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Previous Level:- Lasers: Everything is Switched (Level $5$)

Here is my next level for the continuation of this game. And I got some doubled bridges !!

Rules :

  • There will be lasers which are shaped like an arrow. The arrows pointing in the respective direction shows where the laser goes and the colour shows the colour which it gives out.
  • There will be boxes which are respectively coloured and these boxes need to get touched by the lasers in order to find a solution. In order to find the solution, you can make a move by rotating or moving the lasers or the mirrors, or moving the grey tiles or the coloured boxes 90° clockwise.
  • A mirror reflects a laser's path in exactly 90° angle clockwise or counter-clockwise, depending on the path. A mirror would only reflect the laser it is facing, or else it will block other laser paths.
  • Each box should receive the light of one single laser. In future levels a box may receive the light of 2 or more lasers.
  • Lasers(the arrows of the lasers) and Grey Tiles, along with the sides of the Mirrors would block other lasers' paths.
  • Bridges have a specific colour to allow lasers to cross through a box from a particular direction from the same colour, or else it would block lasers from coming through other directions as well as lasers with different colours. Bridges cannot be rotated but in future levels they may be rotated.
  • You can move objects (like lasers, mirrors, grey-tiles, bridges, etc.) such that they move as far as possible in the grid in a particular direction, until they reach the edge, or they collide with another piece. Brown objects cannot be moved.

What's New :-

  • From now on, we will have doubled Bridges. They perform the same rule exactly like a bridge does, but they allow to pass 2 lasers of 2 specific colours from 2 particular directions unlike a normal bridge. Examples are given here :-

  • Bridges, also be rotated. This includes both 1-sided and 2-sided bridges.

Here is the puzzle for today, can you solve it? (Level 6)

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    $\begingroup$ It seems like solving the puzzle should be pretty easy if you've done sliding puzzles before, given how much empty space there is - it's just writing out the solution that would actually be tedious. The final configuration is pretty obvious, and because of how much empty space there is, it wouldn't be too hard to move things into their proper places. And why the [optimization] tag? Are you asking people to find a minimal-move solution, and if so, is there any 'nice' way to show minimality? $\endgroup$
    – Deusovi
    Nov 12 '20 at 9:24
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    $\begingroup$ @Deusovi, if you think this is easy and you have found the solution, let me know. As for the tag, I had asked before to count the no. of moves (min.) it takes to go to the solution, but I think it's not needed now. (Removed tag) . $\endgroup$
    – Anonymous
    Nov 12 '20 at 9:35
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    $\begingroup$ The final placement of the pieces is easy (just place the bridges in the middle 3×3 and the blocks on the outside); finding the way to get to that arrangement is going to be painfully long but not in any way challenging. 3 blocks of extra space is more than enough to rearrange the pieces however you'd like. (I'd post an answer if I had the time for a full writeup -- if you're fine with an answer just giving the final configuration, then that would be easy.) $\endgroup$
    – Deusovi
    Nov 12 '20 at 10:09
  • $\begingroup$ How do you know, for example, that you will be able to place the bridges such that they allow passing the light of the lasers? Wouldn't that be challenging? Also I don't like this downvote here. $\endgroup$
    – Anonymous
    Nov 12 '20 at 10:34
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    $\begingroup$ No, it's not challenging? Every combination of horizontal and vertical colors is there once. So you can just place each of the bridges in the proper row and column - the "blue horizontal, red vertical" bridge goes in the top left of the middle 3×3, and so on. $\endgroup$
    – Deusovi
    Nov 12 '20 at 10:47
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Since each block can only receive one laser, all 6 of them (A, C, F, H, Q and S) should go to the outer ring. Specifically, R1,C2 to C4 and C5, R2 to R4. Given how bridges 1, 2, 3, 4, 5 and 6 are orientated, this can be solved WITHOUT rotations. Move directions: d=down r=right u=up l=left

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My solution: Bd Aru Gl Rdl Bur Ar Cu Rrd Grd 2dr Fu Ql 3d Fr Qu 2l 3u Gu 4l Hur Ru 5lu Sr Rdl 5r 3dr 2rd 5lul 4r Gl Bl Ad Gd 46 moves in total

Ends with:

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31 moves, because R doesn't need to move either:

Using Hakdo's notation, Gl 4l 5ul Cru Hru Qdr Gd 3d 2rd 5d 4d Bd Aur Fur Su Gl 3d 5l 4u 3r 2d 5u Gur
enter image description here

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