Masters Exam. Fall 2004 Department of Computer Science University of Arizona

Transcription

1 Masters Exam Fall 2004 Department of Computer Science University of Arizona This exam consists of ten problems, from which you will do six questions. The questions are in three areas: Theory (CSc 520, 545, 573), Systems (CSc 552, 553, 576), and Applications (CSc 522, 525, 533, 560). Answer two questions from each area. If more than two questions are attempted in an area, the two highest scores are used. You have two hours for the exam. Books or notes are not permitted. At the end of the exam, submit only your solutions in the envelope provided; do not include scrap paper. Write your answers on the paper provided separately. Start the answer to each question on a new sheet of paper. On each page you hand in, write the problem number in the upper-left corner, and the last four digits of your social security number in the upper-right corner. Results will be posted using the four-digit numbers. Some problems may ask you to write an algorithm or a program. Unless directed otherwise, use informal pseudocode. You may use programmming constructs that help present your solution provided their meaning is clear and they do not make the problem trivial. For example, when describing code that iterates over the elements of a set S, you may use a construct such as for x S do statement s

2 Theory 1 CSc 520 (Programming Language Theory) The following questions concern scoping of identifiers. (a) (3 points) Compare and contrast static and dynamic scoping. Explain how identifier bindings are resolved at compile-time or run-time. (b) (3 points) Briefly, why has dynamic scoping fallen out of favor? Give two distinct reasons. (c) (4 points) Write a single short program that prints the word dynamic if the language implementation uses dynamic scoping, and prints the word static if the implementation uses static scoping. ( by Todd Proebsting.) (a) Static scoping resolves all references at compile-time using lexical scoping. No run-time properties are necessary to determine bindings. Dynamic scoping requires binding resolution of free variables that is dependent on the execution context. An instance of the same identifier could be bound to different variables depending on execution context (i.e. the calling procedure). (b) Dynamic scoping requires identifier binding at run-time, which is a costly operation. Dynamic scoping makes programs hard to reason about because the programmer must anticipate all possible calling contexts for a procedure s free variables. (c) One solution is the following: var x : integer := 1; procedure foo() if (x==0) then print "dynamic" else print "static"; procedure main() var x : integer := 0; call foo() 1

3 Theory 2 CSc 545 (Algorithms) Suppose you have a pile of nuts and bolts, and you want to find the pairs of nuts and bolts that match. Specifically, there are n nuts and n bolts, and each nut matches exactly one bolt. When comparing a nut to a bolt, if they don t match you learn which one is bigger than the other, but you cannot directly compare two nuts or two bolts. (a) (3 points) How fast can you match the nuts and bolts in the worst case? More precisely, what is the worst-case running time of your algorithm in terms of the number of nut-bolt comparisons? (b) (7 points) How fast can you match the nuts and bolts in the expected case? You may use randomization to speed up your algorithm from Part (a). ( by Stephen Kobourov.) (a) Here is a trivial Θ(n 2 ) time algorithm. Compare each nut to each bolt. Since there are n nuts and n bolts, this takes Θ(n 2 ) comparisons in the worst case. (b) We can do better. A simple modification of QuickSort shows there is a randomized algorithm whose expected number of comparisons (and running time) is O(n log n). Just pick a random bolt, compare it to all nuts, find its matching nut, and compare it to all bolts, splitting the problem into two subproblems, one consisting of the nuts and bolts smaller than the matched pair and one consisting of the larger ones. Repeating in this manner yields an algorithm whose expected running time is O(n log n). The analysis is similar to that for QuickSort. 2

4 Theory 3 CSc 573 (Theory of Computation) The following questions concern closure properties of languages. (a) (4 points) Let A be a regular language and B be some other language that we know nothing about. Suppose we find that A B is not regular. Which of the following can we say about B? (1) B is regular. (2) B is not regular. (3) B might or might not be regular. Justify your answer carefully. (b) (6 points) Prove or disprove the following: If languages A and B are both context-free, then their intersection A B is context-free. ( by Stephen Kobourov.) (a) Answer (2) is correct: B is not regular. To see this, suppose B was regular. Since the class of regular languages is closed under union, this implies A B is regular, which is a contradiction. (b) This is false. Both and A := B := { } a i b i c j : i, j 1 { } a i b j c j : i, j 1 are context-free, but their intersection is { } A B = a i b i c i : i 1, which can be shown to be non-context-free using the pumping lemma. 3

5 Systems 1 CSc 552 (Operating Systems) The following questions concern remote procedure calls. (a) (3 points) What is the difference between idempotent and non-idempotent operations? Give an example of each. (b) (4 points) What is the difference between at-least-once and at-most-once remoteprocedure-call semantics? Briefly describe how each might be implemented. (c) (3 points) Why is the distinction between idempotent and non-idempotent operations important with respect to the different remote-procedure-call semantics? ( by John Hartman.) (a) An idempotent operation is one that has the same effect no matter how many times it is performed. Examples are reading a block of a file, or setting an account balance to a particular value. A non-idempotent operation has the opposite quality: its effect depends on how many times it is performed. Examples are appending to a file, or adding to an account balance. (b) At-least-once semantics guarantee that a remote-procedure call (RPC) is performed by the server at least once, even in the face of server failures. This can be implemented by having the client retransmit the RPC request until a reply is received. Since the server may fail after performing the RPC but before sending the reply, it may redo the RPC when it reboots. At-most-once semantics guarantee that an RPC is performed at most once by the server, even in the face of server failures. This can be accomplished by maintaining an epoch number on the server that the clients must include in the RPC requests. The server increments the epoch number when it reboots and refuses to service RPCs from the previous epoch. Clients eventually time-out when a reply is not received. (c) At-least-once RPC semantics require idempotent operations since the RPC may be performed multiple times. At-most-once RPC semantics can use either semantics, although it is left up to the client as to how to proceed if an RPC fails. 4

6 Systems 2 CSc 553 (Compilers) Briefly describe in one or two sentences what each of the following terms means in compilation, and what it is used for: register allocation, instruction scheduling, peephole optimization, liveness analysis, and control flow graphs. (Register allocation) This refers to the process of determining which variables are kept in registers at any given point in a program, and which are kept in memory. It is used to improve the performance of the compiled code by keeping frequently accessed variables in registers. (Instruction scheduling) This refers to the process of determining an ordering for the instructions in a program that preserves program semantics, but which may be more efficient than the original instruction order for the program. It is used to hide the latencies of expensive operations where possible. (Peephole optimization) This is an optimization that uses a sliding window over the instruction sequences to look for specific patterns of instructions, and replace them with equivalent sequences that are shorter or faster. This is a simple and cheap approach to effecting local improvements in the code generated by the compiler. (Liveness analysis) Liveness analysis is used to determine which variables in the program are live at each point in the program, i.e. which may be used at a later program point prior to being redefined. Variable liveness information is used for a variety of optimizations, e.g. in register allocation to determine whether the value of a variable in a register can be overwritten by that of some other variable. (Control flow graphs) A control flow graph is a directed graph where each node is a basic block and where there is an edge from node A to node B if it is possible for control to leave A and go immediately to B. It is used as the basic internal low-level program representation for a wide variety of program analyses and optimizations. 5

7 Systems 3 CSc 576 (Architecture) The following questions concern instruction scheduling and memory caches. (a) (5 points) Scoreboarding and Tomasulo s algorithm are commonly-used algorithms for dynamic instruction-scheduling in processors that support out-of-order execution. Explain how these algorithms differ in handling WAW and WAR hazards. (b) (5 points) The average memory-access time is determined by three factors. List these factors, and for each, name a technique that reduces the cache-hit time by affecting that factor. (a) Through register renaming, Tomasulo s algorithm effectively eliminates these hazards. Each instruction is assigned a unique destination register to which it writes its result by the register renaming mechanism, thereby eliminating these hazards. In scoreboarding, the execution of an instruction is delayed until the hazards are resolved. (b) The factors are: (1) cache miss-rate, affected by using a victim cache, (2) cache hit-time, affected by using a virtually-indexed and physically-tagged cache, and (3) cache miss-penalty, affected by giving priorities to reads over writes. 6

8 Applications 1 CSc 522 (Parallel and Distributed Programming) The following questions concern three mechanisms for creating and terminating processes: co/oc (also called cobegin/coend), fork/join/quit, and static process declaration. (a) (5 points) Informally describe the semantics of each of the above mechanisms. (b) (5 points) Discuss the advantages and disadvantages of each. In particular, describe which of the mechanisms above can simulate the other mechanisms. (a) Their semantics are: co/oc: These primitives explicitly specify a sequence of program statements (or processes) that may be executed concurrently. fork/join/quit: If process p executes fork x, it creates a new process q that starts executing the instruction labeled x. Processes p and q can execute concurrently. The command join(t, y) has the following effect: t := t-1 if t = 0 goto y; else quit; The command quit causes the process to terminate. Static process declaration: With this mechanism, process types are declared in the definition of a process p. When p is activated, the processes declared within the definition of p are also activated. (b) Their advantages and disadvantages are: co/oc: This can describe any process flow graph that is a series-parallel graph, but nothing more general. fork/join/quit: These can describe arbitrary process flow graphs and can simulate the two other mechanisms. The disadvantage is they are very unstructured. Static process declaration: This allows explicitly designating code segments as separate processes. The disadvantage is there is no mechanism for joining processes. 7

9 Applications 2 CSc 525 (Networking) The Internet is built on an infrastructure that is prone to failures. Suppose we found a way to build an infrastructure that did not fail. In particular, network links that connect hosts and routers never drop packets, never have bit errors, and every packet transmitted over a link gets to its destination. Furthermore, routers that connect networks never fail: every packet that gets to a router gets sent over a link to another router along a path to its destination. (a) (5 points) Consider the IP layer. What are the main features that IP provides in the traditional Internet? Of these features, what features would still be present in the never-fail Internet? (b) (5 points) Consider the TCP layer. What are the main features that TCP provides in the traditional Internet? Of these features, what features would still be present in the never-fail Internet? In your answers to Parts (a) and (b), for each feature still present in the never-fail Internet, give a very brief explanation why that feature is needed. ( by Patrick Homer.) (a) IP in the traditional Internet provides: host-to-host communication, datagram delivery, packet formatting, fragmentation and reassembly, IP addressing (or naming of hosts), and packet routing. In the never-fail Internet, IP would still provide these, but some would change in service: host-to-host communication: same, datagram delivery: same, but all datagrams (packets) get through, packet formatting: same, fragmentation and reassembly: same, as it s still needed to accomodate various sizes of packets or frames accepted by the underlying network hardware, IP addressing: same, packet routing: same, as we still need to consider the best (fastest or leastcongested) path for a packet to follow. (b) TCP in the traditional Internet provides: application-to-application communication, 8

10 byte streams, flow control, reliable delivery, ordered delivery of data, and virtual circuits. In the never-fail Internet, TCP would still provide the following: application-to-application communication: still needed to multiplex connections, byte streams: still a useful abstraction on top of packets, flow control: still needed (a key point!) since congestion is still possible, reliable delivery: not needed, since the underlying network is reliable, ordered delivery: still needed, since packets can arrive out-of-order by taking different paths through the Internet, virtual circuits: can be argued either way. They can be useful to establish a route for data to follow. One can also argue they are not necessary since the underlying network will deliver every packet. (TCP uses virtual circuits in part to avoid the how-long-do-i-wait-for-an-answer problem that UDP has.) 9

11 Applications 3 CSc 533 (Graphics) The following questions concern projecting points onto a viewing plane. (a) (1 point) Sketch a right-hand coordinate system with a point at (x, y, z) and a camera plane at z = f. Now illustrate how a point (x, y, z) with z > f gets projected onto the camera plane to become point (x, y, z ). (b) (1 point) Determine an expression for (x, y, z ) from Part (a). The derivation need not be long, but full credit requires some indication of how you got your answer. (c) (2 points) Is there a 3-by-3 matrix which maps (x, y, z) to (x, y, z )? If yes, what is it? If no, why not? (d) (2 points) Is there a 4-by-4 matrix which maps a homogeneous coordinate representation for (x, y, z) to a homogeneous coordinate representation for (x, y, z )? If yes, what is it? If no, why not? (e) (1 point) Regardless of whether the matrices in Parts (c) and (d) exist, state two independent properties of a matrix A that maps points in a three-dimensional world onto a plane. (f) (2 points) Suppose the camera is now at an arbitrary location as specified by the center of its image plane, and is pointing in an arbitrary direction. You can assume you know a camera-centric right-hand coordinate system. The focal length is still f. Describe how to compute a single matrix that can be used to map points onto the camera plane. (A complete derivation is not needed, just an outline of how it can be done.) (g) (1 point) Suppose you have an implementation that speeds up your answer for Part (f) by a factor of five. In rough terms, what is the anticipated computational improvement of the overall graphics pipeline when rendering a complex scene? ( by Kobus Barnard.) (a) See the following figure. 10

12 (b) See the above figure. (c) There is no such matrix, because the mapping derived in Part (b) is not a linear transformation. We would need to be able to express (x, y, z ) as a linear combination of x, y, and z. (d) It can be done in homogeneous coordinates because we can use the fourth coordinate w, to encode the division. In particular, the 4-by-4 matrix with ones for the first three diagonal entries, 1/f for entry (4, 3), and zero everywhere else does the job. (e) Two independent properties are: (1) A 2 = A, and (2) det(a) = 0, or inv(a) does not exist. (f) We can tackle this problem by changing coordinate systems, making it so that the objects are now expressed in terms of the the coordinate system drawn in Part (a). We can then follow this by the matrix operation for Part (d) to get a single matrix. We can do the change of coordinates with a sequence of translations and rotations. We can change coordinates by a translation which takes the camera center to the origin, a rotation to align the two right-hand coordinate systems, and then a final translation to put the focal point at the center. Having found a sequence of matrix operations, which when followed by the matrix in Part (d) does the projection, we get a single matrix by multiplying them together in the right order (right to left if we list them in the order that they would apply to a point). (g) The impact is minimal because the computation of the above matrix is only done once, and the single matrix is then applied to potentially millions of objects. It already does not take much time relative to other parts of the application (if done by any remotely reasonable method), so speeding it up has almost no effect. 11

13 Applications 4 CSc 560 (Databases) One method for finding nearest neighbors in a multidimensional database, where Euclidean distance is used to measure the nearness of data objects, is the branch-and-bound algorithm with an R-tree index. The key feature of the branch-and-bound algorithm is that it takes advantage of the Minimum- Bounding-Region Face-Property to prune the search space. Explain what this property is, and how it helps reduce query-processing costs. ( by Bongki Moon.) The Minimum-Bounding-Region (MBR) Face-Property is that every face of an MBR must touch at least one object enclosed by the MBR. This property implies the following: Given any query point q and any set of data objects S, there exists an object x S such that dist(q, x) MinMaxDist(q, MBR(S)). If an object y is encountered during the tree search and dist(q, y) > MinMaxDist(q, M) for any MBR M that has already been encountered, then object y cannot be the nearest neighbor of q and can be discarded. This is because the MBR M contains at least one object closer to q than y by the above. 12