THE PROBLEfIJIIST

SUPPLEfTIEHTISSUE 21

JANUARY 1996

B D Stephenson, 9 Roydfield Drive, Waterthorpe, SHEFFIELD, 519 6NDM McDowell, 23A Harcourt Avenue, SOUTHEND-ON-SEA,Essex, SS2 6HTJ R Coward, 25 Elmwood Avenue, HARROW, Middlesex, HA3 8AJ

EDITOR:SOLUTIONS EDITOR:SELECTIONS EDITOR:

All originals printed here take part in the normalProblemist tourneys, so that publication in this supple-ment is equivalent to publication in the main magazine.

For this supplement, I am looking for straightforwardoriginals of all types, including endgame studies. ldeally,they should be pointed, well constructed, and haveentertainment value. Please submit originals to me at theabove address. I would appreciate it if composers wouldsubmit problems clearly drawn or stamped on diagrams,please. I would also be happy if composers couldsomehow hide the solution so that I may have a chanceof solving their originals. Many thanks!

Comoosers should note that their contributions havebeen accepted unless or until they get a rejection fromme. I regret that I do not have time to acknowledge eachand every contribution. lf acknowledgements are espe-cially required, please enclose a stamped addressenvelope or equivalent. Composers should also note thatI have a very large folder of material ready to publish andthat for fairies and helpmates especially, a year's delaybefore publication will not be unusual! lf composers getimpatient waiting, they should inform me please beforesending their material elsewhere. Email contributions arewelcome and will be acknowledged. My address is1 O0 4 47 .20 43 @cf.m puserve. com.

The supplement has its own unitied solving ladder andall are eligible to enter it. Prizes are in line with the mainmagazlne.

Sadly, we don't have a SELECrTOTVS column in thisissue because I don't have one currently ready forpublication. All contributions would be most welcome!There are an extra four pages this issue to completecompensation for the missing issue in 1995. The extraoriginals also help to speed up publication of manyoroblems that I have had on hand for some time.

Paul Valois always mentions the important birthdaysof important people in the world of chess composition inhis commentary at the beginning of each issue of lheProblemist. Sadly, I have to tell you that he has beenvery remiss in this issue because he has ommitted tomention an important birthday of one of the mostimportant people of alll But then he would, because he isa very modest man! Paul himself will be 50 on 6 March1996. Congratulations to our editor and President! lnorder to celebrate this occasion I have put together anarticle containing three of Paul's compositions, most ofwhich will, I hope, be new to readers.

BDS

nFI'htf

soLUTroNs (SEPTEMBER 1 995)

PS353 (Sp6re) a) 1.S94 (2.R93/Sf2) 1..e1Q/e1S 2.R93/Sf2 b) 1.Se2 (2.S93/Rf2) 1..f1Q/f1S 2.S93/Rf2 c) 1.Sf3(2.Rh2/Rgl) 1..exf1Q/exf1S 2.Rh2/Rg1 "Promotions toqueen and knight separate the threats in each phase,and the white rook mates form a cross pattern." (MM)"Curious. I don't know what to make of it." (B.P.Barnes)"Well worth the Zeroposihon stipulation.' (C.Vaughan)

PS354 (Onoda) a) 1.Rd4 Kxes 2.897 b) 1.Kf4 Kxe62.Ke4 KfG 3.RgB Ke6/fO 4.R96/Bg5'Easy, but interestingfor the radical change such a simple subslitution makes."(BPB)

PS355 (Nicolson & Stephenson) 1.8e4 (2.Qd5) 1..Kxe4ifxe4/Sb6 or e7 2.Qe6/Rxg5/R(x)e7 "Good flighfgivingkey, needed to provide for 1-.d5.' (C.J.Morse) 'Nearly fellfor 1.Qf/? Se7! More please!" (BPB)

P3S56 (Lewis) 1.8e5 (2.595) 1..Kxe5/Sxe5/Bxe5 2.Re2lRf4/Qxb7 "lt doesn't seem logical to close the Q+Sbattery." (BPB) 'Unfortunately this is completely antici-pated by K.S.Howard, Observer 1926 8lltb'lQ2KJ44s3l8l2B1k1P1l3s4l3P1RlB/8 #2 1.8e5 (pointed out byStephen Emmerson). Howard uses a second black knightto provide an extra variation, and a black pawn to preventthe dual mate 2.Sc5 after loose moves of the d3 knight."(MM)

PS357 (Shire) Set 1..Qc6+/Beo 2.Qxc6/Qd3 1.Sc4?(2.Se3) 1..Qc6/Be6 2.Qf5/Qg2 1..Sd6! 1.Qfs! (2.Rd4)1..Qc6+/KdG 2.Sc4/Qd7 'Delicate confection, but notquite one thing nor the other." (BPB)'Unexpected flight.'(cv)

PS358 (Lincoln) 1Sd-? (2.Rd4) 1..R94 or Bdo 2.Qa81..Rd6l 1.Sb7? (2.Rd4) 1..R94! 1.Se8? (2.Rd4) 1..Rd6/Rg4 2.Sc7/Sfo 1..8d6! 1.Sc8! (2.Rd4) 1..Rd6/Bd6/Rg4/Bes 2.Se7/Sxb6/Sxb6/Rxe5 "A random knight moveopens a queen guard on c5, allowing White to threaten2.Rd4, cutting the bishop guard, but fails to provide for1..Rd6, closing the queen line again. 1.Sb7? puls a directguard on c5, but closes the a8-d5 line, allowing thedefence 1..R94!. 1.Se8? provides a mate to exploit theinterference after 1..Rd6, but prevents the potentialqueen male after the interference 1..8d6! The keyprovides a new male after 1..8d6, plus a flne self-interference-after-selfblock male 2.Se7. Excellent com-plex line effects from a near-Meredith setting." (MM)"Never easy to place Bob's whirling knights." (BPB)

PS359 (Pachl & Weissaue0 1.Se4? (2.Qf2) 1..Qb5/Qb8/Qh2 or 93 2.Sc5/Sd6/Sg3 1..Qxh5! 1.Se6! (2.Qe2)1..Qbs/Qb8/Qh2 2.Sc5/Sc7/Sef4 "Took me back 30years when choice of unpin was de rigueur - but nicelydone." (BPB)

PS360 (Chandrasekaran) 1.Sd2? (2.Sxf3) SeM.Sf2?(2.S94) Se3! 1.S93? (2.Sxc6) Sb4! 1.Sd6? (2.Sxco) Sas!1.Rd6? (2.Re6) Sc5! 1.S95! (2.Sxf3) 1..Sel/Sd2/Kf42.Qh2/Qb8/Sfi "Set mates at h2 and b8 are provided forBS moves, and the five thematic tries interfere with thesemates. Only the flight-giving key preserves them." (MM)"Obvious tries (as they are meant to be). Wonderful keyswitches attack (and BS defences) to f3. Top classl(BPB)

PS361 (Petite) 1.8f6? A (2.Sc3) 1..Bxbs b 2.exf7 C1..Sb1 2.Qxf3 1..Bxd3! a 1.895? B (2.Sc5) 1..Bxd3 a2.exf7 C 1..Sb3 2.Qxf3 1..Bxb5! b 1.exf7! C (2.8f5)1..Bxd3 a 2.895 B 1..Bxbs b 2.8f6 A 1..8e6 2.Rxe6 "Eggon faces time. A repeat of PS211 from the May 1994issue. Still, as SE says, "l enjoyed this one the first time I

solved it!'. Over to you Brian!' (MM) 'Well, l'll go to thefoot of our stairs! How did that happen? My face is veryred!!'(BDS)

PS362 (Caillaud & Rotenberg) Set 1..Rc1 2.Bxf3 1.Sxfs?(2.Qc4) 1.. Kc6/Kxe6iQco 2.Qe4iSd4/Se3 1.. RcM.Sxf3l(2.Oc4) 1..Kc6/Kxe6/Qc6iRc1 2.Sd4/Qxf5/Qxf5/Sd4"Makes the most of the Paisley switchback theme, but itdoesn't quite add up, and the play after the try is better.'(BPB)

PS363 (Russ) 1.Rb1/Rb4/Rb8/Ra2lRd2tRf2lRh2lKo4lKd3? Sg3/Sf4/Shffi/Sc7/Sd6/Sef6/Sef6/Sd6+/Sf4+ ! 1.Kd4!eS-/hS- 2.Rb8+/Rh2+ and 3.RxS "The WR musl notabandon its distiant ambush and the WK has only onesafe square to move to." (CJM) "Very pretty, and a greatkey. \Mthout knowledge of the composer, I would haveguessed Wuzburg.' (BPB) ?s the position is a block,does the symmetry not make the key obvious?" (MM)

PS364 (Beasley & McDowell) Set 1..c2 2.Qb4+ Rxb43.Ra5 1.Qh4! (2.Ra5+ QxeT/Qd6 3.Qxe7/Bxd6) 1..Qxh42.Sb4 (3.Sc2) Kxb4iRxb4 3.R95/Ra5 1..Qxe7 2.Qxe7and 3.Ra5 1..QdG 2.BxdG and 3.Ra5'Thoroughly en-joyed this spectacular sequence." (BPB) 'The problemwas developed from J.Jespersen, 2nd Prize, SouthernTrades Gazette 1883 1b293lp4b2l2R94h1p1p1Q1 lk1p4ilPlp3p1/K5p1/5s1q #3 1.Qhs (2.Ras+/Qd1+) 1..Bxh5/Rxhs 2.Sb5/Se4. lt was hoped that the set queensacrifice would catch the eye but unfortunately the massof black in the NE corner looks susoicious. I tried it out onGraham Lee, who solved it in about a minute flat (!), but itbemused some of the solvers in this year's FrenchChampionship. John and I are in agreement that theproblem should have been captioned 'J.Jespersen,version by JDB and MM'.' (MM) "The composers are toomodest! The original had an unfortunate double threat,and they have done well to eliminate it and they haveadded a worthwhile line of set play, both feats ihatJespersen couldn't perform even though he composedthe problem twice and won two prizes with it! To explainthat last sentence I need to mention a piece otserendipity. When the disk bearing this column from MMarrived, I luckily didn't work on it until a day later, but thatnight, my bedtime reading was lre Problemist, Septem-ber 1936. \Mat should | find, but the following short pieceby C S Kipping: 'lt is rather a doubtful point how muchyou must vary your proNems for them to rank asindependent. If a composer is working on some pafticulartheme he is evidently justified in showing a number of

different settings, but to gain prizes these must presum-ably differ considerahly. Rev. J Jespersen (d. 1914) wasa veyprolific composer and it is just possible that hemay have re-composed the second problem below'ithas happened in our own case somefimes! At the sametine, had the woild collection been in existence thesecond version would undoubtedly have been disqualifiedsince the play is identical. This problen during the hotsummer of 1906 caused our retirement to the cellar withwet towels before it was solved. Now a similar problem

would perhaps take a minute!'The first problem given byCSK is the one quoted by MM, the second, which gained2nd Pize, Tidskrift for Schack, 1905, was 3B3s/1pb4b/45R2/1Q3p1r/r2p1p1klp4p1P17Kl2s5 #3. 1.Qa5(2.Rh6+,Qe1+) 1...Rxas/Bxas 2.Sxd4/S95. PresumablyGraham Lee worked his way through lhe wet towel phasesome years ago! we've all been there! The worldCollection referred to by CSK was the White-HumeCollection of which he was the main curator in 1936."(BDS)

PS365 (Wllmott) 1.Sb2 Kd6 2.Sxc4+ Kd5 3.Sb6+ Kd64.Sc8+ KdS 5.c4+ Kxc4 6.Sb6 "BK batted to and fro asthe WS proceeds with purpose via a freeing 2.Sxc4+."(BPB)

PS366 (Bell) 1.8e3 Ra3+ 2.Kd2 Rxd3+ 3.Ke1 Rxdl+4.Kt2 Rf1+ 5.K93 Rf3+ 6.Kh4 Rh3+ 7.K95 Rh5+ LKf6Rfs+ 9.Ke7 Rr/+ 10.Kd6 Rd7+ 11.Kxc5 Flxds+ 12.Kb4Rbs+ 13.Kc3 Rds 14.Re1+ Rd1 1s.Rxdl 'A long WKrundlauf, and a brilliant and subtle construclion. Con-gratulations! (BPB) 'lnteresting king's stroll with switch-backs' (G.Lucenti) "Mediaeval funl (SE)

PS367 (Andzej) 1.gxh6? Kxh6 2.8e4 h2 wins. 1.8e4!dxe4 2.gxh6 KxhG 3.e6! fxe6 4.K94! h2 5.Rxg2 h1Q6.Rh2+ Qxh2 stalemate. 'Pleasant stalemate, bul thestudy solves itself as soon as the possibility of WKg4 is

realised.' (J.D.Beasley)

PS368 (Bakcsi & Zoltan) a) 1.Kds RaG 2.8d4 Sf4: b)1.Sc3 b4 2.Sd5 Sf5 "A magical net of mates. Veryrefined.' (BPB) "lt struck me as being rather ordinary forthese composers.' (MM)

PS369 (Koludrovic) 1.Sd4 Qb3 2.Sf4 Re3; 1.Se5 Rd32.Se3 Rd4 'These eDaulette mates cannot be donebetter." (BPB) "Excellent control of potential cooks.' (SE)

PS370 (Jonsson) 1.Re6 f3 2.Qb5+ Rxb5; 1.Se6 f42.8b7+ BxbT "A black piece stands idle each time. Don'tlike it." (BPB) 'Well matched, but are the checks spoof?"(SE) "No. They force the black move orders." (MM)

PS37{ (Manhart) 1.KeO e4 A 2.R7 Sxcs B; 1.Kf5 Sxc5B 2.RgO e4 A; 1.Ke4 c4 C 2.Rf5 Sd2 D; 1.Kxd5 Sd2 D

2.ReO c4 C'Heavy-handed clue, but it's nice to havedouble-crossing friends! A great construction all thebetter for two pairs of white sequence reversals." (BPB)

PS372 (Grigorian) a) 1.Rxb3 Qxa2 2.Rb2 Qds; b) 1.BxcsQxc3 2.891 Qc6; c) 1.Kxg2 Qb1 2.Kh1 Oe4 "Laboured inthe extreme." (BPB)

PS373 (Shire) a) 1.Rc€+ Re7 2.Kd6 Rd7; b) 1.8d3+ Bg42.Ke4 Bf3'Sharp as ever from DJS" (BPB)

PS374 (Turner) 1.8f5 c8R 2.Ke6 RxeS 3.Qd8 exdSQ "Atricky underpromotion that is not easily spotted. Thosewho solved failed to comment, and those who commentfailed to solve!" (MM)

PS375 (Fasher) 1..Kb4 2.RdB Kc5 3.Kd7 Bh3; 1.R97 d7

1672.BgO d8Q 3.Kfi Bxc4 "Two strong lines of play.Enjoyable." (BPB)

PS376 (Durham) 1.Kd6 Bxbl 2.Ke7 Sc6+ 3.Ke8 896;1.Kxe5 Sd7+ 2.K14 93+ 3.K94 Ses; 1.Kd4 Bxe2 2.Kc3Se4+ 3.Kc2 Bd1; 1.Kxc5 Bfs 2.Kb6 Sc4+ 3.Ka6 Bc8"Dull, without any semblance of a subsidiary theme.' (SE)"The absence of a model mate in the 1.Kd4 line is alsodisappointing." (MM)

PS377 (Vitale) 1.Qe4 Kd6 2.Qf4+ Ke7 3.Ke5 f3 4.fS gxf4"A little mid-board ideal-mate with kings and pawns andthat's all! (MM)

PS378 (Stein) 1.Q93 a3 2.2.Qc3+ bxc3 3.Rxa3 c2;1..axb3 2.Qf4+ Kc3 3.Qd2+ Rxd2 "Sweetly done, withtwo queen sacrifices.' (BPB)

PS379 (Nymian) 1.896 es 2.8h5 Ke6 3.894+ Kfo 4.8f5Sd7 "Not difiicult, but a pointed and good bishoprundlauf.' (BPB) 'A lovely thing.' (JDB)

PS380 (Petkov) Set 1..Kb1=; 1.RPc2 Ka2 2.c'lRS+ Kb1+3.RSb3+ Kc2+ 4.RSa1+ Kc3= "Good sequence.' (SE) "Asplendid piece of humour from the Grandmaster.Platzwechsel (exchange of squares) with minimal force!'(MM)

PS38l (Aistleitner) 1.Kd7 Kd2 2.KcG Kc3 3.Kb5 Rhs+4.Ka41a5'l was half expecting O-O!" (SE)

PS382 (Geissler) a) 1.f18 Ke1 2.Ba6 Kf2 3.Bb7 a8Q4.K91 Qa7 5.8f3+ Kf1; b) 1.e18 Ke2 2.893 Kf1+ 3.Kh2Kf2 4.BbB a8Q 5.893 Qh1 "ln a) 5.8f3+ (note - checkfrom the black king!) is a subtle choice. s.Bhl would notdo, as 6.Kh2 would then be possible, and 5.892 wouldmake 5..Kf1 iflegal. In the final position Kf2 and Kg2 arcillegal because the bishop provides support for KxQ, andKh1 and Kh2 are illegal because the king would not beadjacent to a piece. The point of the underpromotion isonly revealed on move 5, as 5..Kf1 would be illegal with aqueen on f3. In b) the reason for the underpromotion iseven more subtle; in the final position the bishop cannotmove away from 93 to make QxK illegal as it cannot finda support, however a queen could play to e3 or f3!Intriguing stuff!" (MM)

PS383 (Mantha) a) 1.Kb4 Kb1 2.Ka3 Bc5; 1.Ka2 Sb32.a3 Sc1; b) 1.Kc3 Ka22.Kb4 a3 3.Kc3 Ka1 4.Kb3'Thesolulions have little meril in themselves, but the possibil-ity of such different stipulations to the one diagram isworth showing." (MM)

PS384 (Sikdar) 5.c1S 6.a1R 7.Ra4 9.Sf4 11.Ke3 'l2.Re416.a18 17.8d4 Rg3 but Rolf Sieberg indicates manycooks in 15, with BRb7, BBaT and BKbS for Rg8 mate.Black AUW has been shown in serieshelDmate formmany times, the pioneer being M.Myllyniemi, 3rd Prize,feenschach 1966 8/5S2/p7lP7lp1k1P3lP7lppP5l7Ksh#16 1.b1S 2.a1Q 3.Qxa3 4.Qb4 7.alR 9.Rc5 14.a1B15.8d4 16.Sc3 Sd6.

PS385 (Lubin) 1.gO Bhs 2.gxc2 Kh3 3.Ra2 Bd1; 1.KflKg3 2.Rd7 Kf3 3.Rd3 f2'ln the first solution the bishopbecomes an 'amazon' (Q+S) with powers gained from theblack pawn. This pawn is pinned, as after cxdl the d2pawn would be checking with powers gained from therook! In the second solution the c2 pawn prevents therook from checking and leaving the d-file. The e2 pawnmust move to allow the bishop to take the rook's powers,but must stop specifically at f2 to prevent Black fromplaying 4.Rxc2!. As the composer says, the solutionsmay not be related, but with light material they present

many features of the genre." (MM)

PS386 (Aistleitner) 1.8e7+ Kg3 2.8d8 Bh3 3.Se7 Kh44.Se5 or Rf8 Sd6'More unils on home squhres - and stillno O-O!' (SE)

PS387 (Grigorian) 5.clS 7.Sxe5 8.Sd7 13.e18 14.8x9315.8c7 18.glR 20.RdB Bb7'Many near tries in 21.'(SE)"Easy solving for those who had already found theinlenlion of No.384." (MM)

PS388 (Ruppin) 1.8h5 0 1..b6/b5/c6/fs/R- 2.Kb2lKb1lKc2lKd2lKdl "Five anti-circe battery mates with K+8.Extremely well done.' (SE) "The problem is an incom-plete block, the key providing for moves of the f-rook.The a-rook cannot move because the rebirth square hB isoccupied. 1..Kxds is illegal because the king would returnto eB (however it would be legal if dl was occupied, asthen QxeS would be illegall). A move of the \A/K, byvacating the b4 bishop's rebirth square, threatens check,but as the king's move will occupy another piece's rebirthsquare he has to wait for a black error. ll for example, hetried to mate by Kb2, Black could reply with Kb6 as axb6would be illegal (b2 being occupied), however 1..b6selfblocks, allowing what is essentially a strange form ofself-interference! 1..b5 and 1..c6 are similar selfblocks.1..f5 and 1..R- open new guards on d4 and e8respectively, allowing the WK to interfere with the existingguards. A very clear-cut example of Anti-circe strategy.'(MM)

lan Shanahan responds to my observation on the Daltontheme in lhe September issue by sending one of his ownproblems which features a Dalton try and key. l.S. (afterJ.Coombe-Tennant), USPB July 1 994 2Ql S3/K2sl rp1l8/3P1k1Pl2S5l5P2l8l2B5 #2 1.Qb7? Re7! 1.Qc7! Doreaders know of any other examples? \Nho will be stirredinto composing a setting showing changed play? Nowlhere's a challenge!

J Buglos sends a much improved version of his PSl59from November 1993 (lssue 9). This is 5Q2/1p3S2/4p1 p1 | 1 K3b1 il2P 1 P 1 Pl /3S 1 kP1 /4r1 pP l3B2B1 #2 1...B.x-e4lBxg4+ 2.Sfe5/Sg5 # 1.Qa3 (2.Sf2 #) 1...8xe4+/Bxg4+iKxe4 2.Sde5/Sc5/Sc5 #.

MM

ORIGINALS

The '#2' under PS407 to PS413 means 'Vvhite to playand mate in two moves against any defence.' Suchoroblems are called two-movers. lf the word 'setappears under the diagram, this means that the solvershould be on the look out for set play. Set play is whatwould happen if it were Black to play in the diagramposition. After one or more black moves Vvhite will beable to mate in one. lf the words 'try' or 'tries' appearunder the diagram, the solver should look for tries, whichare white moves which nearly mate in two, but fail to justone black defence. This try play is often called virtualplay. Of course, there is still that which actually happensafter the key, the actual play. Each of these parts of aproblem, the set play, the virtual play and the actual playare called phases. Many modern two-move themescover more than one phase, and you will find suchmulti-phase problems among the twGmovers in thisissue.

. We start of with the miniature (seven pieces or less)PS407 from our solutions editor. This problem is anexample of a twin. As well as the diagram position, thereis also a twin position for solution. To get to the twin

168position move the wB from d3 to f4. Newman Guttman,the Rrst of several American contributors this issue, firstcomposed PS409 in 1981, but has left it unpublished untilnow. PS4l2 is an aftempt to show with the white piecesan effect more usually seen with the black ones, and wehope that it has come off. In PS4l3 David Lynn shows usthe Urania theme, where the same white move appearsin three separate phases as key, threat and mate.

No three-movers so straight on to the more-moversin which, in all cases in this issue, White is to play andforce mate in five moves. PS4l4 and PS415 areminiatures and PS416 nearly is. PS4'16 is in fact aMeredith, which is the term used for problems withbetween 8 and 12 men.

PS417 from Andrew Miller in an endgame study inwhich Vvhite is to play and draw. There is no limitation onthe number of moves, so treat it just like a bit ofadjournment analysis from a game.

PS418 to PS425 are helpmates in which Black,playing first, co-operates with White so that White canmate Black in the number of moves specified. PS418,PS419. PS42l. PS422 and P5424 arc all twins in whichpart (b) can be reached by following the instructionsunder the diagrams. PS420 has four solutions and PS423has two.

PS426 to PS429 are selfmates in which White playsfirst and forces an unwilling Black to mate him in thenumber of moves soecified. The first three of these arefrom the United States, but the fourth is home-grown.

PS430, by our regular contribuior Brian Edwards, is areflexmate, which is like a selfmate but with the addedcondition that either side must mate on ihe move if it ispossible to do so.

The last twelve problems are fairies. That is, they areunorthodox in either stipulation, pieces involved, or rulesapplying, and in some cases, all three!

First the unorthodox stioulations. PS434 and PS438are serieshelpmates in which Black plays a sequence ofconsecutive moves (without Vvhite playing at all) to reacha oosition where Vvhite can mate in one. PS43l andPS437 are serieshelpstalemates that are similar toserieshelpmates except that the aim is the stalemate ofBlack. Note that part (b) of PSa3l is a serieshelpmate!PS441 is a direct stalemate where White plays andforces the stalemate of Black in two moves.

The princess on cB in PS433 is the first unorthodoxpiece this issue. lt is a bishop and knight combined, socan move like a bishop to b7, a6, e7 etc. and like a knightto a7, b6+, d6 and e7+. A very welcome problem fromHelmut zajic, PS439, features Lions. They must hopover one man (the hurdle) of either colour to move orcapture, but may land on any square on the line beyondthat man but not past a further man or the board edge.

We have five unorthodox conditions this issue. In theMaximummer, PS432, Black can only play his geometrFcally longest move, or if there are several of equal length,choose between them. Measurements are made fromsquare centre to square centre.

In Andernach Chess pieces (not kings) changecolour after capturlng. Thus in PS434 1.Bxg6(=wB) wouldbe illegal self-check.

The Circe condition in PS440 is defined as follows.When captured, a piece (not a king) is immediatelyreplaced on its square of origin (game array square) ifthat square is empty; in the case of R, B or S on thesouare of the same colour as that on which it wascaptured. Thus a wR captured on h8 (a black square)would be replaced on a1, provided al was empty; if a1

was occupied the R would be removed from the board asin a normal c€lpture. Pawns go lo the initial square of thefile they were captured on. A replaced R is deemed notto have moved for castling purposes.

Part (a) of PS436 is orthodox, but part (b) featuresMonochrome Circe. This is normal Circe plus the extracondition thal each and every move (including thatcapturing the king) must be between squares of likecolour. Thus in the diagram 'l.KhO is not legal.

ln the Platzwechsel Circe of PS442, captor andcaptive exchange squares but check is normal.

Send you solutions and comments to Michael McDow-ell (note his new address within 2months of receiving this issue. Enioy your solving!

BDS

THE WORK OF VACLAV CISAR

(A lecture by G W Chandler presented to the BritishChess Problem Society on 26th October 1956, that, asfar as we can tell, has never appeared in print before)

The subject of this paper is one of the veterancomoosers of the Bohemian School. He was born atPilsen in October 1879, so is now 77, but he is stillcomposing, and a problem of his was honoured in therecent Bohemian National Tourney, the award of whichappeared a few months back.

He learned chess from his uncle Josef Kesl, who wasa well known Bohemian composer round about 1890, andhis first problem appeared when he was 15. Two yearslater he won second prize in a three-move tourneysponsored by the ltalian A/ew Chess Revlew, and herapidly became a first class composer. Three of theproblems I am showing this evening were published

before he was 20, and his outstanding achievement wasto win second prize in the tourney of the BohemianChess Club of Prague at the age of 19. From that time henever looked back, and his successes in one importanttourney after another placed him in the front rank ofCzech comoosers.

He has composed some 300 problems, nearly allthree-movers, and his many tourney succ€sses includeat least 10 first prizewinners. His style is typicallyBohemian, many of his problems showing the interplay ofa White force limited to a few pieces and with no pawns,

but he generally has something to say, and is not contentwith the "check and mate here, check and mate there"type of three'er, which lesser composers have turned outrather too often. Many of his problems show fine keysand good strategic ideas, with quiet play in lhe mainlines; others are forcible and brilliant, but none are everdull.

His other hobby is music, and he would probably havecomposed many more problems had he not been sooccuoied with administrative work in the Czech musicalworld. Accountancy is his profession, and for many yearshe held the post of Chief Borough Accountant in hisnative town of Pilsen.

In 1940 he collaborated with two other Pilsen compos-ers, Jan Berkovec and Frantisek Hladik, to produce acollection of problems by the three of them, and 104 ofhis problems appear in that collection.

(Continued on p. 173)

PS4O7 M MGDOWELL(Southend+n-Sea)

PS4IO L ULANOV(Latvia)

PS4I3 D P LYNN(usA)

PS4Il RA LINCOLN(usA)

PS414 R TURNBULL(Gatelawbridge)

PS4O9 N GUTTMAN(usA)

PS4I2SEMMERSON&BDS(Reading & Shettield)

PS4I5 W W NIKITIN

#2 (tty)

#2 (seutry)

#2 (tries)

PS4l6CAHRUSS(Heme Bay)

PS4{9 T GARAI(usA)

PS422 A A GRIGORIAN(Armenia)

PS4ISHBFBOUMEESTER(Netherlands)

PS42I A ZHURAVLEV(Russia)

PS423 L VITALE(ttaty)

H#2 (b) bRa2 -> c2

%, ,/r, '%.

.%,

H#2 (b) wKfS -> h8 H#2 4 solutions H#2 (b) 900 clockwise

H#3 (b) wsg3->h5 H#3 2 solutions H#a (b) h1=ba

PS425 A C REEVES(St Agnes)

PS428 R J BALES(usA)

PS431 A A GRIGORIAN(Arnenia)

171

PS426 I L STEIN(usA)

PS432 T LINSS(Gernany)

PS427 I L STEIN(usA)

PS43O B EDWARDS(Paisley)

PS433 A WILLMOTT(Australia)

H#2 Princess c8 (b) wKc3->c7(c) bKds->el (d) bP6->e2

SH=7 (b) wBhs->ho & SH#7 S#15 Maximummer: 2 solutions

PS434 J M RICE(Surbiton)

PS437 V RAMASWAMY(lndia)

PS'S40 G BAKCSI(Hungary)

172

PS435 E A DUGAS(usA)

PS438 T KARDOS(Hungary)

PS441 G MALEIKA(Gernany)

PS439 H ZAJIC(Austria)

PS442 V NEBOTOV(Ukraine)

SH#6 Andernach Chess H#2 (b) Monochrome Circe

SH=6 (b) bKal->b1

#5 Circe H#2 (b) Platzwechsel Circe

| /JV CISAR

1st Prize, Hampstead andHighgate Express,1909

#3

At flrst sight one might well wonder how a position likethis could offer anything at all. A lone bK with only onesquare to move to and White with a Q and four pawns.What could any composer do with that? But the solver'sdoubt turns to amazement when he realises lhat the keyis 1.Qf8! For how could any composer arrange such akey, and get the problem sound, without using any Blackforce except the bK? Those of us who have tried willrealise what an achievement this is, and so did thejudges who gave it a first prize. The play is 1.QfB 01...Kxes 2.Qc5+ Ke4 3.f3 #:1...Ke4 2.Ke6 Kd4 3.Qb4 #;1...K95 2.Qb4 0 Kh6,Khs/Kfs 3.Qh4/Qf4 #; 1...K94 2.Qh6Kf5 3.Qf4 #.

V CISARTidskrift f6r Schack, 1 896

lf White could get on the fourth rank with his Q itwould be mate, but 1.Qh8 is met by 1...Kxd4 2.RxcO andthere is no mate for 2...a4: again 1.Qe6 looks over-whelming, but 1...Kxd4 defeats it. Even when onedecides that the Q must come over to the other side,1.Qb8 looks stronger than the actual key, which is 1 .Qa8.The reason forthis becomes apparentwhen 1...Kxd4 etc.The echoed pin-model is very charming, and the play isquiet throughout. The probleqn was published in 1896,and the composer was not 17 until the 4th October in thatyear, so he probably composed it at the age of 16. Thisand the following problem show that he had the skill of amature composer at a very early age. The full solution is1.Qa8 0 1...Kxd4 2.RxcG 0 Ke4la4le4 3.Rc4/Qxa4/Qh8#; 1...cS 2.Qh8 (3.Oh4 #) Kxd4/exd4 3.Rf4/QeB #;1..a42.Qb8 (3.Qxe5 #) Kxd4/exd4 3.Qb4/Qe8,Qf4,Re6,Rf4 #.

(Diagram top of next column)1.Sb6 (2.Rc4+ Ke5 3.Sxd7 # and 2.Sd3 & 3.Rc4,c3 #)

1...9xf4 2.Rc4+ Kes 3.d4 #; 1...Kxcs 2.Sd3+ Kd4/Kb53.c3lc4 #; 1...Sxc5 2.Se2+ Kes 3.d4 #; 1...Bxc5 2.c3+Kes 3.Sd3 #; 1...d5 2.Se6+ Kes 3.Sc4 #. There is a

V CISARCeske Listy Sachove, 1896

second threat 2.Sd3 which is never forced, and thisadded to the composer's difficulty, because he could noteradicate it, and had to make every black defenceprevent it as well as the checking threat. Note, forexample, how he does it when 1...d5 2.Sd3 Sf5 and nomate. There is point in the key because when one movesa S in that position it is invariably to a square where itcan check. lf the problem contains no quiet play, it has asurprising wealth of beautiful variations, combined with adexterity that is truly remarkable for so young acomposer. He uses two white pawns and gets the utmostout of them. They guard vital squares in the K's field, andgive four out of the six model mates. The wQ nevermoves in the forced play, but she guards eight squareson rank and diagonal and fully justifies her use. Thesetting is elegant and the construction leaves nothing tobe desired. Had it competed in a tourney it could hardlyhave failed to win a Drize.

V CISAR2nd Ptize, Casopis Ceskych

Sact,.stu, 1907

#3

Here we have the same white pieces, but in a typicalBohemian setting with no white pawns. lt opens with amoderate key 1.SdO (2.Sxd3+ Kxa4 3.Qa2 #) 1-..Bf72.Se4 (3.Sxc5 #) cxb2lc2 3.Qd1/Qa2 #: 1...cxb2 2.Ra2 &3.Qxb2 #; 1...c2 2.Qa2+ Kc3 3.Se4 #; 1...Sb5 2.Sbc4+Kxa4 3.Qxb5 #. For a second prizewinner there isnothing startling in this, but I quote it as typical of manyproblems by Cisar, in which quiet play, model mates andclean construction are all much in evidence, and thisperfection of finish evidently appealed to the judges.

(Diagran top of next page)

This has a rather unlikely key because it puts the Ron a square where a S could check, but as a matter offact the S at f2 is not required to move at all. The key is1.Rd3 (2.Qf4+ Kxf4lKds 3.S96/Qd6 # and if 1...8f62.Qe3+ dxe3/Kd5 3.Bd6/Qe6 #. a nice combination of Q

V CISARlstPAze, Das Neue lllustierle

BlatuBighton and Hove Soclety 1903

sacrifices, but the real beauty of the problem lies in thedefence 1...f4, when White continues 2.Rxd4 threatening3.Qxf4,Rd5 # and if 2...cxd4 3.Qa5 #. There is no otherplay of note. lf 1...R91 2.Qxe2+ and 1...Kd5 2.Of4 (3.Od6#) Bes 3.Qf3 # land the dual 2.Qe3 (3.Qe6 #) Bes 3.Qf3# - BDSI. One feels a slight disappointment thatsomething better could not be found for this last line,necessitating, as it does, a wP on b3. The other wP at h6is to stop a cook by 1.Qh6, and the bS at a2 is requiredto stop a rather spectacular cook by 1.Rc3 land one byt.Qd3 - 8DSl. The bP at b7 prevents 1.Re3+ Kds 2.Qa5.All these expedients show that it must have been verydifficult to get the problem sound, with a reasonable key.

V CISAR2ndPtize, Cesky Spolek

Sachovnl, 1 898

And now to the problem mentioned in the openingreview, a masterly work for a 19 years old composer. Thekey 1.Re2 is decidedly'give-and-take". ll rescues the Rfrom a P capture and offers it 1o two other pieces instead.ll 1...8xe2 the threat 2.Bb4+ gives a model mate by2..Kf43.8d2. There is a second threat, which is forced by'1...Sb3 and 1...93, but it does not lead to anythingnotable. lt is 2.Qd7 threatening 3.Qd6 or Qfs and 3.8d6.Apart from this threat there is no quiet play in theproblem, but the checking continuations have sparkle. lf1...d3 we have the checking threat repeated by 2.8c5+Kf4 3.8e3 #. But the best line is 1...f5 2.Qb5+ Ke6/Kf43.8d5/Qxf5 #. Note how the key makes possible thecheck at b5. A clever defence to both threats is 1...S95, aself-block which makes it oossible for the B to discovercheck on the other diagonal, and it can and must go tod8, so thatwhen 2...SeO itcan mate by 3.8c7 #. lf 1...Sf8the wB must take it and come in on the other end of thediagonal by 3.Bxh6 #. And if 1...Kf4 2.8d6+ won't do. Wemust play 2.Qb8+, keeping the B for the pin after 2...K95

1743.Qe5 #.

The next four problems have the same White force,Q, 2 Ss and a B. Let us see how Cisar handles them.

V CISAR1st Prize, Hampstead andHighgate Express, 1 908

#3

The last problem showed the composer's mastery ofbattery play, and this one, published ten years later, mayhave been inspired by it, though the play is entirelydifferent. The key sets up the battery by 1.Qf1 threaten-ing 2.891+ KxeS/Kg3 3.Sc4/Sf5,Sde4 #. The black Sscan readily defend, and we have a pair of echoedself-blocks when 1...Sd2 or d4, While sacrificing the B one3 and mating at c4 or f5. But what if 1...Sa5? The replyis 2.Sf5, and if the other S to d2 a model mate is forcedby 3.893, and if 2..lGe5 3.8d4. lf 1...Kxe5 2.Sc4+ Kf4and the wB can discover at e1 or h4, just spoiling a sixthmodel mate, which must have been very annoying to thecomposer. The key is remarkable because White has a

very strong move 1.Qe2, threatening mate on the moveat e3, e4 and Shs. But 1...Rxe5 and \Mite has nocontinuation! This strong Black move is well used in theminor variation 1...Rxes 2.8d4+, and 1...Rc7 2.Qe2. In allrespects a masterly work, well-keyed, well varied, artisticand fairly ditficult to solve.

V CISARTyden Rozhlasu, 1942

#3

This is one of Cisa/s later problems, and there aretwo rather queer points about it. Firstly, it is a waiter withall the appearance of a threat problem, and secondly thekey is 1.Qc7, for which there appears to be no particularneed, as Vvhite could easily prevent the bK going to d3by playing 1.Qa3 or 1.Q93, both of which look muchstronger. But 1.Qa3? fails after 1...dxe6, land 1...cr3! -8Dfl and 1.Q93? is surprisingly met by 1...Sf31. So thewQ has to step backwards so as to be able to reach h7

after 1...dxe6. The solution is 1.Qc7 0 1...Sxc7 2.Sc5+Kes/Ke3 3.Sfl7/Sg4 #; 1...dxe6 2.Qxh7+ Ke3/Kes 3.594

/SfZ #; 1...Sh3 2.Re2+ Kf3/Kd3 3.Sxd4/Qc2 #; 1...d32.Sxg5+ Kd4 3.Sf5 #, a pure mate in which the wR takesno part; 2...Ke3 3.Qf4 #; 1...94 2.Qf4+. After the key it isall fairly straightforward, in fact a typical Bohemianproblem showing three analogous model mates, so alikethat they almost constitute a triple echo, and one othermodel very nicely combined.

V CISAR1st Prize, Ceske Spo/ek

Sachovni, 1916

#3

One may well be surprised that such a simple lookingposition, with only 10 pieces, could win first prize in oneof Czechoslovakia's leading tourneys. But when weexamine the solution we have to hand it to the composer.He has done his work with amazlng simplicity, and theeconomy of force is truly remarkable. The key is quite agood one, and again the position gives no hint of waitingplay. 1.Bgl 0 1...Sa6 2.Se5 and Black cannot prevent3.Qd4 #; 1...Sc6 2.Q93 & 3.Sf6. The bP on d7 stops anawkward move of the bS, but gives a fine variation 1...d62.Qe3+ Kds 3.Qf3 #. The bP on c4 blocks a square inthis variation, and when it moves we have 2.Q92+ Kt43.8e3 or 2...Kd3 3.Se5 with a self-block on c3 and a ourebut not economical mate. lf 1...Kd5 2.56+ 2n6 66 qmates on the diagonal. Unavoidable short mates follow1...d5 and 1...95. I think you will agree that the composerhas extracted every bit of play which the position offers,and has done it with extraordinarv refinement.

V CISAR1st HM., Casopls Ceskych

Sachistu,1909

#3

I have always considered that this problem did not getits deserts. Whether the author regarded it as his firststring in the tourney I don't know, but his other entry gotsth Prize and was placed immediately above this one. Asa matter of interest I will show it next so that we can allexpress an opinion as to which problem is the better.This problem appeared in a lecture I gave four years ago,

175but those of you who saw it then will, I am sure, bepleased to renew acquaintance with it. The key 1.Se5moves a S to a square from which it can check, but in thecontinuations we have precisely the reverse process, andthat is what makes it difficult to solve. For example, thethreat is 2.Sc4, with a model after 2...Kc3. A randommove of the bS will prevent this, and Vvhite can continue2.Sxe6+, mating at b2. A black conection 1...Sd6cunningly prevents this, allowing a Q sacrifice at c5. Nowany move of the bR along the second rank can preventthe threatened mate at b2. lf it goes to d2 the threatoperates with another self-block if 2...Rd3. lf to e2, a Qsacrifice follows on c3. lf lo 12, lnen 2.8b2+ Ke3 3.Qc3 #.But what if 1...R92? The deep reply is 2.Sfd3 (threateningboth 3.Qc4 and Qc5) and if 2...exd3 3.Sf3, giving thefourth model mate! 1...Kxes 2.8b2+ with a pin-matecompletes some excellent minor variety. lt is just a pitythat a bP must occupy 97, [e/se 1.-.597! would refute thekey - BD$ because without it there would be anotherquiet though minor variation 1...R98 2.Sd7. As it is thisonly comes in as a dual when 1...R96, but the composer,having achieved a wonderful combination with completesuccess, is not likely to have worried over that.

V CISARsth Prize, Casoprs Ceskyc,

Sachrslu,1909

#3

1.8b6 (2.Qe4+ ds/Kbs 3.Qxd5/Qo4 #) 1...S93 2.Qe6(3.Qds,Qc4 #) dxe6 3.8e8 #; 1...Sf2 2.Sc2 (3.Sd4 #) Kbs3.Qxd7 #; 1...Kbs 2.Qxd7+ Kb4 3.Sc2 #; 1...R95 2.Qxg5& 3.Qd5 #. This is the position which was placed abovethe previous problem, although to my mind it is not sogood a problem. The key takes a flight square and thethreat is forcible. The point lies, of course, in the two bSdefences, which certainly have strategic value, but onlythree lines result in model mates, and two of those havethe white pieces in exactly the same position with the bKon a different square. lt has an attractive setting with nowhite pawns, and it would appear that the judges gaveconsiderable weight to this feature, more so, I think, thanany English judges would have done. An English judgewould have virtually disregarded the fact that the previousproblem has two white pawns which take no part in anyof lhe four main variations, but Bohemian judges maywell have oenalised it on the score that this is a breach oltrue economy.

Note: The general verdict of those present was thatthe 1st HM. winner is by far the better problem.

(Diagram top of next page)

My final selection is another which I quoted four yearsago, but no lecture on Cisar's problems could be withoutit. Instead of quiet play we have a wealth of checkingvariations which are almost bewildering in their complex-ity. lt opens with a splendid key, for White has very

V CISAR1st Prize, Magyar Sakkvilag,1913

strong squares for the Q, c2, aO or even b5+, but he hasto concentrate on neutralising the bR at e6, and 1.Qh3 isthe way he starts 10 do itl The threat is 2.Qxe6+ fxe63.8b7 #, and if 1...KcO we have the same 2.Qxe6+dxe6,Bxe6 3.Se7. Most moves of the bR on the e flle will,of course, give a short mate but 1...Re5 2.8b7+ Kd63.Sbs #, a pin-model. A subtle defence is 1...8f8 or Bh6,opening the line of the bR to defend b7 afier the threat of2.Qxe6+, and White again plays 2.8b7+ Rc6 3.Sxf6 #,another pin-model. This is repeated after 1...S94 2.Qf3+Re4 3.Se7 #, and finally when 1...Sd7 2.Se7+ decoys theR behind his S for 3.8b7 to mate. Even in the minorvariation 1...Ke4 2.Qf3+ Kf5 3.Se7 # the R is pinned,incidentally for the fourth time, and the only poorcontinuation is 1...d3 2.Qxd3+. The theme of the problemis the neutralisation of the bR by pin, decoy or capture,and it is done six times with model mates in every line.Depth, as generally understood, may be somewhatlacking, indeed it all looks easy when one is shown thevariations. But they take a lot of finding, and when theproblem was quoted many years ago, a surprisingnumber of solvers failed to find one or more of thedefences. Viewed as a feat of construction, it is averitable triumph, and one can only marvel at the skillwhich put it logether. lt must surely rank as one of thefinest checking three'ers ever composed.

THREE PROBLEMS BY PAUL VALOISTo celebrate our President's 50th birthday, I am proud

to present three problems composed by him.

P S VALOIS3rd Prize, McWlliam Ty., 1963

The earliest problem by Piul that we can trace is this,composed when he was a teenager. lt features plus-flights, the four orthogonal flights of the black king. Theseare much more difficult to compose than star flights, asthe mating moves have to each cover two of the other

176flights without there being any duals. The solution is1 . Sd3 0 1 ... Kxd3 | Ke4 | KcA I KdS 2. Rg4/Rd6/Rd7/Rf4 #.

P S VALOIS2nd Pnze. Gazeta Czestochowska 1972

s#2

Here the wSf6 and wReS are half-pinned. The key,1.Rf5, seeks to take advantage of this with the threat2.hxg8=Q,B Qxg8 # a pin-mate with the wSf6 pinned.Black's thematic defences unpin this knight and allow it tomake the second move discovering check from the keyrook. Each black move also opens a line to the wRfs andthe wS must be careful to close it again. This gives us

two dual-avoidance variations - 1...Rc3,Rd4 2.Se4+ Sf6# and 1...d4 2.Sd5+ Sf6 #. There is also the variation1...Sd4 that sets up a masked battery against the wK.The answer to this is the self-pin 2.Sxg8+ Sxfs #. Thereis also 1...8xf6+ 2.Rxf6 SxfG #. Quoting this problem inhis wonderful new book Das Maft des weisse, Ktjn€ls,Friedrich Chlubna writes that it is a pity that PSV, throughhis extensive duties as editor of The Problenist and assub-editor of the selfmates and reflexmates section, findsscarcely any time for composing. Just so!

P S VALOISThe Problemist, 1983

In this reflexmate, the key is 1.Sxb4 (2.Bxf3+ Rxf3 #).Black can defend by any move ofthe bsd2, pinning bRZ,but also pinning wPe3. So 1...Sd- 2.Sd5 Qxd4 #. ButBlack can correct against this twice - 1...Se4 2.Re2fxe2# and 1...Sc4 2.8c3 Bxe3 #. Other lines are 1...Re22.Bxf3+ Rxf3 #; 1...8e4 2.Qe2fu'e2 #; 1...Qc6 2.Sb7 Qe4#; 1...Sx95 2.Sdc6 Se6 #; 1...Sxes 2.Qxgo Sxg6 #.

I hope that the above has proved that Paul is not justa very good editor, organiser and President for theBCPS. He is a good composer who would have moresuccess but for all the hard volunteer work that he does.As director of solving events in which Paul has takenoart. I can add that he is also a fine solver! Thanks toMMCD for finding two of these problems.

BDS