Assignment solution

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1 Assignment 04 solution Theodoe S. Novell 689DueOct04 Q0 [4](Read all pats befoe attempting any.) Supposeaisanaayofnumbesoflengthn,andxholdsanumbe. (a)[4]witeacontact(specification)focomputing i {0,..n} a(i) xi intoy. {tue}?{y i {0,..n} a(i) xi } (b)[4] Use the techniue of eplacing a constant by a vaiable to obtain an invaiant. I lleplacethenwithanatualvaiablekwhichangesfom0ton(inclusive) I:k n y a(i) x i i {0,..k} (c)[4] Wite a coect poof outline that solves the poblem. (You may assume that computing x i fo any intege i is an opeation in the pogamming language, although an expensive one.) Don t woy about efficiency at this point. {tue} k:0 y:0 {I:k n y i {0,..k} a(i) xi } whilek ndo y:y+a(k) x k k:k+ {y i {0,..n} a(i) xi } (d)[4] What is the vaiant? 0

2 n k (e)[4] Intoduce a tacking vaiable to impove the efficiency of the algoithm. State the linking invaiantthatelatesthenewvaiabletotheestofthestate. I llintoduceanumbevaiablezwithlinkinginvaiantzx k. (f)[4] Rewite the poof outline fom pat(c) to use the tacking vaiable. {tue} k:0 y:0 z: {I :k n y } i {0,..k} a(i) xi zx k whilek ndo y:y+a(k) z z:z x k:k+ {y } i {0,..n} a(i) xi [Bytheway,itisinteestingtotyadiffeentinvaiant I :k n y a(i) x i k i {k,..n} (I ewote x i as x i 0 befoe eplacing the two occuences of 0 with k.) This invaiant leads staight to a solution with only one multiplication in the loop: an algoithm known as Hone s ule. {tue} k:n y:0 {I :k n y i {k,..n} a(i) xi k } whilek 0do y:x y+a(k ) k:k {y } i {0,..n} a(i) xi ] Q [6](Read all pats befoe attempting any.)

3 (a)[4] Wite a contact(specification) fo computing the intege pat of the suae oot of a natual numbe. (Intege pat means floo, i.e., ounded down.) LetsandpbevaiablesoftypeN. Thecontactis {tue }? {p s } o,euivalently, {tue }? {p s<(p+) } (b)[4] Give a linking invaiant that makes you specification euivalent to the specification { A(m) A(n) m<n}?{ A(p) A(p+)} Fist I m going to ewite my postcondition as p s < (p+), which can be futhe ewitten as ( p >s ) (p+) >s ThissuggestsafunctionA(i) ( i >s ). Weneedtopickanmsmallenoughthatm >sisnot possible. I llpick m0. Wealsoneednlageenoughthatn >s issuetobe tue. I llpick ns+. (Pickingnswillnotdo,as0 >0isnottueandnois >.) Thustheliking invaiant is L:m0 n(s+) i N (A(i) ( i >s )) Assuming this linking invaiant L we have A(m) A(n) m<n ( m >s ) n >s m<n ( 0 >s ) (s+) >s 0<s+ 0 s (s+) >s 0<s+ tue Fo the postcondition we have A(p) A(p+) p >s (p+) >s p s (p+) >s p s So given this linking invaiant holds, the pe- and postconditions fom pat(a) ae euivalent to those given in the uestion.

4 (c)[8] Use you linking invaiant to deive a coect poof outline fo the contact given in pat (a) fom the algoithm given in slide set 4 pages {5,..,8}. Running time should oughly popotionaltothenumbeofbitseuiedtoepesenttheinput. Seeslideset5,foanexempla. I suggest doing this in thee stages: fist, intoduce additional vaiables and the linking invaiant;second,ewitethealgoithmssothata(atleast)isnolongeneeded;thideliminateaand any othe vaiables no longe needed. (i) Intoduce additional vaiable s of type N using the following linking invaiant fom pat(b) L:m0 n(s+) i N (A(i) ( i >s )) Sincem, n, andaaenotchangedbythealgoithm,theeisnoneedtoaddomodifyany code to maintain this invaiant. { A(m) A(n) m<n L} p:m :n {I:m p< n A(p) A() L} while p+do {I p+ L} : p+ {p<< I L} ifa()then:elsep:endif { A(p) A(p+) L} (ii)rewitesothata,m,andnaeonlymentionedinl. { ( 0 >s ) (s+) >s 0<s+ L} p:0 :s+ {I :0 p< s+ ( p >s ) ( >s ) L} while p+do {I p+} : p+ {p<< I } if >sthen:elsep:endif { ( p >s ) (p+) >s L} 3

5 (iii) Eliminate A, m, and n andcleanup. At this point, I m also dopping the euiement that s+fomtheinvaiant,asitisneveused. {tue } p:0 :s+ {I :p< p s< } while p+do {I p+} : p+ {p<< I } if >sthen:elsep:endif {p s } Bonus[5]. Extend the solution fom Q to find an outline that uses no multiplications. Slideset5setsoutaoadmapfothisinthecasethatnisapoweof i.e. sis onelessthanapoweof. Fothesakeoffindingoutwhatwouldhappen,IthoughtIwouldtackle the poblem diectly(i.e. without switching to the c-i epesentation fom the p- epesentation). Tostat,let slookcloselyat,whichisthemultiplicationtoeliminate Whenp+iseven p+ ( p + p + ) 4 Whenp+isodd p+ ( p + p + ) 4 (p+)+ 4 Thissuggeststhatweuseatackingvaiablesfop,p,and. Let L:ppp pp 4

6 DataefiningthesolutiontoQ,weget {tue } p 0 pp p : s+ 0 0 (s+) {I :p< p s< L} while p+do {I p+} ifp+iseventhen [ else [ endif ] [ : (p+) if>sthen : p else p pp : p endif {p s } 4 (pp+ p+) ] [ : (p+ ) 4 (pp+ p+) (p+)+ 4 p ] ] Nowwehavemultiplicationsp and intheloop. Solet sinvestigatethem. Whenp+ isevenwehave Whenp+isoddwehave p+ p ( p +p ) ( p + ) p+ p ( p +p p ) ( p + ) 5

7 Which leads to {tue } p 0 pp p : s+ 0 0 (s+) {I :p< p s< L} while p+do {I p+} ifp+iseventhen else p p : : endif if>sthen : p else p pp : p endif {p s } (p+) 4 (pp+ p+) (pp+p) (p+) p+ 4 (pp+ p+) (p+)+ 4 (pp+p p) (p+ ) p Thee is only one multiplication left and that is the initialization of. Howeve we ve now emoved all multiplications(that can t be epesented by shifts in binay) out of the loop. And I d be pefectly happy to stop hee, having accomplished that much. 6

8 But let s eliminate that last multiplication anyway. The only euiement on the initial value ofisthatitisbigenoughthats<. Sowecouldeplacetheinitializationcodewith {tue } p 0 pp p : 0 0 {0p< pow() L} whiles do[,]:[,4 ]endwhile {I :p< p s< L} Mission accomplished. No multiplications ae left. Howeve. Itiswothnoticingthat,ifweinitializelikethis,then pisapoweoftwoafte thefistloop. And pbeingapoweofispesevedbythebodyofthesecondloop,asthe followingagumentshows: If pisapoweoftwoand p>,then pisevenandsois p+. Thusishalfwaybetweenpand,so p p andisalsopoweoftwo. Sincepowesofgeatethanaealleven,wecanignoethecasewheep+isodd. Usingthenotationpow(x)tomeanthatxisapoweofwehave {tue } p 0 pp p : 0 0 {0p< pow() L} whiles do[,]:[,4 ]endwhile {I :p< p s< pow( p) L} while p+do {I p+} p : (p+) 4 (pp+ p+) (pp+p) (p+) if>sthen : else p p {p s } p pp p : endif Wemightwoyaboutthedivisionsbyand4. Doweneedtowoyaboutnonintegalvalues being assigned to, p o? (Ealie we saw that p+ is even, so no need to woy about.)sincewedesignedthealgoithmsothatp,,andaenatualnumbes,wecanbesuethat, p, and ae all integes and thus, by L, so ae, p, and. But we can also show this diectly, without efeence to the way the algoithm was developed: We can see that p> even(p) even()isaninvaiantofbothloops; thuspp,p,and aealldivisible by4;thustheexpessions 4 (pp+ p+), (pp+p),and (p+)aeallinteges. 7