# Программа для отсутствия тупиков в операционной системе

Опубликовано: 28 Декабря, 2021

Дано: система имеет R идентичных ресурсов, P процессов конкурируют за них, а N - максимальная потребность каждого процесса. Задача - найти минимальное количество требуемых ресурсов, чтобы никогда не возникла тупиковая ситуация.

Формула:

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

Примеры:

``` Ввод: P = 3, N = 4
Выход: R> = 10

Ввод: P = 7, N = 2
Выход: R> = 8```

Подход:

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 resources` `int` `Resources(` `int` `process,` `int` `need)` `{`` ``int` `minResources = 0;`` `` ``// Condition so that deadlock`` ``// will not occuur`` ``minResources = process * (need - 1) + 1;`` `` ``return` `minResources;` `}`` ``// Driver code` `int` `main()` `{`` ``int` `process = 3, need = 4;`` `` ``cout <<` `"R >= "` `<< Resources(process, need);`` ``return` `0;` `}`

## Джава

 `// Java implementation of above program`` ``class` `GFG` `{`` ``// function that calculates` `// the minimum no. of resources` `static` `int` `Resources(` `int` `process,` `int` `need)` `{`` ``int` `minResources =` `0` `;`` `` ``// Condition so that deadlock`` ``// will not occuur`` ``minResources = process * (need -` `1` `) +` `1` `;`` `` ``return` `minResources;` `}`` ``// Driver Code` `public` `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 resources` `def` `Resources(process, need):`` `` ``minResources` `=` `0`` `` ``# Condition so that deadlock`` ``# will not occuur`` ``minResources` `=` `process` `*` `(need` `-` `1` `)` `+` `1`` `` ``return` `minResources`` ``# Driver Code` `if` `__name__` `=` `=` `"__main__"` `:`` `` ``process, need` `=` `3` `,` `4`` `` ``print` `(` `"R >="` `, Resources(process, need))`` ``# This Code is Contributed` `# by Naman_Garg`

## C #

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

Выход:

` R> = 10`

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