Friday, June 26, 2015

Mathematical thinking and how to solve "lowest common ancestor"

It is quite possible on your interview to have questions related to (binary) trees. Here is one that I have on mine a year ago.

Given values of two nodes in a Binary Tree, write a c program to find the Lowest Common Ancestor (LCA).

The solution that I will present is not optimal, but did the job. Most interesting is how I found the solution using technique described in the book
Thinking Mathematically
On interview I implement it in Python but here I will use Common Lisp. Let's start time is up. Here is the scheme from the book that I follow:


It is strange that representation is not given so the first question is: How to represent a tree in Common Lisp?

The easiest way is as nested list. For example:
(1 (2 (6 7)) (3) (4 (9 (12))) (5 (10 11)))

The first element of a list is the root of that tree (or a node) and the rest of the elements are the subtrees (every new open parenthesis is a next level of the tree). For example, the subtree (6 7) is a tree with 2 at its root and the elements 6 7 as its children. So I start digging on this.
Another representation that I found is to use Paul Graham "ANSI Common Lisp" representation:

(defstruct node
  elt (l nil) (r nil))
I usually create a lisp file and start codding, trying and testing. Sometime I receive strange errors.
As you know defstruct in Lisp also defines predicates e.g. node-p. It is possible to clash names easily. Word node is quite used when we talk for trees.
Doing this SBCL returns bizarre message:

caught WARNING:
  Derived type of TREE is
  conflicting with its asserted type
  See also:
    The SBCL Manual, Node "Handling of Types"


DO NOT forget to comment unused code!

Because I have binary tree may be a better way to represent it with nested lists is to use 'a-list' (more readable). For example:
(1 (2 (6 . 7)) (3) (4 (9 (12))) (5 (10 . 11))).
Even better - display empty:
(0 (1 (2 . 3) NIL) (4 (5 (6) NIL) NIL))

What is a node? (elm (left right))

This is not consistent - (left right) could not be treated as 'elm'. Let's fix it:
(0 (1 (2) (3)) (4 (5 (6))))

Great now I have the tree!

;; Format is ugly but more like a tree
(defparameter *bin-tree* '(0
                            (2) (3))


(defun make-bin-leaf (elm)
  "Create a leaf."
  (list elm))

(defun make-bin-node (parent elm1 &optional elm2)
  (list parent elm1 elm2))

;; simple creation test
(deftest test-make-bin-tree ()
    (equal (make-bin-node 0 (make-bin-node 1 (make-bin-leaf 2)
                                           (make-bin-leaf 3))
                          (make-bin-node 4 (make-bin-node 5 (make-bin-leaf 6))))
           '(0 (1 (2) (3)) (4 (5 (6) NIL) NIL)) )))


Define what is node and leafs (not during the interview time is up)

(defun node-elm (node)
  (first node))

(defun node-left (node)
  (first (second node)))

(defun node-right (node)
  (first (third node)))

(deftest test-nodes ()
    (eq (node-left '(1 (2) (3))) 2)
    (eq (node-right '(1 (2) (3 (4) (5)))) 3)
    (eq (node-right 1) nil)
    (eq (node-elm '(1)) 1)
    (eq (node-elm nil) nil)))

;;; Predicates

(defun leaf-p (node)
  "Test if binary tree NODE is a leaf."
  (and (listp node)
       (= (list-length node) 1)))

(defun node-p (node)
  "Test if binary tree node is a node with children"
  (not (leaf-p node)))

;; wishful thinking ;) 
(defun member-p (elm tree)
  (eq (find-anywhere elm tree) elm))


I discover that it very easy to write 'find-anywhere' when we present trees as nested lists this: (0 (1) (2)) 'car' is the current 'cdr's
are the successors and we stop when 'car' is an atom. There is no need to remember DFS.

(defun find-anywhere (item tree)
  "Does ITEM occur anywhere in TREE?
Returns searched element if found else nil."
  (if (atom tree)
      (if (eql item tree) tree)
      (or (find-anywhere item (first tree))
          (find-anywhere item (rest tree)))))

(deftest test-member-p ()
    (eq (member-p 6 *bin-tree*) t)))

Good job! (Cheer up yourself)


Do not make big steps test carefully everything!


Now I have a tree and I can concentrate on solving the problem.


From where to start? As a rule I start from easiest case - one root and two leafs. In this case task is easy. If (root (p) (q)) then 'root' is the parent.


1. Names for the searched leafs: 'p' and 'q'
2. Method signature: (defun lca (root p q) ...)

Concentrate on (elm (left) (right)) and because this is the simplest case apply rule for the rest. From my previous experience I know that binary
trees are recursive structure so recursion will pass perfectly. Of course be careful how to terminate.

Algorithm could be:
- find first in a left sub-tree
- find second in a right sub-tree
- compare to the searched 'p' and 'q'


How to track context? I could find easily element but I have to know its parent. Some how context should be created.

Is it possible to use recursion to create context?
A tree is a collection of smaller trees up to a simple node. In the same way that a list is a collection of smaller list.
We reduce 'elm'(it is tricky to choose proper name could be e.g. 'tree-branch-node-leaf-nil') so it is possible to be node, leaf or null. Wait a minute...

AHA! I could build the recursive algorithm using possible 'elm' values:
(lca-rec nil 1 2)
                         (lca-rec '(1) 1 2)
                         (lca-rec '(0 (1) (2)) 1 2)
And I can even use them also for tests.

Concentrate on the small picture - only one node.

;; if elm is null or leaf
(defun lca-rec (elm p q)
  (cond ((null elm) nil)
        ((or (eq (car elm) p)
             (eq (car elm) q)) elm)

More difficult part - if it is a node (elm (p) (q))

(defun lca-rec (elm p q)
  (cond ((null elm) nil)
        ((or (eq (car elm) p)
             (eq (car elm) q)) elm)
        (t (let ((left (cadr elm))      ; ***
                 (right (caddr elm)))   ; ***
             (when (and left right)     ; ***
                 (car elm))))))         ; ***

Now I have a tree - the same solution recursively and reduce 'elm'

(defun lca-rec (elm p q)
  (cond ((null elm) nil)
        ((or (eq (car elm) p)
             (eq (car elm) q)) elm)
        (t (let ((left (lca-rec (cadr elm) p q))     ; ***
                 (right (lca-rec (caddr elm) p q)))  ; ***
             (when (and left right)
               (car elm))))))

What we have after recursion?
After recursion we are finished with current 'segment' from the big picture.

Is it possible to discover nothing?
YES! Because with recursion we do not traverse the whole tree.

But elements could be on different levels so I somehow to pass trough recursive calls what is found. That is OK, what I return from recursion is passed to the next recursive step.


Now the better question is when it fails? Only when one or none elements are found.
Should I return if only one element is found? YES!

REMEMBER. Look at the recursive procedure as simplest part of something big. Recursive call is the glue to the big picture.

In our case:
.. ((left (lca-rec (cadr elm) p q))
    (right (lca-rec (caddr elm) p q))) ...

finds only 'left' or 'right' of the current node it doesn't search the whole tree. It seems like - but it is not true. That's why I have to return what is fond in current 'node' and the recursion will pass it to the next level until
both are found or not. Finally we have:

(defun lca-rec (elm p q)
  (cond ((null elm) nil)
        ((or (eq (car elm) p)
             (eq (car elm) q)) elm)
        (t (let ((left (lca-rec (cadr elm) p q))
                 (right (lca-rec (caddr elm) p q)))
             (if (and left right) (car elm)
                 ;; return what is found until now and hope other will be found on next iteration
                 (or left right))))))        ;; ***

Done. It seems difficult at first but when practice a while solving such problems are fun.

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