# No.1262 (MC&JML)

 No.1262 Michel Caillaud & Jean-Marc Loustau (Франция) Оригинальные сказочные задачи JF – 10/2017-3/2018: октябрь’2017 – март’2018

Определения: (показать/спрятать)

 No.1262 Michel Caillaud & Jean-Marc Loustau France original – 22.12.2017 Solution: (click to show/hide) white Kf2 Qf3 Sa3 Bb2 Rf4 BHa1h6 RHe8f1 Pc2c6g5 black Kd2 Sa5 Bh4 Gc1d1 RHb7d6e3g3 BHb1e1 Pg6 #2                                             (12+12) Grasshoppers c1,d1 Rookhoppers e8,f1,b7,d6,e3,g3 Bishophoppers a1,h6,b1,e1 a) Anticircé Calvet b) Anticircé Cheylan a) 1.Rf4-f8 ? {(1.Rf~?)} threat: 2.BHh6-f4 # but 1...Gc1-c3 ! {(and if 2 Sxb1+? RGxb1!) So:} 1.Rf4-b4 ! threat: 2.BHh6-f4 # 1...Gd1-d3 2.Kf2*e1 # {A (and not: 2 Sxb1+? Gxb1! Or 2 Qd1+? Gxd1!)} 1...BHb1-d3 2.Qf3*d1 # {B (interference of RGd6)} 1...Gc1-c3 2.Sa3*b1 # {C (and not: 2 Kxe1+? impossible! Or 2 Bc1+? Gxc1!)} 1...BHe1-c3 2.Bb2*c1 # {D (blocking of c3)} 1...RHd6*h6[bRHh6->h1] 2.Qf3*d1 # b) 1.Rf4-f8 ? {(1.Rf~?)} threat: 2.BHh6-f4 # but 1...RHd6*h6[bRHh6->h1] ! { So:} 1.Rf4-f6 ! threat: 2.BHh6-f4 # 1...Gd1-d3 2.Qf3-d1 # {B} 1...BHb1-d3 2.Sa3-b1 # {C} 1...Gc1-c3 2.Bb2-c1 # {D} 1...BHe1-c3 2.Kf2-e1 # {A (C+ by Popeye 4.79)} 4-fold Lacny theme combined with original twinning with 2 pairs of variations on the same squares (d3, c3) White Correction by the Rook f4, in the 2 phases, with different keys In position a): -> Anti-dual couple between variations 1... Gd3 and 1... Gc3 -> The couples of variations on d3 and c3 can be seen as 2 2nd degree couples (arrival correction), the primary harmful effect being respectively the interference of RGd6 and the blocking of c3 In position b): -> 4 specific Umnov variations (Authors)

### 4 комментария: No.1262 (MC&JML)

1. seetharaman пишет:

Superb !

2. Kjell Widlert пишет:

I love subtle twinning, and this is more subtle than most: not a change from one fairy condition to another, but a change between two forms of the same fairy condition! So the changed play (Lacný) is caused solely by the right for a piece to capture on its own rebirth square: legal or not?

The mechanism depends on the fact that b1-c1-d1-e1 are rebirth squares both for white attackers and for black defenders (because they are all fairy pieces). So part b (Cheylan) is fairly automatic: Black leaves a thematic square, so White is able to go there (no capture) but Black is unable to defend by going back (with capture). Part a (Calvet), however, has required constructional tricks: the defences must let in other thematic mates according to the Lacný pattern. This has succeeded very well.

A valuable extra feature is the fact that the two parts have different keys. Naturally, this differentiation is also caused by the difference between Calvet and Cheylan.

3. Kjell Widlert пишет:

(If you read between the lines, you can see that I too find the problem excellent!)

4. Jacques Rotenberg пишет:

Nice
a) better than b) but it does not really matter here .
Can be shown in a less conventional way:
2# anticirce
the solution being split in two
– if you mean “Cheylan”
– if you mean “Calvet”