Determine least common ancestor at compile-time

Question

There are tons of questions about the least common ancestor algorithm, but this one is different because I'm trying to determine the LCA at compile-time, and my tree is neither binary nor a search tree, even though my simplified version may look like one.

Suppose you have a bunch of structures which contain a member typedef parent, which is another similar structure:

struct G
{
    typedef G parent;    // 'root' node has itself as parent
};

struct F
{
    typedef G parent;
};

struct E
{
    typedef G parent;
};

struct D
{
    typedef F parent;
};

struct C
{
     typedef F parent;
};

struct B
{
    typedef E parent;
};

struct A
{
     typedef E parent;
};

which together make a tree such as

A    B    C    D
 \  /      \  /
  E         F
   \       /
    \     /
     \   /
       G

NOTE: There is no inheritance relationship between the structures.

What I would like to do is create a type-trait least_common_ancestor such that:

least_common_ancestor<A, B>::type;    // E
least_common_ancestor<C, D>::type;    // F
least_common_ancestor<A, E>::type;    // E
least_common_ancestor<A, F>::type;    // G

What's the best way to do this?

I'm not concerned with algorithmic complexity particularly since the tree depth is small, rather I'm looking for the simplest meta-program that that will get to the correct answer.

EDIT: I need to be able to build the solution with msvc2013, among other compilers, so answers without constexpr are preferred.

score 11 · Accepted Answer · answered Apr 11 '16 at 15:11

This may probably be improved, but you could first compute the depth of your type and then use this information to go up one branch or the other:

template <typename U, typename = typename U::parent>
struct depth {
    static const int value = depth<typename U::parent>::value + 1;
};

template <typename U>
struct depth<U, U> {
    static const int value = 0;
};

The above will basically compute the depth of your type in your tree.

Then you could use std::enable_if:

template <typename U, typename V, typename Enabler = void>
struct least_common_ancestor;

template <typename U>
struct least_common_ancestor<U, U> {
    using type = U;
};

template <typename U, typename V>
struct least_common_ancestor<U, V,
                             typename std::enable_if<(depth<U>::value < depth<V>::value)>::type> {
    using type = typename least_common_ancestor<U, typename V::parent>::type;
};

template <typename U, typename V>
struct least_common_ancestor<U, V,
                             typename std::enable_if<(depth<V>::value < depth<U>::value)>::type> {
    using type = typename least_common_ancestor<V, typename U::parent>::type;
};

template <typename U, typename V>
struct least_common_ancestor<U, V,
                             typename std::enable_if<!std::is_same<U, V>::value && (depth<V>::value == depth<U>::value)>::type> {
    using type = typename least_common_ancestor<typename V::parent, typename U::parent>::type;
};

Output:

int main(int, char *[]) {

    std::cout << std::is_same<least_common_ancestor<A, B>::type, E>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<C, D>::type, F>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<A, E>::type, E>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<A, F>::type, G>::value << std::endl;
    std::cout << std::is_same<least_common_ancestor<A, A>::type, A>::value << std::endl;

    return 0;
}

Gives:

1 1 1 1 1

This may probably be improved but could be used as a starting point.

Sorry, I should have mentioned I'm building on msvc2013 which doesn't support `constexpr`. — Nicolas Holthaus, Apr 11 '16 at 15:16
@NicolasHolthaus You can replace `constexpr` with `const` in this case, it was just cleaner for me. I have updated the answer. — Holt, Apr 11 '16 at 15:16
Thanks, I'm testing it out. Visual Studio is also weird about how it chooses template specializations since it doesn't do two-phase lookup so I'll probably have to re-factor the `enable_if` statements before I know if it will work. — Nicolas Holthaus, Apr 11 '16 at 15:23
@NicolasHolthaus I cannot test it right now with msvc 2013, but it looks like it works with msvc 2015 (http://rextester.com/l/cpp_online_compiler_visual). — Holt, Apr 11 '16 at 15:29

Piotr Skotnicki · Answer 2 · 2016-04-12T08:58:30.950

template <typename...> struct typelist {};

template <typename T, typename... Ts>
struct path : path<typename T::parent, T, Ts...> {};

template <typename T, typename... Ts>
struct path<T, T, Ts...> { using type = typelist<T, Ts...>; };

template <typename T, typename U>
struct least;

template <typename T, typename... Vs, typename... Us>
struct least<typelist<T, Vs...>, typelist<T, Us...>> { using type = T; };

template <typename T, typename W, typename... Vs, typename... Us>
struct least<typelist<T, W, Vs...>, typelist<T, W, Us...>>
    : least<typelist<W, Vs...>, typelist<W, Us...>> {};

template <typename V, typename U>
using least_common_ancestor = least<typename path<V>::type, typename path<U>::type>;

DEMO

Bottom-up: Form paths (path::type) from both nodes to the root by means of prepending a typelist at each level (path<?, T, Ts...>), until parent equals the currently processed node (<T, T, ?...>). Moving upwards is performed by replacing T by T::parent.
Top-down: Descend the two typelists simultaneously (least), until there's a mismatch at corresponding positions (Vs..., Us...); if so, the last common node is the common ancestor (T); otherwise (<T, W, ?...>, <T, W, ?...>), remove the matching node (T) and step one level down (now W is the last known common node).

score 5 · Answer 3 · answered Apr 11 '16 at 15:17

This is probably not the most algorithmically efficient approach, but it's functional.

First, we're going to create lists out of the ancestors of each type. So for A this would be <A,E,G> and for G this would be <G>:

template <class X>
using parent_t = typename X::parent;

template <class... > struct typelist {}; 
template <class T> struct tag_t { using type = T; };

template <class, class> struct concat;
template <class X, class Y> using concat_t = typename concat<X, Y>::type;

template <class... Xs, class... Ys> 
struct concat<typelist<Xs...>, typelist<Ys...>>
: tag_t<typelist<Xs..., Ys...>>
{ };

template <class X, class = parent_t<X>>
struct ancestors
    : tag_t<concat_t<typelist<X>, typename ancestors<parent_t<X>>::type>>
{ };

template <class X>
struct ancestors<X, X>
    : tag_t<typelist<X>>
{ };

template <class X>
using ancestors_t = typename ancestors<X>::type;

Then the least common ancestor of two nodes is just going to be the first node in one node's ancestors that is contained in the other node's ancestors:

template <class X, class TL> struct contains;
template <class X, class TL> using contains_t = typename contains<X, TL>::type;

template <class X, class... Xs>
struct contains<X, typelist<X, Xs...>> : std::true_type { };

template <class X, class Y, class... Xs>
struct contains<X, typelist<Y, Xs...>> : contains_t<X, typelist<Xs...>> { };

template <class X>
struct contains<X, typelist<>> : std::false_type { };

template <class X, class Y>
struct lca_impl;

template <class X, class Y>
struct lca : lca_impl<ancestors_t<X>, ancestors_t<Y>> { };

template <class X, class... Xs, class TL>
struct lca_impl<typelist<X, Xs...>, TL>
    : tag_t<
        typename std::conditional_t<contains_t<X, TL>::value,
            tag_t<X>,
            lca_impl<typelist<Xs...>, TL>
            >::type
        >
{ };


template <class X, class Y>
using lca_t = typename lca<X, Y>::type;

which has the behavior you'd expect:

static_assert(std::is_same<lca_t<A, E>, E>{}, "!");
static_assert(std::is_same<lca_t<A, D>, G>{}, "!");

Determine least common ancestor at compile-time

3 Answers3