How to Define DP States

A guide to defining Dynamic Programming States

Before you can apply DP to a problem, you need to clearly define the DP states—the variables that represent the subproblems you are solving. Defining these states correctly is crucial for making your DP solution both correct and efficient.

In this blog post, we’ll break down how to define DP states. We'll look at what DP states are, why they are important, and provide a framework to help you define them systematically using some examples.

What are Dynamic Programming States?

In Dynamic Programming, a state refers to a specific subproblem or configuration that represents a partial solution to the overall problem. The idea is that by solving these subproblems and storing their results, we can build the solution to the larger problem efficiently.

In simple terms, a DP state captures a snapshot of the problem at a particular point in time, and the DP table (or memoization cache) stores the solution to that snapshot. The challenge is to identify the right set of states that will allow you to reconstruct the optimal solution.

Key Components of a DP State

1. State Variables: These are the key parameters or indices that define a subproblem. They represent the different "dimensions" of the problem.

2. State Transition: Once you've defined the states, you'll need to figure out how to transition from one state to another (i.e., how to move from solving one subproblem to another).

   
   

Guide to Defining DP States

Step 1: Understand the Problem and Identify the Structure

Before you can define states, it's crucial to understand the problem and figure out how it can be divided into subproblems. Ask yourself questions like:

• What are the key decisions or choices involved in the problem?

• Can the problem be broken down into smaller overlapping subproblems?

• What values or parameters define a subproblem at any point?

For example, in the Knapsack problem, we need to decide which items to include in the knapsack based on their weight and value, while respecting the knapsack's weight limit.

Step 2: Identify the Parameters That Define the Subproblems

For each problem, think about the parameters or dimensions that define a subproblem. These parameters will serve as the state variables in your DP solution. The subproblem should represent a partial solution at some point.

Example 1: Fibonacci Sequence

In the case of the Fibonacci sequence, the subproblem is simply computing the nth Fibonacci number. The state can be defined by the index n, i.e., how far we are in the sequence.

• State variable: dp[n] – the nth Fibonacci number.

• State transition: dp[n] = dp[n-1] + dp[n-2] (Using the results of the previous two numbers to calculate the current one).

Example 2: Knapsack Problem

In the Knapsack problem, we are given a list of items, each with a weight and value. The goal is to maximize the value of the items you can carry without exceeding the weight capacity of the knapsack.

• State variables: dp[i][w], where i represents the first i items considered, and w is the remaining weight capacity of the knapsack.

• State transition: dp[i][w] = max(dp[i-1][w], dp[i-1][w-weight[i]] + value[i]), which compares the maximum value obtained by either including or excluding the item.

Step 3: Break Down the Problem Recursively

Once you’ve identified the parameters of the subproblem, think about how to break the problem down into smaller subproblems. In other words, how can you recursively define the DP state transitions?

Example 3: Longest Common Subsequence (LCS)

For the LCS problem, where we want to find the longest subsequence that two sequences have in common, the state variables are the indices of the two sequences we are comparing.

• State variables: dp[i][j] – the length of the LCS between the first i characters of string A and the first j characters of string B.

• State transition:

  • If A[i-1] == B[j-1], then dp[i][j] = dp[i-1][j-1] + 1 (Include this matching character).

  • Otherwise, dp[i][j] = max(dp[i-1][j], dp[i][j-1]) (Either skip a character from A or B).

Step 4: Decide the Direction of Computation (Top-Down vs. Bottom-Up)

• Top-Down (Memoization): Start solving the problem from the original problem and break it down recursively, caching results as you go. You need to ensure you define the base cases for recursion (e.g., what happens when i == 0 or w == 0).

• Bottom-Up (Tabulation): Start solving the problem from the simplest subproblems and build up to the final solution. This approach often uses an iterative solution to fill a DP table.

Step 5: Define Base Cases

Base cases are the smallest subproblems that have known solutions without any further breakdown. Defining them correctly is important to avoid unnecessary recursion or iteration.

• Example: In the Knapsack problem, if there are no items or the knapsack has zero capacity, the solution is straightforward.

  • dp[0][w] = 0 for all w (No items, so the value is 0).

  • dp[i][0] = 0 for all i (No capacity, so no items can be included).

Examples: Defining States in Common DP Problems

1. Knapsack Problem

• Problem: Given a set of items, each with a weight and value, determine the maximum value that can be placed into a knapsack of capacity W.

• State Variables:

  • dp[i][w]: Maximum value achievable using the first i items and a knapsack capacity of w.

• Base Cases:

  • dp[0][w] = 0 for all w (No items means no value).

  • dp[i][0] = 0 for all i (Zero capacity means no items can be added).

• State Transition:

  • If we exclude the item: dp[i][w] = dp[i-1][w]

  • If we include the item: dp[i][w] = dp[i-1][w-weight[i]] + value[i]

  • The final result is dp[n][W] where n is the number of items.

2. Longest Common Subsequence (LCS)

• Problem: Find the longest subsequence that two strings have in common.

• State Variables:

  • dp[i][j]: Length of the LCS of the first i characters of string A and the first j characters of string B.

• Base Cases:

  • dp[0][j] = 0 for all j (Empty string A).

  • dp[i][0] = 0 for all i (Empty string B).

• State Transition:

  • If A[i-1] == B[j-1]: dp[i][j] = dp[i-1][j-1] + 1

  • Else: dp[i][j] = max(dp[i-1][j], dp[i][j-1])

Once you have the states defined, the rest of the solution—whether top-down or bottom-up—becomes a matter of filling the DP table or cache based on your state transitions. Hopefully this provides a helpful structure and thought process to help define DP States.

Please share your thoughts and feedback in the comments below!