Oh, and it just occurs to me that your solutions also work with 3 TotalInvisibles instead of 2+1 Invisibles. This…
Bonjour René I think that your new version is cooked: 1.-Sg2 2.I(K)xg2 I~ 3.Kh1/Kh2 Kf2 4.dxIc2 Rh8# (wI on b6-e3)…
Georgy, see “December 29, 2020 08:54” above: “…in the I-system an IxI-capture is not valid. For more, ask T.Wakashima, the…
Isn't in the correction possible 1...Kg1 2.I- Re1 3.Ixe1 Qf3 4.d2 Qf1# ?
Georgy, in your sequence 3.Ixf1 is necessarily 3.Ke2xf1, implying that 2.I-- is 2.Kd3-e2 or Ke3-e2 (2 initial squares for the…
And what about 12.Kf2xBg2(+wKe1) Rd1-d2+ 13.Ke1-f2 Rd2-d1+ 14.Kf3xSg4(+wKe1) Bh3-g2+ 15.Kf2-f3 Rd1-d2+ 16.Ke1-f2 Rd2-d1+ 17.Kf2xBg2(+wKe1) Rd1-d2+ 18.Ke1-f2 Rd2-d1+ 19.Ke6×Bd7(+wKe1) Be8-d7+ 20.Ke5-e6…
Manfred, in the I-system an IxI-capture is not valid. For more, ask T.Wakashima, the inventor. Only 1…Kc1 2.I-- Kc2 3.Bb8 Ra1…
No, you are right: square e7 cannot be occupied (a fact I had forgotten). I sent a new correction to…
No, it isn't! In fact it works both ways. Feel free to confuse your own brain by checking why I…
Second additional cook must be 1.- Kf2! ("wasting" 1 Invisible) 2.Bb8 Qe4! 3.Bh2 Ixh2 (the 2nd one) 4.Ixh2 Qh4(8?)#, sorry!…