Pseudoku

For fans of Killer Sudoku, Samurai Sudoku and other variants

Pseudoku

Postby Pyrrhon » Sat Sep 23, 2006 11:10 am

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.

Image

Have fun.

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Postby Jean-Christophe » Sat Sep 23, 2006 12:31 pm

Interesting:)

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

Image
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Postby Pyrrhon » Sat Sep 23, 2006 6:30 pm

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

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Postby Smythe Dakota » Sun Sep 24, 2006 7:52 am

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

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Postby gsf » Sun Sep 24, 2006 8:47 am

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
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Postby Jean-Christophe » Sun Sep 24, 2006 8:48 am

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
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Postby Smythe Dakota » Sun Sep 24, 2006 8:13 pm

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.)

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Postby Smythe Dakota » Tue Sep 26, 2006 9:56 am

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.

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Postby udosuk » Thu Sep 28, 2006 8:45 am

Jean-Christophe wrote: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

:idea:
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Postby Pyrrhon » Thu Sep 28, 2006 1:17 pm

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.

ImagePyrrhon

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

Image
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Postby Jean-Christophe » Thu Sep 28, 2006 1:33 pm

This one is terribly hard to solve
Required ... 5 miliseconds:D
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Postby HATMAN » Fri Sep 29, 2006 7:39 am

The need for symmetry and a single solution makes it obvious.
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Postby tarek » Fri Sep 29, 2006 11:52 am

Pyrrhon wrote: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:!:

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Postby Pyrrhon » Fri Sep 29, 2006 1:16 pm

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

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Postby udosuk » Fri Sep 29, 2006 1:54 pm

Pyrrhon wrote: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:D :

Image

How many milliseconds do you need, JC?:twisted:
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