Ripple effects: A suite of puzzles and conundrums

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.

Puzzles

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

Yellow squares are a number from 1-5.

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

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?

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.

• Could you please explain to me what a ripple puzzle is? – eedrah Oct 22 '17 at 5:45
• @eedrah Oops. I really should think before I ask. Sorry about that – Wen1now Oct 22 '17 at 5:57
• Correction: n 1xn pieces in a row is only bad for n greater than 1 – boboquack Oct 22 '17 at 20:36
• 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)? – Alconja Oct 22 '17 at 22:59
• @Alconja Yes, it does. – Wen1now Oct 22 '17 at 23:01

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

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

Puzzle 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 $$(i,j)$$, $$i. By symmetry, $$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

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.

• Nice work! That nearly led to disaster though... the question mark in the key was not supposed to be filled in. – Wen1now Oct 22 '17 at 23:16
• Should I redo my workings without the extra 1 @Wen1now? – boboquack Oct 22 '17 at 23:17
• @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... – Wen1now Oct 22 '17 at 23:21
• @Wen1now fixed. – boboquack Oct 22 '17 at 23:27
• 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. – Kruga Nov 1 '17 at 10:09