#1




Multiple greater/lesser clues
I've gotten stuck on several puzzles because of greater/lesser values. I understand how these relative values work, but I've often found that clues seem to be missing on one side or the other of the relative values. Am I missing something, or do several puzzles omit key information?

#2




Each puzzle can be solved with the given clues.
The greater than/less than clues can be combined to get false relationships needed to solve the puzzle. There are many ways the clues can be combined, and cerine's post #2 in the following forum thread illustrates one of the more common ways using "fixed distance" or "specific" greater than/less than clues. https://www.logicpuzzles.org/forum/...ead.php?t=1073 
#3




I have a lot of trouble with those kinds of clues, too. After getting that type of hint multiple times, I tried following the logic from each one until the patterns started to click. I still get stuck on them sometimes, but if I hit a wall, I figure that there's a multiple greater/lesser clue somewhere and I can figure it out.

#4




There are many ways the greater than/less than clues can be combined, and the puzzle creator can make the logic pretty interesting with four categories and seven rows to work with.
In the following examples, there are four categories, and the lesser quantity is always on the left. The puzzle creator can make things more interesting by mixing up the lesser and greater quantity presentation order in the clues. I have the normalized the order in the examples to facilitate comparison among the examples. By convention, I am calling the ordering category A (left side of upper left subgrid) and the other three categories: B, C, and D (tops of the upper three subgrids in a 4Xn puzzle). The categories and row numbers will be determined from the clues. The examples just illustrate the logic for a given assignment of categories and row numbers. 1. B1 < C1, C1 < D1 => B1 != D1 If B1 is less than C1, and C1 is less than D1, then B1 cannot be in the same row as D1. 2. B1 > C1, B1 >> D1 => C1 != D1 If B1 is exactly one row above C1, and B1 is exactly two rows above D1, then C1 cannot be in the same row as D1. 3. B1 > C1, B1 > D1 => C1 == D1 If B1 is exactly one row above both C1 and D1, then C1 and D1 must be in the same row. 4. B1 > C1, B2 > D1 => C1 != D1 If an element of category B is exactly one row above C1, and a different element of category B is exactly one row above D1, then C1 cannot be in the same row as D1. This is because the two elements of category B cannot be in the same row, and the B elements are a fixed distance from elements in the other two categories. Example 4 works for any fixed distance, and the common category elements (B1, B2) can be on the right side instead of the left side. C1 >> B1, D1 >> B2 => C1 != D1 C1 and D1 are both exactly two rows above B1 and B2 respectively. Since B1 and B2 cannot be in the same row and since they are a fixed distance from C1 and D1, C1 and D1 cannot be in the same row either. 5. I hesitate to include this next one, but the hint system has confused lots of people when grid state and/or substitution is involved. All of the above examples can be obscured by requiring a substitution before arriving at the canonical representation shown in the examples above. B1 > C1 (from clue) C2 > D3 (from clue) B2 = C2 (from clue or current solution grid state) B2 > D3 (after substituting B2 for C2) C1 != D3 (deduction) If B1 is exactly one row above C1, C2 is exactly one row above D3, and B2 and C2 are in the same row, then C1 cannot be in the same row as D3. This is because B2 is in the same row as C2, and B2 and B1 cannot be in the same row. The fixed distance relationships between the resulting B category elements and C1 and D3 put C1 and D3 in different rows. The full tutorial can be found here: https://www.logicpuzzles.org/howto...gicpuzzle.php 
#5




Thanks for the help, but here's a specific...
Found one! I still can't seem to find out how things are linked up. In this example, it's close, with hints #2 and #8:
Tau Beta Pi > Cedar Street / Sigma Beta > Lawn Street ...therefore Tau Beta Pi is inequal to Lawn Street. All good so far. But... I cannot find a connection between Lawn Street and Theta Delta. How do I deduce that? I can't find any further links/relationships. Help...? What am I missing? 
#6




Ok so, clue 8 tells us that Cedar St has 6 more members than Lawn Street
(or, Lawn St has 6 fewer members than Cedar Street) Clue 11 tells us that Theta Delta has 6 fewer members than York Court So Lawn Street has six fewer members than Cedar St But Theta Delta has six fewer members than York Court ergo Theta Delta can't be on Lawn Street, otherwise York St and Cedar St would need to have the same amount of members 
#7




Ariana's post says it all, and I have annotated the post to use the canonical notation.
Quote:

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