Julia's Fairies


Eric Huber

Original Fairy problems
July – December’2019

Definition: (click to show/hide)

No.1463 Eric Huber

original – 15.12.2019

Solutions: (click to show/hide)
White Ke1 TIf1 TIh8 TIg3 TIa1 TIc3 Black Kg4 Pf4h3 Neutral Qh1 Pg2

h#2.5          2 solutions        (1+3+2)
4 Total Invisibles

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Georgy Evseev
Georgy Evseev
December 17, 2019 07:09

As there are neutral pieces on the board, the TIs may be neutral (why not?) and then second solution does not work, as the final TI may be a neutral rook (not a white one).

December 17, 2019 08:43
Reply to  Georgy Evseev

Perhaps it should be stipulated that as only neutral pawn and queen are there the TI can only be that, not a neutral rook !!

Thomas Maeder
Thomas Maeder
December 17, 2019 20:08

> the TIs may be neutral (why not?)

Obviously because I won’t program them 🙂

December 17, 2019 21:56

I have published other originals with Total Invisibles in Problem Paradise 86, Apr-Jun. 2019 and it was then already mentioned that “none of the Total Invisibles can be neutral.”

My concise definition of a Total Invisible would be the following:

“A white or black orthodox piece that stands somewhere on the board, but whose color, identity and whereabouts are not known. The initial position and sequence of moves of the solution must be legal. After all aspects of a Total Invisible are revealed, it becomes visible and turns into an ordinary piece.”

If you allow the Invisibles to be also neutral, then you need more moves to identify the White Invisibles that can give mate and exclude the neutral Invisibles that can prevent mate.
The solutions would be more diluted and would lose most of their charm.

Longer solutions would also mean more cooks, all the more so as Popeye doesn’t check problems with neutral Invisibles and can’t help you.

Thomas Maeder
Thomas Maeder
January 3, 2021 22:08

Testing problem 1572 with Popeye (and vice versa!) revealed (pun intended) a number of bugs in the Popeye implementation of TotalInvisibles.
For this problem, Popeye now also finds:
 1…0-0-0  2.nPg2*f1=nQ nQf1*f4  3.nQf4-e5 Rd1-d4 #
other TIs: g1(not Q or R), e3 (b[SP] or w[QRBP])
 1…nQh1-g1  2.f4*g3 0-0-0  3.nQg1-c5 Rd1-d4 #
alternative move order for 1st solution

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