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Stuck wth extreme puzzle...
 
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Riveroaaron9@gmail.com



Joined: 5 Nov 2016
Messages: 5
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For your bottom right square, 8 and 2 HAVE to be in the middle. so you can put a seven there on the top left.
knoregs



Joined: 22 Aug 2015
Messages: 5
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capsaicin wrote:For your bottom right square, 8 and 2 HAVE to be in the middle. so you can put a seven there on the top left.


I ran through the solution after the computer 'hint' and the 7 in the bottom right square ends up in the top middle between the 1 and 3.

~knoregs
Hawkinsrn@gmail.com



Joined: 19 Oct 2012
Messages: 7
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I'm not an expert, but I think I can help. In the center square, the 5,6,9 must be a 9. Here's how I know. Starting in the box to its left (5,6), follow the other boxes with 5,6. Going clockwise, there's one straight above it, then two to the right, then four down, then one to the left, then the (5,6,9). They go in pairs, right? With the last one in the chain linking back to the beginning in a loop. They can't all be 5,6 or else there would be two possible solutions to the puzzle. So that one must be 9.
I don't think this is a standard strategy, I just made it up. But I think it works.
knoregs



Joined: 22 Aug 2015
Messages: 5
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Hawkinsrn@gmail.com wrote:
I don't think this is a standard strategy, I just made it up. But I think it works.


I'll have to think on this and try to apply when it shows up again. Thanks.

Cheers!

~knoregs
bspiglejr



Joined: 8 Jan 2017
Messages: 2
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Howkinsrn strategy is called the swordfish strategy. You can find it discussed in detail if you google it...
Hawkinsrn@gmail.com



Joined: 19 Oct 2012
Messages: 7
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I've been trying to make this conform to swordfish somehow, and I realize that I don't fully understand swordfish -but I want to. I just don't see it. Which are the nine squares, and which digit gets eliminated?
bspiglejr



Joined: 8 Jan 2017
Messages: 2
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You're right. I'm obviously not much of an expert either. There's no swordfish in that area.

Now my judgment at this point is pretty suspect, but I also think your original analysis is incorrect. It's true that they all can't be 5/6, but like 5 of them could be 5/6 while one was 5/9.

If we number 1-9 top to bottom and a-i left to right, couldn't you have 5's in 3d, 4e and 7f, with 6's in 3f, 4d and 7e putting the 9 in 5e? That would still work? If so wouldn't that invalidate the original analysis?

There is a swordfish.
9's in 3c, 3g, 3h, 4e, 4g, 4h, 5b, 5e and 5g for the triplets, meaning 8g must be 6...

Sorry about the screw-up.
 
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