The New Sudoku Players' Forum

[Pyrrhon]

We had a discussion about pseudoku here. I will give a new example:

Fill in the grid so that every row/column contains all the numbers from 1 to the length of that particular row/column.



Have fun.

Pyrrhon

[Jean-Christophe]

Interesting

It can also be viewed as a Latin square where the missing triangle is filled with 9s,8s... like this

[Pyrrhon]

Nice observation, it seems to be a consequence of the diagonal symmetry of the grid.

Pyrrhon

[Smythe Dakota]

So, a Latin square is nothing more than a sudoku without the boxes?

Bill Smythe

[gsf]

So, a Latin square is nothing more than a sudoku without the boxes?

yes, but the timeline is the other way: a sudoku is a latin square with box constraints
also, references to QWH (quasigroup with holes) in older forum threads mean "latin square" solving

[Jean-Christophe]

Pyrrhon, I wanted to play with your puzzle using my soft, so I turned it into a Latin square.

Hint : a swordfish will unlock it

Bill, yes a Latin square is like a sudoku without boxes. See also Wikipedia : http://en.wikipedia.org/wiki/Latin_square

[Smythe Dakota]

.... yes, but the timeline is the other way: a sudoku is a latin square with box constraints ....

I figured it would be.

Also, for a short time I was confusing latin squares with magic squares. Is it true that, if you have a 9x9 latin square, you can convert it to a magic square by adding 9*(r-1)+9*(c-1) to each cell? For example, you'd add 0 to r1c1, 9 to r1c2 and r2c1, 18 to r1c3 and r2c2 and r3c1, etc?

(I haven't really given this a whole lot of thought, so maybe I'm just being incredibly stupid.)

Bill Smythe

[Smythe Dakota]

.... Is it true that, if you have a 9x9 latin square, you can convert it to a magic square by adding 9*(r-1)+9*(c-1) to each cell? For example, you'd add 0 to r1c1, 9 to r1c2 and r2c1, 18 to r1c3 and r2c2 and r3c1, etc? ....

Oops, that's dumb. If, for example, r1c4 and r4c1 have the same digit to begin with, they'd still be the same in the resulting not-so-magic square.

But there must be SOME formula along those lines. Since a Sudoku has an average value of 5, whereas a 9x9 magic square has an average value of 41, you'd have to add an average of 36 to each cell. Can anybody come up with a formula?

By the way, do I have the right idea of a magic square? I'm thinking that each value, 1 through N-squared, must appear exactly once, and that the sum of each row and column must be the same (369 in the 9x9 case). I hope the main diagonals aren't also supposed to be the same, that would be a major monkey wrench.

Bill Smythe

[udosuk]

Hint : a swordfish will unlock it

An extra hint:

A swordfish of 5 followed by a naked quad/hidden pair on column 3 will completely solve it...

Here is the solution:

539872614
796283451
968517342
187946523
87265413
6547312
421365
34512
2134

[Pyrrhon]

Here a little addition. This give-away is called Pseudo Fun. It requires no given digits and is a valid Pseudoku. Empty diagrams can't be copyrighted. So this one is freeware.

Pyrrhon

Fill in the grid so that every row/column contains all the numbers from 1 to the length of that particular row/column.

[Jean-Christophe]

This one is terribly hard to solve
Required ... 5 miliseconds

[HATMAN]

The need for symmetry and a single solution makes it obvious.

[tarek]

Here a little addition. This give-away is called Pseudo Fun. It requires no given digits and is a valid Pseudoku. Empty diagrams can't be copyrighted. So this one is freeware

Is this a joke ??? It lives up to its name alright

tarek

[Pyrrhon]

It was planed that jcbonsai can solve it in 5 ms. Only binary sudoku is easier. Isn't it?

Pyrrhon

[udosuk]

It was planed that jcbonsai can solve it in 5 ms. Only binary sudoku is easier. Isn't it?

Not true. The following puzzle should be the undisputed world champion :



How many milliseconds do you need, JC?