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Comment Re:How about (Score 1) 112

Oh, forgot to link the dry density for you: here you go. 341kcal/100g. Aka 3,41kcal/g.

Which, like I said, should be obvious, since they're almost entirely carbs (~4kcal/g) and protein (~4kcal/g), and they're, as noted, dry (12-16% moisture). It would be quite the trick indeed to get something that is dry and and is almost entirely comprised of things that are 4kcal/g to be 1,38kcal/g! ;)

Comment Re:How about (Score 1) 112

Just in case you need help:

Your calculation: 195g (dry weight) × 1.38 kcal/g = 269 calories per pound of cooked beans.
Correction: Because you used 1.38 kcal/g (the cooked density) as if it were the dry density, you essentially diluted the calories twice.
The Actual Math: 195g of dry beans * 3.4 kcal/g (actual dry density) = 663 kcal.

When those 195g of dry beans absorb water to weigh 454g (1 pound), they still contain those same 663 calories (since water has zero calories).

Comment Re:How about (Score 1) 112

Canned beans are ALREADY COOKED. *facepalm*. You can eat them straight out of the can.

which is waaaay more than I would want to eat at a sitting.

I can't think of a single ingredient - any ingredient - that I would want to eat exclusively as my diet, so this is a really stupid argument.

Comment Re: I'm I'm skeptical too. (Score 1) 85

I think a giant context is not going to be the answer. It's just got too many problems. Better will probably be parsing the context into connected pieces, and at a different level assembling the "lemmas" into "theorums". (Yeah, those aren't quite the right words, but I'm not sure the right words exist, and that's the analogy from math proofs. Code library isn't the right concept as the "lemma" will often be quite specific to the current task.)

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