## Fiendish 148 19th May

All about puzzles in newspapers, magazines, and books
This one was the toughest one I've done. Does anyone use the fact that Wayne Gould builds some sort of symmetry into his puzzles (I think he wrote that the clues are distributed symmetrically) to their advantage? How can that fact be employed to figure out where to look for clues?
Last edited by determined on Fri May 20, 2005 9:07 pm, edited 1 time in total.
determined

Posts: 5
Joined: 19 May 2005

### Times 148

Arnie2 wrote:Thanks for the v prompt reply...I'm not quite there...

1. 2 and 8 are "paired" - I see that
2. in column 7 7 must go in r 5 or 6 only
3. I cannot see anything interesting in row 6 nor place a furthernumber in box 4

Thicko here!

I understand and agree with the reply to 1 above but 2 totally escapes me! How is the 7 eliminated from row 1 and row 3? If it is eliminated from col7/row 3, as is suggested by the answer above, then the only possibility for this cell is 9, I think - anyone disagree?
Guest

### Re: Times 148

Steve Smith wrote:
Thicko here!

I understand and agree with the reply to 1 above but 2 totally escapes me! How is the 7 eliminated from row 1 and row 3? If it is eliminated from col7/row 3, as is suggested by the answer above, then the only possibility for this cell is 9, I think - anyone disagree?

Oops! Super thicko here!! I was looking at col 8 rather than col 7! I am now reprogrammed!
Guest

Posts: 312
Joined: 25 November 2005

In you're previous post animator you make the claim:

"Now when you look at row 6 you will see that there are only two cells that can have the number 6 and 9. "

I am thoroughly at a loss to see how this is statement can be logically made. There appears to be some alternatives, and this is not a "must" have 6 or 9 in two cells (as the alternatives lie in blocks 4,5,6)

Can you please elaborate on your choice of logic for making this statement?

-SC
Guest

Ok,

Let's first look at the starting point.

Possibilities for row 6:
r6c1: 3, 6, 7, 9
r6c3: 3, 5, 7
r6c6: 6, 9
r6c7: 5, 7, 9
r6c8: 6, 7, 9

Now, when you follow all steps I previously explained, then you will be able to remove the number 7 as a candidate for r6c8:

Row6: (after excluding 7)
r6c1: 3, 6, 7, 9
r6c3: 3, 5, 7
r6c6: 6, 9
r6c7: 5, 7, 9
r6c8: 6, 9

Now you see that r6c6 and r6c8 has to have a 6 or a 9, if you fill in a 6 or a 9 in any other place in the row then you cannot finish the grid... (you will either end up with two cells that can only be 6, or two cells that has 9 as only candidate, both are incorrect).
Animator

Posts: 469
Joined: 08 April 2005

### Also needing help

Help - I've been having great difficulty with this one too, the 1st fiendish that has had me stumped. I'm on about my tenth iteration and my marriage is in danger.

With the help of Animator's hints, and some serious rethinking, I've got as far as this, but can no longer see a way forward. I think I've used an x-wing with 4s, successfully, and once went off on a trial substitution (that got me nowhere), and found the pairs of 28 and 69. But now I'm all out of ideas, have no unique numbers and can find no more pairs.

Can anyone help?

*58 2** 4**
*94 *71 35*
*1* *4* **6

461 *57 *23
92* *3* ***
*8* 12* **4

549 *1* 68*
176 *8* *3*
832 **5 14*

Thanks
Nigel
nputtick

Posts: 3
Joined: 22 May 2005

Take a close look at number 5 :)

And after that you should look at the number 3 and/or 7.
Animator

Posts: 469
Joined: 08 April 2005

Thank you, thank you. How could I have missed that 5? And there was I thinking some special new skill was needed.

And the rest went down like dominoes.

Cheers
Nigel
nputtick

Posts: 3
Joined: 22 May 2005

My first go at sudoku was on May 19, No 148. Not the best start point I think! I got to this point which didn't quite meet up with the forum discussions. Can you please advise next step.

**8 2** 4**
*9* *** 35*
*1* *** **6

4*1 **7 *23
*2* *3* ***
*8* 1** **4

549 *** 68*
176 *8* *3*
832 **5 14*
Guest

Some notes: you shouldn't start with a findish puzzle and definetly not with the one of 19th May :)

And you are missing a clue in your grid... (is it only here? or also at the grid you use to solve it?)

http://www.timesonline.co.uk/article/0,,18209-1617487,00.html ==> r2c5 has a clue (the number 7)...

Anyhow, fill in the 7 and then start looking at the number 4.

You should be able to complete it to what the person in this topic has: http://forum.enjoysudoku.com/viewtopic.php?t=254
Animator

Posts: 469
Joined: 08 April 2005

Thank you for your very prompt reply. And yes, the 7 is missing in the grid I am using to solve it!!! Novices!!!
Guest

Animator wrote:Then look carefully at box 6... (and remember what you just said about the 7)

It allows you to remove 7 as a candidate from several cells, and when you do that 2 values on row 6 will be 'paired'...

(A small note though, that is the first move I saw, not nessesarily the easiest one)

This is seriously depressing me ... I'm not following any of your hints with regard to Box 6 or 4! I got as far as Arnie's grid above: I have already determined that:

r2c9 and 23c7 had to be 2/8
r1c2 and 23c2 had to be 5/6

But then it all went blank on me! What am I missing?
Any help is appreciated
Guest

Posts: 312
Joined: 25 November 2005

### May 19th Sudoku

Hello, I'm so glad that others have had a problem with this one too, I've been carrying it around with me since the 19th and can't move forward, please help!! I can't even get to the point the previous person who asked for help achieved

**8 2** 4**
*94 *7* 35*
*1* *** **6

4** **7 *23
*2* *3* ***
*8* 1** **4

54* *** 68*
*76 *8* *3*
832 **5 14*

in r8c1 and r7c3 how does everyone manage to work out which one is 9 and which one 1?

Thanks
Sarah, confused, london
Guest

You are missing the number 7 :)

Let's take a look at column 7 and see where we could possible place the number 7. There are 4 empty cells: r3c7, r4c7, r5c7, r7c7.

You said yourself that r3c7 has to be 2 or 8, so number 7 is not possible there.

Now if you look at r7c7, then you will see that row 7 already has the number, so it isn't possible there either.

This leaves us with r4c7 and r5c7. Both of these cells are part of box 6. This means that the only place where number 7 can go in box 6 is r4c7 or r5c7.

Which means that you can remove it from all other cells in box 6.

r6c8 has three candidates: 6, 7, 9. We just said that 7 can be excluded, this leaves us with 6, 9. which are exactly the same numbers as r6c6.

This means that there are only two places where the numbers 6 and 9 can go: r6c6 and r6c8. Now you can remove these two numbers from all other cells on row 6, leaving you with only one place for the number 9 in box 4.
Animator

Posts: 469
Joined: 08 April 2005

### Re: May 19th Sudoku

sarah wrote:Hello, I'm so glad that others have had a problem with this one too, I've been carrying it around with me since the 19th and can't move forward, please help!! I can't even get to the point the previous person who asked for help achieved :-(

You are missing some moves:

* look at column 5 see where number 4 can go.
* look at row 4, pay attention to number 1

If you figured out where the number 1 is row 4, then you can fill in box 7.

Then you are get to the point where the person in this thread is stuck: http://forum.enjoysudoku.com/viewtopic.php?t=254
Animator

Posts: 469
Joined: 08 April 2005

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