Langlands Program (2)

Posted on 2016/03/30, 10:58 PM By admin22

関数型言語Haskellで、OctaveとSwiftでテストして前回と同様のことをしてみました。

環境: GHC 7.6.3, Haskell for Mac 1.0.3 / Mac 10.10.5

import Control.Monad

lsmul :: [Int] -> [Int] ->[[Int]]
lsmul [] ys = []
lsmul xs ys = [(map (\n -> n * (head xs)) ys)] ++ (lsmul (tail xs) ys)

poly xs ys = foldl (\x -> zipWith (+) x) (take len $ repeat 0) (addz (lsmul xs ys))
  where len = length xs + length ys - 1
        addz ns = map (\(n, xs) -> (take n $ repeat 0) ++ xs ++ (take (len-n) $ repeat 0)) (zip [0..] ns)

loopf n ls = when (n > 0) $ do
                let xs = foldl (\x -> poly x) [1] [ls, insz(n-1), insz(n-1), insz(n*11-1), insz(n*11-1)]
                print xs
                loopf (n-1) xs

insz n = [-1] ++ (take n $ repeat 0) ++ [1]

main = do
       loopf 13 [1]

import Control.Monad

lsmul :: [Int] -> [Int] ->[[Int]]

lsmul [] ys = []

lsmul xs ys = [(map (\n -> n * (head xs)) ys)] ++ (lsmul (tail xs) ys)

poly xs ys = foldl (\x -> zipWith (+) x) (take len $ repeat 0) (addz (lsmul xs ys))

where len = length xs + length ys - 1

addz ns = map (\(n, xs) -> (take n $ repeat 0) ++ xs ++ (take (len-n) $ repeat 0)) (zip [0..] ns)

loopf n ls = when (n > 0) $ do

let xs = foldl (\x -> poly x) [1] [ls, insz(n-1), insz(n-1), insz(n*11-1), insz(n*11-1)]

print xs

loopf (n-1) xs

insz n = [-1] ++ (take n $ repeat 0) ++ [1]

main = do

loopf 13 [1]

ghc poly.hs

でコンパイル。
係数の計算は実行速度が遅かったので、13までにしました。
このコードだとOctaveの方が速い。

解の個数はHaskellっぽい手法でシンプルに解を求められます。

./poly

の実行結果

素数3,5,7,9,13まで、係数と解の個数が一致しているのが確認できました。

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