Algorithm Analysis and
Big Oh Notation
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Algorithm Analysis: Measuring Algorithm Efficiency with Big O Notation, Slides of Computer Science
An introduction to algorithm analysis, focusing on big o notation. Algorithm analysis is a subfield of computer science that offers tools for determining the efficiency of different methods for solving a problem. The importance of comparing the efficiency of different algorithms, particularly in relation to the number of elements in an array (n). It also covers the concept of big o notation, which measures an algorithm's time requirement as a function of the problem size. Examples of common growth-rate functions and their differences as n grows larger.
Typology: Slides
2012/2013
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Algorithm Analysis andBig Oh Notation
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Measuring the Efficiency of Algorithms
^
Analysis of algorithms^ ^
Area of computer science ^
Provides tools for
determining efficiency
of different
methods of solving a problem^
E.g., the sorting problem - which sorting method is more efficient ^
Comparing the
efficiency
of different methods of solution.
^ Concerned with
significant
differences
^ E.g.:
^ n
- the number of items to be sorted ^ Is the running time proportional to
n^ or proportional to
(^2) n?
^ Big difference: e.g., for
n^ =100 it results in 100-fold difference; for
n=
1000 it results in 1000-fold difference; for
n^ = …
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The Execution Time of Algorithms
^
Count the number of basic operations ofan algorithm^
Read, write, compare, assign, jump,arithmetic operations (increment, decrement,^ add, subtract, multiply, divide),
open, close,
add, subtract, multiply, divide),
open, close,
logical operations (not/complement, AND, OR,XOR), …
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The Execution Time of Algorithms
^
Counting
an algorithm’s
operations
^
Example: calculating a sum of array elements^ int sum = item[0];int j = 1;while (j <
n
)
{
<- 1 assignment <- 1 assignment <-
n
comparisons
{
sum += item[j];++j; }
<-
n
-1 plus/assignments <-
n
-1 plus/assignments Total: 3n operations
^
Notice: Problem size
n
= number of elements in an array
This problem of size n requires solution with 3n operations
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Common Growth-Rate Functions - I
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Common Growth-Rate Functions - II
^ Differences
among the growth-rate functions
grow with
n
^ See the
differences
growing on the diagram on the previous page
^ The bigger n, the bigger differences -- that’s why algorithm efficiency is “
concern for large problems only”
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Big-Oh Notation
^ Algorithm A is order f(n) —denoted O(f(n))—
if there exist
constants k and n
such that A requires <= k*f(n) time units 0
to solve a problem of size n >= n
0
^ Examples
^ n
^ n
2 /5^ ^
O(n
2 ): k=1/5, n
^ 5*n
^ O(n): k=5, n
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More Examples
^
How about
n
2 -3n+10?
^
It is
O(n
2 ) if
there exist k and n
such that 0
kn
2 ≥ n
2 -3n+10 for all n ≥
n
0
^ We see (fig.) that:
3 n
2 ≥ n
2 -3n+10 for all n ≥
^ So
k=3, n
^ More k
- n^0
pairs could be found, but finding just one is
More k
- n^0
pairs could be found, but finding just one is
enough to prove that
(^2) n -3n+10 is O(n
2 )
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Worst-case vs. Average-case Analyses
^
An algorithm can require different times tosolve different problems of the same size.
^
Worst-case analysis
= find the maximum
number of operations an algorithm canexecute in all situations^ ^
Worst
- case analysis is
easier to calculate
^
Worst
- case analysis is
easier to calculate
^
More common ^
Average-case analysis
= enumerate all
possible situations, find the time of each ofthe m possible cases, total and divide by m^ ^
Average-case analysis is
harder to compute
^
Yields a
more realistic
expected behavior
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Bigger Example: Analysis of Selection Sort
Divides the array into twoparts: already sorted, and notyet sorted.
values [ 0 ]
[ 1 ]
On each pass, finds thesmallest of the unsortedelements, and swap it into itscorrect place,
thereby
increasing the number ofsorted elements by one.
[ 2 ] [ 3 ] [ 4 ]
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Selection Sort: End of Pass One
values [ 0 ]
[ 1 ]
U
SORTED
[ 2 ] [ 3 ] [ 4 ]
U N S O R T E D
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Selection Sort: Pass Two
SORTED
values [ 0 ]
[ 1 ] [ 2 ] [ 3 ]
U N S O R
[ 3 ] [ 4 ]
R T E D
To find the smallest in UNSORTED:indexMin =
1
comp. 1:
check if values[2] = 10 < values[indexMin] = 24 - yes => indexMin =
2
comp. 2:
check if values[3] = 36 < values[indexMin] = 10 - NO
comp. 3:
check if values[4] = 12 < values[indexMin] = 10 - NO
Thus indexMin =
2;^
swap values[1] = 24 with values[indexMin] = 10 – see next slide
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Selection Sort: Pass Three
values [ 0 ]
[ 1 ] [ 2 ]
U N S
SORTED
[ 3 ] [ 4 ]
S O R T E D
To find the smallest in UNSORTED:indexMin =
2
comp. 1:
check if values[3] = 36 < values[indexMin] = 24 - NO
comp. 2:
check if values[4] = 12 < values[indexMin] = 24 -
yes => indexMin =
4
Thus indexMin =
4;^
swap values[2] = 24 with values[indexMin] = 12 – see next slide
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Selection Sort: End of Pass Three
values [ 0 ]
[ 1 ]
S O R
[ 2 ] [ 3 ] [ 4 ]
T E D
UNSORTED
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