This is the puzzle we are going to solve. As mentioned, it’s an easy level 1 puzzle, just enough to get the idea, not to stretch our brainpower to the limit!

As we have a couple of zeros here, we can immediately eliminate the first column. It cannot contain any digits, so we can strike through all the cells in that column.

Let’s move onto the second column. The top number tells us we must place a single digit in this column, and the bottom number tells us that the contents of the column must add up to 7. In other words, we know we have to place a single 7 somewhere in the column. Therefore, we can eliminate any cells where putting in the 7 would bust a row. As we can see, none of the bottom three cells can contain the seven, as they belong to rows that add up to 2, 6, and 5. We can put strike marks in to eliminate them.
The top three cells could all potentially hold the 7, so we’ll have to come back to this column later.

We’ll finish looking at the ‘1’ columns, as they give us more easy eliminations. Here we know we have to place a 4, so we can eliminate the cell in the row that sums to 2…

…and in this one we must place a 6, so we can eliminate the cells in the intersecting 2 and 5 rows.

Now we can turn our attention to the rows. We must place a single 8 in the first row. We’ve already eliminated the first cell, and in fact we can eliminate the next four as to put an 8 in any of them would bust the totals for the intersecting columns. That just leaves the last cell, so we must put the 8 into that.

Let’s take a look at the end column now. The top number tells us we need to place two digits, and the bottom number says they must add up to 15. We’ve placed the 8, so we know the second digit must be a 7 (because 8+7=15). As there’s only one cell that can take a 7, that’s where we must put it. We can eliminate the remaining cells in that column, which will help complete the other digits.

The 7 we just placed means that we have completed the row in the green box, so we can eliminate all the remaining cells in that.
As we only have one free cell in the row in the blue box, we can place the 2 that is required in that row.

The knock-on effect of placing the 2 is that we now know we must place a 5 in this column. That’s because the column requires two digits that add up to a total of 7. There’s only one valid cell to put the 5, so we place it and eliminate the remaining cells.

For completeness, we can eliminate the last cell in this row as we’ve complete it.

Let’s take a look at this row. We can solve it using simple arithmetic. We need two digits, they must add up to 11. From the three available, we must use the 7 and the 4. Placing these digits completes a couple of columns, too.

Only one digit left to place, and it’s easy because it’s the only one left! That’s it, puzzle solved.

This is the puzzle we are going to solve. We’ve deliberately chosen a simple one to keep this tutorial manageable. The same concepts apply whatever the size of the grid, though.

Let’s begin by looking for the easiest patterns. The first of those is the AxCC. There are three of those in this puzzle, all in the vertical plane. We know that the empty cell in this pattern must be the A symbol. If we put a C in there, we’d have three Cs in a row, which isn’t allowed. And if we put a B in there, we would have A, B and C in a row, which also isn’t allowed. So from the left, the first empty cell must be a square, the second also a square, and the third one, on the right edge, must be a circle.

Filling in those three symbols has created a a new AxCC pattern (in yellow), so we know the empty cell must be a square. It’s also given us an ABxB pattern (green). This must solve to A, for exactly the same reason as before. So the empty box in the green area must be a circle.

We’ve exhausted the simple gap patterns for now. Let’s look for intersections. The easiest to spot are the doubles (it’s quick to find two symbols the same on the grid). Here’s one example, right at the top. We have an AAx intersecting with a BAx. We know this must solve to B, because if we tried to put an A in the intersecting cell we’d have AAA which isn’t allowed, and if we tried to put a C in there, we’d have BAC which is also not allowed. So the empty cell must contain a circle.

Here’s another, almost the same. This time it’s ABx / AAx, but it solves to the same solution for the same reason. We have to put a circle in the intersecting cell.

Here’s a different kind of intersection. This is an AxC / ABx. From our table above, we know it resolves to A. That’s because if we tried to put a B in the intersecting cell, we would be creating an ABC run in the green area, and if we put a C in there, we’d be creating an ABC run in the yellow one. A is the only option. The empty cell contains a square.

This is an easy puzzle, and there are lots of intersections and patterns appearing all over the place. Let’s look at this AxCC for no other reason than because it’s a diagonal, and we haven’t done any of those yet! It’s important to remember to check the diagonals, particularly when puzzles get harder, as they add two extra dimensions and can often be the key to unlocking the grid.
Being an AxCC pattern, we know the empty cell solves to A, so it’s a circle.

We can do intersections on diagonals as well. Here’s another AAx / BAx, with the AAx on the diagonal. We know it solves to B, so the empty cell at the intersection must be a square.

Here’s another diagonal, an AxCC. It must be a square.

And here’s an ABxB. As we fill out the cells, the simple patterns keep popping up again, so it’s worth keeping an eye out for them. This must be a circle.

Let’s speed up a bit. We’ve got two patterns here. The blue is another ABxB, so that’s a circle. The yellow and green intersection is an AAx / AxC, so that has to solves to C - a circle in this case.

Nearly there. Here are two more. The yellow and green intersection is an AAx / ABx, so must solve to B. It’s a square.
The blue intersection is an AxA / ABx. It must solve to B, so it’s a circle.

A quick AxCC here. It must be a circle.

Just two more to go. This is a new intersection: ABx / ACx. This must solve to A (square). If we put a B in here, we’d be creating an ACB run on the diagonal. if we a C in there, it would be an ABC on the vertical. So, it’s a square.

That just leaves us with a simple ABxB pattern along the bottom. It must be another circle.

That’s it, we’ve completed the puzzle. And we didn’t even have to draw any triangles! I said it was an easy one. If you want to try some puzzles like this yourself, read on – there are some freebies to download below…

This is the puzzle we are going to work through. We’re using a very simple puzzle here as anything larger would make this walk-through way too long. Being an easy puzzle, we are provided with clue numbers in all the cells. Harder levels have fewer clues.

Before we get started, a quick word on notation. As we’ll be eliminating possible arrow directions, we need a consistent way of doing so. Any given cell has three possible arrow directions: up / down / horizontal for the sides, and left / right / vertical for the top and bottom. Here we have the three possible side arrows shown. As we eliminate possible placements, we can place an X in the relevant position to rule it out. 

The very first thing we can do on any puzzle is eliminate ‘illegal’ arrow positions in the corners. The rules state that arrows must point towards at least one number. That means that the arrows shown here are not acceptable as they don’t point at any numbers….

…so we can put Xs in all those positions to rule them out. It’s not strictly necessary, but it makes things easier later on.

Now let’s look at the 0. No arrow can point to it, so we can add in more Xs as we eliminate all the positions that could point to it. That means no vertical arrows in the column the 0 is in. I’ve eliminated those with the purple Xs.

It also means no horizontal arrows for the row the 0 is in, so we eliminate those with Xs too.

And of course we mustn’t forget the diagonals. That gives us four more arrow positions to eliminate.

With all those eliminations, we’ve got some places to put some arrows on the board. We’ll start at the top. With the vertical arrow and diagonal right arrows not an option, we know we must place an arrow going diagonal left, pointing at the cells highlighted here.

At this point we are going to add another kind of note to the board. Each of the three highlighted cells now has an arrow pointing at them, so we’ll mark each one with a dot. This will help us keep track of which cells are ‘complete’, which will be essential later on.

Working around clockwise, we can place a horizontal arrow pointing to this row. Again, we add a dot to each of the cells it’s pointing to. 

Still going clockwise, we have another placement here. And again, we add dots to all the numbers the arrow is pointing to.

And lastly for now, we can put in this vertical arrow.

So where next? We’ve used up all of the known placements, so how can we eliminate any more? We’re going to have to find some constrained numbers – those with limited placement opportunities that can help us out.

We know that numbers on the main diagonals are already more constrained than the other numbers on the board, as they can only have a maximum of six arrows pointing at them, not eight. The largest number on a main diagonal is this 5. It’s already got two arrows pointing to it, so it needs three more. There are four more potentials, but we don’t know which three are the ones we need to use.

Note: Although we don’t know which of those four arrows will be used, we do know that at least one of the ones on the diagonal will have to be used. Had this 5 been deeper in the board and not in a corner, that information could have allowed us to add a dot to all the other numbers on the same diagonal, which may have helped us move forward. In this puzzle, with the 5 being in the corner, it’s no help, but this is an essential technique when working on harder level puzzles.

The next largest number on a main diagonal is this 4. It’s already got one arrow pointing to it. Out of the remaining five possibles, two have been eliminated (highlighted here in yellow). We need three arrows, and lo and behold there are only three places to put them!

We’ll draw them in one at a time, taking care to add dots to our number tallies as we go. First the top one…

…then this bottom one. That’s ‘completed’ two of the numbers in this column. The dot tallies let us see at a glance that the 2 and the 3 both have the requisite number of arrows pointing at them. We’ll draw circles around those numbers to remind us we can’t point any more arrows at them. That will allow us to eliminate more potential arrow positions in a moment, but first…

…we’ll just draw in that last arrow for the 4. Adding the dots confirms the 4 is now complete, so we can circle it.

Now we can use these three newly circled cells to eliminate more arrow positions. The 2 gives us three new eliminations (purple Xs). I haven’t drawn in the one to the left of the row as that arrow is already placed, so we don’t need it.

The completed 3 also eliminates three arrow positions (purple Xs).

We know the 4 doesn’t eliminate anything because we already used up all the remaining positions.

We’re really motoring now. We’ve got lots more arrows we can place. Going around the board clockwise again, we’ll start with this diagonal. That completes the 1 in the corner, which in turn allows us to eliminate an arrow position on the diagonal (purple X).

This next new arrow puts three more dots on the board…

…and this one completes two numbers, which in turn give us two new eliminations.

We can place all three arrows up in this corner. They complete several new numbers, which in turn eliminate further positions (again, the purple Xs).

We know where to place the final three arrows. I’ve added the dot tallies to the numbers, but there’s no need to circle them, we’re all done!

This is the puzzle we are going to solve. It’s a Level 1 puzzle; 7x7, easy to solve but sufficient to demonstrate the common techniques we use.

We’ll begin by looking for the easy patterns. There aren’t any ‘3 in a row’ patterns here, and neither are there any “2+1” patterns. There are quite a few “XYX”s though, which are shown here in yellow for rows and green for columns. We know that in each of these cases, we can circle the middle digit – it must be kept, because we know we’ll have to eliminate (shade) at least one digit immediately adjacent to it. We don’t know which will be shaded (and indeed it could be both), but it doesn’t matter, at least one will be and as we cannot have two shaded cells next to each other, the middle cell can’t be shaded so must be circled.

Now we can look along each row with circled digits, and see if there are duplicates of those digits. If there are, they can be eliminated. In row 2 we are looking for any other 5s and 2s. There is a 5, so we will be able to eliminate it.
Row 6 also has a duplicate 5 that we can take out.

We’ll do the same for the columns. In column 3, there’s a duplicate 5, so that can be shaded in.

As we know that we can never have two shaded squares orthogonally adjacent, we can circle the digits immediately above, below, to the left, and to the right of those we have shaded in. They must be kept. We’ve circled these here in orange.

With eleven new circles on the board, we can go back through the rows and columns looking for duplicates we will be able to remove. For example, in column 2 we’ve a newly circled 6, so we can eliminate the duplicate 6 right at the top. Between all the rows and columns, that lets us eliminate ten new cells (shown here in yellow).

Now we repeat our earlier check, circling the cells above, below, and to the left and right of the newly eliminated ones. That gives us lots of new circles (shown here in orange)…

…and of course, we can use those to further eliminate doubles in the rows and columns. In fact, despite all those nice new orange circles, we can only eliminate one new cell – the 4 right in the middle.

That 4 lets us circle the 4 immediately to its right.
We’ve reached a dead-end with the rippling of these eliminations; the four newly circled in orange don’t let us eliminate anything new, so where next? Let’s take stock of the board…remember, we only know that we’ve completed the puzzle once we are sure there are no duplicate numbers in any rows and columns.

In fact, there are no more duplicates in any row or column, so we’ve completed the puzzle. I said it was an easy one!

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 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!