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Theorem

Okay so this is not an actual portal puzzle, even though it technically is a portal puzzle. It's just something I thought of today and I thought maybe a selected few of you would like it. Maybe that is not the case, but that doesn't really matter.

Anyway, imagine the following. A long hallway divided by fizzler upon fizzler. There are weighted buttons on the floors in the spaces between the fizzlers, each deactivating the fizzler right in front of it (see thumbnail for examples). All the walls are portalable and the cubes do not auto-respawn. The player's objective is to connect the entrance area and the exit area with portals.
Imagine you are allowed to design this puzzle and you are allowed to hand the player N cubes (where N is a non-negative integer, of course). You may decide where the cube respawn buttons are placed, but no other items are allowed.
Q: How long can the level be? How many rooms can it contain?
A: Arbitrarily many, even with just one cube. Exercise: why is this? Hint: read on.

Okay, so that was boring. But now imagine that we are not allowed to place the cube respawn buttons in the first N rooms, counting from the entrance.
Q: Under this restriction, how long can the level be?
I think this is a really fun and accessible question for anyone to think about and try to see how far they can get. Again, this is an example of a question one might ask in mathematics - combinatorics, to be precise. And often these seemingly simple questions have quite intricate answers.
Hint: play the map and find the solutions. This should give you the right idea.

Spoilers below, so if you actually want to think about the question, don't read beyond this paragraph. I take this time as a warning for everyone that the map that accompanies this description is a bit dull and not especially fun to play. I would rate the rooms as difficulty 0.5/10, 1.5/10 and 3/10 respectively, which can be obtained by rounding (#rooms / pi) to the nearest half-integer.

Theorem: The answer on the second question is (N+1)^2+1.
Proof:
- We first observe that the player can only work from the entrance to the exit. We also observe that there is no benefit to the player to preplace cubes in rooms far ahead, because the player is never allowed to walk through a fizzler after getting a portal in the entrance room, so reaching a preplaced cube could also have been realised by bringing said cube along the whole journey. It is clear that leaving cubes behind is of no benefit either. By contruction the player cannot reach any respawn buttons without portals and with N cubes, starting from the first room. This means that it is most beneficial to the player to have all cubes with them at all times.
- Without respawning cubes, the player can get to room N+2 with all cubes. Namely, the player places cubes on the first N buttons and stands on button N+1 to be able to see room N+2. Then the player collects all cubes in order and returns to room 1, to enter room N+2 with all cubes. The player cannot get further than that, because one deactivated fizzler in the way would prevent the player from being able to collect all cubes, which can only be done in the direction of the entrance.
- So without respawn buttons, the player can have all cubes in room N+2. We observe that respawn buttons being further along will be beneficial for the player, because then the player will have access to them for as long as possible. Thus the player can place all N cubes to reach room 2N+2. Here the player can respawn a cube and by collecting and respawning, reobtain all N cubes in room 2N+2.
Exercise: Describe this respawning and collecting process in detail. Hint: play the map and find it out for yourself.
- The first respawn button could not have been further than room 2N+2. Also, without using another respawn button, the player cannot get all cubes past the 2N+2'th fizzler. Hence being in room 2N+2 with all cubes with one respawn button is optimal. One can perform this optimal respawning procedure a total of N times, when the player will be able to reach room (N+2)+NxN = N^2+N+2 with all cubes and no respawns left. Since the player had all cubes with them in a room as far as possible with a given number of respawn buttons used, this is the farthest the player can go with all cubes. Now the player can use the cubes to advance a final N rooms, ending in room N^2+2N+2 = (N+1)^2+1. Hence, this is the maximal length of the puzzle. QED.

Now I'm just waiting for someone to find a counter-example to my theorem and put me to shame. Truthfully, the above proof is not very rigorous, so there is actually a small chance that the above statement is simply false. Let's see if anyone actually responds to this whole endeavour... but I enjoyed thinking about this problem writing up this description so I can't really lose.
If you made it all the way here, thanks.

Of course I won't copystrike you for the music if you record your run :p
In other news: My puzzle game Sokoban: The RPG has been released on steam! Please consider checking it out!