## Wednesday, August 29, 2012

### 5-phasic KoBul tasks

The condition KoBul kings is a fairy condition, (that is with this condition the problem is not orthodox any more), proposed by the composer Diyan Kostadinov from Bulgaria. (KoBul = Kostadinov Bulgaria). With this condition, the kings change abilities of moving and capturing, acquiring the abilities of the friendly piece most recently captured. The kings return to the abilities of a simple King when a friendly pawn is captured.

The condition has got interesting strategic applications, which become apparent just as the kings acquire super-powers(!), but soon the time for the checkmate comes. In the present post we are interested only for compositions where the king is mated in five of his phases, (in the final pictures of the five checkmates appears as king, as queen, as rook, as bishop, as knight), which we have named as 5-phasic.

Let suppose that bK is going to be mated.
In the initial position he may be in king's phase (bK), or already in another phase (bQK or bRK or bBK or bSK), and we need one move less. We may have Forsberg twins, where for each twin the bK changes phase on its square.
The bK can change phase when a black piece, already present on the board, is captured or a black pawn might have previously been promoted to a black piece (and that means we need one more move). We may have Forsberg twins, where for each twin a black piece changes type (bQ or bR or bB or bS or bP) on its square.
The promoted piece can be captured on the promotion square or after it moves to another square, (and that means we need one more move).

Let us see now some 5-phasic problems by Greek composers.

 8/1p2R3/7P/1P1P3p/3k4/3p3P/p2Pp1p1/1b6 (6 + 8) Problem-611 Emmanuel Manolas 4th Honourable Mention, Thematic Tourney 1, KoBulchess.com, 2012 h#3, KoBul kings (wRKe7 + bKd4) 5 solutions 1.g1=Q h7 2.Qg8 hxg8=Q(bK=bQK) 3.QKa1 Qg7# 1.g1=R h7 2.Rg8 hxg8=Q(bK=bRK) 3.RKh4 Qg4# 1.e1=B RKxe1(bK=bBK) 2.BKa7 RKe8 3.b6 RKa8# 1.e1=S RKxe1(bK=bSK) 2.SKc2+ RKc1+ 3.SKb4 RKc4# 1.Kc5 d6 2.Kb6 d7 3.Kc7 d8=Q# 5-phasic task 2K3b1/1P5s/4P1P1/pP6/3k4/p7/2P1R2q/2r5 (7 + 7) Problem-612 Emmanuel Manolas 2nd Prize KoBulchess.com July 2012 h#2, KoBul kings (wKc8 + bKd4) 5 solutions 1.Qg2 Rxg2(bK=bQK) 2.QKh8 g7# 1.Bxe6+ Rxe6(bK=bBK) 2.BKa7 b6# 1.Rxc2+ Rxc2(bK=bRK) 2.RKa4 Rc4# 1.Kc5 gxh7(bK=bSK) 2.a4 Re5# 1.Kc5 Re5+ 2.Kb6 b8=Q# 5-phasic task s6b/r4p1p/2k5/Pp4P1/1PpR3P/1p1P3K/6P1/8 (8 + 9) Problem-613 Ioannis Garoufalidis original h#2, KoBul kings (wKh3 + bKc6), 5 twins, a) diagram : 1.Sc7 a6 2.Kb6 Rd6# b) with bKQc6 : 1.QKg6 Rd6+ 2.QKh5 g4# c) with bKRc6 : 1.cxd3 Rd5 2.RKc3 Rc5# d) with bKBc6 : 1.Re7 a6 2.Re8 Rd6# e) with bKSc6 : 1.SKe7 Re4+ 2.SKg8 Re8# The phase-changing of bK takes place before the 1st black move. 5-phasic task

paul said...

Problem 613: In four of the five phases is no need of Kobul rules.

Emmanuel Manolas said...

Thank you Paul for your justified comment.
We may suppose that the changing of phase for the bK has happened before the move B1. For example, in the position [s6b/r4p1p/2k5/Pp4P1/1Ppp2RP/1p1P3K/6P1/8] with h#2.5 and with Forsberg changing of the piece on d4, the five solutions are close enough to full use of KoBul rules, including KoBul captures. (But, in two cases we have two transpositions of the moves. Have you any ideas for a version?)
Accepting only the white move Rxd4 as last white move, we observe unique mating of the bK in all his phases, which was the subject of this post.