lefti
Newbie level 3
Hello,
I am a rookie in embedded systems and I am trying to implement Huffman encoding using Microblaze in Spartan 3E starter kit. I use the following code in C:
and the hyperterminal out is the following:
Characters with their codes and the encoded text are not appeared. Could you help me to do this? Thank you.
I am a rookie in embedded systems and I am trying to implement Huffman encoding using Microblaze in Spartan 3E starter kit. I use the following code in C:
Code:
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#include <string.h>
#include <xuartlite_l.h>
#include <xutil.h>
#include <xparameters.h>
/*STRUCTURES*/
struct tree //The structure is used to build the binary tree.
{ int data; //Variable 'data' is used to represent the frequency of each character
int trav_val; //trav_value takes 0 if the node is traversed to the right or takes 1 if it is traversed to the left
char ckey; //This is the character assigned to the leaf node
struct tree *rlink; //Right child
struct tree *llink; //Left child
}**root=NULL,*temp;
struct calc_sum //The structure is used to store the sum of two min values
{ int sum; //Sum is the total value
int right; //right is the biggest of the two min values
int left; //left is the smallest of the two min values
}*total;
struct string //The structure is used to assign the frequency of each character
{ char alph; //alph is the character
int freq; //freq is the frequency of the character
}*ori_string,*pre_string; /*ori_string is the original string(containing freq and char of each character entered by user)
and pre_string is the preserved string*/
struct encode //The structure is used to assign the Huffman's code to different characters
{ char letter; //This is the character assigned to the leaf node
int *code; //It contains the Huffman's code for the character
int length; //Contains the length of the Huffman's code
}*enc;
/*FUNCTION DECLARATIIONS*/
void sort(struct string [],int);
void insert(struct calc_sum [],int);
void display(struct tree *);
//void encode(struct encode[],char *);
int next_char,str_len;
static int *trav,m,n,root_check,i,j,k;
char ch,input[50];
int main(void)
{
do
{
int flag,char_freq;
xil_printf ("Enter the string and terminate it with backspace:\n\r");
for(n=0;n<50;n++)
{
input[n]= XUartLite_RecvByte (0x84000000);
if (input[n]==0x8)
break;
}
input[n]='\0';
str_len=n;
xil_printf("\n\r");
xil_printf("Length of the string:%d",str_len);
for(i=32,next_char=0;i<127;i++)
{
flag=0,char_freq=0,j=0;
while(input[j]!='\0')
{
if(input[j]==i)
{
flag=1;
ori_string[next_char].alph=pre_string[next_char].alph=input[j];
char_freq++;
}
j++;
}
if(flag)
{
ori_string[next_char].freq=pre_string[next_char].freq=char_freq;
next_char++;
}
}
if(next_char==1)
xil_printf("\nEncoding is not possible. You have entered same set of characters");
else
{
xil_printf("\n\r");
xil_printf("Character\tFrequency\n\r");
for(i=0;i<next_char;i++)
xil_printf("%c\t\t %d\n\r",pre_string[i].alph,pre_string[i].freq);
sort(ori_string,next_char);
}
xil_printf("\n\nDo you want to continue?(Y/N)\n");
ch= XUartLite_RecvByte (0x84000000);
} while(ch=='y'||ch=='Y');
}
/*
Name of Routine : Sort
Type/Classified as : Sorting
Function Call : sort() from main
Parameters Passed : ori_string,next_char
Return Type : Void
Purpose : Sorts the frequencies in decreasing order
Swap is used for swapping values
k stores the no. of sum values
*/
void sort(struct string ori_string[],int next_char)
{ k=next_char-1;
int swap;
xil_printf("\n\r");
xil_printf("Left\tRight\tTotal\n\r");
while(k>0)
{ for(i=0;i<k+1;i++)
for(j=i+1;j<k+1;j++)
{ if(ori_string[i].freq<ori_string[j].freq)
{ swap=ori_string[i].freq;
ori_string[i].freq=ori_string[j].freq;
ori_string[j].freq=swap;
}
}
total[next_char-1-k].left=ori_string[i-1].freq;
total[next_char-1-k].right=ori_string[i-2].freq;
total[next_char-1-k].sum=total[next_char-1-k].left+total[next_char-1-k].right;
ori_string[k-1].freq=total[next_char-1-k].sum;
xil_printf("%d \t%d \t%d\n\r",total[next_char-1-k].left,total[next_char-1-k].right,total[next_char-1-k].sum);
k--;
}
insert(total,next_char);
}
/*
Name of Routine : Insert
Type/Classified as : Insertion
Function Call : insert() from sort()
Parameters Passed : total,next_char;
Return Type : Void
Purpose : Constructs a binary tree using the sum and the frequency values
Variables rflag and lflag are used to identify a new root
Variables lchange and r change are used to compress two roots into 1
tot_roots calcultes the no of roots possible and integer value is assigned to roots using the power function
roots for max roots that a tree can contain
count gives the no. of roots at any time
swap for swapping two characters and k stores the no of sum values
*/
void insert(struct calc_sum total[],int next_char)
{ int roots,count=0,lflag,rflag,rchange=0,lchange=0;
k=0;
char swap;
float tot_roots;
tot_roots=log(2*next_char-1)/log(2);
roots=pow(2,(int)(tot_roots-1));
for(i=0;i<roots;i++)
root[i]=NULL;
struct tree *r;
struct tree *l;
while(k<(next_char-1))
{ lflag=rflag=1;
temp->data=total[k].sum;
if(k==0)
{
r->data=total[k].right;
r->trav_val=1;
r->rlink=NULL;
r->llink=NULL;
for(i=next_char-1;i>=0;i--)
{ if(pre_string[i].freq==total[k].right)
{ r->ckey=pre_string[i].alph;
pre_string[i].freq=0;
break;
}
}
temp->rlink=r;
l->data=total[k].left;
l->trav_val=0;
l->rlink=NULL;
l->llink=NULL;
for(i=next_char-1;i>=0;i--)
{ if(pre_string[i].freq==total[k].left)
{ l->ckey=pre_string[i].alph;
pre_string[i].freq=0;
break;
}
}
temp->llink=l;
root[0]=temp;
}
else
{ for(i=0;i<roots;i++) // check for right
{ if(total[k].right==root[i]->data)
{
root[i]->trav_val=1;
temp->rlink=root[i];
rflag=0;
rchange=i;
root[rchange]=temp;
}
if(!rflag)
break;
}
if(i>=roots)
{
r->data=total[k].right;
r->trav_val=1;
r->rlink=NULL;
r->llink=NULL;
for(i=next_char-1;i>=0;i--)
{ if(pre_string[i].freq==total[k].right)
{ r->ckey=pre_string[i].alph;
pre_string[i].freq=0;
break;
}
}
temp->rlink=r;
}
for(i=0;i<roots;i++) //check for left
{ if(total[k].left==root[i]->data)
{
root[i]->trav_val=0;
temp->llink=root[i];
lflag=0;
lchange=i;
root[lchange]=temp;
}
if(!lflag)
break;
}
if(i>=roots)
{
l->data=total[k].left;
l->trav_val=0;
l->rlink=NULL;
l->llink=NULL;
for(i=next_char-1;i>=0;i--)
{ if(pre_string[i].freq==total[k].left)
{ l->ckey=pre_string[i].alph;
pre_string[i].freq=0;
break;
}
}
temp->llink=l;
}
if(rflag&&lflag)
{ root[++count]=temp;
if(root[count]->llink->data==root[count]->rlink->data)
{
swap=root[count]->llink->ckey;
root[count]->llink->ckey=root[count]->rlink->ckey;
root[count]->rlink->ckey=swap;
}
}
if(!rflag&&!lflag)
root[rchange]=NULL;
}
k++;
}
xil_printf("HUFFMAN CODES\n");
m=0,i=0,j=0,k=0;
display(root[lchange]);
xil_printf("GET THE ENCODED STRING\n");
// encode(enc,input);
for(i=0;i<str_len;i++)
{ j=0;
while(j<next_char)
{ if(enc[j].letter==input[i])
for(k=0;k<enc[j].length;k++)
xil_printf("%d",enc[j].code[k]);
j++;
}
}
}
/*
Name of Routine : Display
Type/Classified as : I/O
Function Call : disp() from insert()
Parameters Passed : root[lchange]
Return Type : Void
Purpose : Displays the Huffcode for each character
root_check is used to check if the current node is the root
trav[] is an array which stores the huffman code for each character
*/
void display(struct tree *disp)
{ if(disp!=NULL)
{ if(root_check>=1)
trav[m++]=disp->trav_val;
root_check++;
display(disp->llink);
display(disp->rlink);
if(disp->llink==NULL&&disp->rlink==NULL)
{
xil_printf("%c-",disp->ckey);
enc[i].letter=disp->ckey;
enc[i].length=0;
for(m=0;m<root_check-1;m++)
{ xil_printf("%d",trav[m]);
enc[i].code[j]=trav[m];
enc[i].length++;
j++;
}
i++;
j=0;
xil_printf("\n\r");
}
m--;
root_check--;
}
}and the hyperterminal out is the following:
Code:
Enter the string and terminate it with backspace:
Iron Maiden
Length of the string:11
Character Frequency
1
I 1
M 1
a 1
d 1
e 1
i 1
n 2
o 1
r 1
Left Right Total
1 1 2
1 1 2
1 1 2
2 1 2
2 2 4
2 2 4
2 2 4
4 4 8
4 8 12
HUFFMAN CODES
GET THE ENCODED STRING
Do you want to continue?(Y/N)Characters with their codes and the encoded text are not appeared. Could you help me to do this? Thank you.
