Midterm 2 - Solutions 1

Transcription

1 COS 26 Gnral Computr Scinc Spring 999 Midtrm 2 - Solutions. Writ a C function int count(char s[ ]) that taks as input a \ trminatd string and outputs th numbr of charactrs in th string (not including th \ ). Do not us any library functions or pointr arithmtic. int count (char s[]) { int i = ; whil (s[i]!= \ ) i++; rturn i; } S also String Exrcis in th cours packt. 2. Considr th following rcursiv C program. void mystry(int N) { if (N < ) rturn; printf("%d ", N); mystry(n-2); mystry(n-3); printf("%d ", N); } Giv th rsults of mystry(6). Circl th corrct answr. (a) 2463 (b) 2436 (c) 6423 (d) () (f) (g) (h) sgmntation fault Th vry first and vry last thing that mystry(n) dos is print th intgr N. So th answr must start and nd with 6. This liminats vrything but (), (f), and (g). Just aftr printing th first 6, mystry(6) calls mystry(4). Th vry first thing that mystry(4) dos is print 4, so th scond numbr printd is 4. This lavs only (). Copyright 999, COS 26.

2 3. What dos th following TOY program print out? Assum that th following numbrs ar loadd into mmory and that th machin is startd with th PC st to. : B4 R <- 4 : B R <- 2: 3 R <- R * R 3: 4 print R 4: 72 R--; if (R > ) goto 2 5: 4 C 8 8 Th program is a singl loop which dcrmnts R in ach itration. In ach itration R is multiplid by R andr is printd. Thus th function prints n, n (n ),..., n (n )... 2, n (n )... 2, whr n is th initial valu assignd to R. Not th last two trms ar both n!. Also, b carful to do all arithmtic in hx. 4. What dos th following TOY program print out? Assum that th following numbrs ar loadd into mmory and that th machin is startd with th PC st to. : B R <- 4: : B R <- 4: 46 2: B24 R2 <- 4 42: 2 3: 9B2 R3 <- mm[r2 + ] 43: 48 4: 43 print R3 44: 3 5: 9A2 R2 <- mm[r2 + ] 45: 6: 623 if (R2 > ) goto 3 46: 4 7: halt 47: 42 48: 5 49: Th PC will nvr b st to 4-49 so you can think of ths mmory locations as storing data. Th guts of th program is in lins -7. Throughout th computation R = and R =. Indxd addrssing is usd in instructions 9B2 and 9A2 sinc th scond hx digit is gratr than or qual to 8. Lin 3 rads in a valu from mmory location R2 and lin 4 prints it out. Think of R2 as a pointr - it is th addrss in mmory whr som data rsids. Lin 5 updats R2 to b th contnts of mmory addrss R2 +. Th mmory addrss pair 4, 4 contains two pics of information: th first is th data and is printd, th scond is th mmory addrss of th nxt pic of data. This procss is rpatd until th mmory addrss is (i.., NULL). Th data in lins 4-49 is rally a linkd list! 2

3 5. Giv th contnts of th stack aftr th givn PostScript program is xcutd dup add mul add dup add mul add 73 Hr ar th stack contnts along th way, whr th top of th stack is at th lft Which strings ar gnratd by th rgular xprssion? ( + ) Circl on or mor of th following. (a) (b) (c) (d) () (a) First obsrv that th last bit must b : this liminats (b) and (). Scond obsrv that thr must b at last on : this liminats (c). Third, obsrv that thr can t b mor than two conscutiv s: this liminats (b) and (d). 7. Considr th languag gnratd by th following grammar. trminals, nontrminals A, B start A Ruls A B B A A Is th languag rgular? If ys, giv a rgular xprssion that gnrats xactly th sam languag. If no, xplain why not. ys, ()* or ()* Th languag is Typ III (rgular). This implis that thr must xist a corrsponding rgular xprssion. Th production rul A B must b immdiatly followd by B A, so you could think of having th compound rul A A. Thus, xprssions of th following form could b gnratd: A, A,A,A,A. Ultimatly, w must nd up with strings. This is accomplishd as soon as w apply th rul A. 3

4 8. Considr th following FSA. 2 Circl all of th statmnts blow that ar tru. (a) Th FSA accpts. (b) Th FSA accpts. 3 (c) Th FSA rjcts. (d) Th FSA accpts all bit strings with an odd numbr of s. () Th FSA is nondtrministic. (f) Thr xists a dtrministic Turing machin that rcognizs th sam languag as th FSA. (g) Thr xists a nondtrministic Turing machin that rcognizs th sam languag as th FSA. b,, f, g Th FSA is nondtrministic for many rasons: if you ar in stat and th nxt bit is, you can transition to stat or 3. Th n-fsa accpts sinc it could choos th following transitions: It accpts bcaus of 3 2. Th n-fsa will not accpt or - ithr of ths liminats (d). Also, th n-fsa dos not accpt any strings nding in (including ) sincthonlywaytogtto stat 3 aftr rading a is to com from stat, but thr is no way to gt to stat aftr rading a. Statmnt (f) is tru sinc dtrministic Turing machins ar mor powrful than FSA s or n-fsa s. Dtrministic Turing machins can rcogniz xactly th sam languags as nondtrministic ons, so (g) is also tru. 9. Construct a dtrministic finit stat automaton (FSA) that rcognizs all bit strings with a multipl of thr s. (For xampl, th following strings ar in th languag:,,,,, but not,,,.) Us as fw stats as possibl. 2 4

5 a/a/r x/x/r sarch b a/a/l b/b/l b/x/l a/x/r x/x/l x/x/r lft skip x ys no a/x/l b/x/r sarch a x/x/r b/b/r. Considr th Turing machin abov. Th alphabt consists of th four charactrs: a, b, x, and #. Suppos th Turing machin is startd with th input tap and initial cursor as shown blow. Circl all of th inputs blow that will b accptd. # a b # (a) # a b # (b) # a b b a # # a b b a # (c) # a b b b b b b a # # a b b b b b b a # (d) Charactriz in plain English which inputs (i.., charactrs btwn th two dlimiting # s) ar accptd by th Turing machin. 5

6 (a), (b), all strings with an qual numbr of a s and b s Th TM will accpt all strings with an qual numbr of a s and b s. To do this it first movs to th lft nd of th tap. If th first charactr (othr than #) isana, it will dlt it by ovrwriting it with an x. Thn th TM will sarch for th lftmost b. If it finds a b, it dlts it by ovrwriting it with an x. Othrwis, th TM concluds that thr ar mor a s than b s and procds to stat no. (Th procss is analogous if th first charactr is a b. Thn it dlts th b and sarchs for an a to dlt.) This whol procdur is rpatd, until th string contains no mor a s or b s. In this cas, th TM will go to stat ys. It works bcaus vry tim it dlts an a, it also dlts a corrsponding b, thrby rducing th siz of th problm. Clarly th nw smallr input has an qual numbr of a s and b s if and only if th original input did. Hr s a dscription of ach stat. ys: th accpt stat no: th rjct stat lft: This stat rpatdly movs th cursor on position to th lft, unlss th cursor alrady points to #. Thatis,lft movs th cursor to th lft #. If instad, th cursor is alrady pointing to #, thn it transitions to skip x and movs th cursor on position to th right, i.., to th bginning of th intrsting portion of th tap. skip x: This stat rpatdly movs th cursor on position to th right, until it rachs th first charactr that is not x. If th first such charactr is #, thn it accpts th string; if it is a, th TM gos to stat sarch b to find a b to dlt ; if it is b, th TM gos to stat sarch a to find an a to dlt. sarch b: This stat skips ovr all a s and x s. If it finds a b, it dlts it by ovrwriting it with an x and gos back to stat lft. If it dosn t find a b (it rachs th right #) thnin concluds thr wr mor a s than b s, and rjcts th input. sarch a: analogous to sarch b 6

7 . For ach of th dscription on th lft, choos th bst matching machin on th right. d c d c c d corrsponds with intuitiv notion of algorithm pushs and pops charactrs using a stack rads and writs charactrs to and from an array can rcogniz any languag last powrful machins that can rcogniz languag of all valid C programs last powrful machins that can rcogniz languag of all C programs that don t go into an infinit loop ths nondtrministic machins rcogniz mor languags than thir dtrministic countrparts rcognizs fwr languags than nondtrministic FSA s namd aftr Alan Turing (a) dtrministic FSA s (b) nondtrministic FSA s (c) nondtrministic PDA s (d) dtrministic Turing machins () non of th abov Th machins in th right column ar listd in nondcrasing ordr of powr, as in th Chomsky hirarchy. Not that dtrministic and nondtrministic FSA s rcogniz xactly th sam (rgular) languags. Nondtrminism maks th PDA mor powrful. (S Lctur Not 5.2). PDA s ar FSA s with a stack; Turing machins ar FSA s with an infinit tap or array. No machin in th Chomsky hirarchy can rcogniz vry languag. Th most famous of ths xampls is th Halting problm, i., rcogniz th languag of all C programs that don t go into an infinit loop. Th languag of all lgal C programs (programs that compil without syntax rrors) is dscribd by a contxt-fr grammar (as in on of th xrciss and K+R Appndix A3). Th Chomsky hirarchy says that n-pda s ar quivalnt to contxt-fr grammars, so PDA s ar usd by compilrs to find syntax rrors. 2. Th following squntial circuit dscribs a mastr-slav flip flop. Fill in th timing diagram for th output ( OUTPUT ), givn th input ( D ) and clock signal ( CLOCK ) shown blow. Rcall that a D-flip flop stors a bit. Th bit can only chang whn its Cl input is. In this cas, it sts th bit to if its D input is, and to if its D input is. S Lctur Not 2.6. Th important fatur of a mastr-slav flip flop is that th output can only chang just whn th clock gos from on to off. Th OUTPUT is th valu of D at this tim. It rmains th sam until just aftr th nxt clock tick. 7