## Program to implement deadlock prevention technique

Concept:(Bankers algorithm) When a new process enters a system, it must declare the maximum number of instances of each resource type it needed. This number may exceed the total number of resources in the system. When the user request a set of resources, the system must determine whether the allocation of each resources will leave the system in safe state. If it will the resources are allocation; otherwise the processmust wait until some other process release the resources.

Data structures
n-Number of process, m-number of resource types.

Available: Available[j]=k, k – instance of resource type Rj isavailable.

Max: If max[i, j]=k, Pi may request at most k instancesresource Rj.

Allocation: If Allocation [i, j]=k, Pi allocated to k instances of resource Rj

Need: If Need[I, j]=k, Pi may need k more instances of resource type Rj, Need[I, j]=Max[I, j]-Allocation[I, j];

Work and Finish be the vector of length m and n respectively,
Work=Available and Finish[i] =False.

Find an i such that both
Finish[i] =False
Need<=Work
If no such I exists go to step 4.
work=work+Allocation, Finish[i] =True;
if Finish=True for all I, then the system is in safe state

ALGORITHM:
1. Start.
2. Get the values of resources and processes.
3. Get the avail value.
4. After allocation find the need value.
5. Check whether its possible to allocate.
6. If it is possible then the system is in safe state.
7. Else system is not in safety state
8. Stop

				
#include< stdio.h>
#include< conio.h>
void main()
{
int allocated,max,need,avail,tres,work,flag;
int pno,rno,i,j,prc,count,t,total;
count=0;
clrscr();
printf("\n Enter number of process:");
scanf("%d",&pno);
printf("\n Enter number of resources:");
scanf("%d",&rno);
for(i=1;i< =pno;i++)
{
flag[i]=0;
}
printf("\n Enter total numbers of each resources:");
for(i=1;i<= rno;i++)
scanf("%d",&tres[i]);
printf("\n Enter Max resources for each process:");
for(i=1;i<= pno;i++)
{
printf("\n for process %d:",i);
for(j=1;j<= rno;j++)
scanf("%d",&max[i][j]);
}
printf("\n Enter allocated resources for each process:");
for(i=1;i<= pno;i++)
{
printf("\n for process %d:",i);
for(j=1;j<= rno;j++)
scanf("%d",&allocated[i][j]);
}

printf("\n available resources:\n");
for(j=1;j<= rno;j++)
{
avail[j]=0;
total=0;
for(i=1;i<= pno;i++)
{
total+=allocated[i][j];
}
avail[j]=tres[j]-total;
work[j]=avail[j];
printf(" %d \t",work[j]);
}
do
{
for(i=1;i<= pno;i++)
{
for(j=1;j<= rno;j++)
{
need[i][j]=max[i][j]-allocated[i][j];
}
}
printf("\n Allocated matrix Max need");
for(i=1;i<= pno;i++)
{
printf("\n");
for(j=1;j<= rno;j++)
{
printf("%4d",allocated[i][j]);
}
printf("|");
for(j=1;j<= rno;j++)
{
printf("%4d",max[i][j]);
}
printf("|");
for(j=1;j<= rno;j++)
{
printf("%4d",need[i][j]);
}

}
prc=0;
for(i=1;i<= pno;i++)
{
if(flag[i]==0)
{
prc=i;
for(j=1;j<= rno;j++)
{
if(work[j]< need[i][j])
{
prc=0;
break;
}
}
}
if(prc!=0)
break;
}
if(prc!=0)
{
printf("\n Process %d completed",i);
count++;
printf("\n Available matrix:");
for(j=1;j<= rno;j++)
{
work[j]+=allocated[prc][j];
allocated[prc][j]=0;
max[prc][j]=0;
flag[prc]=1;
printf(" %d",work[j]);
}
}
}while(count!=pno&&prc!=0);
if(count==pno)
printf("\nThe system is in a safe state!!");
else

printf("\nThe system is in an unsafe state!!");
getch();
}



OUTPUT:
Enter number of process:5
Enter number of resources:3
Enter total numbers of each resources:10 5 7
Enter Max resources for each process:
for process 1:7 5 3
for process 2:3 2 2
for process 3:9 0 2
for process 4:2 2 2
for process 5:4 3 3
Enter allocated resources for each process:
for process 1:0 1 0
for process 2:3 0 2
for process 3:3 0 2
for process 4:2 1 1
for process 5:0 0 2
available resources:
2 3 0
Allocated matrix Max need
0 1 0| 7 5 3| 7 4 3
3 0 2| 3 2 2| 0 2 0
3 0 2| 9 0 2| 6 0 0
2 1 1| 2 2 2| 0 1 1
0 0 2| 4 3 3| 4 3 1
Process 2 completed
Available matrix: 5 3 2
Allocated matrix Max need
0 1 0| 7 5 3| 7 4 3
0 0 0| 0 0 0| 0 0 0
3 0 2| 9 0 2| 6 0 0
2 1 1| 2 2 2| 0 1 1
0 0 2| 4 3 3| 4 3 1
Process 4 completed
Available matrix: 7 4 3
Allocated matrix Max need
0 1 0| 7 5 3| 7 4 3
0 0 0| 0 0 0| 0 0 0
3 0 2| 9 0 2| 6 0 0
0 0 0| 0 0 0| 0 0 0
0 0 2| 4 3 3| 4 3 1
Process 1 completed
Available matrix: 7 5 3
Allocated matrix Max need
0 0 0| 0 0 0| 0 0 0
0 0 0| 0 0 0| 0 0 0
3 0 2| 9 0 2| 6 0 0
0 0 0| 0 0 0| 0 0 0
0 0 2| 4 3 3| 4 3 1
Process 3 completed
Available matrix: 10 5 5
Allocated matrix Max need
0 0 0| 0 0 0| 0 0 0
0 0 0| 0 0 0| 0 0 0
0 0 0| 0 0 0| 0 0 0
0 0 0| 0 0 0| 0 0 0
0 0 2| 4 3 3| 4 3 1
Process 5 completed
Available matrix: 10 5 7
The system is in a safe state!!