Divide and Conquer

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Divide-and-Conquer

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Divide and Conquer

  1. 1. Divide-and-ConquerRepublic of YemenTHAMAR UNIVERSITYFaculty of Computer Science&Information System1 Eng: Mohammed Hussein
  2. 2. Divide-and-Conquer2 Eng: Mohammed Hussein
  3. 3. Divide-and-Conquer The whole problem we want to solve may too big to understand or solve atonce. We break it up into smaller pieces, solve the pieces separately, andcombine the separate pieces together. Divide-and conquer is a general algorithm design paradigm:1. Divide: divide the input data S in two or more disjoint subsets S1,S2, …2. Recur: solve the subproblems recursively3. Conquer: combine the solutions for S1, S2, …, into a solution for S The base case for the recursion are subproblems of constant size Analysis can be done using recurrence equations3 Eng: Mohammed Hussein
  4. 4. Recurrence Equation Analysis The conquer step of merge-sort consists of merging two sorted sequences, each withn/2 elements and implemented by means of a doubly linked list, takes at most bn steps,for some constant b. Likewise, the basis case (n < 2) will take at b most steps. Therefore, if we let T(n) denote the running time of merge-sort: We can therefore analyze the running time of merge-sort by finding a closed formsolution to the above equation. That is, a solution that has T(n) only on the left-hand side.2if)2/(22if)(nbnnTnbnT4 Eng: Mohammed Hussein
  5. 5. Solving recurrences Recurrences are like solving integrals, differential equations, etc. Should learn a few tricks. Three ways of solving recurrences1. Guess-and-Test Method(If we know the answer)2. RecursionTree Method (Very useful !)3. MasterTheorem (Save our effort)5 Eng: Mohammed Hussein
  6. 6. Guess-and-Test Method In the guess-and-test method, we guess a closed form solution and then try to prove it istrue by induction: Guess that:T(n) < c n log n. Wrong: we cannot make this last line be less than cn log nncnnbncnncnnbnncnnbnnncnbnnTnTloglogloglog)2log(loglog))2/log()2/((2log)2/(2)(2iflog)2/(22if)(nnbnnTnbnT6 Eng: Mohammed HusseinLog(x/y)= log x – log yLog 2 =1‫بالتعويض‬
  7. 7. Guess-and-Test Method, Part 2 Recall the recurrence equation: Guess #2 that:T(n) < cn log2 n. if c > b. So,T(n) is O(n log2 n). In general, to use this method, you need to have a good guess and you need to be good atinduction proofs.ncnnbncnncnncnnbnnncnnbnncnnbnnncnbnnTnT222222logloglog2loglog))2(log2loglog2(loglog)2log(loglog))2/(log)2/((2log)2/(2)(2iflog)2/(22if)(nnbnnTnbnT7 Eng: Mohammed Hussein(x-y) 2= x2 -2xy +y2Log 2 =1nj <= nd if j <= dBig-Oh of highest degree
  8. 8. Recurrences In Mathematic a recurrence is a function that is defined interms of one or more base cases, and itself, with smallerarguments Examples:8 Eng: Mohammed Hussein
  9. 9. Master Method (tree) The master theorem concerns the recurrence a≥1 , b> 1 n is the size of the problem. a is the number of subproblems in the recursion. (‫)الفزوع‬ n/b is the size of each subproblem. (Here it is assumed thatall subproblems are essentially the same size.) f (n) is the cost of the work done outside the recursive calls)()/()( nfbnaTnT 9 Eng: Mohammed Husseinmerge-sort ‫خوارسمية‬‫االستدعاء‬ ‫عمليات‬ ‫عند‬‫الذاتي‬‫الدمج‬ ‫عملية‬ ‫تنفيذ‬ ‫مستويات‬ ‫مختلف‬ ‫في‬
  10. 10. The Tree Definition The root of this tree is node A. Definitions: Parent (A) Children (E, F) Siblings (C, D) Root (A) Leaf / Leaves K, L, F, G, M, I, J… The degree of a node is the number of sub-trees of the node. The level of a node: Initially letting the root be at level one the level of the node’s parent plus one. Level H=level(Parent (H)+1)); The height or depth of a tree is the maximum level of any node in the tree. height =Max(level(tree))10 Eng: Mohammed Hussein
  11. 11. Divide and conquer (tree) We analyze this in some generality: suppose we have a pieces, each of size n/b and merging takes time f(n). (In this example a=b=2 and f(n)=O(log n) but it will not always be true thata=b sometimes the pieces will overlap.) For simplicity, lets assume n is a power of b, and that the recursion stopswhen n is 1. Notice that the size of a node depends only on its level: size(i) = n/(bi). What is time taken by a node at level i? : time(i) = f(n/ bi) How many levels can we have before we get down to n=1? For bottom level, n/bi =1, so n=bi and i=(log n)/(log b). How many items at level i?: ai.11 Eng: Mohammed Hussein)()/()( nfbnaTnT time(i) = f(n/ bi)
  12. 12. The Recursion Tree Draw the recursion tree for the recurrence relation and look for a pattern: At each successive level of recursion the sub problems get halved in size (n/2). At the (log n) th level the sub problems get down to size 1, and so the recursion ends.Level/depthT’s size0 1 n1 2 n/2k 2k n/2k… … …2if)2/(22if)(nbnnTnbnTtimebnbnbn…Total time = bn + bn log n(last level plus all previous levels)2k =n  k= log n for each level.log nNotice that the size of a node depends only on its level12 Eng: Mohammed Husseintime=b x size = bn
  13. 13. Steps using recursion tree. Height = k=log n Levels = log n + 1 Cost/level = cn Bottom node number n = 2log n forheight log n dividing problem by 2each time. Recall that a balanced binary tree ofheight k has k + 1 levels. Recall: alog an = n so 2log 2n = 2log n =n Bottom cost = T(1) * n = cn Note that T(1) is problem size n=1,there are n subproblems at bottom.13 Eng: Mohammed Hussein 2k =n  k= log n for each level.
  14. 14. Merge Sort - Finding O Recall that a balanced binary tree of height k has k + 1 levels. A problem of size n=16 divided as described by the following recurrence equationwould then be represented by the tree. Use the implied coefficient c for arithmetic purposes:14 Eng: Mohammed Hussein16=24
  15. 15. Master Method Many divide-and-conquer recurrence equations have the form: The MasterTheorem:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb15 Eng: Mohammed Husseinvery small f(n)moderate f(n)very large f(n)
  16. 16. Analysis of Merge SortAnalysis of Merge Sort Divide:Trivial. Conquer:Recursively sort 2subarrays. Combine:Linear-time merge. Using Master theorem16 Eng: Mohammed Hussein
  17. 17. Analysis of Binary Search Recurrence for the worst case
  18. 18. Master Method, Example 1 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)()2/(4)( nOnTnT Solution: logba =log24=2, so case 1 says T(n) is Θ(n2).18 Eng: Mohammed Hussein
  19. 19. Master Method, Example 2 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)log()2/(2)( nnnTnT Solution: logba=log22=1, so case 2 says T(n) is Θ(n log2 n).19 Eng: Mohammed Hussein
  20. 20. Master Method, Example 3 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)log()3/()( nnnTnT Solution: logba=0, so case 3 says T(n) is Θ(n log n).20 Eng: Mohammed Hussein
  21. 21. Master Method, Example 4 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)()2/(8)( 2nnTnT Solution: logba=3, so case 1 says T(n) is Θ(n3).21 Eng: Mohammed Hussein
  22. 22. Master Method, Example 5 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)()3/(9)( 3nnTnT Solution: logba=2, so case 3 says T(n) is Θ(n3).22 Eng: Mohammed Hussein
  23. 23. Master Method, Example 6 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)1()2/()(  nTnTSolution: logba=0, so case 2 says T(n) is Θ(log n).(binary search)23 Eng: Mohammed Hussein
  24. 24. Master Method, Example 7 The form: The MasterTheorem: Example:dnnfbnaTdncnTif)()/(if)(.1somefor)()/(provided)),((is)(then),(is)(if3.)log(is)(then),log(is)(if2.)(is)(then),(is)(if1.log1loglogloglognfbnafnfnTnnfnnnTnnnfnnTnOnfakakaaabbbbb)(log)2/(2)( nnTnT Solution: logba=1, so case 1 says T(n) is Θ(n).(heap construction)24 Eng: Mohammed Hussein
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