Double Dutch Sudoku Advent (14) - Renban vs. Point to Next
(Published on 14. 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.
Renban
Place the digits from 1 to 9 in every row, column and 3x3-block. Digits in coloured areas form Renban groups. These groups contain consecutive digits, in any order.
Point to Next
If a digit N is placed in a cell containing an arrow, then the digit N+1 must be placed in a cell pointed at by the arrow.
Solving hints
9 in R5C5
R5C2 must be 1
1 in R6 not in R6C45; 1 in R6C7 or R6C8
R5C4 must be 4; R5C3 must be 5; 6 in C3 in R23C3
R5C6 = 7; R5C89 = pair {68}
R23C5 = pair {47}
R4C5 not 3
R6C4 = 2 or 3; because of arrow in R6C4 the 3 cannot be in R6C5 or R6C6. :=> R6C5 = 6
3 in C5 locked in R78C5
R6C6 must be 5
Maximum for R6C7 = 4; R7C6 = 6; R6C7 = 4; R7C7 = 5; R6C4 = 2; R6C8 = 1
R4C7 = 2; R4C8 = 5
R3C6 = 3; R3C7 = 1
C5 and middle block can be filled now
Renban group: R4C3 must be 7; R3C3 minimum 6
Renban group: R6C3 is not 9 since 7 can't be in the group. :=>R6C3 = 3; R7C4 = 1; R7C3 = 2
1 in C9 can only go in R9C9
Solution code: Row 2, followed by Row 8
