dataset_name: college_computer_science
description: 'The following are multiple choice questions (with answers) about college
  computer science.


  Q: Which of the following regular expressions is equivalent to (describes the same
  set of strings as) (a* + b)*(c + d)?

  (A) a*(c + d)+ b(c + d)

  (B) a*(c + d)* + b(c + d)*

  (C) a*(c + d)+ b*(c + d)

  (D) (a + b)*c +(a + b)*d

  A: Let''s think step by step. We know that:

  1. (X* + Y)* = (X + Y)*

  2. X(Y + Z)? = XY + XZ

  Using equation 1 we can rewrite (a* + b)*(c + d)? as:

  3. (a + b)*(c + d)?

  Using equation 2 we can rewrite equation 3 as:

  (a + b)*c + (a + b)*d The answer is (D).


  Q: The Singleton design pattern is used to guarantee that only a single instance
  of a class may be instantiated. Which of the following is (are) true of this design
  pattern?

  I. The Singleton class has a static factory method to provide its instance.

  II. The Singleton class can be a subclass of another class.

  III. The Singleton class has a private constructor.

  (A) I only

  (B) II only

  (C) III only

  (D) I, II, and III

  A: Let''s think step by step. Statement I is a correct statement about a Singleton,
  because a Singleton restricts instantiation to a single, static method. Statement
  II is also correct, because there is no inherent restriction regarding the inheritance
  of a Singleton. Statement III is also correct, because a Singletons must be instantiated
  only once, so its constructor is made private to prevent any construction except
  via its static factory method.

  Given these facts, statements I, II, and III are all correct. The answer is (D).


  Q: A certain pipelined RISC machine has 8 general-purpose registers R0, R1, . .
  . , R7 and supports the following operations:

  ADD Rs1, Rs2, Rd (Add Rs1 to Rs2 and put the sum in Rd)

  MUL Rs1, Rs2, Rd (Multiply Rs1 by Rs2 and put the product in Rd)

  An operation normally takes one cycle; however, an operation takes two cycles if
  it produces a result required by the immediately following operation in an operation
  sequence.

  Consider the expression AB + ABC + BC, where variables A, B, C are located in registers
  R0, R1, R2. If the contents of these three registers must not be modified, what
  is the minimum number of clock cycles required for an operation sequence that computes
  the value of AB + ABC + BC?

  (A) 5 (B) 6 (C) 7 (D) 8

  A: Let''s think step by step. First, we are given that A is in R0, B is in R1, and
  C is in R2.

  Next, we can see that we must compute three multiplies (AB, BC, and ABC) and two
  adds (AB + ABC, (AB + ABC) + BC) to compute our final answer, resulting in a minimum
  of five clock cycles.

  Next, we can see that there is no way to avoid at least one pipeline stall when
  computing our final answer, because to compute our final sum we must wait at least
  one cycle for the results from the previous stage to be ready. Thus, our minimum
  number of cycles must be 6.

  We can verify that we can create a solution that requires only six cycles as follows:

  compute AB: MUL R0, R1, R3

  compute BC: MUL R1, R2, R4

  compute ABC: MUL R3, R4, R5

  compute AB + BC: ADD R3, R4, R6

  STALL

  compute AB + ABC + BC: ADD R5, R6, R7

  So there are 6 cycles. The answer is (B).


  Q: A compiler generates code for the following assignment statement.

  G := (A + B) * C - (D + E) * F

  The target machine has a single accumulator and a single-address instruction set
  consisting of instructions load, store, add, subtract, and multiply. For the arithmetic
  operations, the left operand is taken from the accumulator and the result appears
  in the accumulator. The smallest possible number of instructions in the resulting
  code is

  (A) 5 (B) 6 (C) 7 (D) 9

  A: Let''s think step by step. We can compute the final answer with the following
  sequence of operations:

  1. LOAD D  (accumulator = D)

  2. ADD E  (accumulator = D+E)

  3. MUL F  (accumulator = (D+E)*F)

  4. STORE X (X = (D+E)*F)

  5. LOAD A  (accumulator = A)

  6. ADD B  (accumulator = A+B)

  7. MUL C  (accumulator = (A+B)*C)

  8. SUB X  (accumulator = (A+B)*C - (D+E)*F)

  9. STORE G (G = (A+B)*C - (D+E)*F)

  This sequence takes 9 instructions. The answer is (D).


  Q: Consider a computer design in which multiple processors, each with a private
  cache memory, share global memory using a single bus. This bus is the critical system
  resource. Each processor can execute one instruction every 500 nanoseconds as long
  as memory references are satisfied by its local cache. When a cache miss occurs,
  the processor is delayed for an additional 2,000 nanoseconds. During half of this
  additional delay, the bus is dedicated to serving the cache miss. During the other
  half, the processor cannot continue, but the bus is free to service requests from
  other processors. On average, each instruction requires 2 memory references. On
  average, cache misses occur on 1 percent of references. What proportion of the capacity
  of the bus would a single processor consume, ignoring delays due to competition
  from other processors?

  (A) 1/50 (B) 1/27 (C) 1/25 (D) 2/27

  A: Let''s think step by step. We know that each instruction requires two memory
  references per instruction, and that there is an average cache miss rate of one
  percent.

  Thus a given processor has:

  (1 cache miss / 100 references) * (2 references / instruction) =

  (2 cache misses / 100 instructions), so:

  misses_per_instruction = 1 cache miss / 50 instructions.

  Next, we know that each instruction requires 500 nanoseconds when there is no cache
  miss, and 500 + 2000 = 2500 nanoseconds when there is a cache miss. Thus:

  50 instructions / (49 * 500) + (1 * 2500) nanoseconds, so:

  instructions_per_ns = 50 instructions / 27000 nanoseconds.

  Now, we know that each cache miss locks the bus for half of the 2000 nanosecond
  cache miss delay, or 1000 nanoseconds, so:

  lock_ns_per_miss = 1000 nanoseconds / cache miss.

  Thus we can see that on average a single processor will lock the bus for:

  lock_ns_per_miss * misses_per_instruction * instructions_per_ns =

  (1000 nanoseconds / cache miss) * (1 cache miss / 50 instructions) * (50 instructions
  / 27000 nanoseconds) = 1000 * (1/50) * (50/27000) = 1000/27000 = 1/27. The answer
  is (B).'
include: _mmlu_flan_cot_fewshot_template_yaml
task: mmlu_flan_cot_fewshot_college_computer_science
