Transcript Document 7659645
Analysis of Algorithms
Estimate the running time Estimate the memory space required. Time and space depend on the input size.
Analysis of Algorithms 1
Running Time (
§
3.1)
Most algorithms transform input objects into output objects.
The running time of an algorithm typically grows with the input size.
Average case time is often difficult to determine.
We focus on the worst case running time.
Easier to analyze Crucial to applications such as games, finance and robotics Analysis of Algorithms
120 100 80 60 40 20 0 1000
best case average case worst case
2000 3000 Input Size 4000
2
Experimental Studies
Write a program implementing the algorithm Run the program with inputs of varying size and composition Use a method like System.currentTimeMillis() to get an accurate measure of the actual running time Plot the results
9000 8000 7000 6000 5000 4000 3000 2000 1000 0 0
Analysis of Algorithms
50 Input Size
3
100
Limitations of Experiments
It is necessary to implement the algorithm, which may be difficult Results may not be indicative of the running time on other inputs not included in the experiment. In order to compare two algorithms, the same hardware and software environments must be used Analysis of Algorithms 4
Theoretical Analysis
Uses a high-level description of the algorithm instead of an implementation Characterizes running time as a function of the input size,
n
.
Takes into account all possible inputs Allows us to evaluate the speed of an algorithm independent of the hardware/software environment Analysis of Algorithms 5
Pseudocode (
§
3.2)
High-level description of an algorithm More structured than English prose Less detailed than a program Preferred notation for describing algorithms Hides program design issues
Algorithm
arrayMax
(
A
,
n
)
Input
array
A
of
n
integers
Output
maximum element of
A currentMax
for if
i
Example: find max element of an array
A
[
i
] 1
to
n
A
[0] 1
do
currentMax currentMax
A
[
i
]
return
currentMax
then
Analysis of Algorithms 6
Pseudocode Details
Control flow
if
…
then
… [
else
…]
while
…
do
…
repeat
…
until
…
for
…
do
… Indentation replaces braces Method declaration
Algorithm
method
(
arg
[,
arg
…])
Input
…
Output
… Expressions Assignment (like in Java)
n
2 Equality testing (like in Java) Superscripts and other mathematical formatting allowed Analysis of Algorithms 7
Primitive Operations ( time unit ) Basic computations performed by an algorithm Identifiable in pseudocode Largely independent from the programming language Exact definition not important (we will see why later) Assumed to take a constant amount of time in the RAM model Examples: Evaluating an expression Assigning a value to a variable Indexing into an array Calling a method Returning from a method Comparison x==y x>Y Analysis of Algorithms 8
Counting Primitive Operations (
§
3.4)
By inspecting the pseudocode, we can determine the maximum number of primitive operations executed by an algorithm, as a function of the input size
Algorithm
arrayMax
(
A
,
n
)
currentMax
A
[0]
for
(
i
=1; i
n
if
A
[
i
] (i=1 once, i
currentMax currentMax
then
A
[
i
] 2( 2(
n n
1) 1)
return
currentMax
1 Total 6
n
1 Analysis of Algorithms 9
Estimating Running Time
Algorithm
arrayMax
executes 6
n
operations in the worst case. 1 primitive Define:
a
= Time taken by the fastest primitive operation
b
= Time taken by the slowest primitive operation Let
T
(
n
) be worst-case time of
arrayMax.
a
(8
n
2)
T
(
n
)
b
(8
n
2) Hence, the running time
T
(
n
) linear functions Then is bounded by two Analysis of Algorithms 10
Growth Rate of Running Time
Changing the hardware/ software environment Affects
T
(
n
) by a constant factor, but Does not alter the growth rate of
T
(
n
) The linear growth rate of the running time
T
(
n
) is an intrinsic property of algorithm
arrayMax
Analysis of Algorithms 11
n
log
n n n
log
n n
2
n
3 2
n
4 8 16 32 64 128 256 512 1024 2 3 4 5 6 7 8 9 10 4 8 16 32 64 128 256 512 1,024 8 24 64 160 384 896 2,048 4,608 10,240 16 64 256 1,024 4,094 16,384 65,536 262,144 1,048,576 64 512 4,096 32,768 262,144 2,097,152 16,777,216 134,217,728 1,073,741,824 16 256 65,536 4,294,967,296 1.84 * 10 19 3.40 * 10 38 1.15 * 10 77 1.34 * 10 154 1.79 * 10 308 The Growth Rate of the Six Popular functions Analysis of Algorithms 12
Big-Oh Notation
To simplify the running time estimation, for a function
f(n),
we ignore the constants and lower order terms.
Example:
10n
3
+4n
2
-4n+5
is
O(n
3
).
Analysis of Algorithms 13
Big-Oh Notation
(Formal Definition) 10,000 Given functions
f
(
n
) and
g
(
n
) , we say that
f
(
n
) is 1,000
O
(
g
(
n
)) if there are positive constants
c
and
n
0
such that 100
f
(
n
)
cg
( Example:
n
2 )
n
for +
n
10
n
0
is
O
(
n
) ( 2
n c
+ 10 2)
n
cn
10
n
10 / (
c
2) Pick
c
3 and
n
0
10 10 1 1 3n 2n+10 n 10
n
100 Analysis of Algorithms 14 1,000
Big-Oh Example
Example: the function
n
2 is not
O
(
n
) 1,000,000 100,000
n
2
n
c cn
The above inequality cannot be satisfied since
c
must be a constant n 2 is O(n 2 ).
10,000 1,000 100 10 1 1 n^2 100n 10n n 10
n
100 Analysis of Algorithms 1,000 15
More Big-Oh Examples
7n-2 7n-2 is O(n) need c > 0 and n 0 1 such that 7n-2 this is true for c = 7 and n 0 = 1 c•n for n n 0 3n 3 3n 3 + 20n + 20n 2 2 + 5 + 5 is O(n 3 ) need c > 0 and n 0 1 such that 3n this is true for c = 4 and n 0 = 21 3 3 log n + 5 + 20n 2 + 5 c•n 3 for n n 0 3 log n + 5 is O(log n) need c > 0 and n 0 1 such that 3 log n + 5 this is true for c = 8 and n 0 = 2 Analysis of Algorithms c•log n for n n 0 16
Big-Oh and Growth Rate
The big-Oh notation gives an upper bound on the growth rate of a function The statement “
f
(
n
) is
O
(
g
(
n
)) ” means that the growth rate of
f
(
n
) is no more than the growth rate of
g
(
n
) We can use the big-Oh notation to rank functions according to their growth rate Analysis of Algorithms 17
Big-Oh Rules
If
f
(
n
) is a polynomial of degree
d
, then
f
(
n
)
O
(
n d
) , i.e., 1.
2.
Drop lower-order terms Drop constant factors Use the smallest possible class of functions Say “ 2
n
is
O
(
n
) ” instead of “ 2
n
is
O
(
n
2 ) ” Use the simplest expression of the class Say “ 3
n
+ 5 is
O
(
n
) ” instead of “ 3
n
+ 5 is
O
(3
n
) ” is Analysis of Algorithms 18
Growth Rate of Running Time Consider a program with time complexity O(n 2 ).
For the input of size n, it takes 5 seconds.
If the input size is doubled (2n), then it takes 20 seconds.
Consider a program with time complexity O(n).
For the input of size n, it takes 5 seconds.
If the input size is doubled (2n), then it takes 10 seconds.
Consider a program with time complexity O(n 3 ).
For the input of size n, it takes 5 seconds.
If the input size is doubled (2n), then it takes 40 seconds.
Analysis of Algorithms 19
Asymptotic Algorithm Analysis
The asymptotic analysis of an algorithm determines the running time in big-Oh notation To perform the asymptotic analysis We find the worst-case number of primitive operations executed as a function of the input size We express this function with big-Oh notation Example: We determine that algorithm
arrayMax
6
n
1 primitive operations We say that algorithm
arrayMax
executes at most “runs in
O
(
n
) time” Since constant factors and lower-order terms are eventually dropped anyhow, we can disregard them when counting primitive operations Analysis of Algorithms 20
Computing Prefix Averages
( We further illustrate asymptotic analysis with two algorithms for prefix averages The
i
-th prefix average of an array
X i
+ 1) is average of the first elements of
X:
A
[
i
] (
X
[0] +
X
[1] + … +
X
[
i
])/(
i
+1) Computing the array prefix averages of another array
X
has applications to financial analysis
A
of 10 5 0 35 30 25 20 15
X A
1 2 3 4 5 6 7 Analysis of Algorithms 21
Prefix Averages (Quadratic)
The following algorithm computes prefix averages in quadratic time by applying the definition
Algorithm
prefixAverages1
(
X, n
)
Input
array
X
of
n
integers
Output
A
for
i
array
A
of prefix averages of new array of
n
to
n
1
do
integers
X
{
s A
[ 0
for
i
]
j X
[0]
s s
1
to
/
i
(
i s
+
do
+
X
[
j
] 1) } 1 1 + + 2 2 #operations + +
n n n
… …
n
+ + ( (
n n
1) 1)
return
A
1 Analysis of Algorithms 22
Arithmetic Progression
The running time of
prefixAverages1
is
O
(1 + 2 + … +
n
) The sum of the first
n
integers is
n
(
n
+ 1) / 2 There is a simple visual proof of this fact Thus, algorithm
prefixAverages1
runs in
O
(
n
2 ) time Analysis of Algorithms 23
Prefix Averages (Linear)
The following algorithm computes prefix averages in linear time by keeping a running sum
Algorithm
prefixAverages2
(
X, n
)
Input
array
X
of
n
integers
Output
A s
for
i
array
A
of prefix averages of new array of
n
integers 0 {
s
0
A
[
i
]
to
s
+
s n
1 /
X
[
i
] (
i
+
do
1) }
X
return
A
Algorithm
prefixAverages2
runs in
O
(
n
) time #operations
n
1
n n n
1 Analysis of Algorithms 24
Exercise: Give a big-Oh characterization
Algorithm
Ex1 (A
, n
)
Input
an array
X
of
n
integers
Output
the sum of the elements in A
s
for
i
A[0]
s
0
s
to
+
n
A[
i
] 1
do return
s Analysis of Algorithms 25
Exercise: Give a big-Oh characterization
Algorithm
Ex2 (A
, n
)
Input
an array
X
of
n
integers
Output
the sum of the elements at even cells in A
s
for
i
A[0]
s
2
s
to
+
n
A[
i
] 1 by increments of 2
do return
s
Analysis of Algorithms 26
Exercise: Give a big-Oh characterization
Algorithm
Ex1 (A
, n
)
Input
an array
X
of
n
integers
Output
s
for
i
0 { the sum of the prefix sums A 0
s
to
for
s
n s
j
+
1
s
1
A[0]
to
+
do
i
do
A[j]
}
return
s
Analysis of Algorithms 27
Remarks: In the first tutorial, ask the students to try programs with running time O(n), O(n log n), O(n 2 ), O(n 2 log n), O(2 n ) with various inputs.
They will get intuitive ideas about those functions.
for (i=1; i<=n; i++) for (j=1; j<=n; j++) { x=x+1; delay(1 second); } Analysis of Algorithms 28