assignment1_sol - else L=1; R=m; while (L<=R)...

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Unformatted text preview: else L=1; R=m; while (L<=R) i=floor((L+R)/2); j=k-i; if (j=n or A[i]<=B[j+1]) and (i=m or B[j]<=A[i+1]) then return max(A[i],B[j]); /* A[1..i], B[1..j] contains the smallest k items */ else if A[i]>B[j+1] then R=i-1; /* rank(A[i]) > k, pick a smaller i */ else if B[j]>A[i+1] then L=i+1; /* rank(B[j]) > k, pick a smaller j and a bigger i */ end if end while end if end 2. Let X be a list of n objects X 1 ,X 2 , ··· ,X n . Construct a list Y of at most n 2 objects such that X 2 i − 1 ∈ Y if and only if X 2 i − 1 = X 2 i for i = 1 , 2 , ··· , ⌊ n 2 ⌋ . Note that Y does not cover X n when n is odd. Suppose more than half of the the objects in X are equal to x . (a) When n is even, the number of x in X is bigger than n 2 = ⌊ n 2 ⌋ . When n is odd and x negationslash = X n , the number of x in X is bigger than n − 1 2 = ⌊ n 2 ⌋ In both cases, there must exists i such that X 2 i − 1 = X 2 i = x . So x ∈ Y . (b) Consider any object y in Y such that the number of y in Y is ≤ | Y | 2 . Since for each pair of objects X 2 i − 1 ,X 2 i such that X 2 i − 1 negationslash = X 2 i , at most one of them can be equal...
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assignment1_sol - else L=1; R=m; while (L<=R)...

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