Putting a fraction into lowest terms

[Brutus]

Hey everyone, I'm new here so I'm not really sure how to post, but here goes. I am making a program that is supposed to calculate slpoe of a line segment. I can get the slope, but the problem is getting it into lowest terms. Any one know how I would achieve this? here is my program so far:

code:

%Analytic Geometry program opening var coor1 : int coor1 := 0 var coor2 : int coor2 := 0 var coor3 : int coor3 := 0 var coor4 : int coor4 := 0 var answer : int put "Welcome to the Analytic Geometry program. This program is designed to calculate the slope, length and equasion of a line." delay (8000) cls loop     put "Please enter the 'x' co-ordinate for the first endpoint of your line."     get coor1     put "Now the 'y' co-ordinate for the first endpoint of your line."     get coor2     put "Ok thanks. Now I need the 'x' co-ordinate for the second endpoint."     get coor3     put "Got it. Now the 'y' co-ordinate for the second endpoint"     get coor4     delay (5000)     cls     put "Alright, I've got it all worked out now, so go ahead and choose an option."     loop         put "For the slope of you line, press 1."         put "For the length of your line, press 2."         put "For the equasion of your line, press 3."         put "To re-enter your numbers, press 4."         get answer         exit when answer = 4         if answer = 1 then             put "The slope of your line is ", coor2 - coor4, "/", coor1 - coor3, "."         elsif answer = 2 then             put "The length of your line is  "         elsif answer = 3 then             put "The equasion of your line is "         else             put "Hey! Come on now buddy. I was nice enough to bail you out by doing all the    hard stuff for you, so do me a favour and choose one of the options I gave you."         end if         delay (10000)         cls     end loop end loop

thanks in advance to whoever helps me

[Tony]

to get a fraction in its lowest term, you find the factors of both nominator and denominator and divide them both by their common factors.

[Bacchus]

i sugest using mod to see the largest common denometer

code:

var num1,num2,highest:int num1:=6 num2:=9 if num1<num2 then for decreasing i:num1..1 if num1 mod i =0 & num2 mod i =0 then highest:=i exit end if end for elsif num1<num2 then for decreasing i:num2..1 if num1 mod i =0 & num2 mod i =0 then highest:=i exit end if end for else highest:=num1 end if

not the bset but u get the idea

[wtd]

Indentation is a good thing.

[AsianSensation]

check up on Euclid's Algorithm. It's a very good way to find the GCD between two numbers and there are surprisingly quite alot of computer programming code out there regarding Euclid's Algorithm.

[Bacchus]

wtd wrote:

Indentation is a good thing.

ya i kno but i didnt wrote it in turing and im lazy

[[Gandalf]]

Here's a quick tip on making your program more 'efficient'.

code:

var coor4 : int coor4 := 0

turns into

code:

var coor4 := 0   %or if you prefer:   var coor4 : int := 0

If anything, I would use the mod function - its probably more simple than some Eulcid's Algorithm ( lol, but it might not work as well).

[AsianSensation]

Euclid's Algorithm is an algorithm, that is to say, a set of instructions that you can follow and produce some result, that could be used over and over again until the desired answer is found.

Mod in turing is a function, and it only returns only 1 value. They are not the samething.

Besides, Euclid's algorithm, when typed into CS code, will need the mod function. As the math logic behind it even calls for the use of mod.

[Brutus]

Ok, thanks very much to all that posted.
Bacchus, I tryed to use your idea, but i had to add the abs thing due to negative number. So here's what I have now.

code:

procedure lowestterms (var rise, run: int) if abs (rise) > abs (run) then for i: decreasing abs run .. 2 if rise mod i = 0 and run mod i - 0 then rise:= rise/i run:= run/i end if end for else for i: decreasing abs (rise) .. 2 if run mod i = 0 and rise mod i = 0 then rise := rise/i run:= run/i end if end for end if end lowestterms

Problem is now, that it won't run with the "i" and decreasing things. Any idea why?

[scottyrush13]

i think the other i's inside the for statement need to be in brackets

[Bacchus]

no but the decreasing part has to come b4 the i ex) for decreasing i:5..1

[Brutus]

Thanks scotty and Bacchus. Hahaha, ya, the decreasing comes before the "i". Also, the "/" had to be changed to div. I hate it when the hardest answer to find is the simple one.... Anyways, thanks again.

[phuong]

if you're still interested... (don't think so but anyway)
here's a program that CAN reduce fractions. check it out.
here's the datafile for it too.

[[Gandalf]]

Well, I don't know - this was quite a while ago. I would just post something in the source code section - a new topic.

Also, when you submit something, include all the files neccessary.