So Doku

I have just been sent a little puzzle called a So Doku. Not a mathematical puzzle dispite it's appearance. It is quite addictive. There is an example of such a puzzle from here. Scroll down, there is an easy, medium, hard and very hard one.

The rules are easy, each row has to contain the digits 1-9 once and once only, each column 1-9 once and once only and each 3x3 box 1-9 omce and once only.

I hope you all waste as much time on this as I did.

Ha...looks like some fun for the lunch break. Good stuff.

[edit]Wow...I'm hooked. Because I'm lazy and don't like to go see if the solution on the web matches my solution, do you think that the following statement is true? (I'm sure someone could mathematically prove or disprove it, but the logic required for that eludes me.)

The So Doku is complete if and only if the sum of every row, the sum of every column, and the sum of every square is 45.

That's what I've used to check my solutions so far. When I do go back to the website to check, they're right. I'm just wondering if I can get a wrong solution that still satisfies the above condition.

[edit] Never mind...the counterexample of every square is 5 tosses that one. Now I'm wondering if the sums conditions and the conditions that the product of every row, every column, and every square is 362880 is enough.

Fun puzzles!

If you are systematic, those problems aren't hard. Begin by writing in every square every possible number for that square, then start erasing numbers and you'll pretty soon have the solution.

Originally Posted by adrianxw

Not a mathematical puzzle

I think this is a typical mathematical puzzle. Hardly everything in mathematics has to do with calculations.

Originally Posted by pianorain

is complete if and only if the sum of every row, the sum of every column, and the sum of every square is 45.

That is a necessary condition, but it is not sufficient.

Originally Posted by pianorain

Now I'm wondering if the sums conditions and the conditions that the product of every row, every column, and every square is 362880 is enough.

Let's see: 362880 has the following prime factors:
2 2 2 2 2 2 2 3 3 3 3 5 7

If we want to combine these into nine number we can do it in many different ways, for example:
9 9 7 5 4 4 2 2 2

Is it possible to contruct the square with these numbers? Yes! Just take a completly filled square, substitute
9 -> 9
8 -> 9
7 -> 7
6 -> 5
5 -> 4
4 -> 4
3 -> 2
2 -> 2
1 -> 2
Now you will have a square with every product of 362880, but not a valid solution to the problem.

So no, having each product equal to 362880 is not sufficient.

>>> I think this is a typical mathematical puzzle.

I don't think it is mathematical at all, purely logical. It uses numbers, but it could use a collection of nine different fruits for example such that each row, column and box had to have a complete fruit salad.

It's annoyingly addictive, I have wasted copious amounts of time on this.

Originally Posted by adrianxw

I don't think it is mathematical at all, purely logical. It uses numbers, but it could use a collection of nine different fruits for example such that each row, column and box had to have a complete fruit salad.

Yes, and that doesn't make it any less mathematical.
Combinatorics is branch of mathematics.

Originally Posted by Sang-drax

So no, having each product equal to 362880 is not sufficient.

Thanks for the explanation on exactly how to figure that out. However, you didn't really answer my question. That's probably due to the ill-stated nature of it. Here's my conjecture.

Code:

The So Doku is complete if and only if the following conditions are met:
1) The sum of every row, the sum of every column, and the sum of every square is 45.
2) The product of every row, the product of every column, and the product of every square is 362880

[edit]Once again, that's not sufficient either. However, there are only two possibilities: 1 2 3 4 5 6 7 8 9 and 1 2 4 4 4 5 7 9 9.