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(from TD 100) | by Nils Adrian Bakke | ![]() |
Chameleon echo models in S# max. and R#
by Nils Adrian Bakke, FIDE-master in composition 1990
What are echo model mates and chameleon echo model mates? We find the
answer by studying diagram 1. After the try 1. Qc4? (1. - h3!) Black may
play 1. - Kxh1 answered by 2. Qf1+ Kh2 3. Bf4#. In the solution 1. Qf5!, 1. - Kxh1 is
answered by 2. Qh3+ Kg1 3. Be3#. These two model mates are so-called echo model
mates.*)
What is the difference, then? In echo model mates the black king occupies squares of the
same color, while in chameleon echo model mates the black king changes from a dark to a
light square (or vice versa!). In this example by Fossum we note that in the first
chameleon model wQ occupies a dark square and bK a light one. In the second chameleon
model this is switched: wQ mates on a light square the bK which now stands on a dark
one.
In ordinary selfmate problems this is an everyday occurrence. The Czechs have
investigated the possibilities here thoroughly. In selfmate maximummers and reflex-mates,
however, this is virgin soil. Diagram 2 is a most acceptable showing.
Probably this is the only ordinary selfmate maximummer with triple chameleon models.
Solutions to diagram 2 and the following diagrams are placed at the end of the
article.
Diagram 1 André Fossum 14. All-round 1973/74 ![]()
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Diagram 2 Nils Adrian Bakke Norsk Sjakkblad 1999 Tilegnet Jiri Jelinek ![]()
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Diagram 3 Nils Adrian Bakke The Problemist 1999 Tilegnet John Beasley ![]()
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Diagram 4 Nils Adrian Bakke Original ![]()
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Then on to the reflex-mate problems. As early as 1975 I made Diagram 5,
a reflex-mate miniature with chameleon echo. Unfortunately it had just two echo mates,
but it was an easy task to remodel it, so it now shows three chameleon model mates.
Diagram 6 was a sensation when it was published in THE PROBLEMIST. The
Russian composer, who is not quite unknown, managed to show three chameleon mates in
a miniature. Note that wK, wRR, and bP all participate in the chameleon echo model
mates.
Diagram 5 Nils Adrian Bakke V. Feenschach 1975 ![]()
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Diagram 6 Viktor Chepizhny The Problemist 1997 ![]()
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Diagram 7 Nils Adrian Bakke Norsk Sjakkblad 1998 ![]()
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In Diagram 7 I had to employ two promoted wSS, but the position is legal! I wouldn't exactly claim that it was easy to compose this problem! |
Solutions:
Diagram 1.
Try: 1 Qc4? Kxh1 2 Qf1+ Kh2 3 Bf4#.
1 - Kg1 2 Qxh4 Kf1 3 Qe1#, but 1 - h3!
Solution: 1 Qf5! Kxh1 2 Qh3+ Kg1 3 Be3#.
1 - Kg1 2 Qh3 Kf2 3 Qg2#;
1 - h3 2 Be3 Kxh1 3 Qxh3#.
Diagram 2.
1 Rf6! a5 2 Qe1 Qxe1 3 Kb5 Qb4#.
1 - b5 2 Qf1 Qxf1 3 Rcf2 Qc4#;
1 - e5 2 Qb2+ Qxb2 3 Kd5 Qd4#.