Double Dutch Sudoku Advent (8) - Arrow vs. Battenburg
(Published on 8. December 2014, 12:00 by Eisbär)
Richard 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.
Arrow
Place the digits from 1 to 9 in every row, column and 3x3-block. The digit in the point of the arrow is the sum of the other digits on the same arrow.
Battenburg
Wherever 2 odd and 2 even digits form a 2x2 checkerboard pattern, a Battenburg marking is
given. If there is no marking, the above pattern is not allowed.
Solving hints
R46C5 = pair {46}; R5C46 = pair {28}
R2C5 = even, R2C6 = odd. R2C4 must be even, because of no BB-sign.
R8C5 = even, R8C4 = odd. R8C6 must be even, because of no BB-sign.
Two pairs of {28} in R2 and R8.
BB-combination in R23C23: R2C2 and R3C3 are odd. They can’t be even since of missing BB-signs. :=>R2C3 = 6; R3C2 = 8; R2C1 = 9; R4C3 = 1.
BB-combination in R23C78; R2C7 and R3C8 are even. :=> R2C7 = 4; R3C8 = 2.
One of the arrows in C5 = 1 + 8 = 9. R5C5 not 1/9.
1 in R5 in R5C89. R5C3 = 9 R6C3 must be odd
.
R7C6 = 9; R8C7 = 6; R9C8 = 3
R7C2 = 6; R8C3 = 4; R8C1 = 1; R6C3 = 7
Solution code: Row 4, followed by Column 6
Comments
on 24. October 2021, 11:31 by uvo_mod
Labels ergänzt.
on 12. December 2014, 18:19 by cornuto
Die Varianten gefallen mir immer besser. Vielen Dank dafür. Und die Lösungscodes sind ausgezeichnet ausgewählt, denn die entsprechenden Zeilen/Spalten ergeben sich bei mir immer zuletzt.
