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| | | (1) Posted by Hauke Reddmann [Monday, Oct 5, 2009 10:37] | Style
Does problem chess allow individual style such that
you can easily tell a Mr. Foo from a Mr. Bar?
Well, here is some anecdotal evidence. When I once judged
the Schwalbe 2#, of course I tried to guess the composers.
I guessed prize 1, 2 and 3 correct, so much for neutralization :-)
Of course there were several facts in my favor:
- I am a keen pattern spotter anyway.
- The Schwalbe, even though an international mix, naturally gets
more German stuff, reducing the number of usual suspects.
- The themes are a strong giveaway.
- My favorites were of course top-notch problems from likewise
top-notch composers (otherwiese I'll be a lousy judge :-)
It would be interesting to hear if other judges (especially from
other areas - can you tell a Keym from a Caillaud from a Plaksin
retro?) can repeat my little trick.
Hauke
| | | (2) Posted by Frank Richter [Monday, Oct 5, 2009 13:12] |
I think, I could distinguish a selfmate by Gamnitzer from a selfmate by Petkov.
But in generally for this task some background information (as you described) is needed, I would say.
| | | (3) Posted by Kevin Begley [Monday, Oct 5, 2009 22:58]; edited by Kevin Begley [09-10-05] |
Sometimes, it seems only a matter of who landed first.
The following is perhaps not a fair survey for this audience, because it contains some highly memorable problems of four exceptional composers. Nevertheless, I have always found a certain shared-style streaming through many of their problems. One may argue that the shared-pattern is more about refinement than style, per se; but refinement can be a style too, and certainly is a component of author-recognition.
That said, match the following list of Authors to the list of Problems (for those who do not remember, good luck!):
Authors (sorted alphabetically, by last name):
a) Feather, Chris
b) Jonsson, Christer
c) Meinking, Dan
d) Wiehagen, Rolf
Problems (sorted by year):
w) 1994
 (= 5+7 )
h#3*
x) 1995
 (= 3+2 )
h#3.5 2.1.1...
y) 1999
 (= 3+9 )
h#4 2.1.1...
z) 2001
 (= 3+12 )
h#3 4.1.1...
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