Question

The main question: How many digits?

Let me explain. I have a number in binary system: 11000000 and in decimal is 192.

After converting to decimal, how many digits it will have (in dicimal)? In my example, it’s 3 digits. But, it isn’t a problem. I’ve searched over internet and found one algorithm for integral part and one for fractional part. I’m not quite understand them, but (I think) they works.

When converting from binary to octal, it’s more easy: each 3 bits give you 1 digit in octal. Same for hex: each 4 bits = 1 hex digit.

But, I’m very curious, what to do, if I have a number in P numeral system and want to convert it to the Q numeral system? I know how to do it (I think, I know :)), but, 1st of all, I want to know how many digits in Q system it will take (u no, I must preallocate space).

Answer 1

There was an error in my previous answer: look at the comment by Ben Schwehn.
Sorry for the confusion, I found and explain the error I made in my previous answer below.

Please use the answer provided by Paul Tomblin. (rewritten to use P, Q and n)

Y = ln(P^n) / ln(Q)
Y = n * ln(P) / ln(Q)

So Y (rounded up) is the number of characters you need in system Q to express the highest number you can encode in n characters in system P.

I have no answer (that wouldn’t convert the number already and take up that many space in a temporary variable) to get the bare minimum for a given number 1000(bin) = 8(dec) while you would reserve 2 decimal positions using this formula.

If a temporary memory usage isn’t a problem, you might cheat and use (Python):

len(str(int(otherBaseStr,P)))

This will give you the number of decimals needed to convert a number in base P, cast as a string (otherBaseStr), into decimals.

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Old WRONG answer:

If you have a number in P numeral system of length n
Then you can calculate the highest number that is possible in n characters:

P^(n-1)

To express this highest number in number system Q you need to use logarithms (because they are the inverse to exponentiation):

log((P^(n-1))/log(Q)
(n-1)*log(P) / log(Q)

For example
11000000 in binary is 8 characters.
To get it in Decimal you would need:

(8-1)*log(2) / log(10) = 2.1 digits (round up to 3)

Reason it was wrong:

The highest number that is possible in n characters is

(P^n) - 1

not

P^(n-1)