How to neatly get the sum of symmetric elements in a list?

This is a simple though somewhat wasteful solution:

symmSum[l_List] := Take[l + Reverse[l], Ceiling[Length[l]/2]]

It will give you the middle element twice, not once. Is it important that you only get c (and not 2c) when applying this function to {a,b,c,d,e}? That's easy to do (avoiding computing elements twice is also easy), but will make the function slightly longer. These solutions all use Length though.

---

Here's a pattern-based solution which avoids Length:

iter[{result___}, {s_, mid___, e_}] := iter[{result, e + s}, {mid}]
iter[{result___}, {}] := {result}
iter[{result___}, {mid_}] := {result, mid}

symmSum[l_List] := iter[{}, l]

You may want to modify this as

symmSum[l_List] := Block[{$IterationLimit = Infinity}, iter[{}, l]]

to make it work for arbitrarily long lists.

Playing with patterns:

{a, b, c, d, e} //. {h_, b___, t : Except[_List]} :> {h + t, {b}} // Flatten

This can be written more efficiently, without the full rescanning inherent in //., using recursion:

f1 = # /. {h_, b___, t_} :> Prepend[f1 @ {b}, h + t] &;

Also as a DownValues definition which is a bit more efficient still:

f2[{h_, b___, t_}] := Prepend[f2 @ {b}, h + t]
f2[x_] := x

f2 @ {a, b, c, d, e}
f2 @ {a, b, c, d, e, f}

{a + e, b + d, c}
{a + f, b + e, c + d}

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Disregarding elegance, this is the fastest method I could come up with for packed arrays:

Module[{ln = Length@#, x},
  x  = #[[ ;; ⌈ln/2`⌉ ]];
  x += #[[ -1 ;; ⌊ln/2`⌋ + 1 ;; -1 ]];
  If[OddQ @ ln, x[[-1]] /= 2 ];
  x
] &

I imagine it can be bested by compile-to-C in v8, but I don't have that.

What's wrong with Length? It's a very efficient function to callculate (since lists are vectors internally, i.e. arrays of fixed size, whose length is precisely known to the program at all times).

symSum = Module[{result, length = Length[#]},
        result = (# + Reverse@#) [[1 ;; Ceiling[length/2]]];
        If[OddQ[length], result[[-1]] /= 2];
        result
    ] &;

symSum@{a, b, c, d, e, f}
symSum@{a, b, c, d, e}

{a + f, b + e, c + d}
{a + e, b + d, c}