How many bits to a trit?

There are \(\log_2(3) \approx 1.585\) bits to a Trit. This can be in­ter­preted a few differ­ent ways:

  1. If you mul­ti­ply the num­ber of mes­sages you might want to send by 3, then the cost of en­cod­ing the mes­sage will go up by 1.58 bits on av­er­age. See Marginal mes­sage cost for more on this in­ter­pre­ta­tion.

  2. If you pack \(n\) of in­de­pen­dent and equally likely 3-mes­sages to­gether into one gi­ant \(3^n\) mes­sage, then the cost (in bits) per in­di­vi­d­ual 3-mes­sage drops as \(n\) grows, ul­ti­mately con­verg­ing to \(\log_2(3)\) bits per 3-mes­sage as \(n\) gets very large. For more on this, see Aver­age mes­sage cost and the GalCom ex­am­ple of en­cod­ing trits us­ing bits.

  3. The in­finite ex­pan­sion of \(\log_2(3) = 1.58496250072\ldots\) tells us not just how many bits it takes to send one 3-mes­sage \((\approx \lceil 1.585 \rceil = 2)\) but also how long it takes to send any num­ber of 3-mes­sages put to­gether. For ex­am­ple, it costs 2 bits to send one 3-mes­sage; 16 bits to send 10; 159 bits to send 1000; 1585 to send 10,000; 15850 to send 100,000; 158497 to send 1,000,000; and so on. For more on this in­ter­pre­ta­tion, see the “se­ries of ceilings” in­ter­pre­ta­tion of log­a­r­ithms.