Overlap
June 8, 2018
A little bit of thinking (and a few trials when I thought backwards) gives this solution:
(define (overlap xlo xhi ylo yhi)
(if (or (< xhi ylo) (< yhi xlo)) '()
(list (max xlo ylo) (min xhi yhi))))
If an overlap exists, it starts at the maximum of the lo values and ends at the minimum of the hi values. Here are some examples:
> (overlap 17 25 12 19) (17 19) > (overlap 12 17 19 25) () > (overlap 19 25 12 17) () > (overlap 19 25 22 30) (22 25)
You can run the program at https://ideone.com/c7a0XU.
Programming Praxis 
In Ruby. Using Ranges (built-in)
Outout:
22..25
nil
Here’s a solution in C.
#include <stdio.h> #include <stdlib.h> void print_usage_and_exit(char* prog) { fprintf(stderr, "Usage: %s min1 max1 min2 max2\n", prog); exit(EXIT_FAILURE); } int main(int argc, char* argv[]) { if (argc != 5) { print_usage_and_exit(argv[0]); } int min1 = atoi(argv[1]); int max1 = atoi(argv[2]); int min2 = atoi(argv[3]); int max2 = atoi(argv[4]); if (max1 < min1 || max2 < min2) { print_usage_and_exit(argv[0]); } if (min1 <= max2 && max1 >= min2) { int min3 = min1 > min2 ? min1 : min2; int max3 = max1 < max2 ? max1 : max2; printf("%d %d\n", min3, max3); } return EXIT_SUCCESS; }Example:
A Haskell version. For clarity we create a type for ranges.
data Range a = R a a | Empty deriving (Show) mkRange :: Ord a => a -> a -> Range a mkRange x y | x <= y = R x y | otherwise = Empty overlap :: Ord a => Range a -> Range a -> Range a overlap (R x1 y1) (R x2 y2) = let (x, y) = (max x1 x2, min y1 y2) in if x <= y then R x y else Empty overlap _ _ = Empty main :: IO () main = do print $ overlap (mkRange 19 25) (mkRange 22 30) print $ overlap (mkRange 19 25) (mkRange 12 17) print $ overlap (mkRange 'k' 'o') (mkRange 'a' 'm')Python
def overlap(a_min, a_max, b_min, b_max): '' return list(sorted(set(range(a_min, a_max+1)) & set(range(b_min, b_max+1)))) if __name__ == "__main__": '' print (overlap(19, 25, 22, 30)) print (overlap(12, 17, 19, 25))[22, 23, 24, 25]
[]
> def does_overlap(a, b, c, d): overlap = set(range(a, b+1)).intersection(set(range(c, d+1))) print(sorted(overlap)) if overlap else print("No overlap")Here’s a solution in Python.
def overlap(min1, max1, min2, max2): if min1 <= max2 and max1 >= min2: return max(min1, min2), min(max1, max2) return None print(overlap(17, 25, 12, 19)) print(overlap(12, 17, 19, 25)) print(overlap(19, 25, 12, 17)) print(overlap(19, 25, 22, 30))Output:
Mumps version
Klong version
pairs::{[a b]; a::(x@0)|(y@0); b::(x@1)&(y@1); :[b<a; []; a,b]} :dyad pairs([1 18]; [19 25]) [] pairs([1 19]; [19 25]) [19 19] pairs([1 20]; [19 25]) [19 20] pairs([19 20]; [19 25]) [19 20] pairs([20 20]; [19 25]) [20 20] pairs([20 24]; [19 25]) [20 24] pairs([20 25]; [19 25]) [20 25] pairs([20 26]; [19 25]) [20 25] pairs([24 26]; [19 25]) [24 25] pairs([25 26]; [19 25]) [25 25] pairs([26 26]; [19 25]) []racket version
(define (overlap a b) (cond [(and (empty? a) (empty? b)) empty] [(or (empty? a) (empty? b)) 'NoOverlap] [else (let* ((x (if (< (car a) (car b)) a b)) (y (if (equal? x a) b a))) (if (> (car y) (cadr x)) 'NoOverlap (list (car y) (min (cadr x) (cadr y)))))]))Testing