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Today we're going to explore the problem of finding the longest common subwords between two sequences. Can anyone tell me what a subword is?
Is it a part of a word, like a segment?
Exactly! A subword is indeed a segment that can appear in both strings. For example, in the words 'secret' and 'secretary', the segment 'secret' is a common subword.
What about the lengths of these common subwords?
Good question! Our focus today will also cover how to compute the lengths of these common subwords efficiently.
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We can solve the longest common subword problem using brute force, but that can be slow. Can anyone suggest why?
Because it checks every possible combination?
Exactly! This approach has a high time complexity. That's why we look towards a more efficient solution—dynamic programming. It helps us break the problem into smaller, manageable subproblems.
How does that work?
We build a table to track the lengths of common subwords at various positions, allowing us to compute the solution progressively.
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Now, let's discuss an essential concept called inductive observation. If we find a common letter at two positions, what can we infer about the lengths of common subwords?
That the length of the common subword increases by one?
That's correct! If the letters match, then we can say there is a common subword of length k at those positions, which implies there’s also a common subword of length k-1 starting one position forward.
But what happens when they don't match?
Exactly! In that case, we conclude that there cannot be a common subword at those positions.
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Finally, let's look at how we fill out the dynamic programming table with the common subwords. How do we start filling in the values?
By initializing the last row and column to zero?
Yes! We initialize to zero since an empty substring has no matching segments. Then we fill the table by checking character matches and using our inductive observations.
And where do we find the length of the longest common subword?
Great question! We look for the highest value in the table once it's completed.
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In this section, the concept of longest common subwords is introduced, where the problem involves finding segments that appear in both sequences. Various methods to solve this problem are discussed, including brute force and dynamic programming approaches.
This section delves into the problem of finding the longest common subwords and subsequences between two sequences. With the aid of examples like 'secret' and 'secretary' or 'bisect' and 'trisect', we illustrate how common segments can be identified within two strings. The aim is to compute the length of these longest common subwords rather than the subwords themselves.
The chapter discusses a brute force algorithm that considers every combination of positions within the two strings. However, the efficiency of this method is poor due to its O(m*n^2) time complexity.
To improve the efficiency, the section introduces an inductive observation, which relies on establishing commonalities between subwords. It forms the basis for a more efficient dynamic programming solution. In this approach, we define subproblems that relate the lengths of the common subwords at different positions in the strings using recursion.
The discussion further covers boundary conditions, where if one word is empty, the length of the common subword is 0. The section also presents a code example and emphasizes the dynamic programming table's construction for tracking lengths of common subwords, culminating in an efficient algorithm to solve the longest common subword problem.
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Let us suppose we are given two words for the moment, let us just assume they're 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 this introduction, we are discussing the problem of finding the longest common subword between two given strings. A subword is defined as a segment of the strings. For example, if we have the words 'secret' and 'secretary', the longest common subword is 'secret'. The focus here is not to identify the subword itself but to calculate its length.
Think of it like finding the longest shared sequence of letters in two words, similar to identifying the common part of two overlapping mountain ranges where the peaks represent letters in the words.
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So, let us look at this thing more formally, so we have two words a0 to am and b0 to bn. If I write out a0, a1, then have a position ai, ai + 1 and so on, then am...
We can represent the two strings algebraically. For string A, the letters are indexed from 0 to m, and for string B, they are indexed from 0 to n. If we take an index position i from A and j from B, and read k letters from both strings starting at these positions, we can identify if they match. If they do, then there is a common subword of length k.
Imagine you are trying to find the longest route on a two-way street. If you start from point A (i) and point B (j) and proceed down the street (k), what you want is to see if you can drive both cars the same distance without turning, and if they do match, that's your common path.
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So, here is a Brute force algorithm, you just try out every position, you look at every ai and bj...
The brute force method involves checking every possible starting position in both strings. By comparing characters one by one, we can track how far we can go without mismatching. This approach, however, has a time complexity of O(m * n^2). We match characters until we encounter the first mismatch, and we keep a record of the longest match we've found so far.
It's similar to trying to solve a jigsaw puzzle. You start fitting pieces together one by one from every combination until you either find a match or realize that it's a wrong fit, and then you restart the other combinations.
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The first observation in the inductive observation is that if I have ai ai + 1...
The principle of induction we use here states that a common subword of length k exists at positions i and j only if a common subword of length k-1 exists at the next positions, i+1 and j+1. If the characters at these positions are equal, we can say that we can extend our common subword by one character.
Consider it as building a tower: if you successfully stack two blocks on top of each other (matching subword), you can only add another block (extend the common subword) if the current blocks fit together.
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For convenience, although our words go from a0 to am and from b0 to bn...
To manage our comparisons effectively, we extend both strings to include an additional position, which allows us to check scenarios where one string might be exhausted. If we reach the end of either string while trying to match, we conclude that there cannot be any common subword beyond these points.
Imagine two runners on a track. If one runner reaches the finish line (the end of the string) first, they can’t continue running, meaning the other runner has no one to match with beyond that point.
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So, we could write a memorized recursive function for that...
Rather than computing values repeatedly, dynamic programming allows us to store results of already computed subproblems to avoid unnecessary recalculations. We define a recursive function that builds upon previously calculated results, making the whole process efficient.
This is akin to having a cookbook. Instead of figuring out each ingredient quantity afresh for a dish, you can reference previously made dishes and simply adjust quantities as needed.
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What we have to do is find the entry with a largest value...
After populating our table with lengths of common subwords, we identify the maximum value, which corresponds to the length of the longest common subword between the two strings. We can retrieve this length and the actual subword from our table.
Think of it as searching for the tallest building in a city skyline where each building's height is represented by the common subword lengths. The tallest building indicates the length of the longest common subword.
Learn essential terms and foundational ideas that form the basis of the topic.
Key Concepts
Longest Common Subword: The longest segment that appears in both sequences.
Brute Force Algorithm: A straightforward approach that checks every possible segment, often inefficient.
Dynamic Programming: A more efficient method that divides the problem into subproblems and uses their solutions.
Inductive Observation: A way to reason about subword lengths by establishing conditions based on character matches.
See how the concepts apply in real-world scenarios to understand their practical implications.
For the strings 'director' and 'secretary', the common subwords 'ec' and 're' are identified, each being of length 2.
In the words 'bisect' and 'trisect', the longest common subword is 'isect', which has a length of 4.
Use mnemonics, acronyms, or visual cues to help remember key information more easily.
To find a common word, look for the letters that sound the same; the correct segments are part of the game.
Once, two words named 'director' and 'secretary' met at a park. They discovered that by examining their letters closely, they had hidden segments that made them closer, revealing shared stories of 'ec' and 're'.
Use 'DIP' for dynamic programming: Decompose, Identify, and Process.
Review key concepts with flashcards.
Review the Definitions for terms.
Term: Subword
Definition:
A segment of a word that occurs in another word.
Term: Longest Common Subword
Definition:
The longest segment that appears in both strings.
Term: Dynamic Programming
Definition:
A method for solving complex problems by breaking them down into simpler subproblems.
Term: Inductive Observation
Definition:
Method of reasoning where the conclusion is based on observing patterns in specific cases.
Term: Brute Force Algorithm
Definition:
A straightforward method that tries all possible combinations to find a solution.