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-   -   I see the solution, but what's the logic? (http://www.logic-puzzles.org/forum/showthread.php?t=299)

 BillsBayou 06-17-2013 08:34 PM

I see the solution, but what's the logic?

2 Attachment(s)
Here's the blank:
Attachment 86

And here's the bottleneck:
Attachment 87

If you replace mushroom with Fusilli, clues 1 and 5 reduce to just two categories:
Yolanda plus \$1 is Capellini
Fusilli plus \$1 is Angie

With:
Angie not Yolanda
Fusilli not Capellini
Angie not Capellini
Yolanda not Capellini

I worked it out that this cannot happen:
Angie is \$9.99
Fusilli is \$8.99
Leaving Yolanda to be either:
Yolanda is \$7.99, but Capellini cannot be \$8.99 (already taken)
Yolanda is \$8.99, but Capellini cannot be \$9.99 (Angie cannot be Capellini)

So, Angie must be \$8.99, and she is, and everything falls into place.

BUT WHAT IS THE LOGIC beyond the bottleneck? I don't see it. How do I frame what's left into a logical statement?

 shriekingviolet 06-17-2013 09:30 PM

Yikes, this one is tricky. Hopefully this will make sense.

Clue 5 - Angie is \$1 more than the mushroom sauce. So either \$7.99=mushroom/\$8.99=Angie -OR- \$8.99=mushroom/\$9.99=Angie. So either way, \$8.99 must be Angie or whoever had the mushroom sauce (either Wade or Yolanda), so you can eliminate Margie from \$8.99.

Therefore you can eliminate \$8.99 from capellini

After that, go back to clue 1 and you can now eliminate \$7.99 from Yolanda

So now, from clue 1 again, but substituting Margie for capellini: \$8.99=Yolanda/\$9.99=Margie or \$9.99=Yolanda/\$10.99=Margie. So \$9.99 must be either Margie or Yolanda, which leaves Angie at \$8.99

 zenobia43 06-18-2013 12:57 AM

And here is yet another method using the two ordering clues with a substitution.

Lay out clues 1 and 5 with the substitution for capellini.

Yolanda -> capellini = Margie

mushroom -> Angie

Notice that Margie and Angie are in the same category, same side of the relation, and a fixed distance from the left side.

That means Yolanda cannot be mushroom. Wade must be mushroom.

Shriekingviolet has a more interesting approach, and it demonstrates how the one-sided "Of A and B, one is C and the other is D" clue can yield extra Xs beside the obvious one.

"A is either C or D" is very similar to its two-sided cousin. Just find out what C and D can never be, and apply those exclusions to A.

As SV points out, from the partial solution grid, \$8.99 is either Angie or mushroom. From the grid, we see that the intersection of Angie and mushroom also has four Xs that can be applied to \$8.99. One of these Xs is "\$8.99 cannot be Margie."

 BillsBayou 06-18-2013 03:52 PM

Thanks to ShriekingViolet and Zenobia43 for laying this out. Yet another technique learned, here.

I'm working up a graphic to explain it all for anyone who doesn't see it. I'll link to it here in a few minutes.

Zenobia43, it looks like I still need work on my approach to multiple fixed-relational clues that use the same category. That would have solved it for me.

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