3.5. Code Implementation of Dynamic Programming Algorithm

Let us suppose we are given two words for the moment, let us just assume there going in English or a language like that, we given two strings and we want to find the length of the longest common subword between them. In a subword, it is just a segment.

This chunk introduces the concept of the longest common subword problem, which involves finding the length of the longest segment (subword) that appears in both of two given strings. A subword is simply defined as a continuous part of the string.

Imagine you are trying to find common phrases in two different articles. If both articles contain the phrase 'deep learning', then that phrase is like the common subword that connects them.

So, the focus at the moment is to get the length of the subword... my main goal at the moment is to compute not the subword itself, but the length.

This chunk highlights that the primary objective is not to actually identify the common subword, but rather to determine the length of that subword. Length gives us a quantifiable measure of similarity between two strings.

Think of it as measuring the height of trees in two forests. You may not care about which trees are alike, but you want to know how similar the forests are based on tree height.

Here is a Brute force algorithm... So, therefore, this will be O of m times n into n or O m n square.

This chunk discusses the brute force approach to find the longest common subword. It encompasses comparing all possible positions in both strings. The computational complexity of this method is analyzed, indicating it can become slow for longer strings since it operates in O(m * n²) time.

Imagine trying to find matching items in two stores by checking each item one by one. If there are many items, this method can take an incredibly long time.

If I have a i equal to b j, I need to look at a i plus 1, b j plus 1... which gives us a handle on the inductive structure.

The focus here is on leveraging an inductive approach. If characters from both strings match, you can build on that by checking the next characters. If they don't match, the search for a longer common subword must revert. This creates a recursive dependency in the calculation.

Consider matching pairs of dominoes: when two dominos connect (match), you can keep extending the chain. If they don’t connect, you must start a new sequence.

So, we will allow a m plus one position and an n plus one position... then we can say that there cannot be a common subword, because there are no letters to use.

This chunk discusses boundary conditions in the algorithm. When one of the words is exhausted (reaches the end), it is impossible to have a common subword, and thus the length of common subword is 0 under these conditions.

Think of a race where one runner cannot continue past the finish line. Once a runner stops, they cannot match pace with the other: that race ends for them.

So, we could write a memorized recursive function for that, but we will directly try to compute a dynamic programming solution for this problem.

This section introduces the dynamic programming approach for solving the longest common subword problem. It emphasizes building solutions from previously computed values (subproblems), rather than recalculating existing solutions, thus saving time.

Imagine solving a jigsaw puzzle: once you find that two pieces fit together, you remember that connection rather than forgetting it and searching for it again each time.

Here is some code for this particular algorithm... the maximum common subword of length 0.

This chunk provides practical code examples of the dynamic programming algorithm discussed, explaining how the array is filled out and the logic behind it. It emphasizes column-wise processing of the problem.

If you think of a painting by numbers, every area corresponds to a number that will dictate what color goes there. By filling one section at a time, you gradually see the complete image, just like filling out the matrix yields the answer.

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