\chapter{Sweep line algorithms} \index{sweep line} Many geometric problems can be solved using \key{sweep line} algorithms. The idea in such algorithms is to represent the problem as a set of events that correspond to points in the plane. The events are processed in increasing order according to their x or y coordinate. As an example, let us consider a problem where there is a company that has $n$ employees, and we know for each employee their arrival and leaving times on a certain day. Our task is to calculate the maximum number of employees that were in the office at the same time. The problem can be solved by modelling the situation so that each employee is assigned two events that corresponds to their arrival and leaving times. After sorting the events, we can go through them and keep track of the number of people in the office. For example, the table \begin{center} \begin{tabular}{ccc} person & arrival time & leaving time \\ \hline Uolevi & 10 & 15 \\ Maija & 6 & 12 \\ Kaaleppi & 14 & 16 \\ Liisa & 5 & 13 \\ \end{tabular} \end{center} corresponds to the following events: \begin{center} \begin{tikzpicture}[scale=0.6] \draw (0,0) rectangle (17,-6.5); \path[draw,thick,-] (10,-1) -- (15,-1); \path[draw,thick,-] (6,-2.5) -- (12,-2.5); \path[draw,thick,-] (14,-4) -- (16,-4); \path[draw,thick,-] (5,-5.5) -- (13,-5.5); \draw[fill] (10,-1) circle [radius=0.05]; \draw[fill] (15,-1) circle [radius=0.05]; \draw[fill] (6,-2.5) circle [radius=0.05]; \draw[fill] (12,-2.5) circle [radius=0.05]; \draw[fill] (14,-4) circle [radius=0.05]; \draw[fill] (16,-4) circle [radius=0.05]; \draw[fill] (5,-5.5) circle [radius=0.05]; \draw[fill] (13,-5.5) circle [radius=0.05]; \node at (2,-1) {Uolevi}; \node at (2,-2.5) {Maija}; \node at (2,-4) {Kaaleppi}; \node at (2,-5.5) {Liisa}; \end{tikzpicture} \end{center} We go through the events from left to right and maintain a counter. Always when a person arrives, we increase the value of the counter by one, and when a person leaves, we decrease the value of the counter by one. The answer to the problem is the maximum value of the counter during the algorithm. In the example, the events are processed as follows: \begin{center} \begin{tikzpicture}[scale=0.6] \path[draw,thick,->] (0.5,0.5) -- (16.5,0.5); \draw (0,0) rectangle (17,-6.5); \path[draw,thick,-] (10,-1) -- (15,-1); \path[draw,thick,-] (6,-2.5) -- (12,-2.5); \path[draw,thick,-] (14,-4) -- (16,-4); \path[draw,thick,-] (5,-5.5) -- (13,-5.5); \draw[fill] (10,-1) circle [radius=0.05]; \draw[fill] (15,-1) circle [radius=0.05]; \draw[fill] (6,-2.5) circle [radius=0.05]; \draw[fill] (12,-2.5) circle [radius=0.05]; \draw[fill] (14,-4) circle [radius=0.05]; \draw[fill] (16,-4) circle [radius=0.05]; \draw[fill] (5,-5.5) circle [radius=0.05]; \draw[fill] (13,-5.5) circle [radius=0.05]; \node at (2,-1) {Uolevi}; \node at (2,-2.5) {Maija}; \node at (2,-4) {Kaaleppi}; \node at (2,-5.5) {Liisa}; \path[draw,dashed] (10,0)--(10,-6.5); \path[draw,dashed] (15,0)--(15,-6.5); \path[draw,dashed] (6,0)--(6,-6.5); \path[draw,dashed] (12,0)--(12,-6.5); \path[draw,dashed] (14,0)--(14,-6.5); \path[draw,dashed] (16,0)--(16,-6.5); \path[draw,dashed] (5,0)--(5,-6.5); \path[draw,dashed] (13,0)--(13,-6.5); \node at (10,-7) {$+$}; \node at (15,-7) {$-$}; \node at (6,-7) {$+$}; \node at (12,-7) {$-$}; \node at (14,-7) {$+$}; \node at (16,-7) {$-$}; \node at (5,-7) {$+$}; \node at (13,-7) {$-$}; \node at (10,-8) {$3$}; \node at (15,-8) {$1$}; \node at (6,-8) {$2$}; \node at (12,-8) {$2$}; \node at (14,-8) {$2$}; \node at (16,-8) {$0$}; \node at (5,-8) {$1$}; \node at (13,-8) {$1$}; \end{tikzpicture} \end{center} The symbols $+$ and $-$ indicate whether the value of the counter increases or decreases, and the value of the counter is shown below. The maximum value of the counter is 3 between Uolevi's arrival time and Maija's leaving time. The running time of the algorithm is $O(n \log n)$, because sorting the events takes $O(n \log n)$ time and the rest of the algorithm takes $O(n)$ time. \section{Intersection points} \index{intersection point} Given a set of $n$ line segments, each of them being either horizontal or vertical, consider the problem of counting the total number of intersection points. For example, when the line segments are \begin{center} \begin{tikzpicture}[scale=0.5] \path[draw,thick,-] (0,2) -- (5,2); \path[draw,thick,-] (1,4) -- (6,4); \path[draw,thick,-] (6,3) -- (10,3); \path[draw,thick,-] (2,1) -- (2,6); \path[draw,thick,-] (8,2) -- (8,5); \end{tikzpicture} \end{center} there are three intersection points: \begin{center} \begin{tikzpicture}[scale=0.5] \path[draw,thick,-] (0,2) -- (5,2); \path[draw,thick,-] (1,4) -- (6,4); \path[draw,thick,-] (6,3) -- (10,3); \path[draw,thick,-] (2,1) -- (2,6); \path[draw,thick,-] (8,2) -- (8,5); \draw[fill] (2,2) circle [radius=0.15]; \draw[fill] (2,4) circle [radius=0.15]; \draw[fill] (8,3) circle [radius=0.15]; \end{tikzpicture} \end{center} It is easy to solve the problem in $O(n^2)$ time, because we can go through all possible pairs of segments and check if they intersect. However, we can solve the problem more efficiently in $O(n \log n)$ time using a sweep line algorithm. The idea is to generate two types of events: \begin{enumerate}[noitemsep] \item[(1)] horizontal segment begins \item[(2)] horizontal segment ends \item[(3)] vertical segment \end{enumerate} The following events correspond to the example: \begin{center} \begin{tikzpicture}[scale=0.6] \path[draw,dashed] (0,2) -- (5,2); \path[draw,dashed] (1,4) -- (6,4); \path[draw,dashed] (6,3) -- (10,3); \path[draw,dashed] (2,1) -- (2,6); \path[draw,dashed] (8,2) -- (8,5); \node at (0,2) {$1$}; \node at (5,2) {$2$}; \node at (1,4) {$1$}; \node at (6,4) {$2$}; \node at (6,3) {$1$}; \node at (10,3) {$2$}; \node at (2,3.5) {$3$}; \node at (8,3.5) {$3$}; \end{tikzpicture} \end{center} We go through the events from left to right and use a data structure that maintains a set of y coordinates where there is an active horizontal segment. At event 1, we add the y coordinate of the segment to the set, and at event 2, we remove the y coordinate from the set. Intersection points are calculated at event 3. When there is a vertical segment between points $y_1$ and $y_2$, we count the number of active horizontal segments whose y coordinate is between $y_1$ and $y_2$, and add this number to the total number of intersection points. An appropriate data structure for y coordinates of horizontal segments is either a binary indexed tree or a segment tree, possibly with index compression. Using such structures, processing each event takes $O(\log n)$ time, so the total running time of the algorithm is $O(n \log n)$. \section{Nearest points} \index{nearest points} Given a set of $n$ points, our next problem is to find two points whose distance is minimum. For example, if the points are \begin{center} \begin{tikzpicture}[scale=0.7] \draw (0,0)--(12,0)--(12,4)--(0,4)--(0,0); \draw (1,2) circle [radius=0.1]; \draw (3,1) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5.5,1.5) circle [radius=0.1]; \draw (6,2.5) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (9,1.5) circle [radius=0.1]; \draw (10,2) circle [radius=0.1]; \draw (1.5,3.5) circle [radius=0.1]; \draw (1.5,1) circle [radius=0.1]; \draw (2.5,3) circle [radius=0.1]; \draw (4.5,1.5) circle [radius=0.1]; \draw (5.25,0.5) circle [radius=0.1]; \draw (6.5,2) circle [radius=0.1]; \end{tikzpicture} \end{center} \begin{samepage} we should find the following points: \begin{center} \begin{tikzpicture}[scale=0.7] \draw (0,0)--(12,0)--(12,4)--(0,4)--(0,0); \draw (1,2) circle [radius=0.1]; \draw (3,1) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5.5,1.5) circle [radius=0.1]; \draw[fill] (6,2.5) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (9,1.5) circle [radius=0.1]; \draw (10,2) circle [radius=0.1]; \draw (1.5,3.5) circle [radius=0.1]; \draw (1.5,1) circle [radius=0.1]; \draw (2.5,3) circle [radius=0.1]; \draw (4.5,1.5) circle [radius=0.1]; \draw (5.25,0.5) circle [radius=0.1]; \draw[fill] (6.5,2) circle [radius=0.1]; \end{tikzpicture} \end{center} \end{samepage} This is another example of a problem that can be also solved in $O(n \log n)$ time using a sweep line algorithm. We go through the points from left to right and maintain a value $d$: the minimum distance between two points seen so far. At each point, we find the nearest point to the left. If the distance is less than $d$, it is the new minimum distance and we update the value of $d$. If the current point is $(x,y)$ and there is a point to the left within a distance of less than $d$, the x coordinate of such a point must be between $[x-d,x]$ and the y coordinate must be between $[y-d,y+d]$. Thus, it suffices to only consider points that are located in those ranges, which makes the algorithm efficient. For example, in the following picture the region marked with dashed lines contains the points that can be within a distance of $d$ from the active point: \begin{center} \begin{tikzpicture}[scale=0.7] \draw (0,0)--(12,0)--(12,4)--(0,4)--(0,0); \draw (1,2) circle [radius=0.1]; \draw (3,1) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5.5,1.5) circle [radius=0.1]; \draw (6,2.5) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (9,1.5) circle [radius=0.1]; \draw (10,2) circle [radius=0.1]; \draw (1.5,3.5) circle [radius=0.1]; \draw (1.5,1) circle [radius=0.1]; \draw (2.5,3) circle [radius=0.1]; \draw (4.5,1.5) circle [radius=0.1]; \draw (5.25,0.5) circle [radius=0.1]; \draw[fill] (6.5,2) circle [radius=0.1]; \draw[dashed] (6.5,0.75)--(6.5,3.25); \draw[dashed] (5.25,0.75)--(5.25,3.25); \draw[dashed] (5.25,0.75)--(6.5,0.75); \draw[dashed] (5.25,3.25)--(6.5,3.25); \draw [decoration={brace}, decorate, line width=0.3mm] (5.25,3.5) -- (6.5,3.5); \node at (5.875,4) {$d$}; \draw [decoration={brace}, decorate, line width=0.3mm] (6.75,3.25) -- (6.75,2); \node at (7.25,2.625) {$d$}; \end{tikzpicture} \end{center} The efficiency of the algorithm is based on the fact that such a region always contains only $O(1)$ points. We can go through those points in $O(\log n)$ time by maintaining a set of points whose x coordinate is between $[x-d,x]$, in increasing order according to their y coordinates. The time complexity of the algorithm is $O(n \log n)$, because we go through $n$ points and find for each point the nearest point to the left in $O(\log n)$ time. \section{Convex hull} A \key{convex hull} is the smallest convex polygon that contains all points of a given set. Convexity means that a line segment between any two vertices of the polygon is completely inside the polygon. An intuitive definition for a convex hull is that it surrounds the given points using a tight rope. \begin{samepage} For example, for the points \begin{center} \begin{tikzpicture}[scale=0.7] \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \end{tikzpicture} \end{center} \end{samepage} the convex hull is as follows: \begin{center} \begin{tikzpicture}[scale=0.7] \draw (0,0)--(4,-1)--(7,1)--(6,3)--(2,4)--(0,2)--(0,0); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \end{tikzpicture} \end{center} \index{Andrew's algorithm} \key{Andrew's algorithm} is an easy algorithm that can be used to construct the convex hull for a set of points in $O(n \log n)$ time. The algorithm constructs the convex hull in two steps: first the upper hull and then the lower hull. Both steps are similar, so we can focus on constructing the upper hull. We sort the points primarily according to x coordinates and secondarily according to y coordinates. After this, we go through the points and always add the new point to the hull. After adding a point we check using cross products whether the tree last point in the hull turn left. If this holds, we remove the middle point from the hull. After this we keep checking again the three last points and removing points, until the three last points do not turn left. The following pictures show how Andrew's algorithm works: \\ \begin{tabular}{ccccccc} \\ \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(1,1); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(1,1)--(2,2); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,2); \end{tikzpicture} \\ 1 & & 2 & & 3 & & 4 \\ \end{tabular} \\ \begin{tabular}{ccccccc} \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,2)--(2,4); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(3,2); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(3,2)--(4,-1); \end{tikzpicture} \\ 5 & & 6 & & 7 & & 8 \\ \end{tabular} \\ \begin{tabular}{ccccccc} \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(3,2)--(4,-1)--(4,0); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(3,2)--(4,0); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(3,2)--(4,0)--(4,3); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(3,2)--(4,3); \end{tikzpicture} \\ 9 & & 10 & & 11 & & 12 \\ \end{tabular} \\ \begin{tabular}{ccccccc} \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(4,3); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(4,3)--(5,2); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(4,3)--(5,2)--(6,1); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(4,3)--(5,2)--(6,1)--(6,3); \end{tikzpicture} \\ 13 & & 14 & & 15 & & 16 \\ \end{tabular} \\ \begin{tabular}{ccccccc} \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(4,3)--(5,2)--(6,3); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(4,3)--(6,3); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(6,3); \end{tikzpicture} & \hspace{0.1cm} & \begin{tikzpicture}[scale=0.3] \draw (-1,-2)--(8,-2)--(8,5)--(-1,5)--(-1,-2); \draw (0,0) circle [radius=0.1]; \draw (4,-1) circle [radius=0.1]; \draw (7,1) circle [radius=0.1]; \draw (6,3) circle [radius=0.1]; \draw (2,4) circle [radius=0.1]; \draw (0,2) circle [radius=0.1]; \draw (1,1) circle [radius=0.1]; \draw (2,2) circle [radius=0.1]; \draw (3,2) circle [radius=0.1]; \draw (4,0) circle [radius=0.1]; \draw (4,3) circle [radius=0.1]; \draw (5,2) circle [radius=0.1]; \draw (6,1) circle [radius=0.1]; \draw (0,0)--(0,2)--(2,4)--(6,3)--(7,1); \end{tikzpicture} \\ 17 & & 18 & & 19 & & 20 \end{tabular}