## help again!! (+img)

Everything about Sudoku that doesn't fit in one of the other sections

### help again!! (+img)

Ugh, I barely got anywhere on this one then got stuck. I'm just getting familiar with all the techniques, but I searched for some x-wings and stuff and could not find ANYTHING! :| maybe i'm just not spotting it, but i am stuck!! :P
Spazactaz

Posts: 6
Joined: 29 September 2005

r8c5 can't have 5 as a candidate

I can see a few more that go in after that, and then I'm stuck as well.
Lardarse

Posts: 106
Joined: 01 July 2005

Lardarse wrote:I can see a few more that go in after that, and then I'm stuck as well.

Requires 2 nishios to complete starting from the orange cell:

angusj

Posts: 306
Joined: 12 June 2005

AJ: The floor is yours to explain those...
Lardarse

Posts: 106
Joined: 01 July 2005

jest one step

Last edited by bennys on Sun Oct 02, 2005 7:50 pm, edited 2 times in total.
bennys

Posts: 156
Joined: 28 September 2005

Lardarse wrote:AJ: The floor is yours to explain those...

Hints:
First Nishio - what happens to candidate 9s in Column 5 when 8 assigned to r5c1? (sorry the colouring isn't overly helpful in that example)
Second Nishio - what happens to candidate 3s in row 5 when 3 is assigned to r3c1.
angusj

Posts: 306
Joined: 12 June 2005

I meant your turn to explain Nishio... I assume that it means "Make an assumption, and then contradict it"
Lardarse

Posts: 106
Joined: 01 July 2005

Lardarse wrote:I meant your turn to explain Nishio... I assume that it means "Make an assumption, and then contradict it"

That pretty much sums it up .

Edit: I consider Nishio to be "Trial and Error that you can do in your head" (ie without having to resort to erasing pencilmarks or solver undos).
angusj

Posts: 306
Joined: 12 June 2005

The two Nishio can be replaced by two forcing chains, if preferred:

Chain#1:
r5c4=8
=> r5c1<>8;
r5c4<>8
=> r5c4=9
=> r4c5<>9
=> r9c5=9
=> r9c1<>9
=> r9c1=8
=> r5c1<>8;
therefore r5c1<>8.
*Or, in alternative notation:
r5c1~8~r5c4~9~r4c5-9-r9c5-9-r9c1~8~r5c1 => r5c1<>8.

Chain#2:
r3c8<>9
=> r3c1=9;
r3c8=9
=> r4c8<>9
=> r4c8=2
=> r5c7<>2
=> r5c3=2
=> r6c1<>2
=> r6c1=3
=> r3c1<>3
=> r3c1=9;
therefore r3c1=9.
*Or, in alternative notation:
r3c1-9-r3c8~9~r4c8~2~r5c7-2-r5c3-2-r6c1~3~r3c1 => r3c1=9.

*EDIT: Added alternative notation.
Last edited by r.e.s. on Sun Oct 02, 2005 8:37 pm, edited 1 time in total.
r.e.s.

Posts: 337
Joined: 31 August 2005

angusj wrote:I consider Nishio to be "Trial and Error that you can do in your head"

As does Pappocom, that's why he calls puzzles that need it "Arguibly unfair"
Lardarse

Posts: 106
Joined: 01 July 2005

Lardarse wrote:
angusj wrote:I consider Nishio to be "Trial and Error that you can do in your head"

As does Pappocom, that's why he calls puzzles that need it "Arguibly unfair"

Which is why I posted the solution using forcing chains -- they are FAIR beyond argument, imho .
r.e.s.

Posts: 337
Joined: 31 August 2005

I don't understand anything in Benny's message.

He said:
"We have 1 in the blue or 8 in the red".

First, is that an exclusive OR or not?

Second, I don't see how anything that is done in either column affects the candidates in the other. Could somebody please explain?

Then he gives a second way to see the same thing:
"Because the 79 in green we have 1 or 8 in the yellow". But to eliminate the 7 and 9 candidates in the yellow we need to have a 79 pair (naked or otherwise) elsewhere in the column, and we don't. Am I missing something here also?

Thanks very much!

RobR
RobR

Posts: 6
Joined: 24 September 2005

Sorry if I was not clear.
Because r1c5 can't be both 1 and 8 we have to have 1 in the blue or 8 in the red (not exlusive)
Thats mean that we cant have a=8(prevent 8 in red) and c=1(prevent 1 in blue).
but a=8 ->b=4->c=1 so we get that if a=8 then both a=8 and c=1
and that cant be and so we get that a<>8.

Regarding the second argument.
If we dont get 1 or 8 in the yellow boxes then the only values that left for them are 7 and 9 so one of them will be 9 and the other 7 but then the green does not any values left as candidates.

So we gut that we have 1 or 8 in the yellow and that mean that a<>8
or c<>1 and from here we continue like the second part of the first argument to show that a<>8.
bennys

Posts: 156
Joined: 28 September 2005

Benny,

Thanks very much! I printed your explanation and held it up next to your original post and understood it.

RobR
RobR

Posts: 6
Joined: 24 September 2005

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