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Chapter 2: Algorithm Analysis. Time Complexity. Big-O, Big- , Big-, Little-O. Running Time Calculation. Analyzing Specific Functions. CS 340. Page 10. Time Complexity.

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Chapter 2: Algorithm Analysis


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slide1

Chapter 2: Algorithm Analysis

  • Time Complexity
  • Big-O, Big-, Big-, Little-O
  • Running Time Calculation
  • Analyzing Specific Functions

CS 340

Page 10

slide2

Time Complexity

A fundamental question when determining the efficiency of an algorithm is the relationship between the size of the problem and the amount of computation needed to solve it.

CS 340

Page 11

slide3

Big-O Notation

Function T(n) is said to be O(f (n)) if there are positive constants c and n0 such that T(n)  c f (n) for any n  n0 (i.e., T(n) is ultimately bounded above by c f (n)).

  • Example: n3 + 3n2 + 6n + 5 is O(n3). (Use c = 15 and n0 = 1.)

Proof: For n ≥ 1, n2 ≥ n and n3 ≥ n2(by mult. property of ineq.)

For n ≥ 1, n3 ≥ n and n3 ≥ 1 (by transitive law of ≥)

For n ≥ 1, 3n3 ≥ 3n2, 6n3 ≥ 6n, and 5n3 ≥ 5 (by mult. property of ineq.)

For n ≥ 1, n3 + 3n3 + 6n3 + 5n3 ≥ n3 + 3n2 + 6n + 5 (by add. property of ineq.)

For n ≥ 1, 15n3 ≥ n3 + 3n2 + 6n + 5 (by distrib. law of * over +)

  • Example: n2 + nlogn is O(n2). (Use c = 2 and n0 = 1.)

Proof: For n ≥ 1, n ≥ logn(by inductive proof)

For n ≥ 1, n2 ≥ nlogn(by mult. property of ineq.)

For n ≥ 1, n2 + n2 ≥ n2 + nlogn(by add. property of ineq.)

For n ≥ 1, 2n2 ≥ n2 + nlogn(by distrib. law of * over +)

CS 340

Page 12

slide4

g(n)

r(n)

ng

nr

An Intuitive Big-O Example

r(n) is O(g(n)) since 1*g(n) exceeds r(n) for all n-values past ng

g(n) is O(r(n)) since 3*r(n) exceeds g(n) for all n-values past nr

CS 340

Page 13

slide5

ALGORITHM

A

B

10

1,110

11,110

Input

Size n

100

1,010,100

2,010,100

1,000

1,001,001,000

1,101,001,000

10,000

1,000,100,010,000

1,010,100,010,000

100,000

1,000,010,000,100,000

1,001,010,000,100,000

1,000,000

1,000,001,000,001,000,000

1,000,101,000,001,000,000

Demonstrating The Big-O Concept

Each of the algorithms below has O(n3) time complexity...

(In fact, the execution time for Algorithm A is n3 + n2 + n, and the execution time for Algorithm B is n3 + 101n2 + n.)

CS 340

Page 14

slide6

ALGORITHM

C

D

Input

Size n

10

123

10,123

100

10,203

110,203

1,000

1,002,003

2,002,003

10,000

100,020,003

110,020,003

100,000

10,000,200,003

10,100,200,003

1,000,000

1,000,002,000,003

1,001,002,000,003

A Second Big-O Demonstration

Each of the algorithms below has O(n2) time complexity...

(In fact, the execution time for Algorithm C is n2 + 2n + 3, and the execution time for Algorithm D is n2 + 1002n + 3.)

CS 340

Page 15

slide7

ALGORITHM

E

F

Input

Size n

10

83

1,083

100

1,164

11,164

1,000

14,966

114,966

10,000

182,877

1,182,877

100,000

2,160,964

12,160,964

1,000,000

24,931,569

124,931,569

One More Big-O Demonstration

Each of the algorithms below has O(n logn) time complexity…

(In fact, the execution time for Algorithm E is nlogn + 5n, and the execution time for Algorithm F is nlogn + 105n. Note that the linear term for Algorithm F will dominate until n reaches 2105.)

CS 340

Page 16

slide8

g(n)

2*r(n)

r(n)

1*g(n)

r(n)

g(n)

Big-O Represents An Upper Bound

If T(n) is O(f(n)), then f(n) is basically a cap on how bad T(n) will behave when n gets big.

Is r(n) O(g(n))?

Is g(n) O(r(n))?

r(n)  1*g(n) for all n≥ 0

g(n)  2*r(n) for all n≥ 0

YES!

Try c=1

starting at n=0

YES!

Try c=2

starting at n=0

CS 340

Page 17

slide9

2*v(n)

v(n)

v(n)

y(n)

1*y(n)

y(n)

n0

n0

Another Example

Is y(n) O(v(n))?

Is v(n) O(y(n))?

v(n)  1*y(n) for all n≥ n0

y(n)  2*v(n) for all n≥ n0

YES!

Try c=1

after 3rd cross

YES!

Try c=2

after 2nd cross

CS 340

Page 18

slide10

c*b(n)

p(n)

p(n)

1*p(n)

b(n)

b(n)

One More Example

Is p(n) O(b(n))?

Is b(n) O(p(n))?

b(n)  1*p(n) for all n≥ 0

NO!

b(n) returns

to 0 for any c!

YES!

Try c=1

starting at n=0

CS 340

Page 19

slide11

g(n)

r(n)

nr

Big- (Big-Omega) Notation

Function T(n) is said to be (g(n)) if there are positive constants c and n0 such that T(n)  c g (n) for any n  n0(i.e., T(n) is ultimately bounded below by c g (n)).

Example: n3 + 3n2 + 6n + 5 is (n3). (Use c = 1 and n0 = 1.)

Example: n2 + nlogn is (n2). (Use c = 1 and n0 = 1.)

r(n) is not(g(n)) since for every positive constant c, c*g(n) ultimately gets bigger than r(n)

g(n) is (r(n)) since g(n) exceeds 1*r(n) for all n-values past nr

CS 340

Page 20

slide12

r(n)

g(n)

n0

Big- (Big-Theta) Notation

Function T(n) is said to be (h(n)) if T(n) is both O(h(n)) and (h(n)).

Example: n3 + 3n2 + 6n + 5 is (n3).

Example: n2 + nlogn is (n2).

r(n) is (g(n)) since r(n) is squeezed between 1*g(n) and 2*g(n) once n exceeds n0

g(n) is (r(n)) since g(n) is squeezed between ½*r(n) and 1*r(n) once n exceeds n0

CS 340

Page 21

slide13

Little-O Notation

Function T(n) is said to be o(p(n)) if T(n) is O(p(n)) but not(p(n)).

Example: n3 + 3n2 + 6n + 5 is O(n4). (Use c = 15 and n0 = 1.) However, n3 + 3n2 + 6n + 5 is not (n4).

Proof (by contradiction):

Assume that there are positive constants c and n0 such that n3 + 3n2 + 6n + 5  c n4 for all n  n0.

Then dividing by n4 on both sides yields the fact that (1/n)+(3/n2)+(6/n3)+(5/n4)  c, for all n  n0.

Since limn((1/n)+(3/n2)+(6/n3)+(5/n4)) = 0, we must conclude that 0  c, which contradicts the fact that c must be a positive constant.

CS 340

Page 22

slide15

Computational Model For Algorithm Analysis

To formally analyze the performance of algorithms, we will use a computational model with a couple of simplifying assumptions:

Each simple instruction (assignment, comparison, addition, multiplication, memory access, etc.) is assumed to execute in a single time unit.

(For example, accessing a one-dimensional array element A[i] requires a multiplication, an addition, and a memory access – three time units in all. Similarly, accessing a two-dimensional array element B[j,k] requires two multiplications, two additions, and one memory access – five time units in all.)

Memory is assumed to be limitless, so there is always room to store whatever data is needed.

The size of the input, n, will normally be used as our main variable, and we’ll primarily be interested in “worst case” scenarios.

CS 340

Page 24

slide16

General Rules For Running Time Calculation

Rule One: Loops

The running time of a loop is at most the running time of the statements inside the loop, multiplied by the number of iterations.

Example:

for (i = 0; i < n; i++) // n iterations,

A[i] = (1-t)*X[i] + t*Y[i]; // 15 time units

// per iteration

(Retrieving X[i] requires one multiplication, one addition, and one memory access, as does retrieving Y[i]; the calculation involves a subtraction, two multiplications, and an addition; assigning A[i] the resulting value requires one multiplication, one addition, and one memory access; and each loop iteration requires a comparison and either an assignment or an increment. This totals fifteen primitive operations.)

Thus, the total running time is 15n time units, i.e., this part of the program is O(n).

CS 340

Page 25

slide17

Rule Two: Nested Loops

The running time of a nested loop is at most the running time of the statements inside the innermost loop, multiplied by the product of the number of iterations of all of the loops.

Example:

for (i = 0; i < n; i++) // n iterations. 2 ops each

for (j = 0; j < n; j++) // n iterations, 2 ops each

C[i,j] = j*A[i] + i*B[j]; // 14 time units/iteration

(3 for retrieving A[i], 3 for retrieving B[j], 3 for the RHS arithmetic, 5 for assigning C[i,j].)

Total running time: ((14+2)n+2)n = 16n2+2ntime units, which is O(n2).

More complex example:

for (i = 0; i < n; i++) // n iter, 2 ops each

for (j = i; j < n; j++) // n-i iter, 2 ops each

C[j,i] = C[i,j] = j*A[i]+i*B[j]; // 19 time units/iter

Total running time:  i=0,n-1(2+ j=i, n-121) =

 i=0,n-1(2+21(n-i)) = 2n+21( i=0,n-1n -  i=0,n-1i) = 2n+21(n2 - ½n(n-1)) = 10.5n2+ 12.5n time units, which is also O(n2).

CS 340

Page 26

slide18

Rule Three: Consecutive Statements

The running time of a sequence of statements is merely the sum of the running times of the individual statements.

Example:

for (i = 0; i < n; i++)

{ // 28n time units

A[i] = (1-t)*X[i] + t*Y[i]; // for this

B[i] = (1-s)*X[i] + s*Y[i]; // entire loop

}

for (i = 0; i < n; i++) // (16n+2)n time

for (j = 0; j < n; j++) // units for this

C[i,j] = j*A[i] + i*B[j]; // nested loop

Total running time: 16n2+30n time units, i.e., this code is O(n2).

CS 340

Page 27

slide19

Rule Four: Conditional Statements

The running time of an if-else statement is at most the running time of the conditional test, added to the maximum of the running times of the if and else blocks of statements.

Example:

if (amt > cost + tax) // 2 time units

{

count = 0; // 1 time unit

while ((count<n) && (amt>cost+tax)) // At most n iter.,

// 4 TUs each, but

{ // (count<n) is

// evaluated when

// the (n+1)-st

// iteration is

// tried

amt -= (cost + tax); // 3 time units

count++; // 1 time unit

}

cout << “CAPACITY:” << count; // 2 time units

}

else

cout << “INSUFFICIENT FUNDS”; // 1 time unit

Total running time: 2 + max(1 + (4 + 3 + 1)n + 1 + 2, 1) = 8n + 6 time units, i.e., this code is O(n).

CS 340

Page 28

slide20

Analysis Of Binary Search Function

intbinsrch(constetype A[], constetype x, constint n)

{

int low = 0, high = n-1; // 3 time units

int middle; // 0 time units

while (low <= high) // 1 time unit per iteration

{

middle = (low + high)/2; // 3 time units

if (A[middle] < x) // 4 TU | <-- Worst Case

low = middle + 1; // 2 TU |

else if (A[middle] > x) // 4 TU | <-- Worst Case

high = middle - 1; // 2 TU | <-- Worst Case

else // 0 TU |

return middle; // 1 TU |

}

return -1; // If search is unsuccessful; 1 time unit.

}

In the worst case, the loop will keep dividing the distance between the low and high indices in half until they are equal, iterating at most logn times. Thus, the total running time is: 17logn+ 4 time units, which is O(logn).

CS 340

Page 29

slide21

Analysis Of Another Function: SuperFreq

etypeSuperFreq(constetype A[], constint n)

{

etypebestElement = A[0]; // 4 time units

intbestFreq = 0; // 1 time unit

intcurrFreq; // 0 time units

for (i = 0; i < n; i++) // n iterations; 2 TUs each

{

currFreq = 0; // 1 time unit

for (j = i; j < n; j++) // n-i iterations; 2 TUs each

if (A[i] == A[j]) // 7 time units

currFreq++; // 1 time unit

if (currFreq > bestFreq) // 1 time unit

bestElement = A[i]; // 4 time units

}

return bestElement; // 1 time unit

}

Note that the function is obviously O(n2) due to its familiar nested loop structure. Specifically, its worst-case running time is 6+ i=0,n-1(8+ j=i, n-110) = 5n2 + 13n + 6.

CS 340

Page 30

slide22

What About Recursion?

humongIntpow(consthumongInt &val, consthumongInt &n)

{

if (n == 0)

return humongInt(0);

if (n == 1)

return val;

if (n % 2 == 0)

return pow(val*val, n/2);

return pow(val*val, n/2) * val;

}

The worst-case running time would require all 3 conditions to be checked, and to fail (taking 4 time units for the comparisons and arithmetic).

The last return statement requires 3 time units for arithmetic operations each time it’s executed, which happens logn – 1 times (since it halves n with each execution, until it reaches a value of 1).

When the parameterized n-value finally reaches 1, three last operations are performed.

Thus, the worst-case running time is 7logn - 4.

CS 340

Page 31

slide23

Recurrence Relations To Evaluate Recursion

int powerOf2(constint &n)

{

if (n == 0)

return 1;

return powerOf2(n-1) + powerOf2(n-1);

}

Assume that there is a function T(n) such that it takes T(k) time to execute powerOf2(k). Examining the code allows us to conclude the following:

T(0) = 2

T(k) = 5 + 2T(k-1) for all k > 0

The second fact tells us that:

T(n) = 5 + 2T(n-1) = 5 + 2(5 + 2T(n-2))

= 5 + 2(5 + 2(5 + 2(T(n-3)))) = …

= 5(1 + 2 + 22 + 23 + … + 2n-1) + 2nT(0)

= 5(2n-1) + 2n(2) = 7(2n) - 5, which is O(2n).

CS 340

Page 32

slide24

Another Recurrence Relation Example

int alternatePowerOf2(constint &n)

{

if (n == 0)

return 1;

return 2*alternatePowerOf2(n-1);

}

Assume that there is a function T(n) such that it takes T(k) time to execute alternatePowerOf2(k). Examining the code allows us to conclude that:

T(0) = 2

T(k) = 4 + T(k-1) for all k > 0

The second fact tells us that:

T(n) = 4 + T(n-1) = 4 + (4 + T(n-2))

= 4 + (4 + (4 + (T(n-3)))) = …

= 4n + T(0) = 4n + 2, which is O(n).

CS 340

Page 33