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SUBMITTED TO-

Santosh Kumar

SUBITTED BY-Kanika Garg4th sem6212/11

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Sorting takes an unordered collectionand makes it an ordered one.

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1 2 3 4 5 65 12 35 42 77 101

1 2 3 4 5 6

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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

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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 1011 2 3 4 5 6

Swap

42 77

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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

512357742 1011 2 3 4 5 6

Swap

35 77

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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

512773542 1011 2 3 4 5 6

Swap

12 77

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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

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No need to swap

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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

577123542 1011 2 3 4 5 6

Swap5 101

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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

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101

Largest value correctly placed

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index A[index + 1]) then

Swap(A[index], A[index + 1])

endifindex

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Notice that only the largest value iscorrectly placed All other values are still out of order So we need to repeat this process

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101

Largest value correctly placed

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If we have N elements And if each time we bubble anelement, we place it in its correct

location Then we repeat the bubble upprocess N 1 times. This guarantees well correctlyplace all N elements.

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77123542 51 2 3 4 5 6

101

5421235 771 2 3 4 5 6

101

4253512 771 2 3 4 5 6

101

4235512 771 2 3 4 5 6

101

4235125 771 2 3 4 5 6

101

N

-1

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12354277 1011 2 3 4 5 6

5

77123542 51 2 3 4 5 6

101

5421235 771 2 3 4 5 6

101

4253512 771 2 3 4 5 6

101

4235512 771 2 3 4 5 6

101

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On the Nthbubble up, we only need todo MAX-N comparisons. For example:

This is the 4thbubble up MAX is 6 Thus we have 2 comparisons to do

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procedure Bubblesort(A isoftype in/out Arr_Type)

to_do, index isoftype Num

to_do A[index + 1]) then

Swap(A[index], A[index + 1])

endif

index

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What if the collection was already sorted? What if only a few elements were out ofplace and after a couple of bubble ups,

the collection was sorted? We want to be able to detect thisand stop early!

4235125 771 2 3 4 5 6101

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We can use a boolean variable to determine ifany swapping occurred during the bubbleup. If no swapping occurred, then we know thatthe collection is already sorted! This boolean flag needs to be reset aftereach bubble up.

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Bubble Up algorithm will move largestvalue 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 wecompare each time one is correctlyplaced

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NOBODY EVER USES BUBBLE SORT NOBODY NOT EVER BECAUSE IT IS EXTREMELY INEFFICIENT

LB

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