SWISH -- SWI-Prolog for SHaring

% Student exercise profile
:- set_prolog_flag(occurs_check, error).        % disallow cyclic terms
:- set_prolog_stack(global, limit(8 000 000)).  % limit term space (8Mb)
:- set_prolog_stack(local,  limit(2 000 000)).  % limit environment space

% given a Lst, compute ...the minimum, in a list.
/* Here is a version which corresponds to our tail-recursive
 * version from a previous exercise
 * 
 * ;min‑helper : number, list‑of‑number ‑> number;
 * ;Return the smallest number out of those in `nums‑remaining` and `min‑so‑far`.;
 * ;Note that nums‑remaining may be empty
 * (define (min‑helper   min‑so‑far  nums‑remaining)  
 *   (cond [(empty? nums‑remaining) min‑so‑far]
 *         [(cons?  nums‑remaining)  (let* {[a (first nums‑remaining)]} 
 *                                     (min‑helper (if (< min‑so‑far a) min‑so‑far a)
 *                                                 (rest nums‑remaining)))])) 
 */

/*
% A first version -- transliterated from above, but not yet idiomized.
myMin( Nums, Answer ) :- Nums=[F|R], minHelper( F, R, Answer ).
minHelper( MinSoFar, NumsRemaining, TheMin ) :- NumsRemaining=[], TheMin=MinSoFar.
minHelper( MinSoFar, NumsRemaining, TheMin ) :- NumsRemaining=[F|R], A=F, MinSoFar < A, minHelper(MinSoFar,R,TheMin).
minHelper( MinSoFar, NumsRemaining, TheMin ) :- NumsRemaining=[F|R], A=F, MinSoFar >= A, minHelper(A,R,TheMin).
*/

% Now, just idiomize by removing "=" from r.h.s.s
myMin( [F|R], Answer ) :- minHelper( F, R, Answer ).
minHelper( TheMin, [], TheMin ).
minHelper( MinSoFar, [F|R], TheMin ) :- MinSoFar< F, minHelper(MinSoFar,R,TheMin).
minHelper( MinSoFar, [F|R], TheMin ) :- MinSoFar>=F, minHelper(F,R,TheMin).

/*
minHelper(2,[],X).
minHelper(2,[3],X).
minHelper(2,[4],X).
minHelper(2,[4,3],X).
minHelper(2,[4,1],X).

myMin([],X).
myMin([4],X).
myMin([3,2,4],X).