Crosswords, Sudoku, etc.

[Shiny Star]

I sometimes liked word searches. Honestly not much for puzzles, but probably should do them more.

Same here. Word searches are better than puzzles. I’ve tried crosswords and sudoku before though.  

Same. Not good at the latter two, especially sudoku. 

Me neither but I’m even worse at most crosswords.

[Detective Osprey]

Same here. Word searches are better than puzzles. I’ve tried crosswords and sudoku before though.  

Same. Not good at the latter two, especially sudoku.  

Me neither but I’m even worse at most crosswords. 

I have a hard time remembering synonyms and words sometimes.

[Kyng]

So, has anyone done any Killer Sudoku?

This is a bit like a regular Sudoku, except you're not given any of the numbers to start. Instead, the grid is divided into groups, each of which must add up to a specific number.

I tried this one... but, after over an hour, I found out that I'd messed up somewhere and had to start again .

[Kyng]

Well, I just had to give that Killer Su Doku another go after work today... and, I didn't get very far .

However, I did make some notes, on how to make a start with it:

CJ's Su Doku notes

Every row/columnn contains the numbers 1 to 9 exactly once. This means that every row and column must add up to 45.
In particular: the top row, the bottom row, the left-most column, and the right-most column, must sum to 4x5 = 180.
The numbers around the perimeter sum to (28 + 12 + 22 + 14 + 23 + 17 + 20 + 10) = 146.
This double-counts the four corners; therefore, the four corners must add up to 34.
The only way in which this can work is if the four corners are two 8s and two 9s.
Therefore, one opposing pair of corners has two 8s, and the other has two 9s

The middle column adds up to 45.
The middle column (minus the two ends) adds up to (10 + 13 + 10) = 33.
Therefore, the two ends of the middle column must add to (45 - 33) = 12.
There are three combinations that make this work: 5 and 7; 4 and 8; and 3 and 9.
However, the last two are impossible, as the 8 and 9 are already used in the corners.
Therefore, the middle top and middle bottom must be a 5 and a 7

If the middle bottom was a 5, then the two cells either side of it would add up to 12.
However, they can't be a 5 and 7, because the 5 is already used in the middle.
They can't be a 4 and 8, because the 8 is already in one corner - and they can't be a 3 and 9, because the 9 is already used in the other corner.
Therefore, there's no combination that makes this work.
This means the middle bottom must be a 7, and the top middle must be a 5

The two cells either side of the top middle add up to 7.
They can't be a 2 and a 5, because the 5 is already used in the middle.
Therefore, the two cells either side of the 5 at the top must be either a 1 and a 6, or a 3 and a 4

The two cells either side of the bottom middle add up to 10.
They can't be a 3 and 7, because the 7 is already used in the middle.
They can't be a 2 and 8, because the 8 is already in one corner - and they can't be a 1 and 9, because the 9 is already used in the other corner.
This just leaves 4 and 6.
Therefore, the two cells either side of the 7 at the bottom must be a 4 and 6

The middle row contains two pairs which add up to 10.
They can't be a 3 and a 7, because the 7 is already used at the bottom.
Therefore, the 3 must be somewhere in the middle bit, which adds up to 13

Now, the remaining numbers in the middle row consist of three pairs, which add up to 10.
These will be: a 1 and a 9; a 2 and an 8; and a 4 and a 6.
The 4 and the 6 cannot be at the top, because there will be either a 4 or a 6 to the side of the 5.
The 4 and the 6 cannot be at the bottom, because there's both a 4 and a 6 to the side of the 7.
Therefore, the 4 and the 6 are in the middle, together with the 3

The four rows on the right add up to (4 x 45) = 180.
These four rows (minus the two cells to the right of the 5 and 7) add up to (22 + 10 + 13 + 23 + 14 + 31 + 19 + 23 + 15) = 170.
Therefore, the cell to the right of the 5, and the the cell to the right of the 7, must add up to 10.
This is only possible if the cell to the right of the 5 is either a 4 or a 6.
Therefore, the number to the right of the 5 is either a 4 or a 6, and the number to the left of the 5 is either a 1 or a 3

Perhaps someone else will pick this up where I left off. Or perhaps I'll do it myself in a few months' time, when I'm either bored or in a masochistic mood .

[Pyrite]

I love crosswords, sudokus and the like. They give me a chance to think and problem solve in a simplistic environment.

[Kyng]

If anybody's interested in playing su doku on Legendary... here's one I found with only four numbers given:



Now, since that's pretty daunting, there are a few special rules to help you (or, I dunno, make it even more daunting ):

• The blue diagonals must contain each digit (from 1 to 9) exactly once;
  
• Squares that are a chess knight's move apart must not contain the same number;
  
• The grey square in the middle must form a magic square. (That is: each row, each column, and the two diagonals must add up to 15).

There is a video with the solution here, but I haven't watched it yet: I'm going to attempt the puzzle myself first !!!!

[Kyng]

Well, I managed to solve it (and I checked my solution against the one in the video afterwards). But it took up much of my evening - and I was up until 1:30am as a result.

I'm not sure it was worth it 😆.

[Kyng]

Well, I found another of these... this time with only two given digits :



Once again, there are a couple of rules to help you along:

• Positive diagonals (i.e. those running from bottom-left to top-right) must contain a different digit in every square;
  
• Adjacent digits on any positive diagonal must differ by at least 4.

I solved this one... and I think it was actually easier than the previous one, with four numbers given . (To solve this, I found a pair of squares which had to have a 6... so I started solving with a 6 in one square, then with a 6 in the other square. I kept these two 'candidate solutions' going for a while, until one of them gave me a contradiction - and then, I kept going with the other one until the end )

[Kyng]

Okay, if you thought four digits was crazy, and two digits was crazier... how about no digits ?



This one does have 'killer' cages, but you're not even given the digits for those . All you're given is the following two rules:

• Within each 3x3 box, all killer cages have the same number;
  
• No two 3x3 boxes have killer cages with the same number as each other.

It's possible, but bloody difficult (if you're interested, the solution is here). I actually took half an hour just solving the first digit on this thing! It became easier after that; however, it still took me about two hours to solve the entire thing.

Now that I've done this, I don't think I'll be doing any more weird sudoku variants for a while. If I start one, I just have to finish it - and then, it takes over my entire evening .