#If ''x'' contains a 1, replace every digit strictly after the first 1 by 0.
#Replace allany remaining 2s with 1s.
#Interpret the result as a binary number. The result is ''c''(''x'').
For example:
* 1/4 becomesis 0.02020202... in base 3. There are no 1s so the next stage is still 0.02020202... This is rewritten as 0.01010101... When read in base 2, this corresponds to 1/3, so ''c''(1/4) = 1/3.
* 1/5 becomesis 0.01210121... in base 3. The digits after the first 1 are replaced by 0s to produce 0.01000000... This is not rewritten since there are no 2s. When read in base 2, this corresponds to 1/4, so ''c''(1/5) = 1/4.
* 200/243 becomesis 0.21102 (or 0.211012222...) in base 3. The digits after the first 1 are replaced by 0s to produce 0.21. This is rewritten as 0.11. When read in base 2, this corresponds to 3/4, so ''c''(200/243) = 3/4.
