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at 0
Implement Dijkstra‘s algorithm to compute the Shortest path thru a given graph.

Program Algorithm/Flowchart:
Begin
Step1: Declare array path [5] [5], min, a [5][5], index, t[5];
Step2: Declare and initialize st=1,ed=5
Step 3: Declare variables i, j, stp, p, edp
Step 4: print “enter the cost “
Step 5: i=1
Step 6: Repeat step (7 to 11) until (i<=5)
Step 7: j=1
Step 8: repeat step (9 to 10) until (j<=5)
Step 9: Read a[i] [j]
Step 10: increment j
Step 11: increment i
Step 12: print “Enter the path”
Step 13: read p
Step 14: print “Enter possible paths”
Step 15: i=1
Step 16: repeat step(17 to 21) until (i<=p)
Step 17: j=1
Step 18: repeat step(19 to 20) until (i<=5)
Step 19: read path[i][j]
Step 20: increment j
Step 21: increment i
Step 22: j=1
Step 23: repeat step(24 to 34) until(i<=p)
Step 24: t[i]=0
Step 25: stp=st
Step 26: j=1
Step 27: repeat step(26 to 34) until(j<=5)
Step 28: edp=path[i][j+1]
Step 29: t[i]= [ti]+a[stp][edp]
Step 30: if (edp==ed) then
Step 31: break;
Step 32: else
Step 33: stp=edp
Step 34: end if
Step 35: min=t[st]
Step 36: index=st
Step 37: repeat step( 38 to 41) until (i<=p)
Step 38: min>t[i]
Step 39: min=t[i]
Step 40: index=i
Step 41: end if
Step 42: print” minimum cost” min
Step 43: print” minimum cost pth”
Step 44: repeat step(45 to 48) until (i<=5)
Step 45: print path[index][i]
Step 46: if(path[idex][i]==ed) then
Step 47: break
Step 48: end if
End

Program Code: Shortest Path for a given graph
#include<stdio.h>
void main()
{
int path[5][5],i,j,min,a[5][5],p,st=1,ed=5,stp,edp,t[5],index;
printf("Enter the cost matrix\n");
for(i=1;i<=5;i++)
for(j=1;j<=5;j++)
scanf("%d",&a[i][j]);
printf("Enter the paths\n");
scanf("%d",&p);
printf("Enter possible paths\n");
for(i=1;i<=p;i++)
for(j=1;j<=5;j++)
scanf("%d",&path[i][j]);
for(i=1;i<=p;i++)
{
t[i]=0;
stp=st;
for(j=1;j<=5;j++)
{
edp=path[i][j+1];
t[i]=t[i]+a[stp][edp];
if(edp==ed)
break;
else
stp=edp;
}
}
min=t[st];index=st;
for(i=1;i<=p;i++)
{
if(min>t[i])
{
min=t[i];
index=i;
}
}
printf("Minimum cost %d",min);
printf("\n Minimum cost path ");
for(i=1;i<=5;i++)
{
printf("--> %d",path[index][i]);
if(path[index][i]==ed)
break;
}
}

Program Output:
Enter the cost matrix
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
1 2 3 4 5
Enter the paths
2
Enter possible paths
1 2 3 4 5
1 2 3 4 5
Minimum cost 14
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at 1
Implement on a data set of characters the three CRC polynomials – CRC 12, CRC 16 and CRC CCIP.

Introduction to Cyclic Redundancy Check:
CRC method can detect a single burst of length n, since only one bit per column will be changed, a burst of length n+1 will pass undetected, if the first bit is inverted, the last bit is inverted and all other bits are correct. If the block is badly garbled by a long burst or by multiple shorter burst, the probability that any of the n columns will have the correct parity that is 0.5. so the probability of a bad block being expected when it should not be 2 power(-n). This scheme sometimes is known as Cyclic Redundancy Code.

Program Algorithm/Flowchart:
Begin
Step 1:Declare I, j , fr[8], dupfr[11], recfr[11], tlen, flag, gen[4], genl, frl, rem[4] as integer
Step 2: initialize frl=8 and genl=4
Step 3: initialize i=0
Step 4: Repeat step(5to7) until i<frl
Step 5: read fr[i]
Step 6: dupfr[i]=fr[i]
Step 7: increment i
Step 8: initialize i=0
Step 9: repeat step(10to11) until i<genl
Step 10: read gen[i]
Step 11: increment i
Step 12: tlen=frl+genl-1
Step 13: initialize i=frl
Step 14: Repeat step(15to16) until i<tlen
Step 15: dupfr[i]=0
Step 16: increment i
Step 17: call the function remainder(dupfr)
Step 18: initialize i=0
Step 19: repeat step(20 to 21) until j<genl
Step 20: recfr[i]=rem[j]
Step 21: increment I and j
Step 22: call the function remainder(dupfr)
Step 23: initialize flag=0 and i=0
Step 24: Repeat step(25to28) until i<4
Step 25: if rem[i]!=0 then
Step 26: increment flag
Step 27: end if
Step 28: increment i
Step 29: if flag=0 then
Step 25: print frame received correctly
Step 25: else
Step 25: print the received frame is wrong
End
Function: Remainder(int fr[])
Begin
Step 1: Declare k,k1,I,j as integer
Step 2: initialize k=0;
Step 3: repeat step(4 to 14) until k< frl
Step 4: if ((fr[k] == 1) then
Step 5: k1=k
Step 6: initialize i=0, j=k
Step 7: repeat step(8 to 9) until i< genl
Step 8: rem[i] =fr[j] exponential gen[i]
Step 9: increment I and j
Step 10: initialize I = 0
Step 11: repeat step(12to13) until I <genl
Step 12: fr[k1] = rem[i]
Step 13: increment k1 and i
Step 14: end if
End

Program Code: // Program for Cyclic Redundency Check
#include<stdio.h>
int gen[4],genl,frl,rem[4];
void main()
{
int i,j,fr[8],dupfr[11],recfr[11],tlen,flag;
frl=8; genl=4;
printf("Enter frame:");
for(i=0;i<frl;i++)
{
scanf("%d",&fr[i]);
dupfr[i]=fr[i];
}
printf("Enter generator:");
for(i=0;i<genl;i++)
scanf("%d",&gen[i]);
tlen=frl+genl-1;
for(i=frl;i<tlen;i++)
{
dupfr[i]=0;
}
remainder(dupfr);
for(i=0;i<frl;i++)
{
recfr[i]=fr[i];
}
for(i=frl,j=1;j<genl;i++,j++)
{
recfr[i]=rem[j];
}
remainder(recfr);
flag=0;
for(i=0;i<4;i++)
{
if(rem[i]!=0)
flag++;
}
if(flag==0)
{
printf("frame received correctly");
}
else
{
printf("the received frame is wrong");
}
}
remainder(int fr[])
{
int k,k1,i,j;
for(k=0;k<frl;k++)
{
if(fr[k]==1)
{
k1=k;
for(i=0,j=k;i<genl;i++,j++)
{
rem[i]=fr[j]^gen[i];
}
for(i=0;i<genl;i++)
{
fr[k1]=rem[i];
k1++;
}
}
}
}

Program Output:
Enter frame: MLRITM
Enter generator: frame received correctly
------------------
(program exited with code: 24)
Press return to continue
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at 1
Implement the data link layer framing methods such as character stuffing.

Introduction to character stuffing method of framing for Data Link Layer: The framing method gets around the problem of resynchronization after an error by having each frame start with the ASCII character sequence DLE STX and the sequence DLE ETX. If the destination ever losses the track of the frame boundaries all it has to do is look for DLE STX or DLE ETX characters to figure out. The data link layer on the receiving end removes the DLE before the data are given to the network layer. This technique is called character stuffing.


Program Algorithm:
Begin
Step 1: Initialize I and j as 0
Step 2: Declare n and pos as integer and a[20],b[50],ch as character
Step 3: read the string a
Step 4: find the length of the string n, i.e n-strlen(a)
Step 5: read the position, pos
Step 6: if pos > n then
Step 7: print invalid position and read again the position, pos
Step 8: endif
Step 9: read the character, ch
Step 10: Initialize the array b, b[0…5] as ’d’, ’l’, ’e’, ’s’, ’t’,’x’ respectively
Step 11: j=6;
Step 12: Repeat step[(13to22) until i<n
Step 13: if i==pos-1 then
Step 14: initialize b array,b[j],b[j+1]…b[j+6] as‘d’, ‘l’, ‘e’ ,’ch, ’d’, ‘l’,‘e’ respectively
Step 15: increment j by 7, i.e j=j+7
Step 16: endif
Step 17: if a[i]==’d’ and a[i+1]==’l’ and a[i+2]==’e’ then
Step 18: initialize array b, b[13…15]=’d’, ‘l’, ‘e’ respectively
Step 19: increment j by 3, i.e j=j+3
Step 20: endif
Step 21: b[j]=a[i]
Step 22: increment I and j;
Step 23: initialize b array,b[j],b[j+1]…b[j+6] as‘d’, ‘l’,‘e’,’e’,‘t’, ‘x’,‘\0’ respectively
Step 24: print frame after stuffing
Step 25: print b
End


Program Code: //Program for Character Stuffing
#include<stdio.h>
#include<string.h>
#include<process.h>
void main()
{
int i=0,j=0,n,pos;
char a[20],b[50],ch;
printf("Enter string\n");
scanf("%s",&a);
n=strlen(a);
printf("Enter position\n");
scanf("%d",&pos);
if(pos>n)
{
printf("invalid position, Enter again :");
scanf("%d",&pos);}
printf("Enter the character\n");
ch=getche();
b[0]='d';
b[1]='l';
b[2]='e';
b[3]='s';
b[4]='t';
b[5]='x';
j=6;
while(i<n)
{
if(i==pos-1)
{
b[j]='d';
b[j+1]='l';
b[j+2]='e';
b[j+3]=ch;
b[j+4]='d';
b[j+5]='l';
b[j+6]='e';
j=j+7;
}
if(a[i]=='d' && a[i+1]=='l' && a[i+2]=='e')
{
b[j]='d';
b[j+1]='l';
b[j+2]='e';
j=j+3;
}
b[j]=a[i];
i++;
j++;
}
b[j]='d';
b[j+1]='l';
b[j+2]='e';
b[j+3]='e';
b[j+4]='t';
b[j+5]='x';
b[j+6]='\0';
printf("\nframe after stuffing:\n");
printf("%s",b);
}


Program Output:
Enter string
MLRITM
Enter position
2
Enter the character
frame after stuffing:
dlestxMdldleLRITMdleetx
------------------
(program exited with code: 0) Press return to continue
Continue reading
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at 5
Implement the data link layer framing methods such as and bit stuffing.

Introduction to bit stuffing framing method used in Data Link layer:
The new technique allows data frames to contain an arbitrary number of bits and allows character codes with an arbitrary no of bits per character. Each frame begins and ends with the special bit pattern, 01111110, called a flag byte. Whenever the sender's data link layer encounters five consecutive ones in the data, it automatically stuffs a 0 bit into the outgoing bit stream. This bit stuffing is analogous to character stuffing, in which a DLE is stuffed into the outgoing character stream before DLE in the data.



Program Algorithm:
Begin
Step 1: Read frame length n
Step 2: Repeat step (3 to 4) until i<n(: Read values into the input frame (0’s and
1’s) i.e.
Step 3: initialize I i=0;
Step 4: read a[i] and increment i
Step 5: Initialize i=0, j=0,count =0
Step 6: repeat step (7 to 22) until i<n
Step 7: If a[i] == 1 then
Step 8: b[j] = a[i]
Step 9: Repeat step (10 to 18) until (a[k] =1 and k<n and count <5)
Step 10: Initialize k=i+1;
Step 11: Increment j and b[j]= a[k];
Step 12: Increment count ;
Step 13: if count =5 then
Step 14: increment j,
Step 15: b[j] =0
Step 16: end if
Step 17: i=k;
Step 18: increment k
Step 19: else
Step 20: b[j] = a[i]
Step 21: end if
Step 22: increment I and j
Step 23: print the frame after bit stuffing
Step 24: repeat step (25 to 26) until i< j
Step 25: print b[i]
Step 26: increment i
End


Program Code: // BIT Stuffing program
#include<stdio.h>
#include<string.h>
void main()
{
int a[20],b[30],i,j,k,count,n;
printf("Enter frame length:");
scanf("%d",&n);
printf("Enter input frame (0's & 1's only):");
for(i=0;i<n;i++)
scanf("%d",&a[i]);
i=0; count=1; j=0;
while(i<n)
{
if(a[i]==1)
{
b[j]=a[i];
for(k=i+1;a[k]==1 && k<n && count<5;k++)
{
j++;
b[j]=a[k];
count++;
if(count==5)
{
j++;
b[j]=0;
}
i=k;
}}
else
{
b[j]=a[i];
}
i++;
j++;
}
printf("After stuffing the frame is:");
for(i=0;i<j;i++)
printf("%d",b[i]);
}



Program Output:
Enter frame length:5
Enter input frame (0's & 1's only):
1
1
1
1
1
After stuffing the frame is:111110
------------------
(program exited with code: 6)
Press return to continue
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