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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
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
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
Now what?
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
Now what?
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
Now what?
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
Now what?
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
Now what?
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
Now what?
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
Now what?
Use Constructive Induction.
Guess S(n) ≤ an lg n for some constant a and n ≥ 1.
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:
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
S(n) =2
n
n−1∑q=1
S(q) + n − 1 where we left off
≤ 2
n
n−1∑q=1
aq lg q + n − 1 by IH
≤ 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
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
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
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
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
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.
Theorem
The expected number of comparisons for Quicksort is ∼ 2n ln n.
NOTE: Could go back and do the Constructive Induction withS(n) ≤ an ln n, which would simplify the algebra.
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.
Theorem
The expected number of comparisons for Quicksort is ∼ 2n ln n.
NOTE: Could go back and do the Constructive Induction withS(n) ≤ an ln n, which would simplify the algebra.
![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://img.pdfslide.us/doc/110x75/5fde54712cef2318e54396f4/analysis-of-quicksort-university-of-the-average-number-of-comparisons-for-quicksort.jpg)
