## 18th March - Fiendish - #95

Post the puzzle or solving technique that's causing you trouble and someone will help
Congrats to the post saying that in column seven, rows 3 and 8 must be 6 or 7. Now that you have pointed it out it seems obvious, but I had been staring for hours and not seen it.

I'll crack on and see where it leads!

Many thanks.
Guest

Wouldn't have got there without you Ric! Personally, I feel that thinking ahead is guessing. To me it counts as T&E if you have to say "If this cell is an X, then...", as opposed to "This combination of cells tells me that cell must be X". But I'm a curmudgeonly old pedant, so I wouldn't worry too much!
Guest

Done it! - and without having to guess at all.

Once again many thanks to 'guest' for that vital info regarding the position of the 6's and 7's in column 8, that was the key I was missing.
Guest

All of the necessary elements have been mentioned, at some stage. It seems as if IJ and |Ric| got there, between them - IJ's discovery about 6 and 7 in column 8 was very important. It also sounds as if Steve has got it.

I hope you all realize that my "sixes and sevens" crack was meant to be a hint.

It's not necessary to guess. It's not even necessary to "look ahead". The idea of "trying out" the 3 and the 8 in r7c8 is not the way to go. That would count as guessing, in my book.

Some of the previous posts may be hard to follow because a few misdescriptions slipped in. For example, column 8 was called column 7 at one stage.

Can I ask posters to check out the "Basic terms" post in the General/Program forum? Talking about "r5c8" (for example) is so much more certain than talking about "the cell to the left of the 4 in the right-middle block" (which is "box 6", of course).

- Wayne
Pappocom

Posts: 599
Joined: 05 March 2005

### Fiendish Friday

Ric you baffled me yesterday but I read your additional explanation this morning and everything dropped into place. You said that a friend had shown you the way out and you have now shown your generosity by helping the rest of the "bemused brigade". It's given me an extra level of thinking that I hadn't been aware of and restored my faith in The Times Su Doku. This is the first time I haven't finished Fiendish Friday without help so I don't feel the sense of achievement that I usually get but my wings have grown with your help. Well done to you.

Wayne, all at 6's and 7's? I think I was back in the cinema in the 60's sitting in the 3/9's!
Guest

This seems to be a crucial part of the learning process - I've certainly found it too: being shown little tricks, or ways of looking at the puzzle, and the relevance of the various options in each cell to other cells in the same row/column/box.

I guess it's the people who can see this and deduce it for themselves that separate them from the rest of us!
shakers

Posts: 93
Joined: 10 March 2005

Blimey, that puzzle really got to me. I wasted 29 minutes 10 seconds of my life on it. :(
Chris

Posts: 12
Joined: 06 March 2005

Well done everyone! I couldn't believe I couldn't finish this. However, once reassured that it could be solved in the normal way and having had my attention directed to the right hand side of the puzzle I could suddenly see what I had not been able to see although it was staring me in the face. That is, that r2c8, r4c8 and r5c8 can only contain the numbers 3, 8 and 9. That gives a definite 1 for r7c8 and the rest just fell into place as normal.

I'm relieved to know that my infallible method is still infallible, for now at least!

Oh dear, I really should get a life, but these puzzles are seriously addictive!
Kate

Posts: 2
Joined: 18 March 2005

Like several others with programs mine also failed at the same point, and I've been staring at it for nearly a week trying to work out why, before finally realising that the answer probably lay here.

Now I just have to work out how to add this additional check into my program (well an excell sheet with macros really).

I normally solve them all by hand first, before checking against my program, I was convinced this one must have had a typo.

Thanks for the tips.
Guest

### Not sure that the question started at the right point

Going all the way back to the original question ... applying basic rules leaves me wondering where the 1 in R2C7 and the 2 in R8C7 come from ...

However, if you remove those values but instead look at the need to have 3 and 5 in R1C5 and R3C5 then this gives you values for R1C7, R1C9 and R3C1; rich reward and enough to work your way through the rest of the puzzle ...

(One day I will bore you with details of my computer program that allows me to do the "boring" rules and leaves it clear as to where the remaining problems lie ... not that those are not also solvable, of course, but man & machine in harmony is fun too!).
Guest

Going all the way back to the original question ... applying basic rules leaves me wondering where the 1 in R2C7 and the 2 in R8C7 come from ...

Consider Row 9. The values 5, 7, 8 and 9 must occupy the cells r9c1, r9c3, r9c7 and r9c9 in some order, which leaves r8c7 as the only candidate in Column 7 for the value 2. Once that's in place, r2c7 is the only candidate in Column 7 for the value 1.
Sue De Coq

Posts: 93
Joined: 01 April 2005

Sue De Coq wrote:Consider Row 9. The values 5, 7, 8 and 9 must occupy the cells r9c1, r9c3, r9c7 and r9c9 in some order, which leaves r8c7 as the only candidate in Column 7 for the value 2. Once that's in place, r2c7 is the only candidate in Column 7 for the value 1.

Of course　- thank you! So that must be an easy puzzle then ... more than one "clever" move possible at this point ;-) The advantage of my solution is that the "clever" move delivers enough numbers to continue with only basic moves. (I recognise that this depends on my currently vague definition of clever and basic but I think I know what I mean ;-)
Guest

### Recap -- how it is solved -- key steps

Thanks IJ. Your March 19, 2005, 1:53 PM, post did it for me. Thanks too to Pappocom's March 19, 2005, 3:24 PM, post confirming it and noting the col. 7/8 mistype. Wayne's remarks are appropriate as to using the correct terminology. Also, posters should register so they can be emailed, or at least so they can edit their errors.

For those still confused, let me say it is solved as follows;

(1) I started with Jane's grid, given at the start of this thread.

(2) I put in all of the pencil marks (candidates) in each cell.

(3) one particularly important set of pencil marks is the 6's at the bottom (r9) of Box 8. They lead us to eliminate the 6s pencil marks at the bottom (r9) of Box 9. (This is basic, but I'll say it anyway: the possible 6s in Box 8 can only go in row 9 of Box 8 [and 6s go all over Box 9], so the 6s of Box 8 (all in row 9) knock out the 6's in row 9 of Box 9.)

(4) Now, the only 6s possibilities (pencil marks) left in column 7 are in the cells in row 4 and 5 . That is, either r4,c7 or r5,c7 must be a 6. Both of these cells are in Box 6. Therefore, the other pencil mark 6s in Box 6 (in column 8 of Box 6) are eliminated.

(5) Now we switch our attention to column 8. All of the remarks in this paragraph are to cells in column 8. 3678 pencil marks are in row 3, and 1679 pencil marks are in row 8. These are the only 6s and 7s left in the column! So we can shorten the pencil marks to 67 pencil marks in row 3 and 67 pencil marks in row 8. Thus, we eliminated 3 and 8 pencil marks in row 3, and 1 and 9 pencil marks in row 8. As we will see in a moment, eliminating that 1 pencil mark from r8,c8 is crucial.

(6) In column 8, there were (once) only two 1 pencil marks -- at r7,c8 and r8,c8. So either cell could be a 1. But in (5), above, we eliminated the 1 from r8,c8, so we have 1 as the sole number in r7,c8. I assume that from here on the puzzle can be easily solved (I hope).
chipe

Posts: 6
Joined: 26 April 2005

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