1. Binhai Zhu Computer Science Department, Montana State University
Linear Time Selection
Binhai Zhu
Computer Science Department, Montana State University
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Consider the amount of time available and prepare to organize your material. Narrow your topic. Divide your presentation into clear segments. Follow a logical progression. Maintain your focus throughout. Close the presentation with a summary, repetition of the key steps, or a logical conclusion.
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2. Definition
Given an unsorted array A[1..n], return the i-th smallest (largest) element of A
(1) How about the smallest and largest elements?
(2) With sorting, it is easy (but costs too much).
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3. Think of Quicksort Quicksort(A,p,r) (1) if p < r
(2) then q ← partition(A,p,r)
(3) Quicksort(A,p,q-1)
(4) Quicksort(A,q+1,r) 1 2 3 4 5 6 7 8 9 10 A 16 14 10 1 7 9 3 2 5 8
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4. What about partition? partition(A,p,r) (1) x ← A[r] (2) i ← p - 1
(3) for j ← p to r-1
(4) if A[j] ≤ x
(5) then i ← i + 1
(6) exchange A[i] ↔ A[j]
(7) exchange A[i+1] ↔ A[r]
(8) return i+1 1 2 3 4 5 6 7 8 9 10 A 16 14 10 1 7 9 3 2 5 8
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5. What about partition? partition(A,p,r)
(1) x ← A[r]
(2) i ← p - 1
(3) for j ← p to r-1
(4) if A[j] ≤ x
(5) then i ← i + 1
(6) exchange A[i] ↔ A[j]
(7) exchange A[i+1] ↔ A[r]
(8) return i+1
The ideal pivot is exactly the n/2-th smallest element! 1 2 3 4 5 6 7 8 9 10 A 16 14 10 1 7 9 3 2 5 8
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6. What about partition? partition(A,p,r)
(1) x ← A[r]
(2) i ← p - 1
(3) for j ← p to r-1
(4) if A[j] ≤ x
(5) then i ← i + 1
(6) exchange A[i] ↔ A[j]
(7) exchange A[i+1] ↔ A[r]
(8) return i+1
The ideal pivot is exactly the └n/2┘-th smallest element!
When n is even, └n/2┘-th
When n is odd, (n+1)/2-th 1 2 3 4 5 6 7 8 9 10 A 16 14 10 1 7 9 3 2 5 8
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7. Select(A,i)
(1) Partition A into groups of 5 elements, sort each group. 1 2 3 4 5 6 7 8 9 10 A 1 7 10 14 16 2 3 5 8 9
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8. Select(A,i)
(1) Partition A into groups of 5 elements, sort each group.
(2) While sorting each group, identify the medians as well. Collect the ┌n/5┐medians into M. 1 2 3 4 5 6 7 8 9 10 A 1 7 10 14 16 2 3 5 8 9
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9. Select(A,i)
(1) Partition A into groups of 5 elements, sort each group.
(2) While sorting each group, identify the medians as well. Collect the ┌n/5┐medians into M.
(3) Recursively call Select(M,└|M|/2┘) to find the median x of M. 1 2 3 4 5 6 7 8 9 10 A 1 7 10 14 16 2 3 5 8 9
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10. Select(A,i)
(1) Partition A into groups of 5 elements, sort each group.
(2) While sorting each group, identify the median as well. Collect the ┌n/5┐medians into M.
(3) Recursively call Select(M,└|M|/2┘) to find the median x of M.
(4) Partition A around the pivot x (index of x is k). 1 2 3 4 5 6 7 8 9 10 A 1 2 3 5 16 7 10 9 8 14
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k=4

11. Select(A,i)
(1) Partition A into groups of 5 elements, sort each group.
(2) While sorting each group, identify the median as well. Collect the ┌n/5┐medians into M.
(3) Recursively call Select(M,└|M|/2┘) to find the median x of M.
(4) Partition A around the pivot x (index of x is k).
(5) if i=k, return x;
if i < k, return Select(A[1..k-1],i);
if i > k, return Select(A[k+1..n],i-k). 1 2 3 4 5 6 7 8 9 10 A 1 2 3 5 16 7 10 9 8 14
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k=4

12. Cost Let T(n) be the running time for select(A,i).
T(n) ≤ O(1), if n<140
T(n) ≤ T(┌n/5┐) + T(7n/10+6) + O(n), if n ≥ 140
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13. Cost Let T(n) be the running time for select(A,i).
T(n) ≤ O(1), if n<140
T(n) ≤ T(┌n/5┐) + T(7n/10+6) + O(n), if n ≥ 140
Why?
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14. Cost Let T(n) be the running time for select(A,i).
T(n) ≤ O(1), if n<140
T(n) ≤ T(┌n/5┐) + T(7n/10+6) + O(n), if n ≥ 140
How to solve this recurrence relation?
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15. Cost Let T(n) be the running time for select(A,i).
T(n) ≤ O(1), if n<140
T(n) ≤ T(┌n/5┐) + T(7n/10+6) + O(n), if n ≥ 140
T(n) ≤ cn = O(n)!
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