Sort Four
April 11, 2017
Sorting is done like this. First, sort the first two items in the input. Second, sort the last two items in the input. Third, compare the two low items of the first two and the last two; the lower of those two items is the lowest in the final output. Fourth, compare the two high items of the first two and the last two; the higher of those two items is the highest in the final output. Fifth, compare the two remaining items, putting them in the correct order. That’s a mouthful to say, and even more complicated to code:
(define (sort4 a b c d)
(let ((ab (if (< a b) (list a b) (list b a)))
(cd (if (< c d) (list c d) (list d c))))
(let ((first #f) (mid1 #f) (mid2 #f) (last #f))
(cond ((< (car ab) (car cd))
(set! first (car ab))
(set! mid1 (car cd)))
(else
(set! first (car cd))
(set! mid1 (car ab))))
(cond ((< (cadr ab) (cadr cd))
(set! last (cadr cd))
(set! mid2 (cadr ab)))
(else
(set! last (cadr ab))
(set! mid2 (cadr cd))))
(if (< mid1 mid2)
(list first mid1 mid2 last)
(list first mid2 mid1 last)))))
Since there are only 4! = 24 possible inputs, it is easy to show that the program works in all cases:
> (for-each
(lambda (xs)
(display xs) (display " => ")
(display (apply sort4 xs)) (newline))
(permutations '(1 2 3 4)))
(4 3 2 1) => (1 2 3 4)
(3 4 2 1) => (1 2 3 4)
(2 4 3 1) => (1 2 3 4)
(4 2 3 1) => (1 2 3 4)
(3 2 4 1) => (1 2 3 4)
(2 3 4 1) => (1 2 3 4)
(1 4 3 2) => (1 2 3 4)
(4 1 3 2) => (1 2 3 4)
(3 1 4 2) => (1 2 3 4)
(1 3 4 2) => (1 2 3 4)
(4 3 1 2) => (1 2 3 4)
(3 4 1 2) => (1 2 3 4)
(2 1 4 3) => (1 2 3 4)
(1 2 4 3) => (1 2 3 4)
(4 2 1 3) => (1 2 3 4)
(2 4 1 3) => (1 2 3 4)
(1 4 2 3) => (1 2 3 4)
(4 1 2 3) => (1 2 3 4)
(3 2 1 4) => (1 2 3 4)
(2 3 1 4) => (1 2 3 4)
(1 3 2 4) => (1 2 3 4)
(3 1 2 4) => (1 2 3 4)
(2 1 3 4) => (1 2 3 4)
(1 2 3 4) => (1 2 3 4)
You can run the program at http://ideone.com/3jmcKZ.
Programming Praxis 
Like this?
Two versions using Clojure:
(defn four-sort1 [a b c d] (sort [a b c d])) (defn swap [x y] [(min x y) (max x y)]) (defn four-sort2 [a b c d] (let [[x y] (swap a b) [z m] (swap c d) [l y2] (swap x z) [z2 h] (swap y m) [i1 i2] (swap y2 z2)] [l i1 i2 h]))Here’s a depth-3 Parallel Fouring Machine program, P, tested with an interpreter written in Python (see below). Parallel Fouring Machine programs are binary trees of commands that act on four registers so that each command sorts both the two indicated registers and the two complementary registers, in parallel, and they continue to the left branch if either pair was swapped, else to right.
from itertools import permutations P = ((0, 1), ((0, 2), ((0, 3), (), ()), ((0, 3), (), ())), ((0, 2), ((0, 3), (), ()), ((0, 3), (), ()))) for D in permutations(range(4)): print(D, '=>', F(P, D)) else: print((3,1,4,1), '=>', F(P, (3,1,4,1)))Clearly this program P always executes the same three parallel comparisons, so it corresponds to six sequential comparisons. Is off by one close enough? (I was not able to generate even all depth-4 Sequential Fouring Machine programs, let alone enough of the required depth-5. It was a combinatorial explosion, and I had no ideas to prune the search space. So I went on to discover the Parallel Fouring Machine instead. For the Journal of Negative Results, maybe.)
Here’s the interpreter, F, in Python.
def F(P, D): while P: P, D = op(P, D) return D # complementary pairs comp = { (0,1) : (2,3), (0,2) : (1,3), (0,3) : (1,2) } def op(P, D): '''Sorts both the pair indicated by the top opcode, and the complementary pair. Continues to the left branch if either pair was swapped, else continues to the right branch.''' O, L, R = P j, k = O u , w = comp[O] S = {} if D[j] > D[k]: S.update({j:k, k:j}) if D[u] > D[w]: S.update({u:w, w:u}) if S: return L, tuple(D[S.get(t, t)] for t in range(4)) else: return R, DThere are only 36 depth-3 Parallel Fouring Machine programs that always sort the four registers. Some might contain the clue to the 5-step sequential sorting problem, if they were investigated more closely.
For such a small number of items to sort, we will avoid memory writes, by expanding the decision tree, and keeping the four elements in registers until it’s time to store them in the resulting sorted vector. Between 4 and 5 tests are needed to reach the leaves of the decision tree (on average for equiprobable permutations: 4.5).
;; Common Lisp (defun sort-4 (lessp a b c d) (if (funcall lessp a b) (if (funcall lessp c d) (if (funcall lessp a c) (if (funcall lessp b d) (if (funcall lessp b c) (vector a b c d) (vector a c b d)) (vector a c d b)) (if (funcall lessp b d) (vector c a b d) (if (funcall lessp a d) (vector c a d b) (vector c d a b)))) (if (funcall lessp a d) (if (funcall lessp b c) (if (funcall lessp b d) (vector a b d c) (vector a d b c)) (vector a d c b)) (if (funcall lessp b c) (vector d a b c) (if (funcall lessp a c) (vector d a c b) (vector d c a b))))) (if (funcall lessp c d) (if (funcall lessp b c) (if (funcall lessp a d) (if (funcall lessp a c) (vector b a c d) (vector b c a d)) (vector b c d a)) (if (funcall lessp a d) (vector c b a d) (if (funcall lessp b d) (vector c b d a) (vector c d b a)))) (if (funcall lessp b d) (if (funcall lessp a c) (if (funcall lessp a d) (vector b a d c) (vector b d a c)) (vector b d c a)) (if (funcall lessp a c) (vector d b a c) (if (funcall lessp b c) (vector d b c a) (vector d c b a))))))) (assert (every (lambda (v) (equalp (sort-4 (function <) (aref v 0) (aref v 1) (aref v 2) (aref v 3)) #(1 2 3 4))) '(#(1 2 3 4) #(2 1 3 4) #(2 3 1 4) #(2 3 4 1) #(1 3 2 4) #(3 1 2 4) #(3 2 1 4) #(3 2 4 1) #(1 3 4 2) #(3 1 4 2) #(3 4 1 2) #(3 4 2 1) #(1 2 4 3) #(2 1 4 3) #(2 4 1 3) #(2 4 3 1) #(1 4 2 3) #(4 1 2 3) #(4 2 1 3) #(4 2 3 1) #(1 4 3 2) #(4 1 3 2) #(4 3 1 2) #(4 3 2 1))))def sort4(a,b,c,d): if b < a: a,b = b,a if d < c: c,d = d,c # at this point, we know the min value is either a or c # and the max value is either b or d if c < a: a,c = c,a if d < b: b,d = d,b # now, the min is a and the max is b # all that remains is to sort the middle two values if c < b: b,c = c,b return a,b,c,d@Mike, I should have had the courage :) Here’s an interpreter for a simpler Fouring Machine whose six opcodes just sort a pair of registers without branching anywhere. A program is a sequence of opcodes.
def F(P, D): for O in P: D = op(O, D) return D def op(O, D): j, k = O S = {j:k, k:j} if D[j] > D[k] else {} return tuple(D[S.get(t, t)] for t in range(4)) if S else DIt’s a combinatorial piece of cake to generate a 5-command solution now. There are 12. Here’s one, which turns out to be precisely your (Mike’s) solution.
from itertools import permutations P = ((0, 1), (2, 3), (0, 2), (1, 3), (1, 2)) for D in permutations(range(4)): print(D, '=>', F(P, D)) else: print((3,1,4,1), '=>', F(P, (3,1,4,1)))One could write a compiler to get a Python (or Scheme (or C (or why not some assembler))) program.
This is Scheme.
;; library - should be imported from a library (define (fold op init args) (if (pair? args) (fold op (op init (car args)) (cdr args)) init)) (define (pipe . ops) (define (pipe op then) (lambda args (call-with-values (lambda () (apply op args)) then))) (fold pipe values ops)) ;; program : 3 1 4 1 => 1 1 3 4 (define sort4 (let-syntax ((on (syntax-rules (=>) ((on cond in => out) (lambda in (if cond (values . out) (values . in))))))) (let ((ab (on (< b a) (a b c d) => (b a c d))) (ac (on (< c a) (a b c d) => (c b a d))) (bc (on (< c b) (a b c d) => (a c b d))) (bd (on (< d b) (a b c d) => (a d c b))) (cd (on (< d c) (a b c d) => (a b d c)))) (pipe ab cd ac bd bc)))) ;; test (call-with-values (lambda () (sort4 3 1 4 1)) (pipe list display)) (newline) (call-with-values (lambda () (sort4 1 4 1 5)) (pipe list display)) (newline) (call-with-values (lambda () (sort4 4 1 5 9)) (pipe list display)) (newline) (call-with-values (lambda () (sort4 1 5 9 2)) (pipe list display)) (newline) (call-with-values (lambda () (sort4 5 9 2 6)) (pipe list display)) (newline)This scheme version is nice, but scheme compilers won’t be able to avoid the fifth comparison, when it’s useless. You will get exactly 5 comparison, why the Common Lisp expanded version has 4.5 comparisons on average.
MUMP V1:
SORT4 ;New routine
1 (N1,N2,N3,N4) ;
N ARR,I
; When the numbers are placed into the array “ARR”, they are automatically placed in sorted order
F I=N1,N2,N3,N4 S ARR(I)=””
; Write the numbers starting at the least
S I=”” F S I=$O(ARR(I)) Q:I=”” W !,I
Q
MCL> D 1^SORT4(99,87,65,7)
7
65
87
99
Here’s a solution in x86 assembly.
Here’s a C program that calls the function.
/* main.c */ #include <stdio.h> void sort4(int* array); static void print_array(int* array) { printf("["); for (int i = 0; i < 4; ++i) { if (i > 0) printf(", "); printf("%d", array[i]); } printf("]"); } int main(int argc, char* argv[]) { int array[] = {4, 7, 1, 6}; printf("array: "); print_array(array); printf("\n"); sort4(array); printf("sorted: "); print_array(array); printf("\n"); }Example usage:
Output: