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2.27 Two consecutive odd integers that are both prime are called twin primes. The twin primes conjecture is that there are infinitely many twin primes. Write a program that finds all the twin primes that are less than 1000. Use the Primes class from Problem 2.22. Your first five lines of output should look like this:

(a) Let x(n) be any bounded input with Ix(n)l c M .Then it follows that the output, y ( n ) = x2(n),may be bounded by I ~ ( n ) = lx(n)12 < M 2 l Therefore, this system is stable,

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public class TestFill { public static void main(String[] args) { Object[] a = new Object[4]; Arrays.fill(a, new Date()); ch02.ex02.DuplicatingArrays.print(a); Arrays.fill(a, 22); ch02.ex02.DuplicatingArrays.print(a); Arrays.fill(a, "Yo!"); ch02.ex02.DuplicatingArrays.print(a); }

(b) This system is clearly not stable. For example, note that the response of the system to a unit sample x ( n ) = S(n) is infinite for all values of n except n = 1.

3 5 11 17 29 5 7 13 19 31

(d) This system corresponds to a digital integrator and is unstable. Consider, for example, the step response of the system. With x ( n ) = u ( n ) we have, for n 2 0 ,

2.2 2.3 2.4

This system may be shown to be stable by using the following inequality:

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2.28 Test the conjecture that there is at least one prime between each pair of consecutive square numbers. (The square numbers are 1, 4, 9, 16, 25, . . .). Use the Primes class from Problem 2.22. Your first five lines of output should look like this:

Specifically, if x ( n ) is bounded, Ix(n)l < M ,

The sequential search runs in linear time, which means that the time is proportional to the number of elements. So an array with twice as many elements would take twice as long to process: 20 ms. The binary search runs in logarithmic time, so squaring the size of the problem should only double its running time. So an array with 10002 elements would take twice as long to process: 10 ms. The interpolation search runs in hyperlogarithmic time, so squaring the size of the problem should have no appreciable effect on its running time. So an array with 1,000,000 elements would also take about 2 ms to process.

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2.29 The Minimite friar Marin Mersenne (1588 1648) undertook in 1644 the study of numbers of the form n = 2 p 1, where p is a prime. He believed that most of these n are also primes, now

( f ) This system is not stable. This may be seen by considering the bounded input x(n) = cos(nrr/l). Specifically, note that the output of the system at time n = 0 is

which is unbounded. Alternatively, because the input-output relation is one of convolution, this is a linear shift-invariant system with a unit sample response

public class TestBinarySearch { private static final int SIZE = 10000; private static final int START = 0; private static final int RANGE = 10000; private static Random random = new Random(); private static int count = 0; public static void main(String[] args) { int[] a = new int[SIZE]; load(a, START, RANGE); Arrays.sort(a); search(a, random.nextInt(10000)); System.out.println(count + " iterations"); } public static void load(int[] a, int start, int range) { for (int i = 0; i < a.length; i++) { a[i] = start + random.nextInt(range); // random 5-digit numbers } } public static int search(int[] a, int x) { int lo = 0; int hi = a.length; while (lo < hi) { ++count; int i = (lo + hi)/2; if (a[i] == x) { return i; } else if (a[i] < x) { lo = i+1; } else { hi = i; } } return -1; } }

CHAP. 2]

Determine which of the following systems are invertible:

To test for invertibility, we may show that a system is invertible by designing an inverse system that uniquely recovers the input from the output, or we may show that a system is not invertible by finding two different inputs that produce the same output. Each system defined above will be tested for invertibility using one of these two methods.

boolean isSorted(int[] a) { if (a.length < 2) { return true; } for (int i = 1; i < a.length; i++) { if (a[i] < a[i-1]) { return false; } } return true; }

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