logic? i need convincing (NZ Herald - 56949)

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logic? i need convincing (NZ Herald - 56949)

Postby jayal » Sun Jun 12, 2005 9:18 am

from NZ Herald - 56949 - this rated HARD - when all options are filled there remains 3 cells with a choice of 2 numbers, 3 & 4 -
column 4 - row 8
column 9 - rows 7&9
take a punt! - if you get lucky and pick column 9, row 7 for a 4 the rest fall into place - pick one of the other 2 and failure
now where's the logic in that, surely it's trial & error?
please prove me wrong!
jayal
 
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Postby Animator » Sun Jun 12, 2005 11:16 am

It would be lots easier if you posted the grid...

Also note that r8c4 is in a different box... if that is the only empty cell in that box then you can fill it in.

My best guess however looking at the solutions page is to see that row 8 and row 9 already have the number 4. This means there is only one place on row 7 where it can go...

(Each 3x3 box/square has to have the numbers 1 to 9 remember?)
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Postby jayal » Sun Jun 12, 2005 10:18 pm

*56 *7* **9
41* 9*3 576
973 5*6 **1

5** *37 *18
**7 6*1 4*5
*2* *5* **7

**5 7** 98*
*8* **5 762
7** *** 15*

the grid with my entries as far as i could go
jayal
 
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Postby shakers » Mon Jun 13, 2005 11:24 am

jayal wrote:*56 *7* **9
41* 9*3 576
973 5*6 **1

5** *37 *18
**7 6*1 4*5
*2* *5* **7

**5 7** 98*
*8* **5 762
7** *** 15*

the grid with my entries as far as i could go


In Row 4 the only possibly places that a 9 could go are in R4C2 or R4C3, this means that the 9 can be elimated as a candidate for R5C2.

There is only one other possible value for R5C2 which, when placed, should get you going again.

Rgds,
Richard

PS. Not that it helps your problem, but you could have got one more number without having to think too hard - R1C4 is the only possible cell for a 1 in box 2.
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Postby Animator » Mon Jun 13, 2005 11:29 am

I really don't understand the sentence: 'when all options are filled'...

My guess what that you were saying that all other cells are filled, but that clearly is not the case in the grid you posted... so what are you talking about?
Animator
 
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Postby shakers » Mon Jun 13, 2005 11:31 am

Animator wrote:I really don't understand the sentence: 'when all options are filled'...

My guess what that you were saying that all other cells are filled, but that clearly is not the case in the grid you posted... so what are you talking about?


I think it's a reference to pencilmarks
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Postby simes » Mon Jun 13, 2005 12:09 pm

From your part completed grid, there is a completely "logical" solution i.e. no T&E required.

If all else fails, put your grid into one of the automatic solvers and see what it throws up.
Last edited by simes on Sun Dec 11, 2011 9:57 am, edited 1 time in total.
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logic

Postby jayal » Tue Jun 14, 2005 2:43 am

ow! - missed that 1 - thanks "shakers"
"all options filled?" yeah, pencilled in options

"Simes" - am i right in saying there are three choices for 3 & 4 in boxes 8 & 9? - that being the case surely T & E, but if i'm wrong guess i'm not as good as i thought - i did solve it but as said using T & E
jayal
 
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Postby simes » Tue Jun 14, 2005 6:25 am

I don't think it's that complicated. Starting with the grid above, my solver gives:
Code: Select all
Note: Cell coordinates are in (row,column) format with (1,1) at top-left and (9,9) at bottom-right.
(1,4) = 1 : only cell in row 1 that can contain 1
blocks 5 and 6 must contain 9 in rows 5 and 6 - removing 9 from candidate for cell(s) (5,2)(6,3)
(5,2) = 3 : only possible value for this cell
(5,1) = 8 : only possible value for this cell
...
Last edited by simes on Sun Dec 11, 2011 9:57 am, edited 1 time in total.
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Postby jayal » Thu Jun 16, 2005 10:51 am

There are some that would argue trial and error is not a logical technique, and is no better than guessing. Although it's not a technique I like to use, I do consider it logical. When further moves seem impossible, trial and error may be the only way forward. Indeed, some puzzles cannot be completed without it.

The technique involves selecting one candidate for a cell - without any particular reason for that selection - and then seeing whether the puzzle can then be completed. If it can, well done. If not, the trial and error move, and any subsequent moves, are undone, and a different choice is made. For some puzzles, it may be necessary to use trial and error several times. For others, it may be required only once.

In order to better manage the complexity, it's usual, if possible, to choose a cell with only two candidates, but that doesn't have to be the case.

It's worth noting, that this technique becomes the equivalent of a brute-force attack

and so, i rest my case
jayal
 
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Postby simes » Thu Jun 16, 2005 11:45 am

Those words look kinda familiar. Oh, no wonder, they're mine.

It's only an opinion though, hardly a solid foundation to rest your case on.

But I thought you were saying T&E was required to solve this puzzle?
Last edited by simes on Sun Dec 11, 2011 9:57 am, edited 1 time in total.
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Postby scrose » Thu Jun 16, 2005 1:28 pm

jayal wrote:There are some that would argue trial and error is not a logical technique

It's just as easy to state the reverse: There are some that would argue that logic is a form of trial-and-error.

Trial-and-error can be viewed as "if I put the number N in cell C, will the grid become invalid?"

Logical techniques of elimination can be viewed as "the grid will immediately become invalid if I put the number N in cell C".

These are very similar statements, which is why the line between logic/trial-and-error is so blurry. The slight distinction between the two being that (when the number N is placed in cell C) in trial-and-error, the player is not sure if the grid will become invalid, whereas by logical techniques of elimination the player can immediately determine if the grid will become invalid.

This puzzle certainly did not require trial-and-error. Once you have placed a 1 at r1c4 and a 3 at r5c2 the rest of the puzzle unravels quite easily.

Fixed: My incorrect usage of converse.
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