The New Sudoku Players' Forum

[stuartn]

This was Sundays Times Sudoku

Code: Select all

.5.1.....
4..6.2...
.6...817.
74.8.....
.........
.....3.59
.137...2.
...4.6..8
.....5.9.


Am I mistaken, or does an offset x-wing on 4's in R6 and R7 force a 9 in R1C7? - Just thought it was a nice example...

stuartn

http://brightonandhove.org/Sudolinks.htm

[Karyobin]

I really should read more. Never even heard of an offset x-wing. How the blummin' heck do you get a 9 up there from 4's down there?!

[stuartn]

Karyobin - Have a look at this ...

[329] 581 [3479] [47] [49] [463] [2346]
4716 [935] 2 [895] [83] [35]
[329] 6 [29] [59] [3459] 817 [2345]
74 [59] 8 [59] 12 [36] [36]
[29] 3 [259] [59] 6 [47] [478] [48] 1
1862 [47] 3 [47] 59
613789 [45] 2 [45]
597426318
824315697

The fridge beckons.......

I've also put it on

http://www.brightonandhove.org/Sudolinks.htm

if you run Excel....

[Doyle]

I don't see it. The 4s of your "Offset X-Wing" could be in r6c5 and r7c9. Therefore no exclusion of 4s in c7 results. Therefore r1c7 could be a 4 as well as a 9. Your "wing" has three possibilities for the 4s, not two.

[stuartn]

But look at R6C7 and R7C7 - both rows have 4's as doubles only. 4's are in the same column (C7). Why can this not eliminate the other 4's above it?.... because it jolly well does! - happy to discuss.

stuartn

http://www.brightonandhove.org/Sudolinks.htm

[Doyle]

Why? Because the 4 in r6 could be in c5, the 4 in r7 in c9. That placement excludes 4s from r6c7 and r7c7. Leaving two other cells in c7 for a 4, including that cell at r1c7 where you are putting the 9. One of us is not seeing things right, hope it's not me.

[stuartn]

If R7C9 =4, then R3C5 =4. Then R6C5 = 4 and R1C7 = 9.

If R7C9 = 5 then R7C7 = 4. Then R1C7 =9

Personally I prefer my way. ( and it works all the time)

stuartn
http://www.brightonandhove.org/sudolinks.htm

[Doyle]

stuartn; Statement 2 in your last post is correct, but statement 1 is internally contradictory, it places two 4s in column 5. Suggest we sleep on it.

[tso]

Yes, you are mistaken.
Doyle is exactly correct.

1) All dogs have legs.
2) All tables are wood.
Therefore, this table has legs.

The truth of your conclusion does not prove either the truth of your premises nor the validity of your logical argument. The fact that you show that r1c7 *is* 9 by essentially other means is not part of the argument.

The four cells r6c57 and r7c79, *by themselves* do not allow you to make any statement about the value of r1c7. As Doyle correctly points out, there are THREE ways that the 4s can be distributed within the "offset x-wing", one of which does NOT have a 4 in the 7th column. These exact conditions CAN exist in a puzzle in which r1c7 IS 4.

On the other hand, the four cells that comprise the normal x-wing at r15c68, *by themselves*, in a vaccuum, elliminate all candidate 4s from rows 1 and 5, as there are the only candidate 4s in columns 6 and 8, leaving exactly two ways to position the fours -- one in each column and one in each row.

[tso]

stuartn; Statement 2 in your last post is correct, but statement 1 is internally contradictory, it places two 4s in column 5. Suggest we sleep on it.

If R7C9 =4, then R3C5 =4. Then R6C5 = 4 and R1C7 = 9.

On the one hand, the fact that the statement contradicts his off-set x-wing argument doesn't matter. He's simply making a valid "forcing chain" argument from r7c9 to r1c7:

1) r7c9=4 => r3c9<>4 => r3c5=4 => r6c5<>4 => r6c7=4 => r1c7<>4
2) r7c9<>4 => r7c7=4 => r1c7=4.
Therefore, r1c7=4.

On the other hand, this has nothing to do with his offset x-wing, which is does not work.

[Scott H]

By the time I could digest this thread I found a Turbot fish in 4's that excludes 4 from r1c7 : r6c7 - r6c5 - r2c5 - r3c9(box3). But I need software that filters on 4s to see such beasts -- very hard for me to see on paper.

Not sure what an offset X-wing would be, but this Turbot fish looks more like one to me than the 4s in rows 6/7. Also, rows 3/6/7 make a swordfish.

[SteveF]

When I first solved this one I was surprised that I couldn't do so using 'normal' eliminations (after all it is from the (Sunday) Times.

Next step, look for the next level of complexity and I soon found an x-wing in the 4's (normal, not offset).

Two points:
- surely you look for the simplest way to solve first, then the progressively more complex?
- as there is a simple w-wing, does it affect any of the points made in this thread? - 'fraid much of it goes over my head.

[stuartn]

I said earlier:

If R7C9 =4, then R3C5 =4. Then R6C5 = 4 and R1C7 = 9.

If R7C9 = 5 then R7C7 = 4. Then R1C7 =9

what I should have said in the first line was R6C5<> 4

stuartn

http://www.brightonandhove.org/sudolinks

[stuartn]

tso wrote:

On the other hand, the four cells that comprise the normal x-wing at r15c68, *by themselves*, in a vaccuum, elliminate all candidate 4s from rows 1 and 5, as there are the only candidate 4s in columns 6 and 8, leaving exactly two ways to position the fours -- one in each column and one in each row.

Yes, I quite agree and if you look at the link you'll see that that x-wing is highlighted aswell - what I was asking (and why the thread was asking 'what is this') was whether we could do it another way - other than forcing chains - which I also pointed out. I gather the answer is no - had the fridge not been so inviting last night I should have spotted it myself.

Condescending comments are out of place on this forum - I'm sure you'll agree.

[russ]

Karyobin - Have a look at this ...

[329] 581 [3479] [47] [49] [463] [2346]
4716 [935] 2 [895] [83] [35]
[329] 6 [29] [59] [3459] 817 [2345]
74 [59] 8 [59] 12 [36] [36]
[29] 3 [259] [59] 6 [47] [478] [48] 1
1862 [47] 3 [47] 59
613789 [45] 2 [45]
597426318
824315697

You could proceed differently: r3c3 can be neither 5 nor 9
and so must be 2.