Nicolas Couture-Grenier
The first step is to find the largest index j in the list for which a[j] < a[j+1]. The pseudocode is simple...
The pseudo-code represents how you would write this algorithm in an imperative language. If you take the whole algorithm and express it naturally in Haskell, that would likely not be the first step at all. But in any case, I might write that version of the first step like this: j xs = maybeLast . filter snd . zip [0..] $ zipWith (<) xs (drop 1 xs) where maybeLast [] = Nothing maybeLast xs = Just (last xs) Notice my use of the Maybe type - there might not be any such index j at all. Regards, Yitz
j:= n-1
while a[j] > a[j+1] j:=j-1
I've coded a haskell function to do this, but it is much uglier than the pseudocode :
j :: Integral a => [a] -> Int j [] = 0 j xs = if (head (tail (reverse xs)) < last xs) then (length xs)-2 else j (take (length xs - 1) xs)
Does anyone has a more elegant solution for this first step?
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