Largest number of empty groups?

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Largest number of empty groups?

Postby tso » Fri Aug 12, 2005 9:41 pm

Rows, columns and boxes number 1, 5 and 9 are empty.
All other cells that are NOT in these 9 groups have clues.
The puzzle has a unique solution and is not difficult.

Code: Select all
 . . . | . . . | . . .
 . . . | 6 . 5 | 1 4 .
 . . . | 4 . 7 | 3 2 .
-------+-------+------
 . 5 1 | . . . | 4 6 .
 . . . | . . . | . . .
 . 8 3 | . . . | 2 1 .
-------+-------+------
 . 6 7 | 5 . 8 | . . .
 . 1 2 | 7 . 6 | . . .
 . . . | . . . | . . .


Is this the maximum number of empty rows, columns and boxes for a valid puzzle?
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Postby richardginn » Wed Aug 17, 2005 11:58 pm

that puzzle just looks plain nasty!!!!
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Postby tso » Thu Aug 18, 2005 3:53 am

Actually, nothing but naked singles and hidden singles required to solve.
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A conjecture: 4 is the maximum number of empty boxes

Postby Big Blue » Fri Aug 19, 2005 7:41 am

I fooled around a bit and I believe that 4 is the maximal number of empty boxes that may lead to a unique solution.

Here are some hints why this conjecture may be true:

1. It appears to be impossible to construct a Sudoku with 5 empty boxes

Reason: The best attempt seems to

0X0
X0X
0X0

where 0 (X) implies an empty (full) box, respectively. By full box I mean 9 clues in that box (although of course 8 are sufficient and in several cases less than 8 might be enough, but let me keep things as simple as possible).

With such a grid of clues it is trivial to construct the group in the centre, but one can show easily that for the 4 corner groups there are always at least 3 options for each cell, in most cases 4, sometimes even more.

This doesn't imply that there is no Sudoku of that type which could exhibit a unique solution, but it seems unlikely to me.

2. The inverted grid to the one above looks promising:

X0X
0X0
X0X

My two attempts by hand led to one Sudoku with 2 bits of ambiguity and to another one with 3 bits of ambiguity.

Thus, either my naive construction at random was too simple, or these ambiguities are a generic feature. I tend to believe that the latter is not the case.

So for all the computer guys here is a nice homework:

A) Show that no Sudoku with 5 empty boxes is possible, no matter how the clues are arranged

B) Construct a Sudoku with 4 empty boxes, possibly (but not necessarily) with the grid of clues indicated above.

PS: Edited, because I focussed on boxes rather than groups - that is maybe a slightly different topic. So I essentially replaced everywhere "group" by "box" - sorry for this notational slip...
Last edited by Big Blue on Fri Aug 19, 2005 8:01 am, edited 1 time in total.
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Re: Largest number of empty groups?

Postby Addlan » Fri Aug 19, 2005 10:18 am

tso wrote:Rows, columns and boxes number 1, 5 and 9 are empty.
All other cells that are NOT in these 9 groups have clues.
The puzzle has a unique solution and is not difficult.

Code: Select all
 . . . | . . . | . . .
 . . . | 6 . 5 | 1 4 .
 . . . | 4 . 7 | 3 2 .
-------+-------+------
 . 5 1 | . . . | 4 6 .
 . . . | . . . | . . .
 . 8 3 | . . . | 2 1 .
-------+-------+------
 . 6 7 | 5 . 8 | . . .
 . 1 2 | 7 . 6 | . . .
 . . . | . . . | . . .


Is this the maximum number of empty rows, columns and boxes for a valid puzzle?


I like this puzzle very much. I would like to copy it to the thread 'my favorate Sudoku puzzle'.
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Re: A conjecture: 4 is the maximum number of empty boxes

Postby tso » Fri Aug 19, 2005 5:39 pm

Big Blue wrote:IB) Construct a Sudoku with 4 empty boxes, possibly (but not necessarily) with the grid of clues indicated above.


Code: Select all
5 8 . . . . . 6 2
7 . 6 . . . 3 . 1
. 9 . . . . . 7 .
. . . 3 4 2 . . .
. . . 9 . 7 . . .
. . . 8 5 1 . . .
. 7 . . . . . 8 .
4 . 3 . . . 9 . 6
1 2 . . . . . 3 4
Last edited by tso on Sat Aug 20, 2005 4:28 pm, edited 1 time in total.
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Postby tso » Sat Aug 20, 2005 8:20 pm

This one doesn't have 5 empty boxes, but all the clues are restricted to 4 separate 3x3 areas -- and there are 6 empty rows and columns. Unfortunately, it is very, very easy:

Code: Select all
. . . . . . . . .
. 2 1 9 . . 7 . .
. . . . . 4 8 5 .
. 1 7 2 . . 9 . .
. . . . . . . . .
. . 3 . . 8 6 4 .
. 4 5 7 . . . . .
. . 9 . . 5 1 3 .
. . . . . . . . .


I've seen this pattern in published puzzles before. Interesting that though there are 6 empty groups and only 22 clues, the puzzle remains elementary. Pretty though.
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Postby udosuk » Sun Aug 21, 2005 8:51 am

It seems puzzles with very few clues are not unusual to be easier to solve. Like Gfoyle's unique 17 clue puzzle with 5 digit appearing once and 4 digit appearing 3 times...

And most fiendish puzzles I see have clues of 27+...
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Postby Nick70 » Sun Aug 21, 2005 9:41 am

tso wrote:Interesting that though there are 6 empty groups and only 22 clues, the puzzle remains elementary.

My theory is that for any given pattern of clues, the majority of puzzles that can be obtained are elementary.
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Postby dukuso » Sun Aug 21, 2005 1:21 pm

Nick70 wrote:
tso wrote:Interesting that though there are 6 empty groups and only 22 clues, the puzzle remains elementary.

My theory is that for any given pattern of clues, the majority of puzzles that can be obtained are elementary.



take a 17-clue sudoku, which isn't elementary.
You can probably not assign other values to the clues
to make a nonequivalent sudoku.
So all sudokus with this pattern are probably non-elementary.
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Postby Nick70 » Sun Aug 21, 2005 8:11 pm

dukuso wrote:take a 17-clue sudoku, which isn't elementary.
You can probably not assign other values to the clues
to make a nonequivalent sudoku.
So all sudokus with this pattern are probably non-elementary.

Maybe. However, it still hasn't been proven that any 17-clues arrangement is unique.

In the meantime, here are some non-trivial puzzles using tso's proposed arrangement.
The last two are much harder than the others.

Code: Select all
. . . . . . . . .
. 9 5 2 . . 4 . .
. . . . . 7 3 8 .
. 2 4 9 . . 5 . .
. . . . . . . . .
. . 3 . . 1 8 6 .
. 6 2 7 . . . . .
. . 1 . . 8 9 7 .
. . . . . . . . .


. . . . . . . . .
. 9 5 2 . . 4 . .
. . . . . 7 3 8 .
. 5 4 9 . . 2 . .
. . . . . . . . .
. . 3 . . 1 8 6 .
. 6 2 7 . . . . .
. . 1 . . 8 9 7 .
. . . . . . . . .


. . . . . . . . .
. 2 5 1 . . 4 . .
. . . . . 7 3 8 .
. 1 4 2 . . 5 . .
. . . . . . . . .
. . 3 . . 5 8 9 .
. 6 2 7 . . . . .
. . 1 . . 8 9 7 .
. . . . . . . . .


. . . . . . . . .
. 9 5 4 . . 7 . .
. . . . . 2 3 4 .
. 7 4 9 . . 1 . .
. . . . . . . . .
. . 3 . . 5 8 2 .
. 6 2 7 . . . . .
. . 1 . . 8 9 7 .
. . . . . . . . .


. . . . . . . . .
. 3 5 2 . . 6 . .
. . . . . 7 8 4 .
. 9 4 6 . . 5 . .
. . . . . . . . .
. . 3 . . 4 1 8 .
. 6 2 7 . . . . .
. . 1 . . 8 9 7 .
. . . . . . . . .


. . . . . . . . .
. 2 5 6 . . 4 . .
. . . . . 4 3 8 .
. 9 4 2 . . 1 . .
. . . . . . . . .
. . 3 . . 1 8 9 .
. 6 2 7 . . . . .
. . 1 . . 8 9 7 .
. . . . . . . . .
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Postby Anette » Sun Aug 21, 2005 9:34 pm

Nick70, do I need forcing chains or T & E to solve those? I started with the first and got to:

{134}{134}{78}{14568}{14569}{459}{267}{259}{5679}
{78}{9}{5}{2}{68}{3}{4}{1}{67}
{2}{14}{6}{145}{1459}{7}{3}{8}{59}
{178}{2}{4}{9}{78}{6}{5}{3}{17}
{6}{1578}{789}{3458}{3457}{245}{27}{249}{1479}
{579}{57}{3}{45}{2457}{1}{8}{6}{2479}
{3459}{6}{2}{7}{3459}{459}{1}{45}{8}
{345}{345}{1}{3456}{23456}{8}{9}{7}{23456}
{34579}{34578}{789}{13456}{134569}{2459}{26}{245}{3456}

Then the only thing I've spotted so far is to follow the 7's for both alternatives in box 1:

A: R1C3 = 7; then R2C9 = 7, then R4C9 =1
B: R2C1 = 7; then R2C9 = 6; then R2C5 = 8, then R4C5 = 7; then R4C9=1

But is there a way to do this without T & E?

/Anette
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Postby Nick70 » Sun Aug 21, 2005 11:05 pm

Anette wrote:Nick70, do I need forcing chains or T & E to solve those?

You can do the first four with xy-wings plus the other usual paraphernalia.
The last two require multiple forcing chains.
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Thanks...

Postby Big Blue » Mon Aug 22, 2005 7:51 am

...for the nice (and prompt) example with 4 empty boxes, tso. You get extra points for removing superfluous clues:)

I think proving that 5 empty boxes are impossible might be a bit tricky - nogo results tend to be more difficult to obtain than examples.

BTW, I am not 100% sure that the suggested pattern,

0X0
X0X
0X0

really is the best attempt. It could be that instead

X0X
000
X0X

provides better possibilities - or some combination of these two patterns, like e.g.

X00
0X0
X0X

(incidentally, this is similar to tso's 4 box-example, with the notable difference that box number 3 is empty as well). Maybe this last example is more promising - perhaps it can be constructed such that you can fill in (in this order) box 4,8,2,6,3
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Re: Thanks...

Postby gfroyle » Mon Aug 22, 2005 10:49 am

Big Blue wrote:..It could be that instead

X0X
000
X0X



Not this one.. there are no valid Sudoku puzzles with this pattern...

gordon
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