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    Yesterday I received one of the strangest cases of my career as an I.S. PI (Insightful Sleuths Private Investigator). I was reclining in my chair, working on the latest crossword puzzle when I received a call out of the blue. "You've got to help us, Investigator!" cried the voice on the other end of the line. "We're fresh out of leads and the culprits have nearly escaped!"
    Almost tearing my newspaper in excitement, I jumped out of my chair and reached out to my contact at the local police department. With their approval, I set off with a laptop and a spring in my step. When I reached the scene of the crime, I was briefed on the case by Deputy Dependable. The local candy shop had been robbed of its entire stock of confectionery (truffles, sugar cubes, macarons, cotton candy, and more). Although the careless criminals had left behind spades of evidence, there was just one problem: none of the police officers could understand what it meant! "There's something about this evidence that just doesn't square up," complained one of them. "This grid makes it look like they're plotting something."
     "We must stop them!" Dep. Dep. exclaimed, "If we follow the right one of these paths, we might be able to get to them in time."
     "Deputy, " I remarked, "this evidence doesn't square up, but it's absolutely enough to identify which way they went."


The criminals had left the following pieces of evidence behind:

1. A dead python, bent into an odd shape:

Sad snek

2. Incomplete math homework with too many unknowns:

$c_{h, r} = 65 +\ S\ \%\ (n+1)$

3. The following schematic:

Bad cop, not a torus

4. A curiously folded note, reading:

Going outside is useless. Better look for clues inside!

5. Chain weapons strewn about outside, smeared with cotton candy:

What a waste of good candy

6. Some sort of weird substitution cipher:

1996 48 1005 82
1996 142 1000 117
1996 286 502
1006, 997, 154, 1005
1998, 153, 498, 1996, 2011

7. A different kind of candy from the store on each possible route.

Which way did the sneaky sweet snatchers go, and why didn't the evidence square up?

Hint:

Two images specify puzzle types (both to be applied as sub-puzzles)

Hint #2:

To be or not to be, that is the question...

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(Thanks to Stiv and M Oehm in chat and the partial answer, I just came in the end)

First-Trichain

(thanks to Stiv) We see that there are three chains in clue 5, hinting towards a Trichain puzzle. Note that the cotton candy are the same colour as 6 of the numbers in clue 3 - they are our Trichain clues. Solving it, we get this grid:

Solved Trichain puzzle
(Please refer to Stiv's partial answer to see how to solve it)

Second-Slitherlink

(thanks to Stiv) Clue 1 seems to indicate the presence of a Slitherlink, showing a snake 'slithering and linking back to its tail'. If we take the whole schematic in clue 3, it is impossible to solve. But what if we use our Trichain? Using the solution from our Trichain puzzle, we try to make a Slitherlink. If we use the unshaded clues, they give us an impossible puzzle, but if we use the shaded clues, they give us a valid Slitherlink:

Slitherlink puzzle
(Please refer to Stiv's partial answer to see how to solve it)

Interpreting

Solved Slitherlink

With the solved Slitherlink, we can see it is a net of a cube with some gray lines inside.

Don't let that trick you, though!

Thanks to clue 4 and the 2nd hint, we realize that if we draw out spaces that are NOT used by the grey lines, we find this:

Solved Slitherlink with orange lines

That's 5=2||, which can be read as $S = 2^{11}$, as confirmed by the OP

Last Steps

With that in mind, we can now progress onto the next step. $S = 2^{11} = 2048$, which we can plug into clue 2. Using S, we find the new formula: $c_{h, r} = 65 + 2048 % (n + 1)$. This had been somewhat puzzling since it was unclear what $c_{h, r}$ was intended to be. However, the penny dropped when Stiv recalled that ASCII characters begin at 65. $c_{h, r}$ must be a letter code!

Using this revelation to our advantage,

we apply the formula from clue 2 to clue 6, which gives us the following numbers:

116 104 101 121
116 111 111 107
116 104 101
99  117 98  101
114 121 117 116 101

Finally,

we can convert these numbers to ASCII, giving us the solution:

They went the cube route

Final thoughts

this evidence doesn't square up: Of course it doesn't! A cube is a 3D shape after all, while a square is a 2D shape. (OP edit: The cube net is what the pun about the evidence not squaring up was all about)

What is the "cube route"?: They took the route of the sugar cubes!
(Additional note from OP: another pun here about taking the cube root :P)

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    $\begingroup$ Aha, well done to you and M Oehm for working through the last bits until the end :) Here's a +1 for your troubles... $\endgroup$
    – Stiv
    Oct 2 at 12:46
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    $\begingroup$ Congratulations! Those nasty nougat-nabbing ne'er-do-wells won't be getting away this time! $\endgroup$
    – Avi
    Oct 2 at 14:00
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Here's a partial answer explaining (in detail) what I believe to be the first few steps of the puzzle's solution. However, beyond that point I am as yet unsure how to proceed...


First off, consider the schematic in 3 and the image in 5 together. We can combine these...

...to solve a trichain puzzle. Note that there are three 'chain weapons' (suggesting 'tri-chain') and the colour of the cotton candy coating them matches the colour highlighting 6 particular cells of the grid - these are the trichain clues...

Step 1:

Black spaces can be shaded beside the diagonally adjacent 2 and 3. Their spaces can then be resolved.

Trichain Step 1

Step 2:

Since R5C4 must be unshaded, this fixes the bottom-left corner by follow-on deductions...

Trichain Steps 2 and 3

Step 3:

R1C3 cannot be shaded, which fixes the 3-space at top-middle, and the rest follows.

It then makes sense to consider clues 1 and 2 together with what we have found.

Clue 1, depicting a snake eating its own tail to form a loop is highly suggestive of a slitherlink puzzle. The whole schematic, as depicted originally in clue 3, does not form the basis of a valid slitherlink puzzle as it stands (for many reasons - most notably the 0-3-3 formation in Row 4 which cannot be satisfied). Instead, we need some way to modify this schematic to enable a valid slitherlink game - how about using the output of our trichain puzzle?

If we consider using all the spaces that were unshaded (left diagram below), we end up with invalid slitherlink clue combinations (e.g. the triplet of 3's in the bottom-right). Meanwhile, if we use all of the clues that were shaded (right diagram below), we produce an alternative option that might be worth investigating...

Shaded or unshaded spaces?

(I initially attached too much significance to the fact that this grid only contains 0, 1 and 2 clues (no 3's), meaning it might contain a coded message in ternary. However, the two 101 triplets (equivalent to '10' in decimal) would clue two J's or K's (depending on whether you translate via A0Z25 or A1Z26) so this didn't look promising.)

(Also, in the very first version of my solution I made the mistake of assuming we had to add 1 to all of these numbers to make it solvable, misinterpreting the intentions of the mathematical note in clue 2 - that was an error, now fixed below!)

Solving this grid-deduction puzzle then...

Step 1:

Mark all the immediately known X's, where the loop cannot pass.

Slitherlink Step 1a

Now consider the two diagonally adjacent 1's in R2C6 and R3C7. The loop must pass to the Top or Left of the lower-right of these two 1's in order to connect with the other. We can therefore add X's to its Right and Bottom. This forces the line to pass to the Right of the nearby 2 clue, else we would end up with a loop end trapped in the upper-right corner.

Slitherlink Step 1b

In fact, (and I only spotted this afterwards) we couldn't put the loop to the Left of this 1-clue either, else it would have ended up needing two sides shading! We actually end up with several immediate knock-on deductions that complete resolve the right half of the grid:

Slitherlink Step 1c

Step 2:

Consider now the 1 in R3C4 - can we pass it to its Left? No - or the loop can never fully form:

Slitherlink Step 2a

How about to its Right? No - for the same reason:

Slitherlink Step 2b

Well, what about to it's Bottom then?! No! Again, the loop gets broken:

Slitherlink Step 2c

In which case, it must be passed to its Top! Simultaneously, to ensure we end up with a loop, the segment to the Top of the 1 clue in R5C3 must be shaded:

Slitherlink Step 2d

Step 3:

Finally, there is only one way to resolve the bottom part of the grid, the 1 clue in R4C2 must be passed via the Bottom (resolving the centre-left), and the top-left loop path can be deduced as a result:

Slitherlink Step 3

What does this give us? Well, it looks a lot like:

the net of a cube! (This explains the pun in the question about the evidence 'squaring up'!)

Overlaying this on the original grid results in this:

Slitherlink result overlaid on original diagram

The question then is where to go next? It seems that clue 4 is the most promising option...

since the talk of 'looking inside' suggests we should look inside the slitherlink loop / cube net for clues.

However, I'm not yet seeing how to interpret this information. Still thinking about that...

Clues so far unused: 2, 6 and 7.

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  • $\begingroup$ If we sum the numbers on each face (excluding those already used), it seems to clue three numbers on the left side of clue 2 (and partial information on the right side). $\endgroup$
    – Bubbler
    Oct 1 at 15:05

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