dynamic_programming.tex

\section{Dynamic Programming Method}

\subsection{Introduction}
\paragraph{Dynamic programming}
Dynamic Programming is a method of solving problems that can be broken down
into subproblems and solved recursively.

\paragraph{CLRS: Four-Step Method}
\begin{enumerate}
  \item Characterize the structure of the optimal solution.
  \item Recursively define the value of an optimal solution.
  \item Compute the value of an optimal solution; typically this is done bottom-up.
  \item Construct an optimal solution from computed information.
\end{enumerate}

\paragraph{Overlapping Subproblems Property}
We must choose subproblems in a way such that the same subproblem
occurs several times in the recursion tree.
When a subproblem is solved, we store the result so that
subsequent instances of the same subproblem can be answered by a lookup table.

\paragraph{Considerations for Putting it Together}
\begin{itemize}
  \item Is the original problem a subproblem or a combination of subproblems.
  \item Order that the subproblems occur in.
  \item How many subproblems are there? How long does each take?
\end{itemize}

\subsection{Longest Increasing Subsequence}

\paragraph{Problem}
Given a sequence of \(n\) real numbers  \(A[1..n]\), find a subsequence
(not necessarily contiguous), where the values are strictly increasing.

\paragraph{Subproblem: Prefixing Lengths}
A natural choice is, for all \(i \in [1, n]\), find the length of the
longest increasing subsequence of \(A[1..i]\) that ends at  \(A[i]\).
Call this \(Q(i)\).

We try to solve \(Q(i)\)  by extending some sequence \(Q(j)\) for
\( j \leq i \).

We assume that we have solved for all \(j < i\). Then, look for a  \(j < i\)
such that \(A[j] < A[i]\). Pick a sequence \(m\) to get a maximal result and
extend it with \(A[i]\).

We stop at a base case of \(i = 1\), which is the trivial case.

The recurrence we define is \[
  Q(i) = \max\{Q(j) : j < i, A[j] < A[i]\} + 1
.\] 

The longest subsequence will correspond to the largest \(Q(i)\).

\paragraph{Time Complexity}
The \(n\)-th entry may require us to look at  \(n - 1\) entries.
We do this \(n\) times, which means the prefixing takes \(O(n^2)\)
time in total. The final solution being extracted takes \(O(n)\) and
is therefore negligible.

\paragraph{Doing Better}
This can be done in \(n \log n\). Finding such an algorithm is left as an
exercise left to the reader.

\paragraph{Correctness}
We claim that truncating the optimal solution for \(Q(i)\)
will produce an optimal solution for \(Q(m)\). 
I can't be bothered writing out why.

\paragraph{Getting Sequence Not Length}
To get the sequence, create an array of \textit{predecessors}, \(P\).
When finding \(Q[i]\), let  \(P[i] = j\) where  \(j\)
is the index chosen according to the definition of \(Q[i]\).

When backtracking through the array, when the max is found, start from
the max, then follow the predecessors until one with no predecessors is found.

\subsection{Actvitiy Selection}
\paragraph{Problem}
Input is a list of \(n\) activities with starting and finishing times  \(s_i, f_i\). 
Find the maximal duration of a subset of activities, where no two activities
take place at the same time.

This is a variation of a problem from the greedy chapter.
The greedy approach will not work for maximal duration.

\paragraph{Setup}
Sort the activities by their finishing times such that \[
  f_1 \leq f_2 \leq \cdots \leq f_n
.\] 

\paragraph{Subproblem: Maximal Duration}
Let \(P(i)\) be the subproblem of finding the duration \(t(i)\)
of a subsequence  \(\sigma(i)\), of the first  \(i\) acitivities that
 \begin{enumerate}
  \item consists of non-overlapping edges,
  \item ends with activity \(i\), 
  \item is of maximal duration.
\end{enumerate}

\(P(i)\) is solved by appending  activity \(i\) to some sequence \[
  t(i) = \max \{
    t(i) : j < i, f_j < s_i
  \}  + f_i - s_i
.\] 

There is a base case \(P(i) = f_1 - s_2\).

\paragraph{Best Solution}
The best time is \(\max(P)\).

\paragraph{Time Complexity}
\begin{enumerate}
  \item Sorting takes \(O(n \log n)\).
  \item We loop through \(n\) elements, searching  \(n - 1\) elements for the
    best  \(t\) which leads to the complexity for all subproblems
    to be \(O(n^2)\).
  \item Finding the best \(t\) is \(O(n)\).
\end{enumerate}
Thus, the overall time is \(O(n^2)\).

\paragraph{Proof of Correctness of Subproblem}
Let the optimal solution for \(P(i)\) be given by \( \{k_j\}_{j=1}^m \),
where \(k_m = i\).
The requirement is that \(\sigma' = \{k_j\}_{1}^{m - 1}\) is optimal solution
to \(P(k_{m - 1})\). This is proved by contradiction.

Suppose that instead \(P(k_{m - 1})\) is solved by a sequence
\(\tau'\) with a duration larger than  \(\sigma'\).

Then, extend \(\tau'\) with activity  \(i\). Therefore,
\(\tau_i\) will trivially have a larger duration than  \(\sigma'\).
This however contradicts the assumption of \(\sigma\) as the sequence
solving \(i\).
Hence, the optimal solution is achieved by extending the \(P(j)\) of largest duration
that supports extension of \(i\), where  \(j < i\).

\paragraph{Sequence of Activities}
To find the largest sequence, we wish not just to store \(t(i)\) with
each  \(P(i)\) but, also the predecessor \(j\) where \(P(i)\)
extends  \(P(j)\).

\subsection{Making change}
\paragraph{Problem}
You are given \(n\) types of coin denominations of integer values \[
  \{v_i\}_{i=1}^n, \text{ where } v_1 < v_2, \ldots, < v_n
.\] 
Suppose \(v_1 = 1\) and there is an unlimited amount of each \(v_i\).
This allows for it to always be possible to create change. 

\paragraph{Task}
Make change for a given integer amount \(C\), with the minimal number of coins.

\paragraph{Greedy Setup: Works}
Greedily take as many coins of the largest possible denomination \(v_n\)
that is possible. Then, do the same for \(v_{n - 1}\) and so on.

This approach works for all real-world currencies but, not for all sequences.
Consider the denominations \(\{1, 10, 15\}\) with \(C = 20\). Greedy would choose
 \(\{15, 1, 1, 1, 1, 1\} \) but the optimal is \(\{10, 10\}\).

\paragraph{Subproblem}
We will work in a bottom-up direction, finding the optimal solution for all
values up to \(C\).

\paragraph{Solving a Subproblem}
Suppose that we have found the solutions for all amounts \(j < i\)
and want to find a solution for \(i\).

Consider all coins \(v_k\) that are a part of the solution for amount \(i\).
Make up the remaining amount \(i - v_k\) with the previously computed optimal solution.

Of all the optimal solutions (being computed recursively), find the pick the one that uses
the fewest number of coins.

Suppose that we choose the coin \(m\). The optimal solution  \(P(i)\) is found
by adding a coin of denomination  \(v_m\) to  \(P(i - v_m)\).

\paragraph{Base Case}
Suppose that \(C = 0\). Then trivially, the solution has no coins.

\paragraph{Time Complexity}
Each of the \(C\) subproblems can be solved in \(O(n)\) time. As such, the time complexity
is just  \(O(nC)\).
This is \textit{not} polynomial time in terms of \(m\), as \(C\) may be unrestricted.

\paragraph{Proving Optimal Solution}
Consider an optimal solution for some \(i\).
Suppose this contains a coin of denomination  \(v_m\), for some  \(1 \leq m \leq n\).
Remobving this coin must leave an optimal solution for \(i - v_m\) by the
same cut and paste argument.

By consider all coins (of value up to \(i\)), we can pick an \(m\)
for which the optimal solution  \(i - v_m\) uses the fewest coins.

\paragraph{Finding Combinations of Coins}
If the solution also finding what combination of coins is used, then each
\(P(i)\) should also store with it, a pointer to  \(\pred(i)\), which is
the optimal solution for \(i - v_k\), alongside the value of  \(v_k\).

Then, when answering the question, we can just do a backtrace.

\paragraph{Notation}
A \(k\) that minimises \(P(i  - v_k)\) is denoted as \[
  \argmin \opt(i - v_k)
.\] 

\subsection{Knapsack Problem}

\subsubsection{Knapsack: Duplicates Allowed}
\paragraph{Problem}
There are \(n\) items. All items of kind  \(i\) are identical with weight  \(w_i\).
All weights are integers. We can take any number of items of each kind,
to fill a Knapsack of capacity \(C\).

The task is to choose a combination of items of maximal total capacity,
which fit the capacity of the Knapsack.

\paragraph{Subproblem}
The problem is effectively  a reflavoured version of the \underline{making change}
problem.

We assume that we have solved the problem for all \(j < i\). Call this value \(P(j)\).
Then, consider all item types where the \(k\)-th element has a weight
\(w_k\). If we add this item, then the rest of the bag can be
fixed with  an allocation matching the optimal solution for \(i - w_k\).
Choose the \(m\) that maximises the total value of \(P(i - w_m)\).
Then, the optimal solution for \(i\) is \(w_m + P(i - w_m)\).

There is a base case where \(\opt(0) = 0\).

\paragraph{Time Complexity}
There are up to \(C\) subproblems, each of which can be solved in \(O(n)\) time.
As such, the total overall time is  \(O(nC)\). Again, this is not polynomial in terms
of the length of the input.

\subsubsection{Knapsack: Duplicates Allowed}

\paragraph{Flaws in Previous Algorithm}
The previous algorithmn is not able to discern if a value \(v_k\) has already been used.
This breaks the rule of duplicates allowed.

\paragraph{Storing Values is not Enough}
Suppose that we follow the previous approach but store the values of what \(v_k\)
was used at that step. This is not enough. There are two main issues:
\begin{enumerate}
  \item The optimal soltion for \(i - w_k\) is not necessarily unique.
    Thus, we may record some set of items, including  \(k\) when, an equal selection of
    items is possible without \(k\). 
  \item Even if all optimal solutions for \(i - w_k\) use  \(k\), its still possible that
    there is a suboptimal solution for \(i - w_k\) that does not use \(k\).
\end{enumerate}
The reason that this fails is due to the lack of the \underline{optimal substructure property}.

\paragraph{New Way to Solve Subproblem}
For all total weights \(i\), find the optimal solution using only the first \(k\) items.
Then,
\begin{itemize}
  \item If \(k\) is used in the solution then, there is \(i - w_k\) weight left
    to fill using the first \( k - 1 \) items.
  \item If \(k\) is not used in the solution then, we must fill all  \(i\)
    units of weight with the first  \(k - 1\) items.
\end{itemize}

\paragraph{New Subproblem}
For \(0 \leq i \leq C\) and \(0 \leq k \leq n\), let  \(P(i, k)\) be the optimal
solution for determining  \(\opt(i, k)\). That is, the maximum value that can be achieved
up to  \(i\) units of weight, only using the first \(k\) items where, \(m(i, k)\) is the
(largest) index of an item in such a collection.

We have the \underline{recurrence} for \(i > 0\) and \(0 \leq k \leq n\), that  \[
  \opt(i, k) = \max (\opt(i, k - 1), opt(i - w_k, k - 1), + v_k),
\] with \(m(i, k) = m(i, k - 1)\) in the first case and \(k\) in the second.

There is a base case with \(\opt(i, k)\) if  \(i\) or \(k\) is zero and  \(m(i, k)\)
is undefined.

\paragraph{Order Matters}
When we get to \(P(i, k)\), there is an expectation that  \(P(i, k - 1)\)
and  \(P(i - w_k, k - 1)\) has already been solved.
This is guaranteed if the subproblems are solved in increasing order of \(k\)
first and then, increasing capacity of \(i\). 

\paragraph{Overall Solution and Time Complexity}
The final solution is just \(\opt(C, n)\). 

Each problem can be solved in constant time and, there are \(O(nC)\)
subproblems. Thus the overall time complexity is  \(O(nC)\)

\subsection{Balanced Partition}

\paragraph{Problem}
The input is a set of \(n\) positive integers  \(x_i \in  S\).
The task is to partition these integers into two subset  \(S_1, S_2\)
with sums \(\sum_1, \sum_2\) respectively, such that \(|\sum_1 - \sum_2|\)
is minimised.

\paragraph{Reframing the Condition}
Without loss of generatlity, assume \(\sum_1 \geq \sum_2\) and let
\(\sum\) be the sum of all elements in the set. Then, \[
  \sum_1 + \sum_2 = \sum
.\] 
As such, it follows \[
  \sum_1 - \sum_2 = 2\left( \frac{\sum}{2} - \sum_2 \right)
.\] 
That is, we want to find a subset \(S_2\) with a sum as close to \(\frac{\sum}{2}\) as possible,
without exceeding it.

\paragraph{Reframing as Knapsack}
For all \(x_i \in  S\), create an item with both weight and value equal to
\(x_i\).
Consider the knapsack problem (with no duplicates) as specified as above
with capacity of \(\frac{\sum}{2}\).

All items that are put in this knapsack can be used as \(S_1\)
while the remaining items can be a part of \(S_2\).

\subsection{Multiplying Chains of Matrices}

\paragraph{Matrix Multiplication}
The product of any two matrices \(A, B\)  \(AB\) exists if the number of
columns in \(A\) is equal to the number of rows of  \(B\).
If \(A\) is  \(m \times n\) and  \(B\) is  \(n\times p\)
then,  \(AB\) is  \(m\times p\).

Each element of \(AB\) \(ab_{i, j}\) is dot product of the \(i\)-th row of \(A\)
with the  \(j\)-th column of \(B\).
Hence, a naive solution would hvae \(m \times n \times p\) multiplications.

The order in which matrices are multiplied can drastically change the amount of
multiplications required.

\paragraph{Problem}
The input is a sequence of matrices \(A_1, A_2, \ldots, A_n\) where
\(A_i\) is of dimension  \(s_{i - 1} \times S_i\).

The task is to find the order of multiplication that minimises the number of
multiplications required.

\paragraph{Brute Force}
The brute force solution is of form \(\Omega(2^n)\).
The number of groupings \(T(n)\) is called the Catalan numbers.

\paragraph{Dynamic Approach}
There are many redundancies in the brute force approach as many of the same
prefixes are solved more than once.

A natural place to start, is just solve for the prefixes. However, this is not enough.
We need to solve for all contiguous subsequences \(A_{i + 1}, \ldots, A_j\).

\paragraph{Recurrences}
The recurrence will consider all of the possible ways to place the outermost
multiplication, splitting the chain into the product of elements
\((A_{i + 1}, \ldots A_k)(A_{k + 1}, \ldots, A_j)\).
We assume that we have solved the shortest way to multiply the sequence on the
left and, on the right.

The base-case is for a subsequence of length one where no multiplications are
required.

\paragraph{Notation for the Subproblems}
Let \(P(i, j)\) be the problem of determining \(\opt(i, j)\), which is the fewest
multiplications needed to compute the product \(A_{i + 1} \times , \ldots , A_j\).
Then, for all \(j - 1 > i\),  \[
  \opt(i,j) = \min \{
    \opt(i, k) + s_i s_k s_j + \opt(k, j): i < k < j
  \} 
.\] 

\paragraph{Order Matters}
Solving \(P(i, j)\) requires solving  \(P(i, k)\) and \(P(k, j)\) for all  \(i < k < j\).
This is guaranteed if, we solve in order of chain length.

\paragraph{Final Solution and Time Complexity}
The overall solution is \(\opt (0, n)\).
There are \(O(n^2)\) subproblems. Each of these requires \(O(n)\)
checks to find the best point of split. Thus, the time complexity is
\(O(n^3)\) as an upper bound.

However, there are many short chains and, very few short chains and,
the time complexity can be theoretically tightened. This is left as
an exercise to the reader.


\paragraph{Finding Actual Bracketing}
To find the actual location of brackets, we need to record the splitting point
that is used to obtain it. Then, we can just backtrack from the final solution.

\subsection{Longest Common Subsequence}

\subsubsection{Two Sequences}

\paragraph{Problem}
Consider two subsequences \[
  S = \langle a_1, a_2, \ldots, a_n \rangle \qquad
  \quad \text{ and, } \quad 
  \qquad S* = \langle b_1, b_2, \ldots, b_m \rangle
\]
The task is to find the longest common subsequence of \(S\) and \(S^*\).

\paragraph{Subsequences}
A subsequence \(s\) of another sequence  \(S\) is a sequence of ordered elements
of \(S\) obtained by removing some elements of  \(S\) while preserving order.

This may be used as a measure of similarity between two sequences. This can be
particularly useful for genetic analysis.

\paragraph{Prefixes}
Consider \(S_i, S^*_j\) which are just the first \(i\) and  \(j\)
elements of  \(S, S^*\).
Consider the last symbol of both,  \(a_i, b_j\).
If \(a_i = b_j = c\) then, the longest common subsequence is found
by appending  \(c\) to the solution for  \(S_{i-1}, S^*_{j - 1}\).

\paragraph{Subproblem}
For all \(0 \leq i \leq n\) and  \(0 \leq j \leq m\), let \(P(i, j)\) be the
subproblem of determinig  \(\opt(i, j)\). That is, the length of the longest
common subsequence of the truncated sequences \[
  S_i = \langle a_1, a_2, \ldots, a_i \rangle
  \quad \text{ and, }
  S^*_j = \langle b_1, b_2, \ldots, b_j \rangle
.\]

We have that \[
  \opt(i, j) = \begin{cases}
    \opt(i - 1, j - 1) + 1 & \text{ if } a_i = b_j \\
    \max(opt(i - 1, j), opt(i, j - 1)) & \text{ if } a_i \neq b_j
  \end{cases}
.\] 
The base cases are \(\opt(i, 0) = \opt(0, j) = 0\).

\paragraph{Order Matters}
Solving \(P(i, j)\) requires having solved  \(P(i, j - 1), P(i - 1, j)\).
As such, we can proceed in a bottom up manner, by solving all
subproblems \(P(i, 0)\), then all \(P(i, 1)\) until we reach  \(P(i, j)\).

\paragraph{Overall Solution and Time Complexity}
The overall solution is \(O(mn)\).
There are  \(mn\) subproblems, each of which an be solved in constant time.

\subsubsection{Three Sequences}

\paragraph{Problem}
Consider three sequences \(S, S^*, S'\) with elements  \(a_i, b_i, c_i\)
of length  \(l, m, n\) respectively.

\paragraph{Subproblems}
For all \(i, j, k\) let  \(P(i, j, k)\) be the subproblem of determining
the longest common sequence of the truncates sequences \[
  S_i = \langle a_1, a_2, \ldots, a_i \rangle
  \quad \text{ and, }
  S^*_j = \langle b_1, b_2, \ldots, b_j \rangle
  \quad \text{ and, }
  S'_k = \langle c_1, c_2, \ldots, c_k \rangle
\]

Then, for all \(i, j, k\)  \[
  \opt(i, j, k) = \begin{cases}
    opt(i - 1, j - 1, k - 1) + 1 & \text{ if } a_i = b_j = c_k \\
    \max \begin{pmatrix} 
      \opt(i - 1, j, k),
      \opt(i, j - 1, k),
      \opt(i, j, k - 1)
    \end{pmatrix}  & \text{ otherwise}.
  \end{cases}
.\] 

\paragraph{Overall Solution and Time Complexity}
We solve \(\opt(l, m, n)\)
There is a total of \(lmn\) subproblems, each of which can be solved in constant time.
The time complexity is thus \(O(lmn)\).

\subsubsection{Shortest Common Supersequence}

\paragraph{Problem}
Consider two sequences \(s, s'\) with \(n, m\) elements of form  \(a_i, b_i\) 
respectively.

The task is to find a shortest common supersequence of \(S\) and \(S'\).
That is, thne shortest possible sequuence of \(S, S'\)
such that \(s, s'\) are subsequences of the supersequence.

\paragraph{Solution}
Find the longer common subsequence of \(s, s'\).
Then, add back differing elements in the right places, in any arbitrary,
compatible order.

\subsection{String Matching}

\paragraph{Problem}
You are given two strings of length  \(A, B\) of length  \(n, m\).
You can perform the following operations on the strings:
\begin{itemize}
  \item insert a character for cost \(c_I\),
  \item delete a character for cost \(c_D\),
  \item replace a character for cost \(c_R\).
\end{itemize}

\paragraph{Task}
The task is to find the lowest cost total transformation of \(A\) into  \(B\).

\paragraph{Levenshtein Distance}
This is calculated when all costs to insert, delete and replace are equal.
As such, the Levenshtein distance is the minimum number of operations needed.

\paragraph{Consider Prefixes}
Consider prefixes \(A_i = A[1..i]\),  \(B_j = B[1..j]\). 
To transform \(A_i \to B_j\), the following options exist
\begin{itemize}
  \item Delete \(A[i]\) and tranform  \(A[1..i-1] \to  B[1..j]\),
  \item Transform \(A[1..i] \to  B[1..j-1]\) and append \(B[j]\)
  \item Transform  \(A[1..i-1] \to  B[1..j - 1]\). If neccesary, replace
    \(A[i]\) with \(B[j]\).
\end{itemize}
If \(i = 0\) or  \(j = 0\), we only insert or delete respectively.

\paragraph{Subproblems}
For all \(0 \leq i \leq n\) and \(0 \leq j \leq m\), let  \(P(i, j)\) be the
problem of determining  \(\opt(i, j)\). That is, the minimium cost of
transforming the squeuce \(A[1..i]\) into the sequence \(B[1..j]\).

Then, \[
  \opt(i, j) = \min \begin{cases}
    \opt(i - 1, j) + c_D \\
    \opt(i, j - 1) + c_I \\
    \begin{cases}
      \opt(i - 1, j - 1)        & \text{ if } A[i] = B[j] \\
      \opt(i - 1, j - 1) + c_R  & \text{ if } A[i] \neq B[j] \\
    \end{cases}
  \end{cases}
.\] 
There are also base cases where \(\opt(i, 0) = ic_D\) and  \(\opt(0, j) = jC_I\). 

\paragraph{Overall Solution}
The overall solution is \(\opt(n, m)\), leading to a total of \(mn\)
subproblems. Each subproblem can be solved in constant time. Thus, the total time complexity
is \(O(mn)\).

\subsection{Maximising an Expression}

\paragraph{Problem}
Consider a seqeunce of numbers with operations \(+, -, \times\) in between.

The task is to place brackets in such a way that the resulting expression
has maximal value.

\paragraph{Link to Matrix Multiplication}
This problem is effectively the same as the \textit{matrix chain multipliatition} problem.
It is not just enough to solve for prefixes \(A[1..i]\) but, all subsequences.

\paragraph{Cases with Operations}
Suppose that the sequence is of form \(A \circ B\) with  \(\circ in \{+, -, \times\} \).
We want to choose between whether \(A, B\) are maximised or minimised,
depending on the value of \(\circ\):
\begin{itemize}
  \item \(\circ = +\) : \(A, B\) are maximised.
  \item \(\circ = \times\) : \(A, B\) are maximised.
  \item \(\circ = -\) : \(A\) is  maximised,  \(B\) is minimised.
\end{itemize}
Therefore, it it not enough to just care for the maximum values but, also
the minimum values.

\paragraph{Solution}
Left as an exercise to the reader.

\subsection{Shortest Path in Directed, Acyclic Graph}
\paragraph{Notation: Direct Acyclic Graphs}
Define a DAG to be a directed acyclic graph.

\paragraph{Definition: Topological Sort}
Recall that a \textit{topological sort} of a DAG is a ordering of vertices
where all the edges point  \textit{left to right}.

Also note that:
\begin{itemize}
  \item This can occur only if, the graph is acyclic.
  \item The solution of a topological sort may not be unique.
  \item A topological sort can be found in linear \(O(|V| + |E|)\) time.
\end{itemize}

\paragraph{Problem}
Given a DAG \(G\), find the shortest path from \(s\) to \(t\).

\paragraph{Shortest Path: Standard Approach}
If all edge weights are positive, we can use Dijkstra's algorithm in \(O(m \log n)\).
For directed acyclic graphs however, this may be done in \(O(n + m)\) time.

\paragraph{Subproblems}
For all \(t \in V\), let \(P(t)\) be the problem of determining  \(\opt(t)\).
That is, the length of the shortest path from  \(s\) to  \(t\).

Then, for all \(t \neq s\),  \[
  \opt(t) = \min \{
    \opt(v) + w(v, t) : (v, t) \in E
  \} 
.\] 
The trivial base case is where \(\opt(s) = 0\).

\paragraph{Overall Solution}
The overall solution is given by \(\opt(t)\) where at each step, a predecessor must be stored.
There are \(n\) total subproblmes. However, we only do this  for all \(m\) edges and,
each edge only gets used once. Thus, the real bound is  \(O(m)\).

\paragraph{Order of Problems}
\(\opt(t)\) relies on solving all  \(\opt(v)\) where there is an edge from \(v\) to  \(t\).
As such, we can solve in the topological order from left to right.

\subsection{Assembly line scheduling}
\paragraph{Problem}
You are given two assembly lines, each of  \(n\) workstations. The  \(k\)-th workstation
performs the \(k\)-th job, out of  \(n\).

\begin{itemize}
  \item To bring a new job to the start of the assembly line \(i\) takes
    \(s_i\) units of time.
  \item To retrieve a finished product from the end of assembly line \(i\)
    takes  \(f_i\) units of time.
  \item On the assembly line \(i\), the  \(k\)-th job takes  \(a_{i, k}\) 
    units of time to complete.
  \item To move a product from station \(k\) on assembly line  \(i\) to station
    \(k + 1\) takes  \(t_{i, k}\) units of time.
  \item There is no time required to continue from station \(k\) to station
     \(k + 1\) on the same assembly line.
\end{itemize}

The task is to find the fastest way to assemble a product, switching lines
as necessary.

\paragraph{Topological Sort}
Denote vertices \(\left( i, j \right) \) to be the \(k\)-th workstation on assembly
line  \(i\). The problem is effectively requires finding the shortest path
from start node to end.
Trivially, the graph is directed and acyclic, as the product only moves
forwards.

The topological sort is trivially \[
  \text{start}, (1, 1), (2, 1), (1, 2), (2, 2), \ldots, (1, n), (2, n), \text{end}
.\] 

Hence, we can use dynamic programming over the topological sort to
find the shortest path.

\paragraph{Time Complexity}
There are \(2n + 2\) vertices and  \(4n\) edges. Thus, the  DP should
take \(O(n)\) time. Compare this to dijkstra's algorithm, which takes
\(O(n \log n)\).

\paragraph{Recurrence}
For \(i \in \{1, 2\} \) and \(1 \leq k \leq n\), let  \(P(i, k)\)
be the problem of determining  \(\opt(i, k)\). That is, the minimal time taken to complete
the first  \(k\) jobs, with the  \(k\)-th job being performed on assembly line \(i\).

Then,
\begin{align*}
  \opt(1, k) &= \min( \opt(1, k - 1), \opt(2, k -1) + t_{2, k - 1} )  + a_{1, k} \\
  \opt(2, k) &= \min( \opt(2, k - 1), \opt(2, k -1) + t_{1, k - 1} )  + a_{2, k}
.\end{align*}

The base cases are where \(\opt(1, 1) = s_1 + a_{1, 1}\) and \(\opt(2, 1 = s_{2} + a_{2, 1})\).

\subsection{Bellman-Form: Single Source Shortest Paths}

\paragraph{Problem}
There is a directed weighted graph \(G = (V, E)\) with edge weights  \(e\).
Edges may have negative weight but, cycles must have positive total weight.
Also, let  \(n = |V|, m = |E|\).

Let \(s \in V\). Find the shortest path from \(s\) to all other vertices  \(t\).

\paragraph{Dijkstra Fails}
Allowing negative weights means that the greedy strategy does not work.
Observe though, that cycles must not have negative weight. This is because
there is otherwise no finite shortest path as it may endlessly loop over the
cycle.

\paragraph{Property: No Cycles and Uniqueness of Vertex}
Observe that for all \(t \in V\), there exists a shortest path \(s \to  t\) without cycles.

This is true as any cycles in the graph are of positive weight as as such, can
be removed to make the weight of the path no worse.

As such, it follows that all shortest paths \(s \to t\) can contain any vertex at most once
and as such, have at most  \(n - 1\) edges.

\paragraph{Bellman-Ford}
For all vertices \(t\), we find the weight of the shortest path  \(s \to t\)
that consists of at \(i\) edges for all  \(i \in [1, n - 1] \in \mathbb{N}\).

Suppose that the path is of form \[
  p = \underbrace{s \to  \ldots \to  v}_{p'} \to  t
.\] 
Then, \(p'\) itself, must be the shortest path from \(s\) to \(v\),
of at msot  \(i - 1\) edges. This is another subproblem.

Obviously, there are base cases where  \(t = s\) or  \(i = 0\).

\paragraph{Suproblems}
For all \(i \in [1, n - 1] \in \mathbb{N}\), and \(t \in  V\), let \(P(i, t)\)
be the subproblem of determing  \(\opt(i, t)\). That is, the length of
the shortest path from  \(s\) to  \(t\), with at most  \(i\) edges.

Then,  \[
  \opt(i, t) = \min \{
    \opt(i - 1, v) + w(v, t) : (v, t) \in E
  \} 
.\] 
There are base cases where \(\opt(i, s) = 0\) and  for \(t \neq s\),  \(\opt(0, t) = \infty\).

\paragraph{Solution and Time Complexity}
The overall solution is \(\opt(n - 1, t)\).
There are \(n\) rounds for \(i \in [1, n - 1]\). In each round,
each edge is considered at most once. Thus, the overall complexity is  \(O(mn)\).

\paragraph{Construcing the Shortest Path}
Store a predecessor \(\pred(i, t)\) that is the immediate predeessor of vertex \(t\)
on a shortest path from \(s\) to \(t\) of at most \(i\) edges.

As such, \[
  \pred(i, t) = \argmin_{v \in V} \{
    \opt(i - 1, v) + w(v, t) : (v, t) \in E
  \} 
.\] 

\paragraph{Improvements}
As described above, the algorithmn builds a table of size \(O(n^2)\),
where each round is a new row.
This can be reduced to \(O(n)\). Using  \(\opt(i - 1, t)\) as a candidate for
\(\opt(i, t)\) does not affect the recurrence. As such, we can overwrite
the table as we go.

\subsection{Floyd-Warshall: All Pair Shortest Paths}

\paragraph{Problem}
The given graph \(G = (V, E)\) has edge weights  \(w(e)\). Edges may have negative
weight but, there are no cycles of negative total weight.

The task is to find the shortest path from all vertices \(s\), to all other vertices
\(t\).
Let \(n = |V|, m = |E|\).


\paragraph{The Idea}
Label the vertices \(V = \{v_1, v_2, \ldots, v_n\} \).
Similarly to Bellman-Ford, we want to think of the immediate vertices allowed
on a path \(s \to t\).

Let \(S\) be the set of vertices allowed as intermediate vertices.
Initially, \(S\) is empty and, we can add \(v_1, v_2, \ldots, v_n\) one at
a time, depending on the current vertex.

\paragraph{Improving Previous Rounds}
The shortest path from \(s \to t\) using the first \(k\) vertices
as intermediates is an improvement is improvement on the value from
the previous round if, there is a shorter path of form \[
  s \to \underbrace{\cdots}_{v_1, \cdots, v_{k-1}} \to \underbrace{\cdots}_{v_1, \cdots, v_{k-1}} \to t
.\]

\paragraph{Subproblems}
For all  \(i, j \in [1, n]\) and \(k \in [0, n]\), let \(P(i, j, k)\)
be the problem of determining \(\opt(i, j, k)\). That is,
the cost of the shortest path from \(v_i\) to  \(v_j\) using only
\(v_1, \ldots v_k\)  as intermediate vertices.

Then, for all \(i, j\),  \[
  \opt(i, j, k) = \min \{
    \opt(i, j, k - 1),
    \opt(i, k, k - 1),
    \opt(k, j, k - 1)
  \} 
.\] 
The base case is \[
  \opt(i, j, 0) = \begin{cases}
    0              & \text{if } i = j, \\
    w(i, j)        &if i \neq j, \\
    \infty         & \text{otherwise}.
  \end{cases}
.\] 

\paragraph{Solution and Time Complexity}
The subproblems rely on smaller \(k\) values. As such, they should be solved
in increasing order of  \(k\).

The overall solution is given as \(\opt(i, j, n)\). All  \(O(n^3)\)
subproblems can be done in constant time. As such, the complexity is  \(O(n^3)\).

As per Bellman-Ford, the space complexity can be improve by over-writing the same table
every round.