Description (from Rosetta Code)
| 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?
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.
Showing the intermediate states
The problem can be generalized to a different number of doors other than 100.
The following variation creates an image with the intermediate results. Horizontal axis (left to right) represents the number of door, vertical axis (top to bottom) is the the number of vist.
The following is the history for 100 doors. Each square is 3x3 pixels.
The next one is the history for 500 doors. Each square is 1x1 pixel.