dailysudoku.com

[TNova]

Well I'm really sorry that Arkietech seems to have stopped posting his puzzles. I just think it's better that we debate the same puzzle.

So let me try this one. The Hodoku rating is 1842 - this compares to about 1,500 for the VH ones on this site. But occasionally Arkietech's were rated at above 2,800 so I think this one isn't too bad.

I hope the format below is OK (it certainly will be for Hodoku users), if not I hope that Marty can correct me.

Tim

.---------.---------.---------.
| . . . | . 7 4 | 8 . . |
| . 7 4 | . . . | 1 6 . |
| . . . | 6 . . | . 7 . |
:---------+---------+---------:
| . . 9 | . 5 . | 4 2 . |
| 2 . . | . . . | . . 9 |
| . 6 8 | . 9 . | 3 . . |
:---------+---------+---------:
| . 3 . | . . 1 | . . . |
| . 9 1 | . . . | 2 3 . |
| . . 2 | 8 3 . | . . . |
'---------'---------'---------'

[arkietech]

[TNova]

Arkietech,

Thanks for your reply, I was previously unaware of that forum that you mention.

Sorry that you gave up on my puzzle, it's rated less difficult than your pattern game 25.104 that you provide on the new forum, but I do agree that it's difficult nonetheless. I do admit that I rather like these difficult puzzles where you can only solve them by gradually eliminating candidates here and there rather than having one killer move.

When I finally understand notation properly I'll try and contribute a bit more.

Tim

[arkietech]

[TNova]

Hi Arkietech, I'll take you up on your challenge.

But two problems

The first is that I'll have to solve this quite difficult puzzle again. It will be obvious that I'm not cheating because my solutions are never the way the computer does it! Also, since I'm in the UK but looking forward to the US Grand Prix on TV in a few minutes I won't re-solve it today.

The second is that I don't fully understand the notation system on these boards, so I'll have to just explain as best I can.

I hope to post again in a day or two.

Tim

[Marty R.]

[tlanglet]

I am on the road but had some time to check on Sudoku and to work on this puzzle. Here is a two step solution using the grid posted by Marty......

AHP[(68)r45c6=(6)r89c6]-(46=2)r87c5-r3c5=r3c6-(2=7)*r6c6-(7=3)r4c4 => *r5c6<>7, r5c6<>3

After clean-up we have

[TNova]

Arkietech asked for details of my eliminations so I've re-solved this difficult puzzle - starting from Randy's diagram, and I've kept notes of what I did. It was an interesting exercise because it forced me to try and rationalise my thought process. I hope I don't get a headache.

But I think it makes sense for me to use those notes - there are a lot of them - in the morning and check that there are no mistakes.

So there will be another post tomorrow.

All the best, Tim

[TNova]

When I posted my message a few minutes ago I wasn't aware of Tlanglet's contribution to this thread. But I'll still post my "English language" solution to this puzzle tomorrow, which I hope some readers will find interesting. I don't think it's anything like Tlanglet's though.

Tim

[peterj]

Most straightforward approach I could find...

[TNova]

This might be the most incomprehensible post in the history of the site, but anyway here are the eliminations that I promised to Arkietech. The starting point is Marty's diagram.

1. If R4C9 is 8, R6C9 is forced to be 7 and R9C9 is forced to be 1. So whatever the value of R4C9, R9C9 cannot be 6. R9C9 also cannot be 5, to avoid a hidden rectangle in R69C89.

2. We can eliminate several 5's based on the fact that if R9C2 is 4, then R5C2 becomes 5, and a naked triple 159 is formed in C8. These result in forced values for R5C8, R4C9 and R5C7, which becomes 7. In turn R5C3 becomes 3, and R3C3 becomes 5, so that R7C3 cannot be 5. Following this chain further, R3C7 is 9, and R1C8 is 5. Therefore R9C8 cannot be 5.

3. Also, with R1C3 and R3C3 known, it is possible to determine most of the values in C1 - in particular that R789C1 will have values 5, 6 and 8. But the 8 will need to go in R7C1 to avoid a 5/6 hidden rectangle in C1 and 7. So 5 can be eliminated as a candidate in that cell.

Also, since we know that R8C9 is 7, we know that either R8C4 or R8C6 must be a 5. So R8C1 cannot be.

4. So I then turned my attention to eliminating 7's and looked at what would happen with R7C3 set to 6. This forces all the values in R1, and also forces R3C5 to be 2. This forces values 4 and 6 into R78C5. and hence R8C46 must be 5 and 7. So the 7 in R8C1 can be eliminated. And with the values of R8C456 now known to be 5, 6 and 7, R9C8 becomes 8, and R8C1 4. The only possible value for R9C1 is therefore 7 and so we can eliminate 7 as a candidate for R7C1.

5. I couldn't see any further eliminations from there, but I was still interested in the 7's,so I had a look what would happen with R6C6 set to 2. This immediately forces 2 into R3C5, and and 5 and 7 into R8C46 as we saw earlier. So the 7 in R9C6 can be eliminated. The 7 in R8C9 can also be eliminated as this cell cannot be 7, irrespective of the value of R6C6.

6. This creates a 568 naked triple in C9, so the 5 in R6C9 can go as well, as can the 5's in R7C78 and R9C7. This in turn creates a Finned X-wing so the 5 in R9C6 can go as well. Also with a 1 forced into R1C4, R456C4 will contain 3, 4 and 7 so R5C6 cannot be 7.

7. Targetting 1's. If R6C9 is 7, then R6C6 is 2. This forces the values in box 2 as seen earlier, so R1C4 is 1. R6C4 cannot be 1, and hence neither can R5C8.

8. The 5 in R3C1 can be eliminated (Finned X wing), and I think it's relatively straightforward from here. If R5C2 is set to 5, a simple chain proves that R9C1 must be 4, so R9C2 cannot be 4. It's clear that R6C6 must be 2 and so on.

Hope anyone made it this far, Tim

[arkietech]