How can I detect if a float has a repeating decimal expansion in C#?

Question

I simply need to know how I can detect repeating decimal expansion in floats.

Example:

0.123456789123456789

The repeating portion of the number would

Answer 1

There's a nice trick for calculating rational approximations to a given float (based on some properties of Euclid's algorithm for GCDs). We can use this to determine whether or not the "best" approximation is of the form A/(2^a 5^b), if it is then the float terminates (in base 10), if not it will have some repeating component. The tricky bit will be determining which of the approximations is the right one (due to floating point precission issues).

So heres how you get approximate rational expressions.

First iterate x = 1/x - floor(1/x) keeping track of int(x)

x = 0.12341234
1/x = 8.102917
x <= 1/x - 8 = 0.102917
1/x = 9.7165
x <= 1/x - 9 = 0.71265277
1/x = 1.3956
x < 1/x - 1 = 0.3956
...

Next stick the int parts of x into the top row of this table, call them k_i. The values A_i = A_{i-2} + k_i * A_{i-1} and the same for B_i.

           ||  8      |  9  | 1   | 2   | 1   |  1  |    8 |    1 |    1
A =    1 0 ||  1      |  9  | 10  | 29  | 39  |  68 |  583 |  651 | 1234
B =    0 1 ||  8      | 73  | 81  | 235 | 316 | 551 | 4724 | 5275 | 9999

The rational approximations are then A_n/B_n.

1/8       = 0.12500000000000000     | e = 1.5e-3
9/73      = 0.12328767123287671     | e = 1.2e-4
10/81     = 0.12345679012345678     | e = 4.4e-5
29/235    = 0.12340425531914893     | e = 8.1e-6
39/316    = 0.12341772151898735     | e = 5.4e-6
68/551    = 0.12341197822141561     | e = 3.6e-7
583/4724  = 0.12341236240474174     | e = 2.2e-8
651/5275  = 0.12341232227488151     | e = 1.8e-8
1234/9999 = 0.12341234123412341     | e = 1.2e-9

So if we decide our error is low enough at the 1234/9999 stage, we note that 9999 can not be written in the form 2^a 5^b, and thus our decimal expansion is repeating.

Note that while this seems to require a lot of steps we can get faster convergence if we use x = 1/x - round(1/x) (and keep track of round round(1/x) instead). In that case the table becomes

     8  10    -4     2      9     -2
1 0  1  10   -39   -68   -651   1234
0 1  8  81  -316  -551  -5275   9999

This gives you a subset of the previous results, in fewer steps.

It is interesting to note that the fraction A_i/B_i is always such that A_i and B_i have no common factors so you dont event need to worry about canceling out factors or anything like that.

For comparison lets look at the expansion for x = 0.123. The table we get is:

      8   8   -3    -5  
 1 0  1   8  -23   123
 0 1  8  65 -187  1000

Then our sequence of approximations is

 1/8      = 0.125       e = 2.0e-3
 8/65     = 0.12307..   e = 7.6e-5
 23/187   = 0.12299..   e = 5.3e-6
 123/1000 = 0.123       e = 0

And we see that 123/1000 is exactly the fraction we want and since 1000 = 10^3 = 2^3 5^3 our fraction is terminating.

If you actually want to find out what the repeating part of the fraction is (what digits and what period) you need to do some additional tricks. This involves factoring the denominator and finding the lowest number (10^k-1) with all those factors (other than 2 and 5), then k will be your period. So for our top case we found A = 9999 = 10^4-1 (and thus 10^4-1 contains all the factors of A - we were kind of lucky here...) so the period of the repeating part is 4. You can find more details about this final part here.

A final and important aspect of not of this algorithm is that it does not require all the digits to mark a decimal expansion as repeating. Consider x = 0.34482, this has the table:

     3 -10 -156
1 0  1 -10   . 
0 1  3 -29   .

We get a very accurate approximation at the second entry and stop there, concluding that our fraction is probably 10/29 (as that gets use within 1e-5) and from the tables in the link above we can discern that its period will be 28 digits. This could never be determined using string searches on the short version of the number, which would require at least 57 digits of the number to be known.