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2.1 2.2 2.3 2.4 Run a test program to see how the Arrays.fill() method handles an array of objects. If the sequential search took 50 ms to run on an array of 10,000 elements, how long would you expect it to take to run on an array of 20,000 elements on the same computer If the binary search took 5 ms to run on an array of 1,000 elements, how long would you expect it to take to run on an array of 1,000,000 elements on the same computer The interpolation search is the same as the binary search except that in step 4 the element si is chosen so that the proportion of elements less than si in the subsequence ss equals the proportion that would be expected in a uniform distribution. For example, looking up the name Byrd in a phone book of 2,600 pages, one would open first near page 200 because one would expect about 2/26 of all the names to precede it. The interpolation search can be shown to run in O(lglgn) time. If it took 5 ms to run on an array of 1,000 elements, how long would you expect it to take to run on an array of 1,000,000 elements on the same computer
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[CHAP. 2
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2.5 2.6
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Run a test driver for the binary search method in Example 2.5 on page 32 on an array of 10,000 elements and count the number of iterations. Write and test this method:
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boolean isSorted(int[] a) // returns true iff a[0] <= a[1] <= ... <= a[a.length-1]
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int minimum(int[] a) // returns the minimum element of a[]
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double mean(double[] a) // returns the average value of all the elements in a[]
Write and test this method:
int[] withoutDuplicates(int[] a) // returns the specified array after removing all duplicates
2.10 Write and test this method:
void reverse(int[] a) // reverses the elements of a[]
2.11 Write and test this method:
Object[] concatenate(Object[] a, Object[] b) // returns an array containing all of a[] followed by all of b[]
2.12 Write and test this method:
void shuffle(Object[] a) // randomly permutes the elements of a[]
2.13 Write and test this method:
int[] tally(String string) // returns an array a[] of 26 integers that count the frequencies // of the (case insensitive) letters in the given string
2.14 Write and test this method:
double innerProduct(double[] x, double[] y) // returns the algebraic inner product (the sum of the component// wise products) of the two given arrays as (algebraic) vectors
2.15 Write and test this method:
double[][] outerProduct(double[] x, double[] y) // returns the algebraic outer product of the two given arrays // as (algebraic) vectors: p[i][j] = a[i]*b[j]
2.16 Write and test this method:
double[][] product(double[][] a, double[][] b) // returns the matrix product of the two given arrays a matrix: // p[i][j] = Sum(a[i][k]*b[k][j]:k)
2.17 Write and test this method:
double[][] transpose(double[][] a) // returns the transpose ta of the specified array as a matrix: // ta[i][j] = a[j][i]
2.18 Write and test this method:
int[][] pascal(int size) // returns Pascal s triangle of the given size
CHAP. 2]
ARRAYS
2.19 The Sieve of Eratosthenes is an array of boolean elements whose ith element is true if and only if i is a prime number. Use the following algorithm to compute and print a sieve of size 1000: (Precondition: p is an array of n bits.) (Postcondition: p[i] is true if and only if i is prime.) 1. Initialize p[0] and p[1] to be false, and all other p[i] to be true. 2. Repeat step 3 for each i from 3 to n, incrementing by 2. 3. If there is a prime the square root of i that divides i, set p[i] false. 2.20 Repeat Problem 2.19 using a java.util.Vector object. 2.21 Repeat Problem 2.19 using a java.util.BitSet object. 2.22 Define and test a Primes class with these methods:
public public public public public public public Use the BitSet static static static static static static static void setLast(int last) void setLast() void sizeSize(int size) void sizeSize() boolean isPrime(int n) int next() void printPrimes() // // // // // // // sets last sets last=1 sets size of bitset sets bitset size=1000 true if n is prime next prime after last prints sieve
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