# The Universe of Discourse

Mon, 19 Mar 2007

Your age as a fraction
Little kids often report their ages as "two and a half" or sometimes even "three and three quarters". These evaluations are usually based on whole months: if you were born on April 2, 1969, then on October 2, 1971 you start reporting your age as "two and a half", and, if you choose to report your age as "three and three quarters", you conventionally may begin on January 2, 1973.

However, these reports are not quite accurate. On January 2, 1973, exactly 3 years and 9 months from your birthday, you would be 1,371 days old, or 3 years plus 275 days. 275/365 = 0.7534. On January 1, you were only 3 + 274/365 days old, which is 3.7507 years, and so January 1 is the day on which you should have been allowed to start reporting your age as "three and three quarters". This slippage between days and months occurs in the other direction as well, so there may be kids wandering around declaring themselves as "three and a half" a full day before they actually reach that age.

Clearly this is one of the major problems facing our society, so I wanted to make up a table showing, for each number of days d from 1 to 365, what is the simplest fraction a/b such that when it is d days after your birthday, you are (some whole number and) a/b years. That is, I wanted a/b such that d/365 ≤ a/b < (d+1)/365.

Then, by consulting the table each day, anyone could find out what new fraction they might have qualified for, and, if they preferred the new fraction to the old, they might start reporting their age with that fraction.

There is a well-developed branch of mathematics that deals with this problem. To find simple fractions that approximate any given rational number, or lie in any range, we first expand the bounds of the range in continued fraction form. For example, suppose it has been 208 days since your birthday. Then today your age will range from y plus 208/365 years up to y plus 209/365 years.

Then we expand 208/365 and 209/365 as continued fractions:

208/365 = [0; 1, 1, 3, 12, 1, 3]
209/365 = [0; 1, 1, 2, 1, 16, 1, 2]
Where [0; 1, 1, 3, 12, 1, 3] is an abbreviation for the typographically horrendous expression:

$$0 + {1\over \displaystyle 1 + {\strut 1\over\displaystyle 1 + {\strut 1\over\displaystyle 3 + {\strut 1\over\displaystyle 12 + {\strut 1\over\displaystyle 1 + {\strut 1\over\displaystyle 3 }}}}}}$$

And similarly the other one. (Oh, the suffering!)

Then you need to find a continued fraction that lies numerically in between these two but is as short as possible. (Shortness of continued fractions corresponds directly to simplicity of the rational numbers they represent.) To do this, take the common initial segment, which is [0; 1, 1], and then apply an appropriate rule for the next place, which depends on whether the numbers in the next place differ by 1 or by more than 1, whether the first difference occurs in an even position or an odd one, mumble mumble mumble; in this case the rules say we should append 3. The result is [0; 1, 1, 3], or, in conventional notation:

$$0 + {1\over \displaystyle 1 + {\strut 1\over\displaystyle 1 + {\strut 1\over\displaystyle 3 }}}$$

which is equal to 4/7. And indeed, 4/7 of a year is 208.57 days, so sometime on the 208th day of the year, you can start reporting your age as (y and) 4/7 years.

Since I already had a library for calculating with continued fractions, I started extending it with functions to handle this problem, to apply all the fussy little rules for truncating the continued fraction in the right place, and so on.

Then I came to my senses, and realized there was a better way, at least for the cases I wanted to calculate. Given d, we want to find the simplest fraction a/b such that d/365 ≤ a/b < (d+1)/365. Equivalently, we want the smallest integer b such that there is some integer a with db/365 ≤ a < (d+1)b/365. But b must be in the range (2 .. 365), so we can easily calculate this just by trying every possible value of b, from 2 on up:

        use POSIX 'ceil', 'floor';

sub approx_frac {
my ($n,$d) = @_;
for my $b (1 ..$d) {
my ($lb,$ub) = ($n*$b/$d, ($n+1)*$b/$d);
if (ceil($lb) < ceil($ub) && ceil($ub) >$ub) {
return (int($ub),$b);
}
}
return ($n,$d);
}

The fussing with ceil() in the main test is to make the ranges open on the upper end: 2/5 is not in the range [3/10, 4/10), but it is in the range [4/10, 5/10). Then we can embed this in a simple report-printing program:

        my $N = shift || 365; for my$i (1..($N-1)) { my ($a, $b) = approx_frac($i, $N); print "$i/$N:$a/$b\n"; }  For tenths, the simplest fractions are:  1/10 ≤ 1/6 < 2/10 (0.1667) 2/10 ≤ 1/4 < 3/10 (0.2500) 3/10 ≤ 1/3 < 4/10 (0.3333) 4/10 ≤ 2/5 < 5/10 (0.4000) 5/10 ≤ 1/2 < 6/10 (0.5000) 6/10 ≤ 2/3 < 7/10 (0.6667) 7/10 ≤ 3/4 < 8/10 (0.7500) 8/10 ≤ 4/5 < 9/10 (0.8000) 9/10 ≤ 9/10 < 10/10 (0.9000) The simplest fractions that are missing from this table are 1/5, which is in the [2/10, 3/10) range and is beaten out by 1/4, and 3/5, which is in the [6/10, 7/10) range and is beaten out by 2/3. This works fine, and it is a heck of a lot simpler than all the continued fraction stuff. The more so because the continued fraction library is written in C. For the application at hand, an alternative algorithm is to go through all fractions, starting with the simplest, placing each one into the appropriate d/365 slot, unless that slot is already filled by a simpler fraction:  my$N = shift || 365;
my $unfilled =$N;

DEN:
for my $d (2 ..$N) {
for my $n (1 ..$d-1) {
my $a = int($n * $N /$d);
unless (defined $simple[$a]) {
$simple[$a] = [$n,$d];
last DEN if --$unfilled == 0; } } } for (1 ..$N-1) {
print "$_/$N: $simple[$_][0]/$simple[$_][1]\n";
}

A while back I wrote an article about using the sawed-off shotgun approach instead of the subtle technique approach. This is another case where the simple algorithm wins big. It is an n2 algorithm, whereas I think the continued fraction one is n log n in the worst case. But unless you're preparing enormous tables, it really doesn't matter much. And the proportionality constant on the O() is surely a lot smaller for the simple algorithms.

(It might also be that you could optimize the algorithms to go faster: you can skip the body of the loop in the slot-filling algorithm whenever $n and$d have a common factor, which means you are executing the body only n log n times. But testing for common factors takes time too...)

I was going to paste in a bunch of tabulations, but once again I remembered that it makes more sense to just let you run the program for yourself. Here is a form that will generate the table for all the fractions 1/N .. (N-1)/N; use N=365 to generate a table of year fractions for common years, and N=366 to generate the table for leap years:

Here's a program that will take your birthday and calculate your age in fractional years. Put in your birthday in ISO standard format: 2 April, 1969 is 19690402.

[ Addendum 20070429: There is a followup to this article. ]