It took me a rewatch to get it, but then I got it.
Largest Possible Remainder problem in racket:
;;lr = largest remainder so far lrd = divisor with largest remainder (define (largest-remainder n d (lr 0) (lrd 0)) (cond [(= d 0) (values lr lrd)] ;;catch cases like n = 6, d = 3 [(> lr d) (values lr lrd)] ;;stop when d is < largest remainder [(> (modulo n d) lr) (largest-remainder n (sub1 d) (modulo n d) d)] ;;save new largest remainder [else (largest-remainder n (sub1 d) lr lrd)])) ;;iterate with d-1
Common Lisp translation:
(defun largest-remainder (n d &optional (lr 0) (lrd 0)) (cond ((<= d 0) (values lr lrd)) ((< d lr) (values lr lrd)) (t (let ((r (rem n d))) (if (> r lr) (largest-remainder n (1- d) r d) (largest-remainder n (1- d) lr lrd))))))
Iterative in python.
def largest_remainder(n,d): lr = 0 lrd = 0 while (d >= lr) and (d > 0): if (n % d > lr): lr = n % d lrd = d d -= 1 return (lr,lrd) print largest_remainder(20,10) print largest_remainder(6,3)
Care of the hyperbolic tiling generator.
Two puzzles, one solved.
Second: Mathematicians discover a new way to tile pentagons (but not The Pentagon.)
Gets really weird about 4:30.
The video is cited in this paper (PDF).