This is the puzzle we will be working on. It’s a Level 1 puzzle – small, and, with almost half the numbers already filled in, easy. There are no complex chains of logic necessary to solve it.

Being a 6x6 grid, we are looking to fill each row and column with the numbers 1 through to 6.

Although it’s an easy puzzle, there are no rows or columns with only one digit missing, so we’re going to have a look a little bit harder to find where to start.

This column is only missing a 3 and a 4. Neither row that crosses it contain a 3 or a 4, so we cannot use the rows to determine which empty cell contains which number. But we can use the fact that every brick must contain an odd and even number.

This brick contains a 6, so it cannot also contain a 4 because that breaks the rule. Therefore this empty cell must contain the 3, leaving the first cell in the column to take the 4.

This top row now only has two empty cells remaining (a consequence of filling in that 4). As such, it’s probably easy to complete. It’s missing a 3 and a 5. Checking the two columns that cross the empty cells, we can see that one of them has a 3 in it already, so must take the 5. That leaves the 3 to go in the top left.

The 3 in the top-left corner doesn’t help get us any further. The 5 we just filled in is more useful though. Now we have this column with only two empty cells, and we need to fill them with a 2 and a 4. There’s a 2 already in the crossing row at the bottom, so that tells us how to complete this column.

The knock-on effect of filling in the 2 in that column is that this row is now only missing a 3 and a 4. The 3 in the first crossing column tells us that the first cell cannot contain a 3 (because that would be two 3s in the same column, which is not allowed). So the 4 goes in the first cell and the 3 in the last one.

Here’s another row with only two missing numbers. We need to find homes for the 3 and the 6. That 3 in the top-right corner again tells us which goes where. Turns out filling in that 3 earlier on was quite useful after all.

Note that we could also have solved row by looking at the brick containing the 4. It could not take the 6 because that would put two even numbers in the same brick, which isn't allowed. As I said, there are lots of ways of reaching the solution!

Here’s a column missing a 3 and a 4. Can you see where they go? It’s pretty easy to work out, there are two different means of determining which number goes where.

Filling in that last 4 makes it possible to complete this column now. It needs a 2 and a 4, and since we already have a 4 in one of the rows, we know which number goes in which cell.

Nearly there now. This row needs a 2 and a 4. The first crossing column already contains a 4, so it will have to go in the last cell, leaving the first one to take the 2.

This row needs a 5 and 6. Again, the first crossing column has a 6 already, so that's all the information we need.

That just leaves these two columns, each of which are missing a single digit – it doesn’t get any easier than that!

And that’s it, we’re all done. I said it was an easy one! Did you get to the end before finishing this example? Ready to try something a little harder? Read on…

This is the puzzle we are going to solve. It’s a level 1 puzzle, so pretty easy. It doesn’t require complex chains of logic to reach the solution.

We’ll start in the top-left corner. This boundary has to form the top and left edges of a square, we just need to find the bottom and right ones. The smallest possible square we could draw, taking in a circle, would be 2x2. And indeed that’s also the largest we can draw, because if we tried to go up one size to 3x3, we’d be including two more circles. So we have to draw in a 2x2 square here, there’s no other option.

We can work outwards from this corner and put in a couple more 2x2 squares. These are, like the first, constrained in that they cannot be any smaller or they wouldn’t contain any circles, or any bigger because then they’d contain too many.

Working along the top, we can put three more 2x2 squares in. These are all constrained in the exact same way as the other squares we’ve drawn so far.

We can carry on going clockwise around the puzzle and work our way down. The next square has to be at least three cells tall to reach the next circle. We can’t go further than those three cells or we’d be including a second circle. So this has to be a 3x3 square. If we tried to make a smaller square around that circle, we would be orphaning the empty cells above it.

Continuing down this edge, the next square is constrained by the circle to its left. It has to be 2x2.

That shows that we have to put in a 3x3 square in the bottom corner as the only possible way to take in that bottom right-hand circle without grabbing any others and without leaving empty space.

That last square has given us half a square here – we just have to draw in the other half.

Drawing that in means we now have three boundaries for this area. There has to be a square here.

We have two more partially created squares here…

…and drawing those in suggests a big 4x4 square here. It has to be this big, because there’s no other way to include all that empty space in any other square along with a circle.

That just leaves these eight cells, which we only need to cut in half to make two squares…

…like this.

And that’s it, we’re all done! Every square has a circle, every circle is in a square. Puzzle solved. 

How did you get on? Did you race ahead and solve it before the end of the explanation? Are you ready to try some yourself? Read on for more.

This is the puzzle we are going to solve. It’s a level 1 puzzle, so it’s quick and easy to complete, without requiring lengthy chains of logical deduction that may be necessary in harder puzzles.

We’ll begin in the top right, because that’s the easiest of easy wins. The 1 in the circle only touches one cell, so we know we have to fill it. We can cross out the 1 in the circle, as we have met its target. We can also cross off the other 1 touching this cell, because we’ve met its target too. Two for the price of one!

I like to put a mark in cells that I know cannot be filled, because it makes it easier to see which ones to fill later. So I’ve put a dot in the cell touching that second 1 at the top right – we know it can’t be filled.

Working down the right hand side of the grid (as I’m going clockwise), the 2s are all easy fills because they each touch two cells. That also allows us to cross out the 1 on that edge. 

We can cross out the 0s because nothing touching those can be filled, and for the same reason we can add dots to the cells that touch them.

Continuing clockwise, along the bottom row we can fill in a couple of cells, which lets us strike off four 1s as complete, and also mark another couple of cells as having to remain empty.

Now working up the left-hand side, the two 2s are easy fills – we have to fill both cells touching them in each case. Filling those cells also lets us cross of a 2 one column further in, and it also lets us cross of a 1 on the edge.

The top two cells both touch 0s so can be marked as empty. For completeness, I’ve struck out the 0s to make it easier to see where we are up to.

There’s only one cell to fill in on the top row, but doing so lets us cross of four 1s. The other cells must all be blank, so we can put dots in them.

We’ve completed a circuit of the grid. Now we can keep working inwards, like peeling an onion.

In a bigger puzzle, it usually makes sense to work methodically inwards one cell at a time. This is a small puzzle though, so we can work our way around doing several easy fills at once. Take these 2s for example. They are easy to fill because we’ve eliminated the cells above them, meaning we have to fill those below…

…like this. That let us cross out four 2s and a 1, and eliminate a several cells as well. 

If we keep working clockwise, we can see that this 3 needs another cell to be filled, and there’s only one candidate remaining, so we have to fill that.

We could keep going down and fill out some more cells using logical deduction, but why use up the brainpower when there are still easy wins on the board? This 3 is another simple one – there’s only one cell we can fill in to complete it.

Completing that 3 let us strike off a couple of 1s, which allowed us to eliminate any other cells touching those 1s, which means it’s now very easy to complete these three circles.

That’s eliminated more cells, so we can complete this 2, which is also going to complete the 1s above it.

Only a few circles left to knock out now. This one is a no-brainer, it only has one valid cell touching it. Completing it will also complete the 2 to its left and the 3 below that.

That just leaves this 1 and 2. At first glance, it looks like there are two ways we could complete this. We could fill the cell to the bottom right of the 1, and the cell to the top left of the 2. This would meet the criteria for the circles, but it would cut the unfilled cell creek into three distinct parts…

…like this. I’ve added the red line to make it easier to see the path of the creek. The rules state it must be continuous, which is not the case. So the solution must be something else.

In fact we only need to fill one cell to complete both the 2 and the 1, and that leaves a continuous creek.

Here’s the creek highlighted with the red line again. All the empty cells can be reached by travelling horizontally and vertically, and all the number targets have been met, so the puzzle is complete. 

How did you get on? Did you get ahead of me and solve the puzzle before the end of the tutorial? Ready to try some more? Read on!

This is the puzzle we are going to solve. It’s an easy Level 1 grid that provides an overview of the basic techniques necessary.

We’ll begin by separating goats and wolves that are in adjacent cells. There are four such cases on this board, so we start by drawing in sections of fence to separate them.

Because we know that fences must originate and terminate at the borders of the field, and that they can only turn at a fence post, we can extend our fence sections to the borders or posts, whichever come first.

To work out where we must go next, we can start to eliminate some possible routes for fences. Consider the fence at the top left corner. From the fence post, it could turn ninety degrees and go to the top of the field, or it could turn ninety degrees the other way and go to the post three rows further down the field, or it could continue straight on and go to the right-hand edge of the field (as shown here). However, were it to go straight on like that, we would be creating an empty area (yellow) which is not allowed.

Therefore we know the fence cannot go straight on. That means we can strike through those borders to remind ourselves that no fence can run there. This will help us see options as we progress.

We can also strike through a potential route one row further down for the same reason – any fence to run along it would create an empty area in the yellow highlighted cell.

In fact, we can cross off quite a few possible fence routes. All these red strike marks show us where we can’t put a fence because we’d be creating empty areas. This helps us see where we must run fences.

Now, let’s look at the wolf highlighted in yellow. We need to separate it from the two goats to its right. There’s only one place we can possibly put a fence that will do that, so we can draw it in…

…like this.

This new fence gives us a new clue. Because it’s connected to a fence post that already has a fence attached to it, we know that no more fences can ever connect to that post (because fences cannot cross at posts). That means we can strike off the two remaining paths to that post.

Now consider the (highlighted) fence post in the bottom right corner. It’s only got one possible path it can take – it has to connect to the post to its left.

Having drawn it in, it’s easy to see that it must now go upwards. We’ve eliminated the path to the left, and if we were to go down, we would be creating a large empty area. Up to the top of the field is the only route possible…

…like this.

We’re making good progress. There are a couple of ways we could continue, but let’s look at the goat and wolf highlighted here. We need to separate them, and there’s only one fence we can put in which will achieve that.

We run the fence from the border up to the first post we encounter. Now we have a choice to make. Do we continue that fence up to the next post above, or do we turn left? See if you can work it out…

Did you do it? To see the answer, let’s consider what would happen if we ran the fence up towards the top of the field.

At first glance, it looks like it might work. But we’d still have to separate the two enemies highlighted in yellow, and there would be no way of doing that without creating an empty area – it’s just not possible.

So that means it’s not a valid option. We have to run that fence to the left instead.

Now we just have the top corner to worry about. There are two fence posts in the yellow box, and each one has two possible directions in which a fence can be run to reach the border or another fence post. You can think through all the permutations, but there’s only one way to complete the two fences that does not leave us with an empty area…

…and this is how it’s done.

That’s it, the puzzle is solved. Every area contains at least one animal, and there are no goats and wolves within the same area. No fences cross at posts. Simple! Ready to try one yourself? Read on…

This is the puzzle we are going to solve. It’s a Level 1 puzzle, an easy 6x6 grid, which will provide an overview of the basic techniques for approaching Yakazu.

We’ll begin with the easiest of easy wins. There are a couple of two-cell regions we can fill in without a second thought (both of them taking a 2). We’ve also got a three-cell region missing just the 1. That cell also happens to complete a row. 

We can fill in this three-cell region. The middle cell cannot contain the 3 because it’s in a column that already has a 3. So the 3 has to go in the last cell, leaving the middle cell for the 1.

Column three is now easy to fill in. The 1 cannot go in the bottom cell because that’s in a row that already has one, so we put it in the top. That leaves the bottom cell for the 2.

Let’s look at another technique. Consider the row highlighted in yellow; it is missing a 3, 4, and a 5. Ignoring any other way of working out what goes where (because we’re demonstrating a method!), we can focus on the 5. There’s only one place in the row it can go, and we can work it out by looking at the three intersecting columns, which happen to correspond to the three available empty cells. 

Looking from right to left: the orange column is four cells high, so cannot take a 5. The blue column is three cells high, so also cannot possibly have a 5 in it. That leaves the green column, which is six cells high, and therefore can take the 5. Indeed it’s the only cell in that row which can do so…

…and so we can write it in. Where next?

Well, the blue column should be easy enough to fill in, based on the two rows that cross it.

The bottom cell can only be a 3 (because it’s in a row that already has a 1 and a 2), and the 2 can only go in the middle cell, leaving the top cell as having to be the 1.

Now we can complete the third row down, because there’s only one free cell…

Putting a 4 in that last free cell means there’s only one place to put the 3 in the intersecting column, and that in turn leaves just the 4 for the top row. This can happen a lot in Yakazu – filling in one cell can lead to a chain-reaction, which is why it’s always worth checking for consequences when you complete a cell.

There are only three cells left to fill in, and they are simple to work out.

It doesn’t really matter whether you look at the six-cell row first or the six-cell column first, their intersections mean there’s only one way to complete them.

All done! Of course, that was the easiest level…the puzzles can get much tougher than this, requiring some deeper logical thinking. Ready to have a go yourself? Read on…

This is the grid we are going to work through. It’s a Level 1 puzzle which is easy enough to solve but should give you a grasp of the basics. There are lots of places to begin, but we’ll start with the easiest of all, which is probably the two rows labelled with the clue number of 2.

Both of those ‘2’ rows have horizontal sections of aquarium that are considerably larger than two cells, therefore they cannot be filled. We can mark them with Xs to exclude them.

By implication, we can also exclude the cell highlighted in yellow – there’s no way that could contain water if those below it are empty.

Now we know how to complete the top row of the puzzle…

…as there are only two cells remaining. That completes the row, so we strike out the clue number.

Of course, that cell in the top right corner can’t be filled on its own – the water cannot simply float there! We have to fill the rest of the aquarium beneath it....

...like so.

Moving on, we know we can fill all the other empty cells in the ‘8’ row.

That’s completed the row. And because we had to back-fill the aquarium at the far left, it’s also meant we’ve completed the ‘2’ row beneath it.

Where next? We have a few options, but let’s go and exclude some more cells.

 Excluding the two yellow areas for the two ‘3’ rows (because they are wider than three cells) lets us knock out a whole load of other cells by implication.

That’s going to let us complete some more rows…

…specifically, the two ‘4’ rows in the middle.

Now, let’s have a look at the ‘6’ row.

Although we don’t know all the cells that need to be populated to complete that row, we do know that the 2-cell region must be filled. There’s no way to complete that row without filling it. So we can fill it, which incidentally lets us strike off the end column as being complete. That means we can also put an X in the bottom-right cell.

That lets us complete the bottom ‘8’ row, which in turn means we complete the first column (and can strike out the two remaining cells in it). We can also strike out the ‘3’ column as complete.

Finishing the puzzle now is easy, as it’s just a case of filling in the last few cells to meet the column and row targets.

And that’s it – all done. Ready to have a go yourself? Read on!

This is the Snake puzzle we are going to solve. At 8x8, and with every row and column labelled, it’s quite easy.

There are no extreme values on this 8x8 grid (no rows or columns labelled with a 0 or an 8). 

We do however, have a ‘1’ row and a ‘1’ column. Because these are one-way crossings, it’s useful to highlight them. Once the snake’s body crosses each of these, there’s no turning back.

This allows us to immediately cross off three cells. The two to the right of the tail cannot be used, because it takes three cells to cross a ‘1’ column, and that would violate the 2 for that row. We can also eliminate the cell to the left of the head, for the same reason. That means the body extends from the head either up or down. Either way will allow us to fulfil the 2 clue for the column containing the head. Which way to go?

We have to go down. Why? Because the row above the head requires three cells to be filled in. Crossing the ‘1’ column would use our full complement. We would have to get back down to the bottom row to fill in the six cells that must be filled there, and that would violate the ‘2’ row, and the ‘3’ row on our way back. The only way to fill in the ‘6’ row is to go there directly from the head. 

Once on the bottom row, we know we fill three cells to cross the one-way ‘1’ column. In fact, we can also deduce that we must place all six required cells in that row because there’s no valid way to leave the row and return. In other words, we couldn’t go across three cells, up a couple, across again, and back down, because that would violate the rows above.

This one move has completed a row and two columns for us. Where do we move next?

We know we have to move up the ‘7’ column, because we’ve filled the ‘6’ row. Going up one cell completes the ‘2’ row (with the head), so we must go up another cell – there’s no choice.

Now we do have a choice though. Because we have a one-way barrier ahead of us (the highlighted ‘1’ row), we know we must complete the ‘3’ row before moving upwards. So do we go left or right?

The one-way barrier means that once we turn upwards, we will have to continue for at least another two cells. Going right would violate the rules in two ways. Firstly, we would be filling three cells in a ‘2’ column. Secondly, we would end up touching the snake’s tail (diagonally). As the snake cannot touch itself, this would be an invalid move. 

So we have to go left…

…then we have to cross our one-way barrier row by going up three cells, completing another column.

We can’t go up any more, so we have to go right…

…and then upwards, because we’ve completed the three cells allowed in this row, and also because we have to complete our ‘7’ column. We’ve knocked off quite a few rows and columns now. 

It’s a clear run to the end. We have to fill four cells in the top row, and there’s only one way to do that…

…which means we can drop down to the tail, completing the puzzle. Easy! Ready to have a go yourself? Read on!

This is the puzzle we will be working on. It’s an easy grid, but sufficient to demonstrate the techniques used to solve Pipelink puzzles.

We can begin by marking out routes we know the pipe cannot take. This will be helpful later when we come to start extending it. Because we cannot add to the given segments, we know our pipe can’t go anywhere we’ve marked with a red cross.

Next we can extend all the given segments into the middle of the adjoining cells. Doing so has connected some sections of the pipeline together – we’re making progress already!

We can continue extending the pipe. The segments in the highlighted cells can only be extended in a single direction…

…so we can extend those.
Where next? Let’s examine the grid further…

The two highlighted cells have only one possible direction to extend. We have to extend them both downwards.

But this leaves us with a problem. We’ve got two ‘islands’ that are isolated from the rest of the pipeline below. We’ll deal with the smaller one first. If we are to connect it to the rest of the pipeline, we can only do so from the highlighted cell. There’s simply no other option. Do we go down, or to the left? The answer, of course, is both! We cannot have a cell with three segments entering/exiting, it’s either two or four (because either the pipeline traverses the cell, or it crosses itself within the cell). So we can draw those extensions in now.

That’s connected the top right-hand island, but it’s given us a two new problems – the highlighted cells are invalid because they have three entries/exits. Of course, these aren’t really problems, they’re opportunities, because we know we have to add a fourth segment to both of them. They must be crossings.

We’re almost done. We just have one gap, and it’s an easy one to connect up…

…which completes the puzzle. It’s fairly easy to see that this solution is valid just by looking. With larger puzzles, it can be harder to be sure your solution is valid. To check a solution, it can be worth tracing the entire pipeline with a different colour to make sure you can cover off the whole thing, and every cell, without lifting the pencil or stylus.

Here is the grid we are going to work through. This is a level 2 puzzle, so though it’s easy to complete, it still presents enough of a challenge to give an overview of the techniques used to solve these puzzles.

We’ll begin by looking for all the easy wins. Remember there are three kinds of easy win. First up, doubles (two adjacent numbers the same).

There are six doubles. We know that the numbers either side of them must be the opposite of the numbers forming the doubles, otherwise we’d be trying to put three numbers in a row, which isn’t allowed. So between them, these doubles allow us to solve seven cells. We’ll put those numbers in now.

Filling those in gave us three new doubles. We’ve also got an almost completed column (another easy win). We can fill in the yellow cell straight away; we know we must have an even number of 0s and 1s in each row and column. That column already has four 0s, but only three 1s. Therefore the yellow cell must be a 1.

Now the top row is almost complete, so we can figure out what the one remaining (yellow) cell must be. We’ve got all the 1s, so it must be a 0. 

We can also work out the two remaining cells in the second column must both be 0s, for the same reason.

We’ve used up all the easy doubles for now, so let’s check on another kind of easy win - empty cells with the same number either side. There are two such opportunities available here, so we can fill out those cells.

That’s opened up two new opportunities: there’s a single remaining cell in column three (yellow), and a new double in column five (green). Even without the double, we could complete that column just based on the count of 0s and 1s.

The penultimate row has a new double (which is also a single cell surrounded by two numbers the same).
Row five has a full complement of 1s, so we know the last two cells must both be 0.

We’ve got one more easy win for now (the green highlight). There are a couple more cells we can solve here too.

First, the blue column. It’s already got three 0s, and as this is an 8x8 grid we know we need one more. Although there are three empty cells, there’s only one into which we can place the missing 0 – the bottom cell marked with the blue X. If we put it in either of the others, we’d be making a row of three.

Next, the yellow row. We know that one needs three more zeros. We can’t put all of them in the last three cells, so one of them must go in the cell with the purple X.

Now we’ve got a double and a single cell, although we don’t need them anyway because as there are four 0s already, both remaining cells in that column must be 1s.

Here’s another single we can fill in.

That gives us a row with one empty cell.

Here’s an easy double.

That gives us an almost complete row.

We’re out of doubles, and empty singles. In fact, we’re left with several pairs of cells and no obvious answers as to which way around to fill them.

But we do have a way to move forward. We can use several pieces of logic to solve the blue and green cells. We can combine our knowledge of what numbers we still need to place, along with the fact that no two columns can be the same. The column with the blue cells is shaping up to look a lot like column two. And the column with the green cells is looking like its neighbour to the left so far. These facts limit our placement options.

Let’s consider, for example, the end column. We must place three 0s somewhere. If we put them both in the top two free cells, that would make the two blue cells at the top of column six both 1s. But that would make it impossible to fill in the remaining free cells in rows two and three correctly:

This is wrong! Clearly that cannot be the solution...

In fact, if we work through the options, there’s only one way we can fill in the blue and green cells in such a way that the columns don’t duplicate others, and that allows us to complete the last two cells correctly. And that’s it – we’ve completed the puzzle.

This is the grid we are starting with. As a level 2 it’s fairly small, and the thermometers only run horizontally and vertically. We’ll start off by looking for the easiest fills. Being a 7x7 grid, those 6s look promising.

The final column is an easy win. Out of seven cells, six need to be filled. It’s easy enough to work out the one that cannot be filled – it’s in a row labelled 2, at the end of a thermometer four cells long. Obviously the end two cells of that thermometer can never be filled. To make life easier, we can cross them out. We know all the other cells in the column are filled. We can strike through the 6 at the top of the grid too, so we know we’ve done that column.

Let’s look at the first row, which is also a 6.

Because we’ve already got one cell filled in that row, we have to fill the whole of the other thermometer. We cannot possibly fill the cell in 1-column, because, well, that column can only contain one filled segment and that thermometer is four cells high! In fact, while we are at it, we will strike out the top three cells of that thermometer because we know they can never be filled. That row is done, so we can strike out its label, too.

Now we know how to complete the first column.

We can fill up the first five cells of the thermometer, reaching our target of 6. We strike out the bottom cell as it can’t be used, and strike out the label because that column is complete.

Now we’ve done the easy ones, let’s do some elimination. We’ll see if there’s anything we know we definitely cannot fill.

We can take out four more cells. Now, consider the row highlighted in yellow. We need to fill five cells, and have already filled two. There are four cells remaining. We know we’ll have to fill the bulbs of the two thermometers that start in that row, even if we don’t yet know where the fifth cell will be.

Putting those in completes the 1-column. Striking out the cell at the bottom of that column leaves us only one place to put our second cell in the 2-row (bottom row), so we can fill that in as well. These last placements have more knock-on effects: we can strike out cells in both the 3-columns because they are impossible to reach now.

Where next? Well, that second row (5) is looking promising.

We needed to fill three cells, and there were only three left. We’ve completed that row. We’ve also completed the 3-column, so we can cross out the last two cells in that one.

The cell we filled in the 4-column is useful, because it’s at the top of a thermometer. We must fill the cells below it.

That’s completed the 4-column, so we cross out the remaining cell. That leaves us only one place to fill in the last cell in the 5-row highlighted...

…and because that’s the top of a thermometer (even if it’s upside down), we have to fill the cells below it. That completes a whole bunch of columns and rows! In fact, we only have one cell left to fill in to complete the puzzle…

That last cell completes the remaining 2-row and 3-column. All done!