cphb/luku15.tex

\chapter{Spanning trees}

\index{spanning tree}

A \key{spanning tree} is a set of edges of a graph
such that there is a path between any two nodes
in the graph using only the edges in the spanning tree.
Like trees in general, a spanning tree is
connected and acyclic.
Usually, there are many ways to construct a spanning tree.

For example, in the graph
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}
one possible spanning tree is as follows:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

The weight of a spanning tree is the sum of the edge weights.
For example, the weight of the above spanning tree is
$3+5+9+3+2=22$.

\index{minimum spanning tree}

A \key{minimum spanning tree}
is a spanning tree whose weight is as small as possible.
The weight of a minimum spanning tree for the above graph
is 20, and a tree can be constructed as follows:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};

\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

\index{maximum spanning tree}

Correspondingly, a \key{maximum spanning tree}
is a spanning tree whose weight is as large as possible.
The weight of a maximum spanning tree for the
above graph is 32:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
%\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
%\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
%\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

Note that there may be several different ways
for constructing a minimum or maximum spanning tree,
so the trees are not unique.

This chapter discusses algorithms that construct
a minimum or maximum spanning tree for a graph.
It turns out that it is easy to find such spanning trees
because many greedy methods produce an optimal solution.

We will learn two algorithms that both construct the
tree by choosing edges ordered by weights.
We will focus on finding a minimum spanning tree,
but the same algorithms can be used for finding a
maximum spanning tree by processing the edges in reverse order.

\section{Kruskal's algorithm}

\index{Kruskal's algorithm}

In \key{Kruskal's algorithm}, the initial spanning tree
is empty and doesn't contain any edges.
Then the algorithm adds edges to the tree
one at a time
in increasing order of their weights.
At each step, the algorithm includes an edge in the tree
if it doesn't create a cycle.

Kruskal's algorithm maintains the components
in the tree.
Initially, each node of the graph
is in its own component,
and each edge added to the tree joins two components.
Finally, all nodes will be in the same component,
and a minimum spanning tree has been found.

\subsubsection{Example}

\begin{samepage}
Let's consider how Kruskal's algorithm processes the
following graph:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}
\end{samepage}

\begin{samepage}
The first step in the algorithm is to sort the
edges in increasing order of their weights.
The result is the following list:

\begin{tabular}{ll}
\\
edge & weight \\
\hline
5--6 & 2 \\
1--2 & 3 \\
3--6 & 3 \\
1--5 & 5 \\
2--3 & 5 \\
2--5 & 6 \\
4--6 & 7 \\
3--4 & 9 \\
\\
\end{tabular}
\end{samepage}

After this, the algorithm goes through the list
and adds an edge to the tree if it joins
two separate components.

Initially, each node is in its own component:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
%\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
%\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
%\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
%\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
%\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}
The first edge to be added to the tree is
edge 5--6 that joins components
$\{5\}$ and $\{6\}$ into component $\{5,6\}$:

\begin{center}
\begin{tikzpicture}
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};

%\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
%\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
%\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
%\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}
After this, edges 1--2, 3--6 and 1--5 are added in a similar way:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};

\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
%\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

After those steps, many components have been joined
and there are two components in the tree:
$\{1,2,3,5,6\}$ and $\{4\}$.

The next edge in the list is edge 2--3,
but it will not be included in the tree because
nodes 2 and 3 are already in the same component.
For the same reason, edge 2--5 will not be added
to the tree.

\begin{samepage}
Finally, edge 4--6 will be included in the tree:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};

\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}
\end{samepage}

After this, the algorithm terminates because
there is a path between any two nodes and
the graph is connected.
The resulting graph is a minimum spanning tree
with weight $2+3+3+5+7=20$.

\subsubsection{Why does this work?}

It's a good question why Kruskal's algorithm works.
Why does the greedy strategy guarantee that we
will find a minimum spanning tree?

Let's see what happens if the lightest edge in
the graph is not included in the minimum spanning tree.
For example, assume that a minimum spanning tree
for the above graph would not contain the edge
between nodes 5 and 6 with weight 2.
We don't know exactly how the new minimum spanning tree
would look like, but still it has to contain some edges.
Assume that the tree would be as follows:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};

\path[draw,thick,-,dashed] (1) -- (2);
\path[draw,thick,-,dashed] (2) -- (5);
\path[draw,thick,-,dashed] (2) -- (3);
\path[draw,thick,-,dashed] (3) -- (4);
\path[draw,thick,-,dashed] (4) -- (6);
\end{tikzpicture}
\end{center}

However, it's not possible that the above tree
would be a real minimum spanning tree for the graph.
The reason for this is that we can remove an edge
from it and replace it with the edge with weight 2.
This produces a spanning tree whose weight is
\emph{smaller}:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};

\path[draw,thick,-,dashed] (1) -- (2);
\path[draw,thick,-,dashed] (2) -- (5);
\path[draw,thick,-,dashed] (3) -- (4);
\path[draw,thick,-,dashed] (4) -- (6);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\end{tikzpicture}
\end{center}

For this reason, it is always optimal to include the lightest edge
in the minimum spanning tree.
Using a similar argument, we can show that we
can also add the second lightest edge to the tree, and so on.
Thus, Kruskal's algorithm works correctly and
always produces a minimum spanning tree.

\subsubsection{Implementation}

Kruskal's algorithm can be conveniently
implemented using an edge list.
The first phase of the algorithm sorts the
edges in $O(m \log m)$ time.
After this, the second phase of the algorithm
builds the minimum spanning tree. 

The second phase of the algorithm looks as follows:

\begin{lstlisting}
for (...) {
  if (!same(a,b)) union(a,b);
}
\end{lstlisting}

The loop goes through the edges in the list
and always processes an edge $a$--$b$
where $a$ and $b$ are two nodes.
The code uses two functions:
the function \texttt{same} determines
if the nodes are in the same component,
and the function \texttt{unite}
joins two components into a single component.

The problem is how to efficiently implement
the functions \texttt{same} and \texttt{unite}.
One possibility is to maintain the graph
in a usual way and implement the function
\texttt{same} as graph traversal.
However, using this technique,
the running time of the function \texttt{same} would be $O(n+m)$,
and this would be slow because the function will be
called for each edge in the graph.

We will solve the problem using a union-find structure
that implements both the functions in $O(\log n)$ time.
Thus, the time complexity of Kruskal's algorithm
will be only $O(m \log n)$ after sorting the edge list.

\section{Union-find-rakenne}

\index{union-find-rakenne}

\key{Union-find-rakenne} pitää yllä
alkiojoukkoja.
Joukot ovat erillisiä,
eli tietty alkio on tarkalleen
yhdessä joukossa.
Rakenne tarjoaa kaksi operaatiota,
jotka toimivat ajassa $O(\log n)$.
Ensimmäinen operaatio tarkistaa,
ovatko kaksi alkiota samassa joukossa.
Toinen operaatio yhdistää kaksi
joukkoa toisiinsa.

\subsubsection{Rakenne}

Union-find-rakenteessa jokaisella
joukolla on edustaja-alkio.
Kaikki muut joukon alkiot osoittavat
edustajaan joko suoraan tai
muiden alkioiden kautta.

Esimerkiksi jos joukot ovat
$\{1,4,7\}$, $\{5\}$ ja $\{2,3,6,8\}$,
tilanne voisi olla:
\begin{center}
\begin{tikzpicture}
\node[draw, circle] (1) at (0,-1) {$1$};
\node[draw, circle] (2) at (7,0) {$2$};
\node[draw, circle] (3) at (7,-1.5) {$3$};
\node[draw, circle] (4) at (1,0) {$4$};
\node[draw, circle] (5) at (4,0) {$5$};
\node[draw, circle] (6) at (6,-2.5) {$6$};
\node[draw, circle] (7) at (2,-1) {$7$};
\node[draw, circle] (8) at (8,-2.5) {$8$};

\path[draw,thick,->] (1) -- (4);
\path[draw,thick,->] (7) -- (4);

\path[draw,thick,->] (3) -- (2);
\path[draw,thick,->] (6) -- (3);
\path[draw,thick,->] (8) -- (3);

\end{tikzpicture}
\end{center}
Tässä tapauksessa alkiot 4, 5 ja 2
ovat joukkojen edustajat.
Minkä tahansa alkion edustaja
löytyy kulkemalla alkiosta lähtevää polkua
eteenpäin niin kauan, kunnes polku päättyy.
Esimerkiksi alkion 6 edustaja on 2,
koska alkiosta 6 lähtevä
polku on $6 \rightarrow 3 \rightarrow 2$.
Tämän avulla voi selvittää,
ovatko kaksi alkiota samassa joukossa:
jos kummankin alkion edustaja on sama,
alkiot ovat samassa joukossa,
ja muuten ne ovat eri joukoissa.

Kahden joukon yhdistäminen tapahtuu
valitsemalla toinen edustaja
joukkojen yhteiseksi edustajaksi
ja kytkemällä toinen edustaja siihen.
Esimerkiksi joukot $\{1,4,7\}$ ja $\{2,3,6,8\}$
voi yhdistää näin joukoksi $\{1,2,3,4,6,7,8\}$:
\begin{center}
\begin{tikzpicture}
\node[draw, circle] (1) at (2,-1) {$1$};
\node[draw, circle] (2) at (7,0) {$2$};
\node[draw, circle] (3) at (7,-1.5) {$3$};
\node[draw, circle] (4) at (3,0) {$4$};
\node[draw, circle] (6) at (6,-2.5) {$6$};
\node[draw, circle] (7) at (4,-1) {$7$};
\node[draw, circle] (8) at (8,-2.5) {$8$};

\path[draw,thick,->] (1) -- (4);
\path[draw,thick,->] (7) -- (4);

\path[draw,thick,->] (3) -- (2);
\path[draw,thick,->] (6) -- (3);
\path[draw,thick,->] (8) -- (3);

\path[draw,thick,->] (4) -- (2);
\end{tikzpicture}
\end{center}

Joukkojen yhteiseksi edustajaksi valitaan alkio 2,
minkä vuoksi alkio 4 yhdistetään siihen.
Tästä lähtien alkio 2 edustaa kaikkia joukon alkioita.

Tehokkuuden kannalta oleellista on,
miten yhdistäminen tapahtuu.
Osoittautuu, että ratkaisu on yksinkertainen:
riittää yhdistää aina pienempi joukko suurempaan,
tai kummin päin tahansa,
jos joukot ovat yhtä suuret.
Tällöin pisin ketju
alkiosta edustajaan on aina luokkaa $O(\log n)$,
koska jokainen askel eteenpäin
ketjussa kaksinkertaistaa
vastaavan joukon koon.

\subsubsection{Toteutus}

Union-find-rakenne on kätevää toteuttaa
taulukoiden avulla.
Seuraavassa toteutuksessa taulukko \texttt{k}
viittaa seuraavaan alkioon ketjussa
tai alkioon itseensä, jos alkio on edustaja.
Taulukko \texttt{s} taas kertoo jokaiselle edustajalle,
kuinka monta alkiota niiden joukossa on.

Aluksi jokainen alkio on omassa joukossaan,
jonka koko on 1:

\begin{lstlisting}
for (int i = 1; i <= n; i++) k[i] = i;
for (int i = 1; i <= n; i++) s[i] = 1;
\end{lstlisting}

Funktio \texttt{id} kertoo alkion $x$
joukon edustajan. Alkion edustaja löytyy
käymällä ketju läpi alkiosta $x$ alkaen.

\begin{lstlisting}
int id(int x) {
    while (x != k[x]) x = k[x];
    return x;
}
\end{lstlisting}

Funktio \texttt{sama} kertoo,
ovatko alkiot $a$ ja $b$ samassa joukossa.
Tämä onnistuu helposti funktion
\texttt{id} avulla.

\begin{lstlisting}
bool sama(int a, int b) {
    return id(a) == id(b);
}
\end{lstlisting}

\begin{samepage}
Funktio \texttt{liita} yhdistää
puolestaan alkioiden $a$ ja $b$ osoittamat
joukot yhdeksi joukoksi.
Funktio etsii ensin joukkojen edustajat
ja yhdistää sitten pienemmän joukon suurempaan.

\begin{lstlisting}
void liita(int a, int b) {
    a = id(a);
    b = id(b);
    if (s[b] > s[a]) swap(a,b);
    s[a] += s[b];
    k[b] = a;
}
\end{lstlisting}
\end{samepage}

Funktion \texttt{id} aikavaativuus on $O(\log n)$
olettaen, että ketjun pituus on luokkaa $O(\log n)$.
Niinpä myös funktioiden \texttt{sama} ja \texttt{liita}
aikavaativuus on $O(\log n)$.
Funktio \texttt{liita} varmistaa,
että ketjun pituus on luokkaa $O(\log n)$
yhdistämällä pienemmän joukon suurempaan.

% Funktiota \texttt{id} on mahdollista vielä tehostaa
% seuraavasti:
% 
% \begin{lstlisting}
% int id(int x) {
%   if (x == k[x]) return x;
%   return k[x] = id(x);
% }
% \end{lstlisting}
% 
% Nyt joukon edustajan etsimisen yhteydessä kaikki ketjun
% alkiot laitetaan osoittamaan suoraan edustajaan.
% On mahdollista osoittaa, että tämän avulla
% funktioiden \texttt{sama} ja \texttt{liita}
% aikavaativuus on tasoitetusti
% vain $O(\alpha(n))$, missä $\alpha(n)$ on
% hyvin hitaasti kasvava käänteinen Ackermannin funktio.

\section{Primin algoritmi}

\index{Primin algoritmi@Primin algoritmi}

\key{Primin algoritmi} on vaihtoehtoinen menetelmä
verkon pienimmän virittävän puun muodostamiseen.
Algoritmi aloittaa puun muodostamisen jostakin
verkon solmusta ja lisää puuhun aina kaaren,
joka on mahdollisimman kevyt ja joka
liittää puuhun uuden solmun.
Lopulta kaikki solmut on lisätty puuhun
ja pienin virittävä puu on valmis.

Primin algoritmin toiminta on lähellä
Dijkstran algoritmia.
Erona on, että Dijkstran algoritmissa valitaan
kaari, jonka kautta syntyy lyhin polku alkusolmusta
uuteen solmuun, mutta Primin algoritmissa
valitaan vain kevein kaari, joka johtaa uuteen solmuun.

\subsubsection{Esimerkki}

Tarkastellaan Primin algoritmin toimintaa
seuraavassa verkossa:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);

%\path[draw=red,thick,-,line width=2pt] (5) -- (6);
\end{tikzpicture}
\end{center}

Aluksi solmujen välillä ei ole mitään kaaria:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
%\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
%\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
%\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
%\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
%\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

Puun muodostuksen voi aloittaa mistä tahansa solmusta,
ja aloitetaan se nyt solmusta 1.
Kevein kaari on painoltaan 3 ja se johtaa solmuun 2:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
%\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
%\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
%\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
%\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
%\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

Nyt kevein uuteen solmuun johtavan
kaaren paino on 5,
ja voimme laajentaa joko solmuun 3 tai 5.
Valitaan solmu 3:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
%\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
%\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
%\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
%\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}

\begin{samepage}
Sama jatkuu, kunnes kaikki solmut ovat mukana puussa:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1.5,2) {$1$};
\node[draw, circle] (2) at (3,3) {$2$};
\node[draw, circle] (3) at (5,3) {$3$};
\node[draw, circle] (4) at (6.5,2) {$4$};
\node[draw, circle] (5) at (3,1) {$5$};
\node[draw, circle] (6) at (5,1) {$6$};
\path[draw,thick,-] (1) -- node[font=\small,label=above:3] {} (2);
\path[draw,thick,-] (2) -- node[font=\small,label=above:5] {} (3);
%\path[draw,thick,-] (3) -- node[font=\small,label=above:9] {} (4);
%\path[draw,thick,-] (1) -- node[font=\small,label=below:5] {} (5);
\path[draw,thick,-] (5) -- node[font=\small,label=below:2] {} (6);
\path[draw,thick,-] (6) -- node[font=\small,label=below:7] {} (4);
%\path[draw,thick,-] (2) -- node[font=\small,label=left:6] {} (5);
\path[draw,thick,-] (3) -- node[font=\small,label=left:3] {} (6);
\end{tikzpicture}
\end{center}
\end{samepage}

\subsubsection{Toteutus}

Dijkstran algoritmin tavoin Primin algoritmin voi toteuttaa
tehokkaasti käyttämällä prioriteettijonoa.
Primin algoritmin tapauksessa jono sisältää kaikki solmut,
jotka voi yhdistää nykyiseen komponentiin kaarella,
järjestyksessä kaaren painon mukaan kevyimmästä raskaimpaan.

Primin algoritmin aikavaativuus on $O(n + m \log m)$
eli sama kuin Dijkstran algoritmissa.
Käytännössä Primin algoritmi on suunnilleen
yhtä nopea kuin Kruskalin algoritmi,
ja onkin makuasia, kumpaa algoritmia käyttää.
Useimmat kisakoodarit käyttävät kuitenkin Kruskalin algoritmia.