Notation: (P,Q,R,S,T) denotes that form P is in Switch 1, Q in Switch 2, etc.
If Switch 1 is wrong, then
If Switch 2 is wrong, then
If Switch 3 is wrong, then
If Switch 4 is wrong, then
If Switch 5 is wrong, then
If I can resurrect this one, I think I've spotted the links between all the previous partial solutions, worked out an element that had been missed by all, and can explain the additional hints in the text. Your girlfriend wants you to buy her:
Puzzle 1 (solved by @PeregrineRook):
Puzzle 2 (solved by @ThirupathiThangavel):
Puzzle 3 (worked on by @Techidiot ...
Initial information, re-parsed
We have a sequence of length 2012, in which each term is either N (nicely) or A (always mad) or C (mad converted from nicely). Among terms 2009, 2010, 2011, 2012, three are N and the fourth is not, and that's never happened with any previous set of four consecutive terms.
If any term is C, then
If we ...
That particular puzzle is meant to be solved by trial and error. The Super Hint for that puzzle says:
"Thankfully, it doesn't matter if you get it wrong and make the porters buckle under the strain."
If you input an incorrect configuration, it shows which of the two porters is carrying too much and lets you try again. The solution is not deducible ...
This is a more graphical answer with 1's for mad frogs and 0's for nice frogs.
Here are the four possibilities for 2009, 2010, 2011 and 2012
No we can go backward bit by bit for every sequences, using the two given rules and the knowledge that an ashamed frog never happened before. We see that there is a recurring pattern for each possibilities.
Since 1, ...
Clarification of the Problem and Very Partial Answer
Transforming the story into the expression. A through I are distinct digits from 1 to 9, X and Y different digits within 1 to 9, and m and n any integer.
AB can be
CD can be
m and n
EFG can then only be
Exhaustive list of possible values of EFG: