@St_Rev @themattsimpson it seems like the hard part is showing you're on the "same river" as before
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Replying to @sarahdoingthing
@St_Rev @themattsimpson (once you complete the portage)1 reply 0 retweets 1 like -
Replying to @sarahdoingthing
@sarahdoingthing @themattsimpson Well, going one way is frequently easy, lifting back to the original problem not so much.1 reply 0 retweets 0 likes -
Replying to @St_Rev
@sarahdoingthing Here's an example: the idea of 'invariants'. Say you have two colossally tangled masses of string, and you want to know...1 reply 0 retweets 1 like -
Replying to @St_Rev
@sarahdoingthing ...if they're the 'same knot', topologically. This is super hard! But there are things called knot invariants.1 reply 0 retweets 2 likes -
Replying to @St_Rev
@sarahdoingthing A knot invariant is a recipe for turning a horribly complicated object (a tangled ball of string) into a simple one...2 replies 0 retweets 1 like -
Replying to @St_Rev
@sarahdoingthing ...like a number or a polynomial, that always gives you the same value if you start with the same object.1 reply 0 retweets 1 like -
Replying to @St_Rev
@sarahdoingthing So maybe you apply the Linking Number invariant https://en.wikipedia.org/wiki/Linking_number … to your two balls of string.2 replies 0 retweets 0 likes -
Replying to @sarahdoingthing
@sarahdoingthing That's pretty much right. Frictionless elastic yarn, can they be yanked into the exact same configuration w/o cutting?1 reply 0 retweets 1 like
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