Double Dutch Sudoku Advent (5) - Clone vs. Pointing Evens
(Published on 5. December 2014, 12:00 by Richard)
Eisbär (Arvid) and I have created a fresh sudoku advent calendar in which we combine a well known variant with a relatively unknown variant every day. Combining variants leads to interesting and surprising new solving techniques.
Logically solvable
All puzzles can be solved completely logically although the logic is sometimes well hidden and is inherent to the combination of restrictions that the different types offer. For that reason we have written some solving hints for most of the puzzles, published in a very tiny font. If you want to read the hints, simply copy these in a text editor and enlarge the font size.
Clone
Place the digits from 1 to 9 in every row, column and 3x3-block. The grid contains four different shapes. Each shape is cloned once. Cloned shapes may be rotated, but the position of the digits within them remains fixed. Within a single shape, no two digits are the same.
Pointing Evens
Digits outside the grid indicate the number of even digits in the direction of the diagonal arrow.
Solving hints
R1C9 and R9C1 are both even; R1C9 = R4C6 = 8; R9C1 = R6C4 = 6
R4C5 and R5C6 are odd; R1C8 and R2C9 are odd; R3C7 must be even.
All arrows together point to 36 even digits. This means that the diagonal top left – bottom right contains only odd digits and also R8C1 and R9C2 are odd. :=> R8C1 = 9; R2C2 = 9; R1C1 and R3C3 = pair {13}
R7C3 = even
The fourth arrow on the left points to 4 even digits. Only one digit is odd. :=> R6C2 = R8C4 and must be even; both are 8. 8 in R9 in R9C7. R5C8 = 8
R9C5 can’t be even, so R5C1 = even = 4. R9C3 = 4; 4 in C2 in R3C2; R3C7 = 2; R3C1 = 8; R7C3 = 8
Solution code: Row 5, followed by column 7.
Comments
on 29. March 2022, 10:15 by Nick Smirnov
Penpa:
https://tinyurl.com/ya6ae9ja
on 18. June 2017, 10:23 by Saskia
sehr schoen ! :)
on 6. December 2014, 12:33 by pin7guin
Schöne Kombination, hat mir gut gefallen!
