How to Get the Coefficients Of A Linear Equation

Question

In this program, I need to get the solutions of a linear system and this system have n variables and n equations. I need to extract the coefficients of the equations in a text file (.txt) and put them on a matrix. The first line of the text file has a number, which is the number of equations in the system. The other lines, have the equations.

For example, if I have the following text file:

3
2x - 3y = 0
4x + z = 12
5y - 6z = 10

The matrix with the coefficients will be:

|2 -3  0  0|
|4  0  1 12|
|0  5 -6 10|

I know how to get the solutions of the linear system, but how can I get only the coefficients of the system (without a library)? I tried functions that only read numbers from a string(vector of char) but they won't work if the coefficient is a letter (return 0) or doesn't exist.

Rules:

The equations of the system need to be LINEAR, otherwise the program will be closed.

If the coefficient and the variable of the equation doesn't exist, the coefficient is 0.

If the coefficient doesn't exist, but the variable exists, (x, y, z for example), the coefficient is 1.

The variable can be any letter, it doesn't need to be x, y and z.

The code:

#include <stdio.h>
#include <string.h>
#include <ctype.h>

#define MAXCHAR 1000
#include "LinkedList.c"

//in the future use STRTOK e SSCANF (SPRINTF maybe)

int main() {
FILE *fp;
char str[MAXCHAR];
int character;
int c = 0;
char fileaddress[50];
char *cp;
char *variaveis;
char *p;

int quant = 0;
int numDeEquacoes = 0;

printf("Enter the address of the file you want to open: ");

gets(fileaddress);
printf(fileaddress);


//int coeficientes[3][3];

fp = fopen(fileaddress, "r");
if (fp == NULL){
    printf("\n");
    printf("Cannot open file %s",fileaddress);
    return 1;
}
    printf("\n");

while (fgets(str, MAXCHAR, fp) != NULL)
{
    quant++;

    if(quant==1)
    {
        numDeEquacoes = (str[0] - '0');
    }

    printf("%s", str);

    //gets(str, sizeof(str), stdin);
        printf("Variables of equation: ");

        for(cp=str; *cp; ++cp)
            if(isalpha(*cp))
            {
               printf("%c", *cp, "\n");
               //scanf("%c", )
            }

            //THE CODE THAT RESULTS IN ONLY NUMBERS
            while (*p)
            {
                if ( isdigit(*p) || ( (*p=='-'||*p=='+') && isdigit(*(p+1))) )
                {
                    long val = strtol(p, &p, 10); 
                    printf("%ld\n Coefficients:", val, "\n");
                }
                else
                {
                    p++;
                }
            }//ENDS HERE


        //printf("Variaveis: ", variaveis);
        //printf("\n");

    //variaveis = strstr(str);
    //printf(variaveis);


}

fclose(fp);

if(quant-1!=numDeEquacoes)
{
    printf("The number of equations is wrong!\n");
    printf("The number of equations is: %i", numDeEquacoes,"\n");
    printf("The variable quant is: %i", quant,"\n");

    return 0;
}

//int coeficientes[numDeEquacoes][numDeEquacoes];

//printf("coef:",coeficientes);

return 0;
}

Answer 1

You should read the signals (+ or -) and multiply the coefficients. If the signal is +, multiply by +1, else, by -1. If there isn't a variable (or letter), the coefficient is 0.

Answer 2

I don't have time right now to code it in C, but I have this python approach, for if you want to check the algorithm. The output I got is

[[2, -3, 0, 0],
 [4, 0, 1, 12],
 [0, 5, -6, 10]]

You have to check if there is a sign (+ or -) or not (assume +), get the number (1 if it not exist), and store the letter (variable). Then restart the process at each sign (+,- or =) and space.

For the matrix, start a column counter and an array for column indexes, and each time you find a variable search in the array for the index, and if it doesn't exist, assign that counter as the column index for that variable and increment it.

Here the code:

# - For reading the coefficients:
# 3
# 2x - 3y = 0
# 4x + z = 12
# 5y - 6z = 10

def get_index(d, c):
    if c in d.keys():
        return d[c] 
    else:
        return -1
s=1 #sign
n_fg = 0 #number exist
eq_fg = 0 #equal flag
s_fg = 0 #sign flag
letter = '' # store current letter (variable)
coef =''
curr_idx = 0

N = int(input())

letter_idx = {}

mat = [[0 for i in range(N+1)] for i in range(N)]

for i in range(N):

    l = input()
    coef = ''
    eq_fg = 0
    s_fg = 0
    s = 1
    ls = len(l)
    k = 0 # position in line str

    for c in l:

        if c == '-':
            s_fg = 1
            s=-1
        elif c == '+':
            s_fg = 1
            s = 1
        elif c.isalpha():
            if n_fg == 0:
                coef = 1
            letter = c
        elif c.isdigit():
            n_fg = 1
            coef += c

            if k == ls-1:
                if coef is '':
                    coef = 1
                coef = s*int(coef)
                if eq_fg == 0:
                    j = get_index(letter_idx,letter)
                    if j == -1:
                        j = curr_idx
                        letter_idx[letter] = j
                        curr_idx+=1
                else:
                    j = N
                mat[i][j] = coef

        elif (c == ' ' and s_fg != 1) :
            if coef is '':
                coef = 1
            coef = s*int(coef)
            if eq_fg == 0:
                j = get_index(letter_idx,letter)
                if j == -1:
                    j = curr_idx
                    letter_idx[letter] = j
                    curr_idx+=1
            else:
                j = N
            mat[i][j] = coef
            coef = ''
            n_fg = 0
        elif c == ' ' and s_fg == 1:
            s_fg = 0
        elif c == '=':
            eq_fg = 1
            s_fg = 1
            s = 1
        k+=1

print(mat)