Program for Deadlock free condition in Operating System

• Difficulty Level : Easy
• Last Updated : 30 Sep, 2021

Given: A system has R identical resources, P processes competing for them and N is the maximum need of each process. The task is to find the minimum number of Resources required So that deadlock will never occur.

Formula:

R >= P * (N - 1) + 1

Examples:

Input : P = 3, N = 4
Output : R >= 10

Input : P = 7, N = 2
Output : R >= 8

Approach:

Consider, 3 process A, B and C.
Let, Need of each process is 4
Therefore, The maximum resources require will be 3 * 4 = 12 i.e, Give 4 resources to each Process.
And, The minimum resources required will be 3 * (4 – 1) + 1 = 10.
i.e, Give 3 Resources to each of the Process, and we are left out with 1 Resource.
That 1 resource will be given to any of the Process A, B or C.
So that after using that resource by any one of the Process, It left the resources and that resources will be used by any other Process and thus Deadlock will Never Occur.

C++

 // C++ implementation of above program.#include using namespace std; // function that calculates// the minimum no. of resourcesint Resources(int process, int need){    int minResources = 0;     // Condition so that deadlock    // will not occur    minResources = process * (need - 1) + 1;     return minResources;} // Driver codeint main(){    int process = 3, need = 4;     cout << "R >= " << Resources(process, need);    return 0;}

Java

 // Java implementation of above program class GFG{ // function that calculates// the minimum no. of resourcesstatic int Resources(int process, int need){    int minResources = 0;     // Condition so that deadlock    // will not occur    minResources = process * (need - 1) + 1;     return minResources;} // Driver Codepublic static void main(String args[]){    int process = 3, need = 4;         System.out.print("R >= ");    System.out.print(Resources(process, need));}}

Python3

 # Python 3 implementation of# above program  # function that calculates# the minimum no. of resourcesdef Resources(process, need):     minResources = 0     # Condition so that deadlock    # will not occur    minResources = process * (need - 1) + 1     return minResources # Driver Codeif __name__ == "__main__" :     process, need = 3, 4     print("R >=", Resources(process, need)) # This Code is Contributed# by Naman_Garg

C#

 // C# implementation of above programusing System; class GFG{ // function that calculates// the minimum no. of resourcesstatic int Resources(int process, int need){    int minResources = 0;     // Condition so that deadlock    // will not occur    minResources = process * (need - 1) + 1;     return minResources;} // Driver Codepublic static void Main(){    int process = 3, need = 4;         Console.Write("R >= ");    Console.Write(Resources(process, need));}} // This code is contributed// by Sanjit_Prasad

Output:

R >= 10

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