courtesy of mit press. used with permission. lecture 3 of mit press. used with permission. menu •...
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6.006- Introduction to Algorithms
Lecture 3 1
Courtesy of MIT Press. Used with permission.
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Menu
• Sorting! – Insertion Sort – Merge Sort
• Solving Recurrences
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The problem of sorting
Input: array A[1…n] of numbers.
Output: permutation B[1…n] of A such that B[1] ≤ B[2] ≤ … ≤ B[n] .
e.g. A = [7, 2, 5, 5, 9.6] → B = [2, 5, 5, 7, 9.6]
How can we do it efficiently ?
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Why Sorting? • Obvious applications
– Organize an MP3 library – Maintain a telephone directory
• Problems that become easy once items are in sorted order – Find a median, or find closest pairs – Binary search, identify statistical outliers
• Non-obvious applications – Data compression: sorting finds duplicates – Computer graphics: rendering scenes front to back
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Insertion sort
INSERTION-SORT (A, n) [ A[1 . . n] for j ← 2 to n
insert key A[j] into the (already sorted) sub-array A[1 .. j-1]. by pairwise key-swaps down to its right position
Illustration of iteration j
1 i j n A:
keysorted new location of key
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 4 9 3 68 2
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Example of insertion sort 9 38 2 4
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Example of insertion sort 9 38 2 4
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 8 2 4 9 3 6
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Example of insertion sort 8 4 9 3 62
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2 3 4 6 8 9 done Running time? Θ(n2) because Θ(n2) compares and Θ(n2) swaps
e.g. when input is A = [n, n − 1, n − 2, . . . , 2, 1] 16
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Binary Insertion sort
BINARY-INSERTION-SORT (A, n) [ A[1 . . n] for j ← 2 to n
insert key A[j] into the (already sorted) sub-array A[1 .. j-1]. Use binary search to find the right position
Binary search with take Θ(log n) time. However, shifting the elements after insertion will still take Θ(n) time.
Complexity: Θ(n log n) comparisons
Θ
(n2) swaps 17
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Meet Merge Sort
MERGE-SORT A[1 . . n] 1. If n = 1, done (nothing to sort).
divide and 2. Otherwise, recursively sort A[ 1 . . n/2 ]conquer and A[ n/2+1 . . n ] .
3. “Merge” the two sorted sub-arrays.
Key subroutine: MERGE 18
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Merging two sorted arrays
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Merging two sorted arrays
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Merging two sorted arrays
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Time = Θ(n) to merge a total of n elements (linear time).
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Analyzing merge sort
MERGE-SORT A[1 . . n] 1. If n = 1, done. 2. Recursively sort A[ 1 . . ⎡n/2⎤ ]
and A[ ⎡n/2⎤+1 . . n ] . 3. “Merge” the two sorted lists
Θ(1) if n = 1;T(n) =
2T(n/2) + Θ(n) if n > 1. T(n) = ?
T(n) Θ(1) 2T(n/2)
Θ(n)
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Recurrence solving Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
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Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
T(n)
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T(n/2) T(n/2)
Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
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cn/2
cn
cn/2
T(n/4) T(n/4) T(n/4) T(n/4)
Recursion tree olve T(n) = 2T(n/2) + cn, where c > 0 is constant. S
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Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
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Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
h = 1 + lg n
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Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
cn
h = 1 + lg n
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cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
cnh = 1 + lg n
Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
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cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
cnh = 1 + lg n
cn
…
cn
Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
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cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
cn
cn
#leaves = n Θ(n)
…
h = 1 + lg n
Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
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cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
cn
cn
#leaves = n Θ(n) Total ?
…
h = 1 + lg n
Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn cn
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cn/4 cn/4 cn/4 cn/4
cn/2 cn/2
Θ(1)
= 1 + lg n cn
cn
#leaves = n Θ(n)
…
Total = Θ(n lg n)
Recursion tree Solve T(n) = 2T(n/2) + cn, where c > 0 is constant.
cn
h
cn
Equal amount of work done at each level 44
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…
c c c c
c c
Θ(1)
2c
4c
#leaves = n Θ(n) n/2 c
h = 1 + lg n
Tree for different recurrence Solve T(n) = 2T(n/2) + c, where c > 0 is constant.
c c
Note that 1 + ½ + ¼ + … < 2 = ΘTotal (n) All the work done at the leaves
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cn2/2h = 1 + lg n
Θ(n)
…
cn2/16 cn2/16 cn2/16 cn2/16
cn2/4 cn2/4
Θ(1) #leaves = n
cn2/4
Tree for yet another recurrence Solve T(n) = 2T(n/2) + cn2, c > 0 is constant.
cn2 cn2
Note that 1 + ½ + ¼ + … < 2 Total = Θ(n2) All the work done at the root
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