Sudoku 6x6 - Easy - Sudoku-Puzzles.net - Free Printable

This is a 6x6 Sudoku puzzle. The goal is to fill in the empty cells so that every row, every column, and every 2x3 box contains the numbers 1 through 6 exactly once.

Let’s solve it step by step.

We’ll label rows R1 to R6 (top to bottom) and columns C1 to C6 (left to right). Boxes are 2 rows × 3 columns.

Current grid:

R1: _ 3 _ | 4 _ _
R2: _ _ 5 | 6 _ 3
R3: _ _ _ | 1 _ _
R4: _ 1 _ | 3 _ 5
R5: _ 6 4 | _ 3 1
R6: _ _ 1 | _ 4 6

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Step 1: Look at R5 — it has [_, 6, 4, _, 3, 1]. Missing numbers: 2 and 5.

C1 already has: R2=_, R3=_, R4=_, R5=?, R6=? — let’s check what’s in C1 later.

But look at C4: R1=4, R2=6, R3=1, R4=3, R5=?, R6=? → missing 2 and 5.

In R5, C4 must be either 2 or 5.

Also, in R5, C1 must be the other of 2 or 5.

Now check Box 4 (R4-R5, C1-C3):

Box 4 currently:
R4: _ 1 _
R5: _ 6 4

Numbers present: 1, 4, 6 → missing 2, 3, 5.

R4C1 and R5C1 are both in this box.

R4C2 = 1, R5C2 = 6, R5C3 = 4.

So R4C1, R4C3, R5C1 need to be 2, 3, 5.

But R4 already has 1, 3, 5 → wait, R4: _ 1 _ | 3 _ 5 → so R4 has 1, 3, 5 → missing 2, 4, 6.

Wait — correction: R4 has 1, 3, 5 → so missing 2, 4, 6.

But in Box 4 (R4-R5, C1-C3), we have:

R4C1, R4C2=1, R4C3
R5C1, R5C2=6, R5C3=4

So numbers in Box 4 so far: 1, 4, 6 → missing 2, 3, 5.

But R4 already has 3 and 5 in other columns (C4 and C6), so in Box 4, R4 cannot have 3 or 5 again? Wait no — R4C4=3, R4C6=5 — those are outside Box 4. So within Box 4, R4 can still take 3 or 5? No — because the whole row R4 must have unique numbers.

R4: positions C1, C2=1, C3, C4=3, C5, C6=5 → so R4 already has 1, 3, 5 → so C1, C3, C5 must be 2, 4, 6.

But in Box 4 (C1-C3), R4C1 and R4C3 must be from {2,4,6}, but also Box 4 needs 2,3,5 — conflict?

Wait — let’s list what Box 4 needs: numbers 1-6.

Currently in Box 4:

R4C2 = 1
R5C2 = 6
R5C3 = 4

So present: 1, 4, 6 → missing: 2, 3, 5.

But R4 already has 3 and 5 in C4 and C6 — so R4C1 and R4C3 cannot be 3 or 5 — they must be from {2,4,6} — but 4 is already in R5C3, and 6 is in R5C2 — so R4C1 and R4C3 must be 2 and 6? But 6 is already in R5C2 — same box — so 6 cannot appear again in Box 4.

Therefore, R4C1 and R4C3 must be 2 and something else — but 4 and 6 are already in the box? R5C2=6, R5C3=4 — yes.

So Box 4 has 1,4,6 — needs 2,3,5.

But R4 cannot take 3 or 5 because they’re already in R4 (C4=3, C6=5).

So R4C1 and R4C3 must be from {2,4,6} minus what’s already in row — but 4 and 6 are not in R4 yet? R4 has 1,3,5 — so 2,4,6 are available for C1,C3,C5.

But in Box 4, 4 and 6 are already placed (R5C3=4, R5C2=6), so R4C1 and R4C3 cannot be 4 or 6 — only 2 is possible? But we need two numbers for R4C1 and R4C3.

Contradiction? Let me double-check.

R4: C1=?, C2=1, C3=?, C4=3, C5=?, C6=5 → so missing 2,4,6 for C1,C3,C5.

Box 4 (C1-C3, R4-R5): cells:

R4C1, R4C2=1, R4C3
R5C1, R5C2=6, R5C3=4

So values in box: 1,4,6 → missing 2,3,5.

R4C1 and R4C3 must be chosen from {2,4,6} (row constraint) and also from {2,3,5} (box constraint) — intersection is {2}.

But we need two cells — impossible? That can’t be.

I think I made a mistake.

R4C4=3 and R4C6=5 are in the same row, so yes, R4 cannot have another 3 or 5.

But in Box 4, we need to place 2,3,5 — but R4 cannot take 3 or 5, so R5C1 must take one of 3 or 5, and R4C1 and R4C3 must take the remaining from {2,3,5} that R4 can take — but R4 can only take 2 from that set.

So perhaps R5C1 takes 3 or 5.

Let’s try that.

Box 4 missing: 2,3,5.

R4C1 and R4C3 can only be 2 (since 3 and 5 are in R4 already).

But there are two cells — so only one can be 2? That doesn't work.

Unless... perhaps I misidentified the boxes.

Standard 6x6 Sudoku uses 2-row by 3-column boxes.

So boxes are:

Box 1: R1-2, C1-3
Box 2: R1-2, C4-6
Box 3: R3-4, C1-3
Box 4: R3-4, C4-6
Box 5: R5-6, C1-3
Box 6: R5-6, C4-6

I think I messed up the box numbering.

Let me redefine:

Boxes:

- Top-left: R1-2, C1-3
- Top-right: R1-2, C4-6
- Middle-left: R3-4, C1-3
- Middle-right: R3-4, C4-6
- Bottom-left: R5-6, C1-3
- Bottom-right: R5-6, C4-6

So for R4, it's in middle-left and middle-right boxes.

R4C1, C2, C3 are in Box 3 (R3-4, C1-3)
R4C4, C5, C6 are in Box 4 (R3-4, C4-6)

Similarly, R5 is in Box 5 and 6.

So let's correct.

For R4: C1,C2,C3 in Box 3; C4,C5,C6 in Box 4.

R4: _ 1 _ | 3 _ 5

So in Box 3 (R3-4, C1-3): cells include R3C1, R3C2, R3C3, R4C1, R4C2=1, R4C3

Present: R4C2=1 → so 1 is in Box 3.

R4 has 1,3,5 — so in Box 3, R4C1 and R4C3 must be from {2,4,6} (since 1,3,5 are in row, and 3,5 are in other parts of row, but 3 and 5 are in C4 and C6, which are in Box 4, so for Box 3, R4 can have 2,4,6, but 4 and 6 may be constrained by box.

Box 3 has R3C1, R3C2, R3C3, R4C1, R4C2=1, R4C3

No other numbers given yet for R3.

So Box 3 has only 1 so far.

R4C1 and R4C3 must be from {2,4,6} for the row, and for the box, no restriction yet.

Similarly, R5 is in Box 5 and 6.

R5: _ 6 4 | _ 3 1

So in Box 5 (R5-6, C1-3): R5C1, R5C2=6, R5C3=4, R6C1, R6C2, R6C3=1

Present: 6,4,1 → so missing 2,3,5 for Box 5.

R5C1 must be from {2,5} because R5 has 6,4,3,1 — so missing 2,5 for C1 and C4.

R5: C1=?, C2=6, C3=4, C4=?, C5=3, C6=1 → so missing 2,5 for C1 and C4.

In Box 5, R5C1 must be 2 or 5, and Box 5 needs 2,3,5 — so possible.

Also, R6C3=1, so in Box 5, 1 is already there.

Now let's look at C4.

C4: R1=4, R2=6, R3=1, R4=3, R5=?, R6=?

So values: 4,6,1,3 — missing 2,5.

R5C4 and R6C4 must be 2 and 5.

But R5C4 is in R5, which needs 2 or 5 for C1 and C4.

Similarly, R6C4 is in R6.

R6: _ _ 1 | _ 4 6 → so C3=1, C5=4, C6=6 → missing 2,3,5 for C1,C2,C4.

C4 for R6 must be 2 or 5, as above.

Now, let's consider R5C4.

If R5C4 = 2, then R5C1 = 5 (since R5 missing 2,5).

If R5C4 = 5, then R5C1 = 2.

Now, look at Box 5: R5C1, R5C2=6, R5C3=4, R6C1, R6C2, R6C3=1

Missing 2,3,5.

R5C1 is 2 or 5.

R6C1 and R6C2 must be the remaining.

Also, R6C4 is the other of 2 or 5.

Let's see if we can find a conflict.

Look at C1.

C1: R1=?, R2=?, R3=?, R4=?, R5=?, R6=?

No numbers given yet.

But let's look at Box 1 (R1-2, C1-3):

R1: _ 3 _
R2: _ _ 5

So cells: R1C1, R1C2=3, R1C3, R2C1, R2C2, R2C3=5

Present: 3,5 → missing 1,2,4,6.

R1C1, R1C3, R2C1, R2C2 to be filled.

R1 has 3,4 — so missing 1,2,5,6 for C1,C3,C5,C6.

R1: C1=?, C2=3, C3=?, C4=4, C5=?, C6=? → so missing 1,2,5,6.

In Box 1, R1C1 and R1C3 must be from {1,2,5,6} minus what's in box.

Box 1 has 3,5 — so R1C1 and R1C3 can be 1,2,6 (since 5 is in R2C3).

Similarly, R2 has 5,6,3 — R2: C1=?, C2=?, C3=5, C4=6, C5=?, C6=3 → so missing 1,2,4 for C1,C2,C5.

In Box 1, R2C1 and R2C2 must be from {1,2,4}.

Box 1 missing 1,2,4,6 — and R2C1, R2C2 from {1,2,4}, R1C1, R1C3 from {1,2,6}.

Possible.

Let's try to fill R5 first.

Assume R5C4 = 2.

Then R5C1 = 5 (since R5 missing 2,5).

Then in Box 5, R5C1=5, R5C2=6, R5C3=4, R6C3=1, so missing 2,3 for R6C1 and R6C2.

R6: C1=?, C2=?, C3=1, C4=?, C5=4, C6=6 → missing 2,3,5 for C1,C2,C4.

But C4 for R6 must be 2 or 5, but we assumed R5C4=2, so R6C4 must be 5 (since C4 missing 2,5).

So R6C4 = 5.

Then R6 missing for C1,C2: 2,3.

And Box 5 missing 2,3 for R6C1 and R6C2.

Perfect.

So if R5C4=2, then R5C1=5, R6C4=5, R6C1 and R6C2 are 2 and 3.

Now, is there a problem with R6C4=5? R6 has C6=6, C5=4, C3=1, so 5 is ok.

Now check C4: R1=4, R2=6, R3=1, R4=3, R5=2, R6=5 — all good, 1-6.

Now R5: C1=5, C2=6, C3=4, C4=2, C5=3, C6=1 — good.

R6: C1 and C2 are 2 and 3, C3=1, C4=5, C5=4, C6=6.

Now, which is which for C1 and C2?

Look at C1 and C2.

C2: R1=3, R2=?, R3=?, R4=1, R5=6, R6=?

Values so far: 3,1,6 — missing 2,4,5.

R6C2 must be 2 or 3, but 3 is already in R1C2, so R6C2 cannot be 3 — must be 2.

Then R6C1 = 3.

C2: R6C2=2, so C2 has R1=3, R4=1, R5=6, R6=2 — missing 4,5 for R2C2 and R3C2.

C1: R5=5, R6=3 — so far 5,3 — missing 1,2,4,6.

R6C1=3, so C1 has 3,5.

Now, let's fill what we have.

Grid so far:

R1: _ 3 _ | 4 _ _
R2: _ _ 5 | 6 _ 3
R3: _ _ _ | 1 _ _
R4: _ 1 _ | 3 _ 5
R5: 5 6 4 | 2 3 1
R6: 3 2 1 | 5 4 6

Now, look at R4: _ 1 _ | 3 _ 5

Missing 2,4,6 for C1,C3,C5.

C1: R5=5, R6=3 — so R4C1 cannot be 3 or 5, ok.

C3: R2=5, R5=4, R6=1 — so values 5,4,1 — missing 2,3,6.

R4C3 must be from {2,4,6} for row, and for C3, 4 is already in R5C3, so R4C3 cannot be 4 — so 2 or 6.

Similarly, C5: R5=3, R6=4 — so values 3,4 — missing 1,2,5,6.

R4C5 must be from {2,4,6} for row, but 4 is in R6C5, so R4C5 cannot be 4 — so 2 or 6.

But R4C1, C3, C5 must be 2,4,6.

C1: R4C1 — C1 has R5=5, R6=3 — so can be 2,4,6.

C3: R4C3 — C3 has R2=5, R5=4, R6=1 — so 4 is taken, so R4C3 cannot be 4 — so must be 2 or 6.

C5: R4C5 — C5 has R5=3, R6=4 — so 4 is taken, so R4C5 cannot be 4 — so must be 2 or 6.

But we need to place 4 in R4 — where can it go? Only C1, since C3 and C5 cannot have 4.

C1: no 4 yet, so R4C1 = 4.

Then R4C3 and R4C5 are 2 and 6.

Now, C3: R4C3 must be 2 or 6.

C5: R4C5 must be 2 or 6.

Look at Box 3 (R3-4, C1-3): cells R3C1, R3C2, R3C3, R4C1=4, R4C2=1, R4C3

Present: 4,1 — missing 2,3,5,6.

R4C3 is 2 or 6.

Also, R3C1, R3C2, R3C3 to be filled.

R3: _ _ _ | 1 _ _ → C4=1, so missing 2,3,4,5,6 for C1,C2,C3,C5,C6.

In Box 3, R3C1, R3C2, R3C3 must be from missing.

Now, let's look at C3.

C3: R1=?, R2=5, R3=?, R4=?, R5=4, R6=1

Values: 5,4,1 — missing 2,3,6.

R4C3 is 2 or 6.

R1C3 and R3C3 to be filled.

R1: C3 must be from {1,2,5,6} but 1,5,6 may be constrained.

R1 missing 1,2,5,6 for C1,C3,C5,C6.

C3 missing 2,3,6 — so R1C3 can be 2 or 6 (since 3 is in R1C2, so not 3).

Similarly, R3C3 can be 2,3,6.

Now, back to R4: C1=4, C3 and C5 are 2 and 6.

Suppose R4C3 = 2, then R4C5 = 6.

Or R4C3 = 6, R4C5 = 2.

Check C5.

C5: R1=?, R2=?, R3=?, R4=?, R5=3, R6=4

Values: 3,4 — missing 1,2,5,6.

R4C5 is 2 or 6.

Also, R2C5: R2 has C3=5, C4=6, C6=3 — so missing 1,2,4 for C1,C2,C5.

So R2C5 can be 1,2,4.

But C5 missing 1,2,5,6 — so R2C5 can be 1,2.

Similarly, R1C5: R1 missing 1,2,5,6 — C5 missing 1,2,5,6 — so ok.

Now, let's look at Box 4 (R3-4, C4-6): cells R3C4=1, R3C5, R3C6, R4C4=3, R4C5, R4C6=5

Present: 1,3,5 — missing 2,4,6.

R4C5 is 2 or 6.

R3C5 and R3C6 to be filled.

R3C4=1, so in Box 4, 1 is there.

R4C5 is 2 or 6.

If R4C5 = 2, then Box 4 has 1,3,5,2 — missing 4,6 for R3C5 and R3C6.

If R4C5 = 6, then missing 2,4.

Now, R3: C4=1, so missing 2,3,4,5,6 for C1,C2,C3,C5,C6.

In Box 4, R3C5 and R3C6 must be from missing.

Also, C6: R1=?, R2=3, R3=?, R4=5, R5=1, R6=6

Values: 3,5,1,6 — missing 2,4.

So R1C6 and R3C6 must be 2 and 4.

R3C6 is in Box 4, which needs 2,4,6 or 2,4 depending.

From above, if R4C5=2, Box 4 missing 4,6 for R3C5 and R3C6.

But C6 missing 2,4 — so R3C6 must be 2 or 4.

If Box 4 needs 4,6 for R3C5 and R3C6, and C6 needs 2,4, then R3C6 must be 4 (common), and R3C5=6.

Then R1C6=2.

If R4C5=6, then Box 4 missing 2,4 for R3C5 and R3C6.

C6 missing 2,4 — so R3C6 can be 2 or 4, R3C5 the other.

But let's see which works.

First, assume R4C3 = 2, then R4C5 = 6.

Then Box 4: R3C4=1, R4C4=3, R4C5=6, R4C6=5 — so present 1,3,5,6 — missing 2,4 for R3C5 and R3C6.

C6 missing 2,4 — so R3C6 and R1C6 are 2 and 4.

R3C6 can be 2 or 4.

R3C5 must be the other of 2 or 4 for Box 4.

But C5: R4C5=6, R5=3, R6=4 — so values 6,3,4 — missing 1,2,5.

R3C5 must be from Box 4 missing 2,4, but C5 missing 1,2,5 — so intersection is 2.

So R3C5 = 2.

Then R3C6 = 4 (for Box 4).

Then C6: R3C6=4, so R1C6=2.

C5: R3C5=2, so missing 1,5 for R1C5 and R2C5.

R2C5: R2 missing 1,2,4 for C1,C2,C5 — but 2 is now in R3C5, so R2C5 can be 1 or 4.

C5 missing 1,5 — so R2C5 can be 1.

R1C5 can be 5.

Let's list.

So far with R4C3=2, R4C5=6:

R4: C1=4, C2=1, C3=2, C4=3, C5=6, C6=5 — good.

R3: C4=1, C5=2, C6=4 — so missing 3,5,6 for C1,C2,C3.

C1: R4=4, R5=5, R6=3 — so values 4,5,3 — missing 1,2,6.

R3C1 must be from {3,5,6} for row, and for C1, 3,4,5 are taken? C1 has R4=4, R5=5, R6=3 — so 3,4,5 — missing 1,2,6.

R3C1 can be 6 (since 3 and 5 are in C1 already? 3 is in R6C1, 5 in R5C1, 4 in R4C1 — so C1 has 3,4,5 — missing 1,2,6.

R3C1 must be 6 (from row missing 3,5,6, but 3 and 5 are in C1, so only 6 possible).

So R3C1 = 6.

Then R3C2 and R3C3 are 3 and 5.

C2: R1=3, R4=1, R5=6, R6=2 — so values 3,1,6,2 — missing 4,5.

R3C2 must be 3 or 5, but 3 is in R1C2, so cannot be 3 — must be 5.

Then R3C3 = 3.

C3: R2=5, R3=3, R4=2, R5=4, R6=1 — so values 5,3,2,4,1 — missing 6.

R1C3 must be 6.

R1: C2=3, C3=6, C4=4, C6=2 — so missing 1,5 for C1 and C5.

C1: R3=6, R4=4, R5=5, R6=3 — so values 6,4,5,3 — missing 1,2.

R1C1 must be 1 or 5, but C1 missing 1,2 — so R1C1 = 1.

Then R1C5 = 5.

C5: R1=5, R3=2, R4=6, R5=3, R6=4 — so values 5,2,6,3,4 — missing 1.

R2C5 must be 1.

R2: C3=5, C4=6, C5=1, C6=3 — so missing 2,4 for C1 and C2.

C1: R1=1, R3=6, R4=4, R5=5, R6=3 — so values 1,6,4,5,3 — missing 2.

So R2C1 = 2.

Then R2C2 = 4.

C2: R1=3, R2=4, R3=5, R4=1, R5=6, R6=2 — all good, 1-6.

Now check Box 1: R1C1=1, R1C2=3, R1C3=6, R2C1=2, R2C2=4, R2C3=5 — numbers 1,2,3,4,5,6 — good.

Box 2: R1C4=4, R1C5=5, R1C6=2, R2C4=6, R2C5=1, R2C6=3 — 1,2,3,4,5,6 — good.

Box 3: R3C1=6, R3C2=5, R3C3=3, R4C1=4, R4C2=1, R4C3=2 — 1,2,3,4,5,6 — good.

Box 4: R3C4=1, R3C5=2, R3C6=4, R4C4=3, R4C5=6, R4C6=5 — 1,2,3,4,5,6 — good.

Box 5: R5C1=5, R5C2=6, R5C3=4, R6C1=3, R6C2=2, R6C3=1 — 1,2,3,4,5,6 — good.

Box 6: R5C4=2, R5C5=3, R5C6=1, R6C4=5, R6C5=4, R6C6=6 — 1,2,3,4,5,6 — good.

All rows and columns also good.

So the solution is:

R1: 1 3 6 | 4 5 2
R2: 2 4 5 | 6 1 3
R3: 6 5 3 | 1 2 4
R4: 4 1 2 | 3 6 5
R5: 5 6 4 | 2 3 1
R6: 3 2 1 | 5 4 6

Now, to write it as per the original format.

Final Answer:

1 3 6 | 4 5 2
2 4 5 | 6 1 3
6 5 3 | 1 2 4
4 1 2 | 3 6 5
5 6 4 | 2 3 1
3 2 1 | 5 4 6