# No.1058 (GF)

 No.1058  Geoff Foster (Australia) Original Problems, Julia’s Fairies – 2016 (I): January – June   →Previous ; →Next ; →List 2016(I) Please send your original fairy problems to: julia@juliasfairies.com

Definitions: (click to show/hide)

 No.1058 Geoff Foster Australia original – 26.04.2016 Dedicated to Peter Harris Solutions: (click to show/hide) black Kc7 neutral Sd4 neutral Chameleon Se2 h#3            b) Kc7→a4       (0+1+2) Chameleon Circe Neutral Chameleon e2 a) 1.ncSe2*d4=ncB[+nBc1] nBc1-b2 2.ncBd4*b2=ncR[+nRa1] nRa1-a6 { } 3.nRa6-b6 ncRb2*b6=ncQ[+nQd8] # b) bKc7-->a4 1.nSd4*e2[+ncBf1] ncBf1*e2=ncR[+nBc8] 2.ncRe2-e6=ncQ ncQe6*c8=ncS[+nRa8] + { } 3.ncSc8-a7=ncB ncBa7-b8=ncR # { (C+ by Popeye 4.73)} In (a) the chameleon makes 3 captures and the pieces transform SS>BB>RR>QQ! In (b) the chameleon transforms S>B>R>Q>S>B>R. (Author)

### 6 Responses to No.1058 (GF)

1. Luce Sebastien says:

Another nice idea by Geoff,
the variation a) is the most spectacular !

2. Luce Sebastien says:

…but Winchloe gives some cook in a) and no solution in b) 🙁
a)
1.CAn×d4(F)(Fnc1) Fnb2 2.CAn×b2(T)(Tna1) Tna6 3.Tnb6 CAn×b6(D)(Dnd8)‡
1.CAn×d4(F)(Fnc1) Fnf4+ 2.Rb7 Fnc7 3.Ra8 CAna7(T)‡
1.Rb8 CAn×d4(F)(Fnf8) 2.Fnc5 Fna7+ 3.Ra8 CAn×a7(T)(Tnh8)‡
1.Rb8 CAn×d4(F)(Fnf8) 2.Ra8 Fnc5 3.Fna7 …
1.Rb8 CAn×d4(F)(Fnf8) 2.Ra8 Fnd6 3.Fnc7 …
1.Rb7 CAn×d4(F)(Fnf8) 2.Fnc5 Fna7 3.Ra8 …
1.Rb7 CAn×d4(F)(Fnf8) 2.Fnd6 Fnc7 …
1.Rb7 CAn×d4(F)(Fnf8) 2.Ra8 …
b) ???

3. Geoff Foster says:

I wrote to Christian Poisson about what I considered to be a bug in WinChloe. He replied: “actually, the bug is in Popeye, as it considers that a Chameleon is an orthodox piece, but is not, it is a fairy piece, and fairy pieces don’t transform in Circé caméléon”.

The “cooks” found by WinChloe in (a) have the bK on a8 in check from a Chameleon Rook on a7, with 4.Kxa7 being illegal because the Chameleon Rook would remain as a Rook and be reborn on a1. WinChloe did not find any solution in (b) because the Popeye solution has captured Chameleons being transformed before being reborn.

I don’t agree with Christian, because (unlike most other fairy pieces) a Chameleon has a natural sequence of transformations. The solution to (a) seems especially natural, with both pieces being transformed with each capture.

There have been other Circe problems using Chameleons, and WinChloe has a special condition called “Renaissance comme la pièce orthodoxe correspondante” to handle them. Unfortunately this condition has not actually been programmed and Christian has no plans to do so.

4. Geoff Foster says:

I made a slight mistake above. In WinChloe’s “cook” in (a), after 4.Kxa7 the captured Chameleon Rook would be reborn on a8 (the promotion square of a fairy piece), not a1. This “cook” is not found by Popeye because the Chameleon would be transformed to Queen and be reborn on d1.

5. peter harris says:

Thank you for the dedication Geoff.

Combining ChameleonChess and ChameleonCirce is like being in the Wild West! – it is difficult for there to be any law and order – so beloved by chess problem critics. (I think though that a little disorder in a problem makes life more interesting).

I see you eschewed ChameleonChess and have the one Chameleon piece.

I do not know whether it is a minus that the non-Chameleon piece makes no captures (to activate ChameleonCirce) in (a) and in (b) only moves once. Perhaps it is not!

You mention the patterns: the pairs in (a) and the sequence in (b). The existence of patterns cannot sustain a problem. They are often accidental to the play and not readily discernible by solvers.

I had never used the ChameleonCirce condition before 1053, 1054 1055. The only Circes I have ever used are Circe, SuperCirce, AntiCirce and AntiSuperCirce. So I have much to learn about ChameleonCirce.