Given two nodes of a tree, find their most recent common ancestor.

Input. The input consist of one tree. The first line of input is a pair of integers x and y separated by a space. x and y are the nodes that you will consider. The second line of input is a single integer n which is the count of edges in the tree. Each one of the next n lines consist of a pair of integers a and b separated by a space. The pair a b corresponds to an edge from a and b. The following is an example input.

5 7
6
1 2
1 3
3 4
4 5
4 6
6 7

Output. The output consist of a single integer, the most recent common ancestor. The following output corresponds to the example input.

4

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2017-10-08 The Book of Problems currently supports Ruby, Python and now Elixir. More languages will come soon. If there is a language you’d like to use, drop me a line in the comments section.

Solution

We find the most recent common ancestor in the following way.

We load the graph into a data structure such that there is a reference from each node to its parent. The following diagram illustrates the data structure for the example input.

   1
  / \
 2   3
      \
       4
      / \
     6   5
    /
   7

We place two pebbles A and B on top of the nodes we consider, like so.

   1
  / \
 2   3
      \
       4
      / \
     6   5 A
    /
   7 B

We move the pebble that is farther from the root to the same distance than the other. In our example, we move A from 7 to 6.

   1
  / \
 2   3
      \
       4
      / \
   B 6   5 A
    /
   7

We move both pebbles closer to the root one step at a time until they meet. In our example the pebbles meet at 4.

   1
  / \
 2   3
      \
     B 4 A
      / \
     6   5
    /
   7

The node where the pebbles meet is the most recent common ancestor.

Implementation

Here is a Ruby implementation.

#!/usr/bin/env ruby

def depth n, p
  i = 0
  while !!p[n]
    n = p[n]
    i += 1
  end
  return i
end

def most_recent_common_ancestor a, b, p
  da = depth a, p
  db = depth b, p
  if da < db
    diff = db - da
    while diff > 0
      b = p[b]
      diff -= 1
    end
  elsif db < da
    diff = da - db
    while diff > 0
      a = p[a]
      diff -= 1
    end
  end
  while a != b
    a = p[a]
    b = p[b]
  end
  return a
end

def main
  x, y = readline.strip.split(' ').map(&:to_i)
  n = readline.strip.to_i
  p = {}
  while n > 0
    a, b = readline.strip.split(' ').map(&:to_i)
    p[b] = a
    n -= 1
  end
  puts most_recent_common_ancestor x, y, p
end

main

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