## spent ages on this one (Australia - D.Telegraph)

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### spent ages on this one (Australia - D.Telegraph)

...from the friday paper.

The original is below. I have only found 3 more numbers (2 of them were giveaways). I'm interested if anyone can post a few more numbers *and* why they are sure of them. There's tons of squares that I have narrowed to 2 possibilities but no further...

1** 5** **4
**9 7** 8**
*** *6* *13

4** 2** ***
*8* *** *6*
*** **6 **8

65* *2* ***
**1 **7 9**
8** **4 **1

------

I have only figured out 3 definites =/ It now looks like this:

1** 5** *[9]4
**9 7** 8**
*** *6* *13

4** 2** ***
*8* *** *6*
*** **6 **8

65* *2* *[8]*
**1 **7 9**
8[9] **4 **1
pizza

Posts: 8
Joined: 22 May 2005

You definetly should take a look at column 9, you can fill in a number...

Also I'm not sure yet about the 8 in r7c8...
Last edited by Animator on Sun May 22, 2005 8:53 am, edited 2 times in total.
Animator

Posts: 469
Joined: 08 April 2005

This is a somewhat harder hint:

Take a look at colum 5.

You will find three cells that can only have the numers 3, 5 and 8. This allows you to exclude those numbers from the other cells, and that allows you to fill in the 8 in box 5 (there is only one candidate left after excluding the 3, 5 and 8 from the correct cells)
Animator

Posts: 469
Joined: 08 April 2005

The logic for the "8" in r7c8 is not straight-forward. Explanation:

1. Look at r7, by applying simple deduction, you will find that r7c4 and r7c6 must be "1" or "9".

1** 5** *[9]4
**9 7** 8**
*** *6* *13

4** 2** ***
*8* *** *6*
*** **6 **8

65* [1,9]2[1,9] ***
**1 **7 9**
8[9] **4 **1

2. Due to the "8" in r9c1, the "8" for box8 must be on r8.
ie: **7 becomes [8,?][8,?]7
The important realisation is that there must be an 8 in either of those two squares.

3. Now looking at box9, you'll see the only place for a "8" is r7c8 after you apply the known "8"s and the fact that step2 accounts for an "8" on row8.

Animator wrote:You definetly should take a look at column 9, you can fill in a number...

The missing numbers from c9 are 2,5,7,9 and i can see no way of being sure where any of them go based on current certainties.
pizza

Posts: 8
Joined: 22 May 2005

Animator wrote:This is a somewhat harder hint:

Take a look at colum 5.

You will find three cells that can only have the numers 3, 5 and 8. This allows you to exclude those numbers from the other cells, and that allows you to fill in the 8 in box 5 (there is only one candidate left after excluding the 3, 5 and 8 from the correct cells)

Yes i missed this freebie. ty!

Edit: Nope, on double check i cant see any certainties from this reply either =/
Last edited by pizza on Sun May 22, 2005 8:54 am, edited 1 time in total.
pizza

Posts: 8
Joined: 22 May 2005

pizza wrote:The missing numbers from c9 are 2,5,7,9 and i can see no way of being sure where any of them go based on current certainties.

Take a look at row 7...
r7c9:
* 2 is impossible due to r7c5
* 5 is impossible due to r7c2
* 9 is impossible due to r8c7
Animator

Posts: 469
Joined: 08 April 2005

ah ha! lol ty
pizza

Posts: 8
Joined: 22 May 2005

pizza wrote:
Animator wrote:This is a somewhat harder hint:

Take a look at colum 5.

You will find three cells that can only have the numers 3, 5 and 8. This allows you to exclude those numbers from the other cells, and that allows you to fill in the 8 in box 5 (there is only one candidate left after excluding the 3, 5 and 8 from the correct cells)

Yes i missed this freebie. ty!

Edit: Nope, on double check i cant see any certainties from this reply either =/

And I cant see any way to finish it without a look-ahead/guess method. I got this far
Code: Select all
`1.. 5.. .94..9 7.. 8.. ... .6. .13 4.. 2.. ...  .8. ... .6.  ... ..6 ..8  65. .2. .87..1 ..7 9..  897 ..4 ..1`
greengrass

Posts: 5
Joined: 01 June 2005

Do what I said earlier, take a look at column 5.

If you do things properly, then you can fill in the number 8 and the number 5 in box 5. and after that could can fill in some numbers in column 4...
Animator

Posts: 469
Joined: 08 April 2005

Ok here is my potentials
Code: Select all
`.........|.23..67..|.23..6.8.|.........|..3....8.|.23....8.|.2...67..|.........|..........23.5....|.234.6...|.........|.........|1.34.....|123......|.........|.2..5....|.2..56....2..5.7..|.2.4..7..|.2.45..8.|...4...89|.........|.2.....89|.2..5.7..|.........|..................|1.3..67..|..3.56...|.........|1.3...789|1.3.5..8.|1.3.5....|..3.5.7..|....5...9.23.5.7.9|.........|.23.5....|1.34.....|1.34..7.9|1.3.5....|12345....|.........|.2..5...9.23.5.7.9|123...7..|.23.5....|1.34.....|1.34..7.9|.........|12345....|.2345.7..|..................|.........|..34.....|1.......9|.........|1.......9|..34.....|.........|..........23......|.234.....|.........|..3..6.8.|..3.5..8.|.........|.........|..345....|....56............|.........|.........|..3..6...|..3.5....|.........|.23.56...|.23.5....|.........`

Firstly I should ask, if you agree with the possible solutions for the center block.
If you don't, where have I gone wrong.
If you agree, I still can't see the logic for locating the 8 in column 5. To my thought process the 8 in column 5 can be in row 1, or 4, or 8 or is there somethimg I am missing here?
greengrass

Posts: 5
Joined: 01 June 2005

I disagree... (also note that I said filling in 8 in box 5, not column 5)

The key is column 5...

There are three cells that can have the numbers 3, 5 and/or 8:

r1c5: 3, 8
r8c5: 3, 5, 8
r9c5: 3, 5

Only those three cells can have the numbers 3, 5 and/or 8, if you fill it in somewhere else in the column then you end up with an invalid grid (try if you don't belive my explenation).

I know it can be a problem seeing it, so think of the candidates as this:

r1c5: 3, 5, 8
r8c5: 3, 5, 8
r9c5: 3, 5, 8

Would you agree with me that based on that you can remove it from all other cells in that column? If yes, then you can start removing candidates from these cells and end up with the ones you have now...
Animator

Posts: 469
Joined: 08 April 2005

Thanks for the detailed explination. Once over this, it solved in 13 generations. If I had seen 3,5,8 in all three locations, I would have recognized a "disjointed subset" and knocked out all other 3,5,8's from the column.

Is there a name for this type of configuration?
greengrass

Posts: 5
Joined: 01 June 2005

I'm not sure how this is normally called...

some people call it a triple, perhaps you can call it a 'disjointed subset' aswell... usually I only mention the row/column/box, and the numbers, without naming it...
Animator

Posts: 469
Joined: 08 April 2005

I'm new to this and maybe I'm a bit dumb but I've worked this one out three times using the grid you've supplied and it breaks down at the same point with just a few spaces to fill.

I think there's a typo in the grid you've posted and as it is, I think it's unsolvable. R4 C4 shows a 2 in your grid but I've found that you can only complete this one if this number is changed to a 3, solution then as follows:

172 583 694
369 741 852
548 962 713

426 398 175
783 415 269
915 276 438

654 129 387
231 857 946
897 634 521

I do the Sudoku puzzles published in the London Evening Standard. They're pretty easy with the exception of last Friday's (3rd May) which is a real pig (so far) but no doubt there's some six year old Indian kid who's done it for breakfast!!
Nobby1

Posts: 2
Joined: 05 June 2005

Nobby1 wrote:I think there's a typo in the grid you've posted and as it is, I think it's unsolvable. R4 C4 shows a 2 in your grid but I've found that you can only complete this one if this number is changed to a 3......

There is a solution in 12 generations with a 2 in R4,C4
greengrass

Posts: 5
Joined: 01 June 2005

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