Free Printable Alphabet Sudoku Puzzles - Free Printable

Let's solve this Sudoku-like puzzle using letters instead of numbers. The goal is to fill the 6×6 grid so that each row, each column, and each 2×3 block (or "region") contains all six distinct letters: A, B, C, D, E, F, with no repeats.

---

Step 1: Understand the Grid

The grid is 6×6, divided into six 2×3 blocks (not the usual 3×3). Each block must contain all six letters A–F exactly once.

We’ll label the rows from top to bottom as R1 to R6 and columns from left to right as C1 to C6.

Here’s the initial grid:

```
R1: _ C _ | _ A _
R2: D B _ | E C F
R3: B F _ | _ E A
R4: _ _ _ | _ _ _
R5: _ E _ | _ D _
R6: C _ _ | F B _
```

And the blocks are:

- Block 1: R1C1-R2C3
- Block 2: R1C4-R2C6
- Block 3: R3C1-R4C3
- Block 4: R3C4-R4C6
- Block 5: R5C1-R6C3
- Block 6: R5C4-R6C6

---

Step 2: Fill in Missing Letters Using Logic

We'll go step by step.

#### Let's look at Block 1: R1C1 to R2C3
Current:
```
R1: _ C _
R2: D B _
```
Used: C, D, B → missing: A, E, F

So R1C1, R1C3, R2C3 must be A, E, F in some order.

Now check column constraints.

---

Column C1:

R1C1: ?
R2C1: D
R3C1: B
R4C1: ?
R5C1: ?
R6C1: C

Used: D, B, C → missing: A, E, F

So R1C1, R4C1, R5C1 ∈ {A,E,F}

But from Block 1, R1C1 ∈ {A,E,F} — consistent.

---

Column C2:

R1C2: C
R2C2: B
R3C2: F
R4C2: ?
R5C2: E
R6C2: ?

Used: C, B, F, E → missing: A, D

So R4C2 and R6C2 ∈ {A, D}

---

Column C3:

R1C3: ?
R2C3: ?
R3C3: ?
R4C3: ?
R5C3: ?
R6C3: ?

Only R2C3 has a value? Wait, no — let's list:

R1C3: ?
R2C3: ?
R3C3: ?
R4C3: ?
R5C3: ?
R6C3: ?

Wait — only R2C3 is empty? No — actually, only R2C3 is filled? No — R2C3 is blank.

Wait — looking back:

From original:
- R1C3: blank
- R2C3: blank
- R3C3: blank
- R4C3: blank
- R5C3: blank
- R6C3: blank

No values in C3 yet. So we can't deduce much yet.

---

Focus on Block 2: R1C4 to R2C6

```
R1: _ A _
R2: E C F
```

Used: A, E, C, F → missing: B, D

So R1C4 and R1C6 must be B and D in some order.

But R1C4 is already A — wait, no:

Wait! R1C4 is blank, R1C5 is A, R1C6 is blank.

So:
- R1C4: ?
- R1C5: A
- R1C6: ?

R2C4: E
R2C5: C
R2C6: F

So used in block: A, C, E, F → missing: B, D

So R1C4 and R1C6 must be B and D.

Now, look at Column C4:

R1C4: ? → must be B or D
R2C4: E
R3C4: ?
R4C4: ?
R5C4: ?
R6C4: F

Used: E, F → missing: A, B, C, D

So R1C4 ∈ {B,D} — okay.

Similarly, Column C6:
R1C6: ? → B or D
R2C6: F
R3C6: A
R4C6: ?
R5C6: ?
R6C6: ?

Used: F, A → missing: B, C, D, E

So R1C6 ∈ {B,D} — okay.

Now, can we determine which?

Let’s look at Row 1:

R1: ?, C, ?, ?, A, ?

Letters used: C, A → missing: B, D, E, F

But from earlier, R1C4 and R1C6 are B and D → so R1C1 and R1C3 must be E and F.

Recall from Block 1: R1C1, R1C3, R2C3 ∈ {A,E,F}, but A is already used in R1C5, so R1C1 and R1C3 ∈ {E,F}

Yes.

So for Row 1:
- R1C1: E or F
- R1C3: F or E
- R1C4: B or D
- R1C6: D or B

Also, R1C2 = C, R1C5 = A

So R1 needs: B, D, E, F

We have:
- R1C1 and R1C3: E and F
- R1C4 and R1C6: B and D

So that fits.

Now, can we figure out R1C4?

Look at Column C4:
R1C4: ?
R2C4: E
R3C4: ?
R4C4: ?
R5C4: ?
R6C4: F

Used: E, F → need A, B, C, D

But R1C4 is either B or D.

Can it be B?

Try both options later.

---

Now look at Row 2:

R2: D, B, ?, E, C, F

Used: D, B, E, C, F → missing: A

So R2C3 must be A!

That’s key!

So R2C3 = A

Now update Block 1:

Block 1: R1C1, R1C2=C, R1C3; R2C1=D, R2C2=B, R2C3=A

Used: C, D, B, A → missing: E, F

So R1C1 and R1C3 must be E and F.

Earlier we had that.

Now, Row 1:
- R1C1: E or F
- R1C3: F or E
- R1C4: B or D
- R1C6: D or B

But now, Column C3:
R1C3: E or F
R2C3: A
R3C3: ?
R4C3: ?
R5C3: ?
R6C3: ?

Used: A → need B,C,D,E,F

So no constraint yet.

But look at Row 3:
R3: B, F, ?, ?, E, A

Used: B, F, E, A → missing: C, D

So R3C3 and R3C4 must be C and D.

Now, Column C3:
R1C3: E or F
R2C3: A
R3C3: C or D
R4C3: ?
R5C3: ?
R6C3: ?

So far, no duplicates.

But Block 3: R3C1 to R4C3

R3: B, F, ?
R4: ?, ?, ?

Used: B, F → need A, C, D, E

So R3C3 ∈ {C,D} — consistent.

Now, R3C4: must be C or D (from row)

But R3C4 is in Block 4: R3C4 to R4C6

R3C4: C or D
R3C5: E
R3C6: A
R4C4: ?
R4C5: ?
R4C6: ?

Used: E, A → need B, C, D, F

So R3C4 ∈ {C,D} — okay.

Now, Column C4:
R1C4: B or D
R2C4: E
R3C4: C or D
R4C4: ?
R5C4: ?
R6C4: F

Used: E, F → need A, B, C, D

So R1C4 ∈ {B,D}, R3C4 ∈ {C,D}

Can they both be D? Only if not same column.

But column C4: R1C4 and R3C4 both could be D?

Possibly.

But let’s see if we can eliminate.

Back to Row 1:

We know:
- R1C1 and R1C3: E and F
- R1C4 and R1C6: B and D

Now, Column C1:
R1C1: E or F
R2C1: D
R3C1: B
R4C1: ?
R5C1: ?
R6C1: C

Used: D, B, C → missing: A, E, F

So R1C1 ∈ {E,F} — good.

But R1C1 cannot be D or B or C — already used.

So R1C1 ∈ {A,E,F}, but A is used in R1C5 → so R1C1 ∈ {E,F} — correct.

Now, suppose R1C1 = E → then R1C3 = F

Or vice versa.

Let’s try to use Block 5: R5C1 to R6C3

R5: ?, E, ?, ?, D, ?
R6: C, ?, ?, F, B, ?

So Block 5:
R5C1: ?
R5C2: E
R5C3: ?
R6C1: C
R6C2: ?
R6C3: ?

Used: E, C, D, F, B → wait, D is in R5C5, F in R6C4, B in R6C5 — not in this block.

Block 5: R5C1, R5C2=E, R5C3; R6C1=C, R6C2=?, R6C3=?

So used: E, C → missing: A, B, D, F

But R6C2 and R6C3 are in this block.

Now, Column C2:
R1C2: C
R2C2: B
R3C2: F
R4C2: ?
R5C2: E
R6C2: ?

Used: C, B, F, E → missing: A, D

So R4C2 and R6C2 ∈ {A, D}

Similarly, Column C3:
R1C3: E or F
R2C3: A
R3C3: C or D
R4C3: ?
R5C3: ?
R6C3: ?

Used: A → need B,C,D,E,F

Now, back to Row 6:
R6: C, ?, ?, F, B, ?

Used: C, F, B → missing: A, D, E

So R6C2, R6C3, R6C6 ∈ {A,D,E}

But from Column C2, R6C2 ∈ {A,D}

So R6C2 = A or D

From above, R6C2 ∈ {A,D,E} — so possible.

Now, Column C6:
R1C6: B or D
R2C6: F
R3C6: A
R4C6: ?
R5C6: ?
R6C6: ?

Used: F, A → missing: B, C, D, E

So R1C6 ∈ {B,D} — okay.

Now, let’s go back to Row 1.

We have two possibilities:

Option 1: R1C1 = E, R1C3 = F
Option 2: R1C1 = F, R1C3 = E

Let’s test Option 1: R1C1 = E

Then R1C3 = F

Now, Column C1:
R1C1 = E
R2C1 = D
R3C1 = B
R4C1 = ?
R5C1 = ?
R6C1 = C

Used: E, D, B, C → missing: A, F

So R4C1 and R5C1 ∈ {A,F}

Now, Column C3:
R1C3 = F
R2C3 = A
R3C3 = C or D
R4C3 = ?
R5C3 = ?
R6C3 = ?

Used: F, A → need B,C,D,E

Now, Row 3:
R3: B, F, ?, ?, E, A

So R3C3 and R3C4 ∈ {C,D}

Suppose R3C3 = C → then R3C4 = D

Or vice versa.

Now, Column C4:
R1C4: B or D
R2C4: E
R3C4: C or D
R4C4: ?
R5C4: ?
R6C4: F

Used: E, F → need A,B,C,D

Now, Block 4: R3C4 to R4C6

R3C4: C or D
R3C5: E
R3C6: A
R4C4: ?
R4C5: ?
R4C6: ?

Used: E, A → need B,C,D,F

Now, Row 4 is completely blank — hard to start.

Let’s look at Row 5:
R5: ?, E, ?, ?, D, ?

Used: E, D → missing: A, B, C, F

So R5C1, R5C3, R5C4, R5C6 ∈ {A,B,C,F}

Now, Column C1: R5C1 ∈ {A,F} (from earlier, since R4C1 and R5C1 ∈ {A,F})

So R5C1 = A or F

Similarly, Column C3: R5C3 ∈ {B,C,D,E} — but D and E are used in row 5? Not necessarily.

Wait, R5C3 is part of row 5, which has E and D already.

So R5C3 ∈ {A,B,C,F} — but A may be available.

But Column C3 has R1C3=F, R2C3=A → so A and F already used in column C3 → so R5C3 ≠ A, F

So R5C3 ∈ {B,C,D,E}

But row 5 has E and D → so R5C3 ∈ {B,C}

Similarly, Column C4:
R1C4: B or D
R2C4: E
R3C4: C or D
R4C4: ?
R5C4: ?
R6C4: F

Used: E, F → need A,B,C,D

So R5C4 ∈ {A,B,C,D}

But row 5 has D → so R5C4 ∈ {A,B,C}

Now, Column C6:
R1C6: B or D
R2C6: F
R3C6: A
R4C6: ?
R5C6: ?
R6C6: ?

Used: F, A → need B,C,D,E

So R5C6 ∈ {B,C,D,E}

But row 5 has D → so R5C6 ∈ {B,C,E}

Now, let’s go back to Block 1.

We had R1C1 = E (assumption), R1C3 = F

Now, Column C1:
R1C1 = E
R2C1 = D
R3C1 = B
R4C1 = ?
R5C1 = ?
R6C1 = C

Used: E, D, B, C → missing: A, F

So R4C1 and R5C1 ∈ {A,F}

Now, Column C3:
R1C3 = F
R2C3 = A
R3C3 = ?
R4C3 = ?
R5C3 = ?
R6C3 = ?

Used: F, A → need B,C,D,E

But R5C3 ∈ {B,C} (from earlier)

Now, Row 3: R3C3 and R3C4 ∈ {C,D}

Suppose R3C3 = C → then R3C4 = D

Then Column C3 has R3C3 = C

Now, Column C4:
R3C4 = D

R1C4: B or D → but D is now in R3C4 → so R1C4 ≠ D → so R1C4 = B

Then R1C6 = D (since R1C4 and R1C6 are B and D)

So now:

- R1C4 = B
- R1C6 = D

Now, Row 1:
R1: E, C, F, B, A, D

Check: E,C,F,B,A,D → all letters present — good.

Now, Column C4:
R1C4 = B
R2C4 = E
R3C4 = D
R4C4 = ?
R5C4 = ?
R6C4 = F

Used: B, E, D, F → missing: A, C

So R4C4 and R5C4 ∈ {A,C}

But earlier, R5C4 ∈ {A,B,C} → now ∈ {A,C}

Now, Column C6:
R1C6 = D
R2C6 = F
R3C6 = A
R4C6 = ?
R5C6 = ?
R6C6 = ?

Used: D, F, A → missing: B, C, E

So R4C6, R5C6, R6C6 ∈ {B,C,E}

Now, Row 4:
R4: ?, ?, ?, ?, ?, ?

But Block 3: R3C1 to R4C3

R3: B, F, C
R4: ?, ?, ?

Used: B, F, C → need A, D, E

So R4C1, R4C2, R4C3 ∈ {A,D,E}

But Column C1: R4C1 ∈ {A,F} → but F is used in R4C1? No, R4C1 ∈ {A,F} from earlier

But in Block 3, R4C1 ∈ {A,D,E}

So intersection: R4C1 ∈ {A} (since A is common)

So R4C1 = A

Then R5C1 = F (since R4C1 and R5C1 ∈ {A,F})

Now, Column C1:
R4C1 = A
R5C1 = F
R6C1 = C
R3C1 = B
R2C1 = D
R1C1 = E

All filled: A,B,C,D,E,F — perfect.

Now, Row 4:
R4C1 = A

Now, Column C2:
R1C2 = C
R2C2 = B
R3C2 = F
R4C2 = ?
R5C2 = E
R6C2 = ?

Used: C,B,F,E → missing: A,D

So R4C2 and R6C2 ∈ {A,D}

But Block 3: R4C2 ∈ {D,E} (since R4C1=A, R4C2, R4C3 ∈ {D,E} — because used: A, and need D,E)

Wait, Block 3: R3C1=B, R3C2=F, R3C3=C → used: B,F,C → need A,D,E

R4C1=A → so R4C2 and R4C3 ∈ {D,E}

So R4C2 ∈ {D,E}

But from Column C2, R4C2 ∈ {A,D}

Intersection: R4C2 ∈ {D}

So R4C2 = D

Then R6C2 = A (from column C2: missing A,D, R4C2=D → R6C2=A)

Now, Row 4:
R4C1 = A
R4C2 = D
R4C3 = ? → must be E (since R4C3 ∈ {D,E}, but D used → so E)

So R4C3 = E

Now, Row 4: A, D, E, ?, ?, ?

Now, Column C3:
R1C3 = F
R2C3 = A
R3C3 = C
R4C3 = E
R5C3 = ?
R6C3 = ?

Used: F,A,C,E → missing: B,D

So R5C3 and R6C3 ∈ {B,D}

But earlier, R5C3 ∈ {B,C} → so R5C3 = B

Then R6C3 = D

Now, Row 5:
R5: F, E, B, ?, D, ?

Used: F,E,B,D → missing: A,C

So R5C4 and R5C6 ∈ {A,C}

But earlier, R5C4 ∈ {A,C} — good.

Now, Column C4:
R1C4 = B
R2C4 = E
R3C4 = D
R4C4 = ?
R5C4 = ?
R6C4 = F

Used: B,E,D,F → missing: A,C

So R4C4 and R5C4 ∈ {A,C}

R5C4 ∈ {A,C} — good.

Now, Block 4: R3C4 to R4C6

R3: D, E, A
R4: ?, ?, ?

Used: D,E,A → need B,C,F

R4C4, R4C5, R4C6 ∈ {B,C,F}

But Row 4:
R4: A, D, E, ?, ?, ?

So R4C4, R4C5, R4C6 ∈ {B,C,F}

Now, Column C4:
R4C4 ∈ {A,C} → but A is used in R4C1 → so R4C4 ≠ A → so R4C4 = C

Then R5C4 = A (since R4C4 and R5C4 ∈ {A,C}, R4C4=C → R5C4=A)

Now, Row 5:
R5C4 = A

Then R5C6 = C (since R5C4 and R5C6 ∈ {A,C})

Now, Row 5: F, E, B, A, D, C

All letters: F,E,B,A,D,C — good.

Now, Column C6:
R1C6 = D
R2C6 = F
R3C6 = A
R4C6 = ?
R5C6 = C
R6C6 = ?

Used: D,F,A,C → missing: B,E

So R4C6 and R6C6 ∈ {B,E}

Now, Row 4:
R4: A, D, E, C, ?, ?

So R4C5 and R4C6 ∈ {B,F} — but F is used in R2C6, but not in row 4 yet.

Wait, Row 4 has: A,D,E,C → used: A,D,E,C → missing: B,F

So R4C5 and R4C6 ∈ {B,F}

But Column C6 says R4C6 ∈ {B,E}

Intersection: R4C6 ∈ {B}

So R4C6 = B

Then R4C5 = F

Now, Column C6:
R4C6 = B
R5C6 = C
R6C6 = ? → must be E (since missing B,E, B used → E)

So R6C6 = E

Now, Row 6:
R6: C, A, D, F, B, E

Used: C,A,D,F,B,E — all letters — perfect.

Now, Row 4:
R4: A, D, E, C, F, B

All good.

Now, Row 3:
R3: B, F, C, D, E, A

All good.

Now, Row 5:
R5: F, E, B, A, D, C

Good.

Now, Row 6:
R6: C, A, D, F, B, E

Good.

Now, Row 1:
R1: E, C, F, B, A, D

Good.

Row 2:
R2: D, B, A, E, C, F

Good.

Now, check columns:

- C1: E,D,B,A,F,C → all different — good
- C2: C,B,F,D,E,A → good
- C3: F,A,C,E,B,D → good
- C4: B,E,D,C,A,F → good
- C5: A,C,E,F,D,B → good
- C6: D,F,A,B,C,E → good

Now, check blocks:

- Block 1: R1C1-E, R1C2-C, R1C3-F; R2C1-D, R2C2-B, R2C3-A → E,C,F,D,B,A → all good
- Block 2: R1C4-B, R1C5-A, R1C6-D; R2C4-E, R2C5-C, R2C6-F → B,A,D,E,C,F → good
- Block 3: R3C1-B, R3C2-F, R3C3-C; R4C1-A, R4C2-D, R4C3-E → B,F,C,A,D,E → good
- Block 4: R3C4-D, R3C5-E, R3C6-A; R4C4-C, R4C5-F, R4C6-B → D,E,A,C,F,B → good
- Block 5: R5C1-F, R5C2-E, R5C3-B; R6C1-C, R6C2-A, R6C3-D → F,E,B,C,A,D → good
- Block 6: R5C4-A, R5C5-D, R5C6-C; R6C4-F, R6C5-B, R6C6-E → A,D,C,F,B,E → good

All blocks are valid.

---

Final Answer:

```
R1: E C F | B A D
R2: D B A | E C F
R3: B F C | D E A
R4: A D E | C F B
R5: F E B | A D C
R6: C A D | F B E
```

---

Solution Grid:

| | | | | | |
|-----|-----|-----|-----|-----|-----|
| E | C | F | B | A | D |
| D | B | A | E | C | F |
| B | F | C | D | E | A |
| A | D | E | C | F | B |
| F | E | B | A | D | C |
| C | A | D | F | B | E |

✔ All rows, columns, and blocks contain A–F exactly once.

This is the unique solution.