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|(1) Posted by Evgeni Bourd [Thursday, Oct 12, 2006 22:56]|
Some of my latest ...
Hey all , I will show few of my latest works
Harmonie 2005? ( not sure , it might be 2004 )
(= 3+9 )
H#2 Tranmuted Kings , b)-Bf8
This one is from the times I composed some transmutingkings problems , I went for a "clear" grimshaw . You can see black has the option to move the piece but he chooses to use the Grimshaw option , dont get fooled by the meredith position it might give you a headache :)
1. Bf3! Kf6+ 2. Ka6 c8=B#
( 1. Rf3 Kd5+ 2. Ka6 c8=B#?? Rb7!
1. Ra2 Kf6+ 2. Ka6 c8=B#?? Bb7! )
1. Rf3! Kd5+ 2. Kg8 c8=R#
( 1. Bf3 Kf6+ 2. Kg8 c8=R#?? Qd8!
1. Bf1 Kd5+ 2. Kg8 c8=R#?? Rf8! )
The next is one of my many good recent Co works with Arie
Evgeni Bourd & Arie Grinblat
(= 12+13 )
Try: 1. Sac3? [2. Q:d5 [A] #]
But 1... b:c6 [a] !
Try: 1. Kf6? [2. Q:e5 [B] #]
But 1... Bf4 !
Solution: 1. Sc5! [2. Qf7! [3. Se6#]
1... b:c6 [a] 2. Q:d5 [A] +
2... c:d5 3. Se6#
2... K:d5 3. Rd8#
1... Bf4 2. Q:e5 [B] +
2... K:e5 3. Bg7#
2... B:e5 3. Se6#
1... c:b6 2. Kd6 [3. Q:e5, Q:d5#]
2... Bf4 3. Q:d5 [A] #
2... b:c6 3. Q:e5 [B] #
1... Bg6 2. Q:g6 [3. Se6#]
1... Bg8 2. Q:g8 [3. Se6#]
1... f:g4 2. Q:g4+
Dombrovsis , Keller , Rudenko , pretty much everything you want a "thematic" problem to have . There is even a sort of reciprocal changes between 2nd white moves and 3rd white moves in the Kd6 variation .
Rudenko School tourney 2004
(= 14+7 )
Try: 1. Bb2 [F] ? [2. Sf7 [B] #]
1... R:f2 [a] 2. Sf3 [A] #
1... Q:d6 2. f4 [D] #
1... Q:d4 [c] 2. Qd5 [C] #
1... Q:c4 2. S:c4#
But 1... R:d4!
Try: 1. b8=Q/B [E] ? [2. Sf3 [A] #]
1... R:f2 [a] 2. Sf7 [B] #
1... Q:d6 2. Qd5 [C] #
1... Q:d4 [c] 2. f4 [D] #
But 1... R:e3!
Solution: 1. Qf7! [2. Qf6#]
1... R:f2 [a] 2. Qe7#
1... Q:d6 2. Sf3 [A] #
1... K:d4 2. Bb2 [F] #
1... K:d6 2. b8=Q/B [E] #
This one is inspired by Gvozdjak's article about anti-cyclone , reciprocal changes , le grand and an almost 3x3 zaguruiko . if only there was an option to complete the zaguruiko...
The next is one of my favourites , not
because its good but because its funny !
(= 9+11 )
H#2 15?/5? sol
1. Ke5 Bg2
2. Qf5 Sd7, Sc6, S:f7, B:d6, d4#
2. Rf5 Sd7, Sc6, S:f7, B:d6#
2. Bf5 Sd7, Sc6, S:f7#
2. Sf5 Sd7, Sc6#
2. f5 Sd7#
very paradoxal and amusing problem , the stronger the piece less mates appear .
Hope you enjoyed and I would
ld like to hear some opinions !
|(2) Posted by Frank Richter [Friday, Oct 13, 2006 08:02]; edited by Frank Richter [06-10-13]|
the first problem was published in harmonie 83, September 2005. A very clear and clever composition, that will be highly awarded!
You should have received the issue, doesn't you?
|(3) Posted by Evgeni Bourd [Friday, Oct 13, 2006 10:27]|
Yes , I recieved the copy but forgot to write the year in my problemiste collection :(
|(4) Posted by Hauke Reddmann [Friday, Oct 13, 2006 17:06]|
Regarding your "amusing" h#, the first halfmove pair isn't
particularly funny :-) (and save the fact that the K unblocks
the block field, not even thematic).
I would have left it out altogether. And instead constructed
a sixth mate after K moves to block field...if that is
possible. (Since, as I see it, it must be a mate that
doesn't utilize the block, Black musn't have neutral moves,
and that will be a heinous construction task.)
|(5) Posted by Miodrag Mladenović [Friday, Oct 13, 2006 19:23]|
Yes, I do agree with Hauke about helpmate. It will be better problem as H#1. First move does not add any value. Idea is great and I never saw helpmate like this one.
I like faries and threemover a lot. #3 is an excellent problem and I think it'll win some prize.
About #2. The matrix for reciprocal changes where black defense selfpins and blocks a square next to the king is very known. However there are much better settings without flight taking try moves. In my opinion try move in twomover should not be taking flight from black king. It's so huge defect for me that I would say that tries should not be even mentioned with this problem. Yes I know that you do need them to complete some letter themes but as problemists we should not try to compose cycle at any price. There should be some basic principles that everyone follow up. Of course this is just my opinion. As a beginner I learned that in twomover try and key moves should never take a flight. It's sounds so logical that I never questioned that principle neither I ever tried to compose #2 with flight taking moves. Of course I know that there are many problemists that do not care about basic principles if problem is showing some complex letter theme. Sorry, I do like your problems a lot but not this one.
|(6) Posted by Harry Fougiaxis [Saturday, Oct 14, 2006 12:24]|
imho, you should have left out the last one. Certainly experimental and I smiled at first with the idea; if I were asked to guess the composer, I would bet to György Bakcsi at once! Yet, this is not a helpmate and at the end I felt quite uneasy with it.
I would love you had quoted the following joint with Shamir, instead. I was the judge in that section of the match and I think this masterpiece belongs to the top-10 of the modern h#2's composed in the last decade, or so.
1 Pl Israel-Macedonia 2005-06
(= 8+9 )
H#2 b) Kd4 to c4 (-pc4)
a) 1.Sd5! (Sc~?) Sd6! (Sd2?) 2.Qxe5 Scb5#
b) 1.Sc5! (Sb~?) Sb5! (Se2?) 2.Qxb4 Sed6#
The white pawn on c4 is, of course, absolutely necessary (the solutions work even without it), since otherwise 1.Sd5 in part (a) would be a square-block too, as Christopher Jones mentions in his very thorough analysis in the latest "The Problemist", Sep 2006, p.506.
|(7) Posted by David Knezevic [Sunday, Oct 22, 2006 16:09]|
RE h#2 with progressive separation:
Just for comparison, with no further comment by me
Il Due Mosse 1954
(= 10+9 )
|1.Be6! ~ 2.Sf7#|
1... Sd6: 2.Sg6/ef4/d4/Sd7:/Rc5#
1... ed6 2.Sg6/ef4/d4/Sd7:#
1... cd6 2.Sg6/ef4/d4#
1... Rdd6: 2.Sg6/ef4#
1... Rbd6: 2.Sg6#
1... Kd6: 2.Bf4:#
|(8) Posted by Hauke Reddmann [Sunday, Oct 22, 2006 20:27]|
Well, at least you could have commented that
it's ten times harder to do in a direct mate.
Stroke my ego as 2# specialist, dammit :-)
|(9) Posted by Siegfried Hornecker [Sunday, May 20, 2007 00:09]; edited by Siegfried Hornecker [07-05-20]|
The one given above by Evgeni Bourd (Shahmat 2004 in the first posting) reminded me of the following problem that is a classic (even by now):
(= 10+11 )
Ralf Krätschmer, 2001 (source?)
Mate in how many moves?
(I don't know if this was the original stipulation, so sorry if it isn't)
|(10) Posted by Hauke Reddmann [Sunday, May 20, 2007 20:08]|
Source must be 1st Price, Keym Theme Tournament!? (Should be in a
fairly recent SCHWALBE)
|(11) Posted by Kostas Prentos [Sunday, May 20, 2007 23:20]|
According to WinChloe, it's the 182 T.T. of Die Schwalbe 1999-2000, 2nd prize.
The judge was Werner Keym and the award appeared in Die Schwalbe 185 (Oct.2000).
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MatPlus.Net Forum Misc Some of my latest ...