I realized that I had done the same calculation when I wrote an article On Password Strength, not thinking that there were other applications for this. I thought it was interesting and merited a short post.
So, to determine how many bits are needed for an integer of length 1, i.e, the integers
0 - 9, take the base-2 logarithm of 10:
log2(10) = 3.3219280949
This shows the mathematical relationship between base-2 and base-10. It’s saying, “How many times do we need to raise the number
2 to equal
To store a base-10 number we’ll need four bits. We round up, because obviously we can’t have a fraction of a bit.
To see that this is true, let’s count to
0 - 9 in binary:
0000 0001 0010 0011 0100 0101 0110 0111 1000 1001
This demonstrates that the integers
9 do indeed need four bits to be stored in memory.
Take the number 91. Since the number is two digits long, we can calculate the bits needed to store the number thusly:
- 2 * log2(10)
- 2 * 3.3219280949
- 7 bits
That looks right, after all 7 bits is decimal 128 which encompasses the number 91.
Let’s try another one! For instance, decimal 9119. The number is 4 digits long, so:
- 4 * log2(10)
- 4 * 3.3219280949
- 14 bits
Is that right? Let’s ask our trusty little friend asbits!
$ asbits 9119 0010 0011 1001 1111
Yes, it’s 14 bits long! Also, we can see that 14 bits is decimal 16,384, which encompasses our number 9119 (13 bits is less than 9119):
$ bc <<< 2^14 16384 $ bc <<< 2^13 8192
ASCII is a good example for encoding both numerical and alphabetical characters from bits. Recall that ASCII is a character encoding that can fit in 7 bits, and the 8th bit was used by many, many other encoding schemes to add additional characters and symbols that were incompatible with each other. This was a true nightmare before Unicode came to the rescue and solved all of our problems.
This is a good time to mention a classic Joel Spolsky article on the topic that is just as relevant now as it was when it was written in 2003. Read it, if you haven’t before! And if you have, read it again!