The earlier examples of sorting and searching have a striking property: to solve a problem of size n, we have to break it in half and then work on one or more problems of size n/2. A common way to implement such methods uses recursion. We define a function f, which simplifies the problem, and calls itself to solve one or more easier

instances of the same problem. It then combines the results into a solution for the original problem.

For example, suppose we have a set of n words, and want to calculate how many different ways they can be combined to make a sequence of words. If we have only one word (n=1), there is just one way to make it into a sequence. If we have a set of two words, there are two ways to put them into a sequence. For three words there are six possibilities. In general, for n words, there are n x n-1 x ... x 2 x 1 ways (i.e., the factorial of n). We can code this up as follows:

>>> def factorial1(n): ... result = 1 ... for i in range(n): ... result *= (i+1)

... return result

However, there is also a recursive algorithm for solving this problem, based on the following observation. Suppose we have a way to construct all orderings for n-1 distinct words. Then for each such ordering, there are n places where we can insert a new word: at the start, the end, or any of the n-2 boundaries between the words. Thus we simply multiply the number of solutions found for n-1 by the value of n. We also need the base case, to say that if we have a single word, there's just one ordering. We can code this up as follows:

>>> def factorial2(n): ... if n == 1: ... return 1

These two algorithms solve the same problem. One uses iteration while the other uses recursion. We can use recursion to navigate a deeply nested object, such as the WordNet hypernym hierarchy. Let's count the size of the hypernym hierarchy rooted at a given synset s. We'll do this by finding the size of each hyponym of s, then adding these together (we will also add 1 for the synset itself). The following function sizel() does this work; notice that the body of the function includes a recursive call to sizel(): >>> def sizel(s):

... return l + sum(sizel(child) for child in s.hyponyms())

We can also design an iterative solution to this problem which processes the hierarchy in layers. The first layer is the synset itself , then all the hyponyms of the synset, then all the hyponyms of the hyponyms. Each time through the loop it computes the next layer by finding the hyponyms of everything in the last layer . It also maintains a total of the number of synsets encountered so far ©.

... layer = [h for c in layer for h in c.hyponyms()] 0

Not only is the iterative solution much longer, it is harder to interpret. It forces us to think procedurally, and keep track of what is happening with the layer and total variables through time. Let's satisfy ourselves that both solutions give the same result. We'll use a new form of the import statement, allowing us to abbreviate the name wordnet to wn:

>>> from nltk.corpus import wordnet as wn >>> dog = wn.synset('dog.n.0l') >>> sizel(dog) l90

As a final example of recursion, let's use it to construct a deeply nested object. A letter trie is a data structure that can be used for indexing a lexicon, one letter at a time. (The name is based on the word retrieval.) For example, if trie contained a letter trie, then trie['c'] would be a smaller trie which held all words starting with c. Example 4-6 demonstrates the recursive process of building a trie, using Python dictionaries (Section 5.3). To insert the word chien (French for dog), we split off the c and recursively insert hien into the sub-trie trie['c']. The recursion continues until there are no letters remaining in the word, when we store the intended value (in this case, the word dog).

Example 4-6. Building a letter trie: A recursive function that builds a nested dictionary structure; each level of nesting contains all words with a given prefix, and a sub-trie containing all possible continuations.

def insert(trie, key, value): if key:

first, rest = key[0], key[l:] if first not in trie: trie[first] = {} insert(trie[first], rest, value) else:

>>> trie = nltk.defaultdict(dict) >>> insert(trie, 'chat', 'cat') >>> insert(trie, 'chien', 'dog') >>> insert(trie, 'chair', 'flesh') >>> insert(trie, 'chic', 'stylish')

>>> trie = dict(trie) # for nicer printing

{'i': {'r': {'value': 'flesh'}}}, 'i': {'e': {'n': {'value': 'dog'}}} {'c': {'value': 'stylish'}}}}}

Caution!

Despite the simplicity of recursive programming, it comes with a cost. Each time a function is called, some state information needs to be pushed on a stack, so that once the function has completed, execution can continue from where it left off. For this reason, iterative solutions are often more efficient than recursive solutions.

Was this article helpful?

## Post a comment