# No.656 (PH)

 No.656 Peter Harris(South Africa) Original Problems, Julia’s Fairies – 2014 (III): September – December →Previous ; →Next ; →List 2014(III) Please send your original fairy problems to: julia@juliasfairies.com

No.656 by Peter Harris – Five solutions after an interesting addition to No.649! But might it be considered as an independent problem? (JV)

Definitions:

Imitator(I): Every time a piece moves an Imitator (or a set of Imitators) moves simultaneously in an identical manner. An Imitatorcannot move of itself. If an Imitator cannot imitate the move of a piece, the move is illegal. An Imitator may only pass through or enter an unoccupied square and cannot move off the board. Castling is imitated by decomposing into a King move followed by a Rook move.

Maximummer: Black must play the geometrically longest move or may choose from among longest moves of equal length, distances being measured from the center of each square. Diagonal and oblique distances are measured from the orthogonal coordinates by using Pythagora’s theorem (take the square root of the sum of the squares of the orthogonal distances). All other orthodox chess rules apply. White Maximummer: Only White must play the geometrically longest moves, Black plays orthodoxal.

 No.656 Peter HarrisSouth Africaoriginal – 30.11.2014Dedicated to Pierre Tritten. INTRODUCTION. white kc2 rg8 bd1 black kb5 pb4e2 neutral If3 h#3                                     (3+3+1i)b) Rg8→a3; c) Rg8→f4;d) Rg8→b7; e) Rg8→d7Imitator f3White Maximummer This is your 649 position Pierre. Maxi [White or Black] adds something magical to the Imitator. White has to mate with a Maxi move. (Author) Solutions: (click to show/hide) a) 1.Kb5-c6[Ig4] Rg8-c8[Ic4] 2.e2-e1[Ic3]=S + Bd1-f3[Ie5] + 3.Kc6-c5[Ie4] Bf3-b7[Ia8] # b) wRg8-->a3 1.e2*d1[Ie2]=B + Kc2-d3[If3] 2.Bd1-a4[Ic6] Ra3-c3[Ie6] 3.Kb5-a5[Id6] Rc3-c5[Id8] # c) wRg8-->f4 1.e2*d1[Ie2]=R Rf4-f8[Ie6] 2.Rd1-d2[Ie7] + Rf8-b8[Ia7] + 3.Rd2-e2[Ib7] Kc2-b3[Ia8] # d) wRg8-->b7 1.e2-e1[If2]=Q Bd1-f3[Ih4] + 2.Kb5-a4[Ig3] Bf3-c6[Id6] 3.Qe1-c3[Ib8] Rb7-h7[Ih8] # e) wRg8-->d7 1.Kb5-a5[Ie3] Rd7-a7[Ib3] + { 2.e2-e1[Ib2]=I Bd1-g4[Ie5,h4] + 3.Ka5-a4[Ie4,h3] Bg4-c8[Ia8,d7] # (Animation is not supported for promotion into Imitator (JV)) (C+ by Popeye 4.69)}

### 5 Responses to No.656 (PH)

1. peter harris says:

With regard the question implying that Tritten should perhaps be named as co-author:

The following changes were made to Tritten’s problem:

(1) A piece was removed – bPh4
(2) The position of another piece was changed – wRc3
(3) The stipulation was changed – h#2 > h#3
(4) A condition was added – + Whitemaxi

HUNDREDS of other problems could be made by making 4 changes similar to the above, for example,

(1) Remove wBd1
(2) Move bPb4
(3) Make stipulation hs#4

Should all such problems have Tritten as co-author? – of course not.

One should not be influenced by my dedicating my problem to him. This was by way of a goodwill gesture.

2. Pierre Tritten says:

Well, I’m definitely not a co-author of this problem!
I like the unity of the twinning and the unusual battery mates.
Showing Super-AUW in this way is certainly a great achievement, thank you Peter for the dedication!

• Kjell Widlert says:

Agreed: the famous Many Fathers problem (Vielväter) has already shown that content can be totally different even when the position is exactly the same. There is much similarity in the positions of 649 and 656, but the play is completely different. So 656 is definitely an independent problem (and the dedication seems to me quite appropriate).

3. luce says:

Congratulations Peter,
it is a great problem !
SL

4. seetharaman says:

Very ingenious ! To produce something totally different from a known position takes great imagination and skill. Peter Harris works magic with the position of 649 to produce something quite different! Great problem !