From ba57c235addf1bd2f734df7339c3ce507eabc731 Mon Sep 17 00:00:00 2001 From: Camil Staps Date: Thu, 10 Dec 2015 23:09:29 +0000 Subject: Start report practical 2 --- Practical2/report/algorithm.tex | 107 ++++++++++++++++++++++++++++++++++++++++ Practical2/report/notation.tex | 11 +++++ Practical2/report/report.tex | 81 ++++++++++++++++++++++++++++++ 3 files changed, 199 insertions(+) create mode 100644 Practical2/report/algorithm.tex create mode 100644 Practical2/report/notation.tex create mode 100644 Practical2/report/report.tex (limited to 'Practical2') diff --git a/Practical2/report/algorithm.tex b/Practical2/report/algorithm.tex new file mode 100644 index 0000000..73780f0 --- /dev/null +++ b/Practical2/report/algorithm.tex @@ -0,0 +1,107 @@ +\section{Algorithm} +\label{sec:algorithm} + +I will propose a solution that uses dynamic programming. First, I will describe the general structure of the algorithm, and simultaneously argue its correctness. After this, we will discuss some optimisations. + +\subsection{General program flow} +\label{sec:algorithm:general} + +\begin{thmproof}[P_0_k]{lemma}{For any $\bar p, k$, we have $P_0^k=0$.} + There is only one way to partition $\bar p$ (using $k$ empty partitions). The profit of every member of the partitioning is zero, and therefore the overall profit of the partitioning is zero as well. +\end{thmproof} + +\begin{thmproof}[P_p_0]{lemma}{For any $\bar p, p\le|\bar p|$, we have $P_p^0(\bar p)=\PR(S_0^p(\bar p))$.} + In this case there is only one way to partition $\bar p$ (using one complete partition). The profit is the difference between the sum of the elements and their rounded sum. +\end{thmproof} + +\begin{thmproof}[P_p_k]{lemma}{$P_p^k(\bar p) = \max\left(P_p^{k-1}(\bar p), \max\limits_{i=0}^{p-1}\left(P_i^{k-1}(\bar p) + \PR(S_i^p)\right)\right)$.} + Consider the $p$th element of $\bar p$ and a partitioning that gives the maximal profit in the substring until $\bar p_p$, $P_p^k(\bar p)$. We have two possibilities: + + \begin{itemize}[itemsep=0pt] + \item $\bar p_p$ is in one partition with $\bar p_i, \bar p_{i+1}, \dots, \bar p_{p-1}$. In this case we would have $P_p^k(\bar p) = P_i^{k-1}(\bar p) + \PR(S_i^p)$, since we can divide the rest of $\bar p$ over $k-1$ independent partitions and gain a profit of $\PR(S_i^p)$ over that last partition. + \item It is in fact more worthwhile to create $k-1$ partitions rather than $k$. In this case, we have $P_p^k(\bar p)=P_p^{k-1}(\bar p)$. + \end{itemize} + + To get the actual profit, we take the maximum of the two possibilities. +\end{thmproof} + +This suggests the general flow of our algorithm: we maintain an $(n+1)\times (k+1)$-table of values for $P_p^d$ with $0\le d\le k, 0\le p\le n$. Using the recurrences from \autoref{lem:P_p_0}, \ref{lem:P_0_k} and \ref{lem:P_p_k}, that only rely on `smaller values' ($P_{p'}^{d'}$ with $p'\le p$ or $d'\le d$), we generate the whole table. The solution to our problem is then $$\PM_k(\bar p) = \sum\bar p-\PP_k(\bar p) = \sum\bar p - P_{|\bar p|}^k(\bar p).$$ + +\begin{algorithm}[h] + \footnotesize + \caption{Finding the maximum profit $\PP_k(\bar p)$ of sequence $\bar p$ with $k$ partitions} + \label{alg:maximum-profit} + \begin{algorithmic}[1] + \REQUIRE $\bar p, k$ + \STATE $n\gets |\bar p|$ + \STATE $P[n+1][k+1]$ + \FOR{$0 \le p \le n$} + \FOR{$0 \le d \le k$} + \IF{$p = 0$} + \STATE $P[p][d] = 0$ \COMMENT{\autoref{lem:P_0_k}} + \ELSIF{$d = 0$} + \STATE $P[p][d] = \PR(S_0^p(\bar p))$ \COMMENT{\autoref{lem:P_p_0}} + \ELSE + \STATE $P[p][d] = \max\left(P[p][d-1], \max\limits_{i=0}^{p-1}\left(P[i][d-1] + \PR(S_i^p(\bar p))\right)\right)$ \COMMENT{\autoref{lem:P_p_k}} + \ENDIF + \ENDFOR + \ENDFOR + \RETURN $P[n][k]$ + \end{algorithmic} +\end{algorithm} + +\begin{thmproof}[maximum-profit]{theorem}{\autoref{alg:maximum-profit} is a correct implementation of the $\PP$ function.} + This follows from \autoref{lem:P_0_k} through \ref{lem:P_p_k}. +\end{thmproof} + +\subsection{Optimisations} +\label{sec:algorithm:optimisations} + +A first observation we should make is that values in the input sequence $\bar p$ that are divisible by $5$ do not affect the maximum profit of the substring they're in. + +\begin{thmproof}[divide-by-5]{lemma}{For any $\bar p$, $\PP(\bar p) = \PP\left(\seq{p_i \in\bar p \mid 5\nmid p_i}\right)$.} + No matter in which substring, a $p_i$ which is divisible by $5$ will have no influence on the rounded sum of that substring and thus not on the profit. +\end{thmproof} + +On average, \autoref{lem:divide-by-5} allows us to apply \autoref{alg:maximum-profit} on a input $\frac54$ times as small as the original input. While not decreasing the Big-O complexity of the algorithm, this does give us faster running times. + +Another optimisation concerns the $S$-function. In \autoref{alg:maximum-profit} above, it is called once for every $0\le i\le p$ in every iteration over $p$. Usually we would compute $S$ in linear time, iterating over $\bar p$. Computing \emph{all} the values for $S$ we need would then take quadratic time. However, we may as well use dynamic programming to compute all values for $S$ we're going to need in the beginning of the iteration over $p$, which can be done in linear time. This is demonstrated in \autoref{alg:maximum-profit-optimised}. In this algorithm, \autoref{alg:maximum-profit} has been optimised using \autoref{lem:divide-by-5} and the optimisation just discussed. + +\begin{algorithm}[h] + \footnotesize + \caption{Optimised version of \autoref{alg:maximum-profit}, to compute $\PP_k(\bar p)$} + \label{alg:maximum-profit-optimised} + \begin{algorithmic}[1] + \REQUIRE $\bar p, k$ + \STATE $\bar q\gets\seq{p_i\in\bar p\mid 5\nmid p_i}$ \COMMENT{\autoref{lem:divide-by-5}} + \STATE $n\gets |\bar q|$ + \STATE Initialise $P[n+1][k+1]$ + \FOR{$0 \le p \le n$} + \STATE Initialise $S[p]$ \COMMENT{Here we exploit dynamic programming to compute $S$} + \IF{$p=0$} + \STATE $S[0] = 0$ + \ELSE + \STATE $S[p-1] = q_{p-1}$ + \FOR{$i=p-2$ through $0$} + \STATE $S[i] = p_i + S[i+1]$ + \ENDFOR + \ENDIF + + \FOR{$0 \le d \le k$} + \IF{$p = 0$} + \STATE $P[p][d] = 0$ \COMMENT{\autoref{lem:P_0_k}} + \ELSIF{$d = 0$} + \STATE $P[p][d] = \PR(S[0])$ \COMMENT{\autoref{lem:P_p_0}} + \ELSE + \STATE $P[p][d] = \max\left(P[p][d-1], \max\limits_{i=0}^{p-1}\left(P[i][d-1] + \PR(S[i])\right)\right)$ \COMMENT{\autoref{lem:P_p_k}} + \ENDIF + \ENDFOR + \ENDFOR + \RETURN $P[n][k]$ + \end{algorithmic} +\end{algorithm} + +The correctness of \autoref{alg:maximum-profit-optimised} depends on the correctness of \autoref{alg:maximum-profit} which has been proven in \autoref{thm:maximum-profit}. + +We can now easily find $\PM_k(\bar p)$ using the equation we saw in \autoref{sec:notation}: $$\PM_k(\bar p) = \sum\bar p - \PP_k(\bar p).$$ + diff --git a/Practical2/report/notation.tex b/Practical2/report/notation.tex new file mode 100644 index 0000000..d21295e --- /dev/null +++ b/Practical2/report/notation.tex @@ -0,0 +1,11 @@ +\section{Notation} +\label{sec:notation} + +Given a list of $n$ numbers $p_0,p_1,\dots,p_{n-1}$, we consider one of its partitions in $k$ (possibly empty) pairwise disjoint substrings such that the sum of the sums of the substrings, rounded to multiples of $5$, is minimal. Output that minimal sum. + +I will write $\bar p = \seq{p_i \mid i