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COMPUTER ORGANIZATION 3.1 Write the number 229 in base-2. 11100101 3.2 What is the base-10 value of 11100101 229 3.3 What are the units (values) of the first 3 columns in a base-8 (octal) number 64s 82 8s 81 80 units

On the first iteration of the main loop, a[1] is compared with a[0] and interchanged if necessary. So a[1] after the first iteration. If we assume that the loop invariant is true prior to some kth iteration, then it must also be true after that iteration because during it a[k] is inserted between the elements that are less than or equal to it and those that are greater. It follows by mathematical induction that the loop invariant is true for all k.

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The proof is similar to that for the corresponding theorems for the bubble sort and the selection sort. On the first iteration of the outer i loop, the inner j loop iterates once. On the second, it iterates once or twice, depending upon whether a[1] > a[2]. On the third iteration, the inner j loop iterates at most three times, again depending upon how many of the elements on the left of a[3] are greater than a[3]. This pattern continues, so that on the kth iteration of the outer loop, the inner loop iterates at most k times. Thus the maximum total number of iterations is 1 + 2 + + (n 1) = n(n 1)/2. In this case, the inner loop will iterate only once for each iteration of the outer loop. So the total number of iterations of the inner loop is: 1 + 1 + 1 + +1 + 1 = n 1. For the quick sort pivoting on the last element, the only changes needed are in the partition() method:

3.4 What is the base-2 value of the base-8 (octal) number 377 11111111 3.5 Convert the following base-10 numbers to base-2: 37 470 1220 17 99 100101 111010110 10011000100 10001 1100011 3.6 Convert the following base-2 numbers to base-10: 00000111 7 10101010 170 00111001 57 01010101 85 00110011 51 3.7 Assume a 16-bit signed integer data representation where the sign bit is the msb. a b c d a b c d What is the largest positive number that can be represented Write the number 17,440. Write the number 20. What is the largest negative number that can be represented 32,767 0100010000100000 1111111111101100 -32,768

private static int partition(int[] a, int p, int q) { int pivot = a[q-1], i = p-1, j = q-1; while (i < j) { while (i < j && a[++i] <= pivot) ; // empty loop if (i < j) { a[j] = a[i]; } while (j > i && a[--j] >= pivot) ; // empty loop if (j > i) { a[i] = a[j]; } } a[j] = pivot; return j; }

14.25 14.26

3.8 Using ASCII encoding, write the bytes to encode your initials in capital letters. CHR C 67 H 72 R 82 1000011 1001000 1010010

partition() method: private static int partition(int[] a, int p, int q) { int m = (p + q)/2; m = indexOfMedian(a, p, m, q-1); swap(a, p, m); // The rest is the same as lines 20-32 in Example 14.9 on page 264 }

This requires a method for locating the index of three array elements:

3.9 Referring to the list of Intel x86 instructions in this chapter, arrange a set of instructions to add the values stored in memory locations 50 and 51, and then to store the result in memory location 101. You need not show the bit pattern for each instruction; just use the mnemonics listed, followed in each case by the appropriate operand(s). MOV 50, EDX // copy what s in 50 to the register ADD 51, EDX // add what s in 51 to the register MOV EDX, 101 // store the result in 101 3.10 What Intel x86 instructions would you use to accomplish subtraction using 2 s complement addition This instruction set has a SUB instruction, but don t use that; write your own 2 s complement routine instead. MOV NEG ADD MOV 51, EDX EDX EDX, 50 EDX, 101 // // // // copy what s in 51 to the register take 2s complement of register subtract contents of 51 from 52 store the result in 101

private static int indexOfMedian(int[] a, int i, int j, int k) { // Returns the index of the median of {a[i], a[j], a[k]} if (a[i] <= a[j] && a[j] <= a[k]) return j; if (a[i] <= a[k] && a[k] <= a[j]) return k; if (a[j] <= a[i] && a[i] <= a[k]) return i; if (a[j] <= a[k] && a[k] <= a[i]) return k; if (a[k] <= a[i] && a[i] <= a[j]) return i; return j; }

For the quick sort with reversion to the insertion sort on arrays of size < 8, the only changes needed are in the sort() method:

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