The New Sudoku Players' Forum

[jco]

Hi!

Just an observation: this sudoku-X puzzle (SE = 7.6) proposed by philvo here can be solved without pencilmarks.

Code: Select all

+-------+-------+
| . . . | . . . |
| . . 4 | . . . |
+-------+-------+
| . . . | . . 5 |
| . . . | . . 3 |
+-------+-------+
| . . . | . . . |
| . . . | 6 2 . |
+-------+-------+ X

EDIT (Feb 3, 2025): included hidden partial solution

Hidden Text: Show

Due to (6)d1, we have (6)e3|e4 and (6)f5|f6,
and (6)a6|b5|c4.

If a5=6, then a6=6, and there is no
"6" at c4,a1 or a2, so (6)b2|c2.
We cannot have (6)b2

Code: Select all

+--------+-------+
| A .  . | . . . | 6
| . .  4 | . . A | 5
+--------+-------+
| . .  -A| . . 5 | 4
| . .  . | . . 3 | 3
+--------+-------+
| . A  -A| . . . | 2
| . .  . | 6 2 . | 1
+--------+-------+
  a b  c   d e f   Sudoku-X

because there would be no 6 at box 3.
So, under the assumption a5=6, we must
have (6)c2

Code: Select all

+--------+-------+
| A .  . | . . . | 6
| . .  4 | . . A | 5
+--------+-------+
| . .  -A| . . 5 | 4
| . .  . | . . 3 | 3
+--------+-------+
| . -A A | . . . | 2
| . .  . | 6 2 . | 1
+--------+-------+
  a b  c   d e f   Sudoku-X


however this implies that there is
no 6 at the second diagonal (/).
So, we must have f6=6 and
f5=2.