The solution is 1.Rd4+ Ke5 2.Rxg4 Kd5 3.Rd4 Ke5 4.Rxh4 Kd5 5.Kd4 Ke5 6.Rc4 Kd5 7.Rc3 g1=Q 8.Rd3 Qd4 9.Rxd4…
On Conflictio @ Articles
The problem by O. Schmitt can be done without black Bishop, as follows: 8/5p2/3p4/pK1k1pN1/P1R3pp/1p6/1B2P1p1/4rn1r #15 1.Rd4+
On Conflictio @ Articles
Probably all hs= , ser.s= and auto= can be designated as reaching dead positions.
On No.1784 (JM&AB)
In the try in Phase (b) 2...Rf4+ also impossible due to the fairy condition
On No.1783 (AT)
If I counted the solutions correctly the total is 12. WWW = 1 solution WBW = 2 solutions BBW =…
On No.1765 (NSR)
Many excellent problems. Congratulations!
On Tadashi Wakashima 70 JT Award
Thanks Maryan. At least two phases. stipulation h# direct mate, s# or hs# One phase must be Superguards. Other phase(s)…
On SEETHARAMAN-75 JUBILEE TOURNEY FOR SUPERGUARDS
Superguards 1)are other fairy conditions allowed? 2) any stipulation is allowed?
On SEETHARAMAN-75 JUBILEE TOURNEY FOR SUPERGUARDS
Nice idea. It will be more interesting if the black Pao made critical moves from g5, perhaps to selfblock on…
On No.1778 (HM)
Witty trick with the “neutral” rook that after the first half-move entirely forgets about its neutrality, playing only for White.…
On No.1769 (KW)