Long Division

Hi im trying to write a function, that is basically long division, so i will get a value and then also a remainder. Could anybody give me some starting advice please?

function [q,r]=Divide(x,y)

Thats what my first line of my function shall be, where q is the integer value and r the remainder.

Thanks!

[Jan]

This is a double post. You got two answers to your former question already: http://www.mathworks.com/matlabcentral/answers/31403-writing-a-function-for-division

[Ben Davis]

Hello, the answer you gave me is good but i need to somehow get it to work for when x is very large, so then q doesnt approximate:

For example i need my function to be able to perform commands like this:

[q,r]=Divide(2*ones(1,43),7)

But at the moment i cant get it too :\

got any tips? :)

Because for some reason when i call that on the command line i end up with q as =0 and r as 43 zero's in a line.

When the answer should be :

q = Columns 1 through 22

3 1 7 4 6 0 3 1 7 4 6 0 3 1 7 4 6 0 3 1 7 4

Columns 23 through 42

6 0 3 1 7 4 6 0 3 1 7 4 6 0 3 1 7 4 6 0

r = 2

Answer 1

[Walter Roberson]

When you do division by hand, on a piece of paper, what you see is a vector of digits in the numerator and a vector of digits in the denominator. Humans do not usually perceive a number such as 2893 as a single unit: we see it as "two eight nine three". "long division" deals with the situation.

So, if someone gave you pencil and paper and two vectors of digits (numerator and denominator), and asked you to do the division, how would you proceed?

[Ben Davis]

If i was doing long division the process is that you move along digits of the numerator until you find a pair/amount of digits that are able divisible by the denominator. You then find out how many denominators going into the numerator, and then subtract that value from the digits of the numerator. Once that is done we add the remaining digits of the numerator onto our new value, and the process is repeated until an answer is obtained.

Is that right?

[Walter Roberson]

More or less. I probably would not describe it that way myself, but I get the impression you know what you mean.

[Ben Davis]

I understand the process of how to do long division, i just have no idea on how i am meant to write a function to get it to work. Well to get it to work for big values of 'x' anyway. Really stuck :\

[Ben Davis]

I know i should re-write my whole function, so im guessing i need to do something in my programming about 'x=' as that is where my problem lyes?

[Walter Roberson]

Long division:

First order of business is to check if the denominator vector is all zero; if it is, then exit with an appropriate error result.

Second order of business is to remove all leading 0's from the numerator and denominator vectors.

Third, initialize your result to 0

Loop

Is the numerator vector 0 ? If so then you are finished with a 0 remainder.

Is the denominator vector longer than the numerator vector? If so then you are finished and the numerator vector is your remainder.

Compare the denominator vector to the leading digits of the numerator vector. Is the denominator greater than those leading digits? If it is then the result-digit for that position is 0, and add an extra 0 to the end of the numerator vector to achieve the effect of multiplying by 10, and return to the beginning of the loop.

If the denominator was not greater than those leading digits then.... <etc>

end the loop

return the result and the remainder

Now, does _that_ look like your main problem is how you initialize x ??

[Ben Davis]

Nope.....i'll try and work on it now then.

[Ben Davis]

I give up, ive tried so hard with it all....i understand what to do mentally, i just dont know how to program it. This is causing me so much stress, i cant handle it anymore.

[Walter Roberson]

If you post what you have so far, someone might point out problems.

[Ben Davis]

I tried to follow your steps and program it in that order. I have missed out the sections:

'Compare the denominator vector to the leading digits of the numerator vector. Is the denominator greater than those leading digits? If it is then the result-digit for that position is 0, and add an extra 0 to the end of the numerator vector to achieve the effect of multiplying by 10, and return to the beginning of the loop.'

So ,so far i have this:

function [q,r]=Fivide(x,y)
f=length(x)-length(y);
g=zeros(1,f);
x=[g,x];
y=[g,y];
q=x/y;
q=floor(q);
r=x-(q*y);
p=length(q)
for a=p:-1:1;
 if x==0;
 r=0;
 end
 if length(y)>length(x)
 q=x;
 end
 
while x(1)==0;
 y(1)==0;
 x(1)=[];
 y(1)=[];
end
end
end

I have attempted to what i think many work for the function other than the section that i have missed out, i know i have most likely made many mistakes. But i really am trying my best :\ any feedback would be great.

Answer 2

[Daniel Shub]

In your first comment to Walter, you wrote "and the process is repeated until an answer is obtained." That sounds a lot like a "while" loop to me. Think you could express what you do in terms of a while loop ...

Maybe: While the remainder is less than the denominator DO ...

Then from there think about how to write the do in pseudo code. Finally convert it to MATLAB