CSE 102 – Tutorial 5 – Programming with trees

General instructions

After you are finished with the tutorial, please upload your submission to Moodle by completing “Tutorial 5 assignment” before the indicated deadline.

Binary trees and binary search trees

In the problems below, we will use the following class to represent the nodes of a binary tree:

class Node:
    def __init__(self, value, left=None, right=None):
        self.value = value
        self.left = left
        self.right = right

Each node stores a value, and optionally pointers to its left and right children nodes (otherwise set to None if the node does not have a left/right child). A tree is specified by its root node, with the empty tree represented by None.

For example, the binary tree

is represented by the following Python value:

  T1 = Node(0, Node(1, Node(2), Node(3)), Node(4, Node(5)))

while the pair of binary trees

are represented by the following pair of Python values, respectively:

  T2 = Node(0, Node(1), Node(2, Node(3), Node(4, Node(5))))
  T3 = Node(11, Node(10, Node(9, None, Node(8)), Node(7)), Node(6))

For the following problems we will consider binary search trees (BSTs), which you saw previously in CSE101. Recall that a binary tree has the BST property if for every node of the tree, the value at that node is greater than or equal to every value in its left subtree, and less than or equal to every value in its right subtree.

For example, of the following two trees, the one on the right has the BST property, but the one on the left does not:

Searching and insertion into a binary search tree can be performed by recursively choosing to follow the left or right branch depending on the value being searched/inserted. Let us remind you how to do this in Python:

def bst_search(node, x):
    if node is None:
        return False
    elif node.value == x:
        return True
    elif x < node.value:
        return bst_search(node.left, x)
    else:
        return bst_search(node.right, x)
    
def bst_insert(node, x):
    if node is None:
        return Node(x)
    elif x <= node.value:
        node.left = bst_insert(node.left, x)
    else:
        node.right = bst_insert(node.right, x)
    return node

We are now interested in implementing removal of a value from a BST. The procedure works as follows. We start by looking for the node \(N\) of the tree containing the given value to be deleted. Once it has been found, several cases need to be considered:

• \(N\) is a leaf: since the node has no children, we just remove it.

• \(N\) has exactly one child: we remove \(N\) from the tree and replace it with its child.

• \(N\) has two children: In that case, we do not remove \(N\), but instead we replace its value with the value of its in-order successor \(S\) (= the left-most node of the right-subtree of \(N\)). Then, we remove \(S\): since \(S\) has at most one child, we are back in one of the two previous cases.

If you don’t know the right word to use, trie guessing (optional)

A word square is a square of words that can be read both horizontally and vertically, for example:

===========
| L O V E |
| O V A L |
| V A N S |
| E L S E |
===========

In this case the square contains the same words both horizontally and vertically, but in general a word square does not necessarily have to be symmetric (although apparently symmetry is expected for the French version, carrées magiques). For example, the following is a non-symmetric word square:

===========
| L O V E |
| O M E N |
| S E N D |
| E N D S |
===========

The notion naturally generalizes to word rectangles, such as the following 4 x 6 rectangle:

===============
| P Y T H O N |
| E U R O P E |
| E L O P E S |
| D E T E C T |
===============

In Volume 4B of The Art of Computer Programming, Donald Knuth suggests a procedure for constructing \(m \times n\) word rectangles using backtracking combined with a trie data structure. A trie is a kind of dictionary that supports \(O(k)\) worst-case search and insertion time for any given string of length \(k\) (in particular, the time does not depend on the number of strings in the dictionary). The key property of tries that make them useful for backtracking is that they permit efficiently testing whether a string is a valid prefix of a collection of strings, and of finding out what are its possible suffixes. Before discussing the backtracking algorithm itself, let us take a moment to describe tries in more detail, beginning with how to visualize them.

Here is a picture of a trie (adapted from the Wikipedia article linked above) containing the seven strings 'to', 'tea', 'ted', 'ten', 'A', 'in', and 'inn':

The idea is to organize strings into a tree based on their common prefixes. Each of the eleven labelled nodes logically represents a valid prefix of one of the strings in the dictionary, e.g., node 5 represents the prefix 'te', and has three children corresponding to the possible extensions of the prefix by one letter. A red circle indicates that the trie accepts the current prefix as a string, i.e., that it is a complete word in the dictionary. Observe that all of the leaves are accepting, but so is the internal node 6 in this example, since 'in' is both a valid prefix of 'inn' and also a valid string itself. As usual when talking about trees, we think of the trie itself as represented by its root node, i.e., node 0 representing the empty prefix (a.k.a. ε, which is not a word in our dictionary although it is a prefix of every word!).

Now here is a description of Knuth’s algorithm.

Beginning with an empty rectangle — or perhaps beginning with a particular word on the first row, such as “PYTHON” — we start filling in letters contiguously, going across the rows and then down the columns. At each step, we ensure that all of the rows contain valid prefixes of \(n\)-letter words, and that all of the columns contains valid prefixes of \(m\)-letter words. Indeed, to ensure this we just need to check that the newly introduced letter does not create any invalid prefixes. At the end of this process, if we have managed to fill in all of the \(m \times n\) letters then we have a valid word rectangle.

Let’s illustrate with a simple example for the case \(m = n = 4\). Suppose we have already filled in the first two rows and the first two letters of the third row:

===========
| L O V E |
| O M E N |
| S E ? ? |
| ? ? ? ? |
===========

At this point we consider the letters that can possibly follow “SE” and “VE”. Our dictionary of 4-letter words contains 26 possible suffixes for “SE-” (“-AL”, “-AM”, …, “-WS”, “-XY”) corresponding to 10 distinct initial letters, and 12 possible suffixes for “VE-” (“-AL”, “-ER”, …, “-TO”, “-TS”) corresponding to 7 distinct initial letters. By considering the intersection of the sets of initial letters, we see that there are only 5 possible letters that can go in the third column of the third row: A, E, N, R, or T. Suppose we go with E:

===========
| L O V E |
| O M E N |
| S E E ? |
| ? ? ? ? |
===========

We continue filling in letters, say until we end up with the following square:

===========
| L O V E |
| O M E N |
| S E E D |
| T ? ? ? |
===========

At this point we’re stuck, because although our dictionary of 4-letter words contains 156 possible suffixes for “T-” corresponding to 11 distinct initial letters, it only has one possible suffix for “OME-” (“-N”), and the intersection of these sets is empty.

We can backtrack on the choice of T for the first letter of the fourth row, making more progress by instead choosing E:

===========
| L O V E |
| O M E N |
| S E E D |
| E N ? ? |
===========

However, at this point we get stuck again, since the only possible suffix of “VEE-” is “-R”, and “ENR-” is not a valid prefix of a 4-letter word. Eventually, we’ll backtrack to the point where we chose E as the third letter of the third row and replace it by another letter, say N:

===========
| L O V E |
| O M E N |
| S E N ? |
| ? ? ? ? |
===========

And now we can work our way forwards again, until we arrive at the following (unique) solution:

===========
| L O V E |
| O M E N |
| S E N D |
| E N D S |
===========

Implementing tries

Your first goal is to understand and complete the following partially-completed implementation of tries:

class Trie:
    def __init__(self, words = []):
        '''Creates a trie from a list of words.'''
        # initialize an empty trie
        self.__accept = False
        self.__step = {}
        # insert the list of words into the trie
        for w in words:
            self.insert(w)

    def __bool__(self):
        '''Returns True iff the trie contains any words.'''
        return self.__accept or self.__step

    def __contains__(self, w):
        '''Returns True iff w is in the trie.'''
        node = self.prefix(w)
        return node.__accept if node is not None else False

    def __iter__(self):
        '''Iterates over the strings in the trie.'''
        if self.__accept:
            yield ''
        for (c, t) in self.__step.items():
            yield from [c + w for w in t]

    def follows(self, w):
        '''Returns the list of characters that can possibly follow w
        in the trie, assuming w is a valid prefix.'''
        return [c for c in self.prefix(w).__step.keys()]

    def prefix(self, w):
        '''Returns the node of the trie corresponding to prefix w, 
        or else None if w is not a valid prefix.'''
        pass  # your code here

    def insert(self, w):
        '''Inserts w into the trie.'''
        pass  # your code here

In particular, you have to implement the two methods:

which are themselves used in the implementation of some of the other methods. To define prefix and insert correctly, you should understand the interpretations of the two fields stored at each node of the trie:

• __accept: this is a boolean indicating whether the current prefix is a word of the dictionary (corresponding to the color of the node in the visualization we saw earlier, with red = True, black = False);

• __step: this is a Python dictionary associating single-letter extensions of the current prefix to corresponding child nodes.

(Note that a trie can also be thought of as a simple finite-state machine, where the states correspond to valid prefixes of words. Under this interpretation, the __accept field indicates whether the current state is accepting, and the __step field encodes its outgoing transitions.)

Here is an example of a simple test you can try on your implementation:

def test_trie():
    trie = Trie(["CAT", "CATEGORY", "CATION", "DATA", "DO", "DOG"])
    for w in trie.prefix("CAT"):
        print("CAT-", w)
    for w in trie.prefix("DO"):
        print("DO-", w)
    for c in trie.follows("CAT"):
        print("CAT" + c + "?...")

It should output:

CAT- 
CAT- EGORY
CAT- ION
DO- 
DO- G
CATE?...
CATI?...

(The grader server will also run some automatic tests.)

Generating word rectangles

We will represent a partially-filled \(m \times n\) word rectangle as a two-dimensional matrix rect, where rect[i][j] contains the letter (string of length 1) at row i, column j, or else None if that position is unfilled. We have provided a few utility functions that you can use for constructing and printing partially-filled word rectangles:

def partial_rect(m, n, topstrings = []):
    rect = [[None for _ in range(n)] for _ in range(m)]
    for i in range(len(topstrings)):
        s = topstrings[i]
        for j in range(len(s)):
            rect[i][j] = s[j]
    return rect

def rect_print(m, n, rect):
    print('='*(2*n+3))
    for i in range(m):
        print('| ' + ' '.join([c if c else '?' for c in rect[i]]) + ' |')
    print('='*(2*n+3))

Example:

>>> rect = partial_rect(4, 4, ["LOVE","OMEN","SE"])
>>> rect_print(4, 4, rect)
===========
| L O V E |
| O M E N |
| S E ? ? |
| ? ? ? ? |
===========

We have also provided a few different dictionaries that you can use to test your implementation:

• common100.txt: a list of the 100 most common words in English
• common10000.txt: a list of the 10000 most common words in English
• francais1.txt: a dictionary containing 22739 words in French
• english2.txt: a dictionary containing 65199 words in English
• francais2.txt: a dictionary containing 208916 words in French

On most Unix systems there is also a standard dictionary file, available at /usr/share/dict/words. Or feel free to supply your own dictionary!

You can use the following function to load a dictionary file containing one word per line, and return the list of words all converted to uppercase:

def readdict(name):
    with open(name, 'r', encoding='utf-8') as stream:
        return [x.strip().upper() for x in stream]

Finally you are ready to fill some rectangles!