1 / 15

CS100J Lecture 15

CS100J Lecture 15. Previous Lecture Sorting Loop invariants more Rules of Thumb Reading Lewis & Loftus Section 6.3 Savitch Section 6.4 This Lecture Programming concepts Binary search Application of the “rules of thumb” Asymptotic complexity Java Constructs Conditional Expressions.

ginette
Download Presentation

CS100J Lecture 15

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS100J Lecture 15 • Previous Lecture • Sorting • Loop invariants • more Rules of Thumb • Reading • Lewis & Loftus Section 6.3 • Savitch Section 6.4 • This Lecture • Programming concepts • Binary search • Application of the “rules of thumb” • Asymptotic complexity • Java Constructs • Conditional Expressions Lecture 15

  2. Search in a Sorted Array • Problem. Find a given value in a sorted array, or say it doesn’t occur in the array. • Rule of Thumb. Write a precise specification. /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { . . . } • Rule of Thumb. Find inspiration from experience, e.g., looking up a name in a telephone directory. Lecture 15

  3. 0 1 2 3 4 5=N A 10 12 13 18 19 20 0 1 2 3 4 5=N A 10 12 13 18 19 20 0 1 2 3 4 5=N A 10 12 13 18 19 20 0 1 2 3 4 5=N A 10 12 13 18 19 20 Example • Rule of Thumb. Work sample data by hand. Be introspective. Ask yourself: “What am I doing?” val 19 Lecture 15

  4. Loop Pattern • Rule of Thumb. If you smell an iteration, write it down. • Decide between definite iteration and indefinite iteration • Write down an appropriate pattern for the iteration. • Do not fill in the pattern yet. /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { . . . while ( _______________________) { . . . } . . . return ___________________________; } Lecture 15

  5. 0 L R N A Loop Invariant • Rule of Thumb. • Characterize the state after an arbitrary number of iterations, either in English of in a diagram. • Introduce a variable to record the subscript of each boundary expected to change independently during the iteration. val If val occurs in A[0..N], then val occurs in the shaded region A[L..R]. Lecture 15

  6. 0 L R N A 0 R N A Initial and Final Conditions • Rule of Thumb, continued • Characterize the initial and final states, i.e., the state before the iteration begins and after the iteration is expected to stop. • The characterization of the initial and final states should be special cases of the general characterization. Initial: 0=L N=R A Intermediate: Final: L If val occurs in A[0..N], then val occurs in the shaded region A[L..R]. Lecture 15

  7. Initialization and Termination • Rule of Thumb. • Use the characterization to refine the initialization, termination condition, and increment of the loop. /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { int L = _____________; int R = _____________; while ( _______________________) { . . . } . . . return ___________________________; } Lecture 15

  8. Specify the Body • Rule of Thumb. • Use the characterization to specify what the loop body must accomplish. /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { int L = 0; int R = N; /* Make L==R s.t. if val is in A[0..N], then A[L]== val. */ while ( L != R ) { /* Reduce the size of the interval A[L..R] approximately in half, preserving the property that if val was in the original interval, then it is in the new interval. */ . . . } . . . return ___________________________; } Lecture 15

  9. L M R 11 12 13 14 A L R 11 12 13 14 A L R 11 12 13 14 A Refine the Body: Even length case • Midpoint M = (L+R) / 2; • If the interval has even length: • then select either • or Lecture 15

  10. L M R 11 12 13 14 15 A L R 11 12 13 14 15 A L R 11 12 13 14 15 A Refine the Body: Odd length case • Midpoint M = (L+R) / 2; • If the interval has odd length: • then select either • or Lecture 15

  11. Refine the Body . . . /* Make L==R s.t. if val is in A[0..N], then A[L]== val. */ while ( L != R ) { /* Reduce the size of the interval A[L..R] approximately in half, preserving the property that if val was in the original interval, then it is in the new interval. */ int M = (L + R) / 2; if (____________________________) R = __________________________; else L = __________________________; } . . . • Rule of Thumb. Avoid different treatments of cases if you can treat all cases uniformly. Lecture 15

  12. Conditional Expressions • Conditional Expression expression0 ? expression1 : expression2 Meaning: The value of the conditional expression is expression1ifexpression0istrue, and isexpression2otherwise. /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { int L = 0; int R = N; /* Make L==R s.t. if val is in A[0..N], then A[L]== val. */ . . . return ____________________________________; } Lecture 15

  13. Final Program /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { int L = 0; int R = N; /* Make L==R s.t. if val is in A[0..N], then A[L]== val. */ while ( L != R ) { /* Reduce the size of the interval A[L..R] approximately in half, preserving the property that if val was in the original interval, then it is in the new interval. */ int M = (L + R) / 2; if ( A[M] >= val ) R = M; else L = M+1; } return (A[L]==val) ? L : N+1; } Lecture 15

  14. Asymptotic Complexity • How many iterations does binary search take? N+1 1 2 4 8 16 32 ... #iterations 0 1 2 3 4 5 ... log2(N+1) • In contrast, how many iterations does sequential search take in the worst case? /* Given A[0..N] sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or N+1 otherwise. */ int find(int[] A, int N, int val) { int k = 0; while (k <= N && val != A[k]) k++; return k; } N+1 1 2 4 8 16 32 ... #iterations 1 2 4 8 16 32 ... N+1 Lecture 15

  15. Better Final Program /* Given A sorted in non-decreasing order, return the subscript of an occurrence of val in A (if val occurs in A) or A.length otherwise. */ int find(int[] A, int val) { int N = A.length - 1; int L = 0; int R = N; /* Make L==R s.t. if val is in A[0..N], then A[L]== val. */ while ( L != R ) { /* Reduce the size of the interval A[L..R] approximately in half, preserving the property that if val was in the original interval, then it is in the new interval. */ int M = (L + R) / 2; if ( A[M] >= val ) R = M; else L = M+1; } return (A[L]==val) ? L : N+1; } Lecture 15

More Related