Best Problems 51

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Transcript of Best Problems 51

  • 241

    BEST PROBLEMSRassegna dei migliori problemidiretta da Antonio Garofalo

    Hanno collaborato a questo numero:in redazione - C. J. Feather, G. J. Perrone, V. Rallo, Mr. Veneziano;altri collaboratori - Enzo Minerva, Vlaicu Crian, Eric Huber

    Nuovi partecipanti ai concorsi di BP (welcome): Gabriele BRUNORI, Stefano MARIANI, PascalePIET. In questo fascicolo ancora altri 2 verdetti interessanti che occupano molto spazio a discapitodi alcune rubriche. Per la prima volta anche BP adotta il metodo di pubblicare le soluzioni sullostesso fascicoli degli inediti. un sistema pi pratico per il redattore e per i giudici.

    Abbiamo il piacere di ospitare su queste pagine un'altra rappresentante del gentil sessofrancese, la seconda dopo Odette Baudoin; una rarit nel mondo del problemi, impensabile inItalia. Ma ancora con maggior piacere ospitiamo l'opera prima del giovane Gabriele Brunori diPerugia, al quale auguriamo di continuare la carriera appena iniziata. Infine c' un altro granderitorno, quello di Stefano Mariani. I perugini (Agostini, Mariani e Brunori) si sono risvegliati,tremate gente, tremate. E per il prossimo numero vi preannuncio un altro grande ritorno...

    Marjan Kovacevic, Belgrade Internet Ty. 2008, 1 Pr.

    2 (5+13) C+

    1g5 2.xf5 1d4 2.xb8 1c4 2.xd5 1c4 2.xc41.xf5? [2.xd7] 1gxf5 2.xf5 1c4 2.xd5 ma 1h7!1.xd5? [2.a6] 1b3 2.c3 1c4 2.e3 1c7 2.xc7 1xd5 2.xd5 ma 1c4!1.xd5! [2.xb8] 1b4 2.c4 1c4 2.e4 1c7 2.f8 1xd5 2.xd5

    Anno XIII - n. 513/2009 - July

  • 242

    Inediti2262. V. Lukashv

    Russia2263. V. Lukashv

    Russia2264. V. Lukashv

    Russia2265. J. Pitkanen

    FinlandiaUS1EF1U'Q1D31$QQG1Gdddddddd

    GEF'151$$1131QDSDdddddddd

    GEFS11'UDQ1Q113%UGQD$5dddddddd

    GU13SQEFU11Gdddddddd2 vv (10+8) C+ 2 (6+10) C+ 2 vv (11+10) C+ 2 (7+4) C+

    2266. V. LukashvRussia

    2267. P. L. PlacanicoItalia

    2268. A. DikusarovRussia

    2269. A. ArmeniItalia

    'EF$D1$31SQQUGdddddddd

    DD1Q13QU$QQGS1QQ1QUGdddddddd

    15'EFD1U'GDQQS35%Udddddddd

    'G5UQ131111'1QDDQQSUdddddddd2 (7+6) C+ 2 (14+6) C+ 2 (9+8) C+ 2 v (10+10) C+

    2270. A. ArmeniItalia

    2271. G. BrunoriItalia

    2272. V. Agostini &S. Mariani - Italia

    2273. V. AgostiniItalia

    GG5U5UD3$S1Ddddddddd

    151QD3GSU11Q'dddddddd

    D111Q31EF51QQ$USGdddddddd

    EF5GDUS11'1Q3'$Q$UGdddddddd

    2 vv (7+5) C+ 2 vv (6+7) C+ 2* v (8+8) C+ 2 vv (9+9) C+

    2 - n. 2262-2273 (Judge 2009/2010: NN).

  • 243

    2274. P. MacinaItalia

    2275. V. BarsukovRussia

    2276. A. GrinblatIsraele

    2277. V. Agostini &S. Mariani - Italia

    EFS3QQdddddddd

    DS1Q13D1UEFdddddddd

    U'G111Q3Q1$QQ1Q1QDQ1GSDdddddddd

    5D''1$EF1G1131QUQ$Sdddddddd

    3 (4+1) C+ 3 (6+4) C+ 3 (13+10) C+ 3 vv (7+11) C+

    2278. A. DikusarovRussia

    2279. A. DikusarovRussia

    2280. M. KuliginUcraina

    2281. A. ZarurBrasile

    U%GD31Sdddddddd

    SG5U5Q11%3$'dddddddd

    DSD31GG1$dddddddd

    'S$1$Q1'13Q51U5QDDddddddddH2 (4+3) C+b) f2-a7

    H2 (4+8) C+b) wa3, c) we4,d) wc8, e) wg6

    H2 (5+4) C+2 sol.

    H2 (7+11) C+2 sol.

    2282. A. PankratievRussia

    2283. A. PankratievRussia

    2284. P. PietFrancia

    2285. S. ParzuchPolonia

    'DD11551%131'QQGSdddddddd

    D1131'%5US1D5'dddddddd

    G1'15UQ3QSdddddddd

    5$5'31G'SGUdddddddd

    H2 (6+11) C+2 sol.

    H2 (4+10) C+b) g3

    H2 (5+5) C+2 sol.

    H2 (4+7) C+2 sol.

    3, n. 2274-2277 (Judge 2008-2009: Antonio Garofalo).H2, H=2 - n. 2278-2293 (Judge 2008-2009: Chris J. Feather).

  • 244

    2286. A. DikusarovRussia

    2287. C. JonssonSvezia

    2288. L. M. MartinSpagna

    2289. G. Bakcsi & L.Zoltan - Ungheria

    %U31$15$D15'1GSdddddddd

    $S%1311DD1UQ11Gdddddddd

    ''51111$13GQ1$SU%5dddddddd

    $UD3QS1'5dddddddd

    H2 (4+11) C+b) wd7-e5c) wd7-f3d) =c) {e6-c4

    H2 (6+9) C+4 sol.

    H2 (4+14) C+b) wc5-g7

    H2 (4+5) C+2 sol.b) -e7 2 sol.

    2290. F. SimoniItalia

    2291. M. WitztumIsraele

    2292. S. MarianiItalia

    2293. J. CarfFrancia

    5U5%13G1G$$S'U'dddddddd

    %UEF3QQQSdddddddd

    EF11S15Q''1%D1UD31QQ$Q1Q15$dddddddd

    D11113'SD$15Q$ddddddddH2 (5+10) C+b) zc1-g1

    H2 (6+2) C+3 sol.

    H2 (10+16) C+b) e7-c2

    H2 (4+10) C+2 sol.

    2294. V. BarsukovRussia

    2295. M. Kuligin &V. I. Shevchenko

    Ucraina

    2296. P. TrittenFrancia

    2297. M. Kuligin &V. I. Shevchenko

    Ucraina'531DGUSdddddddd

    S1EF11'13Qdddddddd

    5S5'131D%dddddddd

    5111Q5311$U$'Sdddddddd

    H2.5 (4+4) C+3 sol.

    H2.5 (3+6) C+b) zc4-d4

    H3 (2+7) C+2 sol.

    H3 (3+11) C+2 sol.

    H2, H=2 - n. 2278-2293 (Judge 2008-2009: Chris J. Feather).H3/n, H=3/n - n. 2294-2305 (Judge 2008-2009: Viktor Zaitsev).

  • 245

    2298. V. Rallo &A. Garofalo - Italia

    2299. A. OnkoudFrancia

    2300. C. J. A. JonesInghilterra

    2301. M. ParrinelloItalia

    'SQ1Q131Q15U$Gdddddddd

    D11$S%3G'$Q'dddddddd

    S'%G11QQQ1$UQ3Q$5511dddddddd

    5'SQQ1$1%$11Q'1Q3QUdddddddd

    H3 (6+8) C+2 sol.

    H3 (4+8) C+b) {b2vc2

    H3 (8+12) C+b) {g4-g2

    H3 (7+12) C+2 sol.

    2302. T. IlievskiMacedonia

    2303. C. JonssonSvezia

    2304. M. KuliginUcraina

    2305. T. IlievskiMacedonia

    5G'13SQdddddddd

    '$$SQ1%3dddddddd

    3$1S1'Gdddddddd

    3GQSddddddddH3.5 (3+4) C+2 sol.

    H4 (2+6) C+2 sol.

    H5 (2+5) C+1 sol.

    H6.5 (3+1) C+1 sol.

    2306. H. NieuwhartOlanda

    2307. W.SeehoferGermania

    2308. M. ParrinelloItalia

    2309. V. RalloItalia

    VWG43Sdddddddd

    P1QDS3Qdddddddd

    511Q1SQQ3Q1EFDG%'dddddddd

    13SDdddddddd

    H3 (3+2) C+2 sol.4=GrasshopperVW=Locust

    S5* (5+2) C+P=Empress(Imperatrice)

    HS2.5 (8+8) C+b) h2, c) h2Anticirce (Calvet)

    H3.5 (2+2) C+b) wg3-g1Circe parrain

    H3/n, H=3/n - n. 2294-2305 (Judge 2008-2009: Viktor Zaitsev).Fairies n. 2306-2310 (Judge 2008-2009: Dinu-Ioan Nicula).

  • 246

    2310. S. Galletti - ItaliaDiag. A Diag. B Diag. C Diag. D

    'GG13Sdddddddd

    'GG13QSdddddddd

    'GG131QSdddddddd

    'GG131Sdddddddd

    H3 (3+3) C+Anticirce (Calvet)

    H3 (4+3) C+Circe

    H3 (4+4) C+Circe parrain

    H3 (3+4) C+Diagram circe

    Un curioso problema, 4 gemelli con posizioni e condizioni diverse. Per motivi di comodit tipo-grafica ho usato 4 diagrammi.

    Note agli inediti

    Anticirce: on making a capture, any unit (including K) is reborn on its game-array square (asdetermined according to Circe rules), and the captured unit disappears. Since rebirth is obligatory, acapture is legal only if the relevant rebirth-square is unoccupied. A capture may be made from arebirth-square. Promotion with capture is legal provided the rebirth-square of the promoted unit isunoccupied. In Anticirce type Calvet capture on a rebirth-square is allowed (it is the default type). InAnticirce type Cheylan capture on a rebirth-square is not allowed. [Effettuando una cattura, qualsiasipezzo (incluso il Re) rinasce nella sua casa di partenza intesa come una partita di scacchi, in sintoniacon le regole Circe usuali, mentre l'unit catturata sparisce; siccome la rinascita obbligatoria unacattura legale solo se la relativa casa di rinascita libera. Una cattura pu essere fatta da una casa dirinascita. La promozione con cattura legale se la casa di rinascita del pezzo derivante da promozione libera.]Circe parrain: a captured unit is reborn after the next single move on the square situated in arelationship to the capture-square that is equivalent to the departure and arrival squares of thatsingle move, e.g. capture on e4 followed by move g8-h6: rebirth on f2. If the rebirth square isoccupied, or if it would be beyond the board-edge, the capture is normal. [Un pezzo catturato ri-nasce dopo la successiva singola mossa sulla casa che in relazione alla casa di cattura in modoanalogo alle case di partenza e arrivo della stessa singola mossa. Se avviene, ad esempio, una cat-tura in e4, seguita dalla mossa g8-h6 la rinascita in f2. Se la casa di rinascita risulta occupata oal di fuori della scacchiera, la cattura normale.]Diagram Circe: a captured unit, when reborn (see Circe), reappears on its diagram square. [Il pezzocatturato rinasce (vedere Circe) nella casa in cui era nella posizione iniziale del diagramma.]Locust: moves on Queen-lines but only by capturing an enemy unit, arriving on the squareimmediately beyond that unit. [Muove sulle linee della Donna ma solo se pu catturare; la catturaavviene saltando il pezzo catturato e cadendo in una casa vuota; per esempio se nel diagramma laLocusta fosse in h7: LOh7xb7-a7]Empress: Rook+Springer. [L'Imperatrice unisce i movimenti della e del .]

  • 247

    Soluzioni IneditiFascicolo n. 50

    Commenti degli autori, di Gaspar J. Perrone e del redattore.

    2210. (2, Viktor Lukashv)1.c6! [2.EFf5] 1Uf3 2.exf3 1Uh5 2.Dc3 1Ud5 2.EFxd5 1d5 2.Dc52211. (2, Viktor Lukashv)1.Dh4? tempo, ma 1d2! 1.EFf8? [2.EFxd6] ma 1d5! 1.exd6? [2.e5] 1e5 2.EFe3 ma 1d2!1.De1! tempo 1g3 2.Df3 1d2 2.De2 1dxe5 2.EFe3 1d5 2.exd5 1a5 2.Db52212. (2, Viktor Lukashv)1.EFb4? EFa4? [2.EFxd4] ma 1c5! 1.EFc5! [2.EFxd4] 1Df5 2.Dc6 1dxe3 2.EFxe3 1EFb42.Gd2 1Ub4 2.Dg4 1Ua4 2.Gc12213. (2, Alberto Armeni)1.Ud8? [2.Uxd6] 1Uh5 2.EFa1 (a) ma 1Df4! 1.Dg5! [2.Df3] 1g1=D 2.EFa1 (a) 1Dh42.h8=EFG 1De5 2.De6 1Dxe4 2.Ud8 1Gxd1 2.Dxc6 1Se5 2.d42214. (2, Gerhard Maleika & Rainer Paslack)1.Dc~? [2.Ud5] 1Dc4 2.EFxc4 ma 1Df6! 1.De5! [2.Dc6] 1De~ 2.EFc5 1Dc3 2.EFf22215. (2, Rainer Paslack)1.Df6? [2.Ub5] 1Ge4 2.exf8=EF/G 1EFd5 2.EFxd5 ma 1Ue4! 1.Db6? [2.Ub5] 1EFd52.EFxd5 1Ue4 2.Dd3 ma 1Ge4! 1.Df4! [2.Ub5] 1Ge4 2.De6 1Ue4 2.Dfd3 1EFd52.EFxd5 Anti-Somov A1, matti cambiati.2216. (2, Stephan Dietrich)1.f4! [2.Ue5] 1gxf4 2.Dxf4 1fxe6 2.Gxe6 1exf3 e.p. 2.e4 1Dc7+ 2.Dxc7 1Dd6+2.Uxd6 1Dd3 2.Ud4 1Dxc4 2.bxc4 1Uxb6+ 2.Dxb62217. (2, Andre Dikusarov)1.Dd3? [2.EFd7] ma 1Gh3! 1.Dd7? 1.Da4? [2.Db6] ma 1axb5! 1.Dxa6? [2.Dc7] ma 1Ge5!1.De4! [2.EFd6] 1Gd4 2.c4 1Ge5 2.EFf7 1Dd4 2.Dc32218. (2, Andre Dikusarov)1.EFe2? [2.EFe8] ma 1Gxh1! 1.EFa3? [2.EFf8] ma 1Gxh1! 1.EFc3? [2.EFc8] ma 1a6! 1.EFa1!tempo 1f5 2.EFh8 1Gxh1 2.EFxh1 1Gg2 2.Gxg2 1Gf3 2.Gxf3 1Ge4 2.Gxe4 1Gd52.Gxd5 1Gc6 2.Gxc6 1a5 2.EFxa5 1a6 2.EFxa62219. (2, Valerio Agostini)1.EFg3? tempo 1Ge3 2.EFxe3 1Gf2 2.EFxf2 1exf4 2.EFg7 1Dd3/Dd1 2.EF(x)d3 1Dxc42.EFd3 1Da~ 2.EFc3 ma 1Gb1! 1.EFc8! tempo 1Ge3 2.Df3 1Gf2 2.Gxf2 1exf4 2.EFh81Dd3/Dd1 2.U(x)d3 1Dxc4 2.EFxc4 1Gb1 2.Dxb3 1Da~ 2.EF(x)c5 Matti cambiati conchiavi a tempo, effetti Somov B2. Un bel ritorno per il nostro collega perugino.2220. (2, Efren Petite)1Ge4 2.Dg4 1Gf5 2.Df3 1.Uhf4? [2.Dg4] ma 1Gf5! 1.Uff4? [2.Df3] ma 1Ge4! 1.Ud4?[2.Dg4] 1Gf5 2.Df3 1cxd4 2.EFxd4 ma 1axb5! 1.Gd5? [2.Dg4] 1Gf5 2.Df3 ma 1axb5!1.Dd6! [2.EFxc5] 1Ge4 2.Uxe4 1Gf5 2.Uxf5 1Gxd6 2.EFb2Bikos, matti cambiati, Hannelius.2221. (2, Valery A. Kirillov & Boris Maslov)1.c4? [2.Gd5 (A)] 1c6 2.Dc5 (B) 1De7/Df4 2.Ue5 (C) 1cxd6 2.EFxd6 ma 1Gxc4!1.Uh6? [2.Ue5 (C)] 1Gd5 2.Gxd5 (A) 1c5 2.Dxc5 (B) 1f6 2.hxg8=EFG 1cxd6 2.EFxd6 ma1Gf6! 1.Ge5! [2.Dc5] (B) 1Dxe5 2.Uxe5 (C) 1Db3 2.Gd5 (A) 1f6 2.hxg8=EFG 1cxd62.EFxd6 Pseudo-Le Grand ciclico, matti trasferiti.

  • 248

    2222. (3, Karol Mlynka)1.Ua3? [2.Uh3] ma 1g5! 1.Uxa6? tempo 1g5 2.Gf8 [3.Uh6] 1Sh8 2.Gg6 [3.Ua8] ma 1g6!1.Gf8! tempo 1g5 2.Uxa6 [3.Uh6] 1g6 2.Ua3 [3.Uh3] 1Sh6 2.Uxc5 [3.Uh5] 1Sh8 2.Ua3[3.Uh3] Matti cambiati.2223. (3, Evgeny Bogdanov)1Uxa6 2.EFg5+ Sd4 3.c3 1Uc3 2.Gxc3+ Sxd5 3.Gb7 1.e8=EF? [2.EFh5] ma 1Sxd5! 1.EFf4+?Sxd5 2.Gc4+ Sd4 3.c3 2Uxc4 3.EFxd6 ma 1Sd4! 1.cxd3? [2.EFxe4] 1exd5 2.e8=EF+ Ge63.EFxh8 2Sf6 3.EFh6 1Uc4 2.EFf4+ Sd4 3.EFxe4 2Sxd5 3.EFxd6 ma 1Gh7! 1.Gxd3?[2.EFxe4] 1exd5 2.e8=EF+ Ge6 3.EFxh8 2Sf6 3.EFh6 1Uc4 2.EFf4+ Sd4 3.EFxe4 2Sxd53.EFxd6 ma 1Gh7! 1.Gb7! [2.f4+ (A) Sxd5 3.c4 (B)] 1Sxd5 2.c4+ (B) Se5 3.Gc3 (C) 1Uxc22.Gc3+ (C) Uxc3 3.f4 (A) 1exd5 2.e8=EF+ Ge6 3.EFxh8 2Sf6 3.EFh6 1dxc2 2.EFc3+ Uxc33.Gxc3 2Sxd5 3.Gxc62224. (3, Arieh Grinblat)1.f4! [2.Ue5+ fxe5 3.EFxe5] 1Dc5 2.De3+ Sd4 3.Dgf5 2Sd6 3.Dgf5 1Dd3 2.Ued7+ Se63.Ud6 1Df3 2.De3+ Sd6 3.De4 2Sd4 3.Ue4/Uc4 Uno sfortunato duale. 1Uc5 2.Ucd7+ Sc43.De32225. (3, Eligiusz Zimmer)1Sxa2 2.Sc3 tempo Sa3 3.Ua1 1.Ug1? 1.Uf1? 1.Ue1? 1.Ud1? tempo, ma 1Sa3! 1.b4? tempo, ma1Sa3! 1.Sd2? tempo, ma 1Sa3! 1.Uh4! tempo 1Sa3 2.Sc2 [3.Ua4] 1Sc1 2.Uh2 tempo Sd13.Uh1 1Sa1 2.Sc2 tempo Sxa2 3.Ua4 1Sxa2 2.Sc2 [3.Ua4]2226. (H2, Andre Dikusarov)1.Gg5 Sf2 2.Sh4 Sg2 1.Gd6 Gd3 2.Gh2 Gf5 1.Ge6 Sg1 2.Gg4 Gf1 1.Sh2 Gg3+ 2.Sh1 Gf3Eco camaleonte.2227. (H2, Andre Dikusarov)a) 1.Df6 Sg3 2.Se5 Sf3 b) 1.Dg3 Sf2 2.Df5 Gb7 Eco camaleonte anche qui.2228. (H2, Pierre Tritten)1.EFh3 Gg1 2.EFf5 Gh2 1.Ge4 De3 2.Gf5 Dd52229. (H2, Pierre Tritten)1.e3 Df2 2.EFf6+ Gxf6 1.Gb6 Sxb6 2.Ud6+ Uxd6 Controscacchi, matti modello.2230. (H2, Mikola Kuligin & Roman Zalokotsky)1.g5 Gxf7 2.Uxb4 Ua2 1.f2 Gd1 2.Uxb2 Ua4 Zilahi, matti modello, eco.2231. (H2, Karol Mlynka)1.Uf6 Df7+ 2.Sf5 Dg7 1.Ue6 Dg7 2.Gf6 Df7 Scambio di mosse bianche, matti modello.2232. (H2, Tode Ilievski)1.De4 Db3+ 2.Sxd5 Uc5 1.Dc4 De7+ 2.Sxc5 EFg1Autoinchiodatura preventiva, scacchi doppi, matti modello.2233. (H2, Pierre Tritten & Antonio Garofalo)a) 1.Gb5 Ud8 2.Dxc3 Dxc3 b) 1.Ug3 Gh6 2.Uxf2 Dxf2 Una batteria nera viene interferita ondepermettere al pezzo nero antistante di catturare un pedone che impedisce il matto. Nel contempo la prima mossanera sgombera la via a un pezzo bianco che va a controllare la casa d2 la cui guardia deve essere lasciata dalpezzo mattante.2234. (H2, Pierre Tritten & Antonio Garofalo)a) 1.Gf1 (Gg2?) Gf7 2.Uxe4 Dxe4 b) 1.Ub8 (Uxb6?) Uc8 2.Dxb5 Dxb5 Quasi stessa idea tematicadel precedente: in questo caso la batteria nera si autodistrugge perch muove (con antiduale) il pezzoretrostante. Il pezzo tematico bianco si ritrova con una via sgomberata ma solo dopo aver mosso.2235. (H2, Almiro Zarur)1.Dxf3 Dc3 (Dd4?) 2.De1 Dd2 1.Dxc2 Dd4 (Dc3?) 2.De1 De3 Ottimo lavoro.2236. (H2, Alexandre Pankratiev)a) 1.Dd3 Gd1 2.Ue4 Dg2 b) 1.Dd1 Dd3 2.Ge4 Gg4

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    2237. (H2, Alexandre Pankratiev)a) 1.Gxc3+ Sb5 2.Gb4 Dd2 b) 1.Uxb3+ Sa5 2.Ub4 Ge5 Zilahi, Visserman.2238. (H2, Vadim Vinokurov & Alexandre Semenenko)1.fxe3 Gg1 2.D5f4 Gxe3 1.bxc2 Uc3 2.Ub3 Uxc2Theme of the "Romolo Ravarini-90" JT + annihilation of the white knights, tempo move of the thematic whitefigures, exchange of their functions and the effects "Come-and-go" in black play (Authors).2239. (H2, Valery Barsukov & Antonio Garofalo)a) 1.Sxe5 U3xd6 2.Se4 Ue6 b) 1.Sxd3 Gxd6 2.Sc3 Ge5 Zilahi, switchback nero in a), bianco in b).2240. (H2, Pietro L. Placanico)a) 1.Df5 Gc5+ 2.Sd5 Ue5 b) 1.Db7 Se4 2.Sd2 Gb42241. (H2, Francesco Simoni)1.Gb7 Ue8 (Gd6? Ug8?) 2.Gxe4 Ge3 1.Ub4 Gd6 (Ue8? Ug8?) 2.Uxe4 Uf81.Dxd2 Ug8 (Gd6? Ue8?) 2.Dxe4 Dd3 (Dg6?) 1.Gxc8 d3 2.Gxg4 Dg6 (Dd3?)TF H2. I-II-III mostrano autoblocchi nella medesima casa con doppio duale evitato. In I e II c' lo scambio difunzioni tra il pezzo bianco mattante e quello che muove per primo. III-IV mostrano autoblocchi su casediverse con cattura del pezzo bianco che muove nell'altra soluzione e duale evitato nel matto (Author).2242. (2.5, Pierre Tritten)1Gf1 2.exf1=G Da4 3.Gb5 Dec5 1Db1 2.cxb1=U d6 3.Ub6 Dc7 Molti buoni elementi: sacrifici,sottopromozioni, scambio di funzioni, interferenze, matti modello, ma la seconda soluzione manca di corri-spondenza mossa per mossa. Credo che sarebbe preferibile una perfetta corrispondenza con una costruzionegemellare nella quale un Pedone bianco funga da antiduale. 8/p7/k3S3/q7/6K1/2SP3B/2p1p3/8 h2,5 - b)d3d5 a) 1...Db1 2.cxb1=U Gg2 3.Ub6 Dc7 b) 1...Gf1 2.exf1=G Da4 3.Gb5 Dec5 (GJP).2243. (H2.5 Pierre Tritten)1Gd5 2.Sc5 Se8 3.Sd6 De4 1Df3+ 2.Se4 Dg5+ 3.Sf5 Dh32244. (H3, Andre Dikusarov)1.Df6 c7 2.Sd4 c8=EF 3.Se5 EFc5 1.De5 c7 2.Sd4 c8=EF 3.e3 EFc41.Gd3 c7 2.Ge2 c8=EF 3.Sd3 EFc32245. (H3, Leonid Makaronez)1.EFd3 Da8 2.Sd5 Sb6 3.EFe4 Dc7 1.EFd4 Dd5 2.Ud3 Sd6 3.Df3 Dg32246. (H3, Anatoly Mityushin & Roman Zalokotsky)a) 1.d1=EF Sb7 2.EFxd3 c8=EF 3.EFg6 EFh3 b) 1.d1=U Sb8 2.Ua1 c8=U 3.Ua5 Uc62247. H3, Pierre Tritten)1.Sc6 Gb5+ 2.Sb7 Ga4 3.Sa6 Gc6 1.Sc5 Sb1 2.Sb4 Sa2 3.Sa5 Sb32248. (H3, Evgeny Fomichev)1.Gg1 Gxg1 2.EFf2 Dc5+ 3.Sd4 Gxf2 1.EFa7 Gxa7 2.Gb6 Df2+ 3.Sd4 Gxb6 Una bella soluzionecon Bristol di sacrificio e successivo Bristol con effetto Maslar (o "mouse trap"), l'altra... lo specchio (GJP).2249. (H3, Eligiusz Zimmer)a) 1.Gf6 Gh5 2.Ge7 g6 3.Se8 g7 b) 1.Sg8 Gxe5 2.Gf6 gxf6 3.Sh8 f7 Miniatura Indiana con mattimodello ad eco camaleonte, esce dall'ordinario, sebbene il Re nero sia sotto scacco in b).2250. (H5.5, Pierre Tritten)1d4 2.Db4 d5 3.Dc6 dxc6 4.b6 cxd7 5.Gb7 d8=EF 6.Gg2 EFd11b4 2.Dc7 b5 3.Da6 bxa6 4.b6 a7 5.Ga6 a8=EF 6.Ge2 EFh12251. H8, Jorma Pitkanen)1.g1=D h3 2.Df3 exf3 3.e2 f4 4.e1=G f5 5.Gg3 f6 6.Gc7 bxc7 7.b6 c8=U+ 8.Sb7 a8=EF AUW2252. (S2, Efren Petite)1.Ue5? tempo 1fxg6 2.Ug5 Uh1 1f5 2.Uxf5 Uh1 1f6 2.Uf5 Uh1 ma 1fxe6!1.Uf5? tempo 1fxe6 2.Ue5 Uh1 1fxg6 2.Ug5 Uh1 ma 1f6!1.Ug5? tempo 1fxe6 2.Ue5 Uh1 1f5 2.Uxf5 Uh1 1f6 2.Uf5 Uh1 ma 1fxg6!1.Uxb5! tempo 1fxe6 2.Ge5 Uh1 1fxg6 2.EFxg6 Uh1 1f5 2.EFxf5 Uh1 1f6 2.EFf5 Uh1In tre tentativi lo zugzwang nero sventato con zugzwang bianco. La chiave libera pezzi bianchi. Moltobuono! (GJP). Pickaninny; pseudo-Kiss completo; Vladimirov.

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    2253. (S3, Jorma Pitkanen)1.Dd7? ma 1dxe4! 1.Db5? ma 1d4! 1.exd5? tempo 1exd4 2.Dd7 tempo d3 3.EFa6+ Ga7 ma1dxc5! 1.cxd5! 1exd4 2.Dd7 d3 3.EFa6+ Ga7 1dxc5 2.Db5 c4 3.EFa7+ Gxa72254. (H=4.5, Gyrgy Bakcsi & Lszl Zoltan )10-0-0 2.EFxe6(Dg8) Dxf6 3.EFe4 Dxe4 4.Db1 Ue1+ 5.Sd3 Sxb1=2255. (H2, Eligiusz Zimmer)a) 1.Uxh4 Sc3 2.Ua4 b4 b) 1.Uxb2+ Sc4 2.Ub5 Sxb5=2256. (H2, Oleg Ageev)1.Dxd5(B) Gxb2(N) 2.Gxg7(B) Dc3 1.Gxa1(B) Gb2 2.Dxd5(B) Df6 Un po' confuso, ma anche curiosoprocedimento antiduale nella grande diagonale (GJP).2257. (H2, Pierre Tritten)1.Dxg5(Db8) Dc4 2.Sc8 Db6 1.Gxd6(Gf8) Dh7 2.Se8 Df6 Zilahi, autoblocco preventivo, eco.2258. (Serie H13, Tode Ilievski)1.f5 2.f4 3.f3 4.f2 5.f1=D 6.g1=U 7.Ug3 8.Ud3 9.Se3 10.Sd2 11.Sc1 12.Ud1 13.Dd2 Uc32259. (Serie H=37, Jorma Pitkanen)1.Se6 2.Sf6 3.Sg6 4.Sh6 5.Sh5 6.Sh4 7.Sg4 8.Sxf4 9.Sg4 10.Sg5 11.Sg6 12.Sf6 13.Se6 14.Se515.e6 16.Gg4 17.Gf3 18.Ge2 19.Gd1 20.Gc2 21.Gb3 22.Sf4 23.e5 24.e4 25.e3 26.Sf3 27.Sf2 28.e229.e1=U 30.Ue2 31.Sf1 32.Se1 33.Ud2 34.Sd1 35.Sc1 36.Uc2 37.Sb1 Dxb3=2260. (2, Hans Nieuwhart)1.Dxd4(Dg1)! [2.Df3] 1hxg1=EF(EFd8) 2.EFe6 1hxg1=U(Uh8) 2.EFf6 1hxg1=G(Gf8) 2.EFd41hxg1=D(Db8) 2.EFd6 Composizione chiara, con buon uso della condizione fairy e un classico tema. Laminaccia pu essere impedita soltanto con la cattura in g1 e questo origina il blocco d'una casa di partenza, chelascia senza difesa una casa vicina al Re nero e la Regina bianca pu quindi mattare. Eccellente! (GJP). TemaRopke, cio duale evitato basato sulla promozione di 2 (in questo caso 4!) promozioni dello stesso pedone neronella stessa casa, ottenendo uno splendido AUW-anticirce.2261. (S6, Wilfried Seehofer)1Uxb1 1.Dc3! Ub1+ 2.Dd1 Ub8 3.Sg1 Ub1 4.Sh2 Ub8 5.Sh3 Ub1 6.Db2 Uh1

    Soluzioni del fascicolo n. 51

    2262. (2, Viktor Lukashv)1.EFd6? [2.De6] 1Gf2 2.e6 ma 1hxg5!1.De4? [2.Gg3] 1Gf2 2.EFxh6 1Sxe5 2.EFd6 ma 1h1=D!1.EFxh6! [2.Df7] 1Gf2 2.De4 1Gf8 2.De6 1Ge7 2.De6 1Sxe5 2.EFd62263. (2, Viktor Lukashv)1.Da2! [2.Dc3] 1Uc6 2.EFe5 1Gc6 2.EFh7 1Db7 2.EFxc4Tre difese che sgomberano tre linee bianche.2264. (2, Viktor Lukashv)1.De3? [2.EFd5] 1Dxe3 2.Uxe3 1Sxe5 2.EFe8 ma 1Df4!1.De7? [2.EFd5] 1Df4 2.Ue3 1De3 2.Uxe3 1Gxe7 2.EFxc4 ma 1Sxe5!1.Dxb4! [2.EFd5] 1Df4 2.Ue3 1De3 2.Uxe3 1Gxb4 2.EFxc4 1Sxe5 2.EFe82265. (2, Jorma Pitkanen)1.EFa2! tempo 1Sc6 2.Uc3 1Se4 2.Ub4 1Sc4 2.Ua3 1Se6 2.Ub62266. (2, Viktor Lukashv)1.EFh5! [2.EFf3] 1De~ 2.EFd5 1De3 2.f3 1De7 2.Dd6 1Dd4 2.Gd3 (1De5 2.Dg5)2267. 2, Pietro L. Placanico)1.Dc7? tempo, ma 1Sb6! 1.Sd2! tempo 1Dxc2 2.Sxc2 1Dxd5 2.Sc1 1Dxd3 2.Sxd31Dxa2 2.Se3 1Dxc6 2.Dxc6 Quattro matti di batteria reale.

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    2268. (2, Andre Dikusarov)1.Sb2! [2.D2xf3] 1EFxf2 2.Uxf1 1Uxf2 2.Uxg1 1Ue2 2.EFa1 1Gxd2 2.EFxd2 1Sd12.EFa12269. (2, Alberto Armeni)1.d4? [2.d5] 1Uxe8 2.fxe8=EF 1d5 2.Dxc5 1cxd4 2.Dxd4 ma 1cxd3 e.p.! 1.0-0! [2.Ue1]1Uxe8 2.fxe8=EF 1Ug8+ 2.fxg8=EFG 1Uxh2/Uh3 2.f8=D 1d5 2.Dxc5 Vallado.2270. (2, Alberto Armeni)1.Ge4? [2.Dg4] 1Uf4 2.Ue7 1Ug8 2.Dxf7 1Df2 2.Uc5 ma 1Ug7!1.Uc4? [2.Ue4] 1Uf4 2.Gc7 1Dc5 2.Uxc5 1Df2 2.Uc5 ma 1Ub7!1.Dd2! [2.Dc4] 1Uxc7 2.Gxc7 1Uf4 2.Ue7 1Db2 2.Uc52271. (2, Gabriele Brunori)1.Dxf3? [2.Ud4] 1e2 2.Dd2 ma 1Ua4! 1.De6? [2.Ud4] 1Ua4 2.Dc5 ma 1e2! 1.Db3![2.Ud4] 1e2 2.Dd2 1Ua4 2.Dc5 Notevole per essere un'opera prima!2272. (2, Valerio Agostini & Stefano Mariani)1cxd5 2.Dd7 1D~ 2.f4 1Df4 2.EFg7 1.e4? [2.Dd7] 1exd5 2.EFxf5 1Ua7 2.Dc4 ma1d3! 1.e3! [2.EFg7] 1Dxe3 2.f4 1exd5 2.EFe7 1dxe3 2.Dd7 1Ua7 2.exd4Un altro gradito ritorno quello di Stefano Mariani, con un lavoro in coppia che mostra matti cambiati, avviatidal doppio passo di pedone.2273. (2, Valerio Agostini)1.EFc1? [2.Dc4] 1Gf5 2.Dd5 1Gc7 2.Gxf2 1Gd5 2.Dxd5 1Gxd3 2.exd3 ma 1Uc8!1.EFe6? [2.Dd5] 1Gc7 2.Gxf2 1D~ 2.D(x)c4 1Ge5 2.Gxf2 ma 1Ud8!1.EFh3! [2.Gxf2] 1D~ 2.D(x)c4 1Gf5 2.Dd5 1Gf3 2.exf3 1Gxd3 2.exd3Uno splendido pseudo le Grand ciclico, con le tre chiavi che inchiodano tre diversi pezzi neri.2274. (3, Pasquale Macina)1.EFd5! tempo, 1Sf8 2.EFd8+ Sf7 3.e6 1Se8 2.f6 tempo Sf8 3.EFg82275. (3, Valery Barsukov)1.EFa1! tempo, 1f4 2.EFe1+ fxe1=EF 3.Df5 2Sd4 3.Db3 1Sxd2 2.Uxf2+ Se3 3.EFa7 1Sf42.EFd4+ Sg3 3.EFh4 1Se2 2.Dxf5 Sxd2 3.Uxf22276. (3, Arieh Grinblat)1.h4! [2.Ue7+ De6 3.Dxd3 2Sf4 3.Ue4] 1Dxc6 2.Dxd3+ Sxd5 3.Df4 1Dxf3 2.Dc4+ Se43.Dxd62277. (3, Valerio Agostini & Stefano Mariani)1.exd4? [2.Gd3] ma 1Ub8! (2.EFxe6?) 1.exf4? [2.Gd3] ma 1Dc5! (2.EFf4?) 1.Df7! [2.Dxd6]1Gc5 2.exf4 [3.Gd3] Dxf4 3.EFxf4 2Ub8 3.EFxe6 1Gb8 2.exd4 [3.Gd3] Dc5 3.EFxf4Lavoro tipico della scuola logica tedesca con l'applicazione del piano ausiliario (1.Df7!) per poi reintrodurre ipiani principali non subito funzionanti per difese efficaci (Authors).2278. (H2, Andre Dikusarov)a) 1.b1=G Ga5 2.Gd3 Gc3 b) 1.EFb4 Sa6 2.Sc5 Ge52279. (H2, Andre Dikusarov)a) 1.Sg1 Uxc5 2.Gg2 Uc1 b) 1.Uc4 Gd4 2.Ua4 Gxb2 c) 1.Uc3 Ge3 2.Gf3 Ue5 d) 1.Ud7 Uxb22.Ucc7 Ub8 e) 1.Uh5 Gd4 2.Uh6 Ug52280. (H2, Mikola Kuligin)1.Dxc3 Dg6 2.Dd5 Gc2 1.Dc1 Dd5 2.Dd3 Dd62281. (H2, Almiro Zarur)1.Uxe5 Sd7 (Sf8?/c2~?) 2.Uf5 Df2 1.Gxc2 Sf8 (Sd7?/e6?) 2.Gf5 Ug32282. (H2, Alexandre Pankratiev)1.Gf6 d4 (d3?) 2.Gc6 Dxd6 1.Ge5 d3 (d4?) 2.Uc6 e42283. (H2, Alexandre Pankratiev)a) 1.Gf3 Ug6 2.Ud5 Uc6 b) 1.Gc3 Gh4 2.Gd5 Ge7

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    2284. (H2, Pascale Piet)1.Uf4 exf4+ 2.Sg4 Gc8 1.Gxe3 Uxc6 2.Gf2 Uc32285. (H2, Stefan Parzuch)1.Df5 Sc2 2.Ud4 Gf3 1.Gd3 Gb3 2.Ue5 Uf42286. (H2, Andre Dikusarov)a) 1.Se7 Gg5 2.EFd7 Gxf6 b) 1.Sd4 Ug4 2.Sd3 Df2c) 1.Gd3 Gh6 2.c2 Dd2 d) 1.De7 Ug1 2.f5 Dg52287. (H2, Christer Jonsson)1.fxe4 Gh6 2.Sf6 Gg7 1.EFxd4 Sxe7 2.fxe4 Ug51.Sxd4 Ge3+ 2.Sd3 Df2 1.Sxe4 Ue3+ 2.Sf4 De62288. (H2, Luis Miguel Martin)a) 1.De2 Ug4 2.Dd7 Ge7 interferenze alle due Torri nere. b) 1.De2 Ug4 2.Dd7 Ge7 interferenze aDonna e Alfiere neri. Presumo che molti solutori avranno avuto difficolt a risolvere il gemello b) perch,normalmente, ci si aspetta una soluzione diversa. L'originalit del lavoro, a mia opinione, che le stesse mossehanno effetti uguali ma su pezzi diversi.2289. (H2, Gyrgy Bakcsi & Lszl Zoltan)a) 1.Uf1 Dc8 2.Uxf6 Ue7 1.Gc3 Dg6 2.Gxf6 Df8 b) 1.Uf1 f7 2.Uf6 fxe8EF 1.Gc3 f7 2.Gf6 f8D2290. (H2, Francesco Simoni)a) 1.Sf5 Gc2 2.Dg2 Uc5 b) 1.Sd6 Ud1 2.Df5 Gf4 b/b unpin, b/w pin, b/b interference, pin mate.White aristocratic. Distant pins, i.e. the pinned piece is not adjacent to the black king (Author).2291. (H2, Menachem Witztum)1.EFa4 Uc6 2.EFe4 EFd6 1.Sc5 Ub6 2.EFd4 EFc6 1.Sxe5 Ue6+ 2.Sxf5 EFf62292. (H2, Stefano Mariani)a) 1.EFxd4 Uxb5 2.Sf5+ De3 b) 1.EFxd5 EFa7 2.Se3+ Df5Problema eccellente, molto complesso, forzatamente abbondante di pezzi.2293. (H2, Jean Carf)1.De5 bxc4 2.Uf5 De4 1.Dd5 b4 2.Gf5 Dgf72294. (H2.5, Valery Barsukov)1...Uc2 2.Sd5 Gf2 3.Ud6 Uc5 1...Ud2+ 2.Se5 Ud4 3.Ue6 Df3 1...Df7+ 2.Se6 Ud2 3.Ge5 Dd82295. (H2.5, Mikola Kuligin & Vitaly Ivanovich Shevchenko)a) 1...EFa4 2.Se5 d4+ 3.Sd5 EFd7 b) 1...EFxb4 2.Se5 dxe4 3.f4 EFe72296. (H3, Pierre Tritten1.Ub5+ Sc6 2.Ub3+ Sd6 3.Uc3 Df5 1.Gb5 Sa5 2.e3 Sb4 3.Ue4 Df3Black Grimshaw on b5 (Author).2297. (H3, Mikola Kuligin & Vitaly Ivanovich Shevchenko)1.Def3 Uxf1 2.Uxc4 Ue1 3.Dd4 Ue3 1.b3 Uxb2 2.Sb4 Uxb3+ 3.Sa5 Ub5Due soluzioni interessanti ma non coerenti fra loro.2298. (H3, Vito Rallo & Antonio Garofalo)1.Df3 Ug1 2.Dxe5 Ug5 3.Dc6 Uxd5 1.Ua5 Ua1 2.Uxc5 Ua5 3.Uc6 Uxd5I pezzi neri tematici svolgono due compiti: liberare la quinta riga per permettere il passaggio della mattante epoi interferire l'za8 entrambi nella stessa casa.2299. (H3, Abdelaziz Onkoud)a) 1.EFd2 Gd5 2.Sc3 Gb3 3.Gd4 Dd5 b) 1.Sc4 Dd5 2.Gc5 Df4 3.Dd4 Gd52300. (H3, Christopher J.A. Jones)a) 1.De5 Gb6 2.EFg4 Gc5 3.EFxf3 dxe5 b) 1.De4 Ub3 2.EFg3 Uc3 3.EFf2 dxe4Eccellente problema con molti effetti tematici, fra cui una mossa "tempo".2301. (H3, Mario Parrinello)1.EFxg7+ Sxg7 2.Gb5 e8=EF 3.Gxe2 EFxe2 1.Gxe7+ Sxe7 2.Gf7 g8=EF 3.Gxa2 EFxa2Fantastico doppio sacrificio di pezzi neri. Difficile immaginare, prima, la posizione di matto che ne deriva.

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    2302. (H3.5, Tode Ilievski)1...Gf6 2.Uf8 Ge7 3.Se5 Se3 4.Uf5 d4 1...Se2 2.Se4 Gd4 3.Gf5 Ge3 4.Ue5 d32303. (H4, Christer Jonsson)1.EFh8 f6 2.Dg7 fxg7 3.Gg4 gxh8EF 4.Ge2 EFh1 1.Ge6+ fxe6 2.EFa8 e7 3.Dd8+ exd8=EF 4.EFg2 EFd12304. (H5, Mikola Kuligin)1.Sh8 Se4 2.Dg8 Sf5 3.Ge3 Sg6 4.Gh6 Gxd4+ 5.Gg7 Gxg72305. (H6.5, Tode Ilievski)1Gg2+ 2.Sc7 h4 3.Sd6 h5 4.Se5 h6 5.Sf4 h7 6.Sg3 h8=G 7.Sh2 Ge52306. (H3, Hans Nieuwhart)1.Sf5 Ge7 2.Gf7 LOxf7-g6+ 3.Sf4 Gd6 1.Sh3 Sf3 2.Gg2 Gg7 3.Gg8 LOxg8-h82307. (S5, Wilfried Seehofer)1.Id2? [2.Id8 axb6] ma 1Sc6! 1.Da8? 1.Dc8? [2.Db6 axb6] ma 1Sc4!1axb6 1.Id1! [2.Id8 axb6] 1Sc6 2.Da8 Sc5 3.Id2 Sc6 4.Id8+ Sc5 5.Db6 axb62308. (HS2.5, Mario Parrinello)a) 1...exd1=G(Gc8) 2.g5 Gg4 3.Dxg4(Db1)+ EFxb1(EFd8)b) 1...exd1=U(Ua8) 2.f5 Uab8 3.Gxb8(Gc1)+ EFxc1(EFd8)c) 1...exd1=D(Dg8) 2.hxg6(g2) Dh6 3.Uxh6(Ua1)+ EFxa1(EFd8)Elegante e pieno sfruttamento della condizione anticirce, con gemelli Forsberg; un capolavoro!2309. (H3.5, Vito Rallo)a) 1...Dxe3 2.Sh2(f2) Sxf2 3.Sh1(Df1) Df5 4.Dh2 Dg3b) 1...Sxe3 2.Sh1(f3) Sf2 3.fxg2 Sg3(Dh3) 4.g1=D Df22310. (H3, Stefano Galletti)a) 1.a5 Gxa8(Gf1) 2.Sb7 Gb5 3.Sa8 Gc6 b) 1.Sxb7(Gf1) Sg1 2.Sc8 Gxa6(a7)+ 3.Sxb8(Gc1)Gf4 c) 1.a5 Sg1 2.Gxb7 Ga7(Ga6)+ 3.Sxa6 Gxc5(Gc4) d) 1.Sb5 Se2 2.Sc4 Gd5+ 3.Sd4 Ge5

    Award in the Santi Pirrone Memorial Tourneyby Antonio Garofalo

    Some of the participating problems, curiously enough, did not show the required theme, despiteChris Feathers perfect English translation of the text. The example quoted had a structural defect,in that the rear white battery piece has no function after the Grimshaw, and that defect wasunfortunately shared by many of the problems in the tourney. Nevertheless some of thoseproblems have found their way into the award, as they have compensating merits of other kinds.A few notes on problems not featuring in the award: Togookhuus work (qh1/we3) could be setmore economically, saving bx and b{ by using a b| instead. The position in question will besent to the author. Several problems contained totally unnecessary white pieces, going beyond theabovementioned structural defect in the example problem; others did not show a genuine whiteGrimshaw.

    The competing problems were 32 in number, sent by 19 composers. Their surnames are givenin alphabetical order, followed in brackets by the diagram numbers allocated in order of receipt:Fadil Abdurahmanovic (3), Oscar Bonivento (5), Gennady Chumakov (23), Bruno Fanelli (5),Harald Grubert (4, 9), Jean Haymann (16, 17, 20), Christer Jonsson (11), Leonid Makaronez (21,22), Mikhal Marandyuk (2431), Enzo Minerva (19), Arpd Molnar (32), Dieter Mller (4, 8,9), Emanuel Navon (1318), Mike Prcic (3), Michael Shapiro (18), Gerard Smits (10),Lkhundevin Togookhuu (12), Pierre Tritten (1, 2), Menachem Witztum (6, 7).My thanks to all the participants who kindly made the effort to honour the memory of the greatcomposer Santi Pirrone. Here then is my award, which will become final three months from thedate of publication.

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    1st PrizeA. Molnar

    2nd PrizeJ. Haymann

    3rd PrizeM. Witztum

    1st Hon. Ment.G. Smits

    'EF5U1Q3'G1%Sdddddddd

    G'U'5D13$S111%dddddddd

    '%'151DG13$U1Q11$SQ1U1dddddddd

    GU'11$Q1EF1131S1'5%dddddddd

    H2 (5+7) C+2 sol.

    H2 (4+10) C+b) b3-f7

    H2 (7+15) C+b) wb4-c1

    H2 (5+13) C+2 sol.

    2nd Hon. Ment.F. Abdurahmanovic

    & M. Prcic

    3rd Hon. Ment.E. Navon & M. Shapiro

    Commend.C. Jonsson

    'G5GSU1UQQ13Q15'dddddddd

    G1Q1$1553Q11'1'1GU1SUdddddddd

    '%51113111$5EFSGQ$UddddddddH2 (8+8) C+

    2 sol.H2 (7+14) C+

    b) wf6|g3H2 (5+13) C+

    b) d6-f4

    1st Prize - Arpd Molnar1.d4 e4+ 2.f5 f6 1.d4 e4 2.e3 c1Double line-opening for the w, with perfect deployment of the pieces and excellent economy.2nd Prize - Jean Haymanna) 1.g4 (~? d4?) e6 2.f4 d4 b) 1.e3 (~? g5?) e6 2.f4 g5 It seems asthough the thematic black pieces could be sacrificed on the mating square or simply move away,but the need for a selfblock on the square which is no longer controlled after the white interferencedetermines their moves otherwise. The only defect is the idleness of a white piece in each mate,alternating between the twins. Even with a defect like that, this is an excellent piece of work.3rd Prize - Menachem Witztuma) 1.g6 f4 2.b5 xc6 b) 1.g6 f4 2.d1 xd3It is a most interesting idea to use the white Grimshaw to unpin the black piece which thenselfblocks. A perfect match between the twin positions.1st Hon. Ment. - Gerard Smits1.b2 d7 2.e6 e2 1.b2 d7 2.xd6 b8Elegant despite the mass of black pawns.2nd Hon. Ment. - Fadil Abdurahmanovic & Mike Prcic1.g7 ac4 2.d3 d5 1.g7 c4 2.d4 e5Every white piece plays its part in the solution. Direct and indirect battery mates.

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    Commend. - M. Marandyuk Commend. - E. Minerva Commend. - E. Navon$'EF1113%5'U5GS$dddddddd

    S'EF1511111Q3QU$'5Gdddddddd

    '151311'QSD5QGUU$dddddddd

    H2 (4+11) C+b) c6-b5

    H2 (6+12) C+b) wc4zd3

    H2 (7+10) C+2 sol.

    3rd Hon. Ment. - Emanuel Navon & Michael Shapiroa) 1.db6 f4 2.c6 d6 b) 1.xb7 f4 2.dc6 d3The f1 is poorly used in part a), but becomes necessary in part b). But for that this work wouldhave been more highly placed.

    Commendations in alphabetical order:

    Christer Jonssona) 1.c7 c3 2.d4 xd6 b) 1.c7 c3 2.c4 g8 Simple but pleasant.Mikhal Marandyuka) 1.e4 d4 2.d5 f7 b) 1.e4 d4+ 2.c5 c7The indispensable minimum.Enzo Minervaa) 1.g7 e4 2.d5 xf7 b) 1.g7 e4+ 2.xd4 xf6A good exercise in the required theme, with no defects.Emanuel Navon1.c6 f4 2.e5 c3 1.c6 f4 2.f5 xd6Direct and indirect battery mates.

    Possible claims must be sent to the Editor: Antonio Garofalo, via Collodi n.13 70124 Bari - Italy.E-mail: [email protected]

    Bari, May 2009Int. Judge A. Garofalo

    Revision of the Enzo Minerva 45 Jubilee Tourney 2007 Award

    La SPG 22.0 di Gianni Donati, alla quale era stata assegnata la 2 Menzione Onorevole, risultademolita, come segnalato dal Grande Maestro Michel Caillaud: 1.f3 a5 2.d4 a4 3.c6 d54.xd8 g5 5.Se6 g7 6.f8 g4 7.xh7 f5 8.g5 g6 9.h4 f5 10.h5 f6 11.h6 g8 12.h7 e613.h8= c5 14.8h3 c4 15.a3 c6 16.b3 axb3 17.h3 bxa2 18.g1 0-0-0 19.b3 df8 20.f3c3 21.f2 gxf3 22.e3 h6+ Il Verdetto pertanto modificato nella seguente forma definitiva:1 Premio, R. Osorio & J. J. Lois; 2 Premio, K. Prentos & A. Frolkin; 3 Premio, M. Caillaud(SPG 20.5); Menzione Onorevole, M. Caillaud (SPG 17.5); Menzione Onorevole Speciale, P.Raican. Seguono in ordine alfabetico: Lode, I. Denkovski; Lode, G. Donati (SPG 20.0).

    Lesa, 13 maggio 2009, Enzo Minerva

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    Award Best Problems 2002-2003: H3 or more moves.by Vlaicu Crian & Eric Huber

    When Antonio Garofalo posted a request for judges on the Mat Plus forum at the beginning ofOctober 2008, we were immediately interested. The main danger encountered when judgingproblems published a while ago is to apply todays standards to evaluate the quality of yesterdaysoutput. One does tend to remember problems published in the meantime, but this can be avoidedthanks to database help. By the way, we forward many thanks and kudos to the Chess ProblemDatabase Server (http://www.softdecc.com/pdb/index.pdb?langt=EN&langn=EN) for its inva-luable service. In what follows below, we shall refer to the problems registered in the PDB serverby their identifying number and the reader may browse the page presenting the problem with alink like this one: http://www.softdecc.com/pdb/search.pdb?expression=PROBID='P0004254'.

    We received 130 anonymous problems in the form of a PDF document and decided to split theaward in two parts of approximately the same size: 74 helpmates in three and 56 helpmates in fouror more moves. However, the analysis of the problems soon made us realise two regrettable facts:1. The general level was disappointing. The qualities we expected from helpmate problems were

    a clear and neat presentation allied with sufficiently original contents and too few problemsmet our standards.

    2. Too many problems suffered from various levels of anticipation or had been published twice.We would like to emphasise this second point and direct readers and other judges attentionto it. For that purpose, we devote the following paragraphs to this subject.

    First, we cannot help wondering what prompts some composers to send their problem to more thanone chess review. This is clearly not allowed by our Codex (http://www.saunalahti.fi/~stniekat/pccc/codex.htm) and we wholeheartedly encourage composers to update their files. Here is thelist of double publications: 704 (Lasiy) triple publication, also 15012 in Schach 05/2002 and H0678 in StrateGems 07-09/2002. Bravo, Mr. Lasiy! 708 (Bereshnoy), previously published in Problem-Forum 2001. 755 (Bantush), published in Ideal Mate Review 1999. 756 (Lozek), published in The Problemist, 03/2002. 759 (Grushko), republished in Finland P1006153. 814 (Vitale), P0573130 published in The Problemist in 1997. 969 (Grushko), published by the author in a mirrored version P1005779.The leader in this section is M. Grushko. He was caught in the act in 2004, when Christian Poissonobserved M. Grushkos strange mania of publishing his fairy problems simultaneously inProblemesis and other reviews. We guess that nowadays Mr. Grushko pays closer attention to hisproblem files.

    The second subject of concern to us was the considerable number of anticipations. The mostindulgent interpretation of these anticipations is that some composers are attracted to old-fashioned ideas and are therefore bound to recreate existing problems. Here is the complete list: 612 (Pitknen), anticipated by A. M. Broer (P0509280) as far back as1928! 649 (Di Sarno & Vecu), anticipated by P0567825 817 (Nagnibida), Nil novi sub sole: identical with P0502008 (1948) and P0502036 (1971),mirrored position of P0502037 (1988) ! We are looking forward to the next occurrence of thisposition around 2030 A.D. 858 (Nagnibida), we all love Peter Kniest, but we should not go so far as to mirror his P0502030(feenschach 1971) to express the warmth of our feelings.

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    900 (Nagnibida), completely anticipated by P0563300, which was also more economical. 973 (Grigorian), P0501196 was better. 981 (Cistiakov), this one could have been included in the first section. The author manages toget rid of a pawn in his P0575521 (Schach 1995) and then publishes the 'new' problem in amirrored position as an original in the present review. What is the correct term for this behaviour?The winner in the anticipations section is the unchallenged M. Nagnibida (1939-2005).The rest of the competing problems displayed an amazing variety of defects: repeated moves(many originals), white pieces useless in one solution (707, 814, 850), lack of interplay (813) etc..However, there remained enough good problems.

    Helpmates in 3In the dj vu category, we noticed a few problems which could be related to past items featuringvery similar ideas. 603 (Di Sarno & Vecu) slightly improves P0501087 by presenting 4 solutions as 2x2 instead of4x1 but apart from the technical improvement, everything was already said in 1970. 607 (Ilievski), the same us/zx AUW was composed by John Niemann in 1972 (P0508388)with the difference that there were two thematic bPs. Here only one thematic pawn and one unitless is a worthy technical achievement, hardly anything more. 652 (Drazkowski), partly anticipated by P0532335 with different motivations at B1: here thereare active sacrifices of the bQ but in the better P0532335, a black Grimshaw. 779 (Muralidharan), P0556044 was not as economical, but showed reciprocal batteries as earlyas 1955. 810 (Fomichev), P0577699 is very similar and more economical.

    1st Prize895. C. J. A. Jones

    2nd Prize899. V. Vladimirov &

    V. Nefdov

    1st Hon. Ment.611. A. Onkoud &

    A. Garofalo

    2nd Hon. Ment.703. C. Jonsson

    'G1Q%3Q511QQUSdddddddd

    $511U'$Q31G1QQ1Sdddddddd

    %G'11115$1D31SG5Q$$dddddddd

    '1DGS11QQ1Q315'5dddddddd

    H3 - 2 sol. H3 - 2 sol. H3 - b) f4-d6 H3 - 2 sol.

    1st Prize: 895 (Jones)1.e7 d5 2.d6 xb7 3.d5 xd5 1.b4 d5+ 2.c4 a5 3.d5 xd5Typical Jones material and ODT strategy! The theme is quite interesting compared to thedefinition of the last WCCT theme: a white piece stops on a square before leaving it at the secondmove. Subsequently another white piece delivers mate on the same square. Thus, except theexchanged W1 and W3, it is interesting that Black also plays to the thematic square at his lastmove and this active Pelle sacrifice reminded us of Fadil Abdurahmanovics masterpiece (seeP0506884). The construction seems flawless.

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    2nd Prize: 899 (Vladimirov & Nefyodov)1.e3 e4 2.xe4 d3+ 3.f3 f6 1.f7 e5 2.xe5 d4+ 3.e6 g4Another impressive wealth of strategy: a Zilahi with active white sacrifices on squares just vacatedby Black. This Follow-My-Leader (FML) motivation is particularly difficult to achieve and here itis doubled, in diagonal-orthogonal correspondence; the squares where the bw is mated are squaresinitially occupied by the sacrificed white pieces. Each motivation in this problem has been seenpreviously, but the whole package seems to be original. Readers may think, with the benefit ofhindsight, that this work foreshadowed recent problems by the phenomenal Semenenko Brothers:P1072984 (consecutive white sacrifices on empty squares, 2004) and P1074091 (the bw marchesto squares initially occupied by white pieces, 2006).

    The Honourable Mentions go to problems with interesting strategy, but not quite perfectlyexecuted.

    1st Honourable Mention: 611 (Onkoud & Garofalo)a) 1.e4 xe5+ 2.xe5 a4 3.d4 d3 b) 1.c7 xd5 2.cxd5 xc2 3.c6 b4And here is a special case: 611 has by far the most original content in the competition! It featuresa Zilahi with paradoxical elements: the pinned piece finds itself pinned on another line after W2,so B3 (like B1) must unpin it. It is amazing that the by (in a) and bx (in b) pin both white piecesand unpin them! But there is also quite a price to pay: the three b{s, the rather clumsyconstruction and the poor twinning involving bw movement. In spite of all these defects, we stillconsider this composition worth rewarding.

    3rd Hon. Ment.974. S. Hudak

    1st Commend.605. C. J. A. Jones

    2nd Commend.701. I. Murarasu

    3rd Commend.757. T. Garai

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    %U35Q$$SQdddddddd

    U%515GQ13Q11Q1$Q'1Q11EF'Sdddddddd

    '5$31'11$5DUSdddddddd

    H3 - 2 sol. H3 - 2 sol. H2.5 - 2 sol. H3 - 2 sol.

    2nd Honourable Mention: 703 (Jonsson)1.xb4 c4 2.b2 e3 3.b3 b4 1.xa4 d5 2.c2 f4 3.b3 a4On his first move Black annihilates a wv and then performs two consecutive selfblocks, hence amutual Grimshaw on b3. The white strategy is less unified; in particular, W1 does not match,although there is an exchange of wt/wu functions, ending with a mate on the square vacated byBlack. This rendering of the WCCT-8 theme by means of wv annihilations mixed with Grimshawhas been shown before (see for instance P0562265, P0560076 and P0557852). We also deplore therather weak interplay. Without these defects, the problem would have been placed higher in theaward.

    w r y t { v

  • 259

    4th Commend.648. G. Chumakov

    5th Commend.758. C. J. A. Jones

    6th Commend. 896.K. Muralidharan

    3D1U111'QSQdddddddd

    %USG5$QQ1Q3QQdddddddd

    1SQGDD3dddddddd

    H3 ** - 1 sol. H3 - 2 sol. H3 - b) -d3

    3rd Honourable Mention: 974 (Hudak)1.xb3 b5 2.c2 xc3 3.b1 a3 1.xe5 a5 2.d4 xc2+ 3.c5 b4We liked the dual avoidance on W1: 1b1? and 1c5? simply fail because they occupy thesquare aimed at by the bK. B1 is also good, vacating a square allowing the bw passage to themating square, without being a distant selfblock. The switchbacks on W3 and the model mates arealso nice, but the rest of the play is rather dull.1st Commendation: 605 (Jones)1.a5 bxa5 2.c5 b6 3.d5 b4 1.d2 b5 2.c4 c6 3.e4 bxc4These chameleon echo mates are charming. There are certain similarities with P0567754,featuring the same white material, but the mates are quite different.2nd Commendation: 701 (Murarasu)1...f8 2.xc2 e1 3.e6 xf4 1...g5 2.bxc2 b8 3.f5 e8The rather free long range white pieces are very difficult to handle, hence the heavy constructionand the barely legal position. The solutions ending in pinmates are quite homogenous; perhaps theonly slight drawback being that the wR is not supported by the wB in the second solution.3rd Commendation: 757 (Garai)1.c3 f2 2.d1 xe3 3.c5 e4 1.c3 xe3 2.e1 e4+ 3.d3 f2A neat setting of a white move cyclic shift (ABC-CAB), using a halfpin. The FML effect at B2,unpinning the white piece, is not perfectly balanced: 2.d1 unguards the mating square (e4),while 2.e1 does not. We could not overlook the fact that the black strategy is not unified.4th Commendation: 648 (Chumakov)1...xh7 2.g7 gxf6 3.xh8 xh8 1...:f7+ 2.g8 gxh6 3.xh8 f8 1.xg5 hxg5 2.xg7gxf6+ 3.h6 4.xf7The original change of play between set play and solutions offers the Zilahi theme. The set playalso displays echo mates.5th Commendation: 758 (Jones)1.d6 xb8 2.e4 b6 3.d3 e6 1.xg3 xh6 2.d4 g5 3.d3 f6The strategy of annihilation of black units is rather thin, but the two black selfblocks on the samesquare and the white manoeuvres raise this problem to the level of a commendation.6th Commendation: 896 (Muralidharan)a) 1.d6 b1 2.d5 d3 3.d4 c2 b) 1.d5 c3 2.d4+ b2 3.d3 c3Three black tempi in each solution. Nothing less, but nothing more.

  • 260

    Helpmates in 4 moves or moreThe dj vu files are not empty in the longer helpmates either, but only a few problems areinvolved this time. 654 (Chumakov & Fomichev), P0500733 (Kornilov 1983) presents 5xExcelsior and 5 -promo-tions. 716 (Novomesky), nice chameleon echo mates, partially self-anticipated by P1002819. 859 (Lozek & Garofalo), looks like P0500796 (AUW+1 promotion) but mostly like P0508831with mate in the a1 corner.The remaining problems were not breathtakingly original but some could be singled out of the lot.

    Prize857. C. Jonsson

    Hon. Ment.713. C. Jonsson &

    R. Wiehagen

    1st Commend.855. D. Meinking

    2nd Commend.763. M. Ylijoki

    '$'$1S1DD111355%dddddddd

    $SG35QQ'dddddddd

    %51111'Q113'1Sdddddddd

    $1111S$11311'GddddddddH4 - 2 sol. H4 - b) g6-f3 H4 - b) f5-e5 H5.5 - 1 sol.

    Prize: 857 (Jonsson)1.b3 xe8 2.c4 xb8 3.c3 c6 4.d3 d6 1.c3 xe7 2.d4 xb7 3.d3 d6 4.e3c6 Undoubtedly the best problem of the competition! Two perfectly matched solutions end inchameleon echo mates. The double switchback by the wus is achieved by very simple means (i.e.annihilation of black pieces controlling wu initial squares). The black strategy is also simple three selfblocks and the bw running to the mating square. However, the interplay is ensured by theclever position of the bw, who opens the 3rd rank for the bys. The elegance of the interpretation isthe typical trademark of the distinguished Swedish composer.Honourable Mention: 713 (Jonsson & Wiehagen)a) 1.h2 g8 2.d6 f7 3.d5 b4 4.d4 f6 b) 1.h6 a2 2.d6 b3 3.d5 f4 4.d4 b4The old Indian theme is mixed with an impressive change of roles between the by and bz: eachof them alternately blocks on d4 and d6. Of course the white strategy is not entirely original (seefor instance P0516502), but we could not find a clear predecessor for the whole package. Thisproblem would have certainly received a prize if B3 had not been repeated.1st Commendation: 855 (Meinking)a) 1.f3 fxe6 2.a8 exd7 3.c8 dxc8 4.e4 h3 b) 1.h3 exf6 2.h8 f7 3.g8 fxg84.h4 g2The typical rush for r promotion of the wv has been shown several times before. Here theoriginal value is added to by the black Bristol on the 8th rank. Of course, this Bristol cannot bepure, but the double rendering of this idea and the exchange of roles between the bx and bydeserve recognition. Also interesting is that the bw acts as a barrier against bx and by checks hence forcing the order of moves. A pity that this could not be arranged without twinning!

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    3rd Commend.657. R. Wiehagen

    Commend.683. M. Grushko

    Commend.765. L. Vitale &

    R. Wiehagen3'$11QS'%dddddddd

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    $31'11$1QSddddddddH7 - 1 sol. H4 - b) c4-c2 H7 - 2 sol.

    2nd Commendation: 763 (Ylijoki)1...g2 2.h4 h1 3.f1 g2 4.d3 f1 5.g6 d3 6.h5 f5The content speaks for itself: two wt tempos and an interesting manoeuvre by the bzh3 to reachh5 in four moves. However the first wt tempo is quite artificial (no other white move is possible)and so is the switchback. More subtle is the reason why the bw cannot leave g4 and returnproviding a shortcut for the bzh3 to reach h5. The great number of black spectators seems to beunavoidable.3rd Commendation: 657 (Wiehagen)1.h6 e3! 2.h4 e4 3.g5 e5 4.f7 e6+ 5.g6 e7 6.h5 e8 7.g6 f6An achievement worth quoting: surprising Excelsior-hesitation ending in promotion. This hasbeen shown before even more economically (see for instance P0582675), but here the authormanaged to show four active blocks (compared to only 3 in Bakcsi and Zoltan) by using one extrahalf move.

    Commendations without order: "Everything in the first move".

    683 (Grushko)a) 1.c7 c5 2.c4 d8 3.b5 xe6 4.a6 xc7 b) 1.c4 b8 2.a7+ c7 3.b5 c4+4.a6 b5 This neat miniature receives a commendation for the unexpected bx active sacrificeon B1 in a). The twinning mechanism is also paradoxical. We disliked the repeated bw moves - adrawback which severely diminishes this problems value.765 (Vitale & Wiehagen)1.f3 gxf3 2.e4 fxe4 3.g5 e5 4.f4 e6 5.e3 e7 6.d2 e8 7.c1 e1 1.h1 g3 2.d1g4 3.g5 gxf5 4.f4 fxg6 5.e3 g7 6.d2 g8 7.c1 g5 A long Meredith with twosolutions of dissimilar values: the 1.h1! solution is brilliant, because the key can be understoodonly after Whites last move, while the 1.f3 solution has nothing special. Again, the repeatedbw moves (B3-B7) detract.

    We congratulate the winners and thank all the participants.

    Vlaicu Crian & Eric Huber, March 2009.

    My most sincere thanks to Vlaicu Crian e Eric Huber for their ample and detailed award, whichwill become definitive 3 months after publication. Possible claims must be sent to the Editor:Antonio Garofalo, via Collodi n.13 70124 Bari - Italy. E-mail: [email protected]

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    RicostruzioneD513QU1GEF$DGSdddddddd

    J. Tatalin - (Ricostruzione n. 39, BP50)29 Chess Olympiad - Novi Sad 1990 - Lode (Commend.)2 (8+5) C+1g4 2.e6 1exd4 2.a51.c2? [2.xe4] 1g4 2.f7 1exd4 2.d5 ma 1e3!1.e2? [2.xe4] 1g4 2.xg4 1exd4 2.b5 ma 1e3!1.c2! [2.xe4] 1g4 2.c8 1exd4 2.c51.a8? [2.xe4]1g4 2.f8 2.c8 1exd4 2.a5 2.d5 ma 1c6!Zagorouko 42, Ellerman-Makihovi.

    Un problema che a giudizio dei partecipanti e mio personale avrebbe meritato molto di pidella Lode ricevuta. A mia giustificazione solo il fatto che non conosco la qualit degli altriproblemi premiati. Non difficoltosa la ricostruzione, con la sola "trappola" di possibili duali.Partendo dalla soluzione data subito individuabile la posizione iniziale della Donna bianca, inquanto sono indicate ben 5 sue prime mosse. Da questo a trovare la posizione del Re nero, chedalla Donna riceve 7 matti da case differenti, il tratto e breve. Ovvia la posizione di un pedonenero in 'e5' conoscendo la mossa 1...exd4; facile trovare la posizione dell'b3 bianco conoscendodue sue case di arrivo (e6 ed c2). Individuata la posizione di 4 pezzi, due bianchi e due neri,non resta che aggiungere il "contorno" che l'unica cosa che richiede pi tempo; ma in questosono stati bravi tutti i partecipanti che hanno inviato la loro ricostruzione in tempi relativamentebrevi.

    Hanno partecipato alla gara (con commenti): - Prof. Oscar Bonivento: "Provo a cimentarminella ricostruzione (amore di antichi tempi... !)" - Antonio Garofalo: "Potevo evitare il tentativosupplementare a8, ma non vedo lo scopo di aggiungere un a7 nero." Hans Nieuwhart: "Goodhow the black knight can close the line, after the white Bishop close the line for the Queen. Multi-functional moves!" - Valerio Agostini: "questo esercizio mi ha appassionato non poco e mi hapermesso di conoscere un lavoro davvero eccellente. La cosa mi piaciuta molto! ... Il di pi venuto dal fatto che mi sono dovuto impegnare anche come compositore! Non avrei mai credutoche si potessero fare le due cose insieme!!" - Gerd Prahl: "Clouage direct, dclouage direct,Zagorouko 4x2, Ellerman-Makihovi (Winchloe). I like to reconstruct chessproblems. I alreadytake part in Synthetics of The Problemist and The Macedonian Problemist, also in Reconstructieof Probleemblad. Altri partecipanti: Alberto Armeni, Saverio Ce', Valery Krivenko, Efren Petite ePietro L. Placanico. Un record di partecipazione, ben 10!

    Quasi tutti hanno mandato la ricostruzione corrispondente all'originale con l'eccezione di due:Placanico, la cui ricostruzione (8/7S/6r1/4pk1P/3Rp3/1B5P/Q5sK/2B5) ha il pregio di utilizzareun Pedone bianco anzich un Cavallo bianco (mettendo al posto del h2 il Re bianco eaggiungendo il pedone in h3), ma ha il difetto di avere creato un duale in una variante tematica(1.e2? [2.xe4] 1...g4 2.hxg4/xg4). Posizione identica a questa stata inviata dal Prof.Bonivento, con la sola differenza del Re bianco in h1. Da notare: sono gli unici partecipantisprovvisti - sfortunatamente - di computer, altrimenti un programma avrebbe fatto loro notare ilduale.

    Ricostruzione n.40, soluzione tematica:1.d2? (C) [2.d7] 1...f6 (a) 2.d6 (A) Ma 1...e7! (b)1.f2? (D) [2.d7] 1...e7 (b) 2.f4 (B) Ma 1...f6! (a)

  • 263

    1.xb2! [2.d7] 1...e7 (b) 2.d2 (C) 1...f6 (a) 2.f2 (D)Inviare le vostre proposte a (send your diagram to): Vito Rallo, via Manzoni n.162, 91100Trapani. E-mail: [email protected]

    Vito Rallo

    NUTS (25)di Mr. Veneziano John Nunn, 1986

    5$'%3'$511111111QQQQQQQQUDGEFSGDUdddddddd

    Una partita ha inizio con 1.e4 e termina allaquinta mossa con x scacco matto.Siete in grado di ricostruire una simile partita?

    Questo problema scacchistico venne proposto dal tedesco Frederic Friedel, direttore diChessbase, sia a Garry Kasparov, sia ad Anatoly Karpov, riuniti a Londra per le trattative delloro terzo match per il titolo mondiale. Entrambi i K tentarono di risolvere l'enigma, primadurante un lungo viaggio in auto, poi per due giorni, nelle rispettive stanze d'albergo, senza esito.

    Quando si dovettero salutare, Friedel consegn a Garry una busta sigillata con all'interno lasoluzione del problema, chiedendogli di non aprirla e di scrivere all'esterno della busta, quandol'avesse trovata, la sua soluzione.La busta rimase sempre chiusa. Garry la port con s in quella che ancora si chiamava UnioneSovietica. Sottopose il bizzarro problema a Mikhail Botvinnik, ex-campione mondiale e suomaestro. N Botvinnik n gli allievi della sua famosa scuola, tutti promettenti scacchisti,riuscirono a risolvere il quesito.

    In seguito la lettera and persa. Kasparov, furioso, ormai convinto che il problema fossestato mal formulato, prese il telefono, chiam Friedel ad Amburgo e finalmente ebbe lasoluzione. Sapreste fare meglio di ben tre campioni del mondo?

    Mr. V.Soluzione N. (24)(Formnek) Secondo il pensiero di Sam Loyd in un problema i pezzi vengono posizionati sullascacchiera, quindi in precedenza NON sono stati mossi! Dunque 1.0-0 (!) risolve. L'arroccoviene effettuato correttamente, infatti il Re si muove di due passi verso la T e questa lo salta,ponendosi nella casa adiacente al Re! Un joke divertente.

    Notiziario (News)Valentino Ricci Memorial Tourney (VRMT)Sono richiesti tre mosse diretti (3) con almeno 2 o pi fasi di gioco (per esempio GA/GR,GV/GR, GA/GV/GR). Sono anche richiesti qualsiasi tipo di cambi fra le fasi: cambi di matti,difese, minacce, effetti ecc. Premi: 1 Premio 150 Euro, 2 Premio 100 Euro, 3 Premio 50 Euro,Menzioni Onorevoli, Lodi.

    Sweetest nut hathsourest rind!

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    I problemi che non raggiungono il verdetto tornano a disposizione dei loro autori. Il verdetto sarpubblicato su Best Problems. Prego spedire gli inediti a: Valerio AGOSTINI, Via del giochetto,2/F-2; 06122 PERUGIA - (Italy). E-mail: [email protected] - Data di chiusura 30 giugno2010. Giudice: Antonio Garofalo.[Three moves direct (3) are requested with at least 2 phases or more (for example setplay/solution; try/solution; setplay/try/solution). They are also requested any type of changes amongthe phases: changes of mates, defenses, menaces, effects etc. 1st Prize: 150 Euro, 2nd Prize: 100Euro, 3rd Prize: 50 Euro, Honourable Mentions, Commendations. The problems that don't reachthe award return back to their authors. The award will be published on Best Problems Magazine.Please send the originals to: Valerio AGOSTINI, Via del giochetto, 2/F-2; 06122 PERUGIA -(Italy). E-mail: [email protected] - Closing date: June 30, 2010. Judge: Antonio Garofalo.]

    Correzioni (Corrections)

    Diag. 2258. Non minimummer ma un normale Serie H13 (It's not minimummer).Diag. 2260. Nieuwhart olandese, non tedesco (Netherland, not Germany).

    Compleanni da ricordare: Valerio Agostini 3/8/1954, 55 anni; Mario Albasi 21/7/1949, 60anni; Alberto Armeni 14/9/1944, 65 anni; Prof. Oscar Bonivento 27/11/1914, 95 anni; MarcoCampioli 10/6/1954, 55 anni; Alessandro Fasano 3/2/1939, 70 anni; Giuseppe Ferro 26/7/1934, 75anni; Vito Rallo 3/8/1939, 70 anni; Prof. Pietro Rossi 26/5/1924, 85 anni.Auguri a tutti e mi si perdonino eventuali dimenticanze.

    I nostri concorsi:2 (2009-2010: judge not yet appointed).3 (Judge 2008-2009: Antonio Garofalo).H2 (Judge 2008-2009: Chris J. Feather).H3/n (Judge 2008-2009: Viktor Zaitsev).Sm2/3 (Judge 2009-2010: Sven Trommler).Fairies (Judge 2008-2009: Dinu-Ioan Nicula).

    Pubblicazione trimestrale senza scopo di lucro.Per riceverla, contattare: () Antonio Garofalo,

    via Collodi n.13 70124 Bari - Italy 080/5564025 - s CCP: 17784703 E-mail: [email protected]

    http://web.tiscali.it/best_problems/Index.html

    ContentsInediti (Originals) . . . . . . . . . . . . . . . . . . . . . . . p. 242Soluzioni BP50/51 (Solutions BP50/51) . . . . . . . . . . . . . . p. 247Award S. Pirrone Memorial Tourney by Antonio Garofalo . . . . . . p. 253Revision EM45JT award by Enzo Minerva . . . . . . . . . . . . . . p. 255Award H3/n moves 2002-2003 by Vlaicu Crisan & Eric Huber . . . . p. 256Ricostruzione n.39/40 by Vito Rallo . . . . . . . . . . . . . . . p. 262Nuts by Mr. Veneziano . . . . . . . . . . . . . . . . . . . . . . p. 263Notiziario (News) . . . . . . . . . . . . . . . . . . . . . . . . p. 263Correzioni (Corrections) . . . . . . . . . . . . . . . . . . . . p. 264Compleanni (Birthdays) . . . . . . . . . . . . . . . . . . . . . p. 264

    Sostenitori:Pier Giorgio Soranzo Hans Gruber Pietro Luciano Placanico Roberto CassanoClaudio Cesetti Filiberto Pivirotto Vito Rallo Enzo MinervaDaniele Giacobbe Marco Crucioli Marco Guida Daniele GiacobbeRomano Bellucci Dolf Wissmann Alberto Armeni Valerio AgostiniSaverio Ce' Stefano Galletti Royal Library of Netherlands

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