An algorithm is a set of steps of operations to solve a problem performing calculation, data processing, and automated reasoning tasks. An algorithm is an efficient method that can be expressed within finite amount of time and space.
An algorithm is the best way to represent the solution of a particular problem in a very simple and efficient way. If we have an algorithm for a specific problem, then we can implement it in any programming language, meaning that the algorithm is independent from any programming languages.
The important aspects of algorithm design include creating an efficient algorithm to solve a problem in an efficient way using minimum time and space.
To solve a problem, different approaches can be followed. Some of them can be efficient with respect to time consumption, whereas other approaches may be memory efficient. However, one has to keep in mind that both time consumption and memory usage cannot be optimized simultaneously. If we require an algorithm to run in lesser time, we have to invest in more memory and if we require an algorithm to run with lesser memory, we need to have more time.
The following steps are involved in solving computational problems.
The main characteristics of algorithms are as follows −
Algorithms must have a unique name
Algorithms should have explicitly defined set of inputs and outputs
Algorithms are wellordered with unambiguous operations
Algorithms halt in a finite amount of time. Algorithms should not run for infinity, i.e., an algorithm must end at some point
Pseudocode gives a highlevel description of an algorithm without the ambiguity associated with plain text but also without the need to know the syntax of a particular programming language.
The running time can be estimated in a more general manner by using Pseudocode to represent the algorithm as a set of fundamental operations which can then be counted.
An algorithm is a formal definition with some specific characteristics that describes a process, which could be executed by a Turingcomplete computer machine to perform a specific task. Generally, the word "algorithm" can be used to describe any high level task in computer science.
On the other hand, pseudocode is an informal and (often rudimentary) human readable description of an algorithm leaving many granular details of it. Writing a pseudocode has no restriction of styles and its only objective is to describe the high level steps of algorithm in a much realistic manner in natural language.
For example, following is an algorithm for Insertion Sort.
Algorithm: InsertionSort Input: A list L of integers of length n Output: A sorted list L1 containing those integers present in L Step 1: Keep a sorted list L1 which starts off empty Step 2: Perform Step 3 for each element in the original list L Step 3: Insert it into the correct position in the sorted list L1. Step 4: Return the sorted list Step 5: Stop
Here is a pseudocode which describes how the high level abstract process mentioned above in the algorithm InsertionSort could be described in a more realistic way.
for i < 1 to length(A) x < A[i] j < i while j > 0 and A[j1] > x A[j] < A[j1] j < j  1 A[j] < x
In this tutorial, algorithms will be presented in the form of pseudocode, that is similar in many respects to C, C++, Java, Python, and other programming languages.
In theoretical analysis of algorithms, it is common to estimate their complexity in the asymptotic sense, i.e., to estimate the complexity function for arbitrarily large input. The term "analysis of algorithms" was coined by Donald Knuth.
Algorithm analysis is an important part of computational complexity theory, which provides theoretical estimation for the required resources of an algorithm to solve a specific computational problem. Most algorithms are designed to work with inputs of arbitrary length. Analysis of algorithms is the determination of the amount of time and space resources required to execute it.
Usually, the efficiency or running time of an algorithm is stated as a function relating the input length to the number of steps, known as time complexity, or volume of memory, known as space complexity.
In this chapter, we will discuss the need for analysis of algorithms and how to choose a better algorithm for a particular problem as one computational problem can be solved by different algorithms.
By considering an algorithm for a specific problem, we can begin to develop pattern recognition so that similar types of problems can be solved by the help of this algorithm.
Algorithms are often quite different from one another, though the objective of these algorithms are the same. For example, we know that a set of numbers can be sorted using different algorithms. Number of comparisons performed by one algorithm may vary with others for the same input. Hence, time complexity of those algorithms may differ. At the same time, we need to calculate the memory space required by each algorithm.
Analysis of algorithm is the process of analyzing the problemsolving capability of the algorithm in terms of the time and size required (the size of memory for storage while implementation). However, the main concern of analysis of algorithms is the required time or performance. Generally, we perform the following types of analysis −
Worstcase − The maximum number of steps taken on any instance of size a.
Bestcase − The minimum number of steps taken on any instance of size a.
Average case − An average number of steps taken on any instance of size a.
Amortized − A sequence of operations applied to the input of size a averaged over time.
To solve a problem, we need to consider time as well as space complexity as the program may run on a system where memory is limited but adequate space is available or may be viceversa. In this context, if we compare bubble sort and merge sort. Bubble sort does not require additional memory, but merge sort requires additional space. Though time complexity of bubble sort is higher compared to merge sort, we may need to apply bubble sort if the program needs to run in an environment, where memory is very limited.
To measure resource consumption of an algorithm, different strategies are used as discussed in this chapter.
The asymptotic behavior of a function f(n) refers to the growth of f(n) as n gets large.
We typically ignore small values of n, since we are usually interested in estimating how slow the program will be on large inputs.
A good rule of thumb is that the slower the asymptotic growth rate, the better the algorithm. Though it’s not always true.
For example, a linear algorithm $f(n) = d * n + k$ is always asymptotically better than a quadratic one, $f(n) = c.n^2 + q$.
A recurrence is an equation or inequality that describes a function in terms of its value on smaller inputs. Recurrences are generally used in divideandconquer paradigm.
Let us consider T(n) to be the running time on a problem of size n.
If the problem size is small enough, say n < c where c is a constant, the straightforward solution takes constant time, which is written as θ(1). If the division of the problem yields a number of subproblems with size $\frac{n}{b}$.
To solve the problem, the required time is a.T(n/b). If we consider the time required for division is D(n) and the time required for combining the results of subproblems is C(n), the recurrence relation can be represented as −
$$T(n)=\begin{cases}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\theta(1) & if\:n\leqslant c\\a T(\frac{n}{b})+D(n)+C(n) & otherwise\end{cases}$$
A recurrence relation can be solved using the following methods −
Substitution Method − In this method, we guess a bound and using mathematical induction we prove that our assumption was correct.
Recursion Tree Method − In this method, a recurrence tree is formed where each node represents the cost.
Master’s Theorem − This is another important technique to find the complexity of a recurrence relation.
Amortized analysis is generally used for certain algorithms where a sequence of similar operations are performed.
Amortized analysis provides a bound on the actual cost of the entire sequence, instead of bounding the cost of sequence of operations separately.
Amortized analysis differs from averagecase analysis; probability is not involved in amortized analysis. Amortized analysis guarantees the average performance of each operation in the worst case.
It is not just a tool for analysis, it’s a way of thinking about the design, since designing and analysis are closely related.
The aggregate method gives a global view of a problem. In this method, if n operations takes worstcase time T(n) in total. Then the amortized cost of each operation is T(n)/n. Though different operations may take different time, in this method varying cost is neglected.
In this method, different charges are assigned to different operations according to their actual cost. If the amortized cost of an operation exceeds its actual cost, the difference is assigned to the object as credit. This credit helps to pay for later operations for which the amortized cost less than actual cost.
If the actual cost and the amortized cost of i^{th} operation are $c_{i}$ and $\hat{c_{l}}$, then
$$\displaystyle\sum\limits_{i=1}^n \hat{c_{l}}\geqslant\displaystyle\sum\limits_{i=1}^n c_{i}$$
This method represents the prepaid work as potential energy, instead of considering prepaid work as credit. This energy can be released to pay for future operations.
If we perform n operations starting with an initial data structure D_{0}. Let us consider, c_{i} as the actual cost and D_{i} as data structure of i^{th} operation. The potential function Ф maps to a real number Ф(D_{i}), the associated potential of D_{i}. The amortized cost $\hat{c_{l}}$ can be defined by
$$\hat{c_{l}}=c_{i}+\Phi (D_{i})\Phi (D_{i1})$$
Hence, the total amortized cost is
$$\displaystyle\sum\limits_{i=1}^n \hat{c_{l}}=\displaystyle\sum\limits_{i=1}^n (c_{i}+\Phi (D_{i})\Phi (D_{i1}))=\displaystyle\sum\limits_{i=1}^n c_{i}+\Phi (D_{n})\Phi (D_{0})$$
If the allocated space for the table is not enough, we must copy the table into larger size table. Similarly, if large number of members are erased from the table, it is a good idea to reallocate the table with a smaller size.
Using amortized analysis, we can show that the amortized cost of insertion and deletion is constant and unused space in a dynamic table never exceeds a constant fraction of the total space.
In the next chapter of this tutorial, we will discuss Asymptotic Notations in brief.
In designing of Algorithm, complexity analysis of an algorithm is an essential aspect. Mainly, algorithmic complexity is concerned about its performance, how fast or slow it works.
The complexity of an algorithm describes the efficiency of the algorithm in terms of the amount of the memory required to process the data and the processing time.
Complexity of an algorithm is analyzed in two perspectives: Time and Space.
It’s a function describing the amount of time required to run an algorithm in terms of the size of the input. "Time" can mean the number of memory accesses performed, the number of comparisons between integers, the number of times some inner loop is executed, or some other natural unit related to the amount of real time the algorithm will take.
It’s a function describing the amount of memory an algorithm takes in terms of the size of input to the algorithm. We often speak of "extra" memory needed, not counting the memory needed to store the input itself. Again, we use natural (but fixedlength) units to measure this.
Space complexity is sometimes ignored because the space used is minimal and/or obvious, however sometimes it becomes as important an issue as time.
Execution time of an algorithm depends on the instruction set, processor speed, disk I/O speed, etc. Hence, we estimate the efficiency of an algorithm asymptotically.
Time function of an algorithm is represented by T(n), where n is the input size.
Different types of asymptotic notations are used to represent the complexity of an algorithm. Following asymptotic notations are used to calculate the running time complexity of an algorithm.
O − Big Oh
Ω − Big omega
θ − Big theta
o − Little Oh
ω − Little omega
‘O’ (Big Oh) is the most commonly used notation. A function f(n) can be represented is the order of g(n) that is O(g(n)), if there exists a value of positive integer n as n_{0} and a positive constant c such that −
$f(n)\leqslant c.g(n)$ for $n > n_{0}$ in all case
Hence, function g(n) is an upper bound for function f(n), as g(n) grows faster than f(n).
Let us consider a given function, $f(n) = 4.n^3 + 10.n^2 + 5.n + 1$
Considering $g(n) = n^3$,
$f(n)\leqslant 5.g(n)$ for all the values of $n > 2$
Hence, the complexity of f(n) can be represented as $O(g(n))$, i.e. $O(n^3)$
We say that $f(n) = \Omega (g(n))$ when there exists constant c that $f(n)\geqslant c.g(n)$ for all sufficiently large value of n. Here n is a positive integer. It means function g is a lower bound for function f; after a certain value of n, f will never go below g.
Let us consider a given function, $f(n) = 4.n^3 + 10.n^2 + 5.n + 1$.
Considering $g(n) = n^3$, $f(n)\geqslant 4.g(n)$ for all the values of $n > 0$.
Hence, the complexity of f(n) can be represented as $\Omega (g(n))$, i.e. $\Omega (n^3)$
We say that $f(n) = \theta(g(n))$ when there exist constants c_{1} and c_{2} that $c_{1}.g(n) \leqslant f(n) \leqslant c_{2}.g(n)$ for all sufficiently large value of n. Here n is a positive integer.
This means function g is a tight bound for function f.
Let us consider a given function, $f(n) = 4.n^3 + 10.n^2 + 5.n + 1$
Considering $g(n) = n^3$, $4.g(n) \leqslant f(n) \leqslant 5.g(n)$ for all the large values of n.
Hence, the complexity of f(n) can be represented as $\theta (g(n))$, i.e. $\theta (n^3)$.
The asymptotic upper bound provided by Onotation may or may not be asymptotically tight. The bound $2.n^2 = O(n^2)$ is asymptotically tight, but the bound $2.n = O(n^2)$ is not.
We use onotation to denote an upper bound that is not asymptotically tight.
We formally define o(g(n)) (littleoh of g of n) as the set f(n) = o(g(n)) for any positive constant $c > 0$ and there exists a value $n_{0} > 0$, such that $0 \leqslant f(n) \leqslant c.g(n)$.
Intuitively, in the onotation, the function f(n) becomes insignificant relative to g(n) as n approaches infinity; that is,
$$\lim_{n \rightarrow \infty}\left(\frac{f(n)}{g(n)}\right) = 0$$
Let us consider the same function, $f(n) = 4.n^3 + 10.n^2 + 5.n + 1$
Considering $g(n) = n^{4}$,
$$\lim_{n \rightarrow \infty}\left(\frac{4.n^3 + 10.n^2 + 5.n + 1}{n^4}\right) = 0$$
Hence, the complexity of f(n) can be represented as $o(g(n))$, i.e. $o(n^4)$.
We use ωnotation to denote a lower bound that is not asymptotically tight. Formally, however, we define ω(g(n)) (littleomega of g of n) as the set f(n) = ω(g(n)) for any positive constant C > 0 and there exists a value $n_{0} > 0$, such that $0 \leqslant c.g(n) < f(n)$.
For example, $\frac{n^2}{2} = \omega (n)$, but $\frac{n^2}{2} \neq \omega (n^2)$. The relation $f(n) = \omega (g(n))$ implies that the following limit exists
$$\lim_{n \rightarrow \infty}\left(\frac{f(n)}{g(n)}\right) = \infty$$
That is, f(n) becomes arbitrarily large relative to g(n) as n approaches infinity.
Let us consider same function, $f(n) = 4.n^3 + 10.n^2 + 5.n + 1$
Considering $g(n) = n^2$,
$$\lim_{n \rightarrow \infty}\left(\frac{4.n^3 + 10.n^2 + 5.n + 1}{n^2}\right) = \infty$$
Hence, the complexity of f(n) can be represented as $o(g(n))$, i.e. $\omega (n^2)$.
Apriori analysis means, analysis is performed prior to running it on a specific system. This analysis is a stage where a function is defined using some theoretical model. Hence, we determine the time and space complexity of an algorithm by just looking at the algorithm rather than running it on a particular system with a different memory, processor, and compiler.
Apostiari analysis of an algorithm means we perform analysis of an algorithm only after running it on a system. It directly depends on the system and changes from system to system.
In an industry, we cannot perform Apostiari analysis as the software is generally made for an anonymous user, which runs it on a system different from those present in the industry.
In Apriori, it is the reason that we use asymptotic notations to determine time and space complexity as they change from computer to computer; however, asymptotically they are the same.
In this chapter, we will discuss the complexity of computational problems with respect to the amount of space an algorithm requires.
Space complexity shares many of the features of time complexity and serves as a further way of classifying problems according to their computational difficulties.
Space complexity is a function describing the amount of memory (space) an algorithm takes in terms of the amount of input to the algorithm.
We often speak of extra memory needed, not counting the memory needed to store the input itself. Again, we use natural (but fixedlength) units to measure this.
We can use bytes, but it's easier to use, say, the number of integers used, the number of fixedsized structures, etc.
In the end, the function we come up with will be independent of the actual number of bytes needed to represent the unit.
Space complexity is sometimes ignored because the space used is minimal and/or obvious, however sometimes it becomes as important issue as time complexity
Let M be a deterministic Turing machine (TM) that halts on all inputs. The space complexity of M is the function $f \colon N \rightarrow N$, where f(n) is the maximum number of cells of tape and M scans any input of length M. If the space complexity of M is f(n), we can say that M runs in space f(n).
We estimate the space complexity of Turing machine by using asymptotic notation.
Let $f \colon N \rightarrow R^+$ be a function. The space complexity classes can be defined as follows −
SPACE = {L  L is a language decided by an O(f(n)) space deterministic TM}
SPACE = {L  L is a language decided by an O(f(n)) space nondeterministic TM}
PSPACE is the class of languages that are decidable in polynomial space on a deterministic Turing machine.
In other words, PSPACE = U_{k} SPACE (n^{k})
One of the earliest theorem related to space complexity is Savitch’s theorem. According to this theorem, a deterministic machine can simulate nondeterministic machines by using a small amount of space.
For time complexity, such a simulation seems to require an exponential increase in time. For space complexity, this theorem shows that any nondeterministic Turing machine that uses f(n) space can be converted to a deterministic TM that uses f^{2}(n) space.
Hence, Savitch’s theorem states that, for any function, $f \colon N \rightarrow R^+$, where $f(n) \geqslant n$
NSPACE(f(n)) ⊆ SPACE(f(n))
The following diagram depicts the relationship among different complexity classes.
Till now, we have not discussed P and NP classes in this tutorial. These will be discussed later.
Many algorithms are recursive in nature to solve a given problem recursively dealing with subproblems.
In divide and conquer approach, a problem is divided into smaller problems, then the smaller problems are solved independently, and finally the solutions of smaller problems are combined into a solution for the large problem.
Generally, divideandconquer algorithms have three parts −
Divide the problem into a number of subproblems that are smaller instances of the same problem.
Conquer the subproblems by solving them recursively. If they are small enough, solve the subproblems as base cases.
Combine the solutions to the subproblems into the solution for the original problem.
Divide and conquer approach supports parallelism as subproblems are independent. Hence, an algorithm, which is designed using this technique, can run on the multiprocessor system or in different machines simultaneously.
In this approach, most of the algorithms are designed using recursion, hence memory management is very high. For recursive function stack is used, where function state needs to be stored.
Following are some problems, which are solved using divide and conquer approach.
Let us consider a simple problem that can be solved by divide and conquer technique.
The MaxMin Problem in algorithm analysis is finding the maximum and minimum value in an array.
To find the maximum and minimum numbers in a given array numbers[] of size n, the following algorithm can be used. First we are representing the naive method and then we will present divide and conquer approach.
Naïve method is a basic method to solve any problem. In this method, the maximum and minimum number can be found separately. To find the maximum and minimum numbers, the following straightforward algorithm can be used.
Algorithm: MaxMinElement (numbers[]) max := numbers[1] min := numbers[1] for i = 2 to n do if numbers[i] > max then max := numbers[i] if numbers[i] < min then min := numbers[i] return (max, min)
The number of comparison in Naive method is 2n  2.
The number of comparisons can be reduced using the divide and conquer approach. Following is the technique.
In this approach, the array is divided into two halves. Then using recursive approach maximum and minimum numbers in each halves are found. Later, return the maximum of two maxima of each half and the minimum of two minima of each half.
In this given problem, the number of elements in an array is $y  x + 1$, where y is greater than or equal to x.
$\mathbf{\mathit{Max  Min(x, y)}}$ will return the maximum and minimum values of an array $\mathbf{\mathit{numbers[x...y]}}$.
Algorithm: Max  Min(x, y) if y – x ≤ 1 then return (max(numbers[x], numbers[y]), min((numbers[x], numbers[y])) else (max1, min1):= maxmin(x, ⌊((x + y)/2)⌋) (max2, min2):= maxmin(⌊((x + y)/2) + 1)⌋,y) return (max(max1, max2), min(min1, min2))
Let T(n) be the number of comparisons made by $\mathbf{\mathit{Max  Min(x, y)}}$, where the number of elements $n = y  x + 1$.
If T(n) represents the numbers, then the recurrence relation can be represented as
$$T(n) = \begin{cases}T\left(\lfloor\frac{n}{2}\rfloor\right)+T\left(\lceil\frac{n}{2}\rceil\right)+2 & for\: n>2\\1 & for\:n = 2 \\0 & for\:n = 1\end{cases}$$
Let us assume that n is in the form of power of 2. Hence, n = 2^{k} where k is height of the recursion tree.
So,
$$T(n) = 2.T (\frac{n}{2}) + 2 = 2.\left(\begin{array}{c}2.T(\frac{n}{4}) + 2\end{array}\right) + 2 ..... = \frac{3n}{2}  2$$
Compared to Naïve method, in divide and conquer approach, the number of comparisons is less. However, using the asymptotic notation both of the approaches are represented by O(n).
In this chapter, we will discuss merge sort and analyze its complexity.
The problem of sorting a list of numbers lends itself immediately to a divideandconquer strategy: split the list into two halves, recursively sort each half, and then merge the two sorted sublists.
In this algorithm, the numbers are stored in an array numbers[]. Here, p and q represents the start and end index of a subarray.
Algorithm: MergeSort (numbers[], p, r) if p < r then q = ⌊(p + r) / 2⌋ MergeSort (numbers[], p, q) MergeSort (numbers[], q + 1, r) Merge (numbers[], p, q, r)
Function: Merge (numbers[], p, q, r) n_{1} = q – p + 1 n_{2} = r – q declare leftnums[1…n_{1} + 1] and rightnums[1…n_{2} + 1] temporary arrays for i = 1 to n_{1} leftnums[i] = numbers[p + i  1] for j = 1 to n_{2} rightnums[j] = numbers[q+ j] leftnums[n_{1} + 1] = ∞ rightnums[n_{2} + 1] = ∞ i = 1 j = 1 for k = p to r if leftnums[i] ≤ rightnums[j] numbers[k] = leftnums[i] i = i + 1 else numbers[k] = rightnums[j] j = j + 1
Let us consider, the running time of MergeSort as T(n). Hence,
$T(n)=\begin{cases}c & if\:n\leqslant 1\\2\:x\:T(\frac{n}{2})+d\:x\:n & otherwise\end{cases}$ where c and d are constants
Therefore, using this recurrence relation,
$$T(n) = 2^i T(\frac{n}{2^i}) + i.d.n$$
As, $i = log\:n,\: T(n) = 2^{log\:n} T(\frac{n}{2^{log\:n}}) + log\:n.d.n$
$=\:c.n + d.n.log\:n$
Therefore, $T(n) = O(n\:log\:n)$
In the following example, we have shown MergeSort algorithm step by step. First, every iteration array is divided into two subarrays, until the subarray contains only one element. When these subarrays cannot be divided further, then merge operations are performed.
In this chapter, we will discuss another algorithm based on divide and conquer method.
Binary search can be performed on a sorted array. In this approach, the index of an element x is determined if the element belongs to the list of elements. If the array is unsorted, linear search is used to determine the position.
In this algorithm, we want to find whether element x belongs to a set of numbers stored in an array numbers[]. Where l and r represent the left and right index of a subarray in which searching operation should be performed.
Algorithm: BinarySearch(numbers[], x, l, r) if l = r then return l else m := ⌊(l + r) / 2⌋ if x ≤ numbers[m] then return BinarySearch(numbers[], x, l, m) else return BinarySearch(numbers[], x, m+1, r)
Linear search runs in O(n) time. Whereas binary search produces the result in O(log n) time
Let T(n) be the number of comparisons in worstcase in an array of n elements.
Hence,
$$T(n)=\begin{cases}0 & if\:n= 1\\T(\frac{n}{2})+1 & otherwise\end{cases}$$
Using this recurrence relation $T(n) = log\:n$.
Therefore, binary search uses $O(log\:n)$ time.
In this example, we are going to search element 63.
In this chapter, first we will discuss the general method of matrix multiplication and later we will discuss Strassen’s matrix multiplication algorithm.
Let us consider two matrices X and Y. We want to calculate the resultant matrix Z by multiplying X and Y.
First, we will discuss naïve method and its complexity. Here, we are calculating Z = X × Y. Using Naïve method, two matrices (X and Y) can be multiplied if the order of these matrices are p × q and q × r. Following is the algorithm.
Algorithm: MatrixMultiplication (X, Y, Z) for i = 1 to p do for j = 1 to r do Z[i,j] := 0 for k = 1 to q do Z[i,j] := Z[i,j] + X[i,k] × Y[k,j]
Here, we assume that integer operations take O(1) time. There are three for loops in this algorithm and one is nested in other. Hence, the algorithm takes O(n^{3}) time to execute.
In this context, using Strassen’s Matrix multiplication algorithm, the time consumption can be improved a little bit.
Strassen’s Matrix multiplication can be performed only on square matrices where n is a power of 2. Order of both of the matrices are n × n.
Divide X, Y and Z into four (n/2)×(n/2) matrices as represented below −
$Z = \begin{bmatrix}I & J \\K & L \end{bmatrix}$ $X = \begin{bmatrix}A & B \\C & D \end{bmatrix}$ and $Y = \begin{bmatrix}E & F \\G & H \end{bmatrix}$
Using Strassen’s Algorithm compute the following −
$$M_{1} \: \colon= (A+C) \times (E+F)$$
$$M_{2} \: \colon= (B+D) \times (G+H)$$
$$M_{3} \: \colon= (AD) \times (E+H)$$
$$M_{4} \: \colon= A \times (FH)$$
$$M_{5} \: \colon= (C+D) \times (E)$$
$$M_{6} \: \colon= (A+B) \times (H)$$
$$M_{7} \: \colon= D \times (GE)$$
Then,
$$I \: \colon= M_{2} + M_{3}  M_{6}  M_{7}$$
$$J \: \colon= M_{4} + M_{6}$$
$$K \: \colon= M_{5} + M_{7}$$
$$L \: \colon= M_{1}  M_{3}  M_{4}  M_{5}$$
$T(n)=\begin{cases}c & if\:n= 1\\7\:x\:T(\frac{n}{2})+d\:x\:n^2 & otherwise\end{cases}$ where c and d are constants
Using this recurrence relation, we get $T(n) = O(n^{log7})$
Hence, the complexity of Strassen’s matrix multiplication algorithm is $O(n^{log7})$.
Among all the algorithmic approaches, the simplest and straightforward approach is the Greedy method. In this approach, the decision is taken on the basis of current available information without worrying about the effect of the current decision in future.
Greedy algorithms build a solution part by part, choosing the next part in such a way, that it gives an immediate benefit. This approach never reconsiders the choices taken previously. This approach is mainly used to solve optimization problems. Greedy method is easy to implement and quite efficient in most of the cases. Hence, we can say that Greedy algorithm is an algorithmic paradigm based on heuristic that follows local optimal choice at each step with the hope of finding global optimal solution.
In many problems, it does not produce an optimal solution though it gives an approximate (near optimal) solution in a reasonable time.
Greedy algorithms have the following five components −
A candidate set − A solution is created from this set.
A selection function − Used to choose the best candidate to be added to the solution.
A feasibility function − Used to determine whether a candidate can be used to contribute to the solution.
An objective function − Used to assign a value to a solution or a partial solution.
A solution function − Used to indicate whether a complete solution has been reached.
Greedy approach is used to solve many problems, such as
Finding the shortest path between two vertices using Dijkstra’s algorithm.
Finding the minimal spanning tree in a graph using Prim’s /Kruskal’s algorithm, etc.
In many problems, Greedy algorithm fails to find an optimal solution, moreover it may produce a worst solution. Problems like Travelling Salesman and Knapsack cannot be solved using this approach.
The Greedy algorithm could be understood very well with a wellknown problem referred to as Knapsack problem. Although the same problem could be solved by employing other algorithmic approaches, Greedy approach solves Fractional Knapsack problem reasonably in a good time. Let us discuss the Knapsack problem in detail.
Given a set of items, each with a weight and a value, determine a subset of items to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible.
The knapsack problem is in combinatorial optimization problem. It appears as a subproblem in many, more complex mathematical models of realworld problems. One general approach to difficult problems is to identify the most restrictive constraint, ignore the others, solve a knapsack problem, and somehow adjust the solution to satisfy the ignored constraints.
In many cases of resource allocation along with some constraint, the problem can be derived in a similar way of Knapsack problem. Following is a set of example.
A thief is robbing a store and can carry a maximal weight of W into his knapsack. There are n items available in the store and weight of i^{th} item is w_{i} and its profit is p_{i}. What items should the thief take?
In this context, the items should be selected in such a way that the thief will carry those items for which he will gain maximum profit. Hence, the objective of the thief is to maximize the profit.
Based on the nature of the items, Knapsack problems are categorized as
In this case, items can be broken into smaller pieces, hence the thief can select fractions of items.
According to the problem statement,
There are n items in the store
Weight of i^{th} item $w_{i} > 0$
Profit for i^{th} item $p_{i} > 0$ and
Capacity of the Knapsack is W
In this version of Knapsack problem, items can be broken into smaller pieces. So, the thief may take only a fraction x_{i} of i^{th} item.
$$0 \leqslant x_{i} \leqslant 1$$
The i^{th} item contributes the weight $x_{i}.w_{i}$ to the total weight in the knapsack and profit $x_{i}.p_{i}$ to the total profit.
Hence, the objective of this algorithm is to
$$maximize\:\displaystyle\sum\limits_{n=1}^n (x_{i}.p_{}i)$$
subject to constraint,
$$\displaystyle\sum\limits_{n=1}^n (x_{i}.w_{}i) \leqslant W$$
It is clear that an optimal solution must fill the knapsack exactly, otherwise we could add a fraction of one of the remaining items and increase the overall profit.
Thus, an optimal solution can be obtained by
$$\displaystyle\sum\limits_{n=1}^n (x_{i}.w_{}i) = W$$
In this context, first we need to sort those items according to the value of $\frac{p_{i}}{w_{i}}$, so that $\frac{p_{i}+1}{w_{i}+1}$ ≤ $\frac{p_{i}}{w_{i}}$ . Here, x is an array to store the fraction of items.
Algorithm: GreedyFractionalKnapsack (w[1..n], p[1..n], W) for i = 1 to n do x[i] = 0 weight = 0 for i = 1 to n if weight + w[i] ≤ W then x[i] = 1 weight = weight + w[i] else x[i] = (W  weight) / w[i] weight = W break return x
If the provided items are already sorted into a decreasing order of $\mathbf{\frac{p_{i}}{w_{i}}}$, then the whileloop takes a time in O(n); Therefore, the total time including the sort is in O(n logn).
Let us consider that the capacity of the knapsack W = 60 and the list of provided items are shown in the following table −
Item  A  B  C  D 

Profit  280  100  120  120 
Weight  40  10  20  24 
Ratio $(\frac{p_{i}}{w_{i}})$  7  10  6  5 
As the provided items are not sorted based on $\mathbf{\frac{p_{i}}{w_{i}}}$. After sorting, the items are as shown in the following table.
Item  B  A  C  D 

Profit  100  280  120  120 
Weight  10  40  20  24 
Ratio $(\frac{p_{i}}{w_{i}})$  10  7  6  5 
After sorting all the items according to $\frac{p_{i}}{w_{i}}$. First all of B is chosen as weight of B is less than the capacity of the knapsack. Next, item A is chosen, as the available capacity of the knapsack is greater than the weight of A. Now, C is chosen as the next item. However, the whole item cannot be chosen as the remaining capacity of the knapsack is less than the weight of C.
Hence, fraction of C (i.e. (60 − 50)/20) is chosen.
Now, the capacity of the Knapsack is equal to the selected items. Hence, no more item can be selected.
The total weight of the selected items is 10 + 40 + 20 * (10/20) = 60
And the total profit is 100 + 280 + 120 * (10/20) = 380 + 60 = 440
This is the optimal solution. We cannot gain more profit selecting any different combination of items.
In job sequencing problem, the objective is to find a sequence of jobs, which is completed within their deadlines and gives maximum profit.
Let us consider, a set of n given jobs which are associated with deadlines and profit is earned, if a job is completed by its deadline. These jobs need to be ordered in such a way that there is maximum profit.
It may happen that all of the given jobs may not be completed within their deadlines.
Assume, deadline of i^{th} job J_{i} is d_{i} and the profit received from this job is p_{i}. Hence, the optimal solution of this algorithm is a feasible solution with maximum profit.
Thus, $D(i) > 0$ for $1 \leqslant i \leqslant n$.
Initially, these jobs are ordered according to profit, i.e. $p_{1} \geqslant p_{2} \geqslant p_{3} \geqslant \:... \: \geqslant p_{n}$.
Algorithm: JobSequencingWithDeadline (D, J, n, k) D(0) := J(0) := 0 k := 1 J(1) := 1 // means first job is selected for i = 2 … n do r := k while D(J(r)) > D(i) and D(J(r)) ≠ r do r := r – 1 if D(J(r)) ≤ D(i) and D(i) > r then for l = k … r + 1 by 1 do J(l + 1) := J(l) J(r + 1) := i k := k + 1
In this algorithm, we are using two loops, one is within another. Hence, the complexity of this algorithm is $O(n^2)$.
Let us consider a set of given jobs as shown in the following table. We have to find a sequence of jobs, which will be completed within their deadlines and will give maximum profit. Each job is associated with a deadline and profit.
Job  J_{1}  J_{2}  J_{3}  J_{4}  J_{5} 

Deadline  2  1  3  2  1 
Profit  60  100  20  40  20 
To solve this problem, the given jobs are sorted according to their profit in a descending order. Hence, after sorting, the jobs are ordered as shown in the following table.
Job  J_{2}  J_{1}  J_{4}  J_{3}  J_{5} 

Deadline  1  2  2  3  1 
Profit  100  60  40  20  20 
From this set of jobs, first we select J_{2}, as it can be completed within its deadline and contributes maximum profit.
Next, J_{1} is selected as it gives more profit compared to J_{4}.
In the next clock, J_{4} cannot be selected as its deadline is over, hence J_{3} is selected as it executes within its deadline.
The job J_{5} is discarded as it cannot be executed within its deadline.
Thus, the solution is the sequence of jobs (J_{2}, J_{1}, J_{3}), which are being executed within their deadline and gives maximum profit.
Total profit of this sequence is 100 + 60 + 20 = 180.
Merge a set of sorted files of different length into a single sorted file. We need to find an optimal solution, where the resultant file will be generated in minimum time.
If the number of sorted files are given, there are many ways to merge them into a single sorted file. This merge can be performed pair wise. Hence, this type of merging is called as 2way merge patterns.
As, different pairings require different amounts of time, in this strategy we want to determine an optimal way of merging many files together. At each step, two shortest sequences are merged.
To merge a precord file and a qrecord file requires possibly p + q record moves, the obvious choice being, merge the two smallest files together at each step.
Twoway merge patterns can be represented by binary merge trees. Let us consider a set of n sorted files {f_{1}, f_{2}, f_{3}, …, f_{n}}. Initially, each element of this is considered as a single node binary tree. To find this optimal solution, the following algorithm is used.
Algorithm: TREE (n) for i := 1 to n – 1 do declare new node node.leftchild := least (list) node.rightchild := least (list) node.weight) := ((node.leftchild).weight) + ((node.rightchild).weight) insert (list, node); return least (list);
At the end of this algorithm, the weight of the root node represents the optimal cost.
Let us consider the given files, f_{1}, f_{2}, f_{3}, f_{4} and f_{5} with 20, 30, 10, 5 and 30 number of elements respectively.
If merge operations are performed according to the provided sequence, then
M_{1} = merge f_{1} and f_{2} => 20 + 30 = 50
M_{2} = merge M_{1} and f_{3} => 50 + 10 = 60
M_{3} = merge M_{2} and f_{4} => 60 + 5 = 65
M_{4} = merge M_{3} and f_{5} => 65 + 30 = 95
Hence, the total number of operations is
50 + 60 + 65 + 95 = 270
Now, the question arises is there any better solution?
Sorting the numbers according to their size in an ascending order, we get the following sequence −
f_{4}, f_{3}, f_{1}, f_{2}, f_{5}
Hence, merge operations can be performed on this sequence
M_{1} = merge f_{4} and f_{3} => 5 + 10 = 15
M_{2} = merge M_{1} and f_{1} => 15 + 20 = 35
M_{3} = merge M_{2} and f_{2} => 35 + 30 = 65
M_{4} = merge M_{3} and f_{5} => 65 + 30 = 95
Therefore, the total number of operations is
15 + 35 + 65 + 95 = 210
Obviously, this is better than the previous one.
In this context, we are now going to solve the problem using this algorithm.
Hence, the solution takes 15 + 35 + 60 + 95 = 205 number of comparisons.
Dynamic Programming is also used in optimization problems. Like divideandconquer method, Dynamic Programming solves problems by combining the solutions of subproblems. Moreover, Dynamic Programming algorithm solves each subproblem just once and then saves its answer in a table, thereby avoiding the work of recomputing the answer every time.
Two main properties of a problem suggest that the given problem can be solved using Dynamic Programming. These properties are overlapping subproblems and optimal substructure.
Similar to DivideandConquer approach, Dynamic Programming also combines solutions to subproblems. It is mainly used where the solution of one subproblem is needed repeatedly. The computed solutions are stored in a table, so that these don’t have to be recomputed. Hence, this technique is needed where overlapping subproblem exists.
For example, Binary Search does not have overlapping subproblem. Whereas recursive program of Fibonacci numbers have many overlapping subproblems.
A given problem has Optimal Substructure Property, if the optimal solution of the given problem can be obtained using optimal solutions of its subproblems.
For example, the Shortest Path problem has the following optimal substructure property −
If a node x lies in the shortest path from a source node u to destination node v, then the shortest path from u to v is the combination of the shortest path from u to x, and the shortest path from x to v.
The standard All Pair Shortest Path algorithms like FloydWarshall and BellmanFord are typical examples of Dynamic Programming.
Dynamic Programming algorithm is designed using the following four steps −
In this tutorial, earlier we have discussed Fractional Knapsack problem using Greedy approach. We have shown that Greedy approach gives an optimal solution for Fractional Knapsack. However, this chapter will cover 01 Knapsack problem and its analysis.
In 01 Knapsack, items cannot be broken which means the thief should take the item as a whole or should leave it. This is reason behind calling it as 01 Knapsack.
Hence, in case of 01 Knapsack, the value of x_{i} can be either 0 or 1, where other constraints remain the same.
01 Knapsack cannot be solved by Greedy approach. Greedy approach does not ensure an optimal solution. In many instances, Greedy approach may give an optimal solution.
The following examples will establish our statement.
Let us consider that the capacity of the knapsack is W = 25 and the items are as shown in the following table.
Item  A  B  C  D 

Profit  24  18  18  10 
Weight  24  10  10  7 
Without considering the profit per unit weight (p_{i}/w_{i}), if we apply Greedy approach to solve this problem, first item A will be selected as it will contribute max_{i}mum profit among all the elements.
After selecting item A, no more item will be selected. Hence, for this given set of items total profit is 24. Whereas, the optimal solution can be achieved by selecting items, B and C, where the total profit is 18 + 18 = 36.
Instead of selecting the items based on the overall benefit, in this example the items are selected based on ratio p_{i}/w_{i}. Let us consider that the capacity of the knapsack is W = 60 and the items are as shown in the following table.
Item  A  B  C 

Price  100  280  120 
Weight  10  40  20 
Ratio  10  7  6 
Using the Greedy approach, first item A is selected. Then, the next item B is chosen. Hence, the total profit is 100 + 280 = 380. However, the optimal solution of this instance can be achieved by selecting items, B and C, where the total profit is 280 + 120 = 400.
Hence, it can be concluded that Greedy approach may not give an optimal solution.
To solve 01 Knapsack, Dynamic Programming approach is required.
A thief is robbing a store and can carry a max_{i}mal weight of W into his knapsack. There are n items and weight of i^{th} item is w_{i} and the profit of selecting this item is p_{i}. What items should the thief take?
Let i be the highestnumbered item in an optimal solution S for W dollars. Then S^{'} = S  {i} is an optimal solution for W  w_{i} dollars and the value to the solution S is V_{i} plus the value of the subproblem.
We can express this fact in the following formula: define c[i, w] to be the solution for items 1,2, … , i and the max_{i}mum weight w.
The algorithm takes the following inputs
The max_{i}mum weight W
The number of items n
The two sequences v = <v_{1}, v_{2}, …, v_{n}> and w = <w_{1}, w_{2}, …, w_{n}>
Dynamic01knapsack (v, w, n, W) for w = 0 to W do c[0, w] = 0 for i = 1 to n do c[i, 0] = 0 for w = 1 to W do if w_{i} ≤ w then if v_{i} + c[i1, ww_{i}] then c[i, w] = v_{i} + c[i1, ww_{i}] else c[i, w] = c[i1, w] else c[i, w] = c[i1, w]
The set of items to take can be deduced from the table, starting at c[n, w] and tracing backwards where the optimal values came from.
If c[i, w] = c[i1, w], then item i is not part of the solution, and we continue tracing with c[i1, w]. Otherwise, item i is part of the solution, and we continue tracing with c[i1, wW].
This algorithm takes θ(n, w) times as table c has (n + 1).(w + 1) entries, where each entry requires θ(1) time to compute.
The longest common subsequence problem is finding the longest sequence which exists in both the given strings.
Let us consider a sequence S = <s_{1}, s_{2}, s_{3}, s_{4}, …,s_{n}>.
A sequence Z = <z_{1}, z_{2}, z_{3}, z_{4}, …,z_{m}> over S is called a subsequence of S, if and only if it can be derived from S deletion of some elements.
Suppose, X and Y are two sequences over a finite set of elements. We can say that Z is a common subsequence of X and Y, if Z is a subsequence of both X and Y.
If a set of sequences are given, the longest common subsequence problem is to find a common subsequence of all the sequences that is of maximal length.
The longest common subsequence problem is a classic computer science problem, the basis of data comparison programs such as the diffutility, and has applications in bioinformatics. It is also widely used by revision control systems, such as SVN and Git, for reconciling multiple changes made to a revisioncontrolled collection of files.
Let X be a sequence of length m and Y a sequence of length n. Check for every subsequence of X whether it is a subsequence of Y, and return the longest common subsequence found.
There are 2^{m} subsequences of X. Testing sequences whether or not it is a subsequence of Y takes O(n) time. Thus, the naïve algorithm would take O(n2^{m}) time.
Let X = < x_{1}, x_{2}, x_{3},…, x_{m} > and Y = < y_{1}, y_{2}, y_{3},…, y_{n} > be the sequences. To compute the length of an element the following algorithm is used.
In this procedure, table C[m, n] is computed in row major order and another table B[m,n] is computed to construct optimal solution.
Algorithm: LCSLengthTableFormulation (X, Y) m := length(X) n := length(Y) for i = 1 to m do C[i, 0] := 0 for j = 1 to n do C[0, j] := 0 for i = 1 to m do for j = 1 to n do if x_{i} = y_{j} C[i, j] := C[i  1, j  1] + 1 B[i, j] := ‘D’ else if C[i 1, j] ≥ C[i, j 1] C[i, j] := C[i  1, j] + 1 B[i, j] := ‘U’ else C[i, j] := C[i, j  1] B[i, j] := ‘L’ return C and B
Algorithm: PrintLCS (B, X, i, j) if i = 0 and j = 0 return if B[i, j] = ‘D’ PrintLCS(B, X, i1, j1) Print(x_{i}) else if B[i, j] = ‘U’ PrintLCS(B, X, i1, j) else PrintLCS(B, X, i, j1)
This algorithm will print the longest common subsequence of X and Y.
To populate the table, the outer for loop iterates m times and the inner for loop iterates n times. Hence, the complexity of the algorithm is O(m, n), where m and n are the length of two strings.
In this example, we have two strings X = BACDB and Y = BDCB to find the longest common subsequence.
Following the algorithm LCSLengthTableFormulation (as stated above), we have calculated table C (shown on the left hand side) and table B (shown on the right hand side).
In table B, instead of ‘D’, ‘L’ and ‘U’, we are using the diagonal arrow, left arrow and up arrow, respectively. After generating table B, the LCS is determined by function LCSPrint. The result is BCB.
A spanning tree is a subset of an undirected Graph that has all the vertices connected by minimum number of edges.
If all the vertices are connected in a graph, then there exists at least one spanning tree. In a graph, there may exist more than one spanning tree.
In the following graph, the highlighted edges form a spanning tree.
A Minimum Spanning Tree (MST) is a subset of edges of a connected weighted undirected graph that connects all the vertices together with the minimum possible total edge weight. To derive an MST, Prim’s algorithm or Kruskal’s algorithm can be used. Hence, we will discuss Prim’s algorithm in this chapter.
As we have discussed, one graph may have more than one spanning tree. If there are n number of vertices, the spanning tree should have n  1 number of edges. In this context, if each edge of the graph is associated with a weight and there exists more than one spanning tree, we need to find the minimum spanning tree of the graph.
Moreover, if there exist any duplicate weighted edges, the graph may have multiple minimum spanning tree.
In the above graph, we have shown a spanning tree though it’s not the minimum spanning tree. The cost of this spanning tree is (5 + 7 + 3 + 3 + 5 + 8 + 3 + 4) = 38.
We will use Prim’s algorithm to find the minimum spanning tree.
Prim’s algorithm is a greedy approach to find the minimum spanning tree. In this algorithm, to form a MST we can start from an arbitrary vertex.
Algorithm: MSTPrim’s (G, w, r) for each u є G.V u.key = ∞ u.∏ = NIL r.key = 0 Q = G.V while Q ≠ Ф u = ExtractMin (Q) for each v є G.adj[u] if each v є Q and w(u, v) < v.key v.∏ = u v.key = w(u, v)
The function ExtractMin returns the vertex with minimum edge cost. This function works on minheap.
Using Prim’s algorithm, we can start from any vertex, let us start from vertex 1.
Vertex 3 is connected to vertex 1 with minimum edge cost, hence edge (1, 2) is added to the spanning tree.
Next, edge (2, 3) is considered as this is the minimum among edges {(1, 2), (2, 3), (3, 4), (3, 7)}.
In the next step, we get edge (3, 4) and (2, 4) with minimum cost. Edge (3, 4) is selected at random.
In a similar way, edges (4, 5), (5, 7), (7, 8), (6, 8) and (6, 9) are selected. As all the vertices are visited, now the algorithm stops.
The cost of the spanning tree is (2 + 2 + 3 + 2 + 5 + 2 + 3 + 4) = 23. There is no more spanning tree in this graph with cost less than 23.
Dijkstra’s algorithm solves the singlesource shortestpaths problem on a directed weighted graph G = (V, E), where all the edges are nonnegative (i.e., w(u, v) ≥ 0 for each edge (u, v) Є E).
In the following algorithm, we will use one function ExtractMin(), which extracts the node with the smallest key.
Algorithm: Dijkstra’sAlgorithm (G, w, s) for each vertex v Є G.V v.d := ∞ v.∏ := NIL s.d := 0 S := Ф Q := G.V while Q ≠ Ф u := ExtractMin (Q) S := S U {u} for each vertex v Є G.adj[u] if v.d > u.d + w(u, v) v.d := u.d + w(u, v) v.∏ := u
The complexity of this algorithm is fully dependent on the implementation of ExtractMin function. If extract min function is implemented using linear search, the complexity of this algorithm is O(V^{2} + E).
In this algorithm, if we use minheap on which ExtractMin() function works to return the node from Q with the smallest key, the complexity of this algorithm can be reduced further.
Let us consider vertex 1 and 9 as the start and destination vertex respectively. Initially, all the vertices except the start vertex are marked by ∞ and the start vertex is marked by 0.
Vertex  Initial  Step1 V_{1}  Step2 V_{3}  Step3 V_{2}  Step4 V_{4}  Step5 V_{5}  Step6 V_{7}  Step7 V_{8}  Step8 V_{6} 

1  0  0  0  0  0  0  0  0  0 
2  ∞  5  4  4  4  4  4  4  4 
3  ∞  2  2  2  2  2  2  2  2 
4  ∞  ∞  ∞  7  7  7  7  7  7 
5  ∞  ∞  ∞  11  9  9  9  9  9 
6  ∞  ∞  ∞  ∞  ∞  17  17  16  16 
7  ∞  ∞  11  11  11  11  11  11  11 
8  ∞  ∞  ∞  ∞  ∞  16  13  13  13 
9  ∞  ∞  ∞  ∞  ∞  ∞  ∞  ∞  20 
Hence, the minimum distance of vertex 9 from vertex 1 is 20. And the path is
1→ 3→ 7→ 8→ 6→ 9
This path is determined based on predecessor information.
This algorithm solves the single source shortest path problem of a directed graph G = (V, E) in which the edge weights may be negative. Moreover, this algorithm can be applied to find the shortest path, if there does not exist any negative weighted cycle.
Algorithm: BellmanFordAlgorithm (G, w, s) for each vertex v Є G.V v.d := ∞ v.∏ := NIL s.d := 0 for i = 1 to G.V  1 for each edge (u, v) Є G.E if v.d > u.d + w(u, v) v.d := u.d +w(u, v) v.∏ := u for each edge (u, v) Є G.E if v.d > u.d + w(u, v) return FALSE return TRUE
The first for loop is used for initialization, which runs in O(V) times. The next for loop runs V  1 passes over the edges, which takes O(E) times.
Hence, BellmanFord algorithm runs in O(V, E) time.
The following example shows how BellmanFord algorithm works step by step. This graph has a negative edge but does not have any negative cycle, hence the problem can be solved using this technique.
At the time of initialization, all the vertices except the source are marked by ∞ and the source is marked by 0.
In the first step, all the vertices which are reachable from the source are updated by minimum cost. Hence, vertices a and h are updated.
In the next step, vertices a, b, f and e are updated.
Following the same logic, in this step vertices b, f, c and g are updated.
Here, vertices c and d are updated.
Hence, the minimum distance between vertex s and vertex d is 20.
Based on the predecessor information, the path is s→ h→ e→ g→ c→ d
A multistage graph G = (V, E) is a directed graph where vertices are partitioned into k (where k > 1) number of disjoint subsets S = {s_{1},s_{2},…,s_{k}} such that edge (u, v) is in E, then u Є s_{i} and v Є s_{1 + 1} for some subsets in the partition and s_{1} = s_{k} = 1.
The vertex s Є s_{1} is called the source and the vertex t Є s_{k} is called sink.
G is usually assumed to be a weighted graph. In this graph, cost of an edge (i, j) is represented by c(i, j). Hence, the cost of path from source s to sink t is the sum of costs of each edges in this path.
The multistage graph problem is finding the path with minimum cost from source s to sink t.
Consider the following example to understand the concept of multistage graph.
According to the formula, we have to calculate the cost (i, j) using the following steps
In this step, three nodes (node 4, 5. 6) are selected as j. Hence, we have three options to choose the minimum cost at this step.
Cost(3, 4) = min {c(4, 7) + Cost(7, 9),c(4, 8) + Cost(8, 9)} = 7
Cost(3, 5) = min {c(5, 7) + Cost(7, 9),c(5, 8) + Cost(8, 9)} = 5
Cost(3, 6) = min {c(6, 7) + Cost(7, 9),c(6, 8) + Cost(8, 9)} = 5
Two nodes are selected as j because at stage k  3 = 2 there are two nodes, 2 and 3. So, the value i = 2 and j = 2 and 3.
Cost(2, 2) = min {c(2, 4) + Cost(4, 8) + Cost(8, 9),c(2, 6) +
Cost(6, 8) + Cost(8, 9)} = 8
Cost(2, 3) = {c(3, 4) + Cost(4, 8) + Cost(8, 9), c(3, 5) + Cost(5, 8)+ Cost(8, 9), c(3, 6) + Cost(6, 8) + Cost(8, 9)} = 10
Cost (1, 1) = {c(1, 2) + Cost(2, 6) + Cost(6, 8) + Cost(8, 9), c(1, 3) + Cost(3, 5) + Cost(5, 8) + Cost(8, 9))} = 12
c(1, 3) + Cost(3, 6) + Cost(6, 8 + Cost(8, 9))} = 13
Hence, the path having the minimum cost is 1→ 3→ 5→ 8→ 9.
A traveler needs to visit all the cities from a list, where distances between all the cities are known and each city should be visited just once. What is the shortest possible route that he visits each city exactly once and returns to the origin city?
Travelling salesman problem is the most notorious computational problem. We can use bruteforce approach to evaluate every possible tour and select the best one. For n number of vertices in a graph, there are (n  1)! number of possibilities.
Instead of bruteforce using dynamic programming approach, the solution can be obtained in lesser time, though there is no polynomial time algorithm.
Let us consider a graph G = (V, E), where V is a set of cities and E is a set of weighted edges. An edge e(u, v) represents that vertices u and v are connected. Distance between vertex u and v is d(u, v), which should be nonnegative.
Suppose we have started at city 1 and after visiting some cities now we are in city j. Hence, this is a partial tour. We certainly need to know j, since this will determine which cities are most convenient to visit next. We also need to know all the cities visited so far, so that we don't repeat any of them. Hence, this is an appropriate subproblem.
For a subset of cities S Є {1, 2, 3, ... , n} that includes 1, and j Є S, let C(S, j) be the length of the shortest path visiting each node in S exactly once, starting at 1 and ending at j.
When S > 1, we define C(S, 1) = ∝ since the path cannot start and end at 1.
Now, let express C(S, j) in terms of smaller subproblems. We need to start at 1 and end at j. We should select the next city in such a way that
$$C(S, j) = min \:C(S  \lbrace j \rbrace, i) + d(i, j)\:where\: i\in S \: and\: i \neq jc(S, j) = minC(s \lbrace j \rbrace, i)+ d(i,j) \:where\: i\in S \: and\: i \neq j $$
Algorithm: TravelingSalesmanProblem C ({1}, 1) = 0 for s = 2 to n do for all subsets S Є {1, 2, 3, … , n} of size s and containing 1 C (S, 1) = ∞ for all j Є S and j ≠ 1 C (S, j) = min {C (S – {j}, i) + d(i, j) for i Є S and i ≠ j} Return minj C ({1, 2, 3, …, n}, j) + d(j, i)
There are at the most $2^n.n$ subproblems and each one takes linear time to solve. Therefore, the total running time is $O(2^n.n^2)$.
In the following example, we will illustrate the steps to solve the travelling salesman problem.
From the above graph, the following table is prepared.
1  2  3  4  
1  0  10  15  20 
2  5  0  9  10 
3  6  13  0  12 
4  8  8  9  0 
$$\small Cost (2,\Phi,1) = d (2,1) = 5\small Cost(2,\Phi,1)=d(2,1)=5$$
$$\small Cost (3,\Phi,1) = d (3,1) = 6\small Cost(3,\Phi,1)=d(3,1)=6$$
$$\small Cost (4,\Phi,1) = d (4,1) = 8\small Cost(4,\Phi,1)=d(4,1)=8$$
$$\small Cost (i,s) = min \lbrace Cost (j,s – (j)) + d [i,j]\rbrace\small Cost (i,s)=min \lbrace Cost (j,s)(j))+ d [i,j]\rbrace$$
$$\small Cost (2,\lbrace 3 \rbrace,1) = d [2,3] + Cost (3,\Phi,1) = 9 + 6 = 15cost(2,\lbrace3 \rbrace,1)=d[2,3]+cost(3,\Phi ,1)=9+6=15$$
$$\small Cost (2,\lbrace 4 \rbrace,1) = d [2,4] + Cost (4,\Phi,1) = 10 + 8 = 18cost(2,\lbrace4 \rbrace,1)=d[2,4]+cost(4,\Phi,1)=10+8=18$$
$$\small Cost (3,\lbrace 2 \rbrace,1) = d [3,2] + Cost (2,\Phi,1) = 13 + 5 = 18cost(3,\lbrace2 \rbrace,1)=d[3,2]+cost(2,\Phi,1)=13+5=18$$
$$\small Cost (3,\lbrace 4 \rbrace,1) = d [3,4] + Cost (4,\Phi,1) = 12 + 8 = 20cost(3,\lbrace4 \rbrace,1)=d[3,4]+cost(4,\Phi,1)=12+8=20$$
$$\small Cost (4,\lbrace 3 \rbrace,1) = d [4,3] + Cost (3,\Phi,1) = 9 + 6 = 15cost(4,\lbrace3 \rbrace,1)=d[4,3]+cost(3,\Phi,1)=9+6=15$$
$$\small Cost (4,\lbrace 2 \rbrace,1) = d [4,2] + Cost (2,\Phi,1) = 8 + 5 = 13cost(4,\lbrace2 \rbrace,1)=d[4,2]+cost(2,\Phi,1)=8+5=13$$
$$\small Cost(2, \lbrace 3, 4 \rbrace, 1)=\begin{cases}d[2, 3] + Cost(3, \lbrace 4 \rbrace, 1) = 9 + 20 = 29\\d[2, 4] + Cost(4, \lbrace 3 \rbrace, 1) = 10 + 15 = 25=25\small Cost (2,\lbrace 3,4 \rbrace,1)\\\lbrace d[2,3]+ \small cost(3,\lbrace4\rbrace,1)=9+20=29d[2,4]+ \small Cost (4,\lbrace 3 \rbrace ,1)=10+15=25\end{cases}= 25$$
$$\small Cost(3, \lbrace 2, 4 \rbrace, 1)=\begin{cases}d[3, 2] + Cost(2, \lbrace 4 \rbrace, 1) = 13 + 18 = 31\\d[3, 4] + Cost(4, \lbrace 2 \rbrace, 1) = 12 + 13 = 25=25\small Cost (3,\lbrace 2,4 \rbrace,1)\\\lbrace d[3,2]+ \small cost(2,\lbrace4\rbrace,1)=13+18=31d[3,4]+ \small Cost (4,\lbrace 2 \rbrace ,1)=12+13=25\end{cases}= 25$$
$$\small Cost(4, \lbrace 2, 3 \rbrace, 1)=\begin{cases}d[4, 2] + Cost(2, \lbrace 3 \rbrace, 1) = 8 + 15 = 23\\d[4, 3] + Cost(3, \lbrace 2 \rbrace, 1) = 9 + 18 = 27=23\small Cost (4,\lbrace 2,3 \rbrace,1)\\\lbrace d[4,2]+ \small cost(2,\lbrace3\rbrace,1)=8+15=23d[4,3]+ \small Cost (3,\lbrace 2 \rbrace ,1)=9+18=27\end{cases}= 23$$
$$\small Cost(1, \lbrace 2, 3, 4 \rbrace, 1)=\begin{cases}d[1, 2] + Cost(2, \lbrace 3, 4 \rbrace, 1) = 10 + 25 = 35\\d[1, 3] + Cost(3, \lbrace 2, 4 \rbrace, 1) = 15 + 25 = 40\\d[1, 4] + Cost(4, \lbrace 2, 3 \rbrace, 1) = 20 + 23 = 43=35 cost(1,\lbrace 2,3,4 \rbrace),1)\\d[1,2]+cost(2,\lbrace 3,4 \rbrace,1)=10+25=35\\d[1,3]+cost(3,\lbrace 2,4 \rbrace,1)=15+25=40\\d[1,4]+cost(4,\lbrace 2,3 \rbrace ,1)=20+23=43=35\end{cases}$$
The minimum cost path is 35.
Start from cost {1, {2, 3, 4}, 1}, we get the minimum value for d [1, 2]. When s = 3, select the path from 1 to 2 (cost is 10) then go backwards. When s = 2, we get the minimum value for d [4, 2]. Select the path from 2 to 4 (cost is 10) then go backwards.
When s = 1, we get the minimum value for d [4, 3]. Selecting path 4 to 3 (cost is 9), then we shall go to then go to s = Φ step. We get the minimum value for d [3, 1] (cost is 6).
A Binary Search Tree (BST) is a tree where the key values are stored in the internal nodes. The external nodes are null nodes. The keys are ordered lexicographically, i.e. for each internal node all the keys in the left subtree are less than the keys in the node, and all the keys in the right subtree are greater.
When we know the frequency of searching each one of the keys, it is quite easy to compute the expected cost of accessing each node in the tree. An optimal binary search tree is a BST, which has minimal expected cost of locating each node
Search time of an element in a BST is O(n), whereas in a BalancedBST search time is O(log n). Again the search time can be improved in Optimal Cost Binary Search Tree, placing the most frequently used data in the root and closer to the root element, while placing the least frequently used data near leaves and in leaves.
Here, the Optimal Binary Search Tree Algorithm is presented. First, we build a BST from a set of provided n number of distinct keys < k_{1}, k_{2}, k_{3}, ... k_{n} >. Here we assume, the probability of accessing a key K_{i} is p_{i}. Some dummy keys (d_{0}, d_{1}, d_{2}, ... d_{n}) are added as some searches may be performed for the values which are not present in the Key set K. We assume, for each dummy key d_{i} probability of access is q_{i}.
OptimalBinarySearchTree(p, q, n) e[1…n + 1, 0…n], w[1…n + 1, 0…n], root[1…n + 1, 0…n] for i = 1 to n + 1 do e[i, i  1] := q_{i}  1 w[i, i  1] := q_{i}  1 for l = 1 to n do for i = 1 to n – l + 1 do j = i + l – 1 e[i, j] := ∞ w[i, i] := w[i, i 1] + p_{j} + q_{j} for r = i to j do t := e[i, r  1] + e[r + 1, j] + w[i, j] if t < e[i, j] e[i, j] := t root[i, j] := r return e and root
The algorithm requires O (n^{3}) time, since three nested for loops are used. Each of these loops takes on at most n values.
Considering the following tree, the cost is 2.80, though this is not an optimal result.
Node  Depth  Probability  Contribution 

k_{1}  1  0.15  0.30 
k_{2}  0  0.10  0.10 
k_{3}  2  0.05  0.15 
k_{4}  1  0.10  0.20 
k_{5}  2  0.20  0.60 
d_{0}  2  0.05  0.15 
d_{1}  2  0.10  0.30 
d_{2}  3  0.05  0.20 
d_{3}  3  0.05  0.20 
d_{4}  3  0.05  0.20 
d_{5}  3  0.10  0.40 
Total  2.80 
To get an optimal solution, using the algorithm discussed in this chapter, the following tables are generated.
In the following tables, column index is i and row index is j.
e  1  2  3  4  5  6 

5  2.75  2.00  1.30  0.90  0.50  0.10 
4  1.75  1.20  0.60  0.30  0.05  
3  1.25  0.70  0.25  0.05  
2  0.90  0.40  0.05  
1  0.45  0.10  
0  0.05 
w  1  2  3  4  5  6 

5  1.00  0.80  0.60  0.50  0.35  0.10 
4  0.70  0.50  0.30  0.20  0.05  
3  0.55  0.35  0.15  0.05  
2  0.45  0.25  0.05  
1  0.30  0.10  
0  0.05 
root  1  2  3  4  5 

5  2  4  5  5  5 
4  2  2  4  4  
3  2  2  3  
2  1  2  
1  1 
From these tables, the optimal tree can be formed.
There are several types of heaps, however in this chapter, we are going to discuss binary heap. A binary heap is a data structure, which looks similar to a complete binary tree. Heap data structure obeys ordering properties discussed below. Generally, a Heap is represented by an array. In this chapter, we are representing a heap by H.
As the elements of a heap is stored in an array, considering the starting index as 1, the position of the parent node of i^{th} element can be found at ⌊ i/2 ⌋ . Left child and right child of i^{th} node is at position 2i and 2i + 1.
A binary heap can be classified further as either a maxheap or a minheap based on the ordering property.
In this heap, the key value of a node is greater than or equal to the key value of the highest child.
Hence, H[Parent(i)] ≥ H[i]
In meanheap, the key value of a node is lesser than or equal to the key value of the lowest child.
Hence, H[Parent(i)] ≤ H[i]
In this context, basic operations are shown below with respect to MaxHeap. Insertion and deletion of elements in and from heaps need rearrangement of elements. Hence, Heapify function needs to be called.
A complete binary tree can be represented by an array, storing its elements using level order traversal.
Let us consider a heap (as shown below) which will be represented by an array H.
Considering the starting index as 0, using level order traversal, the elements are being kept in an array as follows.
Index  0  1  2  3  4  5  6  7  8  ... 
elements  70  30  50  12  20  35  25  4  8  ... 
In this context, operations on heap are being represented with respect to MaxHeap.
To find the index of the parent of an element at index i, the following algorithm Parent (numbers[], i) is used.
Algorithm: Parent (numbers[], i) if i == 1 return NULL else [i / 2]
The index of the left child of an element at index i can be found using the following algorithm, LeftChild (numbers[], i).
Algorithm: LeftChild (numbers[], i) If 2 * i ≤ heapsize return [2 * i] else return NULL
The index of the right child of an element at index i can be found using the following algorithm, RightChild(numbers[], i).
Algorithm: RightChild (numbers[], i) if 2 * i < heapsize return [2 * i + 1] else return NULL
To insert an element in a heap, the new element is initially appended to the end of the heap as the last element of the array.
After inserting this element, heap property may be violated, hence the heap property is repaired by comparing the added element with its parent and moving the added element up a level, swapping positions with the parent. This process is called percolation up.
The comparison is repeated until the parent is larger than or equal to the percolating element.
Algorithm: MaxHeapInsert (numbers[], key) heapsize = heapsize + 1 numbers[heapsize] = ∞ i = heapsize numbers[i] = key while i > 1 and numbers[Parent(numbers[], i)] < numbers[i] exchange(numbers[i], numbers[Parent(numbers[], i)]) i = Parent (numbers[], i)
Initially, an element is being added at the end of the array. If it violates the heap property, the element is exchanged with its parent. The height of the tree is log n. Maximum log n number of operations needs to be performed.
Hence, the complexity of this function is O(log n).
Let us consider a maxheap, as shown below, where a new element 5 needs to be added.
Initially, 55 will be added at the end of this array.
After insertion, it violates the heap property. Hence, the element needs to swap with its parent. After swap, the heap looks like the following.
Again, the element violates the property of heap. Hence, it is swapped with its parent.
Now, we have to stop.
Heapify method rearranges the elements of an array where the left and right subtree of i^{th} element obeys the heap property.
Algorithm: MaxHeapify(numbers[], i) leftchild := numbers[2i] rightchild := numbers [2i + 1] if leftchild ≤ numbers[].size and numbers[leftchild] > numbers[i] largest := leftchild else largest := i if rightchild ≤ numbers[].size and numbers[rightchild] > numbers[largest] largest := rightchild if largest ≠ i swap numbers[i] with numbers[largest] MaxHeapify(numbers, largest)
When the provided array does not obey the heap property, Heap is built based on the following algorithm BuildMaxHeap (numbers[]).
Algorithm: BuildMaxHeap(numbers[]) numbers[].size := numbers[].length fori = ⌊ numbers[].length/2 ⌋ to 1 by 1 MaxHeapify (numbers[], i)
Extract method is used to extract the root element of a Heap. Following is the algorithm.
Algorithm: HeapExtractMax (numbers[]) max = numbers[1] numbers[1] = numbers[heapsize] heapsize = heapsize – 1 MaxHeapify (numbers[], 1) return max
Let us consider the same example discussed previously. Now we want to extract an element. This method will return the root element of the heap.
After deletion of the root element, the last element will be moved to the root position.
Now, Heapify function will be called. After Heapify, the following heap is generated.
Bubble Sort is an elementary sorting algorithm, which works by repeatedly exchanging adjacent elements, if necessary. When no exchanges are required, the file is sorted.
This is the simplest technique among all sorting algorithms.
Algorithm: SequentialBubbleSort (A) fori← 1 to length [A] do for j ← length [A] downto i +1 do if A[A] < A[j  1] then Exchange A[j] ↔ A[j1]
voidbubbleSort(int numbers[], intarray_size) { inti, j, temp; for (i = (array_size  1); i >= 0; i) for (j = 1; j <= i; j++) if (numbers[j  1] > numbers[j]) { temp = numbers[j1]; numbers[j  1] = numbers[j]; numbers[j] = temp; } }
Here, the number of comparisons are
1 + 2 + 3 +...+ (n  1) = n(n  1)/2 = O(n^{2})
Clearly, the graph shows the n^{2} nature of the bubble sort.
In this algorithm, the number of comparison is irrespective of the data set, i.e. whether the provided input elements are in sorted order or in reverse order or at random.
From the algorithm stated above, it is clear that bubble sort does not require extra memory.
Unsorted list: 

5 > 2 swap 


5 > 1 swap 


5 > 4 swap 


5 > 3 swap 


5 < 7 no swap 


7 > 6 swap 

2 > 1 swap 


2 < 4 no swap 


4 > 3 swap 


4 < 5 no swap 


5 < 6 no swap 

There is no change in 3^{rd}, 4^{th}, 5^{th} and 6^{th} iteration.
Finally,
the sorted list is 

Insertion sort is a very simple method to sort numbers in an ascending or descending order. This method follows the incremental method. It can be compared with the technique how cards are sorted at the time of playing a game.
The numbers, which are needed to be sorted, are known as keys. Here is the algorithm of the insertion sort method.
Algorithm: InsertionSort(A) for j = 2 to A.length key = A[j] i = j – 1 while i > 0 and A[i] > key A[i + 1] = A[i] i = i 1 A[i + 1] = key
Run time of this algorithm is very much dependent on the given input.
If the given numbers are sorted, this algorithm runs in O(n) time. If the given numbers are in reverse order, the algorithm runs in O(n^{2}) time.
Unsorted list: 

1^{st} iteration:
Key = a[2] = 13
a[1] = 2 < 13
Swap, no swap 

2^{nd} iteration:
Key = a[3] = 5
a[2] = 13 > 5
Swap 5 and 13 

Next, a[1] = 2 < 13
Swap, no swap 

3^{rd} iteration:
Key = a[4] = 18
a[3] = 13 < 18,
a[2] = 5 < 18,
a[1] = 2 < 18
Swap, no swap 

4^{th} iteration:
Key = a[5] = 14
a[4] = 18 > 14
Swap 18 and 14 

Next, a[3] = 13 < 14,
a[2] = 5 < 14,
a[1] = 2 < 14
So, no swap 

Finally,
the sorted list is 

This type of sorting is called Selection Sort as it works by repeatedly sorting elements. It works as follows: first find the smallest in the array and exchange it with the element in the first position, then find the second smallest element and exchange it with the element in the second position, and continue in this way until the entire array is sorted.
Algorithm: SelectionSort (A) fori ← 1 to n1 do min j ← i; min x ← A[i] for j ←i + 1 to n do if A[j] < min x then min j ← j min x ← A[j] A[min j] ← A [i] A[i] ← min x
Selection sort is among the simplest of sorting techniques and it works very well for small files. It has a quite important application as each item is actually moved at the most once.
Section sort is a method of choice for sorting files with very large objects (records) and small keys. The worst case occurs if the array is already sorted in a descending order and we want to sort them in an ascending order.
Nonetheless, the time required by selection sort algorithm is not very sensitive to the original order of the array to be sorted: the test if A[j] < min x is executed exactly the same number of times in every case.
Selection sort spends most of its time trying to find the minimum element in the unsorted part of the array. It clearly shows the similarity between Selection sort and Bubble sort.
Bubble sort selects the maximum remaining elements at each stage, but wastes some effort imparting some order to an unsorted part of the array.
Selection sort is quadratic in both the worst and the average case, and requires no extra memory.
For each i from 1 to n  1, there is one exchange and n  i comparisons, so there is a total of n  1 exchanges and
(n − 1) + (n − 2) + ...+ 2 + 1 = n(n − 1)/2 comparisons.
These observations hold, no matter what the input data is.
In the worst case, this could be quadratic, but in the average case, this quantity is O(n log n). It implies that the running time of Selection sort is quite insensitive to the input.
Void SelectionSort(int numbers[], int array_size) { int i, j; int min, temp; for (i = 0; I < array_size1; i++) { min = i; for (j = i+1; j < array_size; j++) if (numbers[j] < numbers[min]) min = j; temp = numbers[i]; numbers[i] = numbers[min]; numbers[min] = temp; } }
Unsorted list: 

Smallest = 5
2 < 5, smallest = 2
1 < 2, smallest = 1
4 > 1, smallest = 1
3 > 1, smallest = 1
Swap 5 and 1 

Smallest = 2
2 < 5, smallest = 2
2 < 4, smallest = 2
2 < 3, smallest = 2
No Swap 

Smallest = 5
4 < 5, smallest = 4
3 < 4, smallest = 3
Swap 5 and 3 

Smallest = 4
4 < 5, smallest = 4
No Swap 

Finally,
the sorted list is 

It is used on the principle of divideandconquer. Quick sort is an algorithm of choice in many situations as it is not difficult to implement. It is a good general purpose sort and it consumes relatively fewer resources during execution.
It is inplace since it uses only a small auxiliary stack.
It requires only n (log n) time to sort n items.
It has an extremely short inner loop.
This algorithm has been subjected to a thorough mathematical analysis, a very precise statement can be made about performance issues.
It is recursive. Especially, if recursion is not available, the implementation is extremely complicated.
It requires quadratic (i.e., n2) time in the worstcase.
It is fragile, i.e. a simple mistake in the implementation can go unnoticed and cause it to perform badly.
Quick sort works by partitioning a given array A[p ... r] into two nonempty sub array A[p ... q] and A[q+1 ... r] such that every key in A[p ... q] is less than or equal to every key in A[q+1 ... r].
Then, the two subarrays are sorted by recursive calls to Quick sort. The exact position of the partition depends on the given array and index q is computed as a part of the partitioning procedure.
Algorithm: QuickSort (A, p, r) if p < r then q Partition (A, p, r) QuickSort (A, p, q) QuickSort (A, q + r, r)
Note that to sort the entire array, the initial call should be QuickSort (A, 1, length[A])
As a first step, Quick Sort chooses one of the items in the array to be sorted as pivot. Then, the array is partitioned on either side of the pivot. Elements that are less than or equal to pivot will move towards the left, while the elements that are greater than or equal to pivot will move towards the right.
Partitioning procedure rearranges the subarrays inplace.
Function: Partition (A, p, r) x ← A[p] i ← p1 j ← r+1 while TRUE do Repeat j ← j  1 until A[j] ≤ x Repeat i← i+1 until A[i] ≥ x if i < j then exchange A[i] ↔ A[j] else return j
The worst case complexity of QuickSort algorithm is O(n^{2}). However using this technique, in average cases generally we get the output in O(n log n) time.
Radix sort is a small method that many people intuitively use when alphabetizing a large list of names. Specifically, the list of names is first sorted according to the first letter of each name, that is, the names are arranged in 26 classes.
Intuitively, one might want to sort numbers on their most significant digit. However, Radix sort works counterintuitively by sorting on the least significant digits first. On the first pass, all the numbers are sorted on the least significant digit and combined in an array. Then on the second pass, the entire numbers are sorted again on the second least significant digits and combined in an array and so on.
Algorithm: RadixSort (list, n) shift = 1 for loop = 1 to keysize do for entry = 1 to n do bucketnumber = (list[entry].key / shift) mod 10 append (bucket[bucketnumber], list[entry]) list = combinebuckets() shift = shift * 10
Each key is looked at once for each digit (or letter if the keys are alphabetic) of the longest key. Hence, if the longest key has m digits and there are n keys, radix sort has order O(m.n).
However, if we look at these two values, the size of the keys will be relatively small when compared to the number of keys. For example, if we have sixdigit keys, we could have a million different records.
Here, we see that the size of the keys is not significant, and this algorithm is of linear complexity O(n).
Following example shows how Radix sort operates on seven 3digits number.
Input  1^{st} Pass  2^{nd} Pass  3^{rd} Pass 

329  720  720  329 
457  355  329  355 
657  436  436  436 
839  457  839  457 
436  657  355  657 
720  329  457  720 
355  839  657  839 
In the above example, the first column is the input. The remaining columns show the list after successive sorts on increasingly significant digits position. The code for Radix sort assumes that each element in an array A of n elements has d digits, where digit 1 is the lowestorder digit and d is the highestorder digit.
To understand class P and NP, first we should know the computational model. Hence, in this chapter we will discuss two important computational models.
One of these models is deterministic onetape Turing machine. This machine consists of a finite state control, a readwrite head and a twoway tape with infinite sequence.
Following is the schematic diagram of a deterministic onetape Turing machine.
A program for a deterministic Turing machine specifies the following information −
In algorithmic analysis, if a problem is solvable in polynomial time by a deterministic one tape Turing machine, the problem belongs to P class.
To solve the computational problem, another model is the Nondeterministic Turing Machine (NDTM). The structure of NDTM is similar to DTM, however here we have one additional module known as the guessing module, which is associated with one writeonly head.
Following is the schematic diagram.
If the problem is solvable in polynomial time by a nondeterministic Turing machine, the problem belongs to NP class.
In an undirected graph, a clique is a complete subgraph of the given graph. Complete subgraph means, all the vertices of this subgraph is connected to all other vertices of this subgraph.
The MaxClique problem is the computational problem of finding maximum clique of the graph. Max clique is used in many realworld problems.
Let us consider a social networking application, where vertices represent people’s profile and the edges represent mutual acquaintance in a graph. In this graph, a clique represents a subset of people who all know each other.
To find a maximum clique, one can systematically inspect all subsets, but this sort of bruteforce search is too timeconsuming for networks comprising more than a few dozen vertices.
Algorithm: MaxClique (G, n, k) S := Φ for i = 1 to k do t := choice (1…n) if t Є S then return failure S := S ∪ t for all pairs (i, j) such that i Є S and j Є S and i ≠ j do if (i, j) is not a edge of the graph then return failure return success
MaxClique problem is a nondeterministic algorithm. In this algorithm, first we try to determine a set of k distinct vertices and then we try to test whether these vertices form a complete graph.
There is no polynomial time deterministic algorithm to solve this problem. This problem is NPComplete.
Take a look at the following graph. Here, the subgraph containing vertices 2, 3, 4 and 6 forms a complete graph. Hence, this subgraph is a clique. As this is the maximum complete subgraph of the provided graph, it’s a 4Clique.
A vertexcover of an undirected graph G = (V, E) is a subset of vertices V^{'} ⊆ V such that if edge (u, v) is an edge of G, then either u in V or v in V^{'} or both.
Find a vertexcover of maximum size in a given undirected graph. This optimal vertexcover is the optimization version of an NPcomplete problem. However, it is not too hard to find a vertexcover that is near optimal.
APPROXVERTEX_COVER (G: Graph) c ← { } E^{'} ← E[G] while E^{'} is not empty do Let (u, v) be an arbitrary edge of E^{'} c ← c U {u, v} Remove from E^{'} every edge incident on either u or v return c
The set of edges of the given graph is −
{(1,6),(1,2),(1,4),(2,3),(2,4),(6,7),(4,7),(7,8),(3,8),(3,5),(8,5)}
Now, we start by selecting an arbitrary edge (1,6). We eliminate all the edges, which are either incident to vertex 1 or 6 and we add edge (1,6) to cover.
In the next step, we have chosen another edge (2,3) at random
Now we select another edge (4,7).
We select another edge (8,5).
Hence, the vertex cover of this graph is {1,2,4,5}.
It is easy to see that the running time of this algorithm is O(V + E), using adjacency list to represent E^{'}.
In Computer Science, many problems are solved where the objective is to maximize or minimize some values, whereas in other problems we try to find whether there is a solution or not. Hence, the problems can be categorized as follows −
Optimization problems are those for which the objective is to maximize or minimize some values. For example,
Finding the minimum number of colors needed to color a given graph.
Finding the shortest path between two vertices in a graph.
There are many problems for which the answer is a Yes or a No. These types of problems are known as decision problems. For example,
Whether a given graph can be colored by only 4colors.
Finding Hamiltonian cycle in a graph is not a decision problem, whereas checking a graph is Hamiltonian or not is a decision problem.
Every decision problem can have only two answers, yes or no. Hence, a decision problem may belong to a language if it provides an answer ‘yes’ for a specific input. A language is the totality of inputs for which the answer is Yes. Most of the algorithms discussed in the previous chapters are polynomial time algorithms.
For input size n, if worstcase time complexity of an algorithm is O(n^{k}), where k is a constant, the algorithm is a polynomial time algorithm.
Algorithms such as Matrix Chain Multiplication, Single Source Shortest Path, All Pair Shortest Path, Minimum Spanning Tree, etc. run in polynomial time. However there are many problems, such as traveling salesperson, optimal graph coloring, Hamiltonian cycles, finding the longest path in a graph, and satisfying a Boolean formula, for which no polynomial time algorithms is known. These problems belong to an interesting class of problems, called the NPComplete problems, whose status is unknown.
In this context, we can categorize the problems as follows −
The class P consists of those problems that are solvable in polynomial time, i.e. these problems can be solved in time O(n^{k}) in worstcase, where k is constant.
These problems are called tractable, while others are called intractable or superpolynomial.
Formally, an algorithm is polynomial time algorithm, if there exists a polynomial p(n) such that the algorithm can solve any instance of size n in a time O(p(n)).
Problem requiring Ω(n^{50}) time to solve are essentially intractable for large n. Most known polynomial time algorithm run in time O(n^{k}) for fairly low value of k.
The advantages in considering the class of polynomialtime algorithms is that all reasonable deterministic single processor model of computation can be simulated on each other with at most a polynomial slowd
The class NP consists of those problems that are verifiable in polynomial time. NP is the class of decision problems for which it is easy to check the correctness of a claimed answer, with the aid of a little extra information. Hence, we aren’t asking for a way to find a solution, but only to verify that an alleged solution really is correct.
Every problem in this class can be solved in exponential time using exhaustive search.
Every decision problem that is solvable by a deterministic polynomial time algorithm is also solvable by a polynomial time nondeterministic algorithm.
All problems in P can be solved with polynomial time algorithms, whereas all problems in NP  P are intractable.
It is not known whether P = NP. However, many problems are known in NP with the property that if they belong to P, then it can be proved that P = NP.
If P ≠ NP, there are problems in NP that are neither in P nor in NPComplete.
The problem belongs to class P if it’s easy to find a solution for the problem. The problem belongs to NP, if it’s easy to check a solution that may have been very tedious to find.
Stephen Cook presented four theorems in his paper “The Complexity of Theorem Proving Procedures”. These theorems are stated below. We do understand that many unknown terms are being used in this chapter, but we don’t have any scope to discuss everything in detail.
Following are the four theorems by Stephen Cook −
If a set S of strings is accepted by some nondeterministic Turing machine within polynomial time, then S is Preducible to {DNF tautologies}.
The following sets are Preducible to each other in pairs (and hence each has the same polynomial degree of difficulty): {tautologies}, {DNF tautologies}, D3, {subgraph pairs}.
For any T_{Q}(k) of type Q, $\mathbf{\frac{T_{Q}(k)}{\frac{\sqrt{k}}{(log\:k)^2}}}$ is unbounded
There is a T_{Q}(k) of type Q such that $T_{Q}(k)\leqslant 2^{k(log\:k)^2}$
If the set S of strings is accepted by a nondeterministic machine within time T(n) = 2^{n}, and if T_{Q}(k) is an honest (i.e. realtime countable) function of type Q, then there is a constant K, so S can be recognized by a deterministic machine within time T_{Q}(K8^{n}).
First, he emphasized the significance of polynomial time reducibility. It means that if we have a polynomial time reduction from one problem to another, this ensures that any polynomial time algorithm from the second problem can be converted into a corresponding polynomial time algorithm for the first problem.
Second, he focused attention on the class NP of decision problems that can be solved in polynomial time by a nondeterministic computer. Most of the intractable problems belong to this class, NP.
Third, he proved that one particular problem in NP has the property that every other problem in NP can be polynomially reduced to it. If the satisfiability problem can be solved with a polynomial time algorithm, then every problem in NP can also be solved in polynomial time. If any problem in NP is intractable, then satisfiability problem must be intractable. Thus, satisfiability problem is the hardest problem in NP.
Fourth, Cook suggested that other problems in NP might share with the satisfiability problem this property of being the hardest member of NP.
A problem is in the class NPC if it is in NP and is as hard as any problem in NP. A problem is NPhard if all problems in NP are polynomial time reducible to it, even though it may not be in NP itself.
If a polynomial time algorithm exists for any of these problems, all problems in NP would be polynomial time solvable. These problems are called NPcomplete. The phenomenon of NPcompleteness is important for both theoretical and practical reasons.
A language B is NPcomplete if it satisfies two conditions
B is in NP
Every A in NP is polynomial time reducible to B.
If a language satisfies the second property, but not necessarily the first one, the language B is known as NPHard. Informally, a search problem B is NPHard if there exists some NPComplete problem A that Turing reduces to B.
The problem in NPHard cannot be solved in polynomial time, until P = NP. If a problem is proved to be NPC, there is no need to waste time on trying to find an efficient algorithm for it. Instead, we can focus on design approximation algorithm.
Following are some NPComplete problems, for which no polynomial time algorithm is known.
The following problems are NPHard
In this context, now we will discuss TSP is NPComplete
The traveling salesman problem consists of a salesman and a set of cities. The salesman has to visit each one of the cities starting from a certain one and returning to the same city. The challenge of the problem is that the traveling salesman wants to minimize the total length of the trip
To prove TSP is NPComplete, first we have to prove that TSP belongs to NP. In TSP, we find a tour and check that the tour contains each vertex once. Then the total cost of the edges of the tour is calculated. Finally, we check if the cost is minimum. This can be completed in polynomial time. Thus TSP belongs to NP.
Secondly, we have to prove that TSP is NPhard. To prove this, one way is to show that Hamiltonian cycle ≤_{p} TSP (as we know that the Hamiltonian cycle problem is NPcomplete).
Assume G = (V, E) to be an instance of Hamiltonian cycle.
Hence, an instance of TSP is constructed. We create the complete graph G^{'} = (V, E^{'}), where
$$E^{'}=\lbrace(i, j)\colon i, j \in V \:\:and\:i\neq j$$
Thus, the cost function is defined as follows −
$$t(i,j)=\begin{cases}0 & if\: (i, j)\: \in E\\1 & otherwise\end{cases}$$
Now, suppose that a Hamiltonian cycle h exists in G. It is clear that the cost of each edge in h is 0 in G^{'} as each edge belongs to E. Therefore, h has a cost of 0 in G^{'}. Thus, if graph G has a Hamiltonian cycle, then graph G^{'} has a tour of 0 cost.
Conversely, we assume that G^{'} has a tour h^{'} of cost at most 0. The cost of edges in E^{'} are 0 and 1 by definition. Hence, each edge must have a cost of 0 as the cost of h^{'} is 0. We therefore conclude that h^{'} contains only edges in E.
We have thus proven that G has a Hamiltonian cycle, if and only if G^{'} has a tour of cost at most 0. TSP is NPcomplete.
The algorithms discussed in the previous chapters run systematically. To achieve the goal, one or more previously explored paths toward the solution need to be stored to find the optimal solution.
For many problems, the path to the goal is irrelevant. For example, in NQueens problem, we don’t need to care about the final configuration of the queens as well as in which order the queens are added.
Hill Climbing is a technique to solve certain optimization problems. In this technique, we start with a suboptimal solution and the solution is improved repeatedly until some condition is maximized.
The idea of starting with a suboptimal solution is compared to starting from the base of the hill, improving the solution is compared to walking up the hill, and finally maximizing some condition is compared to reaching the top of the hill.
Hence, the hill climbing technique can be considered as the following phases −
Hill Climbing technique is mainly used for solving computationally hard problems. It looks only at the current state and immediate future state. Hence, this technique is memory efficient as it does not maintain a search tree.
Algorithm: Hill Climbing Evaluate the initial state. Loop until a solution is found or there are no new operators left to be applied:  Select and apply a new operator  Evaluate the new state: goal → quit better than current state → new current state
In iterative improvement method, the optimal solution is achieved by making progress towards an optimal solution in every iteration. However, this technique may encounter local maxima. In this situation, there is no nearby state for a better solution.
This problem can be avoided by different methods. One of these methods is simulated annealing.
This is another method of solving the problem of local optima. This technique conducts a series of searches. Every time, it starts from a randomly generated initial state. Hence, optima or nearly optimal solution can be obtained comparing the solutions of searches performed.
If the heuristic is not convex, Hill Climbing may converge to local maxima, instead of global maxima.
If the target function creates a narrow ridge, then the climber can only ascend the ridge or descend the alley by zigzagging. In this scenario, the climber needs to take very small steps requiring more time to reach the goal.
A plateau is encountered when the search space is flat or sufficiently flat that the value returned by the target function is indistinguishable from the value returned for nearby regions, due to the precision used by the machine to represent its value.
This technique does not suffer from space related issues, as it looks only at the current state. Previously explored paths are not stored.
For most of the problems in Randomrestart Hill Climbing technique, an optimal solution can be achieved in polynomial time. However, for NPComplete problems, computational time can be exponential based on the number of local maxima.
Hill Climbing technique can be used to solve many problems, where the current state allows for an accurate evaluation function, such as NetworkFlow, Travelling Salesman problem, 8Queens problem, Integrated Circuit design, etc.
Hill Climbing is used in inductive learning methods too. This technique is used in robotics for coordination among multiple robots in a team. There are many other problems where this technique is used.
This technique can be applied to solve the travelling salesman problem. First an initial solution is determined that visits all the cities exactly once. Hence, this initial solution is not optimal in most of the cases. Even this solution can be very poor. The Hill Climbing algorithm starts with such an initial solution and makes improvements to it in an iterative way. Eventually, a much shorter route is likely to be obtained.