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**Chapter 10 Algorithm Efficiency**

CS Data Structures Mehmet H Gunes Modified from authors’ slides

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**Contents What Is a Good Solution?**

Measuring the Efficiency of Algorithms Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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Algorithm Efficiency There are often many approaches (algorithms) to solve a problem. How do we choose between them? At the heart of computer program design are two (sometimes conflicting) goals To design an algorithm that is easy to understand, code, debug. makes efficient use of the resources.

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**Algorithm Efficiency (cont)**

Goal (1) is the concern of Software Engineering. Goal (2) is the concern of data structures and algorithm analysis. When goal (2) is important, how do we measure an algorithm’s cost?

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What Is a Good Solution? Criterion A solution is good if the total cost it incurs over all phases of its life is minimal. Keep in mind, efficiency is only one aspect of a solution’s cost Note: Relative importance of various components of a solution’s cost has changed since early days of computing. Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Measuring Efficiency of Algorithms**

Comparison of algorithms should focus on significant differences in efficiency Difficulties with comparing programs instead of algorithms How are the algorithms coded? What computer should you use? What data should the programs use? Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Execution Time of Algorithm**

Traversal of linked nodes – example: Displaying data in linked chain of n nodes requires time proportional to n Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Algorithm Growth Rates**

Measure algorithm’s time requirement as a function of problem size Compare algorithm efficiencies for large problems Look only at significant differences. Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Algorithm Growth Rates**

Time requirements as a function of the problem size n Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Best, Worst, Average Cases**

Not all inputs of a given size take the same time to run. Sequential search for K in an array of n integers: Begin at first element in array and look at each element in turn until K is found Best case: Worst case: Average case: Best: Find at first position. Cost is 1 compare. Worst: Find at last position. Cost is n compares. Average: (n+1)/2 compares IF we assume the element with value K is equally likely to be in any position in the array.

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Time Analysis Provides upper and lower bounds of running time.

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**Worst Case Provides an upper bound on running time.**

An absolute guarantee that the algorithm would not run longer, no matter what the inputs are.

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**Best Case Provides a lower bound on running time.**

Input is the one for which the algorithm runs the fastest.

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**Average Case Provides an estimate of “average” running time.**

Assumes that the input is random. Useful when best/worst cases do not happen very often i.e., few input cases lead to best/worst cases.

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Which Analysis to Use? While average time appears to be the fairest measure, it may be difficult to determine. When is the worst case time important? Average time analysis requires knowledge of distributions. For example, the assumption of distribution used for average case in the last example. Worst-case time is important for real-time algorithms.

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**How to Measure Efficiency?**

Critical resources: Time, memory, battery, bandwidth, programmer effort, user effort Factors affecting running time: For most algorithms, running time depends on “size” of the input. Running time is expressed as T(n) for some function T on input size n. Empirical comparison is difficult to do “fairly” and is time consuming. Critical resources: Time. Space (disk, RAM). Programmers effort. Ease of use (user’s effort). Factors affecting running time: Machine load. OS. Compiler. Problem size. Specific input values for given problem size. 16

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**How do we analyze an algorithm?**

Need to define objective measures. (1) Compare execution times? Empirical comparison (run programs) Not good: times are specific to a particular machine. (2) Count the number of statements? Not good: number of statements varies with programming language and programming style.

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**How do we analyze an algorithm? (cont.)**

(3) Express running time t as a function of problem size n (i.e., t=f(n) ) Asymptotic Algorithm Analysis Given two algorithms having running times f(n) and g(n), find which functions grows faster - Such an analysis is independent of machine time, programming style, etc.

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Comparing algorithms Given two algorithms having running times f(n) and g(n), how do we decide which one is faster? Compare “rates of growth” of f(n) and g(n)

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**Understanding Rate of Growth**

The low order terms of a function are relatively insignificant for large n n n2 + 10n + 50 Approximation: n4 Highest order term determines rate of growth!

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**Visualizing Orders of Growth**

On a graph, as you go to the right, a faster growing function eventually becomes larger... fA(n)=30n+8 Value of function fB(n)=n2+1 Increasing n

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**Rate of growth The graphs of 3 n2 and n2 - 3 n + 10**

Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Rate of Growth ≡ Asymptotic Analysis**

Using rate of growth as a measure to compare different functions implies comparing them asymptotically i.e., as n If f(x) is faster growing than g(x), then f(x) always eventually becomes larger than g(x) in the limit i.e., for large enough values of x

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Complexity Let us assume two algorithms A and B that solve the same class of problems. The time complexity of A is 5,000n, the one for B is 1.1n for an input with n elements For n = 10, A requires 50,000 steps, but B only 3, so B seems to be superior to A. For n = 1000, A requires 5106 steps, while B requires 2.51041 steps.

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**Names of Orders of Magnitude**

O(1) bounded (by a constant) time O(log2N) logarithmic time O(N) linear time O(N*log2N) N*log2N time O(N2) quadratic time O(N3) cubic time O(2N ) exponential time

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**Order of growth of common functions**

Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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Example

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Growth Rate Graph The lower graph corresponds to the box within the dashed lines in the lower left corner of the upper graph.

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**A comparison of growth-rate functions**

Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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N log2N N*log2N N N ,536 *109 *1019 , *1038

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**A comparison of growth-rate functions**

Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Sample Execution Times**

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**The Growth of Functions**

A problem that can be solved with polynomial worst-case complexity is called tractable Problems of higher complexity are called intractable Problems that no algorithm can solve are called unsolvable

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**Analysis and Big O Notation**

Definition: Algorithm A is order f ( n ) Denoted O( f ( n )) If constants k and n0 exist Such that A requires no more than k f ( n ) time units to solve a problem of size n ≥ n0 . Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Asymptotic Notation O notation: asymptotic “less than”:**

f(n)=O(g(n)) implies: f(n) “≤” c g(n) in the limit* c is a constant (used in worst-case analysis) *formal definition in CS477/677

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**Asymptotic Notation notation: asymptotic “greater than”:**

f(n)= (g(n)) implies: f(n) “≥” c g(n) in the limit* c is a constant (used in best-case analysis) *formal definition in CS477/677

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**Asymptotic Notation notation: asymptotic “equality”:**

f(n)= (g(n)) implies: f(n) “=” c g(n) in the limit* c is a constant (provides a tight bound of running time) (best and worst cases are same) *formal definition in CS477/677

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**More on big-O O(g(n)) is a set of functions f(n)**

f(n) ϵ O(g(n)) if “f(n)≤cg(n)”

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**A Common Misunderstanding**

Confusing worst case with upper bound. Upper bound refers to a growth rate. Worst case refers to the worst input from among the choices for possible inputs of a given size.

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**Algorithm speed vs function growth**

An O(n2) algorithm will be slower than an O(n) algorithm (for large n). But an O(n2) function will grow faster than an O(n) function. fA(n)=30n+8 Value of function fB(n)=n2+1 Increasing n

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**Faster Computer or Algorithm?**

Suppose we buy a computer 10 times faster. n: size of input that can be processed in one second on old computer in 1000 computational units n’: size of input that can be processed in one second on new computer T(n) n n’ Change n’/n 10n 100 1,000 n’ = 10n 10 10n2 31.6 n’= 10n 3.16 3 4 n’ = n + 1 1 + 1/n How much speedup? 10 times. More important: How much increase in problem size for same time expended? That depends on the growth rate. n: Size of input that can be processed in one hour (10,000 steps). n’: Size of input that can be processed in one our on the new machine (100,000 steps). Note: for 2n, if n = 1000, then n’ would be 1003.

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**How do we find f(n)? Associate a "cost" with each statement**

(2) Find total number of times each statement is executed (3) Add up the costs

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**Properties of Growth-Rate Functions**

Ignore low-order terms Ignore a multiplicative constant in the high-order term O(f(n)) + O(g(n)) = O(f(n) + g(n)) Be aware of worst case, average case Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Keeping Your Perspective**

Array-based getEntry is O(1) Link-based getEntry is O(n) Consider how frequently particular ADT operations occur in given application Some seldom-used but critical operations must be efficient Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Keeping Your Perspective**

If problem size always small, ignore an algorithm’s efficiency Weigh trade-offs between algorithm’s time and memory requirements Compare algorithms for both style and efficiency Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Efficiency of Searching Algorithms**

Sequential search Worst case O(n) Average case O(n) Best case O(1) Binary search of sorted array Worst case O(log2n) Remember required overhead for keeping array sorted Data Structures and Problem Solving with C++: Walls and Mirrors, Carrano and Henry, © 2013

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**Running Time Examples The body of the while loop: O(N)**

while (i<N) { X=X+Y; // O(1) result = mystery(X); // O(N) i++; // O(1) } The body of the while loop: O(N) Loop is executed: N times N x O(N) = O(N2)

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**Running Time Examples if (i<j) for ( i=0; i<N; i++ ) X = X+i;**

else X=0; Max ( O(N), O(1) ) = O (N) O(N) O(1)

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**Running Time Examples (cont.’d)**

Algorithm 1 Cost arr[0] = 0; c1 arr[1] = 0; c1 arr[2] = 0; c1 ... arr[N-1] = 0; c1 c1+c1+...+c1 = c1 x N Algorithm Cost for(i=0; i<N; i++) c2 arr[i] = 0; c1 (N+1) x c2 + N x c1 = (c2 + c1) x N + c2

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**Running Time Examples (cont.’d)**

Cost sum = 0; c1 for(i=0; i<N; i++) c2 for(j=0; j<N; j++) c2 sum += arr[i][j]; c3 c1 + c2 x (N+1) + c2 x N x (N+1) + c3 x N x N 50

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**Complexity Examples What does the following algorithm compute?**

int who_knows(int a[n]) { int m = 0; for {int i = 0; i<n; i++} for {int j = i+1; j<n; j++} if ( abs(a[i] – a[j]) > m ) m = abs(a[i] – a[j]); return m; } returns the maximum difference between any two numbers in the input array Comparisons: n-1 + n-2 + n-3 + … + 1 = (n-1)n/2 = 0.5n n Time complexity is O(n2)

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**Complexity Examples Another algorithm solving the same problem:**

int max_diff(int a[n]) { int min = a[0]; int max = a[0]; for {int i = 1; i<n; i++} if ( a[i] < min ) min = a[i]; else if ( a[i] > max ) max = a[i]; return max-min; } Comparisons: 2n - 2 Time complexity is O(n).

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**Examples of Growth Rate**

Position of largest value in "A“ */ static int largest(int[] A) { int currlarge = 0; // Position of largest for (int i=1; i<A.length; i++) if (A[currlarge] < A[i]) currlarge = i; // Remember position return currlarge; // Return largest position } As n grows, how does T(n) grow? Cost: T(n) = c1n + c2 steps

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**Examples (cont) sum = 0; for (i=1; i<=n; i++)**

for (j=1; j<=n; j++) sum++; Example 3: Cost: T(n) = c1n2 + c2. Roughly n2 steps, with sum being n2 at the end. Ignore various overhead such as loop counter increments.

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**Time Complexity Examples (1)**

a = b; This assignment takes constant time, so it is (1). sum = 0; for (i=1; i<=n; i++) sum += n; Asymptotic analysis is defined for equations. We need to convert programs to equations to analyze them. The traditional notation is (1), not (c). (n) even though the value of sum is n2.

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**Time Complexity Examples (2)**

sum = 0; for (j=1; j<=n; j++) for (i=1; i<=j; i++) sum++; for (k=0; k<n; k++) A[k] = k; First statement is (1). Double for loop is i = (n2). Final for loop is (n). Result: (n2).

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**Time Complexity Examples (3)**

sum1 = 0; for (i=1; i<=n; i++) for (j=1; j<=n; j++) sum1++; sum2 = 0; for (j=1; j<=i; j++) sum2++; The only difference is j<=n vs. j<= i. First loop, sum is n2. Second loop, sum is (n+1)(n)/2. Both are (n2).

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**Time Complexity Examples (4)**

sum1 = 0; for (k=1; k<=n; k*=2) for (j=1; j<=n; j++) sum1++; sum2 = 0; for (j=1; j<=k; j++) sum2++; First loop is n for k = 1 to log n, or (n log n). Second loop is 2k for k = 0 to log n - 1, or (n).

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Example

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**Analyze the complexity of the following code**

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