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MatPlus.Net Forum Twomovers Miniature task? |
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| | (1) Posted by Michael McDowell [Wednesday, Oct 19, 2011 22:33] | Miniature task? A problem from a recently published award.
M.McDowell
1st HM, 2nd StrateGems Miniature Ty. 2011
(= 5+2 )
#2
1.Bf3
1…Kd3 2.O-O-O
1…d3 2.Bd1
I was surprised that this problem was honoured, as some time after sending it I discovered a problem by Klaus Exner, Badische Neueste Nachrichten 1996, which is identical except for a different key (1.Sf4-d5) and informed the judge, Bob Lincoln. If my version has any right to an independent existence it lies in the accidental feature that technically there are 9 tries, listed below, each of which leads to a different mating move which is forced at least once. Is 11 different forced mating moves a record for a two-move miniature? I doubt it.
1.Ra8? Kc1/d3 2.Rc8 1…Kd3!
1.Ra7? Kc1/d3 2.Rc7 1…Kd3!
1.Ra6? Kc1/d3 2.Rc6 1…Kd3!
1.Ra5? Kc1/d3 2.Rc5 1…Kd3!
1.Ra3? d3 2.Rc3 1…Kc1!
1.Ra2+? Kd3 2.Rd2 1…Kc1!
1.Rab1? d3 2.R4b2 1…Kd3!
1.Rc1+? Kd3 2.Rb3 1…Kxc1!
1.Bh3? Kd3 2.Bf5 1…d3! | | (2) Posted by Ian Shanahan [Thursday, Oct 20, 2011 08:38] | In defining the task, I'd want to see distinct refutations for the tries. (Many tries here are theoretical only.) | | (3) Posted by Hauke Reddmann [Thursday, Oct 20, 2011 14:03] | Miniature, schminiature. I can do 16 tries, all with
unique refutation (overall 10 equal+6 different others,
all moves are tries!) with 5 pieces:
(= 3+2 )
Where is your god now? :-)
Hauke | | (4) Posted by Michael McDowell [Thursday, Oct 20, 2011 17:52] | Ever the charmer.... | | (5) Posted by Kevin Begley [Thursday, Oct 20, 2011 18:01]; edited by Kevin Begley [11-10-20] | >I can do 16 tries, all with unique refutation (overall 10 equal+6 different others....
Good point, using the same accounting, there would be 16 tries here.
This proves Ian's point... in reality, there are only 7 tries with a unique refutation, here... if not less.
1.Rh~? but ...Rxa8! (9 non-unique versions of this try, which proves Ian's point: should count as only 1 try).
1.Rb8? but ...Rxb8!
1.Rc8? but ...Rxc8!
1.Rd8? but ...Rxd8!
1.Re8? but ...Rxe8!
1.Rf8? but ...Rxf8!
1.Rxg8+? but ...Kxg8!
The last "try" would be a dubious claim -- no threat, no extended variations (in fact no alternative but to capture the checking unit).
So, that leaves 6 unique tries.
But, 5 of the 6 may be reduced to a single try: 1.Ra~8? but 1...Rx~8!
That leaves just 2 "human" tries, of the 16 "computer" tries.
[or maybe 1 try: 1.~R~? but 1...Rx!]
Fairly obvious which number a judge should trust.
>Where is your god now? :-)
Went on a 730km journey... somewhere along the way, she must have taken a step backwards in time. | | (6) Posted by Kevin Begley [Thursday, Oct 20, 2011 18:19]; edited by Kevin Begley [11-10-20] | OK, but seriously... how can the judge justify this outrageous award (1st H.M.)??
He was informed of a virtually identical anticipation, with only a slight difference in the key (yielding no substantial improvement).
I think the judge may want to take a giant leap backwards in time. | | No more posts |
MatPlus.Net Forum Twomovers Miniature task? |
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