# Cartological Cats: A Snaky-Loop Grid Puzzle with Cat-Shaped Matchmaking Clues

This is written for calica for a Secret Santa puzzle exchange. The original version in a printable PDF format can be found here.

Dear calica,

Thank you so much for being an amazing person this year! We are grateful to have you in the party and now we are thrilled to give you a small gift of appreciation! We heard you love a geometrical puzzle, d'oh of course! And cats!! Who doesn't love cats?! So here it is, a present for you that might be far from perfect, a present that was so fun to construct, a present that we wish sparks you joy too!

Merry Christmas and Happy New Year!

Cartological Cats

You are given several sets of four identical-shaped pieces each. You must place those pieces into the grid whose possible slots are already given. The pieces can be rotated but cannot be mirrored. It is your task to determine which piece goes where so the grid becomes a valid puzzle with a single solution.

The goal of the puzzle itself is to draw a single snake-like loop by shading some cells without touching itself. In other words:

• If a cell is shaded, then exactly two of its neighboring cells are shaded.
• All shaded cells must be connected.
• There is exactly one connected component of unshaded cells inside the loop.

Note that each piece has some numbers in it. The numbers are your clues. How the clues act strictly differs from a piece to another piece in the same set. It is also your task to determine which piece uses which rule. The four rules are:

1. The clues cannot be shaded and tell how many of its neighboring cells are shaded.
2. The clues cannot be shaded and tell how many of its corners are touching the loop.

Good luck and happy solving!

P.S. Oops, we think we forgot to mention that each number has been replaced by a letter. Different letters refer to different numbers. And yes, it is again your task to determine which letter represents which number.

The Pieces

The Grid

(also available here in Penpa+ with answer check)

• In this context, does "neighboring" include cells which touch only at corners? Jan 12 at 15:10
• @bobble No, neighboring cells are cells which are sharing a side. Jan 12 at 18:09

# Step by Step Solution

## Step 1

We can start by noting that any clue in a triangle cell of type either 1 "no. neighbors shaded" or 2 "no. corners touching loop" can be at most 3: there are only 3 sides and corners. Actually, a clue of type "no. neighbors shaded" in a triangle also cannot be a 3, since shading all 3 adjacent cells would result in having a small closed loop (since the loop may not touch itself).

Now note that all the brown regions contain an I inside a triangle. One of these will be a clue of type "no. neighbors shaded", which implies that I is at most 2. Similarly, all pink regions contain an L clue in a triangle, and all yellow regions contain a C in a triangle. We conclude that C, I and L have to be 0, 1 and 2 in some order, and A is at least 3.

Three of the orange triangles contain an A, which implies that the remaining triangle (containing the L) has to be a "no. neighbors shaded" clue. Also, one of the A clues has to be a "no. corners touching loop" clue, so A has to be 3.

## Step 2

Next consider the red regions. The top three contain an A/3-clue in a triangle, and therefore the bottom regions needs to be the region with the "no. neighbors shaded" rule. Furthermore, the middle two regions cannot have the "no. corners touching loop" clues, since in both cases this would result in a triangle being shaded with only one neighbor which might also be shaded. We conclude that the top cat contains the "no. corners touching loop" clues.

In this top red region, the square above the A clue therefore has to be shaded. This results in the top L clue having at least two corners touching the loop, and the I clue having at least 1 corner touching the loop. Using our previous observations, this implies L = 2 and I = 1. We also get C = 0.

So in conclusion: C = 0, I = 1, L = 2, A = 3.

## Step 3

We stay with the red regions for a while. Remember that the bottom red piece has "no. neighbors shaded" clues. The L=2 clue at the bottom will need two adjacent shaded cells, in particular when placed in the grid it may not touch the boundary.

Also we see that the right square in the piece needs to be shaded, with the top I=1 clue this means that the left square is not shaded. This again implies that also the bottom I clue may not touch the boundary of the grid. This leaves only one possible position for this piece: in the bottom right corner of the grid.

Together with the fact that the loop may not touch itself, this gives quite some progress:

## Step 4

Next focus on the top red piece. Recall that this has "no. corners touching loop" clues. The A=3 clue forces the square above it to be shaded. Since this shaded square should have two shaded neighbors, it may not touch the boundary of the grid. This is only possible if it is placed in the top right of the grid.

We can now also place the remaining red pieces. The third piece cannot go into the bottom left corner of the grid, we quickly get a contradiction if we try to shade either the odd clues or the even clues. We can put the second piece into the bottom left, but only if we shade the even clues. The third piece should go into the top red region, where we shade the odd clues.

## Step 5

Now we look at the yellow pieces. Note that none of the first three yellow pieces can contain the "no. neighbors shaded" clues: in the first piece the C=0 and A=3 clue would result in a shaded triangle between the I and A clues, in the second piece the L=2 and C=0 also result in a shaded triangle with only one neighbor which might be shaded, and the third piece immediately gives a contradiction because there is an A in a triangle. We conclude that the last piece has the "no. neighbors shaded" clues.

Next we check which piece can go in the bottom right yellow region. The fourth piece with the "no. neighbors shaded" clues would isolate a loop end. The third piece would need to contain "odd shaded" clues, but the I and A clues would result in a small loop being formed. Also the second piece cannot go here, since the C has a corner neighboring the loop, and the L and C cannot both be shaded or empty at the same time. We conclude that the first piece goes in this region.

The only way to interpret the clues in this area is by taking them to be "odd shaded" clues. This gives the following progress:

## Step 6

We continue with the yellow pieces. The third piece cannot have the "no. corners touching loop rule, because of the touching A=3 and C=0, so it has to have the "even shaded" rule. The second piece has the "no. corners touching loop rule".

Note that the third piece cannot go into one of the two leftmost yellow regions: the C clue would get closed in. Also the second piece cannot go in the top left corner, since the C=0 clue would block the I=1 clue from being satisfied here. This leaves only one way for placing the yellow pieces:

## Step 7

This is probably the most difficult step. We look at the pink pieces. Note that none of the pink pieces can contain the "no. corners touching loop" clues without going into the pink region in the center of the grid: for most of the cases we would either have a C=0 touching a shaded cell, or have a clue overlapping with a shaded cell; only for the top two pieces and the top left region some extra argument is needed. The top left piece would get into trouble with its I and C clues in this position, and for the top right piece we get a contradiction by looking at its two adjacent L clues.

We conclude that the piece with the "no. corners touching loop" clues goes into the central pink region.

Now consider the bottom left pink piece. This piece cannot contain the "no. corners touching loop" or "no. neighbors shaded" clues because of how the A=3 and C=0 clues interact. In particular, it does not go into the central pink region.

This piece furthermore cannot contain the "even shaded" clues while going in one of the remaining regions, since then the C=0 clue in the left ear of the cat would always get enclosed in some illegal manner by the A=3 clues around it. We conclude that this piece has the "odd shaded" clues (and therefore also cannot go in the right most region).

To continue with this step, we consider the bottom right pink piece. It cannot contain the "no. neighbors shaded" clues, because of the A=3 clues in the triangles. It also cannot contain the "even shaded" clues, because the center L=2 would get enclosed between its two neighboring I=1 clues. We conclude that this piece contains the "no. corners touching loop" clues, and goes into the central region.

## Step 8

Next consider the top right pink piece. This piece cannot go anywhere with the "no. neighbors shaded" clues: at the right the L clue would get shaded, at the bottom we would get a contradiction with the L=2 at the top of the body of the cat, and at the left the clues in the head of the cat would isolate an end of the loop. We conclude that this piece contains the "even shaded" clues.

By elimination, the top left pink piece has to contain the "no. neighbors shaded" clues. This piece cannot go into the left region because one of the L clues would get shaded. It also cannot go in the bottom region, because the C=0 and A=3 clue would together cause a contradiction here. We conclude it needs to go into the right region.

## Step 9

Suppose we put the bottom left pink piece in the left pink region in the grid. We would get the following situation:

Note that the red line crosses the loop trice. This is not allowed: it should cross the loop an even number of times since both its ends are outside of the loop. We conclude that the bottom left piece should go into the bottom region, and the top right piece goes into the left region. By again using the fact that the previously given red line should cross the loop an even number of times, we get the following progress:

## Step 10

Next we look at the blue pieces. All blue pieces contain an I=1 clue in the left ear, but only for the top region shading this clue does not quickly lead to a contradiction. We conclude that this region contains the "odd shaded" clues.

Note that the bottom region can only contain a piece with "even shaded" clues: with all pieces except the bottom left piece we would otherwise quickly get a contradiction with the C=0 clue near the bottom of the head. For the bottom left piece we need to do some more effort, but we also get the contradiction eventually.

The center region now can only contain the "no. neighbors shaded" clues (because of the I in the left ear), which leaves the "no. corners touching loop" clues for the left region.

The central region can only contain the top left piece, this follows by just considering the ears and the triangle between the ears.

## Step 11

We continue with the blue region on the left. Again by looking at the ears and the triangle between them, we conclude that the bottom left piece should go here.

In the bottom region, we cannot put the bottom right piece because of the C=0 in the tail. So we have to put the top right piece here, and put the top right piece in the top region.

## Step 12

We continue with the orange pieces (the triangles). Most of them are currently pretty well hidden, so if you can't find them I advise looking at the original empty grid again.

The top right triangle is shaded, which means it is an A=3 "odd shaded" clue. One of the triangles should have an L=2 "no. neighbors shaded" clue, this can only be the bottom triangle. We also still need an A=3 "no. corners touching loop" clue, which has to be the central triangle, leaving the A=3 "even shaded" clue for the top left triangle.

## Step 13

Only the brown pieces remain to be placed! Note that all these pieces have an I=1 in the right ear. Using this information, we can see that the "no. corners touching loop" piece can only go in the top position, and then it cannot have an L=2 clue.

The bottom right region also cannot have an L=2 clue, so the central region and the bottom left region should have these L clues. The bottom left region now can only contain "no. neighbors shaded" clues, leaving "even shaded" for the bottom right and "odd shaded for the central region. These last clues allow us to complete this beautiful puzzle!

• A huge task, not only to solve this but also to write the step-by-step solution. Jan 15 at 18:17
• Splendid work and very well explained! All steps are on-point --- steps 7 to 9 are indeed the trickiest part here. Great job! :D Jan 16 at 2:15