The New Sudoku Players' Forum

[Guest]

I can usually solve Fiendish puzzles but today's Times puzzle has me stumped. What do I need to do next? Any suggestions much appreciated. I have got this far...

**8 2** 4**
*94 *71 35*
*1* *4* **6

4*1 **7 *23
*2* *3* ***
*8* 12* **4

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

[Animator]

Ok, here is the hint:

First, look at box 3. (take a close look at the numbers 2 and 8)

Second, look where the 7 can go (in column 7)

Third, take a look at row 6, if you done the previous two correctly you should see something intresting

Fourth, look at box 4, you can place a number there

[Guest]

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

[Animator]

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)

[Guest]

Sorted !! Many thanks...

[Guest]

This is the first Times Fiendish I've completed. As I only heard of Su Doku last week I am dead proud and just wanted to add that. Thankyou. Carry on.

[Kites]

I am still stuck at this point on Su Doku 19 May:
(I have finished todays! as I have done every puzzle since December...)

*58 2** 4**
*94 *71 35*
*1* 54* **6

461 *57 *23
93* *3* ***
*8* 12* **4

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

I have: 2+8 in c9/r2 & c7/r3
6+9 in r6 c6 & c8
7 in c7 r5 & r6 (as well as r6 c1 & c3)

Any advise please?

[Animator]

Take a look at column 4

[Animator]

Ignore that previous hint. You are on the wrong track...

Column 2 (and row 5) has the number 3 twice... This is impossible, you made a mistake somewhere...

[Guest]

Still rather confused by Animator's reply. What numbers in row 4 are "paired"

What good does knowing that 7 could be in r5 or r6?

This one has really got me stumped, most fiendishes I can get done within the hour but I've been struggling for 2 on this one.

[Animator]

I never said row 4, I said row 6, and box 4 (but I'll assume it was a typo:)

Well,

If you look at box 6 then you need to know where to fill in the number 7, if you take a look at column 7 then you see that the only place where it can go in box 6.

This allows you to exclude number 7 from r6c8. And this results in the fact that the numbers 6 and 9 are paried.

Therefor number 9 cannot go in r6c1, and this leaves only one place for the number 9 in box 4.

[Guest]

Sorry, Animator, I still don't get it. Starting from here (i.e. as per first post above):

**8 2** 4**
*94 *71 35*
*1* *4* **6

4*1 **7 *23
*2* *3* ***
*8* 12* **4

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

In box 6, I have possibles as follows:
r4c7 589
r5c7 5789
r5c8 1679
r5c9 15789
r6c7 579
r6c8 679

And I don't see how I exclude the 7 from r6c8, or that I have a pairing of 6 and 9 if I do (either in box 6 or row 6)! Or maybe you're starting from somewhere else!?

Help appreciated.

Thanks

[Guest]

look at box 8 - you should be able to place a number there. ..if you can't look down columns 4 and 5...

[Animator]

Well,

First, let us look at box 3, here is the list of possiblities:

r1c8: 1, 7, 9
r1c9: 1, 7, 9
r2c9: 2, 8
r3c7: 2, 7, 8, 9
r3c8: 7, 9

What is intresting in this box is that there are only two cells that can have the numbers 2 and 8, and that is r2c9 and r3c7.

This means that both the 7 and the 9 in r3c7 are impossible (it is has to be a 2 or an 8).

Now let's look at column 7:

r3c7: 2, 7, 8, 9 ==> We just said that 7 and 9 aren't possible, so it really is 2, 8
r4c7: 5, 8, 9
r5c7: 5, 7, 8, 9
r6c7: 5, 7, 9
r7c7: 2, 5, 9


What is intresting here is the number 7, it can only go in r5c7 or r6c7, both of these cells are part of box 6. This allows us to exclude the 7 from all other cells in box 6.

Box 6:

r3c7: 5, 8, 9
r4c7: 5, 7, 8, 9
r5c8: 1, 6, 7, 9 ==> 7 is not possible here, the 7 of box 6 has to be in column 7, so it really is 1, 6, 9
r5c9: 1, 5, 7, 8, 9 ==> no 7 either ==> 1, 5, 8, 9
r6c7: 5, 7, 9
r6c8: 6, 7, 9 ==> the 7 here is impossible(!) ==> 6, 9

Now when you look at row 6 you will see that there are only two cells that can have the number 6 and 9. This allows you to exclude the 9 from r6c1 leaving only one place for the 9 in box 4.

Did this help?

[Guest]

Yes, thanks very much - v.clear explanation. As always, it seems obvious when someone tells you how!