Factoring By Digital Coding

April 1, 2014

A few days ago, an announcement was made in /r/crypto of a new factoring algorithm that runs in time O(log^4 n). A paper describes the algorithm. And even better, coding for the new algorithm is simple:

1)Set mn.
2) Express m as a list of its decimal digits.
3) Express each decimal digit as a list of binary bits.
4) Append the individual lists of binary bits.
5) Convert the resulting list of binary bits into a decimal number. Call it d.
6) If the greatest common divisor of n and d is between 1 and n, it is a factor of n. Report the factor and halt.
7) Set md and go to Step 1.

Steps 1 through 4 form the digital code of a number. For example, 88837 forms the bit-list 1 0 0 0, 1 0 0 0, 1 0 0 0, 1 1, 1 1 1, where we used commas to delineate the decimal digits. That bit-list is 69919 in decimal, which is the digital code of 88837. To find a factor of 88837, compute successive digital codes until a gcd is greater than 1; the chain runs 88837, 69919, 54073, 2847, 2599, 5529, 2921, 333, at which point gcd(88837, 333) = 37 and we have found a factor: 88837 = 74 × 37. Factoring 96577 is even faster: the digital code of 96577 is 40319, and gcd(96577, 40319) = 23, which is a factor of 96577 = 13 × 17 × 19 × 23. But trying to factor 65059 = 17 × 43 × 89 causes an infinite loop. The paper gives a heuristic solution to that problem, but we’ll stop there and accept that sometimes a factorization fails.

Your task is to write a function that factors by digital coding. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

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