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The following is a "Laser Maze."

The goal is to place the tools into the maze so that when the power is turned up to maximum:

  1. The beam will split into two beams
  2. The beams (collectively) will occupy all currently unoccupied squares
  3. Each beam will exit the maze through separate exit holes

Contrary to popular belief, the beams may cross; they will pass each other without interference or direction change (the splitter will have moved them onto separate planes).



Here are the tools along with their allowed rotations and allotted quantities:

 Qty  Name                  Rotations Available

(13) Corner Mirrors :corner1 corner2 corner3 corner4
(2) Double-Sided Mirrors :mirror1 mirror2
(1) Beam Splitter :splitter1 splitter2


As (faintly) indicated, each tool occupies one square.
All (16) tools must be used.


The tools react to the Laser Beam as follows...
enter image description here *Hitting a corner mirror at 0°, 90°, 180° or 360° stops the beam's progress, as indicated on the far right.



...and hitting the wooden blocks is bad.
enter image description here




The maze:


enter image description here
*The laser is currently set to low power, and only penetrates a few squares.

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  • 2
    $\begingroup$ Does "All (16) tools must be used" mean that the laser must touch them all, or simply that all the tools must be placed on the board? $\endgroup$ – Riley Jul 7 '18 at 18:01
  • $\begingroup$ @Riley I had intended "Both," but alternate solutions placing all and utilizing only some would be interesting... $\endgroup$ – Chowzen Jul 7 '18 at 18:19
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    $\begingroup$ @Riley, rule 2 says that the laser(s) must visit every square that's initially empty, which makes it impossible to place a tool so that it wouldn't be touched by laser. Well, inside the maze's outer walls, at least. $\endgroup$ – Bass Jul 7 '18 at 19:15
  • $\begingroup$ @Bass Ah, I missed that. $\endgroup$ – Riley Jul 7 '18 at 20:07
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Here's my go at this "laze". To make the paths easier to distinguish, I used a beam splitter that changes the colour of one of the beams.

enter image description here

(Image was edited to remove the ugliness mentioned in the comments.)

My approach was to plonk down the unavoidable bits:

* each corner must have a corner mirror, so mirrors 1,2,3,5,9,11,12 and 13 are automatic.
* 8 needs a mirror, and it cannot be in any other orientation without creating a dead end. This also forces 7.
* The beam splitter's square has a three-way intersection, where any other piece would create a dead end, and if the square were empty, the square above it would be impossible to visit without hitting a wall

With all these in place, I just put everything else (mirrors 4,6 and 10, and the two-way mirrors) in where they seemed to be useful, and happened to hit a working solution with the correct number of pieces by chance.


On the (non-)uniqueness of the solution

Notice that the lower double-sided mirror is actually a no-op: without it, the beams would just switch roles, and everything would still work out exactly the same.

Since the mirror isn't needed there, it can be removed and placed anywhere where two different beams cross, as long as it is rotated so that the incoming beams hit opposite sides of the mirror. It won't have any effect in any of those places, either:

  • two spaces up, rotated
  • 4 spaces up, three to the right, rotated
  • 4 spaces up, one to the right, same orientation (looks a bit confusing, but works)
  • 4 spaces up, four to the right, rotated

The upper two-sided mirror isn't fixed either, at least these positions work too:

  • same spot, rotated (the direction of the final loop reverses)
  • two spaces up, either orientation (the loop starts at a different spot)
  • one space up, one to the right, either orientation (pink beam does the loop)
  • one space up, two to the left, either orientation (pink beam does the loop)
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  • $\begingroup$ This works. It's almost identical to the solution I had in mind, just uglier. :) $\endgroup$ – Chowzen Jul 7 '18 at 18:25
  • $\begingroup$ Note to self: peer pressure works! $\endgroup$ – Chowzen Jul 7 '18 at 20:17

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