---
date: 11.09.2024
type: theoretical
---


**travelling salesperson**

## Definition
- Approach to design algorithms
- Optimal solution in polynomial time
- Usually used alongside [Divide and Conquer](Divide%20and%20Conquer.md)
- Used to better the time [complexity](Complexity.md), 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**
```pseudo
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**
```javascript
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](https://www.geeksforgeeks.org/problems/rod-cutting0840/1)
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 sequence](https://www.geeksforgeeks.org/longest-common-subsequence-dp-4/)[^1]

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

#### Strategy
![[LCS.png]]

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 set](https://www.geeksforgeeks.org/dominant-set-of-a-graph/)[^2] 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 
 

[^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](https://en.wikipedia.org/wiki/Dominating_set)