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Problem 537 & 537.1: Gabor Tar & Nikola Predrag - Helpmate
gabor.tarnikola.predrag(21.02.2015) Enjoy another nice helpmate from Hungary! There is a traffic jam on 'd5' ! 
 
(24.02.2015) As desired by Gabor Tar, happy to publish the improved version suggested by Nikola Predrag as their joint effort.

 

 

 

 

537. Gabor Tar (Hungary)
21.02.2015
518
   H#2          4 sol.         (8+11)
 
1.Bxd5 Kh5 2.Be4 d4#
1.cxd5 Sd6 2.d4 Sxf7#
1.Rxd5 Sd4 2.Rd6 Sf3#
1.Sxd5 Bd8 2.Sf6 Bxc7#

 

537.1. Gabor Tar (Hungary) &
Nikola Predrag (Croatia)
24.02.2015
537.1
H#2         5 Sol         (8+12)

 

1.Sexd4 Bd7! (Be2?) 2.Sf5 Bxc6#
1.Sbxd4 Be2! (Be7?) 2.Sf3 Bd3#
1.Bxd4 Kh4 2.Be3 d3#
1.cxd4 Sd5 2.d3 Sxf6#
1.Qxd4 Sd3 2.Qd5 Sf2#
 

Comments  

 
+1 #1 Kenneth Solja 2015-02-21 14:33
Hi again,
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.
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+1 #2 Kenneth Solja 2015-02-21 14:53
2nd 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!
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0 #3 Seetharaman Kalyan 2015-02-21 16:22
Good attempts Kenneth Solja. You have indeed eliminated the uneconomical black rook 'h7'.

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.
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+1 #4 Kenneth Solja 2015-02-21 21:26
3rd version (pure version as Mr. Kalyan says)
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!
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+1 #5 Diyan Kostadinov 2015-02-22 00:47
Usually I don't like to make versions, but actually Kenneth is right that the problem can be improved. Here is one possible way:


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.
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0 #6 Seetharaman Kalyan 2015-02-22 03:39
Good versions Kenneth and Diyan! Of course I would go for pure selfblocks if possible.
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+1 #7 Francesco Simoni 2015-02-22 08:06
Since there isn't dual avoidance as possible unity motif of the solutions, the purity of self-blocks isn't necessary, in my opinion. So I prefer the research of economy, like in Diyan version.
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+1 #8 Kenneth Solja 2015-02-22 10:26
I like both Diyan's versions. Usually I don't do versions, but here the only reason was to get rid off the black pieces which are only taking cooks.
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0 #9 Nikola Predrag 2015-02-22 13:33
A rough scheme for increasing the thematic content:
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)
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+1 #10 Vitaly Medintsev 2015-02-22 15:06
The position presented in initial Nikola's version is illegal, but the other one is correct: 8/2pb2r1/2p1nbNp/1np4P/3PkNB1/1pP4K/3P3P/6q1
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0 #11 Seetharaman Kalyan 2015-02-22 15:19
Bravo Nikola! Ingenious idea to add the extra bishop mate and differentiate between the two black knight captures on d4. Quite a task now! Over to Gabor Tar!

Here is the diagram for Nikola's version. The bishop can be at c8.
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+2 #12 Nikola Predrag 2015-02-22 18:05
Well, yes, 9+15 is illegal, sorry.
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.
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+3 #13 Gabor Tar 2015-02-22 19:07
Dear Seetharaman!

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
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+1 #14 eugene david rosner 2015-02-24 14:14
This is "crazy-good" now! I love the waiting 1...Kh4.
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0 #15 Seetharaman Kalyan 2015-02-24 16:37
Quoting eugene david rosner:
This is "crazy-good" now! I love the waiting 1...Kh4.

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!
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+1 #16 Rodolfo Riva 2015-02-24 17:20
Congratulations to both authors!

In my opinion it is better not to emphasize the "unhappy tries" (Be2?/Bd7?).
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0 #17 Seetharaman Kalyan 2015-02-25 06:35
Quoting Rodolfo Riva:
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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