WPCZ /HMl ;:;?uۻr5-X!1cl|7܃ݮ~ÈJn4OL9HC=@:l\OfOѺsap1Bj#}L 69-OH@?LA}6oNXl &cJaUgYhDn/11 drS@xP j[knu}mGGG$נX # pUN{ % 0(w@47KDell AIO Printer A940WINSPOOLPX,..,PX0 (9 Z 6Times New Roman RegularX($ aEbcEdeDfDgEhijύx13|xj !  Xh  +XX CrosticsWithSpecialProperties h X?X+byWilliamR.Smythe,Jr. v      Supposewehavechosenaquotationwithqlettersandanappropriateacrosticwithaletters. N Thenw=q/aistheaverageanswerlengthandamonolengthpuzzleisatleasttheoreticallypossibleif : andonlyifwisaninteger,i.e.,aisanexactdivisorofq.Monolengthpuzzlesareelegantfroma & constructorsviewpoint,butmaybeeithermoreorlessdifficulttosolvethan normalmultilengthcrostics.ffeTherewillprobablybefewerlongsubquotationstocomplete,buttheconstructormayhavebeenforcedtouseseveralrarewordstomaketheanswerlistworkout.eIdontthinkanygeneralconclusionaboutrelativeappealtosolverscanbedrawn,buttoaconstructormonolengthcrosticspresentrareandinterestingchallenges.  Anotchbelowthemonolengthcategoryineleganceandchallengetoconstructorsiswhatwemaycallthebilengthpuzzleoneinwhicheachanswerhasoneofexactlytwospecifiedlengths.In  Z  thecommonsituationswheremonolengthsareimpossible,wewillnowshowthatbilengthsare always   F  possible.Insuchacase,theaverageanswerlength,w,isnotanintegerThatis,whenyoudivideqbya  6  therewillbepositiveintegersnandr,calledthequotientandremainderrespgrghehctively,suchthat  "  q=na+rand1rac cdd1.Withalittlemathematicalsleightofhand,wecanwriteq=na+ras v  q=n(ar+r)+r=n(a``r)+nr+r=n(ar)+(n+1)r,whichsaysthatyoucanformabilength b  puzzlewithaa abbranswersoflengthnandranswersoflengthn+1.Thiscompletestheproof,butwe N shouldnotethatthepairiin,n+1oflengthsisnotnecessarilyunique;theremaywellbeother : possibilitiesasthefollowingexamplewillshow.  Example. Inmypuzzle,bgc,thequotationhasq=253letterswhiletheacrostichas  a=33letters.Dividing253by33,wefindthat253=7'33+22(a=33,r=22)soasshownabove,  wecouldhave117letteranswersand228letteranswers,whichiswhatbgcactuallyhas.jjButwecouldalsohavewritten253=7(22+11)+22=7'22+7'11+2'11=7'22+9'11,soanotherpossibilitywouldhavebeentohave11wordsoflength9and22oflength7.  Ofcourse,wecouldsimilarlydefinetermssuchastrilength,quadlength,etc.,butitsoon " becomesridiculous.Afterall,everycrosticisklengthforsomek.Attheoppositeextremefrom v monolengthwouldbeapuzzleinwhichnotwoanswershavethesamelength2theyallhavedifferent b lengths.Givenanacrosticlengthaandassumingeachanswermusthaveatleast3letters,theleast N possiblequotelengthisq=3+4+5++(a+2)=(a+2)(a+3)/23=a(a+5)/2.Ifanswers : aretohavenomorethan20letters,thena18.Witha=18,youdhavetohaveq=207because & youdhavetoincludeallthelengths3,4,5,...,20.Witha=15,youdhavetohave187q195.   Itseemsthatopportunitiesfortotallymixedlengthcrosticsarerareinpractice.Constructionwouldbedifficultandhardlyseemsworthwhile,thoughasSueremarked,itwouldbetemptingtotrytoconstructoneintriangularform,i.e.,havinganswersofconsecutivelengthsinorderfromthetopdown.+XX?   p  doublecrostic.comJanuary29,2005 $2#& Ѐ