# How to Solve Sudoku—Step By Step

In this article, we’ll go through the process of solving an easy sudoku puzzle. If you’re a sudoku beginner, you might pick up a few tips!

(Incidentally, this is probably the only standard sudoku puzzle you’ll see on this website.) We’ll start with looking at the 1s, and then the 2s and so on. This is most certainly not the most efficient way, but we do not solve sudokus for efficiency’s sake—we do it for fun! And we’re writing this article to illustrate some common sudoku solving techniques. 1: Looking at the 1s, we find that there’s not much to deduce yet, except that there’s just to cells in B7 (the bottom left box) where the number 1 can go. This is worth noting with pencil marks, because if we later find that another number must be in one of those two cells, we know that the other one must be 1. 2: The same is the case with the 2s: We make pencil marks for 2 in B2 (top middle box) R3. 3: Here, we’ve highlighted all the given 3s and shaded all their rows, columns and boxes. Both B2 and B4 has only one empty cell left for 3 (R1C5 and R5C3), so now we can fill in our first solved cells!

That also limits where the 3s can go in the remaining (white) cells, so we add pencil marks for those (R6 and R8, C4 and C6). 3.1: In B2 (top middle), the added 3 creates an interesting situation: The 1 in R2 limits the 1s in B2 to C4 and C5. We have already made pencil marks for 2 in the same cells, so now we know that those two cells can only contain those two numbers.

This makes a pair: If two cells in the same row, column or box are the only cells that can have two numbers, this effectively excludes those numbers from all other cells in the same row, column or box.

If the two numbers are the only candidates for those two cells, it’s called a naked pair. Those two numbers can be excluded from the rest of the row, column and/or box they’re in. Often, as in this case, a naked pair is part of both a box and a row or a column. This is a very useful thing, as we’ll see both now and later in this puzzle.

If two numbers can only go in the same two cells in a row, column or box but there also are other candidates in the two cells, it’s called a hidden pair. They are harder to spot, but once you spot them, they can be stripped naked by removing all other candidates from those two cells. Then you can proceed by applying the naked pair logic from the previous paragraph.

Note that the same logic applies to triplets, that is, three numbers across three cells. 3.2: Applying said logic, this means that in B3 (top right), the number 1 can only be in R1, so we add pencil marks for those too. 3.3: Then, in R2 in B2, we’re left with just 578 – but C4 already has an 8, so R2C4 can only be 57.

We also make a mental note: The pencil marks in B2 are now complete—we’ve pencilled in all the possible candidates. At the same time, we must keep in mind that the pencil marks in most of the other cells are not complete. 4: The given 4s are numerous, which now makes it easy to fill in the missing ones. Again, we’ve highlighted the 4s and shaded all the cells where they can’t go. We start by putting a 4 in B4 (R6C2), which limits 4 in B1 to R3C1, which in turn limits the 4 in B3 to R2C8. 4.1: Whenever we solve one or more cells, it’s worth taking a look at how that might affect what we’ve already deduced.

Look at B1 (top left): After placing the 4, we now see that the 7 in C3 and the 7s in B2R2 only leaves one cell for 7 in B1: R1C1.

That also limits 7 in B7 (bottom right) to C2, which creates another pair of 1 and 7 both in B7 and in C2. 4.2: The only candidates left for the rest of C2, then, are 689—but since we know that 8 in B2 must be in R2, we can exclude 8 from R2C2.

This in turn means that 8 in B1 must be in C3, so we can add pencil marks for those. For the sake of pencil mark completeness, we take a moment to add pencil marks for all the remaining candidates in B1. 4.3: Our latest deductions about 8s means that 8 in C1 must be in B7 (bottom left), but since there’s already an 8 in R9, the only cell left is R7C1. Looking at B9 (bottom right), we see that there’s only one cell left for 8 there: R8C8. 4.4: Now, there are only two cells left in C1, so we pencil in the only two missing numbers: 2 and 6. 5: A quick look at the 5s lets us place one in R9C3.

This also means that the last cell in B7 (bottom left), R8C3, also must have 2 and 6 as candidates, so we add pencil marks for those. 6: Let’s take a closer look at the number 6 in the bottom row of boxes. In There’s a 6 in R7, and the recently added 8 in R8C8 means that the 6 in B9 has to go in R9. This in turn means that there can’t be a 6 in R9C1, so that must be a 2, and thus R8C3 must be a 6. 6.1: For B1 and B4 this means that R2C2 must be a 6, which then determines the positions of the remaining 8 and 9.

C1 and C3 each have only one empty cell left to solve—a 6 and a 2, respectively. 6.2: In R2 there’s an empty cell. We know that the three cells in B2 must contain 578, so the only number left for R2C7 is 9.

This sets of a long chain of events! Looking at B6, we find that 9 must be in R5C9, which in turn places the 8 and 9 in C2.

In B6, the only one place left for 8 now is R6C7, and in B5 we can add pencil marks for the only two candidate cells for 8—R4C5C6.

In B3, the 8 must be in R1C9, which means that the 1 must be in R1C7. 7: So far, we’ve been looking at each number in order, so up next is 7.

B3 is just missing pencil marks for 6, but upon closer inspection we see that the 6 in R5 limits the 6 in B6 to C8, which in turn means that the 6 in B3R3 must be in C9, and consequently the 7 must be in C8. 7.1: Now we can solve the whole of C7: B6 now has only one cell left for 7, in R5C7. The 6 we just placed in C9 places the 6 in B9 in R9C7.

Finally, there’s just one cell left, which must be a 2.

We’ll also add pencil marks for 1 and 7 in both R9 and C9. 8: As it turns out, there’s nothing more to do with 8 at this time, so we take a look at 9 instead and find that it’s easy to place the 9 in B8 first and then the last 9 in B5.

Finally we’re getting rewarded for setting the pencil marks for 3 in those boxes: Placing the last 9 in B5 removed one of the 3s, so we’re left with only one candidate.

This also places the 3 in B8. 9: In R7, there’s only one place for 5—in C6. Again, we set off a chain of events that places the rest of the 5s, the rest of R2 and all the 8s too. 10: Placing the 6 in B5 is easy; it can only go in C5, since C4 and C6 already have 6s. That places the 6 in C8 at R4. 11: After that, filling in the rest of the cells is as trivial as it gets.

Puzzle solved!

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