What’s the point of the “curious will” story? Am I too stupid to see the mistake that I’m supposed to be making?
We have three fractions that don’t add up to 100% - one twelth is missing. And they don’t result in whole numbers when there are 11 cars.
They do happen to add up to whole numbers when there are 12 cars. They still don’t add up to 100%. One twelfth is still missing.
Which part was I supposed to be surprised about, mathematically?
I’ve got that one in a book somewhere. Yup, that’s the answer the book has.
I have loved this gotcha since I was a child (we had both books). I have never done the math before, but you would need at least 10^(9 * 10^8) atoms to encode 3 * 10^10 characters (the size of the online edition of the Encycopedia Britanica). 10^(9 * 10^8) / avogrado’s number would get you the number of moles of atoms you would need (at maximum). It would probably bet better just to store the two sides of the line as amorphous masses than to try to get them to line up in two straight lines, Of course, they would have to be stored in a perfect vacuum to prevent addition or loss of atoms. Wolfram Alpha is currently choking on that division, but once I get a number, all you have to do is multiply it by the mass of one mole of the type of atom you want to store to how much mass you would need.
Some back of the envelope math says avogardo’s number is 6 * 10^23, so 10^(9 * 10^8) / 610^23 is 610^(410^7). The mass of one mole of hydrogen is roughly one gram, so it would take around 610^(4 * 10^7) grams store the encyclopedia. For comparison, the mass of the observable universe is roughly 3.4*10^57. So we would only need a few hundred thousand universes to store the number this way.
Ugh. This is one of those puzzles that makes me angry when I hear the solution, instead of delighted.
I think we’re supposed to use K, rather than r strategy.
The encyclopedia puzzle isn’t about encoding the entire contents onto the gold rod, it’s about using an extremely precise measurement to determine a decimal value that is equivalent to the encoded contents. If “cat” is encoded as “003001020”, then the encoding method puts a decimal in front, giving a value of .003001020. A mark to measure is then placed on the rod .003001020 from the end of the rod representing 0 (the other end of the rod representing 1). When the mark is measured, the measuring device determines that it is .003001020 units from the end. Decode by removing the decimal and then decode the triads, resulting in C-A-T. So, encoding the entire contents isn’t necessary, only unheard-of precision in measurement. At least, that’s my read on the puzzle, and why a single mark is all that is needed.
[Nevermind, got it]
No. Everyone seems to be ignoring the mythical power of exponentiation :-).
Your arithmetic seems to be off.
10^(910^8) / (610^23) is, for all intents of purposes, 10^(9*10^8).
Well, to be exact:
10^(900,000,000) / (6 * 10^23) = 1/6 * 10^(900,000,000 - 23).
And unless that 900,000,000 was an exact figure, we can stop here.
The back of my envelope says:
To store 10^(3 * 10^10) characters, we first need to decide on an encoding. I will assume good compression and just say we can get it down to one bit per character. So, we’ve got 3*10^10 bits to encode.
Now, we want to store that number by marking exactly one place on a rod. That means, we need as many different possible places that we can mark on the rod as there are possible combinations of 3*10^10 bits. (That’s called unary encoding).
We thus need a rod with 2^(310^10) atoms. That is, about 10^(910^9) atoms.
The number of atoms in the universe is estimated at 10^80. That is a lot less.
We don’t just need a few hundred thousand universes to build a rod big enough to store the number this way. That estimate is WAY off.
With all the atom in the universe, we can store about 266 bits with the alien method.
With a hundred thousand universes, we can store 282 bits.
With ten billion universes, we can store 299 bits.
With a googol universes, we can store 598 bits.
On the other hand, if we had a googolplex universes, we could store over three googol bits using unary encoding.
Remember what the bible says:
And Noah said to the animals: “Go forth and multiply!”
But the adders came to Noah and said: “But we cannot multiply, we are adders.”
Then Noah went out and chopped down some trees and built a table of logs. And he showed it to the adders and said: “Look, I have built you a table of logs! Now you adders can multiply!”
But that is the very meaning of “encoding”. You encode a value by mapping it to a another value that you define to be equivalent.
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