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\title{Scheduling Borrels to maximize students attendance and social experience} %TODO Please add

%\titlerunning{Dummy short title} %TODO optional, please use if title is longer than one line

\author{Floris van der Hout}{Utrecht University , 7234080, Netherlands}{f.l.vanderhout@students.uu.nl}{}{}

\author{Thomas van Maaren}{Utrecht University , studentnumber, Netherlands}{t.vanmaaren@students.uu.nl}{}{}

\author{ Rens Versnel}{Utrecht University , 2693852, Netherlands}{r.k.versnel@students.uu.nl}{}{}

\author{Wessel van der Ven}{Utrecht University , 2828189, Netherlands}{w.j.vanderven@students.uu.nl}{}{}

\author{Ids de Vlas}{Utrecht University , studentnumber, Netherlands}{i.devlas@students.uu.nl}{}{}

\authorrunning{Floris van der Hout, Thomas van Maaren, Rens Versnel, Wessel van der Ven and Ids de Vlas} %TODO mandatory. First: Use abbreviated first/middle names. Second (only in severe cases): Use first author plus 'et al.'

\Copyright{Floris van der Hout, Thomas van Maaren, Rens Versnel, Wessel van der Ven and Ids de Vlas} %TODO mandatory, please use full first names. LIPIcs license is "CC-BY";  http://creativecommons.org/licenses/by/3.0/

\ccsdesc[100]{Online algorithms} %TODO mandatory: Please choose ACM 2012 classifications from https://dl.acm.org/ccs/ccs_flat.cfm 

\keywords{Borrels Scheduling, NP-completeness, competitive ratio's} %TODO mandatory; please add comma-separated list of keywords

\category{} %optional, e.g. invited paper

\relatedversion{} %optional, e.g. full version hosted on arXiv, HAL, or other respository/website
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%\supplementdetails[linktext={opt. text shown instead of the URL}, cite=DBLP:books/mk/GrayR93, subcategory={Description, Subcategory}, swhid={Software Heritage Identifier}]{General Classification (e.g. Software, Dataset, Model, ...)}{URL to related version} %linktext, cite, and subcategory are optional

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%Editor-only macros:: begin (do not touch as author)%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\EventEditors{John Q. Open and Joan R. Access}
\EventNoEds{2}
\EventLongTitle{42nd Conference on Very Important Topics (CVIT 2016)}
\EventShortTitle{CVIT 2016}
\EventAcronym{CVIT}
\EventYear{2016}
\EventDate{December 24--27, 2016}
\EventLocation{Little Whinging, United Kingdom}
\EventLogo{}
\SeriesVolume{42}
\ArticleNo{23}
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\begin{document}

\maketitle

%TODO mandatory: add short abstract of the document
\begin{abstract}
Lorem ipsum dolor sit amet, consectetur adipiscing elit. Praesent convallis orci arcu, eu mollis dolor. Aliquam eleifend suscipit lacinia. Maecenas quam mi, porta ut lacinia sed, convallis ac dui. Lorem ipsum dolor sit amet, consectetur adipiscing elit. Suspendisse potenti. 
\end{abstract}

\section{Introduction}
Image you are part of the student organization responsible for organizing "borrels" (social gatherings) for freshmen. However, the freshmen have many obligations such as deadlines, group meetings, exams, ect. These obligations are flexible and can be rescheduled or delayed, but they still need to be completed eventually. Freshmen are eager to attend the borrels and will do so as long as they don't have any conflicting obligations. However, if their schedules clash with their obligations, they will prioritize those obligations over the borrel. Borrels are typically free for students to attend, but they do incur costs for the student organization, such as renting a venue, providing snacks, and other logistical expenses. It's common for borrels to offer the first drink for free, but any additional drinks must be purchased. The revenue generated from drink sales is used to help cover the borrels costs. If students enjoy themselves, they're more likely to buy drinks. One key factor that can enhance their experience is being able to attend the borrels with their friends. Therefore, it's in the student organization's interest to not only focus on attendance but also on maximizing the happiness of the freshman by scheduling borrels in a way that allows groups of friends to participate together. This can boost the overall success of the event both socially and financially. \\

This paper will present two offline algorithms for scheduling borrels: one focused on optimizing student attendance, and the other designed to maximize the overall happiness of the freshman by considering factors like social connections and shared availability. Additionally, we will provide problem competitive ratios for online algorithms which focusing on maximizing the attendance of freshmen at the borrels in multiple scenario's.

\subsection{Literature review}
We will visit several relevant problems in the literature, the first of which is the Social Event Scheduling Problem (SES) \cite{SocialEventPlanning}. In this context, the goal is to schedule a series of events or borrels while maximizing attendance or overall utility, where utility is defined as the expected attendance across all scheduled events. The primary distinction between the two problems lies in how competing events are treated. In the SES problem, competing events are those that have already been scheduled by third parties, resulting in fixed time slots. In contrast, our constraints, which can be viewed as competing events, can often be postponed within their designated time windows. Furthermore, in the SES framework, each event must be assigned a probability representing a user's likelihood of attendance. This approach presents challenges in our scenario since our students are generally eager to attend unless their schedules prevent them from doing so. \\

Another relevant problem is the scheduling of meetings within large organizations, where the objective is to maximize attendance while considering each participant's individual schedule. Since it is not always feasible for everyone to attend, the focus shifts to maximizing the number of attendees at the meeting \cite{distributedOptimizationOfCalendarEvents} \cite{EventSchedulingWithSoftConstraints}. However, solutions derived from this model are not directly applicable to our problem. In these frameworks, a participant is either available or unavailable which is different from our problem.\\

The final relevant problem is the Job Scheduling Problem with Deadlines \cite{JobShopSchedulingWithDeadlines}, which closely parallels the obligation and borrel planning aspects of our study. In this scenario, each obligation can be viewed as a job that needs to be scheduled on a specific machine, with both a deadline and a start time. Similarly, we have borrels that also need to be scheduled on these machines. In our problems context, we receive points for scheduling borrels on different machines simultaneously, the same as maximizing attendance at the events. However, this is different from the objective in the job scheduling problem with deadlines which is to minimize the makespan while ensuring that all tasks remain feasible within their respective constraints. That is why solutions from this problem can't be used for our problem.

\section{Preliminaries}
\subsection{Formal problem definition}
The timeline is partitioned into t integral time slots. The time interval from time
0 (inclusively) to time 1 (exclusively) is time slot 1, and the time interval from
time $i - 1$ (inclusively) to time $i$ (exclusively) is time slot i for all $i \geq 1$.
There are $m$ borrels, and each borrel $h$ has a length of $b_h$ time slots. There are
$n$ students. Each student $i$ has a set of obligations 
$I_i = \{(I_{i,1}, p_{i,1}), (I_{i,2}, p_{i,2}), ... , (I_{i,l_i} , p_{i,l_i} )\}$,
where $I_{i,j} = [r_{i,j} , d_{i,j} ]$ is a time interval from time slot $r_{i,j}$ to time slot $d_{i,j}$ , and
$p_{i,j}$ is the time slots needed by the obligation $j$ that has to be finished within
$I_{i,j}$ . A feasible schedule for a student $i$ of obligation $j$ is a schedule $S_{i,j}$ , which is a set of time slots, such that $S_{i,j} \subseteq I_{i,j}$ and $|S_{i,j}| = p_{i,j}$ . Moreover, for a student
$i$ and two of their obligations $j$ and $j'$, $S_{i,j} \cap S_{i,j'} = \varnothing$. Note that we assume
$p_{i,j}$ are integral for all $i$ and $j$, and we cannot assign a time slot partially to an
obligation.
As a borrels planner, you aim to schedule each borrel $h$ starting at time slot
$s_h$ and lasting until time slot $s_h + b_h - 1$, such that as many students attend
the borrels as possible. More formally, you want to find the start time $s_h$ of
each borrel $h$ such that there exists a set of feasible schedules $S_{i,j}$ for every pair
of student $i$ and obligation $j$ such that the conflict among the borrels and the
students’ schedules, $[s_h, s_h + b_h - 1] \cap S_{i,j} \neq \varnothing$ for some $i$ and $j$, is minimized.
We also assume that borrels start at integral time (that is, the beginning of a
time slot).
\subsubsection{Offline setting} 
In the offline setting, all the parameters are known to
the algorithm from the very beginning. That is, the algorithm knows $t$, $m$, $b_h$
for all $h \in [1, m]$, $n$, and $I_i$ for all $i \in [1, n]$. Once the algorithm receives the
information, it should suggest schedules for all borrels.
\subsubsection{Online setting}
In the online setting, the online algorithm knows the
number of time slots ($t$), the number of students ($n$), the number of borrels ($m$),
and all $b_h$ for each borrel $h$ in the beginning but only learns about $I_{i,j} = [r_{i,j} , d_{i,j} ]$
and $p_{i,j}$ at the beginning of time slot $r_{i,j}$ (that is, at time $r_{i,j} - 1$). The online algorithm has to decide if a borrel will start at time slot $x + 1$ right at time
$x$ (that is, the beginning of time slot $x + 1$). Formally, at time $x$ (that is, the
beginning of time slot $x + 1$), the algorithm first learns the obligations that have
release time at time slot $x + 1$ and then decides if it wants to schedule a borrel
starting from time slot $x + 1$.


\subsection{Formulation as a decision problem}
% This is the outline
In this section we will formulate the decision version of the borrel scheduling problem in which the aim is to decide if there exists a feasible schedule of borrels such that there are at least $k$ attendances for some $k\in\N$. We will use most of the formulation of the original borrel scheduling problem but change the objective.

% This is the actual thing

%In order to prove the NP-completeness of borrel scheduling it needs to be formulated as a decision problem. The aim of the original problem is to schedule each borrel $h$ starting at time slot $sh$ and lasting until time slot $s_h + b_h - 1 $, such that as many students attend the borrels as possible. In the decision version of this problem the aim is then to decide if there exists a assignment of borrels to timeslots such that there are at least $k$ attendances in total for some $k\in \N$.
% maybe move this formal definition below to somewhere far away (or just delete it)

%More formally, the aim is to decide if there exists a start time $s_h$ for each borrel $h$ such that there exists a set of feasible schedules $S_{i,j}$ for every pair of student $i$ and obligation $j$ where the number of attendances is at least some $k\in \N$. The number of attendances can be defined by defining the set of possible attendances $A = \{(h, i) : \forall j, [s_h, s_h+b_h-1] \cap S_{i,j} = \emptyset \}$. The actual attendances of borrels is a subset $G\subseteq A$ where for any two attendances $(h_1,i_1), (h_2,i_2)\in G: i_1 \neq i_2 \vee [s_{h_1}, s_{h_1}+b_{h_1}-1] \cap [s_{h_2}, s_{h_2}+b_{h_2}-1] = \emptyset$. If there exists a set $G$ with a cardinality of at least some $k\in \N$ for some borrel schedule, then the answer to the decision problem should be positive (``yes'' or ``accept'').
% I think the formal formulation of the problem in the borrel.pdf is not complete or not correct, so I made my own shitty one. Maybe I should just stick to the less formal formulation because this is kind of a mess..

\subsection{Set cover}
% This is the outline
In this section we will briefly describe the set cover decision problem.

% This is the actual thing
%We consider the set cover decision problem with a universe $U=\{1,2,...,n\}$ containing all possible elements and the set $S$ containing $m$ subsets of the universe $U$ and a goal $k\in \N$. %(We can require that the union of $S$ is equal to $U$, but this is not strictly necessary when considering the decision problem, I think). 
%The aim in the set cover problem is to decide if there exists a subset $A$ of $S$ such that the union of $A$ is equal to $U$ and $|A| \leq k$. We will construct an instance of the borrel scheduling decision problem whose answer is the same as the answer to this set cover problem.

\section{Theoretical section}
\subsection{Proof of NP-completeness}
% This is the outline, not the actual thing
In this section we will show that the borrel scheduling decision problem is NP-complete. We will start by formulating the decision version of the problem. Then we will show that this decision problem is in NP. Then we will take the set cover decision problem, which is NP-complete \cite{Karp1972}
, and reduce it to the borrel scheduling decision problem. Then we will have shown that the decision version is NP-complete.
% I think we can also conclude that the optimization version of borrel scheduling is NP-hard.


\subsubsection{borrel scheduling is in NP}
% This is the outline
In order to prove that the borrel scheduling decision problem is in NP, we will define a certificate. This certificate is a set of pairs $(h,i)$ where $(h,i)$ is included if and only if borrel $h$ cannot attended by student $i$ because of an obligation. First, we will show that this certificate is polynomial in the input size. Then, we will show that this problem is certifiable in polynomial time, given the certificate. Lastly, we will show that we can construct a valid certificate for every instance resulting in acceptance if and only if the answer to the decision problem should be positive. 

% show that for every instance there is a valid certificate resulting in a yes
% show that you can correctly verify a certificate in polynomial time
% show that the certificate is of polynomial length (compared to input size)

% We can use a witness (h,i) if borrel h is NOT be attended by student i BECAUSE OF AN OBLIGATION. This way we have a witness (h,i) only if a students has an obligation at that time, so the witness is of size O(|I|^z * m^z) for some $z\in\N$.
% I think we cannot use this because it its size not polynomial in the input size: the witness boolean $A_{hi}$ for all students i and borrels h, denoting if student i attends borrel h.
% Another way to ``fix" this is by only considering the scenario where each person has at least one obligation, this way the input size is always at least O(n * m), meaning that the witness can be of size O(n^k * m^k) for some k, which is easy.

\subsubsection{Reduction of set cover problem to borrel scheduling}
% This is the outline
In this section we will reduce the set cover decision problem to the borrel scheduling decision problem. We will do this by constructing a special instance of the borrel scheduling problem given any instance of set cover. Then we will show that the answers to the original set cover instance and the special instance of the borrel scheduling problem are the same. This will show that the borrel scheduling decision problem is NP-hard and therefore, together with the previous proof that it is NP, will show that it is NP-complete.

% RETRY at readable formulation
\paragraph{Construction of the reduction}
% This is the outline
In this paragraph we will describe how we construct a special instance of the borrel scheduling decision problem given any instance of the set cover decision problem. This borrel scheduling instance will consist of one timeslot for each set in the set cover problem. Additionally, it will have one student for every element in the universe of the set cover problem. Given the set cover problem that aims to cover the universe in $k$ or less sets, the borrel scheduling instance will have $k$ borrels. The problem will have an obligation for each student-timeslot combination if the element corresponding to the student is not in the set corresponding to the timeslot. Additionally, each student will have one obligation with an interval that covers all timeslot and a length that makes sure that the student only has room for one borrel. The aim in this borrel scheduling problem is to decide if there exists a schedule where every students attends a borrel.

% This is the actual thing
%Given a universe $U=\{1,2,...,n_1\}$ containing $n_1$ elements, a set $S$ containing $m_1$ subsets of $U$, and a goal $k_1\in \N$. We create an instance of borrel scheduling with the following parameters:
%\begin{itemize}
%    \item $t = |S| = m_1$ \; the number of timeslots is equal to the number of sets in $S$.
%    \item $n_2 = |U| = n_1$ \; the number of students is equal to the size of the universe.
%    \item $m_2 = k_1$ \; the number of borrels is equal to the set cover goal. 
%    \item $b_h = 1$ \; the length of each borrel $h$ is 1.
%\end{itemize}
%Furthermore, for all students $i$ and subsets $S_a \in S$ we create an obligation $j$ with $I_{i,j} = [a, a]$ and $p_{i,j} = 1$ for student $i$ if $i \not\in S_a$. Additionally we add an obligation $x$ to each student $i$ with $I_{i,x} = [1, t]$ and $p_{i,x} = |\{a : i\in S_a\}| - 1$. The borrel decision problem is to determine if there exists a schedule such that there are at least $t = n_1$ attendances.

\paragraph{Proof of the reduction}
% This is the outline
In this paragraph we will show that the necessary conditions are met to claim that we have reduced the set cover decision problem to the borrel scheduling decision problem. We will start by showing that for all instances of set cover: if the specially constructed instance of borrel planning results in the accepting state, then the answer to the original set cover problem is also positive. Then we will show the inverse: that if an instance of set cover results in acceptance, then the special instance of the borrel scheduling based on this instance will also result in acceptance. Together this will show that the answers to both problems are always the same. Finally, we will show that rewriting the instance of set cover to the special instance of borrel scheduling can be done in polynomial time. Together this will show that set cover decision problem can be reduced to the borrel scheduling decision problem.

% This is the actual thing
%We will show that this instance of the borrel scheduling problem is equivalent to the original set cover problem. Note, first, that each person can attend at most one borrel, since there is one timeslot for each set in $S$ and for every person $i$ there is an obligation of length 1 for every set in $S$ that contains $i$ and there is an obligation of length one less that the number of sets in $S$ that do not contain $i$. We can also see that a student $i$ can only attend a borrel at timeslot $j$ if $i$ is in the $j$'th set $S_j$. Then if the borrel scheduling problem decided that there is a feasible borrel schedule such that there are at least $|U|$ attendances with $k$ borrels, then these must be $|U|$ different students and therefore all elements in the universe. This borrel schedule must also use at most $k$ borrels corresponding to sets in $S$ to cover this universe. This proves that if the answer to this constructed borrel is yes, then the original set cover problem is also a yes-instance.

%Then, conversely, if the original set cover instance can be covered in at most $k$ sets, we can show that there exists some assignment of the borrels such that there are at least $|U|$ attendances. Given the $k$ sets that cover the universe, the $k$ borrels can be scheduled at the timeslots corresponding to those sets. Then each person can attend at least one borrel since the sets cover the entire universe, resulting in at least $|U|$ attendances.

%Since we have shown that a yes-instance of the borrel scheduling implies a yes-instance of the original set cover, and we have shown that a yes-instance of the original set cover implies a yes-instance of the borrel scheduling (and by contraposition a no-instance of borrel scheduling implies a no-instance of the set cover) it follows that the constructed borrel scheduling problem and the set cover problem are equivalent and thus set cover can be reduced to borrel scheduling.

%todo % prove that rewriting this problem can be done in polynomial time.

\subsection{Problem competitive ratio}

\subsubsection{Base problem}
\label{base}
To prove the competitive ratio of the base problem we will start by taking in to consideration the following adversary cases: \\
$A := \{t = 2, m = 1, b_1 = 1, \forall i : 1 < i \leq n : I_i = \{([t_1,t_1],1)\}\} $ \\
$OPT(A) = n$ where $b_1$ is helt at $t_2$ \\
$B := \{t = 2, m = 1, b_1 = 1, I_1 = \{([t_2,t_2],1)\}, \forall i : 1 < i \leq n : I_i = \{([t_1,t_1],1),([t_2,t_2]1)\}\} $ \\
$OPT(B) = 1$ where $b_1$ is helt at $t_1$ \\
\\
For both cases every student except for one has an obligation at timeslot $t_1$. Since there exists at least one borrel $b_1$ with length 1 every online algorithm can either plan this borrel on $t_1$ or not, creating two possible cases. \\
\\
If some algorithm $ALG_1$ chooses to host the borrel on $t_1$ then the adversery case is $A$ where $ALG_1(A) = 1$ and $OPT(A) = n$ giving a competitive ratio of $\frac{OPT(A)}{ALG_1(A)} = \frac{n}{1} = n$. \\
If some algorithm $ALG_2$ chooses not to host the borrel on $t_1$ then the adversery case is $B$ where $ALG_2(B) = 0$ and $OPT(B) = 1$ giving a competitive ratio of $\frac{OPT(B)}{ALG_2(B)} = \frac{1}{0} = \infty$ \\
\\
This gives the problem a competitive ratio of $n$ \\
\\
Moreover, every algorithm hoping to attain a competitive ratio $c < \infty$ should always plan a borrel the first chance it gets to avoid having a score of 0.

\subsubsection{Multiple borrels}
We will investigate how the competitive ratio changes when $m>1$.

\subsubsection{k-foresight}
Next we will look at the problem where the algorithm can look k timeslots ahead instead of 1. So in a setting with $k=3$ the algorithm would be able to see obligation information about timeslots $\{t_1,t_2,t_3\}$ at timeslot $t_1$ and timeslots $\{t_2,t_3,t_4\}$ at timeslot $t_2$.\\
\\
If $k \geq t$ the problem is simply the offline problem and thus trivial with a competitive ratio of $c=1$. \\ 
\\
For a smaller $k$ the competitive ratio can be proven similarly to the base problem with adversary cases: \\
$A := \{t = k + 1, m = 1, b_1 = 1, \forall i : 1 < i \leq n : I_i = \{([t_1,t_k],1)\}\} $ \\
$OPT(A) = n$ where $b_1$ is held at $t_{k+1}$ \\
$B := \{t = k + 1, m = 1, b_1 = 1, I_1 = \{([t_{k+1},t_{k+1}],1)\}, \forall i : 1 < i \leq n : I_i = \{([t_1,t_k],1),([t_{k+1},t_{k+1}]1)\}\} $ \\
$OPT(B) = 1$ where $b_1$ is held at $t_1$ \\
\\
At timeslot $t_1$ any algorithm will see a timeslot with 1 free student at $t_1$ and 0 free students at $\{t_2...t_k\}$ creating once again the same choice every algorithm must make at $t_1$ for hosting a borrel at $t_1$. \\
\\
Then like the base problem:
If some algorithm $ALG_1$ chooses to host the borrel on $t_1$ then the adversary case is $A$ where $ALG_1(A) = 1$ and $OPT(A) = n$ giving a competitive ratio of $\frac{OPT(A)}{ALG_1(A)} = \frac{n}{1} = n$. \\
If some algorithm $ALG_2$ chooses not to host the borrel on $t_1$ then the adversary case is $B$ where $ALG_2(B) = 0$ and $OPT(B) = 1$ giving a competitive ratio of $\frac{OPT(B)}{ALG_2(B)} = \frac{1}{0} = \infty$ \\
\\
This gives this problem also a competitive ratio of $n$ 

\subsubsection{Non-overlapping obligations}
This problem has the same argumentation as subsection \ref{base}.

\subsubsection{At most N different timeslots}
This problem has the same argumentation as subsection \ref{base}.

\subsubsection{Combination of k-foresight, non-overlapping obligations and N different timeslots}
This problem has the same argumentation as subsection \ref{base}, with the only changes being making the first obligation $k$ long instead of 1, and $t=k+1$ instead of $t=2$

%possible additional restrictions: Minstens 1 timeslot vrij tussen obligaties (sterkere versie van non-overlapping), alleen niet heel spannend (met behulp van een "trap" kan dit gebroken worden) maar misschien met t<N wel interessant. No duplicate obligation sets, samen met minstens 1 timeslot vrij tussen obligaties kan dit mogelijk erg interessant zijn, anders zelfde argument als base.

\section{Offline Algorithms}

\section{Experiments}

\section{Conclusion}

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\end{document}
