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Another way to derive this decomposition more directly is as follows. First, we note that the decomposition should begin with a unit step, which generates a value of I at index n = 0.Because x(n) increases to a value of 2 at n = 1, we must add a delayed unit step u(n - 1). At n = 2, x(n) again increases in amplitude by 1, so we add the delayed unit step u(n - 2). At this point, we have

99 88 77

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Thus, all that remains is to bring the sequence back to zero for n > 3. This may be done by subtracting the delayed unit step 3u(n - 3), which produces the same decomposition as before.

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This verifies that the first loop invariant is true on every iteration of the outer loop Since swap(a,i,j) simply interchanges a[i] with a[j], the second loop invariant follows from the first The third loop invariant follows from the second and by mathematical induction During the first iteration of the main loop, the inner loop finds a[j] to be the largest element in the array The swap(a,i,j) puts that largest element in the last location a[i], so a[i] must be >= all the a[j] Prior to the ith iteration of the main loop, we have by the inductive hypothesis that the subarray a[i+1.n) is sorted and all the values in the subarray a[0.i] are smaller than a[i+1] Then after the ith iteration, a[i] is one of those smaller elements, so a[i] a[i+1] .. a[n-1].

99 88 88

For each of the systems below, x ( n ) is the input and y(n) is the output. Determine which systems are homogeneous, which systems are additive, and which are linear.

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(a) If the system is homogeneous, y(n) = T[cx(n)] = cT[x(n)] for any input x(n) and for all complex constants c. The system y(n) = log(x(n)) is not homogeneous because the response of the system to xl(n) = cx(n) is

Again, the proof is essentially the same as that for the corresponding theorem for the bubble sort On the first iteration of the outer i loop, the inner j loop iterates n 1 times On the second, it iterates n 2 times This progression continues, for a total of (n 1) + (n 2) + + 2 + 1 = n(n 1)/2..

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which is not equal to c log(x(n)). For the system to be additive, if yl(n) and y2(n) are the responses to the inputs and xz(n), respectively, the response to x(n) = xl(n) x2(n) must be y(n) = yl(n) y2(n). For this system we have T[xl(n) + x h N = log[x~(n) x2(n)l # log[x~(n)l+log[x2(n)l

public static void sort(int[] a, int[] ndx) { for (int i = 1; i < a.length; i++) { int ndxi = ndx[i], j; for (j = i; j > 0 && a[ndx[j-1]] > a[ndxi]; j--) { ndx[j] = ndx[j-1]; } ndx[j] = ndxi; } }

Therefore, the system is not additive. Finally, because the system is neither additive nor homogeneous, the system is nonlinear. (b) Note that if y(n) is the response to x(n).

f. Trace of the merge sort:

which is not the same as y1 (n). Therefore, this system is not homogeneous. Similarly, note that the response to x(n) = x,(n) x2(n) is

CHAP. 14]

which is not equal to yl(n)

a[0] 44 44 a[1] 88 55 a[2] 55 77 a[3] 99 99 a[4] 66 33 22 66 a[5] 33 66 33 77 66 88 77 77 88 88 88 99 a[6] 22 a[7] 88 a[8] 77

The system is clearly, however, not additive and therefore is nonlinear. Let y,(n) and yz(n) be the responses of the system to the inputs x,(n) and x2(n), respectively. The response to the input x(n) = axl(n) bxz(n)

public static void sort(int[] a) { sort(a, a.length); } public static void sort(int[] a, int n) { if (n < 2) { return; } sort(a, n-1); int temp = a[n-1], j; for (j = n-1; j > 0 && a[j-1] > temp; j--) { a[j] = a[j-1]; } a[j] = temp; }

(y) =

CHAP. 14]

[axl(n)

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