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Phénix

No. 208, Jul-Aug. 2011

6362 - P. WASSONG (after Dmitrij BAIBIKOV)

Phénix 208, Jul-Aug 2011

[rnKb4/pBn1p2p/N1r2p2/2PkPP2/3p1p2/P2P4/pP1p3P/qRB2B2]

13+15. Resolve

[rnKb4/pBn1p2p/N1r2p2/2PkPP2/3p1p2/P2P4/pP1p3P/qRB2B2]

Solution

Scheme : wB~×bBf1, bBc4-f1, b4-b3×Qa2, wQ~-a2, wBd7, bQd6-e6-a1, wB~, wQd7, bBe6-c4, wBg8-f7-~, g7-g8=B, etc...

During the solution, this problem shows 5 interceptions, of which a cycle with 4 links : bBc4 intercepts wQb3, wBd7 intercepts bQe6, bQd6 intercepts wQd7, wQd7 intercepts bQe6, bQe6 intercepts wBf7-g8.

The matrix is inspired by the problem of

Dmitrij Baibikov
PDB N°P1084675, 
1st Prize Mat Plus, 2008

[1k6/1P1p2p1/QP2p3/rnK5/pRnP2PP/1BrpP2p/N1P1P1p1/2N5]

(14+12) Resolve?

[1k6/1P1p2p1/QP2p3/rnK5/pRnP2PP/1BrpP2p/N1P1P1p1/2N5]


6363 - P. WASSONG

Phénix 208, Jul-Aug 2011

[qnQRb1B1/rBrN1N1p/n1R5/3k1P2/5p2/2PK1P2/1PPpPp1P/4b3]

15+12. last move?

[qnQRb1B1/rBrN1N1p/n1R5/3k1P2/5p2/2PK1P2/1PPpPp1P/4b3]

Solution

Retro : 1.e3×Bf2.

This problem shows a bQ in a cage further possible from the Pawns.

Here the nearest Pawn to bQa8 is on ç3. The distance is square of 29=square of (5^2+2^2)


6364 - E. PICHOURON

Phénix 208, Jul-Aug 2011

[2b2q1Q/3p1rk1/np2pnp1/5prp/1p6/5P2/PP1P2P1/RNB1KBNR]

13+14. SPG in 19.5 moves (C+ in 19.0)

[2b2q1Q/3p1rk1/np2pnp1/5prp/1p6/5P2/PP1P2P1/RNB1KBNR]

Solution

1.e4 e6 2.e5 Bd6 3.e×d6 b5 4.d×c7 Bb7 5.c8=N b4 6.Nb6 a×b6 7.c4 Ra5 8.c5 Na6 9.c6 Rg5 10.c7 f5 11.c8=N Nf6 12.Ne7 Bc8 13.Ng6 h×g6 14.h4 00 15.h5 Rf7 16.h6 Qf8 17.Qh5 g×h5 18.f3 g6 19.h7+ Kg7 20.h8=Q+


6365 - N. DUPONT & R. OSORIO

Phénix 208, Jul-Aug 2011

[rn2kbnr/1p1pppp1/B1R4P/2p5/1P2P3/B1N1PK2/1PP1NP2/1Q4R1]

15+12. SPG in 20.5 moves (C+)

[rn2kbnr/1p1pppp1/B1R4P/2p5/1P2P3/B1N1PK2/1PP1NP2/1Q4R1]

Solution

1.e4 c5 2.Ke2 Qb6 3.Kf3 Qb3+ 4.a×b3 h5 5.Ra6 h4 6.Rc6 a5 7.Ba6 h3 8.Ne2 h×g2 9.h4 g1=Q 10.h5 Qg5 11.h6 Qe3+ 12.d×e3 a4 13.Bd2 a3 14.Bb4 a2 15.Nbc3 a1=Q 16.Qb1 Qa5 17.R×c8+ Qd8 18.Rc6 Qc7 19.Ba3 Qh2 20.b4 Qg1 21.R×g1


6366 - R. OSORIO & J.J. LOIS

Phénix 208, Jul-Aug 2011

[1nbqkbn1/p1ppp2Q/p4R2/4B2p/4R3/PPPPKPN1/7P/3B1N2]

14+12. SPG in 25.0 moves (C+)

[1nbqkbn1/p1ppp2Q/p4R2/4B2p/4R3/PPPPKPN1/7P/3B1N2]

Solution

1.a4 Nc6 2.a5 Rb8 3.a6 b×a6 4.Ra4 Rb3 5.c×b3 h5 6.Qc2 Rh6 7.Qh7 Rd6 8.Re4 Rd3 9.e×d3 f5 10.Be2 f4 11.Bd1 f3 12.Ne2 f×g2 13.f3 g1=R+ 14.Kf2 Rg5 15.Rg1 Rc5 16.Rg6 Rc3 17.Rf6 g5 18.d×c3 g4 19.Bf4 g3+ 20.Ke3 g2 21.Nd2 g1=R 22.Nf1 Rg5 23.Neg3 Ra5 24.Be5 Ra3 25.b×a3 Nb8


6367 - V. CRISAN

Phénix 208, Jul-Aug 2011
This problem was later found to be cooked

[8/5N2/2r4p/5k2/7K/8/3p2P1/3Q2b1]

4+5. Proca Retractor -16 & s#1 (Anticirce)

[8/5N2/2r4p/5k2/7K/8/3p2P1/3Q2b1]

Solution

Author’s solution :

1.h5×g6ep(g2) g7-g5+ 2.Qg2×Qh1(Qd1) Bh2-g1+ (h2-h1=Q+ is illegal here) 3.Kg3-h4 Bg1-h2+ 4.Kf2-g3 Bh2-g1+ 5.Kf1-f2 Bg1-h2+ 6.Ke1-f1 d3-d2+ 7.Kh4×g4(Ke1) Bh2-g1+ 8.Kg3-h4 Bg1-h2+ 9.Kf2-g3 Bh2-g1+ 10.Ke1-f2+ Bg1-h2+ 11.Kh4×Rg5(Ke1) Bh2-g1+ 12.Kg3-h4 Bg1-h2+ 13.Kf2-g3 Bh2-g1+ 14.Ke1-f2 Bg1-h2+ 15.Kd5×Be4(Ke1) Rc8-c6+ 16.Kd6-d5 and 1.Qf3+ Q×f3(Qd8)#

Cook:

1.Qd1×Rh5(Qd1) Rg5-h5+ 2.Kg3-h4 Rh5-g5+ 3.Kf2-g3 Bh2-g1+ 4.Ke1-f2 d3-d2+ 5.Kd7×Rc7(Ke1) Rc8-c7+ 6.Ke8-d7 Rc7-c8+ 7.f6×Ng7(g2) Ne6-g7+ 8.Kd7-e8 Rc8-c7+ 9.Qe2×Ne1(Qd1) & 1.Qe4+ K×e4(Ke8) or K×f6(Ke8)#