Math Puzzle
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
Well done :)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
Genius!:) Brad Jennings "You're mom is nice. Mind if I go out with her?" - Jörgen Sigvardsson
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0 is simple, so is 1. For the rest: n=-log2[log2(sqrt(sqrt(...n times...(2)...))))] Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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You have a degree in mathematics, don't you? :) -- I am on fire. Do you need a light?
No I don't :) I'm a electrical engineer, but used to be fond of these kind of quizzes when I was younger. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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No I don't :) I'm a electrical engineer, but used to be fond of these kind of quizzes when I was younger. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
Have you read Gödel Escher Bach - The Eternal Golden Braid? Your solution reminded me a lot of the authors (Hofstadter) representation of the natural numbers. He defined it by an axiom 0 (zero) and an operation S (successor). Your log/sqrt solution for simulating S gave me a flashback.. :) -- I am on fire. Do you need a light?
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
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my answer -~-~...-~2 is simpler and requires only one 2.
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
Brilliant! Well done :)
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
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The answer would involve some constructed function f(x) that increases x by 1. Then, applying f(x) to itself would allow all the positive numbers be generated. IE, f(2) = 3, and f(f(2)) = 4, and f^n(2) = 2+n-1. Since 2 is the only number we are allow, f(2) is really the only possibility, unless you want to consider f(22) or f(222) or f(2/2) or etc, but then we wastes our valuable 2s. Some functions f(x) that satisfy, are: f(x) = -(~x) --> bitwise negation followed by arithmetic negation f(x) = combination of logs and sqrts of a prior post so, my solution, the first f(x), can obtain, for example, 5 which is -~-~-~2. Thanks, Wes
Regarding, my answer, -~-~-~ ... -~2, it is can be done with log n operations by appropriate using replacing some of the inner operations with the sqr or factorial functions. Thanks, Wes
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Have you read Gödel Escher Bach - The Eternal Golden Braid? Your solution reminded me a lot of the authors (Hofstadter) representation of the natural numbers. He defined it by an axiom 0 (zero) and an operation S (successor). Your log/sqrt solution for simulating S gave me a flashback.. :) -- I am on fire. Do you need a light?
Yes, that was a nice book. If you like this sort of stuff, a lighter, but equally fun book on logic matters is What is the Name of this Book? by Raymond Smullyan. It has some Gödelian discussions in the last chapters. Joaquín M López Muñoz Telefónica, Investigación y Desarrollo
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Atlantys wrote: The largest number I can do is: 4194304 (2^22) Really?! What about 222! = approx 1.12*10426
Ryan
"Punctuality is only a virtue for those who aren't smart enough to think of good excuses for being late" John Nichol "Point Of Impact"
Doh! I was trying to use an operator that causes the system to grow large very quickly. Hence ^. I complete forget about !. Dammit! I suppose that's what staying up late does to the brain. Of course... you can then do ((222!)!), etc. :~ :~ I prefer to wear gloves when using it, but that's merely a matter of personal hygiene [Roger Wright on VB] Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning. [Rich Cook]