bubble sort - southern methodist university › ... › lecture-5 › lecture-bubble-sort.pdf ·...

2/1/2016

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CS 3353

Lecture 5

Vidroha debroyMaterial adapted courtesy of Prof. Xiangnan Kong and Prof.

Carolina Ruiz at Worcester Polytechnic Institute

Searching for a number

54

68

3

103

73

131

26

88

3

91

19

Find a specific number “91” -

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Searching for a number

54

68

3

103

73

131

26

88

13

91

19

Find a specific number “91” - what about now?

73

38371

17

35

41

85

114

2161

87

9711

49

94

839

57

66

12

22

11163

102

9976

Why Sorting?

• Practical Application

– People by last name

– Countries by population

– Search engine results by relevance

• Fundamental to other algorithms

• Different algorithms have different asymtotic and constant-factor trade-offs

– no single “best” sort for all scenarios

– knowing one way to sort just isn't enough

Keeping data in “order” allows it to be searched

more efficiently

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Sorting

• Sorting: Rearranging the values in an array or collection into a specific order (usually into their "natural ordering").

– one of the fundamental problems in computer science

– it is estimated that 25~50% of all computing power is used for sorting activities.

• Popular Sorting algorithms:

– bubble sort: swap adjacent pairs that are out of order

– selection sort: look for the smallest element, move to front

– insertion sort: build an increasingly large sorted front portion

– merge sort: recursively divide the array in half and sort it

– heap sort: place the values into a sorted tree structure

– quick sort: recursively partition array based on a middle value

Sorting Example

Bubble Sort

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Problem Definition

• Sorting takes an unordered collection and makes it an ordered one.

512354277 101

1 2 3 4 5 6

5 12 35 42 77 101

1 2 3 4 5 6

Input:

Output:

One Idea: Bubble

1

Water

Air

2

3

1

3

2

1

2

3

1

2

3

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"Bubbling Up" the Largest Element

• Traverse a collection of elements

– Move from the front to the end

– “Bubble” the largest value to the end using pair-wise comparisons and swapping

512354277 101

1 2 3 4 5 6





512354277 101

1 2 3 4 5 6

Swap42 77

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512357742 101

1 2 3 4 5 6

Swap35 77





512773542 101

1 2 3 4 5 6

Swap12 77

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577123542 101

1 2 3 4 5 6

No need to swap





577123542 101

1 2 3 4 5 6

Swap5 101

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77123542 5

1 2 3 4 5 6

101

Largest value correctly placed

The “Bubble Up” Algorithm

Bubble-Sort-step1( A ):

for k = 1 … (N – 1):if A[k] > A[k+1]:

Swap( A, k, k+1 )

Swap( A, x, y ):tmp = A[x]A[x] = A[y]A[y] = tmp

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Need More Iterations

• Notice that only the largest value is correctly placed

• All other values are still out of order

• So we need to repeat this process

77123542 5

1 2 3 4 5 6

101

Largest value correctly placed

Repeat “Bubble Up” How Many Times?

• If we have N elements…

• And if each time we bubble an element, we place it in its correct location…

• Then we repeat the “bubble up” process N – 1 times.

• This guarantees we’ll correctly place all N elements.

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“Bubbling” All the Elements

77123542 5

1 2 3 4 5 6

101

5421235 77

1 2 3 4 5 6

101

4253512 77

1 2 3 4 5 6

101

4235512 77

1 2 3 4 5 6

101

4235125 77

1 2 3 4 5 6

101

N -

1


Bubble-Sort( A ):

for k = 1 … (N – 1):if A[k] > A[k+1]:

Swap( A, k, k+1 )

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Bubble-Sort( A ):

for i = 1 … (N – 1):for k = 1 … (N – 1):

if A[k] > A[k+1]:Swap( A, k, k+1 )

Bubble-Sort( A ):

for i = 1 … (N – 1):

for k = 1 … (N – 1):if A[k] > A[k+1]:

Swap( A, k, k+1 )


Inn

er

loo

p

Ou

ter

loo

p

To bubble a value

To do N-1 iterations

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Reducing the Number of Comparisons

12354277 101

1 2 3 4 5 6

5

77123542 5

1 2 3 4 5 6

101

5421235 77

1 2 3 4 5 6

101

4253512 77

1 2 3 4 5 6

101

4235512 77

1 2 3 4 5 6

101

Reducing the Number of Comparisons

• Assume the array size N

• On the ith iteration, we only need to do N-i comparisons.

• For example:

– N = 6

– i = 4 (4th iteration)

– Thus, we have 2 comparisons to do

4253512 77

1 2 3 4 5 6

101

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Bubble-Sort( A ):

for i = 1 … (N – 1):for k = 1 … (N – 1):

if A[k] > A[k+1]:Swap( A, k, k+1 )


Bubble-Sort( A ):

for i = 1 … (N – 1):for k = 1 … (N – i):

if A[k] > A[k+1]:Swap( A, k, k+1 )


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Code Demo

An Animated Example

674523 14 6 3398 42

1 2 3 4 5 6 7 8

i

k

1

N 8

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An Animated Example

674523 14 6 3398 42

1 2 3 4 5 6 7 8

1

1

N 8

i

k

An Animated Example

674523 14 6 3398 42

1 2 3 4 5 6 7 8

1

1

N 8

Swap

i

k

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An Animated Example

674598 14 6 3323 42

1 2 3 4 5 6 7 8

1

1

N 8

Swap

i

k

An Animated Example

674598 14 6 3323 42

1 2 3 4 5 6 7 8

1

2

N 8

i

k

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An Animated Example

674598 14 6 3323 42

1 2 3 4 5 6 7 8

1

2

N 8

Swap

i

k

An Animated Example

679845 14 6 3323 42

1 2 3 4 5 6 7 8

1

2

N 8

Swap

i

k

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An Animated Example

679845 14 6 3323 42

1 2 3 4 5 6 7 8

1

3

N 8

i

k

An Animated Example

679845 14 6 3323 42

1 2 3 4 5 6 7 8

1

3

N 8

Swap

i

k

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An Animated Example

671445 98 6 3323 42

1 2 3 4 5 6 7 8

1

3

N 8

Swap

i

k

An Animated Example

671445 98 6 3323 42

1 2 3 4 5 6 7 8

1

4

N 8

i

k

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An Animated Example

671445 98 6 3323 42

1 2 3 4 5 6 7 8

1

4

N 8

Swap

i

k

An Animated Example

671445 6 98 3323 42

1 2 3 4 5 6 7 8

1

4

N 8

Swap

i

k

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An Animated Example

671445 6 98 3323 42

1 2 3 4 5 6 7 8

1

5

N 8

i

k

An Animated Example

671445 6 98 3323 42

1 2 3 4 5 6 7 8

1

5

N 8

Swap

i

k

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An Animated Example

981445 6 67 3323 42

1 2 3 4 5 6 7 8

1

5

N 8

Swap

i

k

An Animated Example

981445 6 67 3323 42

1 2 3 4 5 6 7 8

1

6

N 8

i

k

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An Animated Example

981445 6 67 3323 42

1 2 3 4 5 6 7 8

1

6

N 8

Swap

i

k

An Animated Example

331445 6 67 9823 42

1 2 3 4 5 6 7 8

1

6

N 8

Swap

i

k

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An Animated Example

331445 6 67 9823 42

1 2 3 4 5 6 7 8

1

7

N 8

i

k

An Animated Example

331445 6 67 9823 42

1 2 3 4 5 6 7 8

1

7

N 8

Swap

i

k

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An Animated Example

331445 6 67 4223 98

1 2 3 4 5 6 7 8

1

7

N 8

Swap

i

k

After First Pass of Outer Loop

331445 6 67 4223 98

1 2 3 4 5 6 7 8

1

8

N 8

Finished first “Bubble Up”

i

k

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The Second “Bubble Up”

331445 6 67 4223 98

1 2 3 4 5 6 7 8

2

1

N 8

i

k


331445 6 67 4223 98

1 2 3 4 5 6 7 8

2

1

N 8

No Swap

i

k

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331445 6 67 4223 98

1 2 3 4 5 6 7 8

2

2

N 8

i

k


331445 6 67 4223 98

1 2 3 4 5 6 7 8

2

2

N 8

Swap

i

k

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334514 6 67 4223 98

1 2 3 4 5 6 7 8

2

2

N 8

Swap

i

k


334514 6 67 4223 98

1 2 3 4 5 6 7 8

2

3

N 8

i

k

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334514 6 67 4223 98

1 2 3 4 5 6 7 8

2

3

N 8

Swap

i

k


33614 45 67 4223 98

1 2 3 4 5 6 7 8

2

3

N 8

Swap

i

k

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33614 45 67 4223 98

1 2 3 4 5 6 7 8

2

4

N 8

i

k


33614 45 67 4223 98

1 2 3 4 5 6 7 8

2

4

N 8

No Swap

i

k

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33614 45 67 4223 98

1 2 3 4 5 6 7 8

2

5

N 8

i

k


33614 45 67 4223 98

1 2 3 4 5 6 7 8

2

5

N 8

Swap

i

k

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67614 45 33 4223 98

1 2 3 4 5 6 7 8

2

5

N 8

Swap

i

k


67614 45 33 4223 98

1 2 3 4 5 6 7 8

2

6

N 8

i

k

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67614 45 33 4223 98

1 2 3 4 5 6 7 8

2

6

N 8

Swap

i

k


42614 45 33 6723 98

1 2 3 4 5 6 7 8

2

6

N 8

Swap

i

k

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After Second Pass of Outer Loop

42614 45 33 6723 98

1 2 3 4 5 6 7 8

2

7

N 8

Finished second “Bubble Up”

i

k

The Third “Bubble Up”

42614 45 33 6723 98

1 2 3 4 5 6 7 8

3

1

N 8

i

k

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42614 45 33 6723 98

1 2 3 4 5 6 7 8

3

1

N 8

Swap

i

k


42623 45 33 6714 98

1 2 3 4 5 6 7 8

3

1

N 8

Swap

i

k

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42623 45 33 6714 98

1 2 3 4 5 6 7 8

3

2

N 8

i

k


42623 45 33 6714 98

1 2 3 4 5 6 7 8

3

2

N 8

Swap

i

k

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42236 45 33 6714 98

1 2 3 4 5 6 7 8

3

2

N 8

Swap

i

k


42236 45 33 6714 98

1 2 3 4 5 6 7 8

3

3

N 8

i

k

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42236 45 33 6714 98

1 2 3 4 5 6 7 8

3

3

N 8

No Swap

i

k


42236 45 33 6714 98

1 2 3 4 5 6 7 8

3

4

N 8

i

k

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42236 45 33 6714 98

1 2 3 4 5 6 7 8

3

4

N 8

Swap

i

k


42236 33 45 6714 98

1 2 3 4 5 6 7 8

3

4

N 8

Swap

i

k

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42236 33 45 6714 98

1 2 3 4 5 6 7 8

3

5

N 8

i

k


42236 33 45 6714 98

1 2 3 4 5 6 7 8

3

5

N 8

Swap

i

k

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45236 33 42 6714 98

1 2 3 4 5 6 7 8

3

5

N 8

Swap

i

k

After Third Pass of Outer Loop

45236 33 42 6714 98

1 2 3 4 5 6 7 8

3

6

N 8

Finished third “Bubble Up”

i

k

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The Fourth “Bubble Up”

45236 33 42 6714 98

1 2 3 4 5 6 7 8

4

1

N 8

i

k


45236 33 42 6714 98

1 2 3 4 5 6 7 8

4

1

N 8

Swap

i

k

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452314 33 42 676 98

1 2 3 4 5 6 7 8

4

1

N 8

Swap

i

k


452314 33 42 676 98

1 2 3 4 5 6 7 8

4

2

N 8

i

k

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452314 33 42 676 98

1 2 3 4 5 6 7 8

4

2

N 8

No Swap

i

k


452314 33 42 676 98

1 2 3 4 5 6 7 8

4

3

N 8

i

k

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452314 33 42 676 98

1 2 3 4 5 6 7 8

4

3

N 8

No Swap

i

k


452314 33 42 676 98

1 2 3 4 5 6 7 8

4

4

N 8

i

k

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452314 33 42 676 98

1 2 3 4 5 6 7 8

4

4

N 8

No Swap

i

k

After Fourth Pass of Outer Loop

452314 33 42 676 98

1 2 3 4 5 6 7 8

4

5

N 8

Finished fourth “Bubble Up”

i

k

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The Fifth “Bubble Up”

452314 33 42 676 98

1 2 3 4 5 6 7 8

5

1

N 8

i

k


452314 33 42 676 98

1 2 3 4 5 6 7 8

5

1

N 8

No Swap

i

k

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452314 33 42 676 98

1 2 3 4 5 6 7 8

5

2

N 8

i

k


452314 33 42 676 98

1 2 3 4 5 6 7 8

5

2

N 8

No Swap

i

k

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452314 33 42 676 98

1 2 3 4 5 6 7 8

5

3

N 8

i

k


452314 33 42 676 98

1 2 3 4 5 6 7 8

5

3

N 8

No Swap

i

k

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After Fifth Pass of Outer Loop

452314 33 42 676 98

1 2 3 4 5 6 7 8

5

4

N 8

Finished fifth “Bubble Up”

i

k

452314 33 42 676 98

1 2 3 4 5 6 7 8

5

4

N 8

i

k

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Analysis of Bubble Sort and Loop Invariant

Bubble-Sort( A ):

for i = 1 … (N – 1):

for k = 1 … (N – 1):if A[k] > A[k+1]:

Swap( A, k, k+1 )

Bubble Sort Algorithm

Inn

er

loo

p

Ou

ter

loo

p

To bubble a value


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Loop Invariant• It is a condition or property that is guaranteed to be correct with each

iteration in the loop

• Usually used to prove the correctness of the algorithm

for i = 1 … (N – 1):

for k = 1 … (N – 1):if A[k] > A[k+1]:

Swap( A, k, k+1 )

Inn

er

loo

p

Ou

ter

loo

p

To bubble a value


Loop Invariant for Bubble Sort

• By the end of iteration i the right-most i items (largest) are sorted and in place

for i = 1 … (N – 1):

for k = 1 … (N – 1):if A[k] > A[k+1]:

Swap( A, k, k+1 )

Inn

er

loo

p

Ou

ter

loo

p

To bubble a value


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N-1 Iterations

1st

2nd

3rd

4th

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5421235 77

1 2 3 4 5 6

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4253512 77

1 2 3 4 5 6

101

4235512 77

1 2 3 4 5 6

101

4235125 77

1 2 3 4 5 6

101

N -

1

5th

Correctness of Bubble Sort (using Loop Invariant)

• Bubble sort has N-1 Iterations

• Invariant: By the end of iteration i the right-most i items (largest) are sorted and in place

• Then: After the N-1 iterations The right-most N-1 items are sorted

– This implies that all the N items are sorted

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Bubble Sort Time Complexity

• Best-Case Time Complexity

– The scenario under which the algorithm will do the least amount of work (finish the fastest)

– What if we had a pass and nothing was swapped? What does that tell us?

• Worst-Case Time Complexity

– The scenario under which the algorithm will do the largest amount of work (finish the slowest)

Summary

• “Bubble Up” algorithm will move largest value to its correct location (to the right)

• Repeat “Bubble Up” until all elements are correctly placed:

– Maximum of N-1 times

– Can finish early if no swapping occurs

• We reduce the number of elements we compare each time one is correctly placed

• Not the most efficient algorithm around but is easy to understand

bubble sort - southern methodist university › ... › lecture-5 › lecture-bubble-sort.pdf ·...

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