Predicting Number of Decimal Digits in a Binary Number
Date: 03/10/2004 at 11:52:00 From: R.B. Subject: Number of decimal digits from binary number I am trying to write a function that will return the number of decimal digits from a binary number without actually converting from binary to decimal. Here is what I have so far: let d = number of digits d = floor of [(bits - 1)*log2] + 1 However, if (bits - 1)mod 10 = 3,6,or 9 then the number of digits could be: [(bits - 1)*log2] + 1 OR [(bits - 1)*log2] + 2 Say, for instance, the binary number has 4 bits. In decimal that number ranges from 8-15...1 or 2 digits. Is there a way to determine which length is correct without going through a complete conversion?
Date: 03/10/2004 at 12:53:15 From: Doctor Douglas Subject: Re: Number of decimal digits from binary number Hi R.B. Thanks for writing to the Math Forum. I don't think that there is a way to do this with a single formula. For example, consider the two numbers 1111100111 binary = 999 decimal 1111101000 binary = 1000 decimal If you are going to distinguish the fact that one of these numbers gives rise to 3 decimal digits and the other gives rise to 4 decimal digits, you will need to get into the last four digits in the binary expansion--you will need to parse at least the first seven digits of the binary input. To be efficient, as you've realized, you need to implement the direct conversion only when there is a danger of crossing from k decimal digits to k + 1 decimal digits (e.g. 999 to 1000). This is almost, but not quite, identical to your rule using the mod function. I think the smallest number of binary digits for which this differs is 102 bits: 2^103 = 1.01412E+31 has 32 decimal digits Thus a 102 bit number could have either 31 or 32 decimal digits. - Doctor Douglas, The Math Forum http://mathforum.org/dr.math/
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