macOS A little OT

[mdeh]

Hi all,
I am reading Piper/Murphy's Cryptography. There is one sentence, that does not quite make sense. The text is talking about decimal and binary numbers.

They show an example of converting a decimal number to it's binary equivalent, and show how 53 and 86 are converted. "53" is a 6-bit number and "86" is an example of a 7-bit number.

Now the confusing part.

In general, 3.32d gives some idea of the number of bits needed to express d decimal digits in binary

Can anyone elucidate the "3.32d" for example with the digit 1, or 20 etc. It's bugging me, even though in the big picture, it really does not effect anything.

Thanks in advance

 

[kpua]

Just a guess, but is it just multiplication? It sounds like it's suggesting that, for example, to represent a 4-digit decimal number (1000-9999) in binary you would need about 3.32 * 4 = 13.28 bits on average.

 

[mdeh]

Just a guess, but is it just multiplication? It sounds like it's suggesting that, for example, to represent a 4-digit decimal number (1000-9999) in binary you would need about 3.32 * 4 = 13.28 bits on average.

That's what I thought initially too. Thanks

 

[lee1210]

A better way would be:
the ceiling of log base 2(value+1)

so for 86:
log base 2 of 87 is between 6 and 7, so the ceiling is 7. It's a bit more involved, but you can tell for sure how many bits you need.

-Lee

Edit: I didn't see that this was estimating the number of bits needed to estimate some number of digits...
Using the method above you get 10 for 1000, and 14 for 9999
The average is nice, i guess? Everything over 8191 will take 14 bits, everything over 4095 will take 13, so that's 2/5 of the space at 13, and 1/5 at 14... there isn't a lot of the space that can use only 10, 1/10th or so that can use 11, the rest 12... i guess 13.28 is nice to know? I'm not sure what you'd do with it.

 

[mdeh]

A better way would be:
the ceiling of log base 2(value+1)

so for 86:
log base 2 of 87 is between 6 and 7, so the ceiling is 7. It's a bit more involved, but you can tell for sure how many bits you need.

-Lee

Edit: I didn't see that this was estimating the number of bits needed to estimate some number of digits...
Using the method above you get 10 for 1000, and 14 for 9999
The average is nice, i guess? Everything over 8191 will take 14 bits, everything over 4095 will take 13, so that's 2/5 of the space at 13, and 1/5 at 14... there isn't a lot of the space that can use only 10, 1/10th or so that can use 11, the rest 12... i guess 13.28 is nice to know? I'm not sure what you'd do with it.

Thanks Lee