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QuestionHamming algorithm. Pin
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QuestionBignum timing Pin
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Member 419459314-Jan-15 11:20
Member 419459314-Jan-15 11:20 
I have several more questions to ask about bignumb processing, but since you were were the only one to respond to my original question I thought I'd just ask you directly via email, however your email seams to be blocked so I will ask here directly.

When you select the low limit for a potential P value of 40% of the decimal digit count, what would that be? Lets take a simple example - 40% of a 15 digit number would be 6 digits. My assumption is that it would be 100,000, 6 digits and the lowest number in 6 digits. If I wanted to insure that this was an odd number, would that change to 99,999 or to 999,999? 99,999 is not a 6 digit number, but 999,999 is certainly greater than the minimum 6 digit number of 100,000. Which should be used as the limit to match what RSA would generate for encryption?

To calculate this in binary, consider that 10^14 is 100,000,000,000,000 (15 digits) = 0x3,8D7E,A4C6,8000 (50 bits). 40% of 50 bits is 20 bits. The lowest 20 bit number is 0x8,0000 (the lowest number in 20 bits). If I wanted to insure that this was an odd number, would that change to 0x7,FFFF or 0xF,FFFF? 0x7,FFFF is just a 19 bit number, but 0xF,FFFF is certainly is certainly greater than the minimum 20 bit number of 0x8,0000.

The upper limit of 45% is easier to figure out. Just create the maximum number for that digit count (or bit count) and increment the count and create the
number as a base to a power, then decrement the result by 1 (it will always be odd, either 99999... or FFFFF...).

Dave.
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Questionsimple maby for you Pin
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QuestionI'm looking for an algorithm ... Pin
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QuestionAlgorithm to count how many "1" in binary o(1) Pin
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