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MatPlus.Net Forum Helpmates Bror Larsson, Springaren 1947

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The following problem is in the PDB (P0003276). It also appeared in problem 127-132 09/1969.

(= 4+11 )
h#3

1.fxg3 ep.+ Kxe1 2.Rc4 Bxh5 3.Kd3 Bxg6#

However it has the following cooks:
1.Rc4 Bxb3 2.Bc3 Kd1 3.Kd3 Bc2#
1.Rg3 Kxf1 2.Bxg4 Bxb3 3.Kf3 Bd5#
1.Rg3 Bxb3 2.Bxg4+ Kxf1 3.Kf3 Bd5#
1.Bxg4+ Kxf1 2.Rg3 Bxb3 3.Kf3 Bd5#

Is the diagram in the PDB correct? If it is, has a correction ever been published? One way of fixing the cooks would be to add a bSa1 and bPc6.

it seems that with a black Queen c1 to add a black pawn c6 is enough.

The most economic I could get is :

(= 4+11 )

Strictly speaking, a Knight h1 is lighter than a Queen g1, but with the Qg1 there is an interception effect on g3 - a kind of "weasel" effect.

Another attempt:

P0003276v
(= 3+11 )

h#3 (3+11)

1. h:g3+ K:e1 2. g5 B:h3 3. Kd3 Bf5#

yes, Marcel, nice.

Nice problem

I tried to get 13, but did not succeed, just found 2 other ways for 14 :

(= 4+10 )
(4+10)

1.f×g3 e.p.+ K×e1 2.R1h2 B×h3 3.Kd3 Bf5‡

(= 3+11 )
(3+11)

1.e×d3 e.p.+ K×b1 2.d5 B×e3 3.Ka3 Bc5‡

I'm no retro expert, but what about a matrix with Pe4->c4, Bh8, Rh3?

Hauke

The big issue is that the rook needs to free c3 for the black bishop a move earlier, and the white bishop can't come from e3 as well, so Re3 must be (or some other reason). For the rest, it might work with bPc7 (to avoid duals) but I didn't test it.

Stupid me. Also e3xd4 if that field isn't blocked.
P.S. By moving the last diagram two to the right one gets
another version (pawns right fall off, but are needed
to block on the left).

Hauke

EDIT:

(= 4+9 )

13! Correct?

Cooked by 1.Se3 dxe3 2.Bg3 Bxh3 3.Be5 Bg2.

Cooks are unfair, I just looked after the retro :-)
And after Bh4->f4?

Hauke

1. Rg3 Ke1 2. Bg4 Bc4 3. Kf3 Bd5#

Another position with 14 pieces:

(= 3+11 )

H#3 (3+11)

1.cxd3(ep) Kxc1 2.Ka3 Bxh8 3.Qb4 Bb2#

I like corner to corner move by wB.

Can this one be correct?

(= 4+9 )

H#3 (4+9)

1.fxg3(ep) f3+ 2.Kf4 Sf2 3.f5 Sd3#

very nice ideas Miodrag....

the last one is retro cooked : last move may be Kf1-e2

The previous may give this (with promoted force)

(= 3+10 )
(3+10)

1.cxd3(ep) Kxb1 2.Ka3 Bxh8 3.Qb4 Bb2#

The previous play could have been c2xd1R+ Kf1-e2.

This seems to work :

(= 4+9 )
h3# (4+9)

1.f×g3 e.p.+ f3+ 2.Kf4 Sf2 3.g5 Sd3‡

Yes, I oversaw black rook promotion/capturing on d1. I think position you proposed is correct.

(= 4+8 )
h#3
Meredith

Edit: with bRg6 (instead of the bP)