How to resolve the algorithm 100 doors step by step in the REXX programming language
How to resolve the algorithm 100 doors step by step in the REXX programming language
Table of Contents
Problem Statement
There are 100 doors in a row that are all initially closed.
You make 100 passes by the doors.
The first time through, visit every door and toggle the door (if the door is closed, open it; if it is open, close it).
The second time, only visit every 2nd door (door #2, #4, #6, ...), and toggle it.
The third time, visit every 3rd door (door #3, #6, #9, ...), etc, until you only visit the 100th door.
Answer the question: what state are the doors in after the last pass? Which are open, which are closed?
Alternate:
As noted in this page's discussion page, the only doors that remain open are those whose numbers are perfect squares.
Opening only those doors is an optimization that may also be expressed;
however, as should be obvious, this defeats the intent of comparing implementations across programming languages.
Let's start with the solution:
Step by Step solution about How to resolve the algorithm 100 doors step by step in the REXX programming language
Source code in the rexx programming language
/*REXX pgm solves the 100 doors puzzle, doing it the hard way by opening/closing doors.*/
parse arg doors . /*obtain the optional argument from CL.*/
if doors=='' | doors=="," then doors=100 /*not specified? Then assume 100 doors*/
/* 0 = the door is closed. */
/* 1 = " " " open. */
door.=0 /*assume all doors are closed at start.*/
do #=1 for doors /*process a pass─through for all doors.*/
do j=# by # to doors /* ··· every Jth door from this point.*/
door.j= \door.j /*toggle the "openness" of the door. */
end /*j*/
end /*#*/
say 'After ' doors " passes, the following doors are open:"
say
do k=1 for doors
if door.k then say right(k, 20) /*add some indentation for the output. */
end /*k*/ /*stick a fork in it, we're all done. */
/*REXX pgm solves the 100 doors puzzle, doing it the easy way by calculating squares.*/
parse arg doors . /*obtain the optional argument from CL.*/
if doors=='' | doors=="," then doors=100 /*not specified? Then assume 100 doors*/
say 'After ' doors " passes, the following doors are open:"
say
do #=1 while #**2 <= doors /*process easy pass─through (squares).*/
say right(#**2, 20) /*add some indentation for the output. */
end /*#*/ /*stick a fork in it, we're all done. */
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