Problem 537 & 537.1: Gabor Tar & Nikola Predrag - Helpmate |
![]() ![]() (24.02.2015) As desired by Gabor Tar, happy to publish the improved version suggested by Nikola Predrag as their joint effort.
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Comments
I made following version of this problem:
W:Kh4,Lg5,Sf5,Sg7,bc4,bd3,bd5,bh3=8.
B:Ke5,Dg2,Th7,Lf7,Sa8,Sb4 ,bc2,bc6,bc7,be3,bh6=11.
H#2, 4 solutions C+
1.Dxd5 Sd4 2 Dd6 Sf3#
1 Sxd5 Lxe3 2 Sf4 Ld4#
1 cxd5 Sd6 2 d4 Sxf7#
1 Lxd5 Kh5 2 Le4 d4#
But from this position I still would like get rid off Th7 and Sa8,so this is only the first version. I also prefer Ld4-mate instead of Lc7-mate.
In the original there was just black pieces which one want to get off the board, so because of this this version.
W:Kh4,Lg5,Sf5,S g7,bc4,bd3,bd5,bh6=8.
B:Ke5,Dg2,Lf7,Sa8,Sb4,bc2 ,bc6,bc7,be3,bh7=10.
H#2, 4 solutions C+
1.Dxd5 Sd4 2 Dd6 Sf3#
1 Sxd5 Lxe3 2 Sf4 Ld4#
1 cxd5 Sd6 2 d4 Sxf7#
1 Lxd5 Kh5 2 Le4 d4#
Now we have got rid off bTh7!
In the composers versin, none of the mating squares are initially guarded. Therefore all the four self-blocks are pure.
In your version the black queen variation unguards 'f3' disturbing this important unity. This is an avoidable defect. So personally I would prefer the original.
W:Kh4,Lg5,Sf5,Sg7,bc4,bd3,bd5,bh3,bh6=9.
B:Ke5,Ta5,Tg2,Lf7,Sa8,Sb4 ,bc2,bc6, bc7,be3,bh7=11.
H#2, 4 solutions C+
1.Txd5 Sd4 2 Td6 Sf3#
1 Sxd5 Lxe3 2 Sf4 Ld4#
1 cxd5 Sd6 2 d4 Sxf7#
1 Lxd5 Kh5 2 Le4 d4#
But still we don't want bSa8!
Four black pieces (including 2 Rooks) are removed.
Or if we want "pure selfblocks" as Ram said, so this can be like that:
Here if we want to avoid the capture of the bPe3, can be used wPd2 instead it. Actually it is a matter of taste.
W:Sg6,Ph5,Pd4,Sf4,Bg4,Pc3,Kh3,Pd2,Ph2
B:Bc8,Pc7,Pf7,Rg7,Pc6,Se6 ,Bf6,Ph6,Sb5,Pc5,Ra4,Ke4, Ph4,Pb3,Qh1
Stipulation H#2; 5.1.1.1. (9+15)
1.Ra4*d4 Sf4-d3 2.Rd4-d5 Sd3-f2#
1.Sb5*d4 Bg4-e2 2.Sd4-f3 Be2-d3#
1.c5*d4 Sf4-d5 2.d4-d3 Sd5*f6#
1.Se6*d4 Bg4-d7 2.Sd4-f5 Bd7*c6#
1.Bf6*d4 Kh3*h4 2.Bd4-e3 d2-d3#
bQh1 to g1 & bBc8 to d7 would spare bRa4,bPf7,bPh4 (9+12)
Here is the diagram for Nikola's version. The bishop can be at c8.
In the legal example, wPh2 is not needed and bB indeed could be on c8.
Thanks Vitaly and Seetharaman.
I was not careful about the "details", I just wished to encourage Gabor Tar to increase the quantity of the task.
The scheme could be perhaps improved.
I ask Predrag ?r, let there be a co-author
the No.to 537 problems.
Soils me, if he this he accepts it.
Thank You very much:
Gabor Tar
Indeed it is a surprising tempo move! This feature is of course there in the original version. Nikola's idea is to add the second knight variation and to differentiate it beeautifully!
In my opinion it is better not to emphasize the "unhappy tries" (Be2?/Bd7?).
Well, the authors did not mention it. I added it to highlight to dual avoidance effect.
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