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Forum: Math and Practical

Topic: Converting random bytes into a range

Converting random bytes into a range [message #65] 
Wed, 06 April 2011 00:48 
TRNG98
Messages: 21 Registered: January 2011 Location: Lund

Senior Member 


Now, if you wish to convert a random byte (0..255) into a digit (0..9) you can do like this:
 You combine the digits into the maximum that your input can generate (99)
 You discard the rest of the range (100..255)
You can discard less if you have 10 bit input and can generate 3 digits (0..999 and discard only 1000..1023).
This is how to solve for a optimum solution:
 Your input is (0..255) so, in bits it is log (256)/log(2) = 8 (any log will do)
 Your range is log (10)/log(2) = 3.32
 The ratio of a digit to a byte is: 3.32/8=0.415
By adding (to itself) the number 0.415 we see how much we can use of the input, and how large the loss is:
Digits Fill Number of bytes
1 0.415 1
2 0.830 1
3 1.246 2
4 1.661 2
5 2.076 3
6 2.491 3
7 2.907 3
8 3.322 4
9 3.373 4
10 4.152 5
11 4.568 5
12 4.983 5
Since the input comes in bytes, using 3 bytes and convert into 7 digits produce a loss of (3.0002.907)/7=0.013 bytes for each output digit. This obviously get better if we use larger numbers. If you use 5 bytes and generate 12 digits, the loss is only 0.0014 bytes/digit.
A compromise is usually easy to find.
TRNG98 We sell hardware random number generators



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