py_20100608

mat en 2 coups avec Noctambule et Sauterelle

Nil G.G. VAN DIJK
Probleemblad, 1960
1st HM

white Ph7h4g3 Kg8 Rh1 Gg4e6d5b8 Na4c6c8f5 black Bf1d4 Kh3 Qb2 Ph2 Sf3d2 Re1e2

#2
g4, e6, d5, b8: Sauterelles
a4, c6, c8, f5: Noctambules

{

} 1.h7-h8=G + ? but 1...Bd4*h8 !{

} 1.Nf5-d1 + ? but 1...Kh3*g3 !{

} 1.Nf5-e3 + ? but 1...Kh3*g3 !{

} 1.Nf5-e7 + ? but 1...Re2*e6 !{

} 1.Na4*e2 ? but 1...Re1*e2 !{

} 1.Gd5-d3 ! threat: 2.Nf5-d1 #{
} 1...Re2*e6 2.Rh1*h2 #{
} 1...Re2-e5 2.h7-h8=G #{
} 1...Re2-e3 2.Nc8-f2 #{
} 1...Re2-f2 2.Na4-g1 #{
} 1...Bd4-f2 2.Rh1*h2 #{
} 1...Bd4-e3 2.Nf5-e7 #{
} 1...Bd4-e5 2.Nf5-h6 #

aidé 3 coups avec pièces chinoises

Unto HEINONEN
Probleemblad, 1987

white Pe7 Kg8 black VAa4f4 LEa1 MAc6e6e4 Kd5 Pb6d6c4e3g3 PAc7b4

h#3
4 solutions
c7, b4: Pao
a4, f4: Vao
c6, e6, e4: Mao
a1: Leo

Solution

{

} 1.LEa1-e5 e7-e8=VA 2.MAc6-d4 VAe8-h5 3.PAc7-c5 VAh5-f3 #{

} 1.PAb4-b5 e7-e8=LE 2.PAb5-c5 LEe8*a4 3.LEa1-d4 LEa4-b5 #{

} 1.MAe4-c5 e7-e8=MA 2.MAe6-d4 Kg8-f7 3.VAf4-e5 MAe8-f6 #{

} 1.MAc6-d4 e7-e8=PA 2.VAa4-c6 PAe8-a8 3.b6-b5 PAa8-a5 #

aidé 2 coups avec pièces chinoises

Franz PACHL
Jubilé K. Gandev_60
Shakmatna Misl, 2007
1st Prize

white VAg2 NAe2 Kd8 Pf4 Sc1 PAc7h6 black Bf1 NAh2 Kc5 Pa7 Se3 PAa1

h#2
b) Cc1-->c2
c) Cc1-->c3

Solution

{

} a) {

} 1.NAh2-f3 NAe2-b8 2.Kc5-d4 PAh6-c6 #{

} b) wSc1-->c2{

} 1.NAh2-f6 VAg2-b7 2.Kc5-d5 NAe2-c6 #{

} c) wSc1-->c3{

} 1.NAh2-d4 PAh6-a6 2.Kc5-d6 VAg2-c6 #

aidé 2 coups Andernach

Volker GULKE
The Problemist, 2002
1st-2nd Prize

white Pa4e4h4 Kb7 Rg4 black Pd2f6 Ka5 black Chameleon Qg5 Rh1 Sf5 Be1

h#2 Andernach
4 solutions

Solution

{

} 1.cBe1*h4=cR=w cRh4*h1=cQ=b 2.cQh1*e4=cS=w cSe4-c3=cB #{

} 1.cRh1*h4=cQ=w cQh4*g5=cS=b 2.cSg5*e4=cB=w cBe4-d5=cR #{

} 1.cSf5*h4=cB=w cBh4*e1=cR=b 2.cRe1*e4=cQ=w cQe4-c6=cS #{

} 1.cQg5*h4=cS=w cSh4*f5=cB=b 2.cBf5*e4=cR=w cRe4-b4=cQ #

mat en 2 coups Circé

Michel CAILLAUD
Jubilé C. Lytton-70
The Problemist, 2009
2nd HM

white Ba3 Kh4 Qf8 Pd3g5e5 Sc6d6 Rg3d7 black Pg7h3 Sf7 Rh2 Kf4

#2 Circé

Solution

{

} 1.Qf8*g7 ? threat: 2.Qg7-f6 # but 1...Sf7-h6 !{
} 1.Qf8*f7[+bSg8] + ? but 1...Sg8-f6 !{
} 1.Rd7*f7[+bSg8] + ? but 1...Sg8-f6 !{
} 1.Sc6-e7 ? threat: 2.Se7-g6 # but 1...Kf4*e5[+wPe2] !{
} 1.Rg3-f3 + ? but 1...Kf4*f3[+wRh1] !{
} 1.Ba3-c1 + ? but 1...Rh2-d2 !{
} 1.Ba3-c5 ? threat: 2.Bc5-e3 # but 1...Rh2-e2 !{

} 1.Qf8-g8 ! threat: 2.Rd7*f7 #{
} 1...Sf7*d6[+wSg1] 2.Rg3-f3 #{
} 1...Sf7*e5[+wPe2] 2.e2-e3 #{
} 1...Sf7*g5[+wPg2] 2.Ba3-c1 #{
} 1...Sf7-h6 2.Qg8-c4 #{
} 1...Sf7-h8 2.Qg8-c4 #{
} 1...Sf7-d8 2.Qg8-c4 #

mat en 2 coups Rois Transmutés avec Equistoppeurs

Hans Peter REHM
(The Problemist, 2005
1st Prize

white Pd7g7a2 Kc8 Ze4 Sb3 QFh6f4 black Pe5b7f3e2d2c3b4 Sa8 Bh2 Ke6

#2 Rois Transmutés
e4: Zèbre
h6, f4: Equistoppeurs français

Solution

{

} 1.Ze4-b6 + ? but 1...Ke6-c4 !{

} 1.Sb3-c5 + ? but 1...Ke6*c5 !{

} 1.a2-a4 ! threat: 2.Ze4-b6 #{
} 1...Bh2*f4 2.d7-d8=S #{
} 1...Ke6-f6 2.d7-d8=B #{
} 1...Ke6-e7 2.d7-d8=Q #{
} 1...Ke6-d6 2.g7-g8=QF #{
} 1...Ke6-f7 2.d7-d8=QF #{
} 1...Ke6-f5 2.d7-d8=Z #{
} 1...Ke6-d5 2.d7-d8=R #

aidé 3 coups avec Pao et Triton

Chris FEATHER
The Problemist, 2009
3rd Place

white Kd1 PAd4 TRa4 black Pd2d3e3e5c5 Ke4 Rf3 Sd5 Bh3

h#3
b) TRa4-->h4
a4: Triton
d4: Pao

Solution

{

} a) {

} 1.Ke4-f5 PAd4-f4 2.Kf5-e4 TRa4-c4 3.Bh3-f5 PAf4-h4 # {

} b) wTRa4-->h4 {

} 1.Bh3-d7 TRh4-g4 2.c5-c4 TRg4-f4 3.Bd7-a4 + PAd4*a4 #

mat en 2 coups - échecs Fonctionnaires

Jim DUCAK
The Problemist, 2005
3rd HM

white Bc8e7 Kh8 Qf3 Pe3 Sb5b4f7 Rc2h4 black Qb8 Pg3g5 Rb3g4 Ba2g1h7 Ke5

s#9 AntiCircé

Solution

{

} 1.Rc2-c5 ? threat: 2.e3-e4 # but 1...Ba2-b1 !{

} 1.e3-e4 ! threat: 2.Sb4-d3 #{
} 1...Rg4*e4 2.Qf3-f6 #{
} 1...Bh7*e4 2.Rc2-c5 #

inverse de série 15 coups - AntiCircé - Madrasi

Unto HEINONEN
The Problemist, 2010
1st Prize

white Pc5e5a7g7d3 Ka5 black Kd4 Pb2b4b6f6 Bd1 Rh1 Sd6

ser-s#15 AntiCircé Madrasi

Solution

{

} 1.g7-g8=S 2.Ka5-b5 {
} 3.a7-a8=S 4.Sa8-c7 {
} 5.c5-c6 6.Sc7-a6 {
} 7.c6-c7 8.c7-c8=S {
} 9.Sc8-e7 10.e5-e6 {
} 11.Se7-d5 12.e6-e7 {
} 13.e7-e8=S 14.Sg8-e7 15.Se7-c6 + Rh1-g1 #

Aidé 2 coups - Circé Martien

Frantisk SABOL
15è T.T. Chess Composition
Microweb, 2004 - 1st Prize

white Ba8 Kh4 Qc7 Pe7 Sg8 Rd5 black Bg3 Kh3 Qh2g1 Pf2a2b4g5h5 Se3e1 Rg2a1

h#2 Circé Martien
5 solutions

Solution

{

} 1.Qg1*c7 e7-e8=Q 2.Qc7-c3 Sg8*c3 #{

} 1.Qg1*d5 e7-e8=R 2.Qd5-d2 Sg8*d2 #{

} 1.Qg1*a8 e7-e8=B 2.Qa8-a3 Sg8*a3 #{

} 1.Qg1*g8 e7-e8=S 2.Qg8-f8 Se8-g7 #{

} 1.Qg1*e7 Qc7*a1 2.Qe7-g7 Sg8-h6 #

inverse 11 coups - AntiCircé avec Léo et Moa

Klaus WENDA
Die Schwalbe, 2006
1st Prize

white LEg7 Kf8 Pf7f3c5a2b2c2 Sc3 Bc4f2 Rb1 MOe2 black Pd2 Sh1 Ba1 Kd1

s#11 AntiCircé
e2: Moa
g7: Leo

Solution

{

} 1.MOe2-d4! Kd1-c1 {
} 2.Sc3-e2 + Kc1-d1 {
} 3.LEg7-g1 Kd1-c1 {
} 4.LEg1-g5+ Kc1-d1 {
} 5.Se2-f4 Kd1-c1 {
} 6.MOd4-e2 + Kc1-d1 {
} 7.MOe2-c3 Kd1-c1 {
} 8.Sf4-e2+ Kc1-d1 {
} 9.LEg5-g1 Kd1-c1 {
} 10.LEg1-g7 + Kc1-d1 {
} 11.Bf2-e1 d2*e1=Q[bQe1->d8] #

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