data structures and algorithms in java chapter 2 complexity analysis

Data Structures and Algorithms in Java

Chapter 2

Complexity Analysis

Data Structures and Algorithms in Java 2

ObjectivesDiscuss the following topics: • Computational and Asymptotic Complexity• Big-O Notation• Properties of Big-O Notation• Ω and Θ Notations• Examples of Complexities• Finding Asymptotic Complexity: Examples• Amortized Complexity• The Best, Average, and Worst Cases• NP-Completeness


Computational and Asymptotic Complexity

• Computational complexity measures the degree of difficulty of an algorithm

• Indicates how much effort is needed to apply an algorithm or how costly it is

• To evaluate an algorithm’s efficiency, use logical units that express a relationship such as:– The size n of a file or an array

– The amount of time t required to process the data


Computational and Asymptotic Complexity (continued)

• This measure of efficiency is called asymptotic complexity

• It is used when disregarding certain terms of a function– To express the efficiency of an algorithm– When calculating a function is difficult or

impossible and only approximations can be found

f (n) = n2 + 100n + log10n + 1,000


Computational and Asymptotic Complexity (continued)

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


Big-O Notation

• Introduced in 1894, the big-O notation specifies asymptotic complexity, which estimates the rate of function growth

• Definition 1: f (n) is O(g(n)) if there exist positive numbers c and N such that f (n) ≤ cg(n) for all n ≥ N

Figure 2-2 Different values of c and N for function f (n) = 2n2 + 3n + 1 = O(n2) calculated according to the definition of big-O


Big-O Notation (continued)

Figure 2-3 Comparison of functions for different values of c and N from Figure 2-2


Properties of Big-O Notation

• Fact 1 (transitivity) If f (n) is O(g(n)) and g(n) is O(h(n)), then f(n) is O(h(n))

• Fact 2 If f (n) is O(h(n)) and g(n) is O(h(n)), then f(n) + g(n) is O(h(n))

• Fact 3 The function ank is O(nk)


Properties of Big-O Notation (continued)

• Fact 4The function nk is O(nk+j) for any positive j

• Fact 5If f(n) = cg(n), then f(n) is O(g(n))

• Fact 6 The function loga n is O(logb n) for any positive numbers a and b ≠ 1

• Fact 7loga n is O(lg n) for any positive a ≠ 1, where lg n = log2 n


Ω and Θ Notations

• Big-O notation refers to the upper bounds of functions

• There is a symmetrical definition for a lower bound in the definition of big-Ω

• Definition 2: The function f(n) is Ω(g(n)) if there exist positive numbers c and N such that f(n) ≥ cg(n) for all n ≥ N


Ω and Θ Notations (continued)

• The difference between this definition and the definition of big-O notation is the direction of the inequality

• One definition can be turned into the other by replacing “≥” with “≤”

• There is an interconnection between these two notations expressed by the equivalence

f (n) is Ω(g(n)) iff g(n) is O(f (n)) (prove?)


Ω and Θ Notations (continued)

• Definition 3: f(n) is Θ(g(n)) if there exist positive numbers c1, c2, and N such that c1g(n) ≤ f(n) ≤ c2g(n) for all n ≥ N

• When applying any of these notations (big-O,Ω, and Θ), remember they are approximations that hide some detail that in many cases may be considered important


Examples of Complexities

• Algorithms can be classified by their time or space complexities

• An algorithm is called constant if its execution time remains the same for any number of elements

• It is called quadratic if its execution time is O(n2)


Examples of Complexities (continued)

Figure 2-4 Classes of algorithms and their execution times on a computer executing 1 million operations per second (1 sec = 106 μsec = 103 msec)



Figure 2-4 Classes of algorithms and their execution times on a computer executing 1 million operations per second (1 sec = 106 μsec = 103 msec)(continued)



Figure 2-5 Typical functions applied in big-O estimates


Finding Asymptotic Complexity: Examples

• Asymptotic bounds are used to estimate the efficiency of algorithms by assessing the amount of time and memory needed to accomplish the task for which the algorithms were designed

for (i = sum = 0; i < n; i++)sum += a[i]

• Initialize two variables• Execute two assignments

– Update sum– Update iTotal 2+2n assignments for the complete executionAsymptotic complexity is O(n)



• Printing sums of all the sub-arrays that begins with position 0

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

sum += a[j];System.out.println ("sum for subarray 0 through

"+i+" is" + sum);

}

1+3n+ =1+3n+2(1+2+….n-1) =1+3n+n(n-1)=O(n)+ O(n2)= O(n2)

1

1

2n

i

i


Examples Continued• Printing sums of numbers in the last five cells

of the sub-arrays starting in position 0

for (i = 4; i < n; i++) {for (j = i-3, sum = a[i-4]; j <= i; j++)

sum += a[j];System.out.println

("sum for subarray "+(i - 4)+" through "+i+" is"+ sum);

}• n-4 times for outer loop• For each i, inner loop executes only four times

1+8.(n-4)= O(n)



• Finding the length of the longest sub-array with the numbers in increasing order

• For example [1 2 5 ] in [1 8 1 2 5 0 11 12]

for (i = 0, length = 1; i < n-1; i++) {for (i1 = i2 = k = i; k < n-1 && a[k] < a[k+1];

k++, i2++);if (length < i2 - i1 + 1)length = i2 - i1 + 1;

System.out.println ("the length of the longest ordered subarray is" + length);

}


• If all numbers in the array are in decreasing order, the outer loop is executed n-1 times

• But in each iteration, the inner loop executes just one time. The algorithm is O(n)

• If the numbers are in increasing order, the outer loop is executed n- 1 times and the inner loop is executed n-1-i times for each i in {0,…, n-2}. The algorithm is O(n2)



int binarySearch(int[] arr, int key) {int lo = 0, mid, hi = arr.length-1;while (lo <= hi) {

mid = (lo + hi)/2;if (key < arr[mid])

hi = mid - 1;else if (arr[mid] < key)

lo = mid + 1;else return mid; // success

} return -1; // failure

}• O(lg n)


The Best, Average, and Worst Cases

• The worst case is when an algorithm requires a maximum number of steps

• The best case is when the number of steps is the smallest

• The average case falls between these extremes

Cavg = Σip(inputi)steps(inputi)


• The average complexity is established by considering possible inputs to an algorithm,

• determining the number of steps performed by the algorithm for each input,

• adding the number of steps for all the inputs, and dividing by the number of inputs

• This definition assumes that the probability of occurrence of each input is the same. It is not the case always.

• The average complexity is defined as the average over the number of steps executed when processing each input weighted by the probability of occurrence of this input


Consider searching sequentially an unordered arrayto find a number

• The best case is when the number is found in the first cell

• The worst case is when the number is in the last cell or not in the array at all

• The average case?


• Assuming the probability distribution is uniform• The probability equals to 1/n for each position• To find a number in one try is 1/n • To find a number in two tries is 1/n• etc… • The average steps to find a number is

1 + 2 + … + n n

= n + 1 2


• If the probabilities differ, the average case gives a different outcome

• If the probability of finding a number in the first cell is ½ , the probability in the second cell is ¼ and the probability is the same for remaining cells

=

• the average steps

1 - ½ - ¼ n - 2

1 4(n - 2)

8

31

)2(8

6)1(1

)2(4

...3

4

2

2

1

n

n

nn

n

n


Summation Formulas

Let N > 0, let A, B, and C be constants, and let f and g be any functions. Then:

N

k

N

k

kfCkCf11

)()(

S1: factor out constant

N

k

N

k

N

k

kgkfkgkf111

)()())()((

S2: separate summed terms

NCCN

k

1

S3: sum of constant

2

)1(

1

NNk

N

k

S4: sum of k

6

)12)(1(

1

2

NNNk

N

k

S5: sum of k squared

122 1

0

N

N

k

k

S6: sum of 2^k

12)1(21

1

NN

k

k Nk

S7: sum of k2^(k-1)


Logarithms

Let b be a real number, b > 0 and b 1. Then, for any real number x > 0, the logarithm of x to base b is the power to which b must be raised to yield x. That is:

x byx yb ifonly and if )(log

For example:

642 because 6)64(log 62

8/12 because 3)8/1(log 32

12 because 0)1(log 02

If the base is omitted, the standard convention in mathematics is that log base 10 is intended; in computer science the standard convention is that log base 2 is intended.


Logarithms

Let a and b be real numbers, both positive and neither equal to 1. Let x > 0 and y > 0 be real numbers.

0)1(log bL1:

1)(log bbL2:

10 allfor 0)(log xxbL3:

1 xallfor 0)(log xbL4:

)(log)(log)(log yxxy bbb L7:

)(log)(loglog yxy

xbbb

L8:

)(log)(log xyx by

b L9:

yb yb )(logL5:

xb xb )(logL6:)(log

)(log)(log

b

xx

a

ab L10:


Limit of a Function

Definition:

Let f(x) be a function with domain (a, b) and let a < c < b. The limit of f(x) as x approaches c is L if, for every positive real number , there is a positive real number such that whenever |x-c| < then |f(x) – L| < .

The definition being cumbersome, the following theorems on limits are useful. We assume f(x) is a function with domain as described above and that K is a constant.

KKcx

limC1:

cxcx

limC2: 0 allfor lim

rcx rr

cxC3:


Limit of a FunctionHere assume f(x) and g(x) are functions with domain as described above and that K is a constant, and that both the following limits exist (and are finite):

)(lim)(lim xfKxKfcxcx

C4:

)(lim)(lim)()(lim xgxfxgxfcxcxcx

C5:

Axfcx

)(lim Bxgcx

)(lim

Then:

)(lim*)(lim)(*)(lim xgxfxgxfcxcxcx

C6:

0 provided )(lim/)(lim)(/)(lim

Bxgxfxgxfcxcxcx

C7:


Limit as x Approaches Infinity

Definition:

Let f(x) be a function with domain [0, ). The limit of f(x) as x approaches is L if, for every positive real number , there is a positive real number N such that whenever x > N then |f(x) – L| < .

The definition being cumbersome, the following theorems on limits are useful. We assume f(x) is a function with domain [0, ) and that K is a constant.

KKx

limC8:

01

lim xx

C9: 0 allfor 01

lim

rxrx

C10:


Limit of a Rational Function

Given a rational function the last two rules are sufficient if a little algebra is employed:

3

7003

007

5lim

2lim3lim

10lim

5lim7lim

523

1057

lim523

1057lim

2

2

2

2

2

2

xx

xx

xx

xxxx

xx

xxx

xxx

xx

Divide by highest power of x from the denominator.

Take limits term by term.

Apply theorem C3.


Infinite LimitsIn some cases, the limit may be infinite. Mathematically, this means that the limit does not exist.

0 allfor lim

rxr

xC11:

x

xx

x

xx

x

xx

xx

xxx

xx

5lim2lim

10lim5lim7lim

52

1057

lim52

1057lim

2

x

xelimC13:

Example:

xbx

loglimC12:


l'Hôpital's Rule

In some cases, the reduction trick shown for rational functions does not apply:

In such cases, l'Hôpital's Rule is often useful. If f(x) and g(x) are differentiable functions such that

?? 52

10)log(57lim

x

xxx

)(

)(lim

)(

)(lim

xg

xf

xg

xfcxcx

)(lim)(lim xgxfcxcx

then: This also applies if the limit is 0.


l'Hôpital's Rule ExamplesApplying l'Hôpital's Rule:

2

7

2

57

lim 52

10)log(57lim

xx

xxxx

Another example:

06

lim6

lim3

lim 10

lim23

xxxxxxxx ee

x

e

x

e

x

Recall that: )()()( xfDeeD xfxf


Mathematical Induction

Mathematical induction is a technique for proving that a statement is true for all integers in the range from N0 to , where N0 is typically 0 or 1.

Let P(N) be a proposition regarding the integer N, and let S be the set of all integers k for which P(k) is true. If

1) N0 is in S, and

2) whenever N is in S then N+1 is also in S,

then S contains all integers in the range [N0, ).

First (or Weak) Principle of Mathematical Induction

To apply the PMI, we must first establish that a specific integer, N0, is in S (establishing the basis) and then we must establish that if a arbitrary integer, N N0, is in S then its successor, N+1, is also in S.


Induction Example

Theorem: For all integers n 1, n2+n is a multiple of 2.

proof: Let S be the set of all integers for which n2+n is a multiple of 2.

If n = 1, then n2+n = 2, which is obviously a multiple of 2. This establishes the basis, that 1 is in S.

Now suppose that some integer k 1 is an element of S. Then k2+k is a multiple of 2. We need to show that k+1 is an element of S; in other words, we must show that (k+1)2+(k+1) is a multiple of 2. Performing simple algebra:

(k+1)2+(k+1) = (k2 + 2k + 1) + (k + 1) = k2 + 3k + 2

Now we know k2+k is a multiple of 2, and the expression above can be grouped to show:

(k+1)2+(k+1) = (k2 + k) + (2k + 2) = (k2 + k) + 2(k + 1)

The last expression is the sum of two multiples of 2, so it's also a multiple of 2. Therefore, k+1 is an element of S.

Therefore, by PMI, S contains all integers [1, ).

QED


Inadequacy of the First Form of Induction

Theorem: Every integer greater than 3 can be written as a sum of 2's and 5's.

(That is, if N > 3, then there are nonnegative integers x and y such that N = 2x + 5y.)

This is not (easily) provable using the First Principle of Induction. The problem is that the way to write N+1 in terms of 2's and 5's has little to do with the way N is written in terms of 2's and 5's. For example, if we know that

N = 2x + 5y

we can say that

N + 1 = 2x + 5y + 1 = 2x + 5(y – 1) + 5 + 1 = 2(x + 3) + 5(y – 1)

but we have no reason to believe that y – 1 is nonnegative. (Suppose for example that N is 9.)


"Strong" Form of Induction

Let P(N) be a proposition regarding the integer N, and let S be the set of all integers k for which P(k) is true. If

1) N0 is in S, and

2) whenever N0 through N are in S then N+1 is also in S,

then S contains all integers in the range [N0, ).

Second (or Strong) Principle of Mathematical Induction

Interestingly, the "strong" form of induction is logically equivalent to the "weak" form stated earlier; so in principle, anything that can be proved using the "strong" form can also be proved using the "weak" form.

There is a second statement of induction, sometimes called the "strong" form, that is adequate to prove the result on the preceding slide:


Using the Second Form of Induction

Theorem: Every integer greater than 3 can be written as a sum of 2's and 5's.

proof: Let S be the set of all integers n > 3 for which n = 2x + 5y for some nonnegative integers x and y.

If n = 4, then n = 2*2 + 5*0. If n = 5, then n = 2*0 + 5*1. This establishes the basis, that 4 and 5 are in S.

Now suppose that all integers from 4 through k are elements of S, where k 5. We need to show that k+1 is an element of S; in other words, we must show that k+1 = 2r + 5s for some nonnegative integers r and s.

Now k+1 6, so k-1 4. Therefore by our assumption, k-1 = 2x + 5y for some nonnegative integers x and y. Then, simple algebra yields that:

k+1 = k-1 + 2 = 2x + 5y + 2 = 2(x+1) + 5y,

whence k+1 is an element of S.

Therefore, by the Second PMI, S contains all integers [4, ).

QED


Amortized Complexity

• Amortized analysis:– Analyzes sequences of operations

– Can be used to find the average complexity of a worst case sequence of operations

• By analyzing sequences of operations rather than isolated operations, amortized analysis takes into account interdependence between operations and their results


Amortized Complexity (continued)

Worst case:

C(op1, op2, op3, . . .) = Cworst(op1) + Cworst(op2) + Cworst(op3) + . . .

Average case:

C(op1, op2, op3, . . .) = Cavg(op1) + Cavg(op2) + Cavg(op3) + . . .

Amortized:

C(op1, op2, op3, . . .) = C(op1) + C(op2) + C(op3) + . . .

Where C can be worst, average, or best case complexity


Amortized Complexity (continued)

Figure 2-6 Estimating the amortized cost


NP-Completeness

• A deterministic algorithm is a uniquely defined (determined) sequence of steps for a particular input– There is only one way to determine the next step

that the algorithm can make

• A nondeterministic algorithm is an algorithm that can use a special operation that makes a guess when a decision is to be made


NP-Completeness (continued)

• A nondeterministic algorithm is considered polynomial: its running time in the worst case is O(nk) for some k

• Problems that can be solved with such algorithms are called tractable and the algorithms are considered efficient

• A problem is called NP-complete if it is NP (it can be solved efficiently by a nondeterministic polynomial algorithm) and every NP problem can be polynomially reduced to this problem


NP-Completeness (continued)

• The satisfiability problem concerns Boolean expressions in conjunctive normal form (CNF)


Summary

• Computational complexity measures the degree of difficulty of an algorithm.

• This measure of efficiency is called asymptotic complexity.

• To evaluate an algorithm’s efficiency, use logical units that express a relationship.

• This measure of efficiency is called asymptotic complexity.


Summary (continued)

• Introduced in 1894, the big-O notation specifies asymptotic complexity, which estimates the rate of function growth.

• An algorithm is called constant if its execution time remains the same for any number of elements.

• It is called quadratic if its execution time is O(n2).• Amortized analysis analyzes sequences of

operations.


Summary (continued)

• A deterministic algorithm is a uniquely defined (determined) sequence of steps for a particular input.

• A nondeterministic algorithm is an algorithm that can use a special operation that makes a guess when a decision is to be made.

• A nondeterministic algorithm is considered polynomial.

data structures and algorithms in java chapter 2 complexity analysis

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