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CS 213
Data Structures and Algorithms
Second Semester, 2011-2012
Sum of Squares
int sumOfSquares(int n){
int sum = 0;
for(int i=1; i<=n; i++)
sum = sum + i*i;
return sum;
}T(n) = 5n + 4
Selection Sort
Given the following numbers in an array:53 23 10 34 2 17
2 23 10 34 53 17 2 10 23 34 53 17 2 10 17 34 53 23 2 10 17 23 53 34 2 10 17 23 34 53
Still on Selection Sortvoid selectionSort(int A[], int n){
for(int i=0; i<n; i++){ int min = i; for(int j=i+1; j<n; j++){
if(A[j] < A[min]) min = j;}int temp = A[i];A[i] = A[min];A[min] = temp;
}}
What is this function’s T(n)? What we need is to compute for T(n) from the inner loop going outwards.
We need to use the summation notation to solve the T(n).
The Summation Notation
= x1 + x2 + x3 + x4
Still on the summation notation
2)1(
1
nnn
i
i∑4
i=1
i = 1+2+3+4 = 10
∑5
i=1
3 = 3+3+3+3+3 = 15 ∑n
i=1
c =nc =15
Exercise for(int i=1; i<=5; i++) cout<<endl; for(int i=1; i<=n; i++) cout<<endl; for(int i=3; i<=m; i++) cout<<endl; for(int i=1; i<=6; i++)
for(int j=1; j<=8; j++) cout<<endl; for(int i=1; i<=n; i++)
for(int j=1; j<=n; j++) cout<<endl; for(int i=1; i<=m; i++)
for(int j=1; j<=m; j++) cout<<endl; for(int i=1; i<=n; i++)
for(int j=i; j<=n; j++) cout<<endl;
Going back to the Selection Sortvoid selectionSort(int A[], int n){
for(int i=0; i<n; i++){ int min = i; for(int j=i+1; j<n; j++){
if(A[j] < A[min]) min = j;}int temp = A[i];A[i] = A[min];A[min] = temp;
}}
∑n
j=i+1
c∑n
i=1
=∑n
i=1
c(n-i+1-1)
∑n
i=1
c(n-i) = c ∑n
i=1
n-i
∑n
i=1
n-i = ∑n
i=1
n ∑n
i=1
i-
Time Complexity
A function that maps problem size into the time required to solve the problem.
Typically, we are interested in the inherent complexity of computing the solution to problems in a particular class.
Lower Bound
We might want to know how fast we can sort a list of n items, initially in an arbitrary order, regardless of the algorithm used.
What is sought here is the lower bound, L(n), on sorting, a property of the sorting problem and not of any particular algorithm.
This says that no algorithm can do the job in fewer than L(n) time units for arbitrary inputs.
Upper Bound
We might also like to know how long it would take to sort such list using a known algorithm with a worst-case input.
What is sought here is the upper bound, U(n), which says that for arbitrary inputs we can always sort in time at most U(n).
Goal of Time Complexity Analysis
While there are apparently two complexity functions for problems, L(n) and U(n), the ultimate goal is to make these two bounds coincide.
This is the optimal algorithm which has L(n) = U(n). For some of the problems, this goal has not been
realized yet!
Invitation
Consider this, CS 213, as you journey into finding optimal solutions to classes of problems!
Who knows, you might win a million dollars ($$$) from Claymath Foundation!
Upper Bound Complexity
There are two ways in analyzing this bound:Counting instructionsSolving recurrences
Both are used to find the worst case of an algorithm.
Big O-Notation (O(g(n)))
The O-notation is used to describe the worst-case running time of an algorithm.
O(n) means that the growth of the running time of the algorithm is a function of n.
O-notation computes for the upper bound.
O-Notation Defined
O(g(n)) = {f(n): c>0, n0>0 s.t. 0 f(n) cg(n) for all n n0}.
Example: Check if (n2/2) – 3n O(n2)(n2/2) – 3n cn2
½ - 3/n cChoosing c = ½, n0 = 6 proves the claim.
Another Example
3n2 - 100n + 6 O(n2) ?3n2 - 100n + 6 cn2 3 – 100/n + 6/n2 cAt this point, we have to choose a c>0 and an
n0
What values will prove our claim?
Lower Bound Complexity
This is the more difficult of the bounds.There is no algorithm to analyze.(g(n)) is used to describe the lower bound of
the running time of the algorithm or minimum possible running time of the algorithm.
-Notation
(g(n)) = {f(n): c>0, n0>0 s.t. 0 cg(n) f(n) for all n n0}.
Example: Check if (n2/2) – 3n (n2)cn2 (n2/2) – 3n c ½ - 3/nChoosing c = 1/14, n0 = 7 proves the claim.
Another Example
Check if 3n2 - 100n + 6 (n)cn 3n2 - 100n + 6 c 3n – 100 + 6/n
At this point we need to find a c and an n0 that will prove our claim.
What values of c and n0 will suffice the inequality??
-Notation
Used to denote that the lower and upper bounds of the running time of the algorithm is tight, i.e. the growth rate of the upper and lower bounds are the same.
-Notation Defined
(g(n)) = {f(n): c1>0, c2>0, n0>0 s.t. 0 c1g(n) f(n) c2g(n) for all n n0}.
f(n) (g(n)) if f(n) O(g(n)) and f(n) (g(n))
Complexity Classes
Description O-notation
constant O(1)
logarithmic O(log n)
linear O(n)
n log n O(n log n)
quadratic O(n2)
cubic O(n3)
polynomial O(nk), k1
exponential O(an), a>1
Growth rate of complexity classes
class n=2 n=16 n=256 n=1024
1 1 1 1 1
log n 1 4 8 10
n 2 16 256 1024
n log n 2 64 2048 10240
n^2 4 256 65536 1048576
n^3 8 4096 16777216 1.07E+09
2^n 4 65536 1.16E+77 1.8E+308
Graph of the Growth Rates
log n
n
n log n
n 2̂
n 3̂
2 n̂
0
5
10
15
20
25
30
35
40
45
50
log n 0 1 1.58 2 2.32 2.58 2.81 3 3.17 3.32 3.46
n 1 2 3 4 5 6 7 8 9 10 11
n log n 0 2 4.75 8 11.61 15.51 19.65 24 28.53 33.22 38.05
n 2̂ 1 4 9 16 25 36 49 64 81 100 121
n 3̂ 1 8 27 64 125 216 343 512 729 1000 1331
2 n̂ 2 4 8 16 32 64 128 256 512 1024 2048
1 2 3 4 5 6 7 8 9 10 11
Bigger N
log nnn log nn 2̂
n 3̂
2 n̂
0
200
400
600
800
1000
1200
1400
1600
1800
2000
log n 0 1 1.58 2 2.32 2.58 2.81 3 3.17 3.32 3.46
n 1 2 3 4 5 6 7 8 9 10 11
n log n 0 2 4.75 8 11.61 15.51 19.65 24 28.53 33.22 38.05
n 2̂ 1 4 9 16 25 36 49 64 81 100 121
n 3̂ 1 8 27 64 125 216 343 512 729 1000 1331
2 n̂ 2 4 8 16 32 64 128 256 512 1024 2048
1 2 3 4 5 6 7 8 9 10 11
Still on the Graph
100*log n
50*n
20*n log n
10*n 2̂n 3̂
2 n̂
0
200
400
600
800
1000
1200
1400
1600
1800
2000
100*log n 0 100 158.5 200 232.19 258.5 280.74 300 316.99 332.19 345.94
50*n 50 100 150 200 250 300 350 400 450 500 550
20*n log n 0 40 95.1 160 232.19 310.2 393.03 480 570.59 664.39 761.07
10*n 2̂ 10 40 90 160 250 360 490 640 810 1000 1210
n 3̂ 1 8 27 64 125 216 343 512 729 1000 1331
2 n̂ 2 4 8 16 32 64 128 256 512 1024 2048
1 2 3 4 5 6 7 8 9 10 11