"dataset_name": "college_computer_science"
"description": "The following are multiple choice questions (with answers) about college\
  \ computer science.\n\nQ: Which of the following regular expressions is equivalent\
  \ to (describes the same set of strings as) (a* + b)*(c + d)?\n(A) a*(c + d)+ b(c\
  \ + d)\n(B) a*(c + d)* + b(c + d)*\n(C) a*(c + d)+ b*(c + d)\n(D) (a + b)*c +(a\
  \ + b)*d\nA: Let's think step by step. We know that:\n1. (X* + Y)* = (X + Y)*\n\
  2. X(Y + Z)? = XY + XZ\nUsing equation 1 we can rewrite (a* + b)*(c + d)? as:\n\
  3. (a + b)*(c + d)?\nUsing equation 2 we can rewrite equation 3 as:\n(a + b)*c +\
  \ (a + b)*d The answer is (D).\n\nQ: 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?\nI. The Singleton class has a static factory\
  \ method to provide its instance.\nII. The Singleton class can be a subclass of\
  \ another class.\nIII. The Singleton class has a private constructor.\n(A) I only\n\
  (B) II only\n(C) III only\n(D) I, II, and III\nA: 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.\n\
  Given these facts, statements I, II, and III are all correct. The answer is (D).\n\
  \nQ: A certain pipelined RISC machine has 8 general-purpose registers R0, R1, .\
  \ . . , R7 and supports the following operations:\nADD Rs1, Rs2, Rd (Add Rs1 to\
  \ Rs2 and put the sum in Rd)\nMUL Rs1, Rs2, Rd (Multiply Rs1 by Rs2 and put the\
  \ product in Rd)\nAn 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.\nConsider 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?\n(A) 5 (B) 6 (C)\
  \ 7 (D) 8\nA: Let's think step by step. First, we are given that A is in R0, B is\
  \ in R1, and C is in R2.\nNext, 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.\nNext, 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.\nWe can verify that we can create a solution that requires only six cycles\
  \ as follows:\ncompute AB: MUL R0, R1, R3\ncompute BC: MUL R1, R2, R4\ncompute ABC:\
  \ MUL R3, R4, R5\ncompute AB + BC: ADD R3, R4, R6\nSTALL\ncompute AB + ABC + BC:\
  \ ADD R5, R6, R7\nSo there are 6 cycles. The answer is (B).\n\nQ: A compiler generates\
  \ code for the following assignment statement.\nG := (A + B) * C - (D + E) * F\n\
  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\n(A) 5 (B) 6 (C) 7 (D) 9\nA: Let's think step by step. We can compute\
  \ the final answer with the following sequence of operations:\n1. LOAD D  (accumulator\
  \ = D)\n2. ADD E  (accumulator = D+E)\n3. MUL F  (accumulator = (D+E)*F)\n4. STORE\
  \ X (X = (D+E)*F)\n5. LOAD A  (accumulator = A)\n6. ADD B  (accumulator = A+B)\n\
  7. MUL C  (accumulator = (A+B)*C)\n8. SUB X  (accumulator = (A+B)*C - (D+E)*F)\n\
  9. STORE G (G = (A+B)*C - (D+E)*F)\nThis sequence takes 9 instructions. The answer\
  \ is (D).\n\nQ: 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?\n(A) 1/50 (B) 1/27 (C) 1/25 (D)\
  \ 2/27\nA: 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.\nThus a given processor has:\n(1 cache miss / 100 references)\
  \ * (2 references / instruction) =\n(2 cache misses / 100 instructions), so:\nmisses_per_instruction\
  \ = 1 cache miss / 50 instructions.\nNext, 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:\n50 instructions / (49 * 500) + (1 * 2500) nanoseconds,\
  \ so:\ninstructions_per_ns = 50 instructions / 27000 nanoseconds.\nNow, we know\
  \ that each cache miss locks the bus for half of the 2000 nanosecond cache miss\
  \ delay, or 1000 nanoseconds, so:\nlock_ns_per_miss = 1000 nanoseconds / cache miss.\n\
  Thus we can see that on average a single processor will lock the bus for:\nlock_ns_per_miss\
  \ * misses_per_instruction * instructions_per_ns =\n(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).\n\n"
"group": "mmlu_flan_cot_fewshot_stem"
"include": "_mmlu_flan_cot_fewshot_template_yaml"
"task": "mmlu_flan_cot_fewshot_college_computer_science"
