Find an Element x in an Unsorted Array

Transcription

1 Find an Element x in an Unsorted Array What if we try to find a lower bound for the case where the array is not necessarily sorted? J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

2 Merging Two Sorted Arrays Let A[1..n] and B[1..n] be two sorted arrays. We know how to merge A and B into one single sorted array in O(n) time. We need such an algorithm during the merge step of merge sort. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

3 Merging Two Sorted Arrays Let A[1..n] and B[1..n] be two sorted arrays. We know how to merge A and B into one single sorted array in O(n) time. We need such an algorithm during the merge step of merge sort. Can we do better? J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

4 Merging Two Sorted Arrays Let A[1..n] and B[1..n] be two sorted arrays. We know how to merge A and B into one single sorted array in O(n) time. We need such an algorithm during the merge step of merge sort. Can we do better? No. Here is a proof of this fact. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

5 Merging Two Sorted Arrays Let A[1..n] and B[1..n] be two sorted arrays. We know how to merge A and B into one single sorted array in O(n) time. We need such an algorithm during the merge step of merge sort. Can we do better? No. Here is a proof of this fact. Assume that it is possible to merge two sorted arrays faster than in linear time. Then, the recursion equation for merge sort becomes wherre d < 1. T (n) = 2T (n/2) + O(n d ), J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

6 Merging Two Sorted Arrays Let A[1..n] and B[1..n] be two sorted arrays. We know how to merge A and B into one single sorted array in O(n) time. We need such an algorithm during the merge step of merge sort. Can we do better? No. Here is a proof of this fact. Assume that it is possible to merge two sorted arrays faster than in linear time. Then, the recursion equation for merge sort becomes T (n) = 2T (n/2) + O(n d ), wherre d < 1. From the Master Theorem, we get T (n) = O(n). This contradicts the lower bound of Ω(n log(n)) we established for the sorting problem. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

7 Merging Two Sorted Arrays Here is a proof using decision trees. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

8 Merging Two Sorted Arrays Here is a proof using decision trees. First, notice that it is possible to solve this problems using only comparisons (<,, =, > or ). Hence, we can solve this problem with decision trees. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

9 Merging Two Sorted Arrays Here is a proof using decision trees. First, notice that it is possible to solve this problems using only comparisons (<,, =, > or ). Hence, we can solve this problem with decision trees. Given two arrays A[1..n] and B[1..n], how many different outputs are there? J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

10 For instance, if n = 2, we have A[1] < A[2] and B[1] < B[2], from which the possible outputs are as follows. A[1], A[2], B[1], B[2] A[1], B[1], A[2], B[2] A[1], B[1], B[2], A[2] B[1], A[1], A[2], B[2] B[1], A[1], B[2], A[2] B[1], B[2], A[1], A[2] J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

11 For a general n, how many different outputs are there? J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

12 For a general n, how many different outputs are there? We have to fill the 2n positions of the output and make sure that the order in A is satisfied: A[1] < A[2] <... < A[n] as well as the order in B: B[1] < B[2] <... < B[n]. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

13 For a general n, how many different outputs are there? We have to fill the 2n positions of the output and make sure that the order in A is satisfied: A[1] < A[2] <... < A[n] as well as the order in B: B[1] < B[2] <... < B[n]. One way to achieve this is to choose the n positions that will contain the elements of A and then fill the n remaining positions with the elements of B. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

14 For a general n, how many different outputs are there? We have to fill the 2n positions of the output and make sure that the order in A is satisfied: A[1] < A[2] <... < A[n] as well as the order in B: B[1] < B[2] <... < B[n]. One way to achieve this is to choose the n positions that will contain the elements of A and then fill the n remaining positions with the elements of B. How many ways are there to choose n positions among 2n positions? J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

15 For a general n, how many different outputs are there? We have to fill the 2n positions of the output and make sure that the order in A is satisfied: A[1] < A[2] <... < A[n] as well as the order in B: B[1] < B[2] <... < B[n]. One way to achieve this is to choose the n positions that will contain the elements of A and then fill the n remaining positions with the elements of B. How many ways are there to choose n positions among 2n positions? There are ( ) 2n n ways. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

16 For a general n, how many different outputs are there? We have to fill the 2n positions of the output and make sure that the order in A is satisfied: A[1] < A[2] <... < A[n] as well as the order in B: B[1] < B[2] <... < B[n]. One way to achieve this is to choose the n positions that will contain the elements of A and then fill the n remaining positions with the elements of B. How many ways are there to choose n positions among 2n positions? There are ( ) 2n n ways. Therefore, there are at least ( ) 2n n ( leaves in the decision tree. Therefore, the height of the tree is (2n ) ) at least log n. J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

17 From the Stirling formula, we have n! ( n ) n 2πn. e J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

18 From the Stirling formula, we have n! ( n ) n 2πn. e Hence, ( ) 2n = (2n)! n (n!) 2 ( 2π(2n) 2n ) 2n e ( ( 2πn n ) n ) 2 Stirling e = 4πn ( 2n = 4n πn 2πn ( n e ) 2n e ) 2n J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

19 from which we get (( )) 2n log n ( ( )) 4 n = Ω log πn = Ω ( n log(4) log ( π ) log(n) ) = Ω(n). J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

20 from which we get (( )) 2n log n ( ( )) 4 n = Ω log πn = Ω ( n log(4) log ( π ) log(n) ) = Ω(n). Notice that an adversarial argument would me much simpler. Do you see how to proceed? J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17

21 Conclusion J.-L. De Carufel (U. of O.) Design & Analysis of Algorithms Fall / 17