Mike Raiford <"m[raiford]!at"@gmail.com> wrote:
> Warp wrote:

> > 
> >   No offence, but using English as a guide for pronounciation is a joke. ;)
> > 

> It works for those who speak English.

  And who exactly is that? I assume you realize that, for example, people
from Alabama pronounce English rather differently than people from Britain,
Scotland or Australia? The exact same words may be pronounced rather
differently in these English-speaking countries.

> I don't want to get into the IPA debate again, though. I can't 
> understand IPA. I've read the Wikipedia entry on it, and I still can't 
> understand it. It's completely non-intuitive to me.

  English is not much better. It depends on where you are from how you
should pronounce something.

-- 
                                                          - Warp

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Warp wrote:
>   And who exactly is that? I assume you realize that, for example, people
> from Alabama pronounce English rather differently than people from
> Britain, Scotland or Australia? The exact same words may be pronounced
> rather differently in these English-speaking countries.

"The quantity of consonants in the English language is constant. If omitted
in one place, they turn up in another. When a Bostonian 'pahks' his 'cah',
the lost r's migrate southwest, causing a Texan to 'warsh' his car and
invest in 'erl wells'."

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Warp wrote:
>   Inside a 3x3 square, where all the numbers between 1 and 9 are put, the
> number of possible permutations of those numbers is 9! = 362880. Since
> there are 9 such squares, the total number of combined permutations is
> 9^362880. Sure, this is a huge number, but it's a fixed upper bound and
> makes the solver O(1).

Uh, shouldn't that be (9!)^9 instead of 9^(9!)?

Regards,
John
(who does his Sudoku in pen)

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John VanSickle <evi### [at] hotmailcom> wrote:
> Warp wrote:
> >   Inside a 3x3 square, where all the numbers between 1 and 9 are put, the
> > number of possible permutations of those numbers is 9! = 362880. Since
> > there are 9 such squares, the total number of combined permutations is
> > 9^362880. Sure, this is a huge number, but it's a fixed upper bound and
> > makes the solver O(1).

> Uh, shouldn't that be (9!)^9 instead of 9^(9!)?

  Yes. You are right.

  Btw, 9!^9 = 109110688415571316480344899355894085582848000000000, which
would be the total number of possible combinations of those numbers.
(Of course only a small percentage of all these combinations form valid
sudoku solutions, and if equal all symmetric positions, it gets even
smaller.)

-- 
                                                          - Warp

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On Sat, 24 Jan 2009 06:08:51 -0500, Warp wrote:

> Jim Henderson <nos### [at] nospamcom> wrote:
>> I've done several where there were clearly multiple solutions -
>> particularly at easier degrees of difficulty.
> 
>   Did you actually try to solve it in different ways?

Yes.

Jim

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Jim Henderson <nos### [at] nospamcom> wrote:
> On Sat, 24 Jan 2009 06:08:51 -0500, Warp wrote:

> > Jim Henderson <nos### [at] nospamcom> wrote:
> >> I've done several where there were clearly multiple solutions -
> >> particularly at easier degrees of difficulty.
> > 
> >   Did you actually try to solve it in different ways?

> Yes.

  I mean: did you get two different solutions, or did you arrive at the
same solution by two different paths?

-- 
                                                          - Warp

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>  If trial-and-error was acceptable in a sudoku, then you could just as
> well give an empty sudoku grid for someone to solve.

I was assuming (as do most others I think) that sudoku problems only have 1 
unique solution.  What would be the point of an empty grid?

>  The principle in sudoku puzzles is that they can be solved without having
> to guess anything. The chain of required deductions may go very deep in
> the hardest sudokus, but it's always possible to solve it without having
> to guess.

However most people would rather use trial and error than go very deep in 
logical deductions.  ie write a 3 in lightly and fill in a few more squares 
based on that, oops no that doesn't work, original square must have been a 6 
then.  Isn't trial and error a form of deduction anyway?

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>> It works for those who speak English.
>
>  And who exactly is that? I assume you realize that, for example, people
> from Alabama pronounce English rather differently than people from 
> Britain,
> Scotland or Australia? The exact same words may be pronounced rather
> differently in these English-speaking countries.

They probably pronounce sudoku in their local accent though, you would sound 
a bit weird and out-of-place if you tried to sound exactly like a Japanese 
person saying it outside of Japan.

So what you're really trying to explain to people is that the Japanese Su/Ku 
letters rhyme with the English words "do" and "to", and the Japanese Do 
rhymes with a quickly spoken "saw" (OK those might not be the only or most 
accurate English words, but you get the idea).  If you pronounce the words 
to/do/saw differently, then chances are everyone is going to pronounce 
sudoku differently anyway, so you'll be fine.

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Nicolas Alvarez wrote:

> the lost r's migrate southwest, causing a Texan to 'warsh' his car and
> invest in 'erl wells'."

God help me if I ever talk like that.

We need an agreed upon pronunciation standard that A) is easy to 
understand for the majority of folks on here, and B) Won't get Warp's 
hackles up.

-- 
~Mike

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On Sun, 25 Jan 2009 17:03:13 -0500, Warp wrote:

> Jim Henderson <nos### [at] nospamcom> wrote:
>> On Sat, 24 Jan 2009 06:08:51 -0500, Warp wrote:
> 
>> > Jim Henderson <nos### [at] nospamcom> wrote:
>> >> I've done several where there were clearly multiple solutions -
>> >> particularly at easier degrees of difficulty.
>> > 
>> >   Did you actually try to solve it in different ways?
> 
>> Yes.
> 
>   I mean: did you get two different solutions, or did you arrive at the
> same solution by two different paths?

I got two different solutions.

Jim

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