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Best explained here.

Basically there are two rules:

  • fill in each region with the numbers 1 to N, where N is the size of that region
  • two numbers, both K, cannot both be in the same row or column if they are less than or equal to K squares away. e.g the red squares cannot be 3.

demonstration of "x" rule

Puzzles

I was wrapped up in making ripple effects back in the day so here are some:

enter image description here

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Yellow squares are a number from 1-5.

enter image description here

Also, the puzzles are disjoint (e.g each grid is its own puzzle), so I guess you can post partials if you solve, say, 1 puzzle I guess.

There was going to be a meta-puzzle but I decided to scrap it. You might see some artefacts of the meta around.

Conundrums

  1. Show that for n>1, n 1xn pieces in a row is bad. n=4 shown below

enter image description here

  1. Show that on an infinite grid, if all regions are the same size then the ripple effect fails.
  2. What is the smallest number n such that there exists a ripple effect puzzle, which can be extended to infinity, such that all regions are size at most n?
  3. Two 1x2 pieces can be positioned to make a uniquely solvable ripple effect (shown below). How many 1x3 pieces do you need?

enter image description here

  1. (cont.) What about 1x4 pieces? Note: I have not solved this yet

Bonus: Generalise (all of) the above to k dimensions

I did make a bunch more small puzzles and conundrums (~5-10 or so) but my organisation is awful so they kind of... disappeared. Sorry about that. I might post them if they turn up.

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  • 1
    $\begingroup$ Could you please explain to me what a ripple puzzle is? $\endgroup$ – eedrah Oct 22 '17 at 5:45
  • $\begingroup$ @eedrah Oops. I really should think before I ask. Sorry about that $\endgroup$ – Wen1now Oct 22 '17 at 5:57
  • $\begingroup$ Correction: n 1xn pieces in a row is only bad for n greater than 1 $\endgroup$ – boboquack Oct 22 '17 at 20:36
  • $\begingroup$ Does the K not in Ks shadow rule extend over gaps "outside" a puzzle's grid (eg. between the sail and hull of the boat)? $\endgroup$ – Alconja Oct 22 '17 at 22:59
  • $\begingroup$ @Alconja Yes, it does. $\endgroup$ – Wen1now Oct 22 '17 at 23:01
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Partial answer - $\color{red}{\text{Warning - solutions are not spoilered}}$

Order of solve, decomposed into 3 levels:

Sorry to the colourblind

  1. Red, orange, light green, dark green, light blue, dark blue, purple, brown
  2. Thick lines, thin lines
  3. No dot, 1 dot

Solutions to all puzzles

Puzzle 1

#1

NB: the ? should be a 0 to make the word ENCRYPT using AZ26

Puzzle 2

#2

Puzzle 3

#3

Puzzle 4

#4

Puzzle 5

#5

Puzzle 6

#6

Puzzle 7

#7

NB: For puzzles 3, 4 and especially 5, there was quite a bit of casework, which has been omitted from the presented solution, but please feel free to ask me if you need help following a particular step (having done all previous steps).

Solutions to conundrum 1 and 3 and informal solution to 2

Conundrum 1

Label the boxes $1, 2, 3, \dots, n$ and the cells within box $i$ $(i,1), (i,2), (i,3), \dots, (i,n)$.

Observe that if the number $n$ is in $(i,j)$, the number $n$ can't be in $(i,j+1),(i,j+2),\dots,(i,n)$ or $(i+1,1),(i+1,2),\dots,(i+1,j)$ .

So no matter where the number $n$ is in box $1$, it cannot be in $(2,1)$. And no matter where the number $n$ is in box $2$ apart from $(2,1)$, it can't be in either $(3,1)$ or $(3,2)$.

We can extend this argument by induction to say that if the number $n$ is in cell $(i,j)$, $i\leq j$. By symmetry, $i\geq j$.

Therefore the number $n$ can only be in cells which have the form $(x,x)$, for some $1\leq x\leq n$.

Now, consider the number $n-1$. If it is in cell $(i,j)$, it can't be in $(i,j+1),(i,j+2),\dots,(i,n)$ or $(i+1,1),(i+1,2),\dots,(i+1,j-1)$ .

So no matter where $n-1$ is in box 1, since it can't be in $(1,1)$, it can't be in $(2,1)$ either. And no matter where $n-1$ is in box 2, since it can't be in $(2,1)$ or $(2,2)$, it can't be in $(3,1)$, $(3,2)$ or $(3,3)$.

We can extend this argument by induction to say that if the number $n$ is in cell $(i,j)$, $i<j$. By symmetry, $i>j$. But this is absurd, so a ripple effect puzzle with $n$ $1\times n$ pieces in a line is not solvable.

NB: 'By symmetry' applies because we can extend the argument identically, but from the other end of the line.

Conundrum 2 (informal solution)

Suppose the regions are all of size $n$.

The density of the number $n$ in an arbitrary region should be $\frac{1}{n}$, since each region has $1$ $n$ out of $n$ squares.

However, each $n$ eliminates $4n$ squares (with multiplicity) from being $n$, and each square can be eliminated up to $4$ times, so there should be about $n$ non-$n$ squares for every $n$ squares, leading to a density of $\frac{1}{n+1}$.

But $\frac{1}{n}\neq\frac{1}{n+1}$, contradiction.

Obviously, the notion of 'density' needs work on, but this is an intuitive outline to a potential solution.

Conundrum 3

Infinite ripple effect

Solve path:

  • Black - lone 1s
  • Red - 1s can't be anywhere else in the box
  • Blue - Forced infinite ripple - either nothing but 3 can go in the cell, or nothing but 2 can go in the cell
  • Green - Last cell in the box
  • Orange - 2s and 3s can't be anywhere else in the box

Minimality:

Suppose we had only regions of size 1 - then we would have only 1s and two 1s would be adjacent.

Suppose we had only regions of size 1 or 2 - then we would have only 1s and 2s. Then either we have all 1s, all 2s, both are which are bad as above, or a 1 and 2 next to each other. But then neither 1 nor 2 can be in the cell on the opposite side of the 1 to the 2.

So 3 is the smallest n that works.

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  • $\begingroup$ Nice work! That nearly led to disaster though... the question mark in the key was not supposed to be filled in. $\endgroup$ – Wen1now Oct 22 '17 at 23:16
  • $\begingroup$ Should I redo my workings without the extra 1 @Wen1now? $\endgroup$ – boboquack Oct 22 '17 at 23:17
  • $\begingroup$ @Actually, that part can be solved without the extra 1 in exactly the same way. It's up to you if you want to change the picture... $\endgroup$ – Wen1now Oct 22 '17 at 23:21
  • $\begingroup$ @Wen1now fixed. $\endgroup$ – boboquack Oct 22 '17 at 23:27
  • $\begingroup$ For conundrum 2 you can just look at a single row (or column). For each appearance of n, there is another n non-n squares. So there can be max 1/(1+n), which is less than the 1/n that we need. This is true for every row, and therefore the whole grid too. As I was writing this I just realized that this proof also works for all higher dimensions too. $\endgroup$ – Kruga Nov 1 '17 at 10:09

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