2
3
n
Aidés - Helpmates
Inverses - Selfmates
Féeriques - Fairies
Retros
D107 - Bernard DELOBEL
Problemesis 2004
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1.Da6? blocus 1...f×e3 2.Dd6 mais 1...d4! 1.Db2? blocus 1...g×f3 2.Dg7 1...f×e3 2.De5 mais 1...d4! 1.Da4? [2.D×f4] 1...g×f3 2.D×f4 1...f×e3 2.D×g4 mais 1...d4! 1.D×c2? blocus 1...g×f3 2.Dg6 1...f×e3 2.Dc7 mais 1...d4! 1.Da8! blocus 1...g×f3 2.Dg8 1...f×e3 2.Db8 1...d4 2.Tg2 |
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2 | (8+5) C+ |
The scheme is well known, compare to
following 3 twomovers: 1. Karol Mlynka, Hlas Ludu 1971, Kb7 Qh2 Rd1 Be5 Pb5g3b2c2 - Ka2 Pa3b3d2, #2, 5 solutions 1.Qxd2!, 1.Qf2!, 1.Qh8!, 1.Qh6!, 1.Qh4! 2.Ivo Lebloch, 6th Comm Phenix 1994, Kd2 Qa1 Rb8 Be2 Sa4 Pd6e6c5 - Kc6 Pc7d7b6d5, #2 (4 tries) 1.Qe5?, 1.Qc3?, 1.Qh8?, 1.Qb2?, 1.Qg7! 3. Dragan Stojnic, 4th Comm Mat Plus 1997, Kh1 Qa2 Ra3g1 Bb3 Sc1 Pd2e2 - Kf2 Be1 Pe3f3, #2 (5 tries) 1.d4?, 1.Bd5?, 1.Qc2?, 1.Qb2?, 1.Ra8?, 1.Bd1! (Juraj Lörinc) Dieser "Verfuehrungsreigen" ist eine sehr theoretische Angelegenheit, da neben den Mattwechseln teilweise auch die Schlüsselzüge konkurrieren. Das laesst die Praxis, Varianten mit sich wiederholenden Mattzuegen einfach nicht anzugeben, besonders fragwuerdig erscheinen, und auch eine im Prinzip gleichwertige, aber dualistische Verfuehrung wie 1.Da7? hat nicht die geringste Chance, uebersehen zu werden. (Manfred Rittirsch) |
D108 - Alessandro CUPPINI
Problemesis 2004
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1.f8=C! [2.De6] 1...Dh3 2.Ff4 1...Dh6 2.Cf3 1...Fb5 2.Td5 1...Fc4 2.Df5 |
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2 | (12+10) C+ |
Hier gilt Aehnliches: Darf man eindeutig widerlegte Verfuehrungen wie 1.Kd8? (2.e8=D,T#) Lf6! oder 1.Lf5? (2.De6#) Lc4! verleugnen, nur weil sie nicht im Sinne des Autors sind? (Die Letztgenannte hat sogar "nur" noch den Nachteil der verwaessernden Mattwiederholungen.) Dafuer ist diesmal die Loesung mit den 2 differenzierten Linienoeffnungspaaren wenigstens huebsch anzuschauen. (Manfred Rittirsch) |
D109 - Pierre TRITTEN
Problemesis 2004
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1.e8=D? [2.De5,De6] mais 1...Cc5! 1.e8=C? [2.C×f6] mais 1...Cc5! 1.T×e4! [2.Td6] 1...Ch5+ 2.D×h5 1...F×b4 2.C×b4 1...Dd4 2.T×d4 1...Cd4 2.Te5 1...Td7 2.C×b6 (1...Ce8 2.Dh5,Tec4,Td4,T×e3,Tee6 1...C×e4 2.Dh5,F×e4) |
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2 | (11+11) C+ |
D110 - Evgeni BOURD
Problemesis 2004
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1.Dc5? [2.Cd7 A,Cg4 B] mais 1...D×g5! a, Df5! b 1.Dh3? [2.Cd7] 1...Df5 2.D×f5 mais 1...D×g5! 1.De3? [2.Cg4] 1...D×g5 2.D×g5 mais 1...Df5! 1.d5! [2.Cef3,Cc4,C×g6,Cd3,Ce×f7] 1...D×g5 a 2.Cg4 B 1...Df5 b 2.Cd7 A 1...D×h5 2.C×e4 |
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2 | (10+6) C+ |
D111 - Philippe ROBERT
Problemesis 2004
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1.Da1? [2.Cb5] mais 1...C5×e4! 1.Da7? [2.C×e6] mais 1...C3×e4! 1.f5! [2.f4] 1...C3×e4,Cd1,Ce2,Cd5 2.Cb5 1...C5×e4,Cd3 2.C×e6 |
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2 | (11+9) C+ |
Montrer le thème Hannelius sur 2 defenses Schiffmann ne semble pas avoir déjà été fait. C'est difficile à construire. Ce problème montre un des mécanismes possibles : le clouage en essais des 2 PN thématiques. Leur autoclouage dans le JR n'est pas exploité, ce qui est très regrettable, mais n'empèche pas qu'il s'agisse bel et bien de défenses Schiffmann.* (auteur) *Personnellement, je considère que ce ne sont pas des parades Schiffmann quand les auto-clouages ne sont pas utilisés par la suite. (Christian Poisson) Anticipated by Rauf Aliovsadzade, Medzhnun Vagidov, Elhniko oxaki 1981 (74 p.35 FIDE Album 1980-82) : |
T37 - Bernard DELOBEL
Problemesis 2004
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1.Cgf6? blocus 1...f3 2.Db8 f2 3.T×f2 mais 1...Re5! 1.Cef6? blocus 1...f3 2.De3 [3.Ch6,De5] mais 1...R×g5! 1.Tc5? blocus 1...R×e4 2.d6 [3.Dd5] f3 3.De3,Dc4 mais 1...R×g4! 1.Dc3? [2.Cgf2 [3.Df6]] 1...R×g4 2.Tg2+ Rf5/Rh4/Rh5 3.Cd6/Dh8/Dh3,Dh8 mais 1...R×e4! 1.Tf2! blocus 1...R×g4 2.Df3+ Rf5/Rh4 3.D×f4/Th2 1...R×e4 2.Dc4+ Rf5 3.D×f4 |
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3 | (7+3) C+ |
T38 - Vladimir KOZHAKIN
Problemesis 2004
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1.Cc4+! 1...Rc3 2.Ce2+ Rb3 3.Db1 1...Re1 2.Dg2 [3.De2] Rd1 3.Dd2 1...Rc1 2.Dd3 [3.Ce2] |
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3 | (4+3) C+ |
T39 - Andrejs STREBKOVS
Problemesis 2004
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1.Td1! [2.Fe3+ D×e3 3.Td5] 1...Cc3 2.Dd4+ R×d4 3.Fe3 1...Cc7 2.Dd6+ Rb6/R×d6 3.Dd4/F×b4 1...b3 2.Fb4+ R×b4,C×b4/a×b4 3.Dd4/Dd4,Td5 |
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3 | (11+13) C+ |
T40 - Evgeni BOURD
Problemesis 2004
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1.Td7! [2.C×e2+ R×e6 3.Cf4] 1...Cge3 2.Cd×f5+ R×e6 3.Cg7 1...Cce3 2.C×b3+ R×e6 3.Cc5 1...R×f6 2.Tf7+ Re5 3.T×f5 |
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3 | (12+12) C+ |
There are 3 white masked lines aiming at e6 that might be unmasked by white annihilation and the following doublecheck. White uses the linethat is not double masked in threat and 2 defences. Excellent. Compare e.g. to Fernand Joseph, L'Echiquier Belge 1996, Ke1 Rh4c3 Bb7c5 Sh5d4 Pd3 - Ke3 Re7 Bc8 Sd5e2 Pc6, #3, where there is only one such variation 1.Rg4! Sdxc3 2.Sxc6 Kf3 3.Se5#. (Juraj Lörinc) |
M43 - Vladimir KOZHAKIN
Problemesis 2004
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1.Fb3! Rb1 2.Rd1 Ra1 3.Cc4+ Rb1 4.Ca3 | ||
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4 | (5+1) C+ |
Very, very simple. (Baldur Kozdon) |
M44 - Vladimir KOZHAKIN
Problemesis 2004
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1.g6! [2.c8=D+ Rg7 3.f8=D+,f8=F+,Rf5] Rg7,Re7 2.c8=D [3.f8=D+,f8=F+,Rf5] Rf6,R×g6 3.Df5+ Re7,Rg7 4.f8=D | ||
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4 | (4+3) C+ |
Not really a problem. 1.g6 is evident. (Baldur Kozdon) |
M45 - Vladimir KOZHAKIN
Problemesis 2004
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1.Ce6! 1...Rb5 2.Rb7 Ra5 3.a3 Rb5 4.Cc3+ Ra5 5.b4 1...Ra6 2.a4 Ra5 3.Rb8 Ra6 4.Cc5+ Ra5 5.b4 Clé 2×ampliative - The key gives 2 flights |
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5 | (5+1) C+ |
Yes, it's pretty. (Baldur Kozdon) |
M46 - Alessandro CUPPINI
Problemesis 2004
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a) 1.Tf5! [2.T×f7] Tg×f3,Ta×f3 2.Tf1 [3.T×f7+ T×f7 4.T×f7] T×f1 3.T×f1 [4.T×f7] Tf3 4.T×f3 [5.T×f7] f5/f6 5.T×f5/T×f6 2...T×f5 3.T×f5 [4.T×f7] Tf3 4.T×f3 [5.T×f7] f5/f6 5.T×f5/T×f6 b) 1.Tf5! [2.T×f7] 1...Tg×f3 2.T×g5 [3.Tg8] Tg3 3.Tf1 [4.T×f7] Taf3 4.T×a5 [5.Ta8] Ta3 5.T×f7 1...Ta×f3 2.T×a5 [3.Ta8] Ta3 3.Tf1 [4.T×f7] Tgf3 4.T×g5 [5.Tg8] Tg3 5.T×f7 Echange des 2° et 4° coups blancs - Interchange of the 2nd and 4th white moves |
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5 | (11+10) C+ | ||
b) Pe4, Pg5 |
b) is more interesting than a). (Baldur Kozdon) |
M47 - Baldur KOZDON
Problemesis 2004
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1.Fc5! [2.D×e5,De7] C×g3 2.D×e5+ Rh4 3.Cf5+ C×f5 4.Ff2+ Rg5 5.Fe3+ Rh4 6.Dh2 | ||
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6 | (6+13) C+ |
I hope, it's a good problem. (author) |
M48 - K. MURALIDHARAN
Problemesis 2004
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1.Cc1! [2.Cb3+ Re3 3.Df3] Re3 2.Cb3 [3.Df3] Re2 3.Fc4+ Rd1/Re1/Re3 4.Db1/Df1/Df3 | ||
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4 | (6+2) C+ |
It's pretty. (Baldur Kozdon) |
H365 - Stefan MILEWSKI
Problemesis 2004
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1.f1=F T×g5 2.Fh3 Cf3 1.f1=C Tg2 2.Cg3 Th2 |
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h2 | (3+3) C+ | ||
2.1.1.1 |
Selbst mit den Unterverwandlungen muesste es diese bekannte Mustermattkombination schon gegeben haben. (Manfred Rittirsch) |
H366 - Stefan MILEWSKI
Problemesis 2004
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1.Ce3 Tg1 2.Cg2 T×h1 1.Ch2 T×g4 2.Cf2+ C×f2 |
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h2 | (3+4) C+ | ||
2.1.1.1 |
In beiden Loesungen blockt der f-S und der h-S wird geschlagen. Ich kann nur hoffen, dass DAS zur Originalitaet reicht. (Manfred Rittirsch) |
H367 - Stefan MILEWSKI
Problemesis 2004
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1.Fg1 Fh2 2.Fd4 Tg3 3.Re5 Tf3 1.c4 Fb4 2.Fd6 Tf6+ 3.Re5 Fc3 |
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h3 | (3+6) C+ | ||
2.1.1... |
H368 - Stefan MILEWSKI
Problemesis 2004
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1.Te6 d×e6 2.Cd6+ Rd5 3.Ce8 Cg8 1.Td7 Rd3 2.Rd6 Rd4 3.Fe7 Ce8 1.Cf7 C×e4 2.Td8 d6+ 3.Re8 Cf6 |
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h3 | (3+5) C+ | ||
3.1.1... |
H369 - Stefan MILEWSKI
Problemesis 2004
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1.Fc3 T×e6+ 2.R×d3 Te2 3.Td4 Fb5 1.Tc3 Fb3 2.Fg8 Te6+ 3.R×d3 T×d7
Mats modèles - Model mates
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h3 | (5+8) C+ | ||
2.1.1... |
H370 - Stefan MILEWSKI
Problemesis 2004
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1.f3 C×d5 2.Ce7 Tg6 3.Tf4 C×f4 1.Rg4 Cf5 2.Ce7 Th6 3.Rh3 T×h4
Mats modèles - Model mates
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h3 | (3+9) C+ | ||
2.1.1... |
H371 - Stefan MILEWSKI
Problemesis 2004
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a) 1.Rc5 Ce2 2.Fe5 Cc1 3.Rd4 Cb3 b) 1.Re5 Ce6 2.Fc5 Cd8 3.Rd4 Cc6 |
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h3 | (3+6) C+ | ||
b) Fa1®g1 |
H372 - Bernard DELOBEL
Problemesis 2004
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1.C×d3 Td1 2.Ce5 Cde4 1.R×c5 Th5 2.Cd6 Cb3
Boros
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h2 | (5+8) C+ | ||
2.1.1.1 |
H373 - Bernard DELOBEL
Problemesis 2004
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1.Tgb4 c3 2.Fc4 e4 1.Taf4 e3 2.Fe4 c4 1.Rd4 e4 2.Ce3 c3 1.Rd4 c4 2.Cc3 e3
Echange des coups blancs - Interchange of the white moves
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h2 | (6+9) C+ | ||
4.1.1.1 |
Just 2 pairs of symmetrical solutions. (Juraj Lörinc) |
H374 - Michael SHAPIRO
Problemesis 2004
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a) 1.Fb7 C×a4 2.R×c8 Rh2 3.Tc7 Cb6 b) 1.Tb7 C×e7 2.R×b6 Rg2 3.Fc7 Cc8 |
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h3 | (3+11) C+ | ||
b) Dc4®c5 |
Largely anticipated by Vlaicu Crisan, 3rd Prize, Buletin Problemistic 1992-1993, Kf5 Sc5a4 - Kb5 Ra5c4 Bc6a3 Sb7b2 Pf7, h#3, 2 solutions. (Juraj Lörinc) Einzeln hat man alle Elemente schon oft gesehen, aber dass sich die doppelte S-Rueckkehr zum Matt, die differenzierten Tempozuege des wK und besonders der doppelte T/L-Blockwechsel in einer einzigen Aufgabe vereinen lassen, die zudem noch dem Praedikat "Zilahi-Opferminimal" genuegt, raubt mir fast den Atem! Das sollte ueber den Preisbericht hinaus fuer Furore sorgen. (Manfred Rittirsch) |
H375 - Michael SHAPIRO
Problemesis 2004
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1.C×h4 Td7 2.Cf3 Th7 3.Cg1 Ff3 1.C×d4 Fe8 2.Cf3 Fc6 3.Ch2 Td1
Switchbacks
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h3 | (5+4) C+ | ||
2.1.1... |
Auch die Sperrbauern konnten die Herlin-Kritiki nicht lange verhuellen. Dass jene sich in eine symmetrische Anordnung fuegen liessen, ist die eigentliche Ueberraschung. (Manfred Rittirsch) |
H376 - Alessandro CUPPINI
Problemesis 2004
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a) 1.Tc6 Fe2 2.Dd8 Fh5 b) 1.Ff8 Tc6 2.Tf7 Tc8
Boros
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h2 | (6+11) C+ | ||
b) Pd6®f3 |
H377 - Alessandro CUPPINI
Problemesis 2004
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a) 1.Ff2 Ff8 2.F×c5 Tf2+ 3.Re3 F×c5 b) 1.Fh4 Th1 2.Rf2 Th2+ 3.Rg3 Fe5 c) 1.Re3 Te1+ 2.Rf3 Te3+ 3.Rf4 Fh6 |
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h3 | (7+3) C+ | ||
b) +Pf3 c) -Fe1, +Pf5 |
H378 - Viktor SIZONENKO
Problemesis 2004
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1...Rc5 2.Fh7 Rd5 3.Rg6 R×e4 4.d3 Rd5 5.Rf5 Rc5 6.Re4 Rc4 7.Ff5 e×d3
Circuit
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h6,5 | (4+9) C+ |
Interesting matter - is the movement of wK switchback or round trip? As he returns using precisely the same squares as one the way there - for me it is just a switchback. It is just a matter of definition... (Juraj Lörinc) Dass der wK auf seinem Rueckweg sogar den Tempozug beibehaelt, ist eine wunderschoene Bereicherung der in den Zeiten der 3-GHz-Prozessoren gang und gaebe gewordenen h#n-Manoever. (Manfred Rittirsch) |
H379 - Christer JONSSON
Problemesis 2004
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1.T×d7 Tc8 2.Rd6 Tc6 1.F×f6 Ta4 2.Re5 Te4 |
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h2 | (5+4) C+ | ||
2.1.1.1 |
Es muss nicht immer ein Funktionswechsel sein - 2 Turmmatts an der langen Laeuferleine tun es auch! (Manfred Rittirsch) |
H380 - Christer JONSSON
Problemesis 2004
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a) 1.Te5 Rg3 2.Fd5 C×d4 b) 1.Td5 Fb5 2.Fe5 Fd3 |
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h2 | (4+8) C+ | ||
b) -Pe4 |
H381 - Christer JONSSON
Problemesis 2004
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1...T×c4+ 2.R×c4 Fb6 3.d5 Cd6 1...C×d4 2.R×d4 Tf2 3.Re3 Fb6 1...F×b4+ 2.R×b4 Rc1 3.Rb3 Tb2
Zilahi cyclique - Cyclic Zilahi
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h2,5 | (4+8) C+ | ||
3.1.1... |
H382 - Christer JONSSON
Problemesis 2004
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1...Fg6+ 2.Rf8 Fh5 3.Rg8 F×g4 4.Tf8 Fh5 5.Fe6 Fg6 6.Ff7 Fh7 | ||
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h5,5 | (2+9) C+ |
The same question as in H378 - is it switchback or round trip of wB? (Juraj Lörinc) |
H383 - Krzysztof DRAZKOWSKI
Problemesis 2004
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1...Fc5 2.e1=D e7 3.De6 e8=D+ 4.Dc8 D×c8 1...Fd6 2.e1=F e7 3.Ff2 e8=F 4.Fa7 Fc6 |
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h3,5 | (3+3) C+ | ||
2.1.1... |
Bishop or Queen? Fine. (Juraj Lörinc) Mit weniger Material duerfte sich der synchrone UW-Wechsel D/L kaum darstellen lassen. Grandios! (Manfred Rittirsch) |
H384 - Cosme BRULL MAYOL
Problemesis 2004
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1.g3 h×g3 2.Rb1 g4 3.Rc1 g5 4.b1=D g6 5.Fh8 g7 6.Db4 g×h8=D 7.Dd2 Da1 | ||
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h7 | (2+6) C+ |
H385 - K. MURALIDHARAN
Problemesis 2004
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1.T×d6 Fd8 2.Td3 Tb4 3.d5 Fe7 1.T×d6 F×f6 2.Cd4 b3 3.Rd5 Te5 1.Td5 Fd8 2.Cd4 Fc7 3.Cb5 b4
Mats modèles - Model mates
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h3 | (5+8) C+ | ||
1.2.1.1... + 1.1... |
H386 - Evgeni BOURD
Problemesis 2004
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1.Fe5 Fe4 2.Rf4 Dg5 1.Tde5 Te4+ 2.Rd3 Db3
Grimshaw
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h2 | (5+5) C+ | ||
2.1.1.1 |
S27 - Viktor SIZONENKO
Problemesis 2004
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1.Db1+? Tc2 2.Re1 Re3 3.Cf5+ Rd3 4.Fc5 Rc4 5.Tf4+ Rd3 6.Tf2 Rc4 7.Db4+ Rd3 8.Cc1+ T×c1 mais 1...Rd2! 1.Cf4+! 1...Rd4 2.Db2 Re5 3.Tf8 Rd4 4.Fd5 Re5 5.Ta1 Rd4 6.Ta4 Re5 7.Re1 Rd4 8.Fd6+ Re3 9.Dc1+ T×c1 1...Rd2 2.Rg1 Re3 3.Cf5+ Rd2 4.Te1 R×e1 5.Fe4 Rd2 6.Dc2+ Re1 7.Rh1 Rf1 8.Dd1+ Rf2 9.Ch3+ T×h3 |
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s9 | (11+3) |
Quite unorthodox way to attain 3 different model mates - two checking first moves. Will someone prove it is correct? (Juraj Lörinc) |
F320 - Joost DE HEER
Problemesis 2004
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a) 2.a8=N 3.Ng5 4.Na2 5.Nd8 6.Ng2 7.Na5 8.Ng8 9.Nd2 10.Nh4 11.Nb7 12.Ne1 13.Nc5 14.Ng7 15.Nd1 16.Nf5 17.Nh1 18.Nb4 19.Nh7 20.Nd5 21.Nh3 22.Ng1 23.Na4 24.Nc8 25.Nf2 26.Nh6 27.Ng4 28.Na1 29.Ne3 30.Nc2 31.Nf8 32.Nb6 33.Nc4 34.Ng6 35.Ne2 36.Nb8 37.Nh5 38.Nb2 39.Nf4 40.Ne6 41.Nc7 42.Na3 43.Nb1 44.Nc3 45.Ne4 46.Ng3 47.Nf1 48.Nh2 49.Nb5 50.Nh8 51.Nd6 52.Ne8 53.Nf6 54.Nd7 55.Nf3 56.Nd4 57.Nb3 58.Nc1 59.Nf7 60.Nd3 61.Ne5 62.Nc6 63.Ne7 auto= b) 5.b8=N 6.Nh5 7.Nd7 8.Na1 9.Ng4 10.Na7 11.Ne5 12.Ng1 13.Na4 14.Ng7 15.Nd1 16.Nh3 17.Nd5 18.Nh7 19.Ne1 20.Na3 21.Nb1 22.Nh4 23.Nf8 24.Nc2 25.Ne6 26.Na8 27.Nc7 28.Na6 29.Ng3 30.Nc1 31.Na5 32.Ng2 33.Nc4 34.Ne8 35.Nf6 36.Nd2 37.Nf1 38.Ne3 39.Nf5 40.Nb3 41.Nc5 42.Nd3 43.Nh1 44.Nf2 45.Nc8 46.Ne4 47.Na2 48.Nc3 49.Ne2 50.Nh8 51.Nf4 52.Ng6 53.Ne7 54.Ng8 55.Nh6 56.Nd8 57.Nc6 58.Nd4 59.Nh2 60.Nf3 61.Ng5 62.Nf7 63.Nd6 auto= c) 4.b8=N 5.Nh5 6.Nb2 7.Ne8 8.Nh2 9.Nd4 10.Nh6 11.Nd8 12.Ng2 13.Na5 14.Ng8 15.Nd2 16.Nf6 17.Ng4 18.Na1 19.Nd7 20.Ng1 21.Na4 22.Ng7 23.Nd1 24.Nh3 25.Nd5 26.Nf1 27.Ne3 28.Nc2 29.Ne1 30.Nh7 31.Nf3 32.Ne5 33.Na7 34.Nc6 35.Ne7 36.Nh1 37.Nf5 38.Nh4 39.Nf8 40.Ng6 41.Ne2 42.Nf4 43.Ne6 44.Na8 45.Nc7 46.Na6 47.Ng3 48.Nc5 49.Nb3 50.Nc1 51.Nd3 52.Nf2 53.Nc8 54.Ne4 55.Na2 56.Nc3 57.Nb1 58.Na3 59.Nc4 60.Nd6 61.Nh8 62.Nf7 63.Ng5 auto= d) 2.b8=N 3.Ne2 4.Nh8 5.Nb5 6.Nh2 7.Ne8 8.Nb2 9.Nh5 10.Nd7 11.Na1 12.Ng4 13.Na7 14.Ne5 15.Ng1 16.Nh3 17.Nf7 18.Ng5 19.Na8 20.Ne6 21.Na4 22.Nc5 23.Ne1 24.Na3 25.Nc2 26.Na6 27.Nb4 28.Nh1 29.Ne7 30.Nb1 31.Nh4 32.Nf8 33.Ng6 34.Nf4 35.Ng2 36.Na5 37.Nc1 38.Nd3 39.Nf2 40.Nc8 41.Ne4 42.Na2 43.Nc3 44.Nd1 45.Ng7 46.Ne3 47.Nc7 48.Nd5 49.Nh7 50.Nf6 51.Ng8 52.Nh6 53.Nb3 54.Nf1 55.Ng3 56.Nf5 57.Nd6 58.Nc4 59.Nd2 60.Nf3 61.Nd4 62.Nc6 63.Nd8 auto= |
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sd-auto=63 exact | (1+0) C+ | ||
b) pa6®b2 c) pa6®b4 d) pa6®b6 Haan Maximum blanc Cavalier majeur |
F321 - Joost DE HEER
Problemesis 2004
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a) 4.a8=RO 5.ROg8 6.ROg2 7.ROa2 8.ROe6 9.ROg1 10.ROa1 11.ROc6 12.ROh8 13.ROh2 14.ROb2 15.ROb8 16.ROd3 17.ROh7 18.ROb7 19.ROb1 20.ROh1 21.ROf6 22.ROd1 23.ROb6 24.ROh6 25.ROd2 26.ROd8 27.ROf3 28.ROb3 29.ROc5 30.ROf2 31.ROg4 32.ROd7 33.ROe5 34.ROg6 35.ROe7 36.ROa3 37.ROg3 38.ROc7 39.ROh5 40.ROe8 41.ROb5 42.ROg7 43.ROd4 44.ROh4 45.ROf5 46.ROc8 47.ROc4 48.ROf7 49.ROd6 50.ROe4 51.ROg5 52.ROh3 53.ROf4 54.ROc1 55.ROe2 56.ROc3 57.ROd5 58.ROb4 59.ROe1 60.ROc2 61.ROe3 62.ROf1 auto= b) 3.a8=RO 4.ROa2 5.ROg2 6.ROg8 7.ROc4 8.ROh2 9.ROh8 10.ROb8 11.ROb2 12.ROf6 13.ROa4 14.ROf2 15.ROf8 16.ROb4 17.ROh4 18.ROd8 19.ROd2 20.ROb7 21.ROh7 22.ROc5 23.ROg1 24.ROa1 25.ROf3 26.ROc6 27.ROg6 28.ROe1 29.ROe5 30.ROd7 31.ROb6 32.ROd1 33.ROg4 34.ROh6 35.ROf1 36.ROb5 37.ROh5 38.ROc7 39.ROc1 40.ROe6 41.ROb3 42.ROd4 43.ROg7 44.ROg3 45.ROe8 46.ROe4 47.ROb1 48.ROa3 49.ROe7 50.ROc2 51.ROf5 52.ROc8 53.ROd6 54.ROf7 55.ROg5 56.ROh3 57.ROd3 58.ROf4 59.ROe2 60.ROc3 61.ROd5 62.ROe3 auto= c) 1.a8=RO 2.ROa2 3.ROg2 4.ROg8 5.ROc4 6.ROh6 7.ROf1 8.ROb5 9.ROh5 10.ROd1 11.ROd7 12.ROh3 13.ROb3 14.ROf7 15.ROa5 16.ROe1 17.ROe7 18.ROa3 19.ROf5 20.ROc8 21.ROb6 22.ROe3 23.ROg4 24.ROh2 25.ROe5 26.ROb8 27.ROf4 28.ROa6 29.ROc1 30.ROg1 31.ROg7 32.ROe2 33.ROe6 34.ROd4 35.ROa1 36.ROc2 37.ROc6 38.ROd8 39.ROb7 40.ROb1 41.ROh1 42.ROh7 43.ROc5 44.ROf2 45.ROb2 46.ROd3 47.ROb4 48.ROd5 49.ROc7 50.ROe8 51.ROa4 52.ROf6 53.ROc3 54.ROg3 55.ROd6 56.ROd2 57.ROg5 58.ROf3 59.ROh4 60.ROh8 61.ROg6 62.ROf8 auto= d) 4.b8=RO 5.ROb2 6.ROh2 7.ROh8 8.ROf3 9.ROa1 10.ROa7 11.ROe3 12.ROc8 13.ROa3 14.ROd6 15.ROb1 16.ROh1 17.ROh7 18.ROd3 19.ROf8 20.ROa6 21.ROe2 22.ROe8 23.ROa4 24.ROf6 25.ROd1 26.ROg4 27.ROd7 28.ROc5 29.ROg1 30.ROe6 31.ROb3 32.ROc1 33.ROa2 34.ROa8 35.ROc3 36.ROc7 37.ROh5 38.ROd5 39.ROg8 40.ROh6 41.ROf7 42.ROa5 43.ROd8 44.ROd4 45.ROg7 46.ROg3 47.ROf5 48.ROe7 49.ROe1 50.ROg6 51.ROc6 52.ROe5 53.ROc4 54.ROf1 55.ROd2 56.ROg5 57.ROe4 58.ROf2 59.ROh3 60.ROf4 61.ROg2 62.ROh4 auto= e) 2.a8=RO 3.ROg8 4.ROg2 5.ROa2 6.ROe6 7.ROg1 8.ROa1 9.ROc6 10.ROh8 11.ROh2 12.ROb2 13.ROb8 14.ROd3 15.ROh7 16.ROb7 17.ROb1 18.ROh1 19.ROf6 20.ROd1 21.ROb6 22.ROh6 23.ROd2 24.ROd8 25.ROf3 26.ROb3 27.ROc5 28.ROf2 29.ROh3 30.ROg5 31.ROe4 32.ROa4 33.ROc3 34.ROg7 35.ROe2 36.ROh5 37.ROc7 38.ROf4 39.ROb4 40.ROe1 41.ROc2 42.ROc8 43.ROa3 44.ROg3 45.ROb5 46.ROe8 47.ROd6 48.ROf7 49.ROa5 50.ROe5 51.ROd7 52.ROf8 53.ROg6 54.ROe7 55.ROh4 56.ROd4 57.ROf5 58.ROf1 59.ROc4 60.ROg4 61.ROe3 62.ROd5 auto= 2.a8=RO 3.ROa2 4.ROg2 5.ROg8 6.ROc4 7.ROh6 8.ROc8 9.ROc2 10.ROg6 11.ROb4 12.ROe1 13.ROa5 14.ROg5 15.ROc1 16.ROc7 17.ROg3 18.ROa3 19.ROf5 20.ROf1 21.ROb1 22.ROb7 23.ROd2 24.ROh2 25.ROh8 26.ROf3 27.ROa1 28.ROd4 29.ROg7 30.ROe8 31.ROa4 32.ROg4 33.ROb2 34.ROb8 35.ROd3 36.ROh7 37.ROh1 38.ROf6 39.ROd1 40.ROb6 41.ROe3 42.ROe7 43.ROc6 44.ROd8 45.ROe6 46.ROb3 47.ROc5 48.ROf8 49.ROd7 50.ROe5 51.ROf7 52.ROd6 53.ROe4 54.ROf2 55.ROh3 56.ROg1 57.ROe2 58.ROb5 59.ROc3 60.ROd5 61.ROh5 62.ROf4 auto= |
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sd-auto=62 exact | (1+0) C+ | ||
b) pa4®a5 c) pa4®a7 d) pa4®b4 e) pa4®a6, 1.1.2.1.1... Haan Maximum blanc Rose |
F322 - Joost DE HEER
Problemesis 2004
|
1.Nf7 2.Nb5 3.Nh2 4.Ne8 5.Nb2 6.Nd6 7.Nh4 8.Nb1 9.Ne7 10.Nh1 11.Nf5 12.Nb3 13.Nf1 14.Nc7 15.Ne3 16.Na1 17.Nc2 18.Nf8 19.Nb6 20.Nh3 21.Nd1 22.Nf2 23.Nh6 24.Ng4 25.Na7 26.Ne5 27.Na3 28.Nc4 29.Nd2 30.Ng8 31.Ne4 32.Na6 33.Nb4 34.Nh7 35.Ne1 36.Ng5 37.Nf3 38.Ng1 39.Na4 40.Ng7 41.Nc5 42.Nd7 43.Nh5 44.Nf6 45.Nd5 46.Nc3 47.Na2 48.Nc1 49.Ng3 50.Ne2 51.Nb8 52.Nd4 53.Nc6 54.Na5 55.Nb7 56.Nd8 57.Ng2 58.Ne6 59.Nf4 60.Nh8 61.Ng6 auto= 1.Nh1 2.Ne7 3.Nb1 4.Nh4 5.Nb7 6.Nf5 7.Nb3 8.Nd7 9.Ng1 10.Na4 11.Ng7 12.Nc5 13.Ng3 14.Nc1 15.Ne2 16.Nb8 17.Nd4 18.Nf8 19.Ne6 20.Na8 21.Nd2 22.Ng8 23.Ne4 24.Nc8 25.Nb6 26.Nh3 27.Nd5 28.Nf1 29.Nh2 30.Ne8 31.Nb2 32.Nd6 33.Nh8 34.Nf4 35.Ng6 36.Na3 37.Ne5 38.Na7 39.Nd1 40.Nb5 41.Nc3 42.Na2 43.Nd8 44.Nb4 45.Nc6 46.Na5 47.Ng2 48.Nc4 49.Ne3 50.Na1 51.Nc2 52.Ne1 53.Nh7 54.Nf3 55.Ng5 56.Nf7 57.Nh6 58.Nf2 59.Ng4 60.Nf6 61.Nh5 auto= 1.Nb7 2.Nh4 3.Nb1 4.Ne7 5.Nh1 6.Nf5 7.Nb3 8.Nd7 9.Ng1 10.Na4 11.Ng7 12.Nc5 13.Ng3 14.Nc1 15.Ne2 16.Nb8 17.Nd4 18.Nf8 19.Ne6 20.Na8 21.Nd2 22.Ng8 23.Ne4 24.Nc8 25.Nb6 26.Nh3 27.Nd5 28.Nf1 29.Nh2 30.Ne8 31.Nb2 32.Nd6 33.Nh8 34.Nf4 35.Ng6 36.Na3 37.Ne5 38.Na7 39.Nd1 40.Nb5 41.Nc3 42.Na2 43.Nd8 44.Nb4 45.Nc6 46.Na5 47.Ng2 48.Nc4 49.Ne3 50.Na1 51.Nc2 52.Ne1 53.Nh7 54.Nf3 55.Ng5 56.Nf7 57.Nh6 58.Nf2 59.Ng4 60.Nf6 61.Nh5 auto= |
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sd-auto=61 exact | (1+0) C+ | ||
3.1.1... Haan Maximum blanc n =Noctambule |
F323 - Joost DE HEER
Problemesis 2004
|
5.b8=N 6.Ne2 7.Nh8 8.Nd6 9.Nh4 10.Nb1 11.Ne7 12.Nh1 13.Nd3 14.Nh5 15.Nd7 16.Ng1 17.Ne5 18.Na7 19.Nc8 auto= 5.b8=F 6.Fh2 7.Fc7 8.Fg3 9.Fd6 10.Ff8 11.Fh6 12.Fc1 13.Fg5 14.Fd2 15.Ff4 16.Fe3 17.Fg1 18.Ff2 19.Fe1 auto= |
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sd-auto=19 exact | (1+0) C+ | ||
1.1.1.1.2.1.1... Haan Maximum blanc Cavalier majeur |
F324 - Joost DE HEER
Problemesis 2004
|
a) 5.a8=N 6.Nd2 7.Ng8 8.Nc6 9.Na2 10.Ng5 11.Nc7 12.Nf1 13.Nd5 14.Nh7 15.Nf6 16.Nb8 17.Nd7 18.Ng1 19.Nc3 20.Nb1 auto= 5.a8=F 6.Fh1 7.Fb7 8.Fg2 9.Fc6 10.Ff3 11.Fd5 12.Fg8 13.Fe6 14.Fh3 15.Ff5 16.Fb1 17.Fe4 18.Fc2 19.Fb3 20.Fa2 auto= b) 3.a8=N 4.Nd2 5.Ng8 6.Ne4 7.Nc8 8.Na4 9.Ng1 10.Nd7 11.Na1 12.Ng4 13.Nc2 14.Nf8 15.Nd4 16.Nh6 17.Nd8 18.Na2 19.Nc6 20.Nb8 auto= 3.a8=F 4.Fh1 5.Fb7 6.Fg2 7.Fc6 8.Ff3 9.Fh5 10.Fe8 11.Fg6 12.Fb1 13.Ff5 14.Fc2 15.Fe4 16.Fd5 17.Fg8 18.Fe6 19.Fc8 20.Fd7 auto= c) 5.b8=N 6.Ne2 7.Nh8 8.Nd6 9.Nh4 10.Nb1 11.Ne7 12.Nh1 13.Nf5 14.Nd1 15.Nh3 16.Nf7 17.Nc1 18.Ne5 19.Na7 20.Nc8 auto= 5.b8=F 6.Fh2 7.Fc7 8.Fg3 9.Fd6 10.Ff8 11.Fh6 12.Fc1 13.Fg5 14.Fd2 15.Ff4 16.Fe5 17.Fh8 18.Ff6 19.Fd8 20.Fe7 auto= d) 5.d8=N 6.Ng2 7.Na5 8.Ng8 9.Nc6 10.Ng4 11.Na1 12.Nc5 13.Ng3 14.Ne7 15.Nc8 16.Na4 17.Ng1 18.Ne5 19.Nc1 20.Nb3 auto= 5.d8=T 6.Th8 7.Th1 8.Ta1 9.Ta8 10.Ta2 11.Ta7 12.Ta3 13.Th3 14.Tb3 15.Tb8 16.Tb4 17.Tb7 18.Tb5 19.Ta5 20.Ta4 auto= |
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sd-auto=20 exact | (1+0) C+ | ||
1.1.1.1.2.1.1... b) pa3®a5, 1.1.2.1.1... c) pa3®b2 d) pa3®d2 Haan Maximum blanc Cavalier majeur |
F325 - Joost DE HEER
Problemesis 2004
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1.b8=N 2.Ne2 3.Nh8 4.Nb5 5.Nh2 6.Ne8 7.Nb2 8.Nh5 9.Nd3 10.Nf7 11.Nh3 12.Nb6 13.Nf8 14.Nc2 15.Ne6 16.Na4 17.Nc3 18.Ne7 19.Nh1 20.Nf5 21.Nd1 22.Ne3 23.Nc7 24.Na8 auto= 1.b8=F 2.Fh2 3.Fc7 4.Fg3 5.Fd6 6.Fa3 7.Fc5 8.Fg1 9.Fd4 10.Fh8 11.Fe5 12.Fg7 13.Ff8 14.Fe7 15.Fh4 16.Ff6 17.Fg5 18.Fc1 19.Ff4 20.Fd2 21.Fa5 22.Fc3 23.Fa1 24.Fb2 auto= |
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sd-auto=24 exact | (1+0) C+ | ||
2.1.1... Haan Maximum blanc Cavalier majeur |
F326 - Joost DE HEER
Problemesis 2004
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2.a8=N 3.Ng5 4.Na2 5.Nd8 6.Ng2 7.Na5 8.Ng8 9.Nd2 10.Nh4 11.Nb7 12.Ne1 13.Nc5 14.Ng3 15.Ne7 16.Nb1 17.Nd5 18.Nh7 19.Nf8 20.Nc2 21.Ne6 22.Ng7 23.Nd1 24.Nf5 25.Nb3 26.Na1 auto= 2.a8=T 3.Th8 4.Th1 5.Ta1 6.Tg1 7.Tg8 8.Tg2 9.Ta2 10.Tf2 11.Tf8 12.Tf3 13.Ta3 14.Te3 15.Te8 16.Te4 17.Ta4 18.Td4 19.Td8 20.Td5 21.Th5 22.Te5 23.Tg5 24.Tg3 25.Th3 26.Th2 auto= |
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sd-auto=26 exact | (1+0) C+ | ||
1.2.1.1... Haan Maximum blanc Cavalier majeur |
F327 - Joost DE HEER
Problemesis 2004
|
a) 1.Td8 2.Td3 3.Th3 4.Th8 5.Th4 6.Ta4 7.Tg4 8.Tb4 9.Tf4 10.Tf8 11.Tf5 12.Ta5 13.Te5 14.Te1 15.Ta1 16.Td1 17.Tb1 18.Tb3 19.Tb2 20.Tc2 21.Tc8 22.Tc3 23.Tc7 24.Th7 25.Td7 26.Tg7 27.Te7 28.Te8 auto= b) 1.Tc8 2.Tc2 3.Th2 4.Th8 5.Th3 6.Ta3 7.Tg3 8.Tg8 9.Tg4 10.Ta4 11.Tf4 12.Tb4 13.Tb8 14.Tb5 15.Th5 16.Tc5 17.Tg5 18.Td5 19.Td1 20.Th1 21.Te1 22.Te8 23.Te2 24.Te7 25.Te3 26.Tb3 27.Td3 28.Td2 auto= c) 1.Td8 2.Td3 3.Td7 4.Th7 5.Th1 6.Ta1 7.Ta8 8.Ta2 9.Ta7 10.Ta3 11.Ta6 12.Th6 13.Tb6 14.Tg6 15.Tg1 16.Tb1 17.Tb5 18.Th5 19.Tc5 20.Tg5 21.Tg2 22.Te2 23.Te8 24.Te3 25.Te7 26.Te4 27.Ta4 28.Td4 29.Tb4 30.Tb2 31.Tb3 32.Tc3 33.Tc1 34.Tf1 35.Tf8 36.Tf2 37.Tf7 38.Tf3 39.Tf6 40.Tc6 41.Te6 42.Td6 43.Td5 44.Tf5 45.Tf4 46.Th4 47.Th2 48.Th3 49.Tg3 50.Tg4 auto= d) 1.Tc8 2.Tc2 3.Th2 4.Th8 5.Th3 6.Ta3 7.Tg3 8.Tg8 9.Tg4 10.Ta4 11.Tf4 12.Tf8 13.Tf5 14.Ta5 15.Te5 16.Te1 17.Te4 18.Tb4 19.Tb8 20.Tb5 21.Tb7 22.Th7 23.Tc7 24.Tc3 25.Tf3 26.Tf1 27.Th1 28.Tg1 29.Tg2 30.Td2 31.Td8 32.Td3 33.Td7 34.Tg7 35.Tg5 36.Th5 37.Th6 38.Ta6 39.Tg6 40.Tb6 41.Tf6 42.Tc6 43.Tc4 44.Tc5 45.Td5 46.Td6 47.Te6 48.Te8 49.Te7 50.Tf7 auto= |
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sd-auto=28 exact | (1+0) C+ | ||
b) td2®c1 c) sd-auto=50 exact d) =c) td2®c1 Haan Maximum blanc |
F328 - Joost DE HEER
Problemesis 2004
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a) 1.Ne7 2.Nh1 3.Nb4 4.Nh7 5.Ne1 6.Nb7 7.Nh4 8.Nd6 9.Nf2 10.Nh6 11.Nb3 12.Nf1 13.Nc7 14.Ne3 15.Na5 16.Nc6 17.Ne2 18.Nh8 19.Nf4 20.Nd8 21.Ne6 22.Nc2 23.Na1 auto= b) 1.Fh7 2.Fc2 3.Fg6 4.Fd3 5.Fa6 6.Fc4 7.Fg8 8.Fd5 9.Fh1 10.Fe4 11.Fg2 12.Ff1 13.Fe2 14.Fh5 15.Ff3 16.Fg4 17.Fc8 18.Ff5 19.Fd7 20.Fa4 21.Fc6 22.Fa8 23.Fb7 auto= |
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sd-auto=23 exact | (1+0) C+ | ||
b) fb1 Haan Maximum blanc n =Noctambule |
F329 - Alessandro CUPPINI
Problemesis 2004
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a) 1.Nb4 N×d7 2.Na2 Rc2 b) 1.GId2 GId4 2.GIe6 GIe8 3.GIa5 GI×f4 4.GIb1 GIb5 |
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h2 | (4+4) C+ | ||
b) nN=Girafe, h4
nN =Noctambule Girafe |
Deeper strategy is missing, however it is an interesting study of the board geometry - the squares are reachable by different number of movesdepending on the type of the fairy piece involved. (Juraj Lörinc) |
F330 - Alessandro CUPPINI
Problemesis 2004
|
1.a7! [2.a8=S] b2 2.a8=S+ Rb3 3.b8=S+ Rc3 4.Sc8+ Rd2 5.Sd8+ Re2 6.Se8+ R×f1 7.Sf8+ R×g1 8.Sg8+ R×h2 9.Sh8 | ||
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9 | (8+10) C+ | ||
sq =Sauterelle |
Too straightforward. (Juraj Lörinc) |
F331 - Alessandro CUPPINI
Problemesis 2004
|
1.Da4! Df7 2.Db3 D×f1+ 3.Db1 Da6 | ||
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s3 | (3+6) C+ | ||
Maximum |
F332 - Diyan KOSTADINOV
Problemesis 2004
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1...Rnc8 2.e1=Dn Dn×d2(d7)+ 3.d6 Dn×c2(c7) 4.Dn×g6(g2) Dn×d6(d7) 1...Rne8 2.c1=Dn Dn×d2(d7)+ 3.d6 Dn×e2(e7) 4.Dn×a6(a2) Dn×d6(d7)
Captures réciproques - Reciprocal captures
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h3,5 | (0+3+6) C+ | ||
2.1.1... Circé |
Just 2 symmetrical solutions. (Juraj Lörinc) |
F333 - Cosme BRULL MAYOL
Problemesis 2004
|
1...Re2 2.Se3 Rd3 3.Sc3 Rd4 4.Sc4 Rd3 5.Sc7 Rc4 6.Rb5+ Rc5 1...Rf2 2.Sg3 Re3 3.Sd3 Re4 4.Sd4 Re5 5.Rc5 Rd6+ 6.Sd7+ Rd5 |
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h5,5 | (2+3) C+ | ||
2.1.1... Köko sqS =Sauterelle |
F334 - Alberto ARMENI
Problemesis 2004
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1.f4? blocus mais 1...g2! 1.f3! blocus 1...g×h2(Dd1),C×b3(Ff1) 2.Ta5 1...g2 2.Dd6 1...d1=~ 2.Da2 1...Fb2 2.F×b2(Ff8) 1...C×c3,d4 2.Cc4 1...Cc2 2.C×c2(Cg8) |
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2 | (7+7) C+ | ||
Circé |
F335 - Alberto ARMENI
Problemesis 2004
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1.e1=D Oc7 2.Dc1 Cf2 1.e1=T Rd2 2.Tc1 Cf2 1.e1=F Oe5 2.Fc3 Cf2 1.b1=C O×e2 2.Cc3 Cf2 |
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h2 | (6+10) C+ | ||
4.1.1.1
iI =Orphelin |
F336 - Michael GRUSHKO
Problemesis 2004
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a) 1.b6 Cf4 2.Rb7 Cd5 3.Ra8 C×b6(b7) b) 1.R×c7(Fc1) Fb8+ 2.R×b8(Ta1) Fa3 3.Ra8 Fd6 |
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h3 | (2+2) C+ | ||
b) ch3®c7 Circé caméléon |
F337 - Michael GRUSHKO
Problemesis 2004
|
a) 1.Tnc5 b×a4(a2) 2.Tnc2(a1=Dn) Dng7 3.Rn×a2(Rne8) Tnc8(a8=Tn) b) 1...Tnb8 2.a×b3 Tnb7(b2)+ 3.b1=Dn Dn×b3 4.Tnb5(b1=Dn) Dn×b3= |
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h3 | (0+0+4) C+ | ||
a) Circe parrain, Anticirce b) Circe parrain, Eiffel, h=3,5 Circé parrain Eiffel Anticircé |
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1.Fnc4 b7 2.Fng8 b8=Dn 3.Fnh7 g×h7 4.Dne8 h8=Fn 1.Rg7 b7 2.Rh6 b8=Fn 3.Fnd6 g7 4.Fnf8 g8=Cn Mais Éric Huber signale que ce problème est complètement anticipé (position miroir) par Ion Murarasu, Buletin Problemistic 81 juin 2004 (et publié aussi par erreur dans Phénix 128). But Éric Huber indicates that this problem is completely anticipated (mirrored position) by Ion Murarasu, Buletin Problemistic 81 june 2004. |
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h4 | (1+1+3) C+ | ||
2.1.1... |
F339 - Michael GRUSHKO
Problemesis 2004
|
1.Rc5 Sb5 2.Rc6 Sb7 3.R×b7(Sb8) Sb6 1.Re3 Se2 2.Rf3 Se4 3.R×e4(Se8) Se3 1.Rd5 Sc6 2.Rd4 Sc3 3.Rc5 Sc4 |
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h3 | (2+1) C+ | ||
3.1.1... Circé Rois transmutés sq =Sauterelle |
Compare to my own Juraj Lorinc, F1823 The Problemist November 1998, Ge4c1 - Kf6, h#3, Transmuting kings, 4.1... -> yes, although 3 mates are quite mechanic, and less "funny", it shows there are 4 solutions possible. (Juraj Lörinc) |
F340 - Michael GRUSHKO
Problemesis 2004
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1.f1=D Rc2 2.Df5 Cc5 1.f1=T Cd4 2.Tf4 Te6 |
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h2 | (3+2) C+ | ||
2.1.1.1 Rois transmutés |
F341 - Evgeni BOURD
Problemesis 2004
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1.Rb4? blocus 1...T×f5 a 2.Cc6{R«R} A 1...T×g6 b 2.Ced5{R«R} B mais 1...a2! 1.Ra2! blocus 1...T×f5 2.C×f5{R«R} 1...T×g6 2.C×g6{R«R} 1.R×a3! blocus 1...T×f5 a 2.Ced5{R«R} B 1...T×g6 b 2.Cc6{R«R} A |
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2 | (13+9) C+ | ||
2 solutions Swapping Kings |
F342 - Michael GRUSHKO
Problemesis 2004
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a) 1...f8=F+ 2.Rf2 Fh6 3.Rg1 Fe3 1.h1=C f8=F+ 2.Rf2 Fh6 3.Cg3 Fe3 b) 1.Rb6 f8=F 2.c1=D+ Rb4 3.Dc7 Fc5 |
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h3* | (2+2) C+ | ||
b) Ph2®c2 Rois transmutés |
F343 - Joost DE HEER
Problemesis 2004
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1...d8=Fn 2.d1=Dn Dnd5 3.g1=Tn e8=Cn | ||
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h2,5 | (1+1+4) C+ |
Die idealoekonomischen Allumwandlungen scheinen selbst dann nicht zur Neige zu gehen, wenn man sich nur mit Neutralen bewaffnet. (Manfred Rittirsch) |
F344 - Joost DE HEER
Problemesis 2004
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1.a1=Fn c8=Dn 2.Dn×c4(c8=Tn) d×c8=Cn(Tnd7) | ||
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h2 | (1+1+4) C+ | ||
Circé échange |
Die Erkenntnis, dass sich ein orthogonal wirkender Stein auf d7 selbst decken kann, hat die 2 Doppelzuege doch ein wenig laenger werden lassen. Dass es dann nur der Turm sein durfte, kann man im Hinblick auf die AUW als eine glueckliche Fuegung bezeichnen. (Manfred Rittirsch) |
F345 - Michael GRUSHKO
Problemesis 2004
|
1.Rd7 e8=Cn 2.Rd8 c6 3.d5+ Rc5 4.c×d5(d7) Rd6= | ||
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h=4 | (1+1+3) C+ | ||
Anticircé Eiffel |
R173 - Joachim IGLESIAS
Problemesis 2004
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1.c4 g5 2.Dc2 Fg7 3.Dg6 Fh6 4.D×g8+ Ff8 5.Dg6 Fg7 6.Dc6 d×c6 7.g4 D×d2+ 8.R×d2 Fh6 9.Rc3 Ff8 10.Rb4 | ||
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Partie justificative en 9,5 coups | (14+14) C+ | ||
Monochromatique |
Wenn der sL sich hier gleich zweimal zwei Einzelschritte vor und einen Doppelschritt zurueck bewegt, dann hat er zwar bereits zwei Drittel aller schwarzen Zuege verbraucht, ohne von der Stelle gekommen zu sein, aber er hat auch den einzigen Weg gefunden, den Weissen nicht zu stoeren! (Manfred Rittirsch) |
R174 - Joachim IGLESIAS
dédié à D. Antonini
Problemesis 2004
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1.g3 a5 2.g4 Ta6 3.g5 Td6 4.g6 Td3 5.e×d3 Cf6 6.Dh5 Tg8 7.Dh6 g×h6 8.Re2 Fg7 9.Re3 Tf8 10.Rd4 Th8 11.Rc3 Ff8 12.g7 Cc6 13.g8=D Ca7 14.Dg4 Cg8 15.Dd1 | ||
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Partie justificative en 14,5 coups | (15+15) C+ |
Der weisse Bauer, der es ohnehin nicht eilig hatte, muss letztendlich sogar 7 Zuege lang verharren, damit Schwarz Gelegenheit bekommt, das ueberfluessige Tempo zu verlieren. Fuer eine Stellung, die sich so wenig von der PAS entfernt hat, verlaeuft das verblueffend eindeutig. (Manfred Rittirsch) |