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Chapter 2: Complexity AnalysisPowerPoint Presentation

Chapter 2: Complexity Analysis

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Objectives

Looking ahead – in this chapter, we’ll consider:

- Computational and Asymptotic Complexity
- Big-O Notation
- Properties of the Big-O Notation
- Ω and Θ Notations
- Possible Problems with the Notation

Data Structures and Algorithms in C++, Fourth Edition

Objectives (continued)

Examples of Complexities

Finding Asymptotic Complexity

Best, Average, and Worst Cases

Amortized Complexity

NP-Completeness

Data Structures and Algorithms in C++, Fourth Edition

Computational and Asymptotic Complexity

Algorithms are an essential aspect of data structures

Data structures are implemented using algorithms

Some algorithms are more efficient than others

Efficiency is preferred; we need metrics to compare them

An algorithm’s complexity is a function describing the efficiency of the algorithm in terms of the amount of data the algorithm must process

There are two main complexity measures of efficiency

Data Structures and Algorithms in C++, Fourth Edition

Computational and Asymptotic Complexity (continued)

Time complexitydescribes the amount of time an algorithm takes in terms of the amount of input

Space complexitydescribes the amount of memory (space) an algorithm takes in terms of the amount of input

For both measures, we are interested in the algorithm’s asymptotic complexity

This asks: whenn (number of input items) goes to infinity, what happens to the algorithm’s performance?

Data Structures and Algorithms in C++, Fourth Edition

Computational and Asymptotic Complexity (continued)

To illustrate this, consider f(n) = n2 + 100n + log10n + 1000

As the value of n increases, the importance of each term shifts until for large n, only the n2 term is significant

Fig. 2-1The growth rate of all terms of function f (n) = n2 + 100n + log10n + 1,000.

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation

The most commonly used notation for asymptotic complexity used is "big-O" notation

In the previous example we would say n2 + 100n + log10n + 1000 = O(n2) (read "big-oh of n squared")

Definition: Let f(n) and g(n) be functions, wheren ε Z is a positive integer. We write f(n) = O(g(n)) if and only if there exists a real number c and positive integer N satisfying 0 <f(n) <cg(n) for all n>N. (And we say, "f of n is big-oh of g of n.“)

This means that functions like n2 + n, 4n2 - n log n + 12, n2/5 - 100n, n log n, and so forth are all O(n2)

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

Although the definition of big-O is correct, it lacks important information

While cand N exist, it does not tell us how to calculate them or what to do if multiple candidates exist (and they often do)

Consider the function f:

f (n) = 2n2 + 3n + 1

and g:

g(n) = n2

Clearly f (n) is O(n2); possible candidates for c and N are shown in the next slide

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

Fig. 2.2Different values of c and N for function f (n) = 2n2 + 3n + 1 = O(n2) calculated accordingto the definition of big-O

These values are obtained by solving the inequality from the definition of big-O:

f(n) <cg(n)

Substituting for f(n) and g(n) from the previous slide, we have:

2n2 + 3n + 1 <cn2 or 2 + 3/n + 1 / n2 <c

Since n>N, and N is a positive integer, we can start with N = 1 and substitute in either expression to obtain c

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

Generally, we choose an N that allows one term of f to dominate the expression

There are only two terms to consider: 2n2 and 3n, since the last term is a constant

As long as n is greater than 1.5, 2n2 dominates the expression

So N must be 2 or more, and c is greater than 3.75

This illustrates the central point of the definition of big-O, that f(n) <cg(n) relies on the choices of c and N as stated above

The choice of c depends on the choice of N and vice-versa

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

For example, if we wanted the relationship to hold for all positive Nstarting at 1, c would have to be 6

The graphs of f(n) and cg(n)are shown below:

Fig. 2.3Comparison of functions for different values of c and N from Figure 2.2

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

The graph of f(n) crosses the graph of each cg(n) at N

This does not mean other values of c and N are not useful; we can choose to start with any c, provided we choose the correct N

We can also use this information to determine the big-O of a given function

Consider the function 3n2 + 4n - 2

We need a c and N such that:

3n2 + 4n - 2 <cn2

for all n>N

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

Dividing by n2 gives us

3 + 4/n – 2/n2<c

Choosing N = 1, we need to find a c such that

3 + 4 – 2 <c

We can set c = 6, so we have

3n2 + 4n - 2 <6n2

for all n> 1

So our function is O(n2)

Big-O provides a formal method for expressing asymptotic upper bounds, bounding the growth of a function from above

Data Structures and Algorithms in C++, Fourth Edition

Big-O Notation (continued)

Knowing where a function lies within the big-O hierarchy lets us compare it quickly with other functions

Thus we have an idea of which algorithm has the best time performance

Data Structures and Algorithms in C++, Fourth Edition

Properties of Big-O Notation

The big-O notation is awkward to work with all the time

The following is a list of useful theorems you can use to simplify big-O calculations

Big-O is transitive: iff(n) = O(g(n)) and g(n) is O(h(n)), then f(n) = O(h(n))

Iff(n) = O(h(n)) and g(n) is O(h(n)), then f(n) + g(n) = O(h(n))

A function ank = O(nk) for any a > 0

Any kth degree polynomial is O(nk+j) for any j > 0

Data Structures and Algorithms in C++, Fourth Edition

Properties of Big-O Notation (continued)

f(n) = O(g(n)) is true if limn->∞ f(n)/g(n) is a constant. Put another way, if f(n) = cg(n), then f(n) = O(g(n))

logan = O(logbn) for any a, b > 1. This means, except for a few cases, we don’t care what base our logarithms are

Given the preceding, we can use just one base and rewrite the relationship as logan = O(lgn) for positive a ≠ 1 and lg n = log2n

Data Structures and Algorithms in C++, Fourth Edition

Ω and Θ Notations

Big-O only gives us the upper bound of a function

So if we ignore constant factors and let n get big enough, some function will never be bigger than some other function

This can give us too much freedom

Consider that selection sort isO(n3), sincen2 is O(n3) - butO(n2) is a more meaningful upper bound

We need alower bound, a function that always grows more slowly than f(n), and a tight bound, a function that grows at about the same rate asf(n)

Section 2.4 gives a good introduction to these concepts; let’s look at a different way to approach this

Data Structures and Algorithms in C++, Fourth Edition

Ω and Θ Notations (continued)

Big-Ω is for lower bounds what big-O is for upper bounds

Definition: Let f(n) and g(n) be functions, where n is a positive integer. We write f(n) = Ω(g(n)) if and only if g(n) = O(f(n)). We say "f of n is omega of g of n.“

So g is a lower bound for f ; after a certain n, and without regard to multiplicative constants, f will never go below g

Finally, theta notation combines upper bounds with lower bounds to get tight bound

Definition: Let f(n) and g(n) be functions, where n is a positive integer. We write f(n) = Θ(g(n)) if and only if g(n) = O(f(n)) and g(n) = (f(n)). We say "f of n is theta of g of n."

Data Structures and Algorithms in C++, Fourth Edition

Ω and Θ Notations (continued)

There are some additional theorems we can consider when we take Ω and Θ into account; the first four theorems for big-O are also true for Ω and Θ

ReplacingO with Ω and "largest" with "smallest" in the fifth theorem for big-O and it remains true

f(n) = Ω(g(n)) is true if limn->∞ g(n)/f(n) is a constant

f(n) = Θ(g(n)) is true if limn->∞ f(n)/g(n) is a non-zero constant

nk = O((1+ ε) n)) for any positive k and ε

This means any polynomial is bound from above by any exponential

Data Structures and Algorithms in C++, Fourth Edition

Ω and Θ Notations (continued)

So an algorithm that runs in polynomial time is (eventually) preferable to an algorithm that runs in exponential time

(log n)ε = O(nk) for any positive k and ε

This means a logarithm to any power grows more slowly than a polynomial

So an algorithm that runs in logarithmic time is (eventually) preferable to an algorithm that runs in polynomial (or from above, exponential) time

Data Structures and Algorithms in C++, Fourth Edition

Possible Problems

All the notations we’ve considered focus on comparing algorithms designed to solve the same problem

We still have to exercise care; it is possible at first glance to eliminate potentially useful candidate functions

Consider again the definition of big-O

f(n) = O(g(n)) if 0 <f(n) <cg(n)

The number of ns that violate this is finite and can be reduced by proper choice of c

But if c is extremely large, it can cause us to reject a function g even if the function itself is promising

Data Structures and Algorithms in C++, Fourth Edition

Possible Problems (continued)

Consider two algorithms that solve a problem, one requiring 108n steps (O(n)), the other 10n2 (O(n2))

Using big-O alone, we’d reject the second algorithm, because it grows too fast

But that is only true if n > 107 and often it is much smaller, implying the second algorithm would be faster than the first

So we need to consider other factors in our analysis

A “double-O” notation has been proposed in such cases

So f is OO(g(n)) if it is O(g(n)) and the constant c is too large to be useful, implying 108n is OO(g(n))

Data Structures and Algorithms in C++, Fourth Edition

Examples of Complexities

Since we examine algorithms in terms of their time and space complexity, we can classify them this way, too

This is illustrated in the next figure

Fig. 2.4Classes of algorithms and their execution times on a computer executing 1 million operations per second

(1 sec = 106 μsec = 103 msec)

Data Structures and Algorithms in C++, Fourth Edition

Examples of Complexities (continued)

Fig. 2.4(concluded)

The class of an algorithm is the name used to refer to its big-O notation; it is a more convenient way to describe its behavior

For example a linear function is O(n); its time increases in direct proportion to the amount of data processed

Data Structures and Algorithms in C++, Fourth Edition

Examples of Complexities (continued)

This relationship can also be expressed graphically:

Fig. 2.5Typical functions applied in big-O estimates.

This graph, and the previous chart, show that some algorithms have no practical application

Even with today’s supercomputers, cubic order algorithms or higher are impractical for large numbers of elements

Data Structures and Algorithms in C++, Fourth Edition

Finding Asymptotic Complexity

As we have seen, asymptotic bounds are used to determine the time and space efficiency of algorithms

Generally, we are interested in time complexity, which is based on assignments and comparisons in a program

We’ll focus on assignments for the time being

Consider a simple loop:

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

sum = sum + a[i]

Two assignments are executed once (sum = 0 and i = sum) during initialization

In the loop, sum = sum + a[i] is executed ntimes

Data Structures and Algorithms in C++, Fourth Edition

Finding Asymptotic Complexity(continued)

In addition, the i++ in the loop header is executed n times

So there are 2 + 2n assignments in this loop’s execution and it is O(n)

Typically, as loops are nested, the complexity grows by a factor of n, although this isn’t always the case

Consider

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

for (j = 1, sum = a[0]; j <= i; j++)

sum += a[j];

cout << ”sum for subarray 0 through “ << i

<<” is “<<sum<<end1;

}

Data Structures and Algorithms in C++, Fourth Edition

Finding Asymptotic Complexity(continued)

The outer loop initializes i, then executes n times

During each pass through the loop, the variable i is updated, and the inner loop and cout statement are executed

The inner loop initializes j and sum each time, so the number of assignments so far is 1 + 3n

The inner loop executes i times, where i ranges from 1 to n – 1, based on the outer loop (when i is 0, it doesn’t run)

Each time the inner loop executes, it increments j, and assigns a value to sum

So the inner loop executes = 2(1 + 2 + … + n – 1) = 2n(n – 1) assignments

Data Structures and Algorithms in C++, Fourth Edition

Finding Asymptotic Complexity(continued)

The total number of assignments is then 1 + 3n + 2n(n - 1), which is O(1) + O(n) + O(n2) = O(n2)

As mentioned earlier, not all loops increase complexity, so care has to be taken to analyze the processing that takes place

However, additional complexity can be involved if the number of iterations changes during execution

This can be the case in some of the more powerful searching and sorting algorithms

Data Structures and Algorithms in C++, Fourth Edition

Best, Average, and Worst Cases

- If we want to truly get a handle on the complexity of more complicated algorithms, we need to distinguish three cases:
- Worst case– the algorithm takes the maximum number of steps
- Best case – the algorithm takes the fewest number of steps
- Average case – performance falls between the extremes

- For simple situations we can determine the average case by adding together the number of steps required for each input and dividing by the number of inputs
- However, this is based on each input occurring with equal probability, which isn’t always likely

Data Structures and Algorithms in C++, Fourth Edition

Best, Average, and Worst Cases(continued)

- To be more precise, we need to weight the number of steps that occur for a given input by the probability of that input occurring, and sum this over the number of inputs:
- In probability theory, this defines the expected value, which assumes the probabilities can be determined and their distribution known
- Because p is a probability distribution, it satisfies two constraints:
- The function p can never be negative
- The sum of all the probabilities is equal to 1

Data Structures and Algorithms in C++, Fourth Edition

Best, Average, and Worst Cases(continued)

- Consider the example of sequentially searching an unordered array to find a target value
- The best and worst cases are straightforward:
- Best case occurs when we find the target in the first cell
- Worst case occurs when we find the target in the last cell, or not at all (but end up searching the entire array)

- For the average case, we first have to consider the probability of finding the target
- If we assume a uniform distribution of n values, then the probability of finding the target in any one location is

Data Structures and Algorithms in C++, Fourth Edition

Best, Average, and Worst Cases(continued)

So we would find the target in the first location with p = 1/n, in the second location with p = 1/n, etc.

Since the number of steps required to get to each location is the same as the location itself, our sum becomes:

1/n * (1 + 2 + … + n) = (n + 1) / 2

Again, this is based on an equally likely chance of finding the target in any cell

If the probabilities differ, then the computation becomes more involved

Data Structures and Algorithms in C++, Fourth Edition

Amortized Complexity

In many cases, data structures are manipulated by sequences of operations

As a consequence, operations early in the sequence can impact the performance of those later in the sequence

To determine overall performance, we could accumulate the performance for each sequence to determine the result

This can give very inaccurate results however

A more useful approach is to consider the entire sequence of operations of the program

This approach is referred to as amortized analysis

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Amortized Complexity (continued)

This allows us to determine a worst-case bound irrespective of the inputs by looking at all of the operations

The idea is that while some operations may be costly, they do not occur frequently enough to bias the entire program

This is because less costly operations will outnumber the costly ones in the long run, "paying back" the program over a number of iterations

This is particularly useful because rather than making assumptions about the program it guarantees worst-case performance

Data Structures and Algorithms in C++, Fourth Edition

Amortized Complexity (continued)

Consider a dynamic array application where we double the size of the array each time it fills up

Array reallocation may be required, so in the worst case insertion may beO(n)

Yet because remaining insertions are done in constant time, asequence of n insertions can always be done in O(n) time

Consequently, the n insertions can be completed in O(n) time

So theamortized time per operation is O(n) / n = O(1)

Data Structures and Algorithms in C++, Fourth Edition

NP-Completeness

Informally, adeterministic algorithmis one that behaves predictably

Given a particular input, the underlying algorithm has only one way to decide what step to perform at any given point

As opposed to this, a nondeterministic algorithm uses some type of operation to “guess” what to do next when a decision is made

Consequently, it can exhibit different behaviors on different runs

There are several ways this can happen; for example concurrent algorithms may experience race conditions

Data Structures and Algorithms in C++, Fourth Edition

NP-Completeness (continued)

A decision problem is any arbitrary yes-or-no question on an infinite set of inputs

Because of this, it can be defined equivalently as the set of inputs for which the problem returns yes

A nondeterministic algorithm can solve a decision problem if there is a path in the decision tree of the algorithm that leads to a “yes” answer; otherwise it would answer “no”

If the number of steps in the decision tree path to the affirmative answer is O(nk), where n is the size of the specific problem, the algorithm is considered polynomial

A problem that can be solved this way is called tractable

Data Structures and Algorithms in C++, Fourth Edition

NP-Completeness (continued)

Problems that can be solved by a deterministic algorithm in polynomial time belong to a class P of problems

If the problem can be solved in polynomial time by a nondeterministic algorithm, it is of class NP

Class P problems are tractable; NP problems are tractable only if a nondeterministic algorithm is used

Now NP, which simply means that deterministic algorithms are nondeterministic algorithms that don’t use nondeterministic decisions

It is also generally believed P ≠ NP, although this is a famous open problem in computer science

Data Structures and Algorithms in C++, Fourth Edition

NP-Completeness (continued)

The reason for this belief is the existence of NP-complete problems, which is related to the concept of reducibility

A problem is reducible if every instance of the problem can be transformed into instances of another problem using a process referred to as a reduction algorithm

If this transformation can be done efficiently (in polynomial time), then efficient solutions of the second problem can be transformed into efficient solutions of the original problem

A problem is NP-complete if it is NP and every other problem in NP is reducible to the problem in polynomial time

Thus, all NP-complete problems are equivalent, computationally

Data Structures and Algorithms in C++, Fourth Edition

NP-Completeness (continued)

If we can solve an NP-complete problem with a deterministic algorithm, all NP-complete problems can be solved the same way

However, if any NP problem is intractable, so are all NP-complete problems

The reducibility process uses an NP-complete problem to show another problem is NP-complete

However, there has to be at least one problem that can be proven to be NP-complete through a means other than reducibility to make reduction possible

Data Structures and Algorithms in C++, Fourth Edition

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