Solving Slitherlinks - Parity Method

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Slitherlinks are very addictive puzzles. But they will be even more addictive once you've learned the "parity" method of solving. This method is an enhancement of the usual methods to help you solve these much faster.


The rules are simple. In a grid of dots, draw a simple closed loop so that the number of line segments passing through each square matches the number in the square. Here's a fully solved puzzle.

(Click the image for official rules as described by Conceptis Puzzles.)


Each Slitherlink solution is a simple closed loop (also called a Jordan Curve - see below for more details).

Every simple closed loop has ONE inside and ONE outside. You can paint the inside one color, and the outside a different color.

Likewise, as you solve a Slitherlink, you can shade the squares two different colors as you learn whether a square is "inside" or "outside". This is the main idea behind the "parity" method.


Let's solve a puzzle! To get the most of this walkthrough, print yourself a copy and follow along. Or you can follow along in Excel.

Note - this is an actual puzzle published by Conceptis Puzzles.

First, begin by coloring a border around the entire puzzle. This represents the "outside". In this example, pink = outside; and yellow = inside. Note that this is an 8x8 puzzle, so we'll use "chess" coordinates to refer to specific cells. In practice, you don't need to include these coordinates.

Next, we'll apply some "starting techniques". As this website assumes you already know basic solving techniques, I'll refer you to Conceptis for a refresher course.

Go ahead and click that link, and note the following. "Adjacent 0 to 3" gives the lines going around the 3 in e4. "Two adjacent 3's" gives the horizontal lines around the 3's in h6 and h7. "Two diagonal 3's" happens three times in this puzzle, and provides the corner lines drawn in a8, b7, c8, f1, and g2. Finally, "any number in a corner" around 2 in h1 gives us the two lines drawn in g1 and h2.

In practice, applying these "starting techniques" usually takes no more than a minute.

This gives us a good starting place. Now we can start applying "parity". Remember this rule: wherever there's a line separating two squares, those two squares have to be opposite colors. Look at the 3 in a8. It has two lines drawn in the corner. The a8 square must be on the inside, so color it yellow.

Going around the pink border, continue coloring in yellow everwhere there's a line: c8, h2, g1, and f1.

And look at the 0 in a3. It can't have any lines around it. Thus all four squares around it have to be the same color as the 0 square. But look - it's touching the outside. We can color the following squares pink: a2, a3, a4, b3. (Note the 0 in f4. We know that it's the same color as the four squares around it, but we don't have enough information yet to color in those squares.)

Continue changing the color everytime you cross a line. This gives us pink squares at d8, e1, and g2.

Next, look at the 2 in a4. Every 2 has to have two lines touching it. Thus, regardless of the color of a 2 square, it MUST be surrounded by two pinks and two yellows. The 2 in a4 is already surrounded by two pinks: the outside and a3. The other two surrounding squares have to be yellow (a5 and b4). See how this is working?

Do the same for the 2 in a2, and you get two more yellow squares: a1 and b2.

Before we continue, let's update our lines. Remember, wherever two adjacent squares are opposite colors, draw a line between those two squares. From here on out, we'll update the lines at the end of each step.

Next, coloring in opposite colors across lines gives us g3 = yellow.

So far we've seen how to handle 0 and 2. Now, let's learn the rules for 1 and 3.

As each 1 can only have one line, only one square surrounding a 1 can be of the opposite color. Carrying this further, if any two squares surrounding a 1 are the same color, then the 1 square itself must be that color. This means the 1 in e8 has to be pink, as it's surrounded by the outside and by the pink square in d8.

3's are similar. As they have three lines around them, only one square surrounding a 3 can be of the same color. The other three surrounding squares have to be of the opposite color. The 3 in d1 is surrounded by two pink squares. So, the 3 itself has to be the opposite color: yellow.

Before we continue, look at b5. Can it be a pink square? No - as the resulting lines in between the squares would produce a cross, which would ruin the simple closed loop. Checkerboard patterns are never allowed. So, anywhere you have three adjacent squares filled in, and the fourth could possibly make a checkerboard pattern, you must use the other color instead to avoid the checkerboard pattern. Another way to look at it: since you're going around a corner, the color of b5 has to opposite the color of a4 - as it's on the other side of the corner. Either way you look at it, we can color in the following yellow squares: b5, b1, h1, h3.

Don't forget to draw the lines in between opposite colors.

Do you feel like this is a lot to learn? Don't let it bother you too much. Once you do this a few times, you'll quickly get the hang of it and you'll notice that your solving time improves. There's still more to go over.

This next principle is very important, and make sure you understand it well before continuing. Remember that the closed loop has an inside and an outside. Well, more importantly, it has ONLY ONE inside and ONLY ONE outside. In other words, we can't have any lonely pink cells or lonely yellow cells.

With this in mind, look at the pink cell g2 and the as-of-yet uncolored cell f2. Can f2 be yellow? No - as that would cut off the pink g2 from the rest of the pink cells. g2 needs to branch out and meet the other pink cells. So, f2 has to be pink. Got it? This will come up again.

While we're here, e2 then needs to be pink to avoid a checkerboard pattern.

Now look at the 1 in h5. It already has it's line above it. This means that h6 is the one and only opposite color square that surrounds it. The other three squares have to be the same color. But one of those squares is on the outside (pink). So g5, h5, h4 are pink and h6 is yellow.

Applying what we've already learned: the 3 in f8 is yellow because two of the surrounding squares are pink.

The 1 in c1 is surrounded by two yellows, so it also has to be yellow.

Going opposite from h6, we have h7 is pink, and h8 is yellow.

The 1 in e8 already has its opposite color(f8), so e7 has to be pink.

The 3 in d1 already has its same color(c1), so d2 has to be pink.

Before continuing on, I'll introduce a new notation. I don't like to use the x's that most tip websites suggest. As an alternative, I use intermittent lines to show squares that have the same color. This is just a personal preference. X's always work fine, but I think seeing the connecting lines is much more helpful.

All the squares around the 0 in f4 are the same color, so we'll connect all five squares together.

Whatever color the 3 in e4 is, the five squares on the other side of the three lines are of the opposite color, so we'll connect those 5 squares. Likewise, we can connect together the 5 squares around the 3 in b7.

We could probably solve this "easy" puzzle without having to see these connections, but this demonstrates a variety of solving techniques.

All of our hard work is starting to come together. Looking at the puzzle as a whole, you may notice some potentially alone yellow cells. For example, look at h8. That yellow cell needs to reach out to other yellow cells, so g8 is yellow. h6 also needs to branch out, so g6 is yellow.

Next, look at the set of yellow cells in the bottom left hand corner. If c2 is pink, it would cut off that whole set, so c2 has to be yellow. Likewise, c3 has to be yellow. Finally to avoid a checkerboard pattern, c4 has to be yellow.

Look at the cross we drew going through f4. All of those cells have to be the same color. If they're pink, it would cut off all the lower right hand yellow cells. So, all those cells have to be yellow. This forces the cells to the left of that to be pink. We're almost done!

As the 3 in f8 already has its like color(g8), f7 has to be pink. Then to branch out, g7 has to be yellow.

Also the 3 in c5 is surrounded by two yellows, so c5 has to be pink.

As we've colored in those cells in the area of d5 through f3, we can erase those markings we had in there. Other than that, we branch out the yellow cells to pass through the hallway in f6, e6, d6.

Our final connection-marking set has to be pink or yellow. If pink, it would cut off the right-side yellow from the left-side yellow. So all those cells have to be yellow.

We can finish the puzzle now. Opposites give us b7 is pink. Then the pink has to branch out to b8. But what about d7? The 2 in d6 gives us the answer. That 2 already has two surrounding yellow cells, so d7 is pink.

The puzzle is finished. Yeah!


As you solve a Slitherlink, you can combine the traditional methods along with this "parity" method. Use two different colors to keep track of inside vs outside. Between two opposite colored squares, draw a line - and if you draw a line next to a colored square, color the adjacent square the opposite color. Avoid checkerboard patterns. Use the numbers to tell you how many of the surrounding squares are of the opposite color.

0 - all surrounding squares are the same color.
1 - one surrounding square is the opposite color and the other three are the same color.
2 - two surrounding squares are one color and the other two are the opposite color, regardless of what color the 2-square is.
3 - one surrounding square is the same color while the other three are the opposite color.

Finally, remember that the two colors ultimately need to be connected to each other, so they will always branch out to meet each other.

Now, try this out on more puzzles. Conceptis provides four of these free each week. If you're already a seasoned Slitherlink solver, I hope this "parity" method will shorten your solving time. It's definitely helped me solve these much more quickly!


Jordan Curves are "simple" in that you can draw them with a pencil without lifting or crossing the path. They are also "closed" because they form a loop.

No matter how complex you make them, there will always be one inside and one outside. Below, to the left is a really complex Jordan Curve. To the right, you can see that even this complex curve still has only one inside and one outside

(Click the image to learn how to make a Jordan Curve out of any picture.)


Generally, Slitherlink puzzles come on square grids. But there's no reason to stick to this one format. KrazyDad, for one has ventured to use innovative tilings. In these different worlds, some of the traditional solving techniques don't work. But "parity" really comes in handy. You can always rely on shading. Click the picture below to check out these new tilings.



First type the numbers into a grid. If you then highlight the columns and double-click the column width, it will resize the cells into near squares.

Use the paint can (fill color) to shade cells pink/yellow (or your favorite two-color combination).

To draw lines, use "Draw Borders". To get there, click small arrow next to the border icon (usually near the paintcan). In the mini-menu choose "Draw Borders". This will bring up a palette. Click the first icon ("draw border") to activate drawing mode. Then wherever you click, it'll draw a line. To get out of drawing mode, just click the icon again. If you make a mistake, use the 2nd icon to erase the border.


I don't know who first came up with this "parity" method. I discovered it on my own, but anyone who took an elementary topology class (most math majors in college) would come up with the same conclusion. I'm just a little surprised at the lack of websites explaining this method.

Here is where I first tried to introduce the method. It was received like a pile of bricks! Hopefully you - the reader - will be more receptive. Try out the method, and if you like it, spread the word! Parity rules!