Midterm Exam CSC October 2001

Transcription

1 Midterm Exm CSC Otoer 2001 Diretions This exm hs 8 questions, severl of whih hve suprts. Eh question indites its point vlue. The totl is 100 points. Questions 5() nd 6() re optionl; they re not prt of the 100 points, ut will ount for extr redit t the end of the semester. Plese show your work here on the exm, in the spe given. You shouldn t need to write on the ks or in the mrgins; if your nswer won t fit in the spe given then you re trying to write too muh. Put your nme on every pge. Srp pper is ville if you need it, ut the protor will ollet only the exms. I hve tried to mke the questions s ler nd self-explntory s possile. If you do not understnd wht question is sking, mke some resonle ssumption nd write tht ssumption down next to your nswer. The protor hs een instruted not to try to nswer ny questions during the exm. You will hve the entire lss period to work. Good luk! 1. (9 points). Nme nd rieflydesrie(one sentene eh) three differentdt strutures tht ould e used to implement (in three different wys) reltion strtion. rry Eh tuple goes in seprte, ritrry slot of the rry. linked list Tuples re dynmilly lloted nd linked to one nother with pointers or referenes. hsh tle Tuples oupy ukets of lookup tle indexed y some primry key. lned tree Tuples re dynmilly lloted nd orgnized into inry (or n-rry) hierrhy. hrteristi rry Tuples go into slots of therry indited y vlues of some primry key, whih must form dense, ordered, finite set. 1

2 2. Consider the following reltion R: A B C D red 10 lrge x lue 12 lrge y green 6 smll z red 9 smll w yellow 20 smll z () (7 points). Judging from the tuples shown, whih ttriutes or sets of ttriutes might e keys forr? Hown you tell? B, AC, AD re the miniml sets of olumns tht hve unique vlues for ll tuples. They ould e keys, or we might need dditionl olumns, to mke sure tht we will never see ny tuples with identil key vlues. Without semnti informtion (wht do the olumns men), ll we n relly e sure of is tht A, C, D, nd CD re not keys. () (5 points). Give the result of the opertion σ D=z B<10 (R). A B C D green 6 smll z red 9 smll w yellow 20 smll z () (5 points). Give the result of the opertion π C,D (R). C D lrge x lrge y smll z smll w 3. (9 points). Wht does the following progrm print? Explin. #inlude <stdio.h> int min () { hr *s = "string 1"; hr *t = "string 2"; } if (s == t) printf("yes\n"); else printf("no\n"); if (*s == *t) printf("yes\n"); else printf("no\n"); It prints no yes 2

3 Vriles s nd t re pointers to strings (rrys of hrters). The first if sttement heks to see whether s nd t hve the sme vlue s pointers whether they point to the sme lotioninmemory. They do not. The seond if sttement heks to see whether the hrters pointed t y s nd t the first hrters of the two strings re the sme. They re (they re oth s s). If we relly wnted to ompre the ontent of the strings, we d need to write loop or use the strmp lirry funtion: if (!strmp(s, t)) printf("yes\n"); else printf("no\n"); 4. We disussed three possile implementtions of the join opertion: nested loop join, whih itertes over ll pirs of tuples, sort join, whih sorts ll tuples (of oth reltions) y the join ttriute prior to finding mthing pirs, nd n index join, whih itertes over the tuples of one reltion nd performs lookup opertions on the other reltion. Suppose tht we re performing join on reltions R 1 nd R 2 to produe reltion R 3. Suppose further tht reltion R 1 hs N 1 tuples, reltion R 2 hs N 2 tuples, nd reltion R 3 hs N 3 tuples. () (6 points). Give the symptoti omplexity (Big-O running time) of the three join implementtions, in terms of N 1, N 2,ndN 3. nested loop: O(N 1 N 2 ). sort: O(N 1 log N 1 +N 2 log N 2 +N 3 ) or O((N 1 +N 2 )log(n 1 +N 2 )+N 3 ),depending on detils of the implementtion. index: O(N 1 +N 3 ) or O(N 2 +N 3 ),depending on whether you use the R 2 index or the R 1 index, respetively, nd ssuming hsh tle index. With lned tree index, it s O(N 1 log N 2 + N 3 ) or O(N 2 log N 1 + N 3 ). () (5 points). Cn you think of sitution in whih sort join would e preferred over n index join? If you re joining on n ttriute for whih you don t hve n index for either reltion, nd don t wnt to rete one. Alterntively, if R 1 nd R 2 re known to e sorted on the join ttriute lredy, or R 3 needs to e. This ltter se (known to e sorted lredy) might hold if your indies were lned trees indexed y the join ttriute: exploiting the lredy-sorted nture of the tree ould e fster thn repetedly trversing it down from the root. () (5 points). Cn you think of sitution in whih nested loop join would e preferred over sort join? If one of the reltions is muh smller thn the other, so N 1 < log N 2 or N 2 < log N 1,ndthusN 1 N 2 <N 1 log N 1 + N 2 log N 2. Alterntively, if most of the tuples in oth reltions hve the sme vlue for the ttriute on whih you re performing the join, so N 3 N 1 N 2. Note, though, tht you re unlikely to e le to predit this in dvne. NB: I gve prtil redit for if oth reltions re relly smll. Here the nested loop is fster, ut only y smll onstnt ftor. 3

4 5. Consider the lnguge onsisting of ll strings of s, s, nd s, with no two identil onseutive letters. () (7points). Give DFA tht epts this lnguge. () (EXTRA CREDIT, up to 8 points). Give regulr expression tht desries this lnguge. Hint: this is hrd. Strt y reting regulr expression R tht desries strings of lternting s nd s. Then reteregulrexpression tht desries strings of lternting R s nd s. Let R = ()* ( ɛ) ()* ( ɛ) Our nswer is then (R ɛ) (R)* ( ɛ) = (( ()* ( ɛ) ()* ( ɛ)) ɛ) ( ( ()* ( ɛ) ()* ( ɛ)))* ( ɛ). Note tht R must generte non-empty strings, ut our nswer must inlude the empty string. 6. Consider the following NFA: ε ε () (7points). Desrie in English the lnguge epted y this NFA. Note: you will not reeive full redit just for desriing how the NFA works: zero or more s followed y... or zero or more s followed y... isn t eptle. All strings of s nd s in whih either the numer of s is evenly divisile y 2orthe numer of s is evenly divisile y 3. 4

5 () (7 points). Give regulr expression tht desries the sme lnguge. *(*)** *(**)** () (EXTRA CREDIT, up to 8 points). Give DFA tht epts the sme lnguge. 7. (8 points). When performing the suset onstrution to turn n NFA into n equivlent DFA, we need to dopt the onvention tht the NFA epts only if it n end up in n epting stte fter onsuming its entire input. Why?Whtwoulde wrong with onvention tht sys you ept if you n get stuk (no outgoing trnsition) in n epting stte, even if there is some input remining? Suppose there re two NFA sttes A nd B tht n e rehed on the sme input string (prefix) w. Suppose A is epting nd hs no outgoing edges, ut B is non-epting nd hs trnsition on symol to non-epting stte C. A suset stte ontining And Bwill eepting (euse A is), nd will hve n outgoing edge on. Sothe NFA will ept when given input w, utthedfa (suset mhine) will not. 8. Short nswer. () (5 points). A dtse representing demi informtion t the University might hve reltions with the following shemes: student id, student nme, student ddress student id, ourse numer, semester, grde ourse numer, ourse nme, deprtment, redits In priniple we ould store the sme informtion in reltion whose sheme onsists of the nine distint ttriutes from the list ove. Why don t we do this? Beuse it would introdue huge mount of redundnt informtion, wsting spe nd introduing the need to gurd ginst inonsistenies in opies. () (5 points). The serh (find) mehnism in mny text editors llows the user to desrie the trget of the serh with regulrexpression, not just speifi string. The editor onverts the regulr expression to n NFA in order to drive the serh. In mny ses, the editor then emultes the NFA diretly (keeping trk of ll possile urrent sttes), rther thn onverting the NFA to DFA. Why do you suppose it does this? Why not rete the DFA? 5

6 Beuse the onversion to DFA tkes time nd potentilly lot of spe. Sine we re only going to use the mhine one theostof NFA emultion my e less thn the ost of the onversion. The snner in ompiler, y ontrst, is run mny, mny times, nd needs to e relly fst; it definitely wrrnts onversion to deterministi form. () (5 points). The Unix mke utility is designed to mnge olletions of files tht depend on one nother. Progrmmers who uild lrge systems typilly rrnge for mke to ll the ompiler (e.g. g), rther thn lling it themselves. Why? Why not just rete one-line shell ommnd or lis tht invokes g with the right rguments, nd psses it ll the soure files? Beuse typilly when you mke hnge to soure file in lrge system, only smllsuset of the files relly need to e reompiled. By using mke you n rrnge to reompile ll nd only those files, voiding the time required to ompile the others. (d) (5 points). Summrize informlly the generl rule tht determines whether it is profitle to push projet opertion inside join. It s profitle if projeting efore the join would signifintly redue the mount of dt (olumns nd mye rows) tht must e proessed y the join, ut not if it fores us to look t tuples tht we would otherwise hve een le to void inspeting, y performing n index join. 6