How to resolve the algorithm 100 doors step by step in the SheerPower 4GL programming language
How to resolve the algorithm 100 doors step by step in the SheerPower 4GL 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 SheerPower 4GL programming language
Source code in the sheerpower programming language
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
! I n i t i a l i z a t i o n
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
doors% = 100
dim doorArray?(doors%)
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
! M a i n L o g i c A r e a
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
// Initialize Array
for index% = 1 to doors%
doorArray?(index%) = false
next index%
// Execute routine
toggle_doors
// Print results
for index% = 1 to doors%
if doorArray?(index%) = true then print index%, ' is open'
next index%
stop
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
! R o u t i n e s
!%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
routine toggle_doors
for index_outer% = 1 to doors%
for index_inner% = 1 to doors%
if mod(index_inner%, index_outer%) = 0 then
doorArray?(index_inner%) = not doorArray?(index_inner%)
end if
next index_inner%
next index_outer%
end routine
end
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