Notes/Advanced Algorithms/Dynamic Programming.md

date	type
11.09.2024	theoretical

travelling salesperson

Definition

• Approach to design algorithms
• Optimal solution in polynomial time
• Usually used alongside Divide and Conquer
• Used to better the time complexity, but usually worsens the space complexity

How do we create a DP soluton?

• Characterize optimal solution
• Recursively define the value of an optimal solution
• Compute the value
• Construct an optimal solution from the computed information

Problem discussion

Calculating Fibonacci numbers

• Recursion is extremely bad -> PF was wrong
  • Because the branches are too many
  • We compute f(x) where f(x) has already been computed
  • Memoization!

Approaches

• Linear time \mathcal{O}(n) - Bottom to top

F[0] = 1
F[1] = 1
for i in {2, .., n} do
	F[i] = F[i − 1] + F[i − 2]
end for
return F[n]

Compute the entire array and just return the result.

• Memoization \mathcal{O}(n) - Top to bottom

function fibMemo(index, cache) {
  cache = cache || [];
  if (cache[index]) return cache[index];
  else {
    if (index < 3) return 1;
    else {
      cache[index] = fibMemo(index - 1, cache) + fibMemo(index - 2, cache);
    }
  }

  return cache[index];
}

Rod cutting

Given a rod of length N inches and an array of prices, price[]. price[i] denotes the value of a piece of length i. Determine the maximum value obtainable by cutting up the rod and selling the pieces.

Optimal solution

• Let the left part have a length l
• Then R[n] = P[l] + R[n-l]
• Where P[l] is the price
• First cut of length should be the maximal one

Knight Collecting Rewards

Knight Collecting rewards Input: Integers n, m, table of rewards of size n \times m Question: What is the maximum reward that the knight can get for its journey.

Strategy

• We look at the bounds defined by the problem statement
  • We know that the knight can get into the cell (n-1, m-1) either from (n-2,m-3) or (n-3,m-2)
  • Hence, we have to know the most rewarding path to these two points from (0,0)

Approaches

• \mathcal{O}(n*m)

Reward(int n, int m, matrix A)
Create matrix R of size n × m with −∞ values
R[0, 0] = A[0, 0]
for i in {0, .., n − 1} do
	for j in {0, .., m − 1} do
		R[i, j] = A[i, j] + max R[i − 1, j − 2], R[i − 2, j − 1]
	end for
end for
return R[n-1, m-1]

This doesn't tell us how we got there, but it does give us max rewards. We can easily include path memory by making elements be a linked list or create a separate table blah blah.

Longest common non-contiguous sequence¹

LCS Input: Two sequences X = [x1, x2, . . . xm], Y = [y1, y2, . . . , Yn]. Question: Longest common subsequence

Strategy

!

Or this formula:

c[i,j] = \begin{cases}
	0, & \text{if } i=0 \text{ or } j=0\\
	c[i-1,j-1]+1, & \text{if } i,j>0 \text{ and } x_i = y_i\\
	max\{c[i,j-1], c[i-1], j\}, & \text{if } i,j>0 \text{ and } x_i \neq y_i\\

\end{cases}

Dominating set² in a path

Input: A path P with a specified positive cost for each vertex. Output: Choose a subset of vertices S with a minimum cost such that for each v \in P: either v \in S or there is u such that u \in S and u, v are neighbors.

Strategy

• Identify some subproblems
  • A[i] cost of the cheapest set S_i \subset P_i which dominates all vertices in P_i <- not great
  • A[i,0] equals the cost of the cheapest S_i which dominates P_i and v_i \notin S_i
  • A[i,1] equals the cost of the cheapest

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1. contiguous - next or near in time or sequence ↩︎

2. Subset of the vertices of the graph G, such that any vertex of G is in it, or has a neighbor in it ↩︎