‡2
‡n
Aidés - Helpmates
Féeriques - Fairies
Retros


D5 - Abdelaziz ONKOUD
Problemesis 2000
1.Df4? [2.d5‡]
mais 1...c×d4!

1.d5? [2.Df4‡]
mais 1...c×b4!

1.Df3! [2.D×d3‡]
1...g×f3 2.d5‡
1...R×d4 2.Df4‡
1...c×b4 2.T×b4‡
1...c×d4 2.Dc6‡

Reversal
Urania
Zilahi

Re1 Pd4 Pe5 Df7 Tb1 Th4 Fd2 Cb4 Cc7 + Rc4 Pb6 Pc5 Pd3 Pe6 Pg4 Ph5 Fc8
‡2 (9+8) C+

D6 - Abdelaziz ONKOUD
Problemesis 2000
1.f×g3? [2.Df2‡]
mais 1...c2!

1.Da1? [2.Da4‡]
mais 1...g×f2+!

1.Ff1! [2.De3‡]
1...g×f2+ 2.D×f2‡
1...c2 2.Da1‡
Rg1 Pd3 Pe4 Pf2 De1 Fd6 Fe2 + Rd4 Pa5 Pc3 Pg3
‡2 (7+4) C+

D7 - Abdelaziz ONKOUD
Problemesis 2000
1.T×h5? [2.T×f5‡]
mais 1...e×f3!

1.F×c7? [2.F×d6‡]
mais 1...d3!

1.Cf6! [2.Cd7‡]
1...f×g4 2.T×h5‡
1...d5 2.F×c7‡
1...e×f3 2.Te1‡
1...d3 2.Fc3‡
Rf7 Pc4 Pf3 Pg4 Th1 Fa5 Ce8 + Re5 Pc7 Pd6 Pd4 Pe4 Pf5 Pf4 Ph5
‡2 (7+8) C+

D8 - Abdelaziz ONKOUD
Problemesis 2000
1.e×d7? [2.De6‡]
mais 1...Fa2!

1.c×d7? [2.Fc6‡]
mais 1...T×f4!

1.Rd6! [2.Dd5‡]
1...d×c6 2.F×c6‡
1...d×e6 2.D×e6‡
1...T×f4 2.De2‡
1...Fa2 2.Dd3‡
Re7 Pc6 Pe6 Pg3 Dc4 Fb5 Ff4 Ch4 + Re4 Pc5 Pd7 Pd4 Tf2 Fb1
‡2 (8+6) C+


M1 - Ralf KRATSCHMER
Problemesis 2000
1.Te2? [2.T8×e5,T2×e5‡]
mais 1...T×h4‡!

1.Fb1! [2.Fe4‡] c3 2.Fa2+ Tc4 3.Te2 [4.T8×e5,T2×e5‡] T×h4+ 4.Rg6 [5.T8×e5,T2×e5‡] e4 5.f×e4‡

Berlin

Rh5 Pb4 Pf3 Ph4 Te8 Th2 Fa2 Ca7 Cb5 + Rd5 Pc4 Pd2 Pe5 Pf4 Pg5 Ph6 Tc6 Th1 Fe1
‡5 (9+10) C+
A one-liner with obvious key (very few palatable choices for 1st move) (Ryan McCracken)

Weil der Name des Autors ueber dem Diagramm steht, habe ich fuer die Loesung nur 2 Minuten gebraucht. Die Staerke dieser Berlinthema-Darstellung ist sicherlich die direkte Form der thematischen Fluchtfeldgabe, die nur dank der Fesselung funktionieren kann. Schwaechen gibt es allerdings auch, wobei mich die Doppeldrohung am meisten stoert. (Manfred Rittirsch)


H42 - Jacques ROTENBERG
Problemesis 2000
a) 1.a3 Cd3 2.a×b2 Ce5 3.b1=F Tg2 4.Fh7 Cf7‡
b) 1.Rg7 Re2 2.Rf6 Rf3 3.Re5 Ta5+ 4.Rd4 Ce2‡

Mats modèles - Model mates

Rd1 Ta2 Cb2 Cc1 + Rh8 Pa4
h‡4 (4+2) C+
b) -Pa4
Zum huebschen Eckmustermatt mit Umwandlungslaeufer gesellt sich ueberraschend ein respektloses Idealmatt in der Brettmitte. (Manfred Rittirsch)

H43 - Nikolaj ZUJEV
Stepan P. TSIRULIK

Problemesis 2000
1.Fd6 Tg3 2.Ff4 Rc5 3.Ce5 Cd6‡
1.Fc5 Tf7 2.Fe3 Rd6 3.Cd4 Cc5‡

Mats modèles - Model mates
Echo

Rc6 Tg7 Cb7 + Re4 Ff8 Cd2 Cf3
h‡3 (3+4) C+
2.1.1...
very symmetrical play (Ryan McCracken)

H44 - Paul VATARESCU
Emanuel NAVON

Problemesis 2000
1.Td5 C×g4 2.Rd4 Db4‡
1.Rf3 Cf5 2.Tf4 Dh1‡
1.Rf4 Cf7 2.Te4 D×f2‡

Mats modèles - Model mates
Echo

Rc1 De1 Ch6 + Re4 Pd3 Pe2 Pf2 Pg4 Td4
h‡2 (3+6) C+
3.1.1.1
3 nice mates (Ryan McCracken)

Das schoenste Mattbild muss auch in diesem gelungenen Dreispaenner das seltenere bleiben. (Manfred Rittirsch)

H45 - Michel CAILLAUD
Problemesis 2000
1...Ce6 2.Ce5 Cg5‡
1.Cg5 De2+ 2.Rf4 Ch5‡
1.f4 Dc2+ 2.Re3 Cf5‡

Mats modèles - Model mates
Echo caméléon - Chameleon echo

Ra1 Dd2 Cg7 + Re4 Pf5 Cf3
h‡2* (3+3) C+
2.1.1.1
Black's lack of tempo moves causes the mate pattern to shift from set to play...very nice with good economy (Ryan McCracken)

Dreifach (und dann natuerlich auch mit Farbwechsel) kann man dieses Echo (vgl. Kb3 Dg5 Se6 - Kd6 Sb6 Bd7 - h#2* - Wilhelm Kraemer, Schachmatt 1948) nur noch als sensationell bezeichnen! (Manfred Rittirsch)

H46 - Jacques ROTENBERG
Problemesis 2000
1.g4 Cd5 2.Re4 Te8+ 3.Rf5 Te5‡
1.g4 Cd6 2.Re5 Te8+ 3.Rf6 Cce4‡
1.Rd3 Cd5 2.Re4 Te8+ 3.Rf5 Cce3‡

Mats modèles - Model mates

Rh7 Tc8 Cc4 Cc3 + Rd4 Pg5
h‡3 (4+2) C+
1.2.1.1... + 1.1...

H47 - Jacques ROTENBERG
Problemesis 2000
1.Re4 Tf5 2.Cf2 Cc2 3.Cd3 Cg3‡
1.Cg3 Tf6 2.Cf5 Cc3 3.Cd4 Cg4‡

Mats modèles - Model mates
Echo caméléon - Chameleon echo

Rg1 Tf2 Ce3 Ce2 + Re5 Ch1
h‡3 (4+2) C+
2.1.1...

H48 - Jacques ROTENBERG
Problemesis 2000
1.Cd2 Cc2 2.Cb3 Td5+ 3.Rc4 Ce3‡
1.Cc3 Td6 2.Cd5 Cc3 3.Cb4 Ce4‡

Mats modèles - Model mates
Echo caméléon - Chameleon echo

Rf7 Td4 Ca3 Ca2 + Rc5 Cb1
h‡3 (4+2) C+
2.1.1...

H49 - Olivier RONAT
Problemesis 2000
1.Fa8 Tc3 2.Db7 Cc4 3.Dh1 Ce3 4.Fg2 Cg4‡

Turton

Ra2 Pc2 Pe5 Ph4 Tc4 Ca5 + Rh2 Pd4 Pe7 Pe6 Ph7 Da7 Tg1 Fd5
h‡4 (6+8) C+
unexpected and beautiful bristol motivated by need to unpin white pieces and close bR line (Ryan McCracken)

Waere der Loydsche Einleitungszug nicht so verlockend gewesen, haette ich dieses abwegige Mattbild wohl noch sehr, sehr lange gesucht! (Manfred Rittirsch)


F19 - Gianni DONATI
Problemesis 2000
1.d4 3.Rc3 7.d8=D 8.D×h4(Dd8) 9.Df2 14.h8=D 15.Dhd4 17.Re1 18.Dd1+ D×d1‡
Ra5 Pd2 Ph2 + Rb1 Dh4
ss‡18 (3+2) C+
Circé
Mit zwei Damenumwandlungen gelingt der Zweifachexzelsior (Vgl. F15 vom gleichen Autor) sogar in Idealoekonomie. (Manfred Rittirsch)

F20 - Reto ASCHWANDEN
Problemesis 2000
1.CPc5? A [2.LO×b2-c3‡]
1...TP×c5(CPa6) a 2.CP×c5‡ A
1...FPe6 b 2.C×d6‡ B
1...LI×f3 c 2.TLPe6‡ C
mais 1...g4!

1.TLPe6! C [2.TLf5‡]
1...TPc5 a 2.C×d6‡ B
1...FP×e6(TLPg6) b 2.TLP×e6‡ C
1...LI×f3 c 2.CPc5‡ A

Lacny
Reversal

Rg1 Pc7 Pf3 Ta5 Fg8 Ca6 Cd1 Ce8 Wa1 Gc8 Gh2 Yc2 Yd7 Yg6 Yh5 La8 Lh3 + Re4 Pa7 Pa2 Pe5 Pf4 Pg5 Tb5 Ff7 Fh8 Wf8 Wf6 Lb2 Ld6 Ud5
‡2 (17+14) C+
Circé diagramme blanc
# =Pièce Patrouille
g=Girafe
lL=Fou-Lion
U=Lion
wxW=Locuste
y=Tour-Lion
1.Sc5 ? A threat 2.Lxb2-c3 #
Intercepts the observation line wRLc2-c8 to wGIc8. Doublecheck threat by wSc5 and wPf3.
1...Rxc5[+wSa6] a 2.Sxc5 # A
bLId5 is necessarily pinned by the wLLa8 and the wPf3. Ra5 patrols wSc5.
1...LIxf3 b 2.RLe6 # B
2.Lxb2-c3+ ? 2…Bc4 ! Both bRb5 and bBLf7 unpatrolled. d4 is covered in the mate by the wRLd7. This is necessary, because the wGIc8 doesn't guard d4. (Both observations wRLc2-c8 and wBLh3-c8 intercepted).
1...Be6 c 2.Sxd6 # C
2.RLxe6+ ? 2...Kd4! dual avoiding. Intercepts the observation line bBLh3-c8, so there's initially flight d4. The move 2.Sxd6 patrols wGIc8 again, so d4 guarded. Both bLocusts have lost guard of d6.
but: 1…g4 ! Flight d4.

1.RLe6 ! B threat 2.RLf5 #
Intercepts the observation line wBLh3-c8 to wGIc8. Doublecheck threat by wRLf5 and wPf3.
1...Rc5 a 2.Sxd6 # C
2.Sxc5+ ? 2...Kd4! dual avoiding. Intercepts the observation line bRLc2-c8, so there's initially flight d4. The move 2.Sxd6 patrols wGIc8 again, so d4 guarded. Both bLocusts have lost guard of d6.
1...LIxf3 b 2.Sc5 # A
2.RLf5+ ? 2…g4 ! Both bRb5 and bBLf7 unpatrolled. d4 is covered in the mate by the wRLd7. This is necessary, because the wGIc8 doesn't guard d4. (Both observations wRLc2-c8 and wBLh3-c8 intercepted).
1...Bxe6[+wRLg6] c 2.RLxe6 # B
bLId5 is necessarily pinned by the wLLa8 and the wPf3. wBg8 patrols RLe6.

Pioneer example of the above Key Reappearance Theme. Lacný and Reversal 1 included. I've talked about these themes in my lecture in Pula. For more details, read my upcoming feenschach article.
Probably one of the most wonderfull mechanisms I've ever invented. (Author)

F21 - Olivier RONAT
Problemesis 2000
a)
1.GSRd4(+b7) b8=GS 2.GSRh8(+d4) GSe5‡
1.GSRa7(+b7) b8=D+ 2.GSRh1(+a7) a8=GS‡
b)
1.GSRa6(+c7) c8=F+ 2.GSRa7(+a6) Fb7=
1.GSRb5(+c7) c8=D 2.GSRa7(+b5) Dc6=
+ Yb7
h‡2 (0+1) C+
2.1.1.1
b) >Yb7®c7, h=2
Sentinelles en Pion adverse
< =Pièce Royale
Y=Girl-scout

F22 - Olivier RONAT
Problemesis 2000
1.BSRc7(+b4) b5 2.BSRb2(+c7) c8=D 3.BSRa1(+b2) Dc1‡
1.BSRa7(+b4) b5 2.BSRc7(+a7) a8=C+ 3.BSRa7(+c7) b6‡
1.BSRc3(+b4) b5 2.BSRb2(+c3) b6 3.BSRa5(+b2) b4‡
+ Lb4
h‡3 (0+1) C+
3.1.1...
Sentinelles en Pion adverse
< =Pièce Royale
L=Boy-scout


R25 - Michel CAILLAUD
Problemesis 2000
1.f3 e5 2.Rf2 Df6 3.Rg3 Da6 4.Rh4 D×e2 5.g3 Df2 6.Fa6 e4 7.d3 e3 8.Cd2 e2 9.Cb3 e1=D 10.Fg5 De7 11.De2 Dfe3 12.Tf1 Dd8 13.Cc1 Dee7 14.D×e7+

Echange de place de la DN et du PNe7 - bQ and bPe7 interchange their place
Pronkin
Tempo

Rh4 Pa2 Pb2 Pc2 Pd3 Pf3 Pg3 Ph2 De7 Tf1 Th1 Fa6 Fg5 Cc1 Cg1 + Re8 Pa7 Pb7 Pc7 Pd7 Pf7 Pg7 Ph7 Dd8 Ta8 Th8 Fc8 Ff8 Cb8 Cg8
Partie justificative en 13,5 coups (15+15) C+
Die bis auf den Themabauern vollstaendige sPAS und die Tatsache, dass die beiden Themasteine vor dem letzten Einzelzug einem ordentlichen Platzwechsel genuegt haben, verleihen diesem D-Pronkin Letztformcharakter. (Manfred Rittirsch)

R26 - Gianni DONATI
Problemesis 2000
1.h4 h5 2.Th3 Th6 3.Tb3 Tc6 4.Tb6 a×b6 5.Cc3 Ta3 6.Ce4 Tac3 7.d×c3 e6 8.Fh6 D×h4 9.Dd2 Fe7 10.Td1 Ff6 11.Dc1 Re7 12.Cd2 Rd6 13.Cb1+ Rc5 14.Td3 d6 15.Th3 Fd7 16.Th1 Dh3 17.Dd1 h4 18.Fc1

Switchbacks
Circuits

Re1 Pa2 Pb2 Pc3 Pc2 Pe2 Pf2 Pg2 Dd1 Th1 Fc1 Ff1 Cb1 Cg1 + Rc5 Pb7 Pb6 Pc7 Pd6 Pe6 Pf7 Pg7 Ph4 Dh3 Tc6 Fd7 Ff6 Cb8 Cg8
Partie justificative en 17,5 coups (14+15) C+
Besonders gut tarnen konnte sich der a-Turm nicht, aber er kommt sauber durch alle Teamkameraden "hindurch", und als Zugabe findet die S-Rueckkehr ueberraschend in Form eines Rundlaufs statt. (Manfred Rittirsch)