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!