The New Sudoku Players' Forum

[Jean-Christophe]

Here is a puzzle combining both a renban and a killer.



Rules:
Fill in the puzzle such that every row and column contains all the digits from 1 to 9. This is NOT a sudoku, there is no 3x3 boxes / nonets.
The renban groups, framed with a plain thick line, must hold consecutive numbers, in any order. Numbers may not be repeated within a renban group.
eg a renban group of 3 cells may contain {456}, {645}, {123}, {789}... But {124} is not valid since numbers are not consecutive.
The numbers in sum cages, framed with a dashed line, must add up to the sum at the top left. Numbers may not be repeated within a cage.

Enjoy

[Jean-Christophe]

Did anyone solved it ?

Here is a hint:

Search for two "big entities" which almost cover two rows or columns
Findout which values may go (or cannot go) in these two entities.
Think in terms of difference of set of values, similar to Law of Leftovers.
See what set of values works with the rest.


I'll let you search for a day or two and then publish Step 1

[Jean-Christophe]

Here are the first steps :

In C56
Renban/8 in R123C56, R45C6 = all numbers except either 1 or 9
Cage 40/7 in R6789C6+R789C5 = {1456789|2356789}. The complement is 5/2 = {14|23}
LoL on C56 -> Set of numbers in R456C5 = All numbers twice (C56) - (Set of numbers in Renban/8 + Cage 40/7)
-> R456C5 = {(1|9)(14|23)}
Renban/5 in R456C5, R67C6 -> <> {9..} -> {1..} -> {12345}
Renban/8 in R123C56, R45C6 = {9..} = {23456789}, sum = 44

45 on C56 -> R456C5 = 90 - (44+40) = 6 = {123}, R67C6 = {45} (naked pair also within Cage 40/7)
-> Cage 40/7 = {1456789}
Since R78C5 = {6789}, Renban/4 in R78C5, R8C67 -> each in R8C67 >= 4
-> R9C6 = 1 (hidden single in C6)


Remember to use both information from renban groups & sum cages