Theory (CSc 473): Automata, Grammars, and Lanuaes. Let D = fwjw contains an equal number of occurrences of the substrins 0 and 0 Thus 0 2 D because 0

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1 Masters Examination Department of Computer Science October 3, 998 Instructions This examination consists of nine problems. The questions are in three areas:. Theory and Alorithms: CSc 473, 545, and 573; 2. Systems and Architecture: CSc 452, 552, and 576; and 3. Lanuaes and Compilin: CSc 453, 520, and 553. You are to answer any two questions from each area(note: if more than two questions are attempted in any area, the lowest two of their scores will be used). You have three hours to complete the examination. Books or notes are not permitted durin the exam. At the end of the examination, submit just your six solutions. Do not include scrap paper. Write your answers on the tablet paper provided separately. Start the answer to each question on a separate sheet of paper. On each pae that you hand in, write the problem number in the upper left corner and the last four diits of your social security number in the upper riht corner. Hand in your solutions in the envelope provided. Results will be posted usin the four-diit numbers. Some problems will ask you to write an alorithm or a proram. Unless directed otherwise, use an informal pseudo-code. That is, use a lanuae with syntax and semantics similar to that of C or Pascal. You may invent informal constructs to help you present your solutions, provided that the meanin of such constructs is clear and they do not make the problem trivial. For example, when describin code that iterates over the elements of a set, you miht find it helpful to use a construct such as for c 2 S do Stmt where S is a set. SOLUTIONS

2 Theory (CSc 473): Automata, Grammars, and Lanuaes. Let D = fwjw contains an equal number of occurrences of the substrins 0 and 0 Thus 0 2 D because 0 contains a sinle 0 and a sinle 0, but00 62 D because 00 contains two 0s and one 0. Show thatd is a reular lanuae. 2. Let E = fwjw contains an equal number of occurrences of the substrins 0 and Thus E, but E. Show thate is not a reular lanuae. You may use the fact that f0 n n jn 0 is not a reular lanuae.. S is the start state. ffφ 0 n ffφ A ffφ 0??? - Cn B 0-0 ρ S n 6 n D ff0 ο? 0- - nf ΩΨ ΩΨ 0 ΩΨ 2. Proof by contradiction. If E is reular then so is E f0 Λ Λ = f0 n n jn 0.

3 2 Theory 2 (CSc 545): Desin and Analysis of Alorithms A rooted tree is a directed raph G that contains a vertex r (its root) suchthatevery other vertex in G can be reached from r by a unique path. Write an alorithm to determine whether a directed raph, iven as an adjacency list, is in fact a rooted tree, and if so, to find the root. What is the asymptotic runnin time of your alorithm? To receive full credit, your alorithm should take O(n) time, where n is the number of vertices in G.. DFS(G) 2. for each u 2 V 3. DFSfrom(u) 4. DFSfrom(u) 5. if u is unvisited 6. mark u visited 7. for each (u; v) 2 E 8. DFSfrom(v). If the number of edes in the raph G is not n, output Not Rooted Tree". 2. Perform Depth First Search (DFS) on G. DFS produces a sequence of DFS trees, T ;T 2 ;:::;T k, that partition the vertices of G. (Each call to DFSfrom at line 3 visits some (possibly empty) set of vertices in DFS fashion formin a tree. T i is the ith non-empty tree thus created.) Let r i be the root of tree T i then T i consists of all vertices reachable from r i that are not in T [ T 2 [ [T i. 3. Mark all vertices unvisited. 4. Perform DFSfrom(r k ). 5. If the resultin tree contains all vertices in G then output Rooted Tree with root r k ", else output Not Rooted Tree". Runnin Time DFS takes time O(n + m) (m is the numberofedes)ifg is represented as an adjacency list. Thus the alorithm, which performs DFS twice, takes time O(n + m). Since the alorithm checks that m = n before performin DFS, the runnin time is O(n). 2

4 3 Theory 3 (CSc 573): Theory of Computation Let R = fhmi jm is a Turin Machine and L(M) is a reular lanuae: The notation hmi denotes the encodin of the machine M as a strin. The notation L(M) denotes the set of strins accepted by the machine M. Show that R is an undecidable lanuae (i.e. membership in R is undecidable). You may assume that all strins are over the alphabet f0;. You may use the fact that A = fhm;wi jm is a Turin Machine and M accepts w is an undecidable lanuae. You may also use the fact that f0 n n jn 0 is not a reular lanuae. Hint: Use a proof by contradiction. Assume that R is decidable by some Turin Machine S and describe how touses to build a machine that decides A. (From: Introduction to the Theory of Computation by Michael Sipser) Let S be a TM that decides R, and construct a TM T to decide A. Then T works in the followin manner. T = On input hm;wi, where M is a TM and w is a strin:. Construct the followin TM M 2. M 2 = On input x: (a) If x has the form 0 n n, accept. (b) If x does not have this form, run M on input w and accept if M accepts w." 2. Run S on input hm 2 i. 3. If S accepts, accept; ifs rejects, reject." M 2 accepts the nonreular lanuae f0 n n if M does not accept w (so T rejects hm;wi), and accepts the reular lanuae f0; Λ if M accepts w (so T accepts hm;wi). An equally ood answer would be to use Rice's Theorem. 3

5 4 Systems (CSc 452): Principles of Operatin Systems Two types of processes, red and blue, share a resource. Any number of red processes and any number of blue processes many share the resource at the same time, but two processes of different colors are never allowed to share the resource. Also, red processes have priority over blue processes, if a red processes wants to use the resource, no new blue processes should be allowed to access it before the red one. Implement this system usin the followin messae-passin primitives: ffl The blockin function recv(p, &ms), which can be used to receive a messae from an explicitly-named process p. ffl The blockin function recv_any(&p, &ms), which can receive a messae from any process. sender's name is returned in &p.) ffl The non-blockin function send(p,ms), which can be used to send a messae to a named process p. Send queues messaes and those queues are of potentially unbounded lenth. ffl There are no shared data structures; this is a pure messae passin system. Show the code for a sample blue process and fill in the places marked `<your...>' in the server below. The server coordinates the red and blue processes. SERVER = f <your declarations and initializations> while () f recv_any(&p, &ms); switch (ms) f case "start blue": <your code> case "start red": <your code> case "end blue": <your code> case "end red": <your code> (The Blue = f send(server, "start-blue"); recv(server, ms); (Use shared resource here); send(server, "end-blue"); Server = f int red count, blue count = 0; queue red q, blue q = empty; while () f recv any(&p, &ms); switch (ms) f case "start blue": if ((red count == 0) && q empty(red q)) f send(p, "ok"); blue count++; 4

6 else f enqueue(p, blue queue); break; case "start red": if (blue count == 0) f send(p, "ok"); red count++; else f enqueue(p, red queue); break; case "end blue": assert(blue count > 0); blue count--; if ((blue count == 0) &&!q empty(red q)) f for (p = dequeue(red q);!q empty(red q); p = dequeue(red q))f send (p, "ok"); red count++; break; case "end red": assert(red count > 0); red count--; if ((red count == 0) &&!q empty(blue q)) f for (p = dequeue(blue q);!q empty(blue q); p = dequeue(blue q))f send (p, "ok"); blue count++; break; 5

7 5 Systems 2 (CSc 552): Advanced Operatin Systems Distributed shared memory systems (DSMs) and distributed file systems (DFSs) both need to manae distributed caches of shared data. In particular, DSMs cache paes" of shared memory, while DFSs cache files (or pieces of files).. Both DSMs and DFSs use some protocol to coordinate read and write sharin amon the private caches." Describe one such protocol used by DSMs. Use a diaram to aid your description. 2. In file systems, read and writes are done by systems calls and thus are very explicit. In a DSM, the reads and writes are to proram data structures and thus are not the results of explicit calls to the system. Describe two ways in which a DSM miht learn about reads and writes to shared memory. 3. The way in which DSMs and DFSs are used is quite different. This difference leads to some desin decisions bein suitable for one system, but not for another. Give an example of a DFS that handles the manaement of shared data in a way thatwould not be suitable for DSMs. Why is it unsuitable?. Any protocol needs to force exclusive write access and allow shared read access. There are lots of possibilities, but an invalidate protocol is probably the easiest to write down. Each cache line has three possible states, invalid, readable, writable. The diaram should be a FSM with the followin transitions. State invalid : read reads current version of the line (possibly from a cache that has it for write) and transitions to readable; write reads current version of the line, sends invalidate to other caches, transitions to writable; invalidate does nothin; remote reads do nothin. State readable : read does nothin; write already has current version, but must send invalidate and transition to writable; invalidate transitions to invalid; remote reads do nothin. State writable : read does nothin; write does nothin; invalidate must propaate the chaned version to wherever its oin to be read from and then transition to invalid; remote read must propaate the possible chaned version to wherever its oin to be read from and then transition to readable. 2. Two approaches are code instrumentation and VM faultin. In code instrumentation, the code run by the user makes explicit calls to the DSM code when it needs to work with shared data. Those calls miht be added by the prorammer, or by some software that instruments the code behind the prorammers back. With VM faultin, the VM protection is set on paes to force detection of reads and writes. 3. When a file is opened for writin in the Sprite file system, all cachin for that file is disabled and operations must o throuh to the server. Since DSMs need to support fine-rained reads and writes, thus would work very poorly for them. 6

8 6 Systems 3 (CSc 576): Computer Architecture Consider an implementation of the DLX in which the results of a load are not available for use until the third cycle after the load. Furthermore, there are now complex branches that can test for equality and inequality of the values in two reisters (BRE Ra, Rb, loc and BRNE Ra, Rb, loc), not just a reister aainst zero. All branches have one delay slot after the branch and no further branch penalty. Consider the followin loop, which is the (real) inner loop of a hih performance copyin arbae collector. It copies data whose lenth in words is found in R, from the address found in R2 to the address found in R3. You may assume that there is never a possibility that the ranes overlap. LOOP: LW Rtmp, 0(R2) SW Rtmp, 0(R3) ADD R2, R2, # ADD R3, R3, # SUB R, R, # BRNZ R LOOP ADD R0, R0, R0 ; no-op. This loop has a bu in it. What is the bu? How would you fix it? 2. Schedule this loop (i.e., reorder the instructions subject to the constraint that the result computed is uaranteed to be the same as before) so that it executes in as few cycles as possible. Do not eliminate or chane any instructions except to fix any bu(s) you have identified; if you did not find the bu, you may schedule the loop as it appears. Explain why your loop cannot be made any faster. 3. Unroll the loop and schedule it so that the unrolled loop is twice as fast as the current (unscheduled) loop. You may assume no stalls due to cache misses. You may use new temporary reisters, and only the contents of memory need to be the same as for the current loop. You should only write the loop body; you may assume that whatever preamble and/or postamble that is needed to make your loop correct (for example for loops not a multiple of your unrollin) is present. Explain how many times you unrolled and why your loop is twice as fast. Better scores will be iven for usin fewer temp reisters and for unrollin fewer times.. The bu is that it deals with the DLX as if it is word addressed. The #s on the ADDs need to be #4s. 2. LOOP: LW Rtmp, 0(R2) ADD R2, R2, #4 SUB R, R, # SW Rtmp, 0(R3) BRNZ R, LOOP ADD R3, R#, #4 The load stall and the branch delay has been scheduled and there are no more stalls, so it can't be made any faster. 3. SHL R, R, #2 ADD R, R2, R LOOP: LW Rtmp, 0(R2) LW Rtmp2, 4(R2) LW Rtmp3, 8(R2) SW Rtmp, 0(R3) SW Rtmp2, 4(R3) SW Rtmp3, 8(R3) ADD R2, R2, 2 BRNE R, R2, LOOP ADD R3, R3, #2 7

9 The loop has been unrolled three times. This has no stalls and copies a word every 3 instructions, while the oriinal does it every 6. 8

10 7 Lanuaes (CSc 453): Compilers and System Software For each of the statements below, indicate in the space provided the alternative that most accurately completes the statement. Note: Each correct answer is worth point, but each incorrect answer will incur a penalty of point (i.e., count as a score of ). You may wish to take this into account if you are not sure of an answer and want to make a uess.. For a C compiler, the most important requirement for fast compilation speeds is: (a) ood hash functions for fast symbol table lookups; (b) careful I/O buffer manaement and efficient lexical analysis; (c) ood reister allocation. 2. When writin a compiler desined primarily for novice users in an introductory course such as CSc 27, the most important objective should be to: (a) ensure that the enerated code is optimized as much as possible, since novice users can't be relied on to write efficient code; (b) carry out ood dataflow analysis; (c) implement ood error detection, reportin and recovery. 3. LL() parsers are able to handle (a) left-recursive rammars; (b) riht-recursive rammars; (c) both (a) and (b); (d) neither (a) nor (b). 4. LR() parsers are able to handle (a) left-recursive rammars; (b) riht-recursive rammars; (c) both (a) and (b); (d) neither (a) nor (b). 5. An attribute X at a node n in a parse tree is synthesized if (a) the value of X at n is computed at n or one of its descendants; (b) the value of X at n is computed at an ancestor of n or one of its siblins; (c) both (a) and (b); (d) neither (a) nor (b). 6. A rammar G is said to be ambiuous if 9

11 (a) a strin in the lanuae of G has more than one leftmost derivation; (b) it ives rise to a shift/reduce or reduce/reduce conflict within yacc; (c) it cannot be parsed usin a recursive descent parser; (d) all of the above. 7. To implement function calls correctly in a lanuae like C, it is necessary to (a) always pass all aruments and return values on the runtime stack; (b) always pass all aruments and return values in reisters; (c) pass aruments and return values usin any combination of reisters and memory, provided that the caller and the callee aree on where the appropriate values are to be found; (d) none of the above. 8. A C compiler usually carries out dataflow analysis in order to (a) verify that variables are used with types that are consistent with their declarations; (b) ather information for parsin; (c) ather information for code optimization; (d) determine how lare the symbol table ouht to be. 9. In the context of type checkin: (a) two types that are structurally equivalent must be name equivalent; (b) two types that are name equivalent must be structurally equivalent; (c) both (a) and (b); (d) neither (a) nor (b). 0. The followin shows (the structure of) an abstract syntax tree for an expression. We want to label each node with the minimum number of reisters needed to evaluate the subtree rooted at that node without havin to store any intermediate results into memory (this information may then be used, for example, to determine a ood evaluation order for the various subexpressions). The labelled tree that correctly completes the labellin for the tree shown above is: (a) (b) (c) 0

12 . b. 2. c. 3. b. 4. c. 5. a. 6. a. 7. c. 8. c. 9. b. 0. a.

13 8 Lanuaes 2 (CSc 520): Principles of Prorammin Lanuaes Consider a variable x that is a reference parameter to a procedure p: x is referenced n times within the body of p. Suppose that p is such that it can be uaranteed that x can always be passed either by reference or by value-result without affectin the result of the proram. Let the cost of a direct memory reference be C dir, and the cost of an indirect memory reference be C ind (typically, C ind C dir ). You may make the followin assumptions: (a) Scalar aruments (inteers, pointers) can be passed by value in reisters. areates (arrays, records) have to be copied into the callee's stack frame. (b) An address can be computed into a reister without any references to memory. To be passed by value, (c) Copyin a value from one memory location to another requires two direct memory references (a load and a store). (d) Variables local to a procedure may be accessed directly. An element of an array that is local to a procedure can be accessed with a sinle direct memory reference. Answer the followin questions: (i) Suppose that x is an inteer variable. For what values of n is it more efficient topassx by value-result than by reference? (ii) Suppose that x is an array of M inteers. value-result than by reference? For what values of n is it more efficient to pass x by (i) Passin x by value-result involves 2 direct memory references for copyin x: once at entry to p, and once at exit. After this, each reference to the parameter within p is to a local that can be accessed directly. Thus, there are n + 2 direct references in this case, and the cost is (n +2) C dir. If x is passed by reference, no memory references are necessary to compute the address of x into a reister, but the n references to x within the procedure have to access it indirectly. The cost in this case is n C ind. If passin by value-result is to be more efficient, we want (n +2) C dir <n C ind ) ( + 2 n ) <C ind=c dir ) 2 n < (C ind C dir )=C dir ) n>2c dir =(C ind C dir ). (ii) In this case, passin by value-result involves copyin M inteers at entry and at exit: this involves 2M copy operations, each of which incurs two memory references, for a total of 4M direct memory references. After this, any element of the array can be accessed in one direct memory reference, so the total cost is (4M + n) C dir. In passin by reference, there is no initial cost for computin the address of the array into a reister, but then accessin any element takes an indirect reference, as before. Thus the cost in this case is n C ind. If passin by value-result is to be more efficient, then we should have (4M + n) C dir <n C ind ) + 4M n <C ind=c dir ) 4M n < (C ind C dir )=C dir ) n>4mc dir =(C ind C dir ) 2

14 9 Lanuaes 3 (CSc 553): Principles of Compilation Suppose we want to write a proram that will display HTML documents in a browser window: amon the thins we need to do for this is to insert line-breaks (i.e., end the current line and start a new line) at the appropriate places, dependin on the width of the window. In turn, this requires that we be able to determine how wide a iven piece of text is. This question focuses on the problem of determinin the width of text. The rammar we consider is the followin (for simplicity, we'll inore such subtleties as spacin between adjacent letters, proper" display of multiple whitespace characters, etc.): the set of nonterminals is fdocument, Text, Word, the set of terminals is f, , , , Char, the start symbol is Document, and the productions are Document! Text Text! Word j Word Text j Text j Text Word! Char j Char Word Assume the followin:. We have three kinds of fonts: normal, bold, and italic. Initially, text is displayed in normal font; any text within is displayed in bold font, while text within is displayed in italic font. In case of nested and font specifiers, only the mostly deeply nested specifier is considered at each point. 2. The terminal Char has an associated synthesized attribute lexeme that specifies the actual character encountered. The value of this attribute is determined by the lexical analyzer. 3. The width of a character depends on the font bein used (e.., bold characters are somewhat wider than normal ones). You are iven a function chwidth(fnt, chr) that, iven a font fnt and character chr, returns an inteer ivin the width of that letter in that font. For each production of the rammar, specify semantic rules to compute attribute values for the various nonterminals of the rammar so as to determine the width of any strin derived from Text. Specify, for each attribute of each nonterminal, whether it is inherited or synthesized. The nonterminal Document does not have any associated attributes. The nonterminals Text and Word each have an inherited attribute font and a synthesized attribute width. These are determined usin the followin semantic rules: Document! Text Text! Word Text! Word Text Text! Text Text! Text Word! Char Word! Char Word ftext.font = normal; fword.font = Text.font; Text.width = Word.width fword.font = Text.font; Text.font = Text.font; Text:width = Word:width + Text :width ftext.font = bold; Text:width = Text :width ftext.font = italic; Text:width = Text :width f Word.width = chwidth(word.font, Char.lexeme) fword :font = Word:font; Word.width = chwidth(word.font, Char.lexeme) +Word :width 3

1 Theory 1 (CSc 473): Automata, Grammars, and Lanuaes For any strin w = w 1 w 2 :::w n, define w R to be the reversal of w: w R = w n w n 1 :::w 1. Su

1 Theory 1 (CSc 473): Automata, Grammars, and Lanuaes For any strin w = w 1 w 2 :::w n, define w R to be the reversal of w: w R = w n w n 1 :::w 1. Su Masters Examination Department of Computer Science March 27, 1999 Instructions This examination consists of nine problems. The questions are in three areas: 1. Theory and Alorithms: CSc 473, 545, and 573;