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Diffstat (limited to 'Practical2/report/algorithm.tex')
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diff --git a/Practical2/report/algorithm.tex b/Practical2/report/algorithm.tex index 175ad4b..38a9e71 100644 --- a/Practical2/report/algorithm.tex +++ b/Practical2/report/algorithm.tex @@ -7,7 +7,7 @@ I will propose a solution that uses dynamic programming. First, I will describe \label{sec:algorithm:general} \begin{thmproof}[G_0_k]{lemma}{For any $\bar a, k$, we have $G_0^k(\bar a)=0$.} - There is only one way to partition $\bar a$ (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. + There is only one way to partition $\bar a$; using $k+1$ 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}[G_p_0]{lemma}{For any $\bar a, p\le|\bar a|$, we have $G_p^0(\bar a)=\PR(S_0^p(\bar a))$.} @@ -15,21 +15,21 @@ I will propose a solution that uses dynamic programming. First, I will describe \end{thmproof} \begin{thmproof}[G_p_k]{lemma}{$G_p^k(\bar a) = \max\left(G_p^{k-1}(\bar a), \max\limits_{i=0}^{p-1}\left(G_i^{k-1}(\bar a) + \PR(S_i^p(\bar a))\right)\right)$.} - Consider the $p$th element of $\bar a$ and a partitioning that gives the maximal profit in the substring until $\bar a_p$, $G_p^k(\bar a)$. We have two possibilities: + Consider the $p$th element of $\bar a$ ($\bar a_{p-1}$) and a partitioning that gives the maximum gain in the substring until $\bar a_p$, $G_p^k(\bar a)$. We have two possibilities: \begin{itemize}[itemsep=0pt] - \item $\bar a_p$ is in one partition with $\bar a_i, \bar a_{i+1}, \dots, \bar a_{p-1}$. In this case we would have $G_p^k(\bar a) = G_i^{k-1}(\bar a) + \PR(S_i^p(\bar a))$, since we can divide the rest of $\bar a$ over $k-1$ independent partitions and gain a profit of $\PR(S_i^p(\bar a))$ over that last partition. - \item It is in fact more worthwhile to create $k-1$ partitions rather than $k$. In this case, we have $G_p^k(\bar a)=G_p^{k-1}(\bar a)$. + \item $\bar a_{p-1}$ is in one substring with $\bar a_i, \bar a_{i+1}, \dots, \bar a_{p-2}$ (possibly $i=p-1$). In this case we would have $G_p^k(\bar a) = G_i^{k-1}(\bar a) + \PR(S_i^p(\bar a))$, since we can independently divide the rest of $\bar a$ over $k-1$ partitions and gain a profit of $\PR(S_i^p(\bar a))$ over that last partition. + \item It is in fact equally worthwhile to create $k$ partitions rather than $k+1$. In this case, we have $G_p^k(\bar a)=G_p^{k-1}(\bar a)$. \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 $G_p^d$ with $0\le d\le k, 0\le p\le n$. Using the recurrences from \autoref{lem:G_p_0}, \ref{lem:G_0_k} and \ref{lem:G_p_k}, that only rely on `smaller values' ($G_{p'}^{d'}$ with $p'\le p$ or $d'\le d$), we generate the whole table. The solution to our problem is then $$\PM_k(\bar a) = \sum\bar a-\PP_k(\bar a) = \sum\bar a - G_{|\bar a|}^k(\bar a).$$ +This suggests the general flow of our algorithm: we maintain an $(n+1)\times (k+1)$-table of values for $G_p^d$ with $0\le d\le k, 0\le p\le n$. Using the recurrences from \autoref{lem:G_0_k}, \ref{lem:G_p_0} and \ref{lem:G_p_k}, we generate the whole table. \begin{algorithm}[h] \footnotesize - \caption{Finding the maximum profit $\PP_k(\bar a)$ of sequence $\bar a$ with $k$ partitions} + \caption{Finding the maximum profit $\PP_k(\bar a)$ of sequence $\bar a$ with at most $k+1$ partitions} \label{alg:maximum-profit} \begin{algorithmic}[1] \REQUIRE $\bar a, k$ @@ -54,13 +54,15 @@ This suggests the general flow of our algorithm: we maintain an $(n+1)\times (k+ This follows from \autoref{lem:G_0_k} through \ref{lem:G_p_k}. \end{thmproof} +The solution to our problem is then $$\PM_k(\bar a) = \sum\bar a-\PP_k(\bar a) = \sum\bar a - G_{|\bar a|}^k(\bar a).$$ + \subsection{Optimisations} \label{sec:algorithm:optimisations} A first observation we should make is that values in the input sequence $\bar a$ 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 $\bar a$, $\PP(\bar a) = \PP\left(\seq{a_i \in\bar a \mid 5\nmid a_i}\right)$.} - No matter in which substring, an $a_i$ which is divisible by $5$ will have no influence on the rounded sum of that substring and thus not on the profit. +\begin{thmproof}[divide-by-5]{lemma}{For any $\bar a,k$, $\PP_k(\bar a) = \PP_k\left(\seq{a_i \in\bar a \mid 5\nmid a_i}\right)$.} + No matter in which substring (say $\bar a'$), an $a_i$ that is divisible by $5$ will have no influence on $\PR(\bar a')$ and thus not on $\PP_k(\bar a)$. \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. @@ -81,9 +83,9 @@ Another optimisation concerns the $S$-function. In \autoref{alg:maximum-profit} \IF{$p=0$} \STATE $S[0] = 0$ \ELSE - \STATE $S[p-1] = a'_{p-1}$ + \STATE $S[p-1] = \bar a'_{p-1}$ \FOR{$i=p-2$ through $0$} - \STATE $S[i] = a_i + S[i+1]$ + \STATE $S[i] = \bar a'_i + S[i+1]$ \ENDFOR \ENDIF @@ -93,7 +95,7 @@ Another optimisation concerns the $S$-function. In \autoref{alg:maximum-profit} \ELSIF{$d = 0$} \STATE $G[p][d] = \PR(S[0])$ \COMMENT{\autoref{lem:G_p_0}} \ELSE - \STATE $G[p][d] = \max\left(G[p][d-1], \max\limits_{i=0}^{p-1}\left(G[i][d-1] + \PR(S[i])\right)\right)$ \COMMENT{\autoref{lem:G_p_k}} + \STATE $G[p][d] = \max\left(G[p][d-1], \max\limits_{i=0}^{p-1}\left(G[i][d-1] + \PR(S[i])\right)\right)$ \COMMENT{\autoref{lem:G_p_k}} \label{alg:optimised:line:recurrence} \ENDIF \ENDFOR \ENDFOR @@ -103,5 +105,3 @@ Another optimisation concerns the $S$-function. In \autoref{alg:maximum-profit} 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 $\PM_k(\bar a)$ using the equation we saw in \autoref{sec:notation}: $$\PM_k(\bar a) = \sum\bar a - \PP_k(\bar a).$$ - |