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SUBMIT YOUR IDEASecond challenge done...beginner level! #StepByStep
Have always been inspired and jealous of my Prime Minister, who designed his own sudoku solver back in May 2016!
Since then I've always wanted my own design to solving one of my favourite pastimes i.e. Sudoku..
That’s why the wefie with him! Haha!
Advanced Solution:
Logic I used:
If the length of a value is the same as the occurrence of the value in a given set, then the rest of the cells in the same set must not be able to contain that value anymore.
First Example:
if there is only one occurrence of 8 in a row, then the rest of the cells in the row must not be able to have an 8.
Set considered = Rows
Value = 8,
Value Length = 1,
Occurrence = 1,
Second Example:
if there are only two occurrence of 37 (3 and 7 as remaining options in a cell) in a column, then the rest of the cells in the column must not be able to have a 3 or a 7.
Value = 37,
Value Length = 2
Occurrence = 2,
Set considered = Column
Same logic applies for the other sets, ie. Rows(9x1), Columns(1x9) & Blocks(3x3)
Please advise on how I can refine this solution (manually repeated currently), I'd really like to use a reiterative macro to improve this but yet to find a resource that is able to help me use it.
Solution (Beginner)
I'll come back to the Advanced. Promise.