Objec&ves. Review. Divide and conquer algorithms
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1 Objec&ves Divide and conquer algorithms Ø Recurrence rela&ons Ø Coun&ng inversions March 9, 2018 CSCI211 - Sprenkle 1 Review What approach are we learning to solve problems (as of Wednesday)? What is a recurrence rela&on? How can you compute D&C running &mes? March 9, 2018 CSCI211 - Sprenkle 2 1
2 Review: Unrolling Recurrence How many levels are there (assuming n is a power of 2)? How much does each level cost, in terms of the level? What is the total run &me? Number of levels: log 2 n 2 k problems Size: n/2 k c n/2 c n/4 c n/4 cn c n/4 0 c n/2 1 c n/4 2 T(n) Each level takes 2 k * c * (n/2 k ) = cn T(n / 2 k ) O(n log n) c c c c c c c c T(2) March 9, 2018 CSCI211 - Sprenkle 3 Review: Approaches to Solving Recurrences 1. Unroll recursion Ø Look for paverns in run&me at each level Ø Sum up running &mes over all levels 2. Subs&tute guess solu&on into recurrence Ø Check that it works Ø Induc&on on n March 9, 2018 CSCI211 - Sprenkle 4 2
3 Alterna&ve: Proof by Induc&on Claim. If T(n) sa&sfies the recurrence 2 T(n/2) + cn, then T(n) = n log 2 n. Pf. (by induc&on on n) Ø Base case: n = 2 Ø Induc&ve hypothesis: T(n) cn log 2 n Ø Goal: show that T(2n) = 2cn log 2 (2n) Subs&tute 2n in for n into the hypothesis Why doubling n? March 9, 2018 CSCI211 - Sprenkle 5 Proof by Induc&on Claim. If T(n) sa&sfies the recurrence T(n) = 2 T(n/2) + cn, then T(n) = n log 2 n. Pf. (by induc&on on n) Ø Induc&ve hypothesis: T(n) cn log 2 n Ø Goal: show that T(2n) = 2cn log 2 (2n) T(2n) = 2T(n) + c2n Recurrence rela&on = 2cn log 2 n + 2cn Replace T(n) w/ induc&on hypothesis = 2cn (log 2 (2n)-1) + 2cn Log rules: what is the = 2cn log 2 (2n) - 2cn + 2cn difference between = 2cn log 2 (2n) log 2 (2n) and log 2 (n)? March 9, 2018 CSCI211 - Sprenkle 6 3
4 Another Recurrence Rela&on: Binary Search How does binary search work? What is its recurrence rela&on? March 9, 2018 CSCI211 - Sprenkle 7 Analyzing Binary Search BinarySearch( L[1 n], key ): if len(l) == 1 and L[1] == key: return 1 # return the index else: return NOT_FOUND mid = n/2 if L[mid] == key: return mid # return the index if L[mid] < key: return BinarySearch(L[mid+1:], key) else: return BinarySearch(L[:mid], key) What is the recurrence relation? March 9, 2018 CSCI211 - Sprenkle 8 4
5 Analyzing Binary Search BinarySearch( L[1 n], key ): if len(l) == 1 and L[1] == key: return 1 # return the index else: return NOT_FOUND mid = n/2 if L[mid] == key: return mid # return the index if L[mid] < key: return BinarySearch(L[mid+1:], key) else: return BinarySearch(L[:mid], key) What is the recurrence relation? T(n) = T(n/2) + c March 9, 2018 CSCI211 - Sprenkle 9 Unroll the Recurrence T(n) = T(n/2) + c Which makes the run&me? March 9, 2018 CSCI211 - Sprenkle 10 5
6 Unroll the Recurrence T(n) = T(n/2) + c Ø Constant work at each level Ø Number of levels: log n Which makes the run&me? O(log n) March 9, 2018 CSCI211 - Sprenkle 11 Another Recurrence Rela&on Instead of recursively solving 2 problems, solve q problems Ø Size of problems is s&ll n/2 Combining solu&ons is s&ll O(n) Example: q=3: n n/2 n/2 n/2 What is the recurrence relation? March 9, 2018 CSCI211 - Sprenkle 12 6
7 Another Recurrence Rela&on Instead of recursively solving 2 problems, solve q problems Ø Size of problems is s&ll n/2 Combining solu&ons is s&ll O(n) Recurrence rela&on: Ø For some constant c, T(n) q T(n/2) + cn when n > 2 T(2) c Intuition about running time? March 9, 2018 CSCI211 - Sprenkle 13 Unrolling Recurrence, q > 2 T(n) q T(n/2) + cn March 9, 2018 CSCI211 - Sprenkle 14 7
8 Unrolling Recurrence, q > 2 First level: q T(n/2) + cn T(n/2) cn q T(n/2) March 9, 2018 CSCI211 - Sprenkle 15 Unrolling Recurrence, q > 2 Next level: q T(n/4) + c(n/2) c n/2 cn q c n/2 T(n/4) q T(n/4) T(n/4) q T(n/4) March 9, 2018 CSCI211 - Sprenkle 16 8
9 Unrolling Recurrence, q > 2 How much does each level cost, in terms of the level? Number of levels? What is the total run time? c n/2 cn q 0 c n/2 1 q k problems at level k Size: n/2 k Number of levels: log 2 n T(n/4) q T(n/4) T(n/4) q T(n/4) Each level takes q k * c * (n/2 k ) = (q/2) j cn à Total work per level is increasing as level increases March 9, 2018 CSCI211 - Sprenkle 17 Unrolling Recurrence, q > 2 How much does each level cost, in terms of the level? Number of levels? What is the total run time? c n/2 cn q 0 c n/2 1 T(n) Σ j=0,logn (q/2) j cn T(n/4) q T(n/4) T(n/4) q T(n/4) Geometric series: (constant ratio between successive terms) Multiplying previous term by (q/2) O(n log2 q ) March 9, 2018 CSCI211 - Sprenkle 18 9
10 Unrolling the Recurrence Generalize: What are the steps? March 9, 2018 CSCI211 - Sprenkle 19 Summary Use recurrences to analyze the run&me of divide and conquer algorithms Need to figure out Ø Number of sub problems Ø Size of sub problems Ø Number of &mes divided (number of levels) Ø Cost of merging problems March 9, 2018 CSCI211 - Sprenkle 20 10
11 Know Your Recurrence Rela&ons What algorithm has this recurrence relation? What is that algorithm s running time? Recurrence Algorithm Running Time T(n) = T(n/2) + O(1) T(n) = T(n-1) + O(1) T(n) = 2 T(n/2) + O(1) T(n) = T(n-1) + O(n) T(n) = 2 T(n/2) + O(n) March 9, 2018 CSCI211 - Sprenkle 21 Know Your Recurrence Rela&ons What algorithm has this recurrence relation? What is that algorithm s running time? Recurrence Algorithm Running Time T(n) = T(n/2) + O(1) Binary Search O(log n) T(n) = T(n-1) + O(1) T(n) = 2 T(n/2) + O(1) Sequential/ Linear Search Binary Tree Traversal O(n) O(n) T(n) = T(n-1) + O(n) Selection Sort O(n 2 ) T(n) = 2 T(n/2) + O(n) Merge Sort O(n log n) March 9, 2018 CSCI211 - Sprenkle 22 11
12 COUNTING INVERSIONS March 9, 2018 CSCI211 - Sprenkle 23 Comparing Rankings To determine similarity of rankings, need a metric Similarity metric: number of inversions between two rankings Ø My rank: 1, 2,, n Ø Your rank: a 1, a 2,, a n Ø Movies i and j inverted if i < j but a i > a j Me You Movies A B C D E Inversions: 3-2, 4-2 March 9, 2018 CSCI211 - Sprenkle 24 12
13 Color Comparison Test March 9, 2018 CSCI211 - Sprenkle 25 Comparing Rankings To determine similarity of rankings, need a metric Similarity metric: number of inversions between two rankings Ø My rank: 1, 2,, n Ø Your rank: a 1, a 2,, a n Ø Movies i and j inverted if i < j but a i > a j Me You Movies A B C D E Naïve/Brute force solution? Inversions: 3-2, 4-2 March 9, 2018 CSCI211 - Sprenkle 26 13
14 Brute Force Solu&on Look at every pair (i,j) and determine if they are an inversion Requires Θ(n 2 ) &me Ø Note: Already an efficient algorithm but try to improve upon run&me Towards a Better Solution Can t look at each inversion individually March 9, 2018 CSCI211 - Sprenkle 27 Coun&ng Inversions: Divide-and-Conquer Assume number represents where item should be in the list, i.e., where it is in someone else s list Should be at position 5 March 9, 2018 CSCI211 - Sprenkle 28 14
15 Coun&ng Inversions: Divide-and-Conquer Divide: separate list into two pieces Divide: O(1) March 9, 2018 CSCI211 - Sprenkle 29 Coun&ng Inversions: Divide-and-Conquer Divide: separate list into two pieces Conquer: recursively count inversions in each half 5 blue-blue inversions 8 green-green inversions 5-4, 5-2, 4-2, 8-2, , 9-3, 9-7, 12-3, 12-7, 12-11, 11-3, 11-7 Divide: O(1) What is the recurrence relation so far? March 9, 2018 CSCI211 - Sprenkle 30 15
16 Coun&ng Inversions: Divide-and-Conquer Divide: separate list into two pieces Conquer: recursively count inversions in each half Divide: O(1) 5 blue-blue inversions 8 green-green inversions 5-4, 5-2, 4-2, 8-2, , 9-3, 9-7, 12-3, 12-7, 12-11, 11-3, 11-7 Conquer: 2T(n / 2) What do we need to do next? March 9, 2018 CSCI211 - Sprenkle 31 Coun&ng Inversions: Divide-and-Conquer Divide: separate list into two pieces Conquer: recursively count inversions in each half Combine: count inversions where a i and a j are in different halves, and return sum of three quan&&es Divide: O(1) 5 blue-blue inversions 8 green-green inversions 9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, 10-6, 10-9, 10-3, 10-7 Conquer: 2T(n / 2) Combine cost:??? Total = = 22 March 9, 2018 CSCI211 - Sprenkle 32 16
17 Coun&ng Inversions: Divide-and-Conquer Divide: separate list into two pieces Conquer: recursively count inversions in each half Combine: count inversions where a i and a j are in different halves, and return sum of three quan&&es Divide: O(1) 5 blue-blue inversions 8 green-green inversions 9 blue-green inversions 5-3, 4-3, 8-6, 8-3, 8-7, 10-6, 10-9, 10-3, 10-7 Total = = 22 Conquer: 2T(n / 2) Combine: Θ(n 2 ) What would make figuring out blue-green inversions easier? March 9, 2018 CSCI211 - Sprenkle 33 Looking Ahead Wiki: 4.8, Problem Set 7 Ø One of the harder problem sets March 9, 2018 CSCI211 - Sprenkle 34 17
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