# Frog Question Mega Thread (Probably)

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If you haven't already, take a look at this video.

Let's all give it a go for a little bit to try to either disprove or solve some situations of the topic mentioned at the end.

That one guy that's always on

Hi Lucas, awesome challenge, so you want to see if anything is solvable in the waterlogged lillipad situation?

Either any is solvable, any is unsolvable, or just example solutions to the problem.

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I'm trying to think about the best way to display solutions to this puzzle XD, I mean besides getting someone to film me using poker chips 😉

PlanckTime - Amin

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In my answer, I use the term "joint". This is a new idea I am working on to solve this problem, thinking of a waterlogged lily pad as a sort of join that connects two different components of one thing. Also, by trivial and nontrivial, I mean they either do matter to the solving and is a join that connects to another component (nontrivial), or it just dangles on the end, doing nothing (trivial). Without further ado, here is the start of my "joint theory" way of proving/disproving this (tell me what you think on the name).

That one guy that's always on

Just a reminder: this probably isn't going to branch into an actual theory that is studied. I just thought it sounded like a good name.

Sounds awesome ahhah, I see what you've started to do. I had an idea. I think that the use of a graph (as in from graph theory, not x-y axes), to illustrate all the possible places the frogs can jump. Do that for 1 frog, 2 frogs... and where they overlap is possible solutions to the problem.

(This is just a spitball, it may not actually work, or may be unfeasibly long to do)

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Here's a clarification: you use the central joint (or the median joint, as I am now calling it) as a way to figure out how to solve the problem. I will get to even number of joints eventually, which will lead up to what I think is a half-solution of the problem, since it only covers the technique of joint theory to solve it. Here is the "lesson".

That one guy that's always on

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