How is 6 trits equal to 9.5 bits?

Question

This reddit thread says 6 trits ~ 9.5 bits.

How is 6 trits ~ 9.5 bits?

Answer 1

It's talking about information capacity. You can store roughly the same amount of data in 9.5 bits as you can 6 trits because 2^9.5 (724) is approximately 3⁶ (729).

Answer 2

Because 3^6 (729) is approximately 2^9.5 (724).

You can read about the ternary number system for more details.

Answer 3

I suppose, a trit is a "bit with three states" (0, 1 and 2).

Since you can form 3^6=729 different states with 6 trits (every trit can take three different states and you have 6 trits), the above calculation considers how many bits one needs to form 729 different states.

In fact (i.e. reality) you would need ten bits to form 729 different states, but a common calculation to derive how many bits are needed to form a specified number of states is to take the logarithm base 2 of the number. This yields log_2(3^6) = log_2(729) ~ 9.5.

Note: If you want to use the logarithm-base-2 calculation to determine how many bits are needed, you have to take the ceiling of the result.

Answer 4

http://en.wikipedia.org/wiki/Ternary_numeral_system

"One trit contains log23 (about 1.58496) bits of information."

1.58496 * 6 ~ 9.5

Answer 5

log(3**6)/log(2) ~ 9.5097750043269382

Answer 6

Binary are discrete values. You have only whole values, so I guess 9.5 bits are in actuality 10 Bits.

Answer 7

this might all be true, but as computers still works binary (high or low, on or off, open or closed) for a trit you would still need two binary positions. That is not really a good basis to talk about better information density. So as I see it, it takes 100% more space but gives 58% more info? unless ofcourse I completely misunderstand the concept

Answer 8

One trit contains (about 1.58496) bits of information. and hence 6 trit = 9.5 bits (approx)

Some more interesting information in Wikipedia

Answer 9

A trits would be based off of its own discrete transistor with three states rather than two. So it would occupy (hypothetically) the exact same amount of space, but would hold much more data.