|
| Page: [Previous] [Next] 1 2 3 4 |
| (41) Posted by Jacques Rotenberg [Saturday, Jul 7, 2012 21:52]; edited by Jacques Rotenberg [12-07-08] |
This h4# is out of the topic, I agree.
The e.p. is a nice feature that you can't find easily in a helpmate (I could find less than 60 in Winchloe in h4#).
It took me a while to understand that this is no more a retro. So I think that lighter is more important than a precise h2-h3 unique retro move (even if it is tricky).
|
|
| (42) Posted by Jacques Rotenberg [Monday, Jul 9, 2012 15:54] |
Others ways :
with promoted force :
 (= 3+10 )
h#3 (3+10)
1.b×c3 e.p.+ R×g8 2.Sa3 Rc8 3.Sb5 Rc4#
or more "normal" :
 (= 3+10 )
h#2 (3+10)
1.b×c3 e.p.+ h×g8=Q 2.a5 Qb3#
|
|
| (43) Posted by Jacques Rotenberg [Tuesday, Jul 10, 2012 16:12] |
12 !!
 (= 3+9 )
h#4 (3+9)
1.c×d3 e.p.+ K×c1 2.Ka3 Kd2 3.Bc3+ K×c3 4.Qa4 Bb2#
|
|
| (44) Posted by Nikola Predrag [Friday, Jul 13, 2012 05:55]; edited by Nikola Predrag [12-07-15] |
The last one is not only the most economic but has the richest line-play, including the sacrifice on c3 which amends the negative effects of the key. Including the critical retro-move Rc3xe3+, 3 pieces have visited square c3.
A much simplier example, a bit out of topic, shows mutual Zugzwang with virtual mates with wP on e2&e3 and on e4 the real one.
 (= 4+6 )
h#3
1.axb3 ep.+ Ka4 (2.tempo? axb3#; 2.B~? tempo? 3.Bc5 axb3#) - mutual/alternate Zz
2.d3! e4! (2...e3? 3.tempo? axb3#)
3.d4 axb3#
|
|
| (45) Posted by Jacques Rotenberg [Friday, Jul 13, 2012 15:46] |
nice idea, perhaps a little bit better like that :
 (= 4+6 )
with the try 1.Qa5+ Rxc4 2.Qa3/Qd2 ?? 3.Qd3 cxd3#
|
|
| (46) Posted by Miodrag Mladenović [Friday, Jul 13, 2012 15:52] |
This one is incorrect. The last move could be 1.Kb2-c3 and not 1.d2-d4. So there is no solution in this problem.
|
|
| (47) Posted by Jacques Rotenberg [Friday, Jul 13, 2012 16:24]; edited by Jacques Rotenberg [12-07-13] |
oops...
I don't think it worth adding too many things, the ideal mate is also a nice matter
|
|
| (48) Posted by Nikola Predrag [Saturday, Jul 14, 2012 00:01] |
Yes, ideal mate is the best feature, and not only as a formality. All pieces on the diagram play a role in the solution and all that remain participate in the mate. No extra material only to legalize ep. capture. bBc5 is actually a cookstopper, bPc5 would legalize ep. as well, but the order of moves would be dualistic.
|
|
| (49) Posted by Valery Liskovets [Saturday, Jul 14, 2012 19:17] |
Perhaps the following version with W's choice of the first move is some richer:
 (= 4+6 )
h#3
1.axb3 e.p. Ka4! 2.Qf3 e4 3.Qd3 axb3#.
One more modification:
 (= 4+6 )
h#3
1.axb3 e.p. Ka4! 2.Bb1 e4 3.Bd3 axb3#.
|
|
| (50) Posted by seetharaman kalyan [Saturday, Jul 14, 2012 20:47]; edited by seetharaman kalyan [12-07-14] |
Why not 2.Bc2?.
|
|
| (51) Posted by Valery Liskovets [Saturday, Jul 14, 2012 21:15] |
Oh, yes, my blunder. But some other worth modifications are still possible
(say, with bRd5, bQf5 and free sq.c5: 2.Qd3 e4 3.Rc5 axb3#).
|
|
| (52) Posted by Jacques Rotenberg [Saturday, Jul 14, 2012 22:40] |
With Rh5 you mean... ?
 (= 4+6 )
so that you can write :
1.Qb1? Kxa4 2.Rd5 ?? 3.Qb3+ axb3#
1.axb3 e.p. Ka4 2. Qc5? ?? 3.Rd5 axb3#
1.axb3 e.p. Ka4 2. Qd3! e4 3.Rc5 axb3#
but what you lose here is the black zugzwang that you have in the Nikola's position
|
|
| (53) Posted by Valery Liskovets [Sunday, Jul 15, 2012 00:49] |
Why Rh5 instead of d5? But anyway I agree. However, in my first position (with Qd5), losses are less:
seems only the try 2... e3? 3.zz (what looks commensurable with the attained check-avoidance).
|
|
| (54) Posted by Jacques Rotenberg [Sunday, Jul 15, 2012 01:28] |
with Rd5 : cooked in 2
1.Qb1 K×a4 2.Qb3+ a×b3‡
|
|
| (55) Posted by Nikola Predrag [Sunday, Jul 15, 2012 02:04] |
I have added the 'full' solution in post 44 showing alternete Zugzwang in 2nd move and virtual/real mates with 3 positions of white e-Pawn. Such a simple position offers retro-proof and tempo tries.
There are other possibilities, I didn't manage combine the following 2 positions into one where a try and solution would make a Grimshaw on c5. Perhaps more material would be justified.
 (= 3+7 )
h#4
 (= 3+8 )
h#4
|
|
| (56) Posted by Jacques Rotenberg [Sunday, Jul 15, 2012 08:05]; edited by Jacques Rotenberg [12-07-15] |
a simple way to do so would be to forget the e.p. ! example :
 (= 2+8 )
h#5
1.Rc5 Ka3 2.Rd5+ Ka4 3.Bc5 Ka5 4.Bd4+ Ka4 5.c5 cxb3‡
(a bit dry yet)
|
|
| (57) Posted by Frank Richter [Sunday, Jul 15, 2012 11:01]; edited by Frank Richter [12-07-15] |
Without bPb3 and wPb2 you get a second solution with Platzwechsel of the kings.
You saw this?
Yes, the same mating move isn't really nice.
|
|
| (58) Posted by Jacques Rotenberg [Sunday, Jul 15, 2012 11:37] |
yes... I saw it...
|
|
| (59) Posted by Nikola Predrag [Sunday, Jul 15, 2012 12:36]; edited by Nikola Predrag [12-07-15] |
Actually , I tried such Grimshaw in 1 solution in h#6, starting with ep., but the promotions of wPb4 finally conquered me.
Of course, ep. is not necessary for that Grimshaw, so it's probably not worth adding material to avoid cooks just to keep the retro-content.
|
|
| (60) Posted by seetharaman kalyan [Sunday, Jul 15, 2012 13:20] |
Actually this is quite good. Way above many published problems.
|
|
Read more... |
Page: [Previous] [Next] 1 2 3 4
MatPlus.Net  Forum Helpmates Bror Larsson, Springaren 1947 |
ISC 2024
