This is the puzzle we’re going to solve. It’s an easy level 1 puzzle – just enough of a challenge to show some solving techniques in action.

There are a couple of places we could start, but the easiest line to draw in is this one between the 12 and 9. No region can contain two numbers, so there has to be a division here. We don’t know where (or if) that line will extend yet, but it gives us something of a starting point. 

Let’s turn our attention to this 3. Because it’s on an edge, it’s more constrained than numbers in the middle of the board.

There are three ways we can draw a region of three squares that incorporate this number: we could go upwards (as shown by the orange area), or we could go sideways (blue) or downwards (green). To work out which way to go, we need to look at the neighbouring numbers that will affect the squares nearest the 3.

This 9 and 10 are of most interest. Why? Well, consider the possible scenarios: If we extended our 3 region into either the orange or blue squares, that would leave the green ones empty. Every square must be part of a numbered region when the puzzle is complete, so those green squares would have to join another region. The trouble is that neither the 9 region or the 10 region (the two closest) could be drawn in such a way as to include those green squares. 

Therefore, our 3 region must extend into the green squares, because that’s the only way to incorporate them into any region on the board.

We can use the same logic to work out how to complete our partially drawn region around the 9. Given the 3 to the left, the 12 above, and the 10 to the right, we either have to draw around the orange region or the blue one. If we draw around the blue one, that’s going to leave three orphaned squares between the 9 region and the 3 region – squares that cannot be incorporated into any other region.

Therefore, we have to draw around the orange region for the 9.

If we turn our attention to the two 10s, there’s only one way to draw a region around the bottom one because of the constraints imposed by the top 10 and the 20 higher up. We’ll draw that in next.

Now let’s look at the 5. It’s an odd number, and as it’s not divisible by three, it means we have to draw its region as a straight line. That limits the placement options considerably. We can either draw it vertically (the orange or blue regions), or horizontally (the yellow region, which could extend in either direction). 

The answer, once again, comes from working out how we would fill any remaining squares. If we went with the yellow region, or the blue one, we’d be leaving a single empty square above the 5. That square could not be used by the top 12 (because the board is only 9 squares wide), and neither could it be used by the 20 (because regions must be rectangular or square, they cannot be odd shapes that poke out to fill holes!) 

So our only option here is to incorporate that top square into the 5, which means we have to draw around the orange region.

These two 12s are now easy to work out. They have twenty-four squares between them, so we just have to divide them in two. There’s no other way to draw two regions of twelve in this area.

That leaves us with the 10 and the 20, and it should be pretty easy to work out how to divide those.

There, all done! As I said at the start, this was an easy puzzle. But don’t worry, they get a lot more complicated than that. Ready to try some yourself? Read on for some free puzzles to download, and details of where you can get more.

To make things a little easier to understand, this is what the same puzzle would look like if projected into three dimensions, as viewed from above.

And if you were standing at the bottom of the puzzle, it would look like this.

This is the puzzle we are going to solve. It’s a Level 1 puzzle. It’s small, and all the clue numbers are present, making it quite easy to solve. Being a 5x5 puzzle, we’ll be trying to put the numbers 1–5 in each row and column.

We’ve got some easy squares to fill in to start us off; those at the start of rows and columns labelled with a 1 must contain the number 5. It’s the only way to ensure no other building can be seen.

We’ve got another really easy win here in the bottom row. As it’s labelled with a 5 on the right, we have to fill the numbers from 1 to 5 in order, from right to left. It’s the only way to see all five buildings from that position.

Now let’s have a look at this column. We’ve got a clue number of 2 at the top, and we’ve already got a 5 in the bottom of the column. There’s nothing we can put ahead of that 5 to prevent it from being seen from the top. Therefore whatever goes at the top of the column has to be tall enough to prevent the remaining buildings from being seen. In other words, we have to put the 4 there. Anything else would mean we could see at least three buildings from that position.

Next we’ll have a look at the end column (yellow). At first glance, it might look like the missing numbers (2, 3 and 4) could be placed into the three empty squares in any order. However, on closer inspection we can see that’s not the case.

If we take the biggest number (because it’s likely to have the most constraints), the 4, we can see that it cannot go in square A, because that would mean whatever went in square B could not be seen from the bottom of the grid, therefore we wouldn’t reach our target of 4. It could go in square B. What about square C? It cannot go there because there’s already a 4 on that (green) row. So B is the only square into which we can put the 4 in that column.

Let’s try to finish this end column, because there are only two numbers left to place, which means there are only two possible ways of completing it. Either:

A = 2, B = 3

or

A = 3, B = 2

Either way would fulfil the requirements of the clue numbers at the top and bottom of the column. But what about the clue number of 3 at the top right? If we put our 3 in square B, it would become impossible to complete the top row correctly – it would block whatever went to the left of it (a 1 or a 2). So we have to put the 2 in square B, leaving the 3 to go in square A.

We can complete this top row, because there’s only one possible way to fill in the two missing numbers such that they fulfil the clue number of 3 on the right…

…like this.

Let’s look at this yellow column to see another way of working out where to place a number.

We know we need to put a 3 somewhere in the column, and it can’t go in the intersection with the green row because of the 3 already in that row. So we’re limited to two possible squares: A and B. If we put the 3 into square A, it would become impossible to complete that row in a way that respects the left-hand clue number of 2. So the 3 must go into square B.

Now that we’ve got four 3s on the board, it’s easy to work out where the fifth and final one goes. It has to go in the intersection of the only row without a 3 and the only column without a 3.

This is another easy sequence to figure out. We’re missing a 1, 2, 4 and 5. We need to be able to see four buildings from the left-hand side. The only possible way to fit those missing numbers in, in a way that meets the criteria, is to put them in order…

…like this.

Now these two columns are only missing one number each, so they are ‘no-brainers’ to complete.

There are a couple of ways we could work out the last few squares, but let’s keep it simple. We’ve got four 2s on the board, so the only place to put the fifth and final one is the intersection of the only row and column without a 2.

That leaves a single number to complete here…

…which in turn leaves these two rows each with a single square to fill in. Easy!

That’s it, all done. How did you get on? Did you race ahead and complete the puzzle before finishing this tutorial? Don’t worry – we’ve got some much harder ones to keep your brain busy! Read on to find out how to get more…

This is the puzzle we are going to solve. It’s a Level 1 puzzle – pretty small, with some pre-filled numbers – and does not require extended chains of logic to solve.

There are a couple of places we could start, given the pre-filled cells. Let’s begin near the top. This vertical sum needs to add up to 10, and we’ve already got the 2, so this is an easy win – we can enter the 8 to complete the sum.

Having filled in the 8, we now have another simple sum. We’re looking for 13 to complete this horizontal sum, so we can fill in a 5.

That in turn lets us complete this vertical sum. We need it to add up to 8, so we can put in a 3 and we’ll have finished that corner of the puzzle.

There’s another easy win down in this corner.

Now let’s have a look at this horizontal 20 sum. We’ve already got it half filled, so maybe there aren’t too many options left to complete it. A quick look at the Killer Sudoku calculator shows the following number combinations that could be used, bearing in mind we already have a 2 and a 5: 2+4+5+9 and 2+5+6+7.

Of those, we can eliminate the second. Why? Because of the vertically crossing 6 sum, highlighted here in green. Putting either 6 or 7 in the cell where the two sums intersect would not allow us to complete the 6 sum correctly. Therefore the numbers we need to complete the 20 sum must be a 4 and a 9. And by the same reasoning, we cannot put the 9 in that intersecting cell, so it has to be the 4. We have all the information we need to complete this horizontal 20 sum.

Now we know how to complete this vertical 6 sum.

This 20 sum is only missing two digits, so seems to be worthy of closer examination. If we check the Killer Sudoku calculator again (or use the Cheat Sheet), we discover there’s only one set of numbers that can complete the sum and incorporate the 9, 2 and 1 that are already in place: 1+2+3+5+9. So we’re missing a 3 and a 5…

…and the 5 in this crossing sum means that we cannot put the 5 in the intersecting cell (or we’d have two 5s in the same vertical sum, which is not allowed). So there’s only one way to complete the 20 sum correctly.

Now let’s have a look at this vertical 12 sum. There’s a reason I’ve skipped over to this side of the puzzle. A four-cell 12 sum is of interest because it contains two required digits. Which is to say however you complete it, it has to contain a 1 and a 2 because either it’s 1+2+3+6 or it’s 1+2+4+5.

We’ve already got a 1 in there, so we know one of the other cells has to contain a 2. Of the three empty cells, one of them can’t contain a 2 because there’s already one in the horizontal 16 sum that crosses it. So we can write small candidate 2s into the other two cells. These might help us later.

What options do we have for this horizontal 12? It could be any of the following:

3+9, or 4+8, or 5+7 (it cannot be 6+6 because digits cannot be repeated in a sum). Notice anything about these three sums? None of them contain a 2. Therefore that little 2 candidate number we filled in here cannot stay…

…which leaves just one possible place to put the 2 in this sum.

Now let’s turn our attention to this horizontal 16 sum. We’ve only got two cells to fill, so we’ll have a look at the trusty calculator again, to see what combinations of numbers include the 2 we’ve already got. Our options are 2+5+9 or 2+6+8. Can you work out which of this is correct?

These two 5s and the 9 mean there’s no way to fit a 5 and a 9 into our yellow cells together. So the answer must be that these cells are filled with a 6 and an 8. We don’t yet know which goes into which cell, so we’ll just write them in as small candidate numbers for now.

Those candidate numbers might help us with this vertical 23 sum. We can look for combinations of numbers that include the 5 and 9 we’ve already got, and then narrow those combinations down further to only those that also contain either a 6 or an 8, since we know we have to have one of those digits in this sum. These are our options: 1+5+8+9 or 3+5+6+9. That means our top cell in this sum contains either a 1 or a 3…

…and since there’s already a 3 in the crossing 16 sum, we can’t put another one here. So the top cell has to be a 1…

…which means we can now solve these two cells (since the one on the left has to be an 8, to complete the 23 sum).

There are a few places left to fill. We’ll look at this 27 sum next. Given we already have a 3, a 5 and a 6, the only combination of numbers that works for this sum is 3+4+5+6+9. So our two empty cells contain a 4 and a 9. to work out which goes into which, we can again turn to the sum that crosses this one.

If we put the 9 into the intersecting cell, the horizontal 16 sum would add up to 15 with one cell remaining. That would mean we’d have to put a 1 in the last cell of that sum, but that sum already has a 1, so we cannot do that. Therefore we must put the 4 into the intersecting cell, leaving the 9 to go in the top cell…

…like this.
That leaves just two cells to fill, and since each one is a single missing cell for a horizontal sum, they are easy fills.

There, that’s it - all done. How did you get on? Did you rush ahead and finish the puzzle before reading through this example? Are you ready to try some more? Read on!

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!