Finish first version report practical 2

Some changes were made to the code and the Makefile to clean up, make it
more in line with the algorithm as described in the report, etc. No
significant changes.

7 files changed, 110 insertions, 74 deletions

diff --git a/Practical2/Makefile b/Practical2/Makefile
index 0b147dd..b400d34 100644
--- a/Practical2/Makefile
+++ b/Practical2/Makefile
@@ -1,4 +1,5 @@
-CC=gcc
+CC=clang
+CFLAGS=-Wall -Ofast
 OBJS=checkout
 
 .PHONY: all run clean
@@ -6,7 +7,7 @@ OBJS=checkout
 all: $(OBJS)
 
 $(OBJS): % : %.c
-	$(CC) -o $@ $< -Ofast 
+	$(CC) -o $@ $< $(CFLAGS)
 
 clean:
 	rm $(OBJS)
diff --git a/Practical2/checkout.c b/Practical2/checkout.c
index 289a50d..88d962e 100644
--- a/Practical2/checkout.c
+++ b/Practical2/checkout.c
@@ -1,6 +1,8 @@
 #include <stdio.h>
 #include <stdint.h>
 
+#pragma GCC diagnostic push
+#pragma GCC diagnostic ignored "-Wreturn-type"
 int8_t round_profit(uint32_t value) {
     switch (value % 5) {
         case 0: return 0;
@@ -10,61 +12,57 @@ int8_t round_profit(uint32_t value) {
         case 4: return -1;
     }
 }
+#pragma GCC diagnostic pop
 
-int32_t maximum_profit(uint16_t products[], uint16_t n_products, uint8_t dividers) {
-    int8_t profit[n_products+1][dividers+1];
-    uint16_t p, d;
-    for (p = 0; p <= n_products; p++) {
-        uint32_t sums[p];
+int32_t maximum_profit(int seq[], uint16_t seq_len, uint8_t k) {
+    int8_t gain[seq_len+1][k+1];
+    uint16_t p, d, i;
+    for (p = 0; p <= seq_len; p++) {
+        uint32_t sum[p];
         if (p == 0) {
-            sums[0] = 0;
+            sum[0] = 0;
         } else {
-            sums[p-1] = products[p-1];
-            int16_t i;
-            for (i = p-2; i >= 0; i--) {
-                sums[i] = products[i] + sums[i+1];
-            }
+            sum[p-1] = seq[p-1];
+            for (i = p-2; i != UINT16_MAX; i--)
+                sum[i] = seq[i] + sum[i+1];
         }
 
-        for (d = 0; d <= dividers; d++) {
+        for (d = 0; d <= k; d++) {
             if (p == 0) {
-                profit[p][d] = 0;
+                gain[p][d] = 0;
             } else if (d == 0) {
-                profit[p][d] = round_profit(sums[0]);
+                gain[p][d] = round_profit(sum[0]);
             } else {
-                int8_t max = profit[p][d-1];
-                uint16_t i;
+                int8_t max = gain[p][d-1];
                 for (i = 0; i < p; i++) {
-                    int8_t try = profit[i][d-1] + round_profit(sums[i]);
+                    int8_t try = gain[i][d-1] + round_profit(sum[i]);
                     if (try > max)
                         max = try;
                 }
-                profit[p][d] = max;
+                gain[p][d] = max;
             }
         }
     }
 
-    return profit[n_products][dividers];
+    return gain[seq_len][k];
 }
 
 int main(void) {
-    int n_products;
-    int dividers;
+    int seq_len, k;
     uint32_t sum = 0;
+    scanf("%d %d", &seq_len, &k);
+    int seq[seq_len];
 
-    scanf("%d %d", &n_products, &dividers);
-
-    uint16_t products[n_products];
     uint16_t i;
-    for (i = 0; i < n_products; i++) {
-        scanf("%d", &(products[i]));
-        sum += products[i];
-        if (products[i] % 5 == 0) {
-            n_products--;
+    for (i = 0; i < seq_len; i++) {
+        scanf("%d", &(seq[i]));
+        sum += seq[i];
+        if (seq[i] % 5 == 0) {
+            seq_len--;
             i--;
         }
     }
 
-    printf("%d\n", sum - maximum_profit(products, n_products, dividers));
+    printf("%d\n", sum - maximum_profit(seq, seq_len, k));
 }
 
diff --git a/Practical2/report/algorithm.tex b/Practical2/report/algorithm.tex
index 73780f0..175ad4b 100644
--- a/Practical2/report/algorithm.tex
+++ b/Practical2/report/algorithm.tex
@@ -6,102 +6,102 @@ I will propose a solution that uses dynamic programming. First, I will describe
 \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.
+\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.
 \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.
+\begin{thmproof}[G_p_0]{lemma}{For any $\bar a, p\le|\bar a|$, we have $G_p^0(\bar a)=\PR(S_0^p(\bar a))$.}
+    In this case there is only one way to partition $\bar a$ (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{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:
 
     \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 $P_p^k(\bar p)=P_p^{k-1}(\bar p)$.
+        \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 $G_p^k(\bar a)=G_p^{k-1}(\bar a)$.
     \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 $$\PM_k(\bar p) = \sum\bar p-\PP_k(\bar p) = \sum\bar p - P_{|\bar p|}^k(\bar p).$$
+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 $$\PM_k(\bar a) = \sum\bar a-\PP_k(\bar a) = \sum\bar a - G_{|\bar a|}^k(\bar a).$$
 
 \begin{algorithm}[h]
     \footnotesize
-    \caption{Finding the maximum profit $\PP_k(\bar p)$ of sequence $\bar p$ with $k$ partitions}
+    \caption{Finding the maximum profit $\PP_k(\bar a)$ of sequence $\bar a$ with $k$ partitions}
     \label{alg:maximum-profit}
     \begin{algorithmic}[1]
-        \REQUIRE $\bar p, k$
-        \STATE $n\gets |\bar p|$
-        \STATE $P[n+1][k+1]$
+        \REQUIRE $\bar a, k$
+        \STATE $n\gets |\bar a|$
+        \STATE $G[n+1][k+1]$
         \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}}
+                    \STATE $G[p][d] = 0$ \COMMENT{\autoref{lem:G_0_k}}
                 \ELSIF{$d = 0$}
-                    \STATE $P[p][d] = \PR(S_0^p(\bar p))$ \COMMENT{\autoref{lem:P_p_0}}
+                    \STATE $G[p][d] = \PR(S_0^p(\bar a))$ \COMMENT{\autoref{lem:G_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}}
+                \STATE $G[p][d] = \max\left(G[p][d-1], \max\limits_{i=0}^{p-1}\left(G[i][d-1] + \PR(S_i^p(\bar a))\right)\right)$ \COMMENT{\autoref{lem:G_p_k}}
                 \ENDIF
             \ENDFOR
         \ENDFOR
-        \RETURN $P[n][k]$
+        \RETURN $G[n][k]$
     \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}.
+    This follows from \autoref{lem:G_0_k} through \ref{lem:G_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.
+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 $\bar p$, $\PP(\bar p) = \PP\left(\seq{p_i \in\bar p \mid 5\nmid p_i}\right)$.}
-    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.
+\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.
 \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.
+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 $\bar a$. 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 $\PP_k(\bar p)$}
+    \caption{Optimised version of \autoref{alg:maximum-profit}, to compute $\PP_k(\bar a)$}
     \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 $P[n+1][k+1]$
+        \REQUIRE $\bar a, k$
+        \STATE $\bar a'\gets\seq{a_i\in\bar a\mid 5\nmid a_i}$ \COMMENT{\autoref{lem:divide-by-5}}
+        \STATE $n\gets |\bar a'|$
+        \STATE Initialise $G[n+1][k+1]$
         \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 $S[p-1] = q_{p-1}$
+                \STATE $S[p-1] = a'_{p-1}$
                 \FOR{$i=p-2$ through $0$}
-                    \STATE $S[i] = p_i + S[i+1]$
+                    \STATE $S[i] = a_i + S[i+1]$
                 \ENDFOR
             \ENDIF
 
             \FOR{$0 \le d \le k$}
                 \IF{$p = 0$}
-                    \STATE $P[p][d] = 0$ \COMMENT{\autoref{lem:P_0_k}}
+                    \STATE $G[p][d] = 0$ \COMMENT{\autoref{lem:G_0_k}}
                 \ELSIF{$d = 0$}
-                    \STATE $P[p][d] = \PR(S[0])$ \COMMENT{\autoref{lem:P_p_0}}
+                    \STATE $G[p][d] = \PR(S[0])$ \COMMENT{\autoref{lem:G_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}}
+                \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}}
                 \ENDIF
             \ENDFOR
         \ENDFOR
-        \RETURN $P[n][k]$
+        \RETURN $G[n][k]$
     \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}: $$\PM_k(\bar p) = \sum\bar p - \PP_k(\bar p).$$
+We can now easily find $\PM_k(\bar a)$ using the equation we saw in \autoref{sec:notation}: $$\PM_k(\bar a) = \sum\bar a - \PP_k(\bar a).$$
 
diff --git a/Practical2/report/analysis.tex b/Practical2/report/analysis.tex
new file mode 100644
index 0000000..d989fe6
--- /dev/null
+++ b/Practical2/report/analysis.tex
@@ -0,0 +1,18 @@
+\section{Complexity analysis}
+\label{sec:analysis}
+
+We iterate $p$ in $\pazocal O(|\bar a'|) = \pazocal O(|\bar a|)$. In every iteration we do two things:
+
+\begin{itemize}
+    \item We build $S$ in $\pazocal O(p)=\pazocal O(|\bar a|)$.
+    \item We iterate $d$ in $\pazocal O(k)$. Apart from the special cases that $p=0$ or $d=0$, we then have to iterate over $i$ in $\pazocal O(p)=\pazocal O(|\bar a|)$.
+        
+        The computations in this last loop take constant time: $\PR$ can be implemented in constant time as has been seen in \autoref{sec:implementation}, and computing the maximum can be done while iterating over $i$.
+\end{itemize}
+
+This gives a total time complexity of $\pazocal O(k\cdot|\bar a|^2)$.
+
+\medskip
+
+The algorithm uses a two-dimensional array $G$ to store solutions to subproblems. This array uses $\pazocal O(k\cdot|\bar a|)$ space. The auxiliary array $S$ takes only linear space in $|\bar a|$ and can therefore be ignored in the overall space complexity. This gives the algorithm a space complexity of $\pazocal O(k\cdot|\bar a|)$.
+
diff --git a/Practical2/report/implementation.tex b/Practical2/report/implementation.tex
new file mode 100644
index 0000000..cc8a940
--- /dev/null
+++ b/Practical2/report/implementation.tex
@@ -0,0 +1,19 @@
+\section{Implementation}
+\label{sec:implementation}
+
+A C implementation accompanies this report. The $\PR$ function is implemented in $\pazocal O(1)$ in \mintinline{c}{round_profit}, but is not shown here for the sake of brevity.
+
+The implementation of the $\PP$ function, \mintinline{c}{maximum_profit}, follows the flow of \autoref{alg:maximum-profit-optimised}. However, it was easier to take the optimisation of \autoref{lem:divide-by-5} out of this function and apply it when reading in the input sequence.
+
+\inputminted[fontsize=\footnotesize,linenos,firstline=17,lastline=48]{c}{../checkout.c}
+
+In the \mintinline{c}{main} function we see how \autoref{lem:divide-by-5} can be applied. In the last line we also see how the minimal sum may be computed from the maximum profit.
+
+\inputminted[fontsize=\footnotesize,linenos,firstline=50,lastline=67]{c}{../checkout.c}
+
+\subsection{Compilation}
+\label{sec:implementation:compilation}
+Optimal results are achieved with \texttt{clang}, which is on average $25\%$ faster than \texttt{gcc}. In any case, the \texttt{-Ofast} flag should be used. It speeds up the program with a factor $4$ in the case of \texttt{clang} and a factor $3$ in the case of \texttt{gcc}.
+
+A Makefile accompanies the C code. If you don't have \texttt{clang} installed, you should install it. If you don't want to install it, use \mintinline{bash}{make CC=gcc}.
+
diff --git a/Practical2/report/notation.tex b/Practical2/report/notation.tex
index d21295e..cbbbc1d 100644
--- a/Practical2/report/notation.tex
+++ b/Practical2/report/notation.tex
@@ -1,11 +1,9 @@
 \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 $\bar a = \seq{a_i \mid i<n}$ for the sequence of numbers and $k$ for the number of substrings. I redefine $\round{x}$ to mean `the nearest multiple of $5$ to $x$'. We may define the `round profit' as $\PR(x)=x-\round{x}$.
 
-I will write $\bar p = \seq{p_i \mid i<n}$ for the sequence of numbers and $k$ for the number of substrings. I redefine $\round{x}$ to mean `the nearest multiple of $5$ to $x$'. We may define the `round profit' as $\PR(x)=x-\round{x}$.
+Let $\PM_k(\bar a)$ be the minimal sum of rounded sums of $\bar a$ using $k$ partitions. I will use $\PP_k(\bar a)$ as the `maximum profit' of $\bar a$, that is, $\PP_k(\bar a) = \sum\bar a - \PM_k(\bar a)$. This leads to the term `profit' of a particular partitioning of $\bar a$ and of a particular (sub)string $\bar a'$, for which we need not introduce notation. The profit of a string $\bar a'$ is defined as $\PR(\sum\bar a)$. It is not difficult to see that the profit of a partition is equal to the some of the profits of its members.
 
-Let $\PM_k(\bar p)$ be the minimal sum of rounded sums of $\bar p$ using $k$ partitions. I will use $\PP_k(\bar p)$ as the `maximum profit' of $\bar p$, that is, $\PP_k(\bar p) = \sum\bar p - \PM_k(\bar p)$. This leads to the term `profit' of a particular partitioning of $\bar p$ and of a particular (sub)string $\bar q$, for which we need not introduce notation. The profit of a string $\bar q$ is defined as $\PR(\sum\bar q)$. It is not difficult to see that the profit of a partition is equal to the some of the profits of its members.
-
-I will write $P^k_p(\bar p)$ for $\PP_k(\seq{p_i\in\bar p\mid i<p})$ (that is, the maximal profit on $\bar p$'s prefix with length $p$ using $k$ partitions). Lastly, let $S_i^j(\bar p)$ be the sum of all $p_x\in\bar p$ with $i\le x<j$.
+We define the gain $G^k_p(\bar a)$ as $\PP_k(\seq{a_i\in\bar a\mid i<p})$ (that is, the maximal profit on $\bar a$'s prefix with length $p$ using $k$ partitions). Lastly, let $S_i^j(\bar a)$ be the sum of all $p_x\in\bar a$ with $i\le x<j$.
 
diff --git a/Practical2/report/report.tex b/Practical2/report/report.tex
index 4c71eee..2aeb621 100644
--- a/Practical2/report/report.tex
+++ b/Practical2/report/report.tex
@@ -63,7 +63,9 @@
 \thispagestyle{fancy}
 
 \begin{abstract}
-    %todo
+    Given a list of $n$ numbers $a_0, a_1, \dots, a_{n-1}$, 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.
+
+    This paper describes an algorithm to find the minimal sum for such a partition. It can easily be extended to find such a partition as well.
 \end{abstract}
 
 \section{Report organisation}
@@ -73,8 +75,8 @@ When we have seen the algorithm, \autoref{sec:implementation} will go into detai
 
 \input{notation}
 \input{algorithm}
-%\input{implementation}
-%\input{analysis}
+\input{implementation}
+\input{analysis}
 %\input{discussion}
 
 \end{document}