No.626 (PT)

Pierre Tritten 


Original Problems, Julia’s Fairies – 2014 (III): September – December

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baloons7No.626 by Pierre Tritten – Continuation of the theme “one neutral Royal piece on the board”. It is also a Happy Birthday problem for Sébastien Luce! Congratulations! (JV)


Sentinels Pion neutre: When a piece (Pawn excluded) leaves a square outside the first and last rows, a white piece leaves a wP, a black piece leaves a bP and a neutral piece leaves a nP unless 8 Pawns of that color are already on the board.

Royal piece: Piece that executes a function of the King on the board.

No.626 Pierre Tritten

original – 18.10.2014
Dedicated to Sébastien Luce
for his 51st birthday

Hints by author: (click to show/hide)

neutral royal Bh6

h#2                                      (0+0+1n)
Sentinels Pion Neutre
Neutral Royal Bh6
b) Neutral royal Sh6 – h#3,5

Solutions: (click to show/hide)

4 Responses to No.626 (PT)

  1. Sebastien Luce says:

    Dear Pierre,
    what a nice surprise !!
    & thank you also to Julia to publish it the D day !
    I am very happy ! 🙂
    (Remember me your birthdays, I would be pleased to create something special also for you !)

  2. Dominique ForlotDominique Forlot says:

    for the fun, if tou like it, you can add a Twin with two solutions !

    b) Neutral royal Qh6 – h#2 Sentinels Pion Neutre & Einstein

    h#2 (0+0+1n)
    Sentinels Pion Neutre
    Neutral Royal Qh6

    1.nrQh6-f6=R[+nPh6] nrRf6*h6=Q[+nPf6] 2.nrQh6-g5=R[+nPh6] + nrRg5-g7=B[+nPg5] #
    1.nrQh6-g6=R[+nPh6] nrRg6-g7=B[+nPg6]+ 2.nPh6-h5 nrBg7-h8=S[+nPg7] #

    just to add a candle on the cake! Happy birthday to you :o)

  3. Dominique ForlotDominique Forlot says:

    in the same way, with “pat” Rather than “mat” we have another problem: Royal queen in h3

    b) Neutral royal Qh3 : h=2 Sentinels Pion Neutre & Einstein

    h=2 (0+0+1n)
    Sentinels Pion Neutre
    Neutral Royal Qh3

    two solutions:

    a) 1.nrQh3-f1=R[+nPh3] nrRf1-f3=B[+nPf1] 2.nrBf3-h1=S[+nPf3] nrSh1-g3=P[+nPh1] =
    b) 1.nrQh3-h1=R[+nPh3] nrRh1-h2=B[+nPh1] 2.nrBh2-g1=S[+nPh2] + nrSg1-f3=P[+nPg1] =

  4. Dominique ForlotDominique Forlot says:

    the case of “Royal-pawn” is intersting so ( with ReversEinstein here! )

    exemple: Neutral : Pg7 & h#3

    h#3 (0+0+1)
    Sentinelles Pion neutre
    Einstein inversé
    Neutral Royal Pg7

    1.nrg7-g6=S nrSg6-f8=B[+nPg6] 2.nrBf8-h6=T[+nf8] nrTh6×g6=B[+nh6] 3.nf8-f7=S nrBg6-h7=T[+g6]‡

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