1 / 86

Algorithm & Application

Algorithm & Application. Algorithm : A step-by-step procedure for solving a problem Prof. Hyunchul Shin shin@hanyang.ac.kr Hanyang University. Foundations of Algorithms. Richard Neapolitan and Kumarss Naimipour

annick
Download Presentation

Algorithm & Application

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. Algorithm & Application Algorithm : A step-by-step procedure for solving a problem Prof. Hyunchul Shin shin@hanyang.ac.kr Hanyang University

  2. Foundations of Algorithms • Richard Neapolitan and KumarssNaimipour • 3rd Edition. Jones and Bartlett Computer Science, 2004 • Time : CPU cycles • Storage: memory • Instance: Each specific assignment of values to parameters

  3. Problem Is the number x in the list S of n numbers? The answer is yes if x is in S and no if it is not. (ex) S={10,7,11,5,13,8} , n=6 , and x=5 . Solution “yes” Algorithm : search ( S, n, x ) { for ( i=1; i<=n; i++ ) if S[i]==x, return ( “yes” ); return ( “no” ); } /* cf. text P5 */

  4. Exchange Sort Problem : Sort n keys in nondecreasing order Inputs : n, S[1],…,S[n] Outputs : Sorted keys in the array S. Algorithm: Exchange Sort { for( i=1; i<=n; i++ ) for( j=i+1; j<=n; j++ ) if( S[j] < S[i]) exchange S[i] and S[j] }

  5. Algorithm Exchange Sort Algorithm: Exchange Sort (ex) n=4 S=[ 4 3 1 5] { for( i=1; i<=n; i++ ) for( j=i+1; j<=n; j++ ) if( S[j] < S[i]) exchange S[i] and S[j] } Homework Show i , j , S , for exchange sort of S=[ 3 8 5 9 7]. Due 1 week 3 4 1 3 1 5 3 4 3 5 4 5

  6. Matrix Multiplication Cn×n=An×n. Bn×n Cij= aik. Bkj , for i<=n, j <=n. (ex) = Algorithm { /*Matrix multiplication*/ for( i=1; i<=n; i++ ) for( j=1; j<=n; j++ ) { C[i][j]=0; for(k=1;k<=n; k++) C[i][j]= C[i][j] + A[i][k] ×B[k][j]; } }

  7. Fibonacci Sequence f0=0 f1=1 fn= fn-1 + fn-2 for n>=2. (ex) f2 =f1 + f0 =1 + 0=1 f3 =f2 + f1 =1 + 1=2 f4 =f3 + f2 =2 + 1=3 f5 =f4 + f3 =3 + 2=5 …

  8. Fibonacci (Recursive) int fib (int n) { /*divide-and-conquer : chap2 */ if(n<=1) return n; else return( fib(n-1) + fib(n-2) ); } (ex) fib(5) computation

  9. Fibonacci (Iterative) Intfib_iter (int n){/*dynamic programing:chap3*/ Index i; int f[0..n]; f[0]=0; If(n>0){ f[1]=1; for( i=2; i<=n; i++ ) f[i]=f[i-1]+f[i-2]; } Return f[n]; } Complexity (cf text p16) Fib(100) takes 13 days. Fib_iter(100) takes 101 n sec

  10. Complexity: Exchange Sort Algorithm: Exchange Sort { for( i=1; i<=n; i++ ) for( j=i+1; j<=n; j++ ) if( S[j] < S[i]) exchange S[i] and S[j] } Basic operation: Comparison of S[j] with S[i] Input size: n, the number of items to be sorted. Complexity: the number of basic operations T(n)=(n-1)+(n-2)+(n-3)+…+1 =(n-1).n/2 ЄO(n2)

  11. Complexity: Matrix Multiplication Algorithm { /*Matrix multiplication*/ for( i=1; i<=n; i++ ) for( j=1; j<=n; j++ ) { C[i][j]=0; for(k=1;k<=n; k++) C[i][j]= C[i][j] + A[i][k] ×B[k][j]; } } Basic operation: multiplication (innermost for loop) Input size: n, #rows and #columns Complexity: T(n)=n×n×n =n3 ЄO(n3)

  12. Memory Complexity Analysis of algorithm efficiency in terms of memory. Time complexity is usually used. Memory complexity is occasionally useful.

  13. Order : Big O • Definition For a given complexity function f(n), O(f(n)) is the set of complexity functions g(n) for which there exists some positive real constant c and some nonnegative integer N such that for all n ≤ N, . (ex) T1(n)=(n-1).n/2 Є O(n2) T2(n)=n3 Є O(n3) T3(n)=10000++1000 ЄO(n2) (cf. p29)

  14. Divide and Conquer Top-Down Approach (p47) Divide the problem into subproblems Conquer subproblems Obtain the solution from the solutions of subproblems Binary search Problem: Is x in the sorted array S of size n ? Inputs: Sorted array S, a key x. Outputs: Location of x in S (0 if x is not in S)

  15. Binary Search Locationout=location(1,n); Index location (index low, index high) { index mid; if(low>high) return 0 ; else{ mid= ; if (x==S[mid]) return mid; else if (x<S[mid]) return location (low,mid-1); else return location (mid+1,high); } }

  16. Worst-Case Complexity: Binary Search Locationout=location(1,n); Index location (index low, index high){ index mid; if(low>high) return 0 ; else{ mid= (low+high)/2; if (x==S[mid]) return mid; else if (x<S[mid]) return location (low,mid-1); else return location (mid+1,high); } } Basic operation: Comparison of x with S[mid] Input size: n (#items in the array S) W(n)=W(n/2) + 1 recursive top call level

  17. Complexity: Binary Search W(n)=W(n/2)+1 , for n>1 , n a power of 2 W(1)=1 It appears that W(n) = log n + 1 (Induction base) For n=1, t1=1=log1+1 (Induction hypothesis) Assume that W(n)=logn + 1 (Induction step) L=W(2n)=log(2n)+1 R=W(2n)=W(n)+1=(logn + 1)+1 =logn + log2 + 1=log(2n) + 1

  18. Merge Sort (O(nlogn))

  19. Quick Sort Sort by dividing the array into two partitions Using a pivot item. (ex)(first item) Quick sort(index low , index high) {index pivot;/*index of the pivot*/ if(high>low){ partition (low, high, pivot ); quicksort (low, pivot-1); quicksort (pivot+1, high); } }

  20. Homework Given 20 15 25 22 11 20 30 27 (n=8) • Mergesort as in Fig 2.2 P54 • Quicksort as in Fig 2.3 P61 • Partition as in Table 2.2 P62 Due 1 week

  21. Worst-case complexity: Quick sort • Worst case: When the array is already sorted • Time to partition: Tp (x) = n – 1 • Time to sort left subarray = T(0) • Time to sort right subarray = T(n-1) • Quick Sort T(n) = T(0) + T(n-1) + (n-1) for n > 0 T(n) = T(n-1) + (n-1), since T(0) = 0 T(n) = n(n-1)/2 ϵ O() • Average-case complexity: O (nlogn)

  22. Dynamic Programming (Bottom-up) • Dynamic programming • Establish a recursive property • Solve in bottom-up fashion by solving smaller instances first • (ex): Fibonacci (Iterative) • Divide-and-conquer • Divide a problem into smaller instances • Solve these smaller instances (blindly) • Examples: • Fibonacci (Recursive): Instances are related • Merge sort: Instances are unrelated

  23. Binomial coefficient • Frequently, n! is too large to compute directly • Proof:

  24. Binomial coefficients: Divide-and-conquer • Algorithm /* Inefficient */ intbin (intn, intk) { if ( k = = 0 || n = = k) return 1; else returnbin (n-1, k - 1)+bin (n - 1, k); }

  25. Binomial coefficients Figure 3.1: The array B used to compute the binomial coefficient Complexity : O(nK)

  26. Example 3.1: Compute Compute row 0: {This is done only to mimic the algorithm exactly.}   {The value B [0] [0] is not needed in a later computation.} B [0] [0] = 1 Compute row 1: B [1] [0] = 1B [1] [1] = 1 Compute row 2: B [2] [0] = 1B [2] [1] = B [1] [0] + B [1] [1] = 1+1 = 2B [2] [2] = 1 Compute row 3: B [3] [0] = 1B [3] [1] = B [2] [0] + B [2] [1] = 1+2 = 3B [3] [2] = B [2] [1] + B [2] [2] = 2+1 = 3 Compute row 4: B [4] [0] = 1B [4] [1] = B [3] [0] + B [3] [1] = 1+3 = 4B [4] [2] = B [3] [1] + B [3] [2] = 3+3 = 6

  27. Binomial Coefficient: Dynamic Programming • Establish a recursive property • Solve in bottom up fashion Algorithm: intbin2 (intn, intk){  index i, j;intB[0..n][0..k];  for (i = 0; i < = n; i ++)    for (j = 0; j < = minimum(i, k);  j ++)      if (j == 0 || j == i)B[i][j] = 1;      elseB[i][j] = B[i - 1][j - 1] + B[i - 1][j];  return B[n][k];}

  28. HOMEWORK • Use dynamic programming approach to compute B[5][3]. • Draw diagram like figure 3.1 (Page 94) • Due in 1 week

  29. Binary Search Tree • Definition: For a given node n, • Each node contains one key • Key (node in the left subtree of n) <= Key (n) • Key(n) <= Key(node in the right subtree of n) • Optimality depends on the probability

  30. Binary Search Tree • Depth(n): # edges in the unique path from the root to n. • (Depth=level) • Search time = depth(key) + 1 • The root has a depth of 0.

  31. Binary Search Algorithm structnodetype{ Key type key; Nodetype* left; Nodetype* right; }; typeofnodetype* node_pointer; Void search (node_pointer tree, keytypekeyin, node_pointer & p) { {bool found=false; p=tree; while(!found) if (p->key==keyin) found=true; elseif(keyin < p->key) p=p->left; else p=p->right; }

  32. Example

  33. Greedy Approach Start with an empty set and add items to the set until the set represents a solution. Each iteration consists of the following components: A selection procedure A feasibility check A solution check

  34. Spanning Tree A connected subgraph that contains all the vertices and is a tree.

  35. Graph G=(V , E) Where V is a finite set of vertices and E is a set of edges (pairs of vertices in V). (ex) V={v1, v2, v3, v4, v5} E={(v1, v2,), (v1, v3,), (v2, v3,), (v2, v4,), (v3, v4,), (v3, v5,), (v4, v5,), }

  36. Weight of a Graph

  37. Prim’s Algorithm Figure 4.4: A weighted graph (in upper-left corner) and the steps in Prim's algorithm for that graph. The vertices in Y and the edges if F are shaded at each step.

  38. Prim’s Algorithm F = Ø; for (i = 2; i <= n; i++){ // Initialize nearest [i] = 1; // v1 is the nearest distance [i] = W[1] [i] ; // distance is the weight } repeat (n - 1 times){ // Add all n - 1 vertices min = ∞ for (i = 2; i <= n; i++) if (0 ≤ distance [i] < min) { min = distance [i]; vnear = i; } e = edge connecting vnear and nearest [vnear]; F=F ∪{e} //add e to F distance [vnear] = - 1; for (i = 2; i <= n; i++) //update distance if (W[i] [vnear] < distance [i]){ distance = W[i] [vnear]; nearest [i] = vnear; } }

  39. Prim’s Algorithm

  40. Prim’s Spanning Tree Complexity : O(n2) (n-1) iterations of the repeat loop (n-1) iterations in two for loops T(n)= 2(n-1) (n-1) Theorem Prim’s algorithm always produces a minimum Spanning tree.

  41. Dijkstra’s Shortest Paths Figure 4.8: A weighted, directed graph (in upper-left corner) and the steps in Dijkstra's algorithm for that graph. The vertices in Y and the edges in F are shaded in color at each step.

  42. Dijkstra’s Algorithm F = Ø; for (i = 2; i<= n; i++){ //Initialize touch [i] = 1; // paths from V1 length [i] = W[1] [i]; } repeat (n - 1 times){ min = ∞; for (i = 2; i < = n; i++) if ( 0 ≤ length [i] < min) { min = length [i]; vnear = i; } e = edge from from[vnear]to vnear; F=F ∪{e} //add e to F for (i = 2; i < = n; i++) if (length [vnear] + W[vnear] [i] < length [i]){ length[i] = length[vnear] + W[vnear][i]; touch[i] = vnear; } length[vnear] = -1; }

  43. Complexity Prim’s and Dijkstra’s : O (n2) Heap implementation : O (mlogn) Fibonacci heap implementation : O (m + nlogn) 1.Find a minimum spanning tree for the following graph 2.Find the shortest paths from V4 to all the other vertices

  44. Scheduling Minimizing the total time (waiting + service) (ex) Three jobs : t1=5, t2=10, t3=4. Schedule Total Time in the System [1, 2, 3]5+(5+10)+(5+10+4) = 39 [1, 3, 2]5+(5+4)+(5+4+10) = 33 [2, 1, 3]10+(10+5)+(10+5+4) = 44 . . . [3, 1, 2]4+(4+5)+(4+5+10) = 32 3! cases

  45. Optimal Scheduling for Total Time Smallest service time first. // Sort the jobs in nondecreasing order of service time // Schedule in sorted order. Complexity (sorting) w(n)Є O ( nlogn )

  46. Schedule with Deadlines Schedule to maximize the total profit. Each job takes one unit of time to finish. (ex) JobDeadlineProfit 1 2 30 2 1 35 3 2 25 4 1 40 [1,3] : TP=30+25=55 [2,1] : TP=35+30=65 … [4,1] : TP=40+30=70.(optimal) … Is highest profit first optimal?

More Related