quine mccluskey method on C/C++

Reuben Cabrera

hey guys, can I ask for help. can you guys help me with the algorithm/code for quine mccluskey method on C/C++. i kinda having trouble with the algorithm and I need to provide the code on Oct 9 for my CS class. please do help me with my machine problem :) thank you and God bless. :D

David Crow

Reuben Cabrera wrote:

i kinda having trouble with the algorithm...

And that trouble would be?

Reuben Cabrera wrote:

please do help me with my machine problem :)

Help is a two way street. What have you done thus far?

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Lost User

Reuben Cabrera wrote:

I need to provide the code on Oct 9 for my CS class.

And you think submitting someone else's code and claiming it as your own is a good idea?

CPallini

Dear, you are sooooooooooooo old school...

THESE PEOPLE REALLY BOTHER ME!! How can they know what you should do without knowing what you want done?!?! -- C++ FQA Lite

Lost User

Oh you. :-O

Reuben Cabrera

im not submitting someone else's code. i just need help. that's all.

Reuben Cabrera

im already at converting the decimal # to binary. but the problem is the output should be like as a whole. i saw other programs converting decimal to binary and they just printed the remainder in reverse. my plan is to multiply the first digit to 10^0 then the next is by powers of 10 so when I add it i will have a whole number.

Reuben Cabrera

#include<stdio.h>
#include<stdlib.h>
#include<conio.h>

int convert(int);
main()
{
int var,x,y,number,number2;
int minterms[x],minterm[y];

printf("=========QUINE MCCLUSKEY METHOD========");
printf("\nHOW MANY VARIABLES DOES THE FUNCTION HAVE:?");
scanf("%d",&var);

printf("\nHOW MANY MINTERMS?:");
scanf("%d",&number);
number2=number;
for(x=0;number2!=0;number2--)
{
printf("\nENTER MINTERM:");
scanf("%d",&minterms[x]);
x++;
}

//convert each minterm to there binary form//
x=0;
y=0;
while(number!=0)
{
minterm[y]=convert(minterm[x]);
x++;
y++;
}

y=0;
while(number!=0)
{
printf("\n %d",minterm[y]);
number--;
y++;
}

}

int convert(int a)
{
int quot,binarynumber[100],i=1,sum=0;
int multi;
quot=a;
while(quot!=0)
{
binarynumber[i++]=quot%2;
quot= quot/2;
if(i==1)
{
sum=sum + (binarynumber[i++] * 1);
}
else

sum = sum + (binarynumber\[i++\] \* multiplier);
}

return sum;

}

Reuben Cabrera

and quine mccluskey method would require grouping the minterms according to the # of 1's in their binary form. how do you do that? then printing the final output as wxyz w'x' etc.

Lost User

Then you need to explain clearly what help; we cannot guess.

Reuben Cabrera

ok. so the binary form is stored in the string. and the i will make a for loop that will count the number of 1's in the binary form of each entered minterms. my problems is how do I group together the minterms with the same number of 1's cause i need to compare 2 different groups later.