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Old 04-08-2013, 09:18 PM
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Quote:
Originally Posted by zenobia43 View Post
The left side of clue 8 can never be $120 or $124, so that means 105 and 1997 can never be those values either. Now we have the part that really matters, but continue looking at each side of clues 8 and 9 to deduce what the other side can never be and fill in the Xs. I got 9.
Nice. In a 4 part either/or clue, make one side of the argument the overlapping eliminations from a single category.

Follow that logic, and since 1997 cannot be 100, 105, and 120, the overlap is 100 and 120. Eliminate those values from the $145 and Securalert. And that zaps 1994 and Secura by transposing values.

I can follow you along on the rest of processing clues 8 and 9
Quote:
Originally Posted by zenobia43 View Post
(SNIP)
Look at 1997 and $120. They are both in rows 3 and 4. Can they be in the same row? Nope. 1997 cannot be $120. They have to be in different rows.

Clue 5 says that 2007 is one row below $120. If $120 is in row 3, then 2007 is in row 4, but since $120 and 1997 cannot be in the same row, 1997 must be in row 4. Oops! Can't have 1997 and 2007 in the same row.

So $120 must be in the fourth row, 2007 in the fifth row, and 1997 in the third row.
That is quite the BANG getting over that one. But, yeah, it does look like a "What If" step.

You got the solution right on that one. Thanks for the advanced solving technique.

How would we write that up?

Quote:
Technique: In a 4 part either/or clue, make one side of the argument the overlapping eliminations from a single category.

Categories CAT-A, CAT-B... are made up of{A1, A2..., A(n)}, {B1, ..., B(n)}...

Clue type: Of A(x) and B(x), one is C(x) and the other is D(x)

A(x) cannot be B(x); C(x) cannot be D(x) by definition of the clue type.

Find all eliminations on the grid for A(x) in category CAT-B. The result set is all non-A(x) values in CAT-A.

Example: For a given A1 and B1, under CAT-A we see that B1 cannot be A1 and A2. The resulting overlap set is {A2}

Expand the result set by comparing all eliminations in A(x) and B(x) across all the categories in the puzzle.
Example: For a given A1, B1, the matching eliminations are {A2, C2, C3, D4, E1, E5}. Thus, the values for the right side of the equation, C(x) D(x), cannot be any of the values in the result set.

Eliminate all occurrences of the result set from C(x) and D(x).

Repeat for all values of A(x) that cannot exist for B(x). Repeat entire process for C(x) and D(x).
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