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

18 posts
• Page **1** of **2** • **1**, 2

Smythe Dakota wrote: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

- gsf
- 2014 Supporter
**Posts:**7306**Joined:**21 September 2005**Location:**NJ USA

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

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

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

gsf wrote:.... 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
**Posts:**534**Joined:**11 February 2006

I wrote:.... 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

- Smythe Dakota
**Posts:**534**Joined:**11 February 2006

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.

Pyrrhon

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

- Pyrrhon
**Posts:**240**Joined:**26 April 2006

18 posts
• Page **1** of **2** • **1**, 2