# Puzzle 4: Smullyanic Dynasty - The Zotmeister

#### solving the puzzle of life one entry at a time

## Mar. 22nd, 2005

### 06:17 pm - *Puzzle 4: **Smullyanic Dynasty*

*Smullyanic Dynasty*

It started with the name. I have always admired Raymond M. Smullyan, and one day the thought flashed across my mind: there should be a high-quality logic-based pencil puzzle named "*Smullyanic Dynasty*". I sort of let the idea just sit and brew in my head, bouncing around a few ideas as to what such a puzzle would look like, what sort of rules it would employ. I knew it needed to have something to do with either binary logic on a grand scale or functional string manipulation. With nothing else better to concentrate on than the sidewalk when walking from a parking lot to an arcade recently, I gave it careful consideration, and I had the gist of it just before I entered the front door. What I had was obviously inspired by all the Nikoli puzzles I'd been solving recently as well as Smullyan's works; I had a grid of numbers and a binary-logic rule for determining which are to be shaded in. However, when I started experimenting with the puzzle, I found that it didn't really work. It was missing something; it needed another rule to make uniquely solvable puzzles.

I came to realize that my unintended inspiration provided the perfect answer. Nikoli has no fewer than three different puzzles that all have one ruleset in common (no, not the single, uncrossed loop); *Hitori*, *Heyawake*, and (the atrociously-named) *Where is Black Cells* have different mechanics but one unifying concept that they all adhere to. I tried applying it to my puzzle design, and it did the trick - it worked beautifully. It was purely an afterthought that I found 'Dynasty' is an ideal name for this concept, bringing the design process full circle.

The result is part *Hitori*, part Smullyan, part *Minesweeper*, and part something-all-its-own. It is my great pleasure to debut this original puzzle design. I would like to think Mr. Smullyan himself would be honored by this.

You guessed it - before and after. Puzzle on the left, unique solution on the right.

Each cell of the grid is either a "knight" or a "knave". The objective is to determine which for all cells. Knaves never share a side, and all knights are orthogonally contiguous - that's the "Dynasty". A number in a knight is the number of adjacent knaves, including diagonally; a number in a knave is

...Okay,

1) Each cell (unit square) is either a "knight" or a "knave". Mark them all to solve the puzzle. (Rather than colored pencils, I recommend simply shading in knaves and putting a circle in knights. For cells with numbers, either circle the number or shade lightly - you'll probably need that number to stay legible!)

2) The Knaves Stand Alone Rule: Knaves are never orthogonally adjacent - that is, they are never side-by-side. They can touch at a corner, but can never share a side in common. (This means that whenever you find a knave, you know all orthogonally adjacent cells are knights.)

3) The Dynasty Rule: All knights are orthogonally contiguous - just like the water squares in Islands in the Stream [see rule 6].

4) The Smullyanic Rules: Each cell is said to have a "domain", consisting of itself and all adjacent cells, including diagonally. When a cell has a number in it...

- 4a) Knights always tell the truth: ...and the cell is a knight, the number MUST correctly state how many knaves are in its domain.

- 4b) Knaves always lie: ...and the cell is a knave, the number MUST

So how can one even

(Start of sample puzzle solution)

I'll be referring to the cells by coordinates. I'll call the rows A through E, with A at the top; the columns 1 through 5, with 1 on the left. Therefore, the bottom-right-corner cell is E5, which also happens to be where I'll start. It contains a 2. If it were telling the truth, then the two knaves in its domain would have to be D5 and E4, since they can't be side-by-side. But this traps E5, violating the Dynasty Rule. Therefore, E5 is a knave. This makes D5 and E4 knights.

You may want to keep that in mind as a generic rule: a 2 in a corner must be a knave. Here's another: a 3 along a border must be a knave. I'll let you prove that for yourself; it's very similar to the 2-in-a-corner situation. At any rate, we can apply this to E3, which makes D3 and E2 knights.

D4 is next; it must be a knight, and I can give you three separate reasons why: E4 already has its two knaves marked, E4 has no other place for an adjacent knight to keep the Dynasty intact, and E5 would be a knave telling the truth otherwise. Any one of those reasons would be sufficient. With D4 marked, the only undetermined cells in D5's domain are C4 and C5, exactly one of which must be a knave. Well, C5 says 0, so if it's a knight, C4 would have to be a knight as well, which we know cannot be. Therefore, C5 is a knave, making C4 and B5 knights. Checking D4, C3 must be a knight as well.

Now it starts to show a bit of depth. B4 must be a knight, and the proof is up at A2! If B4 were a knave, the Dynasty and Knaves Stand Alone Rules would make A3, A4, A5, and B3 all knights; however, A3 would require A2 to be a knave, while B3 would require it to be a knight! Can't have that. With B4 marked as a knight, C4 makes B3 a knave, which in turn makes A3 and B2 knights. To finish the upper-right, we note from C4 or C5 that between A4 and A5, there is one knight and one knave. If A5 were the knave, A4 would have to be the knight, but it would have two adjacent knaves and its number is 1, so they're the other way around. C3 then gives us A2 as a knight. ...Phew!

The next step bears a striking resemblance to

(End of sample puzzle solution)

Neither of those two rules will help you with the puzzle below. You should be able to find another rule or two to give you a place to start, however. This puzzle is actually rather straightforward, although I've put a little bit of trickiness into its endgame. For this first challenge, I decided to simply be kind and put a number into every cell; don't expect this to always be the case in the future. I hope you enjoy solving this; as always, tell me what you think on the comment page, and email me the solution if you solve it. - ZM

You guessed it - before and after. Puzzle on the left, unique solution on the right.

Each cell of the grid is either a "knight" or a "knave". The objective is to determine which for all cells. Knaves never share a side, and all knights are orthogonally contiguous - that's the "Dynasty". A number in a knight is the number of adjacent knaves, including diagonally; a number in a knave is

*not*the number of adjacent knaves, including diagonally and itself - that's the "Smullyanic"....Okay,

*fine*, be that way - I'll give you a numbered list:1) Each cell (unit square) is either a "knight" or a "knave". Mark them all to solve the puzzle. (Rather than colored pencils, I recommend simply shading in knaves and putting a circle in knights. For cells with numbers, either circle the number or shade lightly - you'll probably need that number to stay legible!)

2) The Knaves Stand Alone Rule: Knaves are never orthogonally adjacent - that is, they are never side-by-side. They can touch at a corner, but can never share a side in common. (This means that whenever you find a knave, you know all orthogonally adjacent cells are knights.)

3) The Dynasty Rule: All knights are orthogonally contiguous - just like the water squares in Islands in the Stream [see rule 6].

4) The Smullyanic Rules: Each cell is said to have a "domain", consisting of itself and all adjacent cells, including diagonally. When a cell has a number in it...

- 4a) Knights always tell the truth: ...and the cell is a knight, the number MUST correctly state how many knaves are in its domain.

- 4b) Knaves always lie: ...and the cell is a knave, the number MUST

**incorrectly**state how many knaves are in its domain. (Remember that this includes the knave itself.)So how can one even

*start*a puzzle where none of the given information can apparently be trusted? Well, there are a few ways. I'll give you two of them in my walkthrough of the sample problem:(Start of sample puzzle solution)

I'll be referring to the cells by coordinates. I'll call the rows A through E, with A at the top; the columns 1 through 5, with 1 on the left. Therefore, the bottom-right-corner cell is E5, which also happens to be where I'll start. It contains a 2. If it were telling the truth, then the two knaves in its domain would have to be D5 and E4, since they can't be side-by-side. But this traps E5, violating the Dynasty Rule. Therefore, E5 is a knave. This makes D5 and E4 knights.

You may want to keep that in mind as a generic rule: a 2 in a corner must be a knave. Here's another: a 3 along a border must be a knave. I'll let you prove that for yourself; it's very similar to the 2-in-a-corner situation. At any rate, we can apply this to E3, which makes D3 and E2 knights.

D4 is next; it must be a knight, and I can give you three separate reasons why: E4 already has its two knaves marked, E4 has no other place for an adjacent knight to keep the Dynasty intact, and E5 would be a knave telling the truth otherwise. Any one of those reasons would be sufficient. With D4 marked, the only undetermined cells in D5's domain are C4 and C5, exactly one of which must be a knave. Well, C5 says 0, so if it's a knight, C4 would have to be a knight as well, which we know cannot be. Therefore, C5 is a knave, making C4 and B5 knights. Checking D4, C3 must be a knight as well.

Now it starts to show a bit of depth. B4 must be a knight, and the proof is up at A2! If B4 were a knave, the Dynasty and Knaves Stand Alone Rules would make A3, A4, A5, and B3 all knights; however, A3 would require A2 to be a knave, while B3 would require it to be a knight! Can't have that. With B4 marked as a knight, C4 makes B3 a knave, which in turn makes A3 and B2 knights. To finish the upper-right, we note from C4 or C5 that between A4 and A5, there is one knight and one knave. If A5 were the knave, A4 would have to be the knight, but it would have two adjacent knaves and its number is 1, so they're the other way around. C3 then gives us A2 as a knight. ...Phew!

The next step bears a striking resemblance to

*Minesweeper*! A2 tells us that there's one knave between A1 and B1; B2 tells us there's one knave between A1, B1, C1, and C2 - but since we know the knave must be one of the first two, C1 and C2 must both be knights! This is a windfall, making finishing easy: C3 forces D2 to be a knave, which in turn forces D1 and E1 to be knights (the former by adjacency, the latter to ensure D2's 3 is a lie or for two other reasons); C2 then marks B1 as a knave, which makes A1 a knight.(End of sample puzzle solution)

Neither of those two rules will help you with the puzzle below. You should be able to find another rule or two to give you a place to start, however. This puzzle is actually rather straightforward, although I've put a little bit of trickiness into its endgame. For this first challenge, I decided to simply be kind and put a number into every cell; don't expect this to always be the case in the future. I hope you enjoy solving this; as always, tell me what you think on the comment page, and email me the solution if you solve it. - ZM