Lecture 03[Insertion Sort]

8/12/2019 Lecture 03[Insertion Sort]

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Design & Analysis of Algorithms

Anwar Ghani

Semester: Spring 2013

Department of Computer Science & Software Engineering,International Islamic University, Islamabad.


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The Sorting Problem

Input:

A sequence of n numbers a 1, a 2 , . . . , a n

Output:

A permutation (reordering) a 1, a2 , . . . , a n of the

input sequence such that a 1 a2 a n

08-Mar-12 InsertionSort/AOA/Spring 12/Shehzad 2


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Insertion Sort like sorting a hand of playing cards

Start with an empty left hand and the cards facingdown on the table.

Remove one card at a time from the table, and insert

compare it with each of the cards already in the hand, fromright to left

The cards held in the left hand are sorted

these cards were originally the top cards of the pile on thetable



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To insert 9, we need to

make room for it by movingfirst 22 and then 15.

Insertion Sort



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Insertion Sort



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Insertion Sort



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Insertion Sort

5 2 4 6 1 3

input array

at each iteration, the array is divided in two sub-arrays :

e su -array r g t su -array

sorted unsorted



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Insertion Sort



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INSERTION-SORTINSERTION-SORT (A)

for j 2 to ndo key A[ j ]

Insert A[ j ] into the sorted sequence A[1 . . j -1]

a 8a 7a 6a 5a 4a 3a 2a 1

1 2 3 4 5 6 7 8

key

-while i > 0 and A[i] > key

do A[i + 1] A[i]

i i 1A[i + 1] key



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Loop Invariant for Insertion SortINSERTION-SORT (A)

for j 2 to n

do key

A[ j ]Insert A[ j ] into the sorted sequence A[1 . . j -1]

i - 1 while i > 0 and A[i] > key

do A[i + 1] A[i]

i

i 1A[i + 1] keyInvariant : at the start of the for loop the elements in A[1 . . j-1]are sorted



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Proof Proving loop invariants works like induction

Initialization (base case): It is true prior to the first iteration of the loop

Maintenance (inductive step): ,

the next iteration

Termination: When the loop terminates, the invariant gives us a useful

property that helps show that the algorithm is correct Stop the induction when the loop terminates



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Loop Invariant for Insertion Sort Initialization:

Just before the first iteration,

= 2:

The sub array A[1 . . j-1] = A[1],

The element originally in A[1] is sorted



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Loop Invariant for Insertion Sort Maintenance: the while inner loop moves A[j -1], A[j -2], A[j -

3], and so on, by one position to the right untilthe proper position for key is found

,position.

So, the sub array A[1..j] is sorted after jth iteration



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Loop Invariant for Insertion Sort Termination: The outer for loop ends when j = n + 1 j-1 = n

Replace n with j-1 in the loop invariant: the sub array A[1 . . n] consists of the elements originally in

A[1 . . n], but in sorted order

The entire array is sorted

-

Invariant : at the start of the for loop the elements in A[1 . . j-1]are in sorted order



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Analysis

cost timesc1 n

c2 n-10 n-1c4 n-1

INSERTION-SORT (A)for j 2 to n

do key A[ j ]Insert A[ j ] into the sorted sequence A[1 . . j -1]

i j - 1

=n

j jt 2 =

n

j jt

2)1(

c5c6c7

c8 n-1

while i > 0 and A[i] > keydo A[i + 1] A[i]

i i 1

A[i + 1]

key

= n

j jt

2)1(

( ) ( ) )1(11)1()1()( 82

72

62

5421 ++++++=

===

nct ct ct cncncncnT n

j j

n

j j

n

j j



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Best Case Analysis Pre-Sorted Array

A[i] keyupon the first time the while loop test is run

(when i = j -1)

t = 1

while i > 0 and A[i] > key

T(n) = c 1n + c 2(n -1) + c 4(n -1) + c 5(n -1) + c 8(n-1)

= (c 1 + c 2 + c 4 + c 5 + c 8)n + (-c 2 - c4 - c5 - c8)

= An + B( ) ( ) )1(11)1()1()( 8

27

26

25421

++++++= ===

nct ct ct cncncncnT n

j j

n

j j

n

j j



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Worst Case Analysis Reverse sorted

Always A[i] > key in while loop test

Have to compare key with all elements to the left of the j th position compare with j-1 elements t j = j

( 1) ( 1) ( 1)1 ( 1)

n n nn n n n n n j j j

+ + = => = => =using we have:

a quadratic function of nWhere

1 2 2 j j j= = =

)1(2

)1(

2

)1(1

2

)1()1()1()( 8765421 +

+

+

+

+++= ncnn

cnn

cnn

cncncncnT

)( 8542 C C C C C +++=

( 2 / )765 C C C A ++=( 2 / )7658421 C C C C C C C B ++++=

C Bn An ++= 2

( 2 / )765 C C C A ++=( 2 / )7658421 C C C C C C C B ++++=

( 2 / )765 C C C A ++=



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Average Case Analysis

INSERTION-SORT (A)

for j

2 to ndo key A[ j ]

Insert A[ j ] into the sorted sequence A[1 . . j -1]

cost times

c1 nc2 n-1

0 n-1

i j - 1

while i > 0 and A[i] > key

do A[i + 1] A[i]i i 1

A[i + 1] key

c4 n-1c5

c6c7

c8 n-1

2 / 2

jn

j =

=

n

j j

2)12 / (

= n

j j

2)12 / (

j/2 comparisons

j/2-1 exchanges



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Average Case Analysis (Cont..) In average case scenario t j = j/2

12 / 122

= == jt n

j

n

j j2 /

22 jt

n

j

n

j j == = and

)1(2 / 2

)1(2 / 2

)1(2 / 12

)1()1()1()( 8765421 +

+

+

++++= ncnncnncnncncncncnT

C Bn An ++= 2 a quadratic function of n

)2 / ( 5842 C C C C C +++=

( 4 / )7658421 C C C C C C C B ++++=( 4 / )765 C C C A ++=Where



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Insertion Sort Advantages

Good running time for Pre-sorted arrays n Disadvantages

n runn ng t me n worst an average case


Lecture 03[Insertion Sort]

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