assignment problem in c

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Is there a greedy algorithm to solve the assignment problem?

The assignment problem is defined as:

There are n people who need to be assigned to n jobs, one person per job. The cost that would accrue if the ith person is assigned to the jth job is a known quantity C[i,j] for each pair i, j = 1, 2, ..., n. The problem is to find an assignment with the minimum total cost.

There is a question asking to design a greedy algorithm to solve the problem. It also asks if the greedy algorithm always yields an optimal solution and for the performance class of the algorithm. Here is my attempt at designing an algorithm:

Am I correct in saying that my algorithm is of O(n^2)? Am I also correct in saying that a greedy algorithm does not always yield an optimal solution? I used my algorithm on the following cost matrix and it is clearly not the optimal solution. Did I Do something wrong?

• greedy-algorithms
• assignment-problem

• $\begingroup$ Algorithms for maximum weight bipartite maximum matching are unfortunately more complicated than that. They don't really follow the greedy paradigm. $\endgroup$ –  Yuval Filmus Commented Apr 7, 2017 at 6:05

The answer of your post question (already given in Yuval comment) is that there is no greedy techniques providing you the optimal answer to an assignment problem.

The commonly used solution is the Hungarian algorithm, see

Harold W. Kuhn, "The Hungarian Method for the assignment problem", Naval Research Logistics Quarterly, 2: 83–97, 1955

for the original paper.

Otherwise your solution seems correct to me, and as usually with greedy algorithms, it will provide you a feasible solution which you may hope to not be "too far" from the global optimal solution.

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Branch and Bound Algorithm

Last updated: March 18, 2024

1. Overview

In computer science, there is a large number of optimization problems which has a finite but extensive number of feasible solutions. Among these, some problems like finding the shortest path in a graph  or  Minimum Spanning Tree  can be solved in polynomial time .

A significant number of optimization problems like production planning , crew scheduling can’t be solved in polynomial time, and they belong to the NP-Hard class . These problems are the example of NP-Hard combinatorial optimization problem .

Branch and bound (B&B) is an algorithm paradigm widely used for solving such problems.

In this tutorial, we’ll discuss the branch and bound method in detail.

2. Basic Idea

Branch and bound algorithms are used to find the optimal solution for combinatory, discrete, and general mathematical optimization problems. In general, given an NP-Hard problem, a branch and bound algorithm explores the entire search space of possible solutions and provides an optimal solution.

A branch and bound algorithm consist of stepwise enumeration of possible candidate solutions by exploring the entire search space. With all the possible solutions, we first build a rooted decision tree. The root node represents the entire search space:

Here, each child node is a partial solution and part of the solution set. Before constructing the rooted decision tree, we set an upper and lower bound for a given problem based on the optimal solution. At each level, we need to make a decision about which node to include in the solution set. At each level, we explore the node with the best bound. In this way, we can find the best and optimal solution fast.

Now it is crucial to find a good upper and lower bound in such cases. We can find an upper bound by using any local optimization method or by picking any point in the search space. On the other hand, we can obtain a lower bound from convex relaxation  or  duality .

In general, we want to partition the solution set into smaller subsets of solution. Then we construct a rooted decision tree, and finally, we choose the best possible subset (node) at each level to find the best possible solution set.

3. When Branch and Bound Is a Good Choice?

We already mentioned some problems where a branch and bound can be an efficient choice over the other algorithms. In this section, we’ll list all such cases where a branch and bound algorithm is a good choice.

If the given problem is a discrete optimization problem, a branch and bound is a good choice. Discrete optimization is a subsection of optimization where the variables in the problem should belong to the discrete set. Examples of such problems are 0-1 Integer Programming  or  Network Flow problem .

Branch and bound work efficiently on the combinatory optimization problems. Given an objective function for an optimization problem, combinatory optimization is a process to find the maxima or minima for the objective function. The domain of the objective function should be discrete and large. Boolean Satisfiability , Integer Linear Programming are examples of the combinatory optimization problems.

4. Branch and Bound Algorithm Example

In this section, we’ll discuss how the job assignment problem can be solved using a branch and bound algorithm.

4.1. Problem Statement

	Job 1	Job 2	Job 3
A	9	3	4
B	7	8	4
C	10	5	2

We can assign any of the available jobs to any worker with the condition that if a job is assigned to a worker, the other workers can’t take that particular job. We should also notice that each job has some cost associated with it, and it differs from one worker to another.

Here the main aim is to complete all the jobs by assigning one job to each worker in such a way that the sum of the cost of all the jobs should be minimized.

4.2. Branch and Bound Algorithm Pseudocode

Now let’s discuss how to solve the job assignment problem using a branch and bound algorithm.

Let’s see the pseudocode first:

In the search space tree, each node contains some information, such as cost, a total number of jobs, as well as a total number of workers.

Now let’s run the algorithm on the sample example we’ve created:

4. Advantages

In a branch and bound algorithm, we don’t explore all the nodes in the tree. That’s why the time complexity of the branch and bound algorithm is less when compared with other algorithms.

If the problem is not large and if we can do the branching in a reasonable amount of time, it finds an optimal solution for a given problem.

The branch and bound algorithm find a minimal path to reach the optimal solution for a given problem. It doesn’t repeat nodes while exploring the tree.

5. Disadvantages

The branch and bound algorithm are time-consuming. Depending on the size of the given problem, the number of nodes in the tree can be too large in the worst case.

Also, parallelization is extremely difficult in the branch and bound algorithm.

6. Conclusion

One of the most popular algorithms used in the optimization problem is the branch and bound algorithm. We’ve discussed it thoroughly in this tutorial.

We’ve explained when a branch and bound algorithm would be the right choice for a user to use. Furthermore, we’ve presented a branch and bound based algorithm for solving the job assignment problem.

Finally, we mentioned some advantages and disadvantages of the branch and bound algorithm.

Press ESC to close

Or check our popular categories..., branch and bound | set 4 (job assignment problem).

Let there be N workers and N jobs. Any worker can be assigned to perform any job, incurring some cost that may vary depending on the work-job assignment. It is required to perform all jobs by assigning exactly one worker to each job and exactly one job to each agent in such a way that the total cost of the assignment is minimized.

Let us explore all approaches for this problem.

Solution 1: Brute Force We generate n! possible job assignments and for each such assignment, we compute its total cost and return the less expensive assignment. Since the solution is a permutation of the n jobs, its complexity is O(n!).

Solution 2: Hungarian Algorithm The optimal assignment can be found using the Hungarian algorithm. The Hungarian algorithm has worst case run-time complexity of O(n^3).

Solution 3: DFS/BFS on state space tree A state space tree is a N-ary tree with property that any path from root to leaf node holds one of many solutions to given problem. We can perform depth-first search on state space tree and but successive moves can take us away from the goal rather than bringing closer. The search of state space tree follows leftmost path from the root regardless of initial state. An answer node may never be found in this approach. We can also perform a Breadth-first search on state space tree. But no matter what the initial state is, the algorithm attempts the same sequence of moves like DFS.

Solution 4: Finding Optimal Solution using Branch and Bound The selection rule for the next node in BFS and DFS is “blind”. i.e. the selection rule does not give any preference to a node that has a very good chance of getting the search to an answer node quickly. The search for an optimal solution can often be speeded by using an “intelligent” ranking function, also called an approximate cost function to avoid searching in sub-trees that do not contain an optimal solution. It is similar to BFS-like search but with one major optimization. Instead of following FIFO order, we choose a live node with least cost. We may not get optimal solution by following node with least promising cost, but it will provide very good chance of getting the search to an answer node quickly.

There are two approaches to calculate the cost function:

• For each worker, we choose job with minimum cost from list of unassigned jobs (take minimum entry from each row).
• For each job, we choose a worker with lowest cost for that job from list of unassigned workers (take minimum entry from each column).

In this article, the first approach is followed.

Let’s take below example and try to calculate promising cost when Job 2 is assigned to worker A.

Since Job 2 is assigned to worker A (marked in green), cost becomes 2 and Job 2 and worker A becomes unavailable (marked in red).

Now we assign job 3 to worker B as it has minimum cost from list of unassigned jobs. Cost becomes 2 + 3 = 5 and Job 3 and worker B also becomes unavailable.

Finally, job 1 gets assigned to worker C as it has minimum cost among unassigned jobs and job 4 gets assigned to worker C as it is only Job left. Total cost becomes 2 + 3 + 5 + 4 = 14.

Below diagram shows complete search space diagram showing optimal solution path in green.

Complete Algorithm:

Below is its C++ implementation.

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assignment-problem

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What is an assignment operator in c.

Assignment Operators in C are used to assign values to the variables. They come under the category of binary operators as they require two operands to operate upon. The left side operand is called a variable and the right side operand is the value. The value on the right side of the "=" is assigned to the variable on the left side of "=". The value on the right side must be of the same data type as the variable on the left side. Hence, the associativity is from right to left.

In this C tutorial , we'll understand the types of C programming assignment operators with examples. To delve deeper you can enroll in our C Programming Course .

Before going in-depth about assignment operators you must know about operators in C. If you haven't visited the Operators in C tutorial, refer to Operators in C: Types of Operators .

Types of Assignment Operators in C

There are two types of assignment operators in C:

1. Simple Assignment Operator (=)

This assigns the value on the right-hand side (RHS) to the variable on the left-hand side (LHS). You can use a literal, another variable, or an expression in the assignment statement.

Example of Simple Assignment Operator

In this example, we have assigned the value 30 to the variable x using the assignment operator (=).

2. Compound Assignment Operators

In addition to the = operator, you can combine arithmetic and bitwise operators with the = symbol to form an augmented or compound assignment operator.

There are five combinations of arithmetic operators with the assignment operator, "=". Let's look at them one by one.

+=	addition assignment	It adds the right operand to the left operand and assigns the result to the left operand.
-=	subtraction assignment	It subtracts the right operand from the left operand and assigns the result to the left operand.
*=	multiplication assignment	It multiplies the right operand with the left operand and assigns the result to the left operand
/=	division assignment	It divides the left operand with the right operand and assigns the result to the left operand.
%=	modulo assignment	It takes modulus using two operands and assigns the result to the left operand.

Example of Augmented Arithmetic and Assignment Operators

There can be five combinations of bitwise operators with the assignment operator, "=". Let's look at them one by one.

&=	bitwise AND assignment	It performs the bitwise AND operation on the variable with the value on the right
|=	bitwise OR assignment	It performs the bitwise OR operation on the variable with the value on the right
^=	bitwise XOR assignment	It performs the bitwise XOR operation on the variable with the value on the right
<<=	bitwise left shift assignment	Shifts the bits of the variable to the left by the value on the right
>>=	bitwise right shift assignment	Shifts the bits of the variable to the right by the value on the right

Example of Augmented Bitwise and Assignment Operators

Practice problems on assignment operators in c, 1. what will the value of "x" be after the execution of the following code.

The correct answer is 52. x starts at 50, increases by 5 to 55, then decreases by 3 to 52.

2. After executing the following code, what is the value of the number variable?

The correct answer is 144. After right-shifting 73 (binary 1001001) by one and then left-shifting the result by two, the value becomes 144 (binary 10010000).

Benefits of Using Assignment Operators

• Simplifies Code: For example, x += 1 is shorter and clearer than x = x + 1.
• Reduces Errors: They break complex expressions into simpler, more manageable parts thus reducing errors.
• Improves Readability: They make the code easier to read and understand by succinctly expressing common operations.
• Enhances Performance: They often operate in place, potentially reducing the need for additional memory or temporary variables.

Best Practices and Tips for Using the Assignment Operator

While performing arithmetic operations with the same variable, use compound assignment operators

• Initialize Variables When Declaring int count = 0 ; // Initialization
• Avoid Complex Expressions in Assignments a = (b + c) * (d - e); // Consider breaking it down: int temp = b + c; a = temp * (d - e);
• Avoid Multiple Assignments in a Single Statement // Instead of this a = b = c = 0 ; // Do this a = 0 ; b = 0 ; c = 0 ;
• Consistent Formatting int result = 0 ; result += 10 ;

When mixing assignments with other operations, use parentheses to ensure the correct order of evaluation.

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One line assignment according condition in c/c++

i would like to know which one of these assignments is faster, safer, better etc. and possibly why:

• conditional-statements
• variable-assignment

• 5 Reading in the character from the file is going to eclipse the time it takes to execute either line. –  Cameron Commented Apr 28, 2012 at 18:41
• 2 Surely 1 is more readable. Whatever the compiler produces in this situation is unlikely to have a noticeable impact on performance. –  user1233508 Commented Apr 28, 2012 at 18:41
• @DCoder: +1 - readability is far more important in this case then minor hypothetical (and premature) micro-optimisations –  Paul R Commented Apr 28, 2012 at 19:01

2 Answers 2

Regarding speed, this would be at least as fast as choice 2:

Whether that would be faster than choice 1 is not immediately obvious to me. However, for tuned performance on an unknown platform, I think that I like the suggested variant on choice 2. The reason is that, at the hardware level, choice 2 (original or variant) does not involve reloading the program counter, but invokes a relatively straightforward shift-register operation, involving relatively few transistors. (Actually, if you want to get really technical about it, I am given to understand that the shift is probably accomplished by multiplexing. To detail that would be too much for the present format, but the point is that the output of (choice == 'y' || choice == 'Y') is effectively piped straight to one of the multiplexer's control lines. Anyway, it's really fast.)

Regarding whether one can safely use the evaluated condition in the manner suggested, ISO/IEC 9899:1999 (E), sect. 6.5.14.3, guarantees that one can safely do this. It reads, "The || operator shall yield 1 if either of its operands compare unequal to 0; otherwise, it yields 0. The result has type int."

(@PaulR rightly observes that electronic-theoretical considerations like the ones this answer offers are not decisive. One must profile actual code on an actual platform to tell for sure which is faster. Nor is this a mere quibble on @PaulR's part. It is all well to argue that choice 2 would be faster, but this does not mean that is is faster. Depending on the CPU in use, branch-predicting or other hardware could promote choice 1, nor would I be extremely surprised if it did.)

• Yeah, i guess like this it would be even faster than choice 2, still is it 100% safe to use evaluated condition as variable ? –  Neko Commented Apr 28, 2012 at 18:47
• -1: you really can't make any general predictions about the efficiency of the above examples, particularly without knowing anything about the specific compiler or platform –  Paul R Commented Apr 28, 2012 at 18:57
• @Neko: It's safe. I have modified the answer above to explain why. –  thb Commented Apr 28, 2012 at 18:57
• 1 @PaulR: Wow, you're a tough audience. Of course you are right, but general electronic principles do come to bear, whether the C standard guarantees their preformance or not. I suspect that multiplexing a bit-shift is cheaper than invoking the CPU's branch-prediction hardware, especially if the brand-predictor guesses wrong and has to flush a pipeline. But what do I know? Nevertheless, your point is well taken (though I'm admittedly a little grumpy about the downvote). –  thb Commented Apr 28, 2012 at 19:04
• @thb: sorry, but it's a pet peeve of mine - making blind assumptions/predictions about performance, particularly without taking compiler and CPU into consideration, is the worst kind of premature optimisation. If performance actually matters in a given case, then profile the different options - evidence in the form of measured performance is infinitely more useful than guesswork. –  Paul R Commented Apr 28, 2012 at 19:08

In my opinion 'Choice 1' is 'faster' because there's only one assignment operation done after the comparison. In 'Choice 2' it does '+' and '<<' along with the comparison part. 'Choice 1' is 'safer' because it is more readable than the other choice so programmer will have less chance to do error in writing it. Choice 1 is 'better' because of the previous two reasons.

• The only problem is that choice 1 is using branching (some compare and jump instruction), where choice 2 is not. –  Neko Commented Apr 28, 2012 at 18:56
• 1 @Neko: You can't know whether the compiler will use branches or not, or what the penalty for branches might be without knowing details of the platform etc –  Paul R Commented Apr 28, 2012 at 18:58
• -1: all predictions are meaningless without taking compiler and platform into consideration –  Paul R Commented Apr 28, 2012 at 19:00
• [sorry,comments don't allow newlines]Dear Neko, I understand what you thought. But look at the code generated by 'gcc' with '-S' option: Choice 1: movl %eax, -4(%rbp) cmpl $121, -4(%rbp) je .L2 cmpl $89, -4(%rbp) jne .L3 .L2: movl $256, %eax jmp .L4 .L3: movl $128, %eax .L4: movl %eax, -8(%rbp) movl $0, %eax leave Choice 2: movl %eax, -4(%rbp) cmpl $121, -4(%rbp) je .L2 cmpl $89, -4(%rbp) jne .L3 .L2: movl $1, %eax jmp .L4 .L3: movl $0, %eax .L4: sall $7, %eax subl $-128, %eax movl %eax, -8(%rbp) movl $0, %eax leave See, the L4 section in both cases. Choice is yours. –  Ayub Commented Apr 28, 2012 at 19:06
• Yes, I agree with Paul R, it is Platform and/or Compiler specific. –  Ayub Commented Apr 28, 2012 at 19:17

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Assignment operators are used for assigning value to a variable. The left side operand of the assignment operator is a variable and right side operand of the assignment operator is a value. The value on the right side must be of the same data-type of the variable on the left side otherwise the compiler will raise an error.

Different types of assignment operators are shown below:

1. “=”: This is the simplest assignment operator. This operator is used to assign the value on the right to the variable on the left. Example:

2. “+=” : This operator is combination of ‘+’ and ‘=’ operators. This operator first adds the current value of the variable on left to the value on the right and then assigns the result to the variable on the left. Example:

If initially value stored in a is 5. Then (a += 6) = 11.

3. “-=” This operator is combination of ‘-‘ and ‘=’ operators. This operator first subtracts the value on the right from the current value of the variable on left and then assigns the result to the variable on the left. Example:

If initially value stored in a is 8. Then (a -= 6) = 2.

4. “*=” This operator is combination of ‘*’ and ‘=’ operators. This operator first multiplies the current value of the variable on left to the value on the right and then assigns the result to the variable on the left. Example:

If initially value stored in a is 5. Then (a *= 6) = 30.

5. “/=” This operator is combination of ‘/’ and ‘=’ operators. This operator first divides the current value of the variable on left by the value on the right and then assigns the result to the variable on the left. Example:

If initially value stored in a is 6. Then (a /= 2) = 3.

Below example illustrates the various Assignment Operators:

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Assignment Operators in C

In C language, the assignment operator stores a certain value in an already declared variable. A variable in C can be assigned the value in the form of a literal, another variable, or an expression.

The value to be assigned forms the right-hand operand, whereas the variable to be assigned should be the operand to the left of the " = " symbol, which is defined as a simple assignment operator in C.

In addition, C has several augmented assignment operators.

The following table lists the assignment operators supported by the C language −

Operator	Description	Example
=	Simple assignment operator. Assigns values from right side operands to left side operand	C = A + B will assign the value of A + B to C
+=	Add AND assignment operator. It adds the right operand to the left operand and assign the result to the left operand.	C += A is equivalent to C = C + A
-=	Subtract AND assignment operator. It subtracts the right operand from the left operand and assigns the result to the left operand.	C -= A is equivalent to C = C - A
*=	Multiply AND assignment operator. It multiplies the right operand with the left operand and assigns the result to the left operand.	C *= A is equivalent to C = C * A
/=	Divide AND assignment operator. It divides the left operand with the right operand and assigns the result to the left operand.	C /= A is equivalent to C = C / A
%=	Modulus AND assignment operator. It takes modulus using two operands and assigns the result to the left operand.	C %= A is equivalent to C = C % A
<<=	Left shift AND assignment operator.	C <<= 2 is same as C = C << 2
>>=	Right shift AND assignment operator.	C >>= 2 is same as C = C >> 2
&=	Bitwise AND assignment operator.	C &= 2 is same as C = C & 2
^=	Bitwise exclusive OR and assignment operator.	C ^= 2 is same as C = C ^ 2
|=	Bitwise inclusive OR and assignment operator.	C |= 2 is same as C = C | 2

Simple Assignment Operator (=)

The = operator is one of the most frequently used operators in C. As per the ANSI C standard, all the variables must be declared in the beginning. Variable declaration after the first processing statement is not allowed.

You can declare a variable to be assigned a value later in the code, or you can initialize it at the time of declaration.

You can use a literal, another variable, or an expression in the assignment statement.

Once a variable of a certain type is declared, it cannot be assigned a value of any other type. In such a case the C compiler reports a type mismatch error.

In C, the expressions that refer to a memory location are called "lvalue" expressions. A lvalue may appear as either the left-hand or right-hand side of an assignment.

On the other hand, the term rvalue refers to a data value that is stored at some address in memory. A rvalue is an expression that cannot have a value assigned to it which means an rvalue may appear on the right-hand side but not on the left-hand side of an assignment.

Variables are lvalues and so they may appear on the left-hand side of an assignment. Numeric literals are rvalues and so they may not be assigned and cannot appear on the left-hand side. Take a look at the following valid and invalid statements −

Augmented Assignment Operators

In addition to the = operator, C allows you to combine arithmetic and bitwise operators with the = symbol to form augmented or compound assignment operator. The augmented operators offer a convenient shortcut for combining arithmetic or bitwise operation with assignment.

For example, the expression "a += b" has the same effect of performing "a + b" first and then assigning the result back to the variable "a".

Run the code and check its output −

Similarly, the expression "a <<= b" has the same effect of performing "a << b" first and then assigning the result back to the variable "a".

Here is a C program that demonstrates the use of assignment operators in C −

When you compile and execute the above program, it will produce the following result −

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There are different kinds of the operators, such as arithmetic, relational, bitwise, assignment, etc., in the C programming language. The assignment operator is used to assign the value, variable and function to another variable. Let's discuss the various types of the assignment operators such as =, +=, -=, /=, *= and %=. It is the operator used to assign the right side operand or variable to the left side variable. Let's create a program to use the simple assignment operator in C. The operator is used to add the left side operand to the left operand and then assign results to the left operand. Let's create a program to use the Plus and assign operator in C. The operator is used to subtract the left operand with the right operand and then assigns the result to the left operand. Let's create a program to use the Subtract and Assign (-=) operator in C. The operator is used to multiply the left operand with the right operand and then assign result to the left operand. Let's create a program to use the multiply and assign operator (*=) in C. An operator is used between the left and right operands, which divides the first number by the second number to return the result in the left operand. Let's create a program to use the divide and assign operator (/=) in C. An operator used between the left operand and the right operand divides the first number (n1) by the second number (n2) and returns the remainder in the left operand. Let's create a program to use the divide and assign operator (%=) in C.

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