Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort...
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The average number of comparisons for quicksort is S (n)= 0 if n =0, 1 ∑ n q=1 1 n [S (q - 1) + S (n - q )] + n - 1 otherwise
Transcript of Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort...
![Page 1: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/1.jpg)
The average number of comparisons for quicksort is
S(n) =
{0 if n = 0, 1∑n
q=11n
[S(q − 1) + S(n − q)] + n − 1 otherwise
Theorem
S(n) ≤ an lg n for some constant a and n ≥ 1.
Proof by Constructive Induction.
Base case: n = 1: S(1) = 0 and a · 1 · lg 1 = 0.
Induction Hypothesis:Assume it holds for all positive integers less than n.So, S(k) ≤ ak lg k for 1 ≤ k ≤ n − 1.
Induction step:
![Page 2: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/2.jpg)
The average number of comparisons for quicksort is
S(n) =
{0 if n = 0, 1∑n
q=11n
[S(q − 1) + S(n − q)] + n − 1 otherwise
Theorem
S(n) ≤ an lg n for some constant a and n ≥ 1.
Proof by Constructive Induction.
Base case: n = 1: S(1) = 0 and a · 1 · lg 1 = 0.
Induction Hypothesis:Assume it holds for all positive integers less than n.So, S(k) ≤ ak lg k for 1 ≤ k ≤ n − 1.
Induction step:
![Page 3: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/3.jpg)
S(n) =n∑
q=1
1
n[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
S(q − 1) +1
n
n∑q=1
S(n − q) + n − 1
=1
n
n−1∑q=0
S(q) +1
n
n−1∑q=0
S(q) + n − 1
=2
n
n−1∑q=0
S(q) + n − 1 =2
n
n−1∑q=1
S(q) + n − 1 since S(0) = 0
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
![Page 4: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/4.jpg)
S(n) =n∑
q=1
1
n[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
S(q − 1) +1
n
n∑q=1
S(n − q) + n − 1
=1
n
n−1∑q=0
S(q) +1
n
n−1∑q=0
S(q) + n − 1
=2
n
n−1∑q=0
S(q) + n − 1 =2
n
n−1∑q=1
S(q) + n − 1 since S(0) = 0
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
![Page 5: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/5.jpg)
S(n) =n∑
q=1
1
n[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
S(q − 1) +1
n
n∑q=1
S(n − q) + n − 1
=1
n
n−1∑q=0
S(q) +1
n
n−1∑q=0
S(q) + n − 1
=2
n
n−1∑q=0
S(q) + n − 1 =2
n
n−1∑q=1
S(q) + n − 1 since S(0) = 0
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
![Page 6: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/6.jpg)
S(n) =n∑
q=1
1
n[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
S(q − 1) +1
n
n∑q=1
S(n − q) + n − 1
=1
n
n−1∑q=0
S(q) +1
n
n−1∑q=0
S(q) + n − 1
=2
n
n−1∑q=0
S(q) + n − 1 =2
n
n−1∑q=1
S(q) + n − 1 since S(0) = 0
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
![Page 7: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/7.jpg)
S(n) =n∑
q=1
1
n[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
S(q − 1) +1
n
n∑q=1
S(n − q) + n − 1
=1
n
n−1∑q=0
S(q) +1
n
n−1∑q=0
S(q) + n − 1
=2
n
n−1∑q=0
S(q) + n − 1 =2
n
n−1∑q=1
S(q) + n − 1 since S(0) = 0
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
![Page 8: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/8.jpg)
S(n) =n∑
q=1
1
n[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
[S(q − 1) + S(n − q)] + n − 1
=1
n
n∑q=1
S(q − 1) +1
n
n∑q=1
S(n − q) + n − 1
=1
n
n−1∑q=0
S(q) +1
n
n−1∑q=0
S(q) + n − 1
=2
n
n−1∑q=0
S(q) + n − 1 =2
n
n−1∑q=1
S(q) + n − 1 since S(0) = 0
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
![Page 9: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/9.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 10: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/10.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 11: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/11.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 12: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/12.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 13: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/13.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 14: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/14.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 15: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/15.jpg)
=2a
n
n−1∑q=1
q lg q + n − 1
≤ 2a
n
∫ n
1
x lg xdx + n − 1 by integral bound
=2a
n
[x2 lg x
2− x2 lg e
4
]|n1 +n − 1
=2a
n
[(n2 lg n
2− n2 lg e
4
)−(
12 lg 1
2− 12 lg e
4
)]+ n − 1
= an lg n − an lg e
2+
a lg e
2n+ n − 1
= an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1
≤ an lg n for the induction to hold
![Page 16: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/16.jpg)
Need
an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1 ≤ an lg n
Need
1− a lg e
2≤ 0 =⇒ a ≥ 2
lg e
Set a = 2lg e≈ 1.39
Need
a lg e
2n− 1 ≤ 0 ⇐⇒ 2 lg e
(lg e)2n− 1 ≤ 0 ⇐⇒ 1
n− 1 ≤ 0
which always holds since n ≥ 1.
So,S(n) ≈ 1.39n lg n
![Page 17: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/17.jpg)
Need
an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1 ≤ an lg n
Need
1− a lg e
2≤ 0 =⇒ a ≥ 2
lg e
Set a = 2lg e≈ 1.39
Need
a lg e
2n− 1 ≤ 0 ⇐⇒ 2 lg e
(lg e)2n− 1 ≤ 0 ⇐⇒ 1
n− 1 ≤ 0
which always holds since n ≥ 1.
So,S(n) ≈ 1.39n lg n
![Page 18: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/18.jpg)
Need
an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1 ≤ an lg n
Need
1− a lg e
2≤ 0 =⇒ a ≥ 2
lg e
Set a = 2lg e≈ 1.39
Need
a lg e
2n− 1 ≤ 0 ⇐⇒ 2 lg e
(lg e)2n− 1 ≤ 0 ⇐⇒ 1
n− 1 ≤ 0
which always holds since n ≥ 1.
So,S(n) ≈ 1.39n lg n
![Page 19: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/19.jpg)
Need
an lg n +
[1− a lg e
2
]n +
a lg e
2n− 1 ≤ an lg n
Need
1− a lg e
2≤ 0 =⇒ a ≥ 2
lg e
Set a = 2lg e≈ 1.39
Need
a lg e
2n− 1 ≤ 0 ⇐⇒ 2 lg e
(lg e)2n− 1 ≤ 0 ⇐⇒ 1
n− 1 ≤ 0
which always holds since n ≥ 1.
So,S(n) ≈ 1.39n lg n
![Page 20: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/20.jpg)
Now that we are finished, we realize that
S(n) ≤ an lg n =2n lg n
lg e= 2n ln n
which is a more natural formula.
Could go back and do the Constructive Induction with S(n) ≤ an ln n,which would simplify the algebra.
![Page 21: Analysis of Quicksort - University Of Maryland · The average number of comparisons for quicksort is S(n) = ˆ 0 ifP n = 0;1 n q=1 1 n [ S( q1) + n)] + 1 otherwise Theorem S(n) anlg](https://reader036.fdocuments.in/reader036/viewer/2022071104/5fde54712cef2318e54396f4/html5/thumbnails/21.jpg)
Now that we are finished, we realize that
S(n) ≤ an lg n =2n lg n
lg e= 2n ln n
which is a more natural formula.
Could go back and do the Constructive Induction with S(n) ≤ an ln n,which would simplify the algebra.
