There’s been a bit of hubbub in the in the math world the last few weeks with Shinichi Mochizuki’s 500 page proof that of the ABC conjecture. Basically, the conjecture states that given three positive coprime integers *a*, *b*, and *c* such that *a* + *b* = *c*, the product of the distinct prime factors of *a*, *b*, and *c* is rarely much smaller than *c*. While this may sound strange, there are a number of interesting consequences that you can read about here.

To make a long story shorter, there was a challenge on Programming Praxis that intrigued me, which was to write code that given a upper bound on *c* would generate a list of all of the triples *(a, b, c)* such that the product is larger.

I’ve been getting more into the hang of Racket, particularly the `for`

family of macros they have, so here’s my Racket code for generating such a list.

First, we could use a set of helper functions:

; keep only unique items in a list (define (unique l) (foldl (lambda (x l) (if (member x l) l (cons x l))) '() l)) ; test if i evenly divides n (define (div? n i) (zero? (remainder n i))) ; test if any item in l satisifies the predicate ? (define (any? ? l) (not (null? (filter ? l)))) ; return the prime factors of n (define (factor n) (let loop ([i 2]) (cond [(> (* i i) n) (list n)] [(div? n i) (append (factor i) (factor (/ n i)))] [else (loop (+ i 1))]))) ; calculate the greatest common demoninator of a and b (define (gcd a b) (cond [(= a b) a] [(< a b) (gcd a (- b a))] [else (gcd (- a b) b)])) ; test if two numbers are co-prime (share no common factors except 1) (define (coprime? a b) (= 1 (gcd a b))) |

Next we need the two functions from the actual proof. The neat thing is that with all of the helper functions above, these have become essentially trivial to write.

; calculate the radical (product of unique prime factors) of a number (define (rad n) (apply * (unique (factor n)))) ; calculate the ration of radicals for a set of coprime a + b = c (define (q a b c) (/ (log c) (log (rad (* a b c))))) |

Finally, the meat of the function. The best part for me was using Racket’s excellent `for*/list`

function to build the list.

; run the test ; for all b, for all a < b; ; if a+b < limit, a and b are coprime, and q > 1, store the result (define (abc n) (sort (for*/list ([b (in-range 1 n)] [a (in-range 1 b)] #:when (and (< (+ a b) n) (coprime? a b) (> (q a b (+ a b)) 1))) (list (q a b (+ a b)) a b (+ a b))) (lambda (x y) (> (car x) (car y))))) |

Now, to test it:

> (abc 1000) '((1.4265653296335432 3 125 128) (1.3175706029138485 1 512 513) (1.311101219926227 1 242 243) (1.292030029884618 1 80 81) (1.2727904629543532 13 243 256) (1.226294385530917 1 8 9) (1.2251805398372944 1 288 289) (1.2039689893561185 49 576 625) (1.198754152359422 169 343 512) (1.1757189916348774 32 49 81) (1.1366732909394883 25 704 729) (1.1126941404922133 1 63 64) (1.1084359145429683 32 343 375) (1.1048460623308174 104 625 729) (1.0921945706283556 1 675 676) (1.0917548251330267 100 243 343) (1.0790468171894105 1 624 625) (1.0458626417646626 1 728 729) (1.0456203807611666 5 507 512) (1.0412424573518235 1 48 49) (1.0370424407259895 81 175 256) (1.0344309549548174 343 625 968) (1.0326159011595437 81 544 625) (1.0326070471078377 7 243 250) (1.0288287974277142 2 243 245) (1.027195810121916 4 121 125) (1.0251241218312794 27 512 539) (1.018975235452531 5 27 32) (1.0129028397298183 1 224 225) (1.0084113092374762 200 529 729) (1.004797211020379 1 960 961))

Which are exactly the results that Programming Praxis got in their solution, so all is well in the world.

Just out of curiosity (since they did something clever in generating a and b while I just brute forced it) I ran timings of both of them on my machine:

> (time (abc 1000)) ; my code cpu time: 8206 real time: 8269 gc time: 139 > (time (abc 1000)) ; their code cpu time: 10280 real time: 10321 gc time: 189 |

Basically, they’re pretty close. Mine is a bit faster, which surprised me slightly, but it’s really not that big of a difference in the grand scheme of things.

In any case, it was a neat little exercise. Perhaps I’ll go and actually read the proof of the conjecture… No, probably not. It’s still an interesting result though.

If you’d like, you can download the entire source of the project here: abc source

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