 5.exf8Q 5.e8Q+? Ka5! (5...Rc6? 6.Qa8+! Kb5 7.Qxb7+ Rb6 8.Qd5+ Ka6 9.Qa8++-) 6.Qxf8 Rc6 7.Qa8+ Kb6= fortress 5...Ra6! 5...Rc6 6.Qa8++- 6.Qc5 Rb6! cyclic zugzwang 1 WTM. 6...Ra5 7.Qc4#

WTM

This analysis uses the term "Cyclic Zugzwang" which is the generalization of the common 3-move (or 5 ply) long triangulation to pass the move to the weaker side. If the stronger side is to move in a "cyclic zugzwang" position, it wins more slowly than with the other side to move and in order to win, the stronger side can be forced to visit the same position but with the other side to move. So now white has to pass the move to the black. There are some R/Q corresponding squares: b6/c5, b5/c7, b4/e7 and it seems like the queen also has the e5 and d6 squares as possible "tempo squares".

7.Ka1!! 7.Qxb6? stalemate; 7.Qe5? Using e5 as a tempo square fails: 7...Kb4!= (After 7...Rb5? white would win by using the corresponding squares 8.Qc7! Rb4 9.Qe7! Rb6 10.Qc5!+-) 7...Rb5 7...Ra6 8.Kb1! (8.Kb2? Rc6! 9.Qa7+ Ra6 10.Qxb7 (10.Qc5 Rc6 perpetual attack) 10...Rb6+! 11.Qxb6 stalemate) 8...Rc6 9.Qa7+ Ra6 10.Qxb7+- 8.Qc7! 8.Qa7+? Kb4 9.Qd4+ Ka5 10.Ka2 Ka6 11.Ka3 Ra5+ 12.Kb3!? Rb5+ (12...b6? 13.Qd7 Rc5 14.Kb4 mutual zugzwang 14...Ra5 15.Qc8+ Ka7 16.Qc7+ Ka6 17.Qb8+-) 13.Kc4 Ra5!= (13...Rh5? 14.Qd3! Ka7 15.Qe3+ Ka6 16.Qe2! b5+ 17.Kb3 Rh3+ 18.Kb2 Rh4 19.Qe6+ Ka5 20.Qe1++-) 8...Rb4 9.Ka2 Rb5 cyclic zugzwang 2 WTM 9...Rb3 10.Qc2+- 10.Qd6! This is the right tempo square! 10.Qe7? Ka5= 10...Rb4 10...Rb3 11.Qd1+-; 10...Ka5 11.Ka3+- 11.Qe7! cyclic zugzwang 3 BTM 11.Qf8?! Ka5 12.Qc5+ Rb5 13.Qa7+ Kb4 14.Qa3+ Kc4 15.Qc1+ Kb4 16.Qc7 Ka4 17.Qd6 Rb4 is only waste of time 11...Rb3 This defence now becames possible because there are no more pins along the a4-d1 diagonal (there is no e0–square on the board)! 11...Rb5 12.Qa3#; 11...Rb6 12.Qc5! cyclic zugzwang 1 BTM. White finally managed to pass the move to the black.

BTM

12...Rb4 13.Qa7++-

Back to the main line we have this position:

12.Qf8! +- This second "Q-promotion on f8" wins! The other "promotion" 12.Qe8+?! Rb5 13.Qd8 Ra5 14.Qb6 Rb5 15.Qc7 Rb4 16.Qe7 Rb3 17.Qf8 is only waste of time.

The c5->d6->e7->f8 "road" is visited both by the white pawn and the promoted white queen!

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