Spin the letter-squares!

Question

Spin the letter-squares!

A 5x3 grid of letters, e.g.,

a b c d e
f g h i j
k l m n o

can be rearranged by spinning either the 3x3 section centered at the second letter of the second row (in this case "g") either clockwise, thus,:

f a b d e
k g c i j
l m h n o

or counterclockwise; or spinning the 3x3 section centered at the fourth letter of the second row (in this case "i") either counterclockwise, thus:

a b d e j
f g c i o
k l h m n

or clockwise. So, if (starting again at the start) the "g" was first rotated clockwise then the "i" rotated counter-, the result would be:

f a d e j
k g b i o
l m c h n

Question 1.

What phrase did this jumbled text start out as?:

a f d s t
i o w ' .
n l w o r

Hint: I gave it

six

spins.

Question 2.

What about this one?

t h o r .
i s _ n t
e x e e i

Hint: I gave it

really I have no idea how many spins. But lots of

spins.

Bonus Question.

Can you be sure?

Answer 1

Question one can be solved with

six rotations of $R$-CWx2, $L$-CCW, $R$-CCW, $L$-CCWx2.

_{There's a GIF below in the detailed answer.}

Question two can be solved with

fourteen rotations of $R$-CCW, $L$-CCW, $R$-CWx3, $L$-CCWx2, $R$-CCWx2, $L$-CW, $R$-CW, $L$-CWx2, $R$-CW.

_{There's a GIF below in the detailed answer.}

---

Getting Started

First, let's assign a letter to each transformable sub-section of the grid; in this case I'll simply use $L$ and $R$ to represent the left and right sides respectively. Next, we begin by analyzing the letters available for plausible solutions. These are:

Question 1: Words won't fail.
Question 2: There is no exit.

_{Identified by @Morris' answer, and confirmed by the author in a comment.}

Solving Question One

Solving question one is simple in that it only requires us to identify the fact that

the period must come after the "L" in fail.

In order to accomplish this, we must

ensure the period follows the L clockwise around the second grid.

To ensure this, we can

move the period to row 3 column 5 with the sequence $R$-CWx2.

This also

moves wor into a position that allows correct placement in the next steps.

Our next move needs to

put the L next to the period, in a way that both squares are moveable in $R$. We accomplish this with $L$-CCW.

Next, we need to

put the wor back where it belongs, which will also put the L. combination in its correct place. This is done with $R$-CCW.

Finally, we

rotate wor into its correct position with $L$-CCWx2 which subsequently solves the entire grid:

So the entire sequence for solving the first question is

$R$-CWx2, $L$-CCW, $R$-CCW, $L$-CCWx2; for a total of six rotations.

You can confirm my answer using an interactive version I made here.

Solving Question Two

This question must be solved by

focusing on the center column and being as efficient as possible with rotations.

To start,

move o into the center square with $R$-CCW then move it up next to the period with $L$-CCW.

The next step took a minute to figure out because

I originally tried chasing the phrase in reverse around the edges (e.g. r, e, ...); however, this was fruitless and maddening.

Ultimately, the next step is to

move the re combo away from the center and move the x back towards the bottom. I focused on the i squares for this since rotating the right square will move the x while moving the left will move the re; since i and i are interchangeable, it didn't matter if they swapped positions, so I moved the i in the right square to the bottom with $R$-CWx3.

Next, we

rotate $L$-CCWx2 then $R$-CCWx2 in order to place the x and prepare for our final rotations.

Afterwards, we need to

put the x back next to the e in the bottom row with $L$-CW.

This

aligns the ex we need for exit, so we rotate $R$-CW to move it into the center column, effectively preparing it for final placement while aligning the r with $L$-CWx2.

Finally,

we can wrap it all up with $R$-CW.

Which gives us a complete sequence of

fourteen rotations; $R$-CCW, $L$-CCW, $R$-CWx3, $L$-CCWx2, $R$-CCWx2, $L$-CW, $R$-CW, $L$-CWx2, $R$-CW

Just-in-case I fudged any of that, here's a GIF:

Feel free to use an interactive version I made here to verify this answer yourself.

Original Idea for Question Two

I originally thought of solving this

in the same manner that I would solve a Rubik's cube; however, I now believe this wouldn't be possible due to the interactions in the center column. However, I did successfully identify one algorithm which results in swapping the top two squares in the final column along with row 2 column 1 and row 3 column 3, while preserving the rest of the grid. The sequence for this is $L$-CCWx2, $R$-CCWx2, $L$-CCW, $R$-CCW, $L$-CW, $R$-CWx3, $L$-CWx2 for a total of 12 turns.

This algorithm swaps the following squares in the grid:



For clarity, this algorithm will swap $i$ with $i$ and $e$ with $o$ in the screenshot above.

Bonus Question

"Can you be sure?"

No; while most of the anagrams that I came up with for the text didn't much sense, they are still possible, and as such they cannot be ruled out without rigorous testing. Until all of the possible algorithms for the grid are known, I don't think we'll ever be sure on this one.

Furthermore, I feel that it's important to note that

with respect to the bonus question posed by the author, we cannot yet rule out the idea that the underscore represents a wild card character of our choosing. This is something I pursued, but due to my lack of sleep, it was not a fruitful adventure, though it was fun.

Answer 2

First one is

"Words won't fail"

Second one looks like

"There is no exit"

but I'm not sure about that...

Answer 3

Here's how I went about solving Question 2.

As with a Rubik's cube, I sought "operators", "transforms", or "tools -- short routines that would change just a couple or few letters at a time while leaving everything else in its original position. Equipped with a few such tools, I should be able to swap around any letters as necessary to get back to start.

Speaking of start: I of course had an advantage, knowing what I'd started from; but I would have happily accepted any solution that brought the grid to anything halfway intelligible. Here are a few scramblings I considered:

 h e r e _
 i s e n t
 i t . x o

 h i r e e
 i s _ n o
 t e x t .

 t o h e x
 i s e n t
 i r e . _
 (Something a witch might say?) 

So, I made a new grid with letters A through O, and wrote a short program that rotated the squares in every possible combination to a depth of about six, and reported back on "interesting" sequences -- those that resulted in fewest changes from start. All the interesting sequences proved to be of the double-swap form; here are a few:

 a b h d e
 f g c i j
 k n m l o
 [L-CCW, R-CW, L-CW, R-CCW]

 a b d c e
 f g h i j
 k m l n o
 [L-CCW, R-CCW, L-CW, R-CW]

 d b h a e
 f g c i j
 k l m n o
 [L-CW, L-CW, R-CCW, L-CCW, R-CW, L-CCW]

 a b c d e
 f g m i j
 n l h k o
 [L-CCW, L-CCW, R-CW, L-CW, R-CCW, L-CW]

This still didn't seem to get me very far; any targeted swap would also inevitably swap two other untargeted letters. Unless ...

If one of the swaps of all my double-swaps was always the same pair, then the damage would be limited: that pair would just swap and swap back repeatedly. Meanwhile I could move other letters around at will.

So I set about devising tools that would always swap a and b. Once I had, for example, this tool:

 b a c d e
 f g n i j
 k l m h o
 [L-CW, L-CW, R-CW, L-CCW, R-CCW, L-CCW]

... I just had to 1) move any two other letters into h's original spot and n's original spot, while keeping a and b in (or returning them to) their spots, 2) run my a/b h/n double-swap, then 3) undo the turns from step 1 (reverse moves in reverse order). For example, to swap j/n instead of h/n was as simple as prepending $R$-CWx2 and appending $R$-CCWx2 to the h/n swap.

Eventually I had a toolbox full enough of a/b double-swaps to allow me to move letters around at will. From there (returning to the puzzle grid), solving was simply a matter of first rotating $R$-CWx2 to get the grid in the best possible starting position, then make three swaps: r/_, o/_, and e/_. This odd number of swaps left t/h swapped, so I also ran an invisible e/e swap. This got the job done in 82 spins. After cutting out redundancies (any $R$-CW immediately followed by a $R$-CCW for example), I had the following 66-spin sequence:

$R$-CCWx2, $L$-CWx2, $R$-CW, $L$-CCW, $R$-CCW, $L$-CCW, $R$-CWx2, $L$-CCW, $R$-CW, $L$-CW, $R$-CCWx3, $L$-CWx2, $R$-CW, $L$-CCW, $R$-CCW, $L$-CCW, $R$-CWx2, $L$-CCW, $R$-CW, $L$-CCWx2, $R$-CW, $L$-CW, $R$-CW, $L$-CWx2, $R$-CCWx4, $L$-CWx2, $R$-CW, $L$-CCW, $R$-CCW, $L$-CCW, $R$-CWx4, $L$-CCWx2, $R$-CCW, $L$-CCW, $R$-CCW, $L$-CWx3, $R$-CCWx2, $L$-CWx2, $R$-CW, $L$-CCW, $R$-CCW, $L$-CCW, $R$-CWx3

Whee!