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View Full Version : Another Day, Another Daisy Chain, Another Stuck Puzzle

BillsBayou
05-30-2013, 03:47 PM
Yes, I pick the hard puzzles on purpose. Under 30% success rate, or over 1000 seconds to solve.

Here I have another stuck puzzle:
48

I can eliminate all but clues 1, 7, and 8:
49

I cannot find any more complementary exclusion sets. Example: The line for \$4.90 in the upper left box, and the "mocha" line in the bottom left box mean that "Mocha" is not \$4.90.

I cannot find any more 2x2 sets. Example: Upper left box for money and men's names.

The daisy-chain for "Bryan < Caffe Latte < (+20 cents) Henry" does not intersect the relationship of "Austin < (+20cents) Fredrick". I've marked that neither Henry nor Bryan can have Caffe Latte, but that's the extent of the exclusions in these relationships.

I know the answer if I ignore logic and start playing "What If": If Fredrick is \$4.80, then Eduardo is \$4.70. Follow that through Toni and Iced Americano is \$4.70. This leaves a violation of clue 8, because Henry is at \$4.90 and Caffe Latte is \$4.60.

Can this be done with logic alone?

If you want it, here's the puzzle with nothing marked:
50

Suefle
05-30-2013, 05:39 PM
I haven't finished the entire puzzle but this may help. Look at clue #7 and clue #8. If Fredrick's coffee is 20 cents more than Austin's and Henry's coffee is 20 cents more than the caffe latte then you know that Austin's coffee cannot be the caffe latte because Fredrick and Henry cannot both drink coffee that is 20 cents more than the caffe latte.

zenobia43
05-30-2013, 05:49 PM
The solution is all laid out neatly in your post.

"The daisy-chain for "Bryan < Caffe Latte < (+20 cents) Henry" does not intersect the relationship of "Austin < (+20cents) Fredrick". I've marked that neither Henry nor Bryan can have Caffe Latte, but that's the extent of the exclusions in these relationships."

caffe latte is 20 cents less than Henry.
Austin is 20 cents less than Fredrick.

Fredrick and Henry are in the same category, they're on the same side of the inequality, and the two inequalities are fixed distance of the same size (20 cents).

Henry cannot be in the same row as Fredrick, so neither can caffe latte and Austin (be in the same row).

cafe latte cannot be Austin.

You did all the work. I just read your summary. Now I'll actually go look at the puzzle to find out what I might have missed with this hasty reply :).

shriekingviolet
05-30-2013, 06:25 PM
The daisy-chain for "Bryan < Caffe Latte < (+20 cents) Henry" does not intersect the relationship of "Austin < (+20cents) Fredrick". I've marked that neither Henry nor Bryan can have Caffe Latte, but that's the extent of the exclusions in these relationships.

I think that this means that Austin could not be the Caffe Latte, right? That might open some things up for you. Because Latte + 20 = Henry and Austin + 20 = Fredrick, so if they were the same, that would be a contradiction.

I got almost as far as you did, but can you explain how you figured out who Patsy goes with? This is my first post here, let me see if I can figure out how to attach where I got stumped

edit - haha, too slow! There weren't any replies when I got started. Y'all are quick!

BillsBayou
05-30-2013, 06:41 PM
Nailed it, guys, thanks. I need to add that one to my toolbox. I've seen it before.

Got to give it a name of some sort. The chains have the same spacing, but because values in the chain are in the same category, we can eliminate the equality of their related links. Both chains begin from different points in the same category and head in the same direction at the same pace. Uneven Parallel Chains?

Ax + c = Jm
Ay + c = Kn
Jm <> Kn

Category-A-index-x, plus some constant, c, equals Category-J-index-m
Category-A-index-y, plus some constant, c, equals Cateogry-K-index-n
Implies that Category-J-index-m cannot be Category-K-index-n
Where c is the same in both equations
Where Category-J or Category-K can equal Category-A (but not both because the problem would be easy to solve)
Where all four values Ax, Ay, Jm, Kn represent four distinct values.

BillsBayou
05-30-2013, 06:55 PM
I think that this means that Austin could not be the Caffe Latte, right? That might open some things up for you. Because Latte + 20 = Henry and Austin + 20 = Fredrick, so if they were the same, that would be a contradiction.

I got almost as far as you did, but can you explain how you figured out who Patsy goes with? This is my first post here, let me see if I can figure out how to attach where I got stumped

edit - haha, too slow! There weren't any replies when I got started. Y'all are quick!

You're almost there. You need to transfer the relationships around the grid. Note that you show that Fredrick cannot be \$4.50. Well, Kristen IS \$4.50. So Kristen cannot be associated with Fredrick. Once you do that, Patsy's value just falls into place.

Also, Austin and Bryan are the only ones who can be \$4.50 and \$4.60. That 2x2 relationship excludes Austin from being \$4.70. Block out that box. Because Fredrick must be 20-cents more than Austin, Fredrick cannot be \$4.90.

shriekingviolet
05-30-2013, 07:03 PM
Gah! I always mess up the transposing step, especially when I'm trying to work quickly. At least this time I just skipped doing it altogether instead of clicking the wrong box ;)

zenobia43
05-30-2013, 08:51 PM
Ax + c = Jm
Ay + c = Kn
Jm <> Kn

This one was relatively straightforward: 2 clues with fixed length, same size inequalities, and, when ordered such that the lesser entity in each clue is on the left, one side of the inequality pair has elements in the same category.

There are tougher situations where a substitution of an existing positive relation has to be made before getting the situation above.

There are others involving more than two ordering/precedence clues, and it is helpful to use the technique shown by the Puzzle Baron in the video. This is the spot in the video where he shows how he uses the notes tab to keep track of the entity ordering.

Often, I find that the puzzles with the longest average solve times are the ones that are cracked with a combination of these ordering clues and other known facts. These are the puzzles that make us doubt that each puzzle can be solved with pure logic.

zenobia43
05-31-2013, 01:20 AM
Here is another example where I found that the ordering clues made the difference between a pure logic solution and a what-if approach.

In the puzzle below, I needed a substitution before I could use clues 6 and 8 to get the key exclusion: Upton is not parakeet.

55

BillsBayou
05-31-2013, 03:39 PM
I get this far when I hit the wall with your puzzle:
56

I've eliminated all but clues 1, 6, 8. How do you do a substitution at this point? I can "What If" and see that Upton cannot be the parakeet, but where's the logic approach?

EDIT: I've managed to push it a bit further. Clue 1 can be restated as "Of the parakeet and the parrot, one was purchased in Upton and the other was bought in May (aka: Quimby)".

Because of clue 8, if one of the birds is in May, the other must be April or June. Thus, Upton cannot be July and, per Clue 6, Bonita cannot be June.

58

shriekingviolet
05-31-2013, 07:55 PM
You've got that Della is the parrot. So from clue 8 Della is one month before the parakeet and from clue six, Bonita is one month before Upton

Della < parakeet
Bonita < Upton

Since Della and Bonita cannot be the same, neither can Upton and the parakeet.

zenobia43
05-31-2013, 08:25 PM
That's it!

By substituting Della for the parrot, you get the relations described by shriekingviolet, and a puzzle that looked like it needed the what-if treatment becomes routine pure logic.

After the substitution, you have two inequalities with entities in the same category on one side and the same distance from the entities on the other side.

As long as the entities on the other side are in different categories, you get another X - hopefully a new one.