sp08-sl-s2 - ECE 3035 A B 4 problems 5 pages Problem 1(3 parts 22 points Computing Mechanisms Exam Two Solutions Spring 2008 9 April 2008 Path Search

# sp08-sl-s2 - ECE 3035 A B 4 problems 5 pages Problem 1(3...

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ECE 3035 A & B Computing Mechanisms Spring 2008 4 problems, 5 pages Exam Two Solutions 9 April 2008 1 Problem 1 (3 parts, 22 points) Path Search Part A (6 points) The diamond search map above list the cost of visiting each square (in black) and the position of the square (in blue). Answer the following questions (show work). Cost to square 55 47 – 38 + 89 – 79 + 60 = 79 Cost to square 48 -75 + 37 = -38 Cost to square 25 -75 + 37 – 90 – 79 – 74 = -281 Part B (11 points) Write a MIPS subroutine that computes (in general) the path cost from the starting square to square 115. Assume \$3 is initialized to the address of square 71. Return the path cost in register \$2. label instruction comment CostTo115: addi \$5, \$0, 4 # count equals run - 1 lw \$2, Array(\$3) # load initial weight Loop: add \$3, \$3, 44 # adjust for next square lw \$1, Array(\$3) # load weight add \$2, \$2, \$1 # add to sum addi \$5, \$5, -1 # next loop? bne \$5, \$0, Loop # loop until done jr \$31 # return to caller Part C (5 points) Explain the flaw in creating a minimum cost path search program with no limits on path length. If the path length is unbound, the search algorithm will enter an infinite loop stepping in and out of negative squares as the cost moves towards negative infinity.
ECE 3035 A & B Computing Mechanisms Spring 2008 4 problems, 5 pages Exam Two Solutions 9 April 2008 2 Problem 2 (2 parts, 30 points) Activation Frames Consider the following C fragment: int main() { int X = 100, Y = 255, Z = 1, A[] = {3, 5, 7}; int Foo(int, int *, int *); Z = Foo(X, &Y, A); printf("X = %d, Y = %d, A[0]=%d, A[1]=%d, A[2]=%d\n", X, Y, A[0], A[1], A[2]); return (Y); } int Foo(int P, int *Q, int V[]) { int Z = 35, W[] = {2, 4};

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