As part of a calculator app, I am trying to implement uses with sigma notation. However, the result it prints out is always a decimal, and the rest isn't important. I simply want to change the decimal to a fraction.
I already have the reduce function, the problem I'm having is getting from a decimal like this: '0.96875' to it's fractional value, '31/32'
Thanks!
PS: I've looked into just about everything, and for the life of me, I can't figure this out. All I need at this point is how to take the decimal out of it, and I can then reduce it.
Here is my reduce method:
-(void)reduce {
int u = numerator;
int v = denominator;
int temp;
while (v != 0) {
temp = u % v;
u = v;
v = temp;
}
numerator /= u;
denominator /= u;
}
Found this out myself. What I did was multiply the numerator and denominator by 1000000 (recalling that the decimal looked like .96875/1) so that it looked like 96875/100000
.
Then, I used this reduce method to bring it into lowest terms:
-(void)reduce {
int u = numerator;
int v = denominator;
int temp;
while (v != 0) {
temp = u % v;
u = v;
v = temp;
}
numerator /= u;
denominator /= u;
}
And finally,I used a print method to get it into fraction form:
//In the .h
@property int numerator, denominator, mixed;
-(void)print;
//In the .m
@synthesize numerator, denominator, mixed;
-(void)print {
if (numerator > denominator) {
//Turn fraction into mixed number
mixed = numerator/denominator;
numerator -= (mixed * denominator);
NSLog(@"= %i %i/%i", mixed, numerator, denominator);
} else if (denominator != 1) {
//Print fraction normally
NSLog(@"= %i/%i", numerator, denominator);
} else {
//Print as integer if it has a denominator of 1
NSLog(@"= %i", numerator);
}
}
And got my desired output:
31/32
I found a fairly good way of doing this a while back, although I don't recall where from. Anyway, it works recursively like this (this is pseudocode, not C):
function getRational(float n)
let i = floor(n); (the integer component of n)
let j = n - i;
if j < 0.0001 (use abritrary precision threshold here), return i/1
let m/n = getRational(1 / j)
return ((i * m) + n) / m
For example, take 3.142857 as a starting point.
i = 3
j = 0.142857
m/n = getRational(7)
i = 7
j = 0
return 7/1
m/n = 7/1
return ((3*7)+1) / 7 = 22/7
Or a more complicated example, 1.55:
i = 1
j = 0.55
m/n = getRational(1.81818181)
i = 1
j = 0.81818181
m/n = getRational(1.22222222)
i = 1
j = 0.22222222
m/n = getRational(4.5)
i = 4
j = 0.5
m/n = getRational(2)
i = 2
j = 0
return 2/1
m/n = 2/1
return ((4*2)+1)/2 = 9/2
m/n = 9/2
return ((1*9)+2)/9 = 11/9
m/n = 11/9
return ((1*11)+9)/11) = 20/11
m/n = 20/11
return ((1*20)+11)/20 = 31/20
I tried this with PI once. It would have gone on a while, but if you set your threshold to 0.01, it only goes down a few recursions before returning 355/113.
There's a bit of a gotcha that you might end up with integers that are too large if it goes down too deep when it returns; I haven't really looked into a good way of allowing for that, except setting the precision threshold to something fairly lax, such as 0.01.
Try this :
-(NSString *)convertToFraction:(CGFloat)floatValue{
double tolerance = 1.0E-6;
CGFloat h1 = 1;
CGFloat h2 = 0;
CGFloat k1 = 0;
CGFloat k2 = 1;
CGFloat b = floatValue;
do{
CGFloat a = floor(b);
CGFloat aux = h1;
h1 = a*h1+h2;
h2 = aux;
aux = k1;
k1 = a*k1+k2;
k2 = aux;
b = 1/(b-a);
}while (ABS(floatValue-h1/k1) > floatValue*tolerance) ;
return k1 > 1 ? [NSString stringWithFormat:@"%.0f/%.0f",h1,k1] : [NSString stringWithFormat:@"%.0f",h1];
}