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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

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

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.

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

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

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.

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

14 posts
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