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# How many valid Soduko grids are there?

## How many valid Soduko grids are there?

(OP)

Hi All,

I'm not sure if this counts as a puzzle, but it's something that I've been trying to exercise my brain on these last few day:

What's the algorithm or formula for determining the number of legal ways of filling a Soduko grid?

Understand, I'm not interested in knowing what the number is (well, I suppose it would be slightly interesting). I would like to know how to go about calculating it.

I've got as far as saying that the first row will have 9! (factoral 9) combinations. For the second row, I'd guess there would be 9 * 2 * 3! (each of the 9 in the first row could go in one of two possible 3 x 3 squares, and for (the second row in) each square there would be 3! combinations).

After that, my brain starts hurting and I can't see how to proceed.

Anyone got any insights?

Mike

__________________________________
Mike Lewis (Edinburgh, Scotland)

My Visual FoxPro site: www.ml-consult.co.uk

Replies continue below

### RE: How many valid Soduko grids are there?

I wrote a Soduko solver in Javascript a while back. It didn't do brute force - which was the fun part

http://code-couch.com/jeff/snippets/sudoku/index.php

It's not perfect - gets stuck on some of the fiendish difficulty level puzzles, but it pauses to let you interact with it when this happens.

Cheers,
Jeff

Jeff's Page @ Code Couch
http://www.codecouch.com/jeff/
http://www.codecouch.com/jeff/blog/

What is Javascript? FAQ216-6094

### RE: How many valid Soduko grids are there?

(OP)

Thanks for both those replies. They didn't answer my question, but the links are interesting -- and worth exploring further.

BabyJeffy: I've toyed with the idea of writing a solver myself, but I doubt I've got the skill to do it (other than by brute force, which as you rightly say wouldn't be fun).

Mike

__________________________________
Mike Lewis (Edinburgh, Scotland)

My Visual FoxPro site: www.ml-consult.co.uk

### RE: How many valid Soduko grids are there?

> What's the algorithm or formula for determining the number of legal ways of filling a Soduko grid?

AFAIK formula is purely deductive. There are 9 blocks:

1 2 3
4 5 6
7 8 9

... each 3x3 in size. Start with #1, #5 and #9, proceed with #2 and #4 (assuming #1/#5/#9 are filled in), then #6 and #8.

Why diagonally? Because blocks within a single step don't affect each other. This simplifies formula a lot. Say, if number of permutations for #1 is N#1, then (N#1)^3 gives count for all 3 blocks in a group (#1/#5/#9) together. Ditto for #2/#4, though this is the toughest part; people unfamiliar with probabilistic math usually do it brute-force. #6/#8 group is trivial. #3/#7 are already uniquely determined with other blocks (multiplier 1 = no need to analyze). All together:

N = (N#1)^3 + (N#2)^2 + (N#6)^2 * 1

or

N = (9!)^3 * (4*9*6*2 + 4*2*2)^2 * (2^3)^2 * 1 = *** gulp! ***

------
select stuff(stuff(replicate('<P> <B> ', 14), 109, 0, '<.'), 112, 0, '/')

### RE: How many valid Soduko grids are there?

(OP)

Vongrunt,

This is fascinating. Your idea of taking the 3 x 3 grids diagonally seems to be the key. I can see that grids 1, 5 and 9 would have (9!)^3 combinations. I haven't got my head round the rest of it yet, but I'll work on it.

By the way, I reckon that:

(9!)^3 * (4*9*6*2 + 4*2*2)^2 * (2^3)^2 * 1

evalutates to approx. 6.14 * 10^23

Mike

__________________________________
Mike Lewis (Edinburgh, Scotland)

My Visual FoxPro site: www.ml-consult.co.uk

### RE: How many valid Soduko grids are there?

That is the Avogadro number ... the universal conspiration is out there :)

Cheers,
Dian

### RE: How many valid Soduko grids are there?

Dan Conspiracy, Well not really.
Avos number is 6.022E23 this is 6.14E23 if you subtract one from the other you get 11.8E20 !!! or a difference of
11,800,000,000,000,000,000,000
Easy to see how these conspiracy theories start though!!

Steve: Delphi a feersum engin indeed.

### RE: How many valid Soduko grids are there?

Yeah.. here is the one I have found long before.. was missing from the favourites. It is with  Excel VBA.
http://www.mrexcel.com/tip109.shtml

________________________________________________________
Zameer Abdulla
Help to find Missing people
Sharp acids corrode their own containers.

### RE: How many valid Soduko grids are there?

(OP)

By the way, I just bought a Sudoku book which includes a 16 x 16 grid. The bill it as "the hardest Sudoku in the world". We'll see.

Mike

__________________________________
Mike Lewis (Edinburgh, Scotland)

My Visual FoxPro site: www.ml-consult.co.uk

### RE: How many valid Soduko grids are there?

My Sudoku book has a 12x12 grid and that was pretty difficult!  I had to create an excel spreadsheet with extra large squares to track all of my candidate numbers!

Leslie

Anything worth doing is a lot more difficult than it's worth - Unknown Induhvidual

Essential reading for anyone working with databases: The Fundamentals of Relational Database Design

### RE: How many valid Soduko grids are there?

(OP)

Leslie,

I've done a few 12 x 12s. I didn't find them particularly difficult, but they do take a lot longer.

I'm saving the 16 x 16 for a long train journey I have to make next month.

Mike

__________________________________
Mike Lewis (Edinburgh, Scotland)

My Visual FoxPro site: www.ml-consult.co.uk

### RE: How many valid Soduko grids are there?

There is an article in Circuit Cellar April issue by Jeff Bachiochi on automating Sudoku puzzle solving.

### RE: How many valid Soduko grids are there?

Hi Mike,

Yuri Rubinov has posted his Sudoku game written in VFP here.

#### Quote:

Play Sudoku: create puzzle on your own, get puzzle, solve it on your own, get the solution, see how the solution works by following it step by step. Written in VFP7, but for the best of my knowledge it should run from VFP command window in versions 5/6/8/9.
I hope it helps.

### RE: How many valid Soduko grids are there?

I was pointed to this thread from another I started in the Squaring the Circle forum.  I enjoy sudoku puzzles and have found a site that has sudoku puzzles combined with cross sums:

http://www.sudoku.org.uk/killersudoku.asp

I posted my original thread trying to get ideas for a solver.  I thought about making an access database and doing quries on the criteria, but with the number of combonations, even doing just the first three rows was to much.

For these puzzles, thought about running through the different combinations of numbers for each sum (for 6: 123 132 213 231 312 321) but would still end up with quite a lot of data to run through.  I also wouldn't even know where to start on entering the criteria.  Guess I am just stuck doing these puzzles which I enjoy the most :)

Blue

If I wasn't Blue, I would just be a Dragon...

### RE: How many valid Soduko grids are there?

Hi Mike,

the number of solutions of course is very high. But if you have one solution, each solution tranlating each of the nine digits to one other would be valid.

for example replace each digit with digit+1 and replace 9 with 1.

If you see all these permutations as similar result I doubt, that there are that many really unique solutions.

Bye, Olaf.

### RE: How many valid Soduko grids are there?

I can understand how a 16x16 grid works (1-16 in each row, column, and 4x4 segment), but how does a 12x12 grid work?

I imagine 1-12 in each row and column, but how are the segments broken up?

### RE: How many valid Soduko grids are there?

here's a link that shows the variant grids.

There's also a GREAT article in the June 2006 issue of Scientific American about sudoku!

Leslie

### RE: How many valid Soduko grids are there?

Free copy of The Sims 2 or GTA: Vice City to the first one to solve this bugger.

JaG

'Very funny, Scotty... Now Beam down my clothes.'

### RE: How many valid Soduko grids are there?

Looks pretty straightforward to me. What are the actual rules? Is there 1 letter missing?

Greg
"Personally, I am always ready to learn, although I do not always like being taught." - Winston Churchill

### RE: How many valid Soduko grids are there?

so is it all letters sans "z"?

As far as solving it.. does putting the info into excel and coding it to find the answer count?

### RE: How many valid Soduko grids are there?

traingamer, one letter missing (Z) because its a 25x25 grid and the inconsiderate buggers who invented the English Alphabet put 26 letters in (Not even thinking that somebody, someday would create a game the required Squared Grids, talk about bad future-proofing)

Skie, if you can solve it in excel VBA, you go for it, but the prize condition is you must show your working (i.e. VBA Code)

Warmest regards,

JaG

'Very funny, Scotty... Now Beam down my clothes.'

### RE: How many valid Soduko grids are there?

Hi!

No, it's a very good thing, that there are 26 letters, you can do a 36x36 grid including the ten digits.

What would that be called? Hexatricosimal system?

Bye, Olaf.

### RE: How many valid Soduko grids are there?

Hi JAG,

solved it (ctrl+A for the spoiler):

#### CODE

etkosudbriwlanxfpycgqmvhj
fgvhlcpyqnjtoisrxembkudwa
dayiphfowkcmbugtqvjnslerx
vluyteimnqfgkcjdbarphxsow
bkmsgpvcjriduxahftownylqe
hxpqodkwlfbsnvycujgetrami
wirfdtuasbqoehlnkmyxgpjvc
cnaejxogyhmwtrpvsliqdkufb
xybgwvnqfeuilmhptksdcjoar
krdpfmhlgwaqxtojcibyvenus
tocuarbpdskjywnelgvhxqmif
sqnjhiytucpfdevaroxmlwgbk
lvemiaxkojscgbrqnwufpdhty
ihtkrlqsboyvpgwmecnujafxd
qdgxmytucpnkifbwosajehrlv
pfoacnerimxujdqlyhkvbtwsg
uwlbefjvkgoahstxiqdrmncyp
nujwbkshpltymaigvrecofxdq
aexlyjrnvuhbqkcodftiwspgm
gpidkqwemyvxsofujbharctnl
omhcqbgfatlnrjdywxpsivkeu
rsftvocixdgewpukmnqlabyjh

Here's the code used (Visual FoxPro). Beware of broken lines.

#### CODE

*#Define cnBase 3
*(3x3)x(3x3) = 9x9 Grid
*#Define ccCharacters "123456789"
*#Define ccSodukoSDF "D:\my\vfp9\soduko\simplesoduko.sdf"

#Define cnBase 5
*(5x5)x(5x5) = 25x25 Grid
#Define ccCharacters "abcdefghijklmnopqrstuvwxy"
#Define ccSodukoSDF "D:\my\vfp9\soduko\abcsoduko.sdf"

Local lnCount, lcFields
Public gcGridFields
gcGridFields=""
lcFields=""
For lnCount = 1 To cnBase*cnBase
lcFields = lcFields + "," + Chr(lnCount+64)+" C(1)"
gcGridFields = gcGridFields + "," + Chr(lnCount+64)
Endfor
gcGridFields = Substr(gcGridFields,2)
lcFields = Substr(lcFields,2)

Create Cursor curGrid (&lcFields, iLevel I)
Append From (ccSodukoSDF) Type Sdf
Replace All iLevel With 1

Local Array laGrid[1]
Select &gcGridFields From curGrid Into Array laGrid

Create Cursor curSteps (iRow I, iCol I, cLetter C(1), iLevel I, iSolvingCase I)
Clear
solvesudoku(@laGrid,2)

Select curGrid
Delete All In curGrid
Append From Array laGrid
Browse

Procedure solvesudoku()
Lparameters taGrid, tnLevel

Local Array laUsable[cnBase*cnBase,cnBase*cnBase]
Local lnCount, lnCount2, lcUsed, lcUsable
Local liRow, liCol

llContinue = .T.
Do While llContinue
llContinue = .F.

* determine used digits/letters of each row, column, block
Local Array laRows[cnBase*cnBase]
Local Array laCols[cnBase*cnBase]
Local Array laBlks[cnBase*cnBase]

For lnCount=1 To cnBase*cnBase
lcRowUsed = ""
lcColUsed = ""
lcBlkUsed = ""
For lnCount2=1 To cnBase*cnBase
lcRowUsed = lcRowUsed + taGrid[lnCount , lnCount2]
lcColUsed = lcColUsed + taGrid[lnCount2, lnCount ]
lcBlkUsed = lcBlkUsed +;
taGrid[Int((lnCount-1)/cnBase)*cnBase +Int((lnCount2-1)/cnBase)+1,;
(lnCount-1)%cnBase*cnBase+(lnCount2-1)%cnBase+1]
Endfor
laRows[lnCount] = lcRowUsed
laCols[lnCount] = lcColUsed
laBlks[lnCount] = lcBlkUsed
Endfor

* determine usable digits/letters of each field
For lnCount=1 To cnBase*cnBase
For lnCount2=1 To cnBase*cnBase
If !Empty(taGrid[lnCount , lnCount2])
Loop
Endif
lcUsed = laRows[lnCount]+laCols[lnCount2]+laBlks[Int((lnCount-1)/cnBase)*cnBase+Int((lnCount2-1)/cnBase)+1]
laUsable[lnCount,lnCount2] = Chrtran(ccCharacters,lcUsed,"")
If Len(laUsable[lnCount,lnCount2])=0
* error, no digit/letter available for a field!
Return .F.
Endif
If Len(laUsable[lnCount,lnCount2])=1
* only one digit/letter available, then take it!
taGrid[lnCount,lnCount2] = laUsable[lnCount,lnCount2]
laUsable[lnCount,lnCount2] = .F.
Insert Into curSteps Values (lnCount,lnCount2, taGrid[lnCount,lnCount2], tnLevel,1)
llContinue = .T.
Exit
Endif
Endfor
Endfor

If llContinue
Loop
Endif

* examine usable digits/letters of rows
For lnCount=1 To cnBase*cnBase
lcUsable = ""
For lnCount2=1 To cnBase*cnBase
If !Empty(taGrid[lnCount , lnCount2])
Loop
Endif
lcUsable = lcUsable + laUsable[lnCount,lnCount2]
Endfor

* any letter only once available?
lcLetter=""
For lnCount2=1 To Len(lcUsable)
If Occurs(Substr(lcUsable,lnCount2,1),lcUsable)=1
* yes!
lcLetter = Substr(lcUsable,lnCount2,1)
Exit
Endif
Endfor

If !Empty(lcLetter)
For lnCount2=1 To cnBase*cnBase
If !Empty(taGrid[lnCount,lnCount2])
Loop
Endif
If lcLetter $laUsable[lnCount,lnCount2] taGrid[lnCount,lnCount2] = lcLetter laUsable[lnCount,lnCount2] = .F. Insert Into curSteps Values (lnCount,lnCount2, taGrid[lnCount,lnCount2], tnLevel,2) llContinue = .T. Exit Endif Endfor Endif If llContinue Exit Endif Endfor If llContinue Loop Endif * examine usable digits/letters of cols For lnCount=1 To cnBase*cnBase lcUsable = "" For lnCount2=1 To cnBase*cnBase If !Empty(taGrid[lnCount2, lnCount]) Loop Endif lcUsable = lcUsable + laUsable[lnCount2,lnCount] Endfor lcLetter="" For lnCount2=1 To Len(lcUsable) If Occurs(Substr(lcUsable,lnCount2,1),lcUsable)=1 lcLetter = Substr(lcUsable,lnCount2,1) Exit Endif Endfor If !Empty(lcLetter) For lnCount2=1 To cnBase*cnBase If !Empty(taGrid[lnCount2,lnCount]) Loop Endif If lcLetter$ laUsable[lnCount2,lnCount]
taGrid[lnCount2,lnCount] = lcLetter

laUsable[lnCount2,lnCount] = .F.
Insert Into curSteps Values (lnCount2, lnCount, taGrid[lnCount2,lnCount], tnLevel,3)
llContinue = .T.
Exit
Endif
Endfor
Endif
If llContinue
Exit
Endif
Endfor

If llContinue
Loop
Endif

* examine usable digits/letters of blocks
For lnCount=1 To cnBase*cnBase
lcUsable = ""
For lnCount2=1 To cnBase*cnBase
liRow = Int((lnCount-1)/cnBase)*cnBase +Int((lnCount2-1)/cnBase)+1
liCol = (lnCount-1)%cnBase*cnBase+(lnCount2-1)%cnBase+1
If !Empty(taGrid[liRow,liCol])
Loop
Endif
lcUsable = lcUsable + laUsable[liRow,liCol]
Endfor

lcLetter=""
For lnCount2=1 To Len(lcUsable)
If Occurs(Substr(lcUsable,lnCount2,1),lcUsable)=1
lcLetter = Substr(lcUsable,lnCount2,1)
Exit
Endif
Endfor

If !Empty(lcLetter)
For lnCount2=1 To cnBase*cnBase
liRow = Int((lnCount-1)/cnBase)*cnBase +Int((lnCount2-1)/cnBase)+1
liCol = (lnCount-1)%cnBase*cnBase+(lnCount2-1)%cnBase+1
If !Empty(taGrid[liRow,liCol])
Loop
Endif
If lcLetter \$ laUsable[liRow,liCol]
taGrid[liRow,liCol] = lcLetter

laUsable[liRow,liCol] = .F.
Insert Into curSteps Values (liRow, liCol, lcLetter, tnLevel,4)
llContinue = .T.
Exit
Endif
Endfor
Endif
If llContinue
Exit
Endif
Endfor

Enddo

* no sure letter, so simply try and error now:
Local lcLeft,lcRight
For lnCount=1 To cnBase*cnBase
For lnCount2=1 To cnBase*cnBase
* two letters available only?
If Empty(taGrid[lnCount,lnCount2]) And Len(laUsable[lnCount,lnCount2])=2
* try both
lcLeft = Left(laUsable[lnCount,lnCount2],1)
lcRight = Right(laUsable[lnCount,lnCount2],1)
Select curGrid
Append From Array taGrid
Replace All iLevel With tnLevel For iLevel=0
laUsable[lnCount,lnCount2] = .F.
taGrid[lnCount,lnCount2] = lcLeft
Insert Into curSteps Values (lnCount, lnCount2, lcLeft, tnLevel,5)

If !solvesudoku(@taGrid, tnLevel+1) &&recurse
* restore Grid from before recursion
Select &gcGridFields From curGrid Where iLevel = tnLevel Into Array taGrid
* restore Grid log
Select * From curGrid Where iLevel<=tnLevel Into cursor curGrid readwrite
* restore steps from before recursion
Select * From curSteps Where iLevel<=tnLevel Into Cursor curSteps Readwrite
* including deletion of the last step
Go Bottom In curSteps
Delete Next 1 In curSteps

taGrid[lnCount,lnCount2] = lcRight
Insert Into curSteps Values (lnCount, lnCount2, lcRight, tnLevel,6)

If !solvesudoku(@taGrid, tnLevel+1) &&recurse
* restore Grid from before recursion
Select &gcGridFields From curGrid Where iLevel = tnLevel Into Array taGrid
* also restore Grid log
Select * From curGrid Where iLevel<tnLevel Into cursor curGrid readwrite

* restore steps from before recursion
Select * From curSteps Where iLevel<=tnLevel Into Cursor curSteps Readwrite
* including deletion of the last step
Go Bottom In curSteps
Delete Next 1 In curSteps

Return .F.
Endif
Endif
Endif
Endfor
Endfor
Endproc

A bit ugly, could have defined some more subroutines, but it's configurable for the normal kind of soduko, just define cnBase, ccCharacters and the initial grid ccSodukoSDF correspondingly.

Additional to the final solution you can see some versions of the grid each time the code went into recursion and curSteps contains all single steps done to find the solution (minus the ones that didn't lead to the solution).

The initial grid is read in using a sdf type format, that is simply an ascii text with 25 lines and the characters without any delimiter, eg for your soduko:

#### CODE

t  s    iwl nxfpy  qmv
fg  lc   njt  s x  bku  a
day  h  wk  b  t vj     x
bq   lj a  vyes u     cn
luyte mn fg  jdb r h so
k   pv  r  uxah t  nyl
x     wl    v  uj  trami
r   u sbqoe  n m      c
a j  g  mwt    liqd
eu  m p  s c oa
kr  fm lg  qx   c byve
toc arb    jy ne   hx mi
n h y   p  evaro m  gbk
ve i x ojs gbrq w  pd ty
krlq b  vpg mecnu af d
g   t   n i b         v
yjsv wa h     m  pf   q o
pf  cneri xuj  ly kv  w
be  v g ahstx   rmnc
u  b sh     a  vr c  xdq
a xl j n  hbqk    t
i k  em vxs   j  arctn
om    gfa  n  d wxpsivke
rs t  c  dg  pu m   ab

Bye, Olaf.

### RE: How many valid Soduko grids are there?

JAG14,

There is no one single solution to that Sudoku puzzle.  After a great amount of work (I was solving it semi-manually), I was able to find multiple solutions that work.

While a brute force approach will find a solution, I was using a "proof by contradiction" approach to try to find the solution, only to find that no such thing exists.

Thanks for the puzzle, though.  It was lots of fun!

### RE: How many valid Soduko grids are there?

I would think "IMHO" that is a standard 9X9 layout
the first block of the first row you can use any of 9 numbers.
the second block of the first row you have one number already dedicated so can choose any of 8 numbers.
and so forth down to the the last block which leaves you only one number to choose from.
First row straight factor of 9.
9*8*7*6*5*4*3*2=362,880

Second row you can not duplicate the number above so you have 8 numbers to choose from.  Next block is where it gets confusing...  Conceivably, you could still have 8 numbers to choose from assuming that you used the number above this block in the first block of this row.  If not then you only have 7 numbers you can choose from.  next block would be only 7 numbers. so I believe it would be:
8*8*7*6*5*4*3*2=322,560

following this logic the next row would be
7*7*7*6*5*4*3*2=246,960
and then:
6*6*6*6*5*4*3*2  and so on and so forth.

But like I said that is only IMHO, so all are more than welcome to call me an idiot and explain it in their own terms.

Shut up and load the Photon Torpedoes...

### RE: How many valid Soduko grids are there?

Hi Tanselm,

You're just considering rows and columns. You also need to satisfy the once per block constraint. But the idea is good. I'd say it comes down to this:

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

And that isn't even considering rotated and mirrored solutions or permutations, which also shouldn't be considered as unique solutions.

Possible permutations are 9*8*7*6*5*4*3*2*1, possible rotations *4, possible mirroring *2.

Which would leave something like 6*5*3*3*2*6*5*4*3*2*3*2*3*2 = 13996800

Still not sure if that takes into account the dependancies of row, column and block correctly.

Bye, Olaf.

### RE: How many valid Soduko grids are there?

Olaf,

You seem to be failing to take into account that there can be some overlap of numbers, so instead of:

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

I venture it would look more like:

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

However, even this solution is oversimplified.  I think vongrunt had a good solution posted above.

### RE: How many valid Soduko grids are there?

Shut up and load the Photon Torpedoes...

### RE: How many valid Soduko grids are there?

Hi KornGeek!

No, I doubt this one:

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

If I fill in from start to end rowwise, if I get to the second row and the second block, there are NOT 6 digits you can choose from, there are 3 digits already used within the 2nd row  and 3 digits within the same block, therefore only in the case the three digits in the same block are identical to the 3 digits of the second row so far, you can choose between any of the 6 remaining digits.

Our schemes each show the most/least possible digits remaining. The factor you need to take should be inbetween, so I fear we are both wrong and the true calculation can't be laid out this way.

Bye, Olaf.

### RE: How many valid Soduko grids are there?

I believe the original question was the maximum possible layouts so yes wou would assume whatever you need to for the maximum.

Shut up and load the Photon Torpedoes...

### RE: How many valid Soduko grids are there?

The original question can be read, can't it. It's about " the number of legal ways of filling a Soduko grid". Not a maximum that is only a rough calculation being a limit to the number of solutions.

Bye, Olaf.

### RE: How many valid Soduko grids are there?

Here you go:

http://en.wikipedia.org/wiki/Mathematics_of_Sudoku

It's much higher as my result, but I said:

#### Quote (myself):

Still not sure if that takes into account the dependancies of row, column and block correctly.

The article says this number was computed through a mixture of logic and brute force computation, so there seems to be no simple formula which you could also apply to the 25x25 grid.

Even taking symmetries into account the number is still higher: 5,472,730,538. As that is not dividable by 9! I'd say permutations are already considered too.

Nevertheless the number of valid sudoku grids is so high, that the industry of sudoku puzzle magazines can thrive quite a long time. It's more probable, that people get tired of these puzzles than that every possible puzzle is ever printed.

Bye, Olaf.

### RE: How many valid Soduko grids are there?

#### Quote:

Our schemes each show the most/least possible digits remaining. The factor you need to take should be inbetween, so I fear we are both wrong and the true calculation can't be laid out this way.

Olaf,
Agreed.  As I stated in my post, I was oversimplifying it because I was going for the maximum (mostly to highlight the possibilities excluded in your scheme).  The bottom line is that the mathematics of this are a little beyond me at the moment.  Thanks for the link.  I'll check it out when I get a chance.

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