How to resolve the algorithm Sorting Algorithms/Circle Sort step by step in the REXX programming language
Published on 12 May 2024 09:40 PM
How to resolve the algorithm Sorting Algorithms/Circle Sort step by step in the REXX programming language
Table of Contents
Problem Statement
Sort an array of integers (of any convenient size) into ascending order using Circlesort. In short, compare the first element to the last element, then the second element to the second last element, etc. Then split the array in two and recurse until there is only one single element in the array, like this: Repeat this procedure until quiescence (i.e. until there are no swaps). Show both the initial, unsorted list and the final sorted list. (Intermediate steps during sorting are optional.) Optimizations (like doing 0.5 log2(n) iterations and then continue with an Insertion sort) are optional. Pseudo code:
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Sorting Algorithms/Circle Sort step by step in the REXX programming language
Source code in the rexx programming language
/*REXX program uses a circle sort algorithm to sort an array (or list) of numbers. */
parse arg x /*obtain optional arguments from the CL*/
if x='' | x="," then x= 6 7 8 9 2 5 3 4 1 /*Not specified? Then use the default.*/
call make_array 'before sort:' /*display the list and make an array. */
call circleSort # /*invoke the circle sort subroutine. */
call make_list ' after sort:' /*make a list and display it to console*/
exit /*stick a fork in it, we're all done. */
/*──────────────────────────────────────────────────────────────────────────────────────*/
circleSort: do while .circleSrt(1, arg(1), 0)\==0; end; return
make_array: #= words(x); do i=1 for #; @.i= word(x, i); end; say arg(1) x; return
make_list: y= @.1; do j=2 for #-1; y= y @.j; end; say arg(1) y; return
.swap: parse arg a,b; parse value @.a @.b with @.b @.a; swaps= swaps+1; return
/*──────────────────────────────────────────────────────────────────────────────────────*/
.circleSrt: procedure expose @.; parse arg LO,HI,swaps /*obtain LO & HI arguments.*/
if LO==HI then return swaps /*1 element? Done with sort.*/
high= HI; low= LO; mid= (HI-LO) % 2 /*assign some indices. */
/* [↓] sort a section of #'s*/
do while LO
if @.LO>@.HI then call .swap LO,HI /*are numbers out of order ? */
LO= LO + 1; HI= HI - 1 /*add to LO; shrink the HI. */
end /*while*/ /*just process one section. */
_= HI + 1 /*point to HI plus one. */
if LO==HI & @.LO>@._ then call .swap LO, _ /*numbers still out of order?*/
swaps= .circleSrt(low, low+mid, swaps) /*sort the lower section. */
swaps= .circleSrt(low+mid+1, high, swaps) /* " " higher " */
return swaps /*the section sorting is done*/
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