50 lines
1.3 KiB
Haskell
50 lines
1.3 KiB
Haskell
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-- It is possible to show that the square root of two
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-- can be expressed as an infinite continued fraction.
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--
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-- √ 2 = 1 + 1/(2 + 1/(2 + 1/(2 + ... ))) = 1.414213...
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--
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-- By expanding this for the first four iterations, we get:
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--
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-- 1 + 1/2 = 3/2 = 1.5
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-- 1 + 1/(2 + 1/2) = 7/5 = 1.4
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-- 1 + 1/(2 + 1/(2 + 1/2)) = 17/12 = 1.41666...
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-- 1 + 1/(2 + 1/(2 + 1/(2 + 1/2))) = 41/29 = 1.41379...
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--
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-- The next three expansions are 99/70, 239/169, and 577/408,
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-- but the eighth expansion, 1393/985,
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-- is the first example where the number of digits in the numerator exceeds
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-- the number of digits in the denominator.
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--
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-- In the first one-thousand expansions,
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-- how many fractions contain a numerator with more digits than denominator?
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--
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-- Thoughts :
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-- 1 + 1/2 = 3/2
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-- 1 + 1/(2 + 1/2) = 1 + 1/(5/2)
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-- = 1 + 2/5
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-- = 5 + 2 / 5
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-- = 7/5
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-- F_n = p/q
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-- F_{n+1}
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--
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-- 1 + 1/( 1 + p/q ) = 1 + 1/( q+p / q)
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-- = 1 + q / q+p
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-- = 2q+p / q+p
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--
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next (p,q) = (2*q + p , p+q)
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nb_digits n = length $ show n
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digits 0 x = []
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digits n x = y:(digits (n-1) y)
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where y = next x
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greater_numerator (p,q) = (nb_digits p) > (nb_digits q)
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nb_greater_numerator n = length $ filter (greater_numerator) (digits n (1,1))
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main = do
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putStrLn $ show $ nb_greater_numerator 1000
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