R J wrote:

Can someone provide a hand calculation of: span (< 0) [-1, -2, -3, 0, 1, 2, -3, -4, -5]? I know the result is ([-1, -2, -3], [0, 1, 2, -3, -4, -5]), but the recursion flummoxes me. Here's the Prelude definition:

First, let's simplify the definition. span _ [] = ([], []) span p xs@(x:xs') | p x = (x:ys, zs) | otherwise = ([], xs) where (ys, zs) = span p xs' === span _ [] = ([], []) span p xs@(x:xs') | p x = let (ys, zs) = span p xs' in (x:ys, zs) | otherwise = ([], xs) === span p xs = case xs of [] -> ([], []) (x:xs') -> if p x then let (ys, zs) = span p xs' in (x:ys, zs) else ([], xs) And finally we can pretty up that recursive call (possibly changing strictness, but otherwise keeping the same meaning). span = \p xs -> case xs of [] -> ([], []) (x:xs') -> if p x then first (x:) (span p xs') else ([], xs) And now, the reduction: span (< 0) (-1:-2:-3:0:1:2:-3:-4:-5:[]) == {delta: unfold the definition of span} (\ p xs -> case xs of [] -> ([], []) (x:xs') -> if p x then first (x:) (span p xs') else ([], xs)) (< 0) (-1:-2:-3:0:1:2:-3:-4:-5:[]) == {beta: reduction of application} (let p = (< 0) in \ xs -> case xs of [] -> ([], []) (x:xs') -> if p x then first (x:) (span p xs') else ([], xs)) (-1:-2:-3:0:1:2:-3:-4:-5:[]) N.B. this next step isn't entirely accurate in Haskell since it looses sharing. == {zeta: reduction of let-binding} (\ xs -> case xs of [] -> ([], []) (x:xs') -> if (< 0) x then first (x:) (span (< 0) xs') else ([], xs)) (-1:-2:-3:0:1:2:-3:-4:-5:[]) == {beta, zeta} case (-1:-2:-3:0:1:2:-3:-4:-5:[]) of [] -> ([], []) (x:xs') -> if (< 0) x then first (x:) (span (< 0) xs') else ([], (-1:-2:-3:0:1:2:-3:-4:-5:[]))) == {iota: reduction of case matching} let x = -1 xs' = (-2:-3:0:1:2:-3:-4:-5:[]) in if (< 0) x then first (x:) (span (< 0) xs') else ([], (-1:-2:-3:0:1:2:-3:-4:-5:[]))) == {zeta, zeta} if (< 0) (-1) then first (-1:) (span (< 0) (-2:-3:0:1:2:-3:-4:-5:[])) else ([], (-1:-2:-3:0:1:2:-3:-4:-5:[]))) == {beta} if -1 < 0 then first (-1:) (span (< 0) (-2:-3:0:1:2:-3:-4:-5:[])) else ([], (-1:-2:-3:0:1:2:-3:-4:-5:[]))) Just to be pedantic :) == {delta} if -1 <# 0 then first (-1:) (span (< 0) (-2:-3:0:1:2:-3:-4:-5:[])) else ([], (-1:-2:-3:0:1:2:-3:-4:-5:[]))) == {primitive} if True then first (-1:) (span (< 0) (-2:-3:0:1:2:-3:-4:-5:[])) else ([], (-1:-2:-3:0:1:2:-3:-4:-5:[]))) == {iota} first (-1:) (span (< 0) (-2:-3:0:1:2:-3:-4:-5:[])) We'll ignore that `first` and evaluate under it for clarity, even though this isn't accurate for Haskell's call-by-need strategy. Technically we'd unfold the definition of `first`, do beta/zeta reduction, and then do the following since we'd need to evaluate the scrutinee of the case match. I'll go a bit faster now since it's the same. == {delta, beta, beta, iota, zeta, zeta, beta, delta, primitive, iota} first (-1:) (first (-2:) (span (< 0) [-3,0,1,2,-3,-4,-5])) == {delta, beta, beta, iota, zeta, zeta, beta, delta, primitive, iota} first (-1:) (first (-2:) (first (-3:) (span (< 0) [0,1,2,-3,-4,-5]))) Now things are a little different, so we'll slow down. == {delta, beta, beta} first (-1:) (first (-2:) (first (-3:) ( case [0,1,2,-3,-4,-5] of [] -> ([], []) (x:xs') -> if (< 0) x then first (x:) (span (< 0) xs') else ([], [0,1,2,-3,-4,-5]) ))) == {iota, zeta, zeta, beta} first (-1:) (first (-2:) (first (-3:) ( if 0 < 0 then first (0:) (span (< 0) [1,2,-3,-4,-5]) else ([], [0,1,2,-3,-4,-5]) ))) == {delta, primitive, iota} first (-1:) (first (-2:) (first (-3:) ([], [0,1,2,-3,-4,-5]))) Now for those `first`s. Again, with pure call-by-need evaluation we'll have already done most of this, and will have done it in reverse order. For clarity we'll do it inside-out a la call-by-value. (Because of strictness analysis, Haskell could perform a hybrid CBN/CBV strategy and could actually follow this order if it determined that `first` was strict in its second argument.) == {delta} first (-1:) (first (-2:) ( (\ f p -> case p of (x,y) -> (f x, y)) (-3:) ([], [0,1,2,-3,-4,-5]))) == {beta, zeta} first (-1:) (first (-2:) ( (\ p -> case p of (x,y) -> ((-3:) x, y)) ([], [0,1,2,-3,-4,-5]))) == {beta, zeta} first (-1:) (first (-2:) ( case ([], [0,1,2,-3,-4,-5]) of (x,y) -> ((-3:) x, y))) == {iota, zeta, zeta} first (-1:) (first (-2:) ((-3:) [], [0,1,2,-3,-4,-5])) Note that since the case match is only head-strict, we don't need to evaluate that application of (-3:) to []. That'll be done later in the call to `show` that's implicit on the ghci REPL. == {delta, beta, zeta, beta, zeta, iota, zeta, zeta, beta} first (-1:) ((-2:) ((-3:) []), [0,1,2,-3,-4,-5]) == {delta, beta, zeta, beta, zeta, iota, zeta, zeta, beta} ((-1:) ((-2:) ((-3:) [])), [0,1,2,-3,-4,-5]) -- Live well, ~wren