dataset_name: college_computer_science
description: The following are multiple choice questions (with answers) about college
  computer science.
fewshot_config:
  sampler: first_n
  samples:
  - question: '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'
    target: '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).'
  - question: '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'
    target: '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).'
  - question: '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'
    target: '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).'
  - question: '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'
    target: '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).'
  - question: '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'
    target: '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).'
group: mmlu_flan_cot_fewshot_stem
include: _mmlu_flan_cot_fewshot_template_yaml
task: mmlu_flan_cot_fewshot_college_computer_science
