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CMP 338

Data Structures and Algorithms I Day 2, 9/1/11. CMP 338. Programming Homework 1. Create a matrix library with the following API Class: Matrix Matrix(int r, int c) Creates a matrix with r rows and c columns Individual entries are random doubles multAdd(Matrix C, Matrix A, Matrix B)

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CMP 338

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  1. Data Structures and Algorithms I Day 2, 9/1/11 CMP 338

  2. Programming Homework 1 Create a matrix library with the following API Class: Matrix Matrix(int r, int c) Creates a matrix with r rows and c columns Individual entries are random doubles multAdd(Matrix C, Matrix A, Matrix B) Throw an exception if matrices are not conformant Otherwise, set C += A*B Note: Real library would have more methods e.g. getters and setters for individual matrix elements

  3. Matrices Implementation: 2-D array of double Multiply-add: for (int i=0; i<M; i++) for (int j=0; j<N; j++) for (int k=0; k<L, k++) c[i][j] += a[i][k]*b[k][j] Analysis: ~ MNL double multiply-adds ~ N3 for square matrices Data structures for sparse matrices

  4. Partition, take 1 int partition(double a[], double p, int lo, int hi) int i = lo; int j = hi while (true) while (a[i++] <= p) while (p < a[--j]) if (i<j) swap(a, i, j) else return j

  5. Partition, take 2 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo; int j = hi while (true) while (a[++i] <= p) while (p < a[--j]) if (i<j) swap(a, i, j) else return j

  6. Partition, take 3 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo-1; int j = hi+1 while (true) while (a[++i] <= p) while (p < a[--j]) if (i<j) swap(a, i, j) else return j

  7. Partition, take 4 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi < a.length int i = lo-1; int j = hi+1 while (true) {a[lo..i] <= p <a[j..hi]} while (a[++i] <= p) {a[lo..i-1]<=p<a[i]} while (p < a[--j]) {a[j]<=p<a[j+1..hi]} if (i<j) swap(a, i, j) else return j{a[lo..j] <= p < a[j+1..hi]}

  8. Partition, take 5 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo-1; int j = hi+1 while (true) {a[lo..i] <= p <a[j..hi]} while (a[++i] <= p) if (i == hi) break {a[lo..i-1]<=p<a[i]} while (p < a[--j]) if (j == lo) break {a[j]<=p<a[j+1..hi]} if (i<j) swap(a, i, j) else return j{a[lo..j] <= p < a[j+1..hi]}

  9. Partition, take 6 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo-1; int j = hi+1 while (true) {a[lo..i] <= p <a[j..hi]} while (a[++i] <= p) if (i == hi) break {a[lo..i-1]<=p<a[i] || i==hi && a[lo..hi]<p} while (p < a[--j]) if (j == lo) break {a[j]<=p<a[j+1..hi] || j==lo && p<a[lo..hi]} if (i<j) swap(a, i, j) else return j{a[lo..j] <= p < a[j+1..hi]}

  10. Partition, take 7 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo-1; int j = hi+1 while (true) {a[lo..i] <= p <a[j..hi]} while (a[i] <= p) if (i++ == hi) break {a[lo..i-1]<=p<a[i] || hi<i && a[lo..hi]<p} while (p < a[j]) if (lo == j++) break {a[j]<=p<a[j+1..hi] || j<lo && p<=a[lo..hi]} if (i<j) swap(a, i, j) else return j{a[lo..j] <= p < a[j+1..hi]}

  11. Partition, take 8 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo; int j = hi while (true) {a[lo..i] <= p <a[j..hi]} while (a[i] <= p) if (i++ == hi) break {a[lo..i-1]<=p<a[i] || hi<i && a[lo..hi]<p} while (p < a[j]) if (lo == j++) break {a[j]<=p<a[j+1..hi] || j<lo && p<=a[lo..hi]} if (i<j) swap(a, i, j) else return j{a[lo..j] <= p < a[j+1..hi]}

  12. Partition, take 9 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo; int j = hi while (true) {a[lo..i-1] <= p <a[j+1..hi]} while (a[i] <= p) if (i++ == hi) break {a[lo..i-1]<=p<a[i] || hi<i && a[lo..hi]<p} while (p < a[j]) if (lo == j++) break {a[j]<=p<a[j+1..hi] || j<lo && p<=a[lo..hi]} if (i<j) swap(a, i, j) else return j{a[lo..j] <= p < a[j+1..hi]}

  13. Partition, take 10 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo; int j = hi while (true) {a[lo..i-1] <= p <a[j+1..hi]} while (a[i] <= p) if (i++ == hi) break {a[lo..i-1]<=p<a[i] || hi<i && a[lo..hi]<p} while (p < a[j]) if (lo == j++) break {a[j]<=p<a[j+1..hi] || j<lo && p<=a[lo..hi]} if (i<j) swap(a, i++, j--) else return j{a[lo..j] <= p < a[j+1..hi]}

  14. Partition, take 10 int partition(double a[], double p, int lo, int hi) assert 0 <= lo <= hi+1 < a.length+1 int i = lo; int j = hi while (true) while (a[i] <= p) if (i++ == hi) break while (p < a[j]) if (lo == j++) break if (i<j) swap(a, i++, j--) else return j

  15. Partition, take 1 int partition(double a[], double p, int lo, int hi) int i = lo; int j = hi while (true) while (a[i++] <= p) while (p < a[j++]) if (i<j) swap(a, i, j) else return j

  16. Programming Homework 2 Implement the following API double pivot(double[] a, int lo, int hi) Assert that 0 <= lo <= hi+1 < ||a||+1 Return p where a[i]<=p<=a[j] for some lo<=j,i<=hi int partition(double[] a, double p, int lo, int hi) Assert that 0 <= lo <= hi+1 < ||a||+1 Return j, lo-1 <= j <= hi,such that a[lo..j] <= p < a[j+1..hi] double select(double[] a, int lo, int hi, int k) Assert that 0<= lo <= hi+1 <= ||a||+1 and 0 <= k <= hi-lo Return the smallest double d such that k elements of a[lo..hi] are less than or equal to d

  17. Selection Algorithm Item select(double[] a, int k, int lo, int hi) if (k == 0 && lo == hi) return a[lo] Item p = pivot(a, lo, hi) // a[lo+(lo+hi)/2] int j =partition(a, p, lo, hi) if (k <= j) return select(a, k, lo, j) else return select(a, k-(j+1), j+1, hi)

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