\chapter{Greedy algorithms}

\index{greedy algorithm}

A \key{greedy algorithm}
constructs a solution for a problem
by always making a choice that looks
the best at the moment.
A greedy algorithm never takes back
its choices, but directly constructs
the final solution.
For this reason, greedy algorithms
are usually very efficient.

The difficulty in designing a greedy algorithm
is to invent a greedy strategy
that always produces an optimal solution
for the problem.
The locally optimal choices in a greedy
algorithm should also be globally optimal.
It's often difficult to argue why
a greedy algorithm works.

\section{Coin problem}

As the first example, we consider a problem
where we are given a set of coin values
and our task is to form a sum of money
using the coins.
The values of the coins are
$\{c_1,c_2,\ldots,c_k\}$,
and each coin can be used as many times we want.
What is the minimum number of coins needed?

For example, if the coins are euro coins (in cents)
\[\{1,2,5,10,20,50,100,200\}\]
and the sum of money is 520,
we need at least four coins.
The optimal solution is to select coins
$200+200+100+20$ whose sum is 520.

\subsubsection{Greedy algorithm}

A natural greedy algorithm for the problem
is to always select the largest possible coin,
until we have constructed the required sum of money.
This algorithm works in the example case,
because we first select two 200 cent coins,
then one 100 cent coin and finally one 20 cent coin.
But does this algorithm always work?

It turns out that, for the set of euro coins,
the greedy algorithm \emph{always} works, i.e.,
it always produces a solution with the fewest
possible number of coins.
The correctness of the algorithm can be
argued as follows:

Each coin 1, 5, 10, 50 and 100 appears
at most once in the optimal solution.
The reason for this is that if the
solution would contain two such coins,
we could replace them by one coin and
obtain a better solution.
For example, if the solution would contain
coins $5+5$, we could replace them by coin $10$.

In the same way, both coins 2 and 20 can appear
at most twice in the optimal solution
because, we could replace
coins $2+2+2$ by coins $5+1$ and
coins $20+20+20$ by coins $50+10$.
Moreover, the optimal solution can't contain
coins $2+2+1$ or $20+20+10$
because we would replace them by coins $5$ and $50$.

Using these observations,
we can show for each coin $x$ that
it is not possible to optimally construct
sum $x$ or any larger sum by only using coins
that are smaller than $x$.
For example, if $x=100$, the largest optimal
sum using the smaller coins is  $5+20+20+5+2+2=99$.
Thus, the greedy algorithm that always selects
the largest coin produces the optimal solution.

This example shows that it can be difficult
to argue why a greedy algorithm works,
even if the algorithm itself is simple.

\subsubsection{General case}

In the general case, the coin set can contain any coins
and the greedy algorithm \emph{not} necessarily produces
an optimal solution.

We can prove that a greedy algorithm doesn't work
by showing a counterexample
where the algorithm gives a wrong answer.
In this problem it's easy to find a counterexample:
if the coins are $\{1,3,4\}$ and the sum of money
is 6, the greedy algorithm produces the solution
$4+1+1$, while the optimal solution is $3+3$.

We don't know if the general coin problem
can be solved using any greedy algorithm.
However, we will revisit the problem in the next chapter
because the general problem can be solved using a dynamic
programming algorithm that always gives the
correct answer.

\section{Scheduling}

Many scheduling problems can be solved
using a greedy strategy.
A classic problem is as follows:
Given $n$ events with their starting and ending
times, our task is to plan a schedule
so that we can join as many events as possible.
It's not possible to join an event partially.
For example, consider the following events:
\begin{center}
\begin{tabular}{lll}
event & starting time & ending time \\
\hline
$A$ & 1 & 3 \\
$B$ & 2 & 5 \\
$C$ & 3 & 9 \\
$D$ & 6 & 8 \\
\end{tabular}
\end{center}
In this case the maximum number of events is two.
For example, we can join events $B$ and $D$
as follows:
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw (2, 0) rectangle (6, -1);
    \draw[fill=lightgray] (4, -1.5) rectangle (10, -2.5);
    \draw (6, -3) rectangle (18, -4);
    \draw[fill=lightgray] (12, -4.5) rectangle (16, -5.5);
    \node at (2.5,-0.5) {$A$};
    \node at (4.5,-2) {$B$};
    \node at (6.5,-3.5) {$C$};
    \node at (12.5,-5) {$D$};
  \end{scope}
\end{tikzpicture}
\end{center}

It is possible to invent several greedy algorithms
for the problem, but which of them works in every case?

\subsubsection*{Algorithm 1}

The first idea is to select as \emph{short}
events as possible.
In the example case this algorithm
selects the following events:
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw[fill=lightgray] (2, 0) rectangle (6, -1);
    \draw (4, -1.5) rectangle (10, -2.5);
    \draw (6, -3) rectangle (18, -4);
    \draw[fill=lightgray] (12, -4.5) rectangle (16, -5.5);
    \node at (2.5,-0.5) {$A$};
    \node at (4.5,-2) {$B$};
    \node at (6.5,-3.5) {$C$};
    \node at (12.5,-5) {$D$};
  \end{scope}
\end{tikzpicture}
\end{center}

However, choosing short events is not always
a correct strategy but the algorithm fails,
for example, in the following case:
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw (1, 0) rectangle (7, -1);
    \draw[fill=lightgray] (6, -1.5) rectangle (9, -2.5);
    \draw (8, -3) rectangle (14, -4);
  \end{scope}
\end{tikzpicture}
\end{center}
If we select the short event, we can only select one event.
However, it would be possible to select both the long events.

\subsubsection*{Algorithm 2}

Another idea is to always select the next possible
event that \emph{begins} as \emph{early} as possible.
This algorithm selects the following events:
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw[fill=lightgray] (2, 0) rectangle (6, -1);
    \draw (4, -1.5) rectangle (10, -2.5);
    \draw[fill=lightgray] (6, -3) rectangle (18, -4);
    \draw (12, -4.5) rectangle (16, -5.5);
    \node at (2.5,-0.5) {$A$};
    \node at (4.5,-2) {$B$};
    \node at (6.5,-3.5) {$C$};
    \node at (12.5,-5) {$D$};
  \end{scope}
\end{tikzpicture}
\end{center}

However, we can find a counterexample for this
algorithm, too.
For example, in the following case,
the algorithm selects only one event:
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw[fill=lightgray] (1, 0) rectangle (14, -1);
    \draw (3, -1.5) rectangle (7, -2.5);
    \draw (8, -3) rectangle (12, -4);
  \end{scope}
\end{tikzpicture}
\end{center}
If we select the first event, it is not possible
to select any other events.
However, it would be possible to join the
other two events.

\subsubsection*{Algorithm 3}

The third idea is to always select the next
possible event that \emph{ends} as \emph{early} as possible.
This algorithm selects the following events: 
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw[fill=lightgray] (2, 0) rectangle (6, -1);
    \draw (4, -1.5) rectangle (10, -2.5);
    \draw (6, -3) rectangle (18, -4);
    \draw[fill=lightgray] (12, -4.5) rectangle (16, -5.5);
    \node at (2.5,-0.5) {$A$};
    \node at (4.5,-2) {$B$};
    \node at (6.5,-3.5) {$C$};
    \node at (12.5,-5) {$D$};
  \end{scope}
\end{tikzpicture}
\end{center}

It turns out that this algorithm
\emph{always} produces an optimal solution.
The algorithm works because 
regarding the final solution, it is
optimal to select an event that
ends as soon as possible.
Then it is optimal to select
the next event using the same strategy, etc.

One way to justify the choice is to think
what happens if we first select some event
that ends later than the event that ends
as soon as possible.
This can never be a better choice
because after an event that ends later,
we will have at most an equal number of
possibilities to select for the next events,
compared to the strategy that we select the
event that ends as soon as possible.

\section{Tasks and deadlines}

We are given $n$ tasks with duration and deadline.
Our task is to choose an order to perform the tasks.
For each task, we get $d-x$ points
where $d$ is the deadline of the task
and $x$ is the moment when we finished the task.
What is the largest possible total score
we can obtain?

For example, if the tasks are
\begin{center}
\begin{tabular}{lll}
task & duration & deadline \\
\hline
$A$ & 4 & 2 \\
$B$ & 3 & 5 \\
$C$ & 2 & 7 \\
$D$ & 4 & 5 \\
\end{tabular}
\end{center}
then the optimal solution is to perform
the tasks as follows:
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw (0, 0) rectangle (4, -1);
    \draw (4, 0) rectangle (10, -1);
    \draw (10, 0) rectangle (18, -1);
    \draw (18, 0) rectangle (26, -1);
    \node at (0.5,-0.5) {$C$};
    \node at (4.5,-0.5) {$B$};
    \node at (10.5,-0.5) {$A$};
    \node at (18.5,-0.5) {$D$};

    \draw (0,1.5) -- (26,1.5);
    \foreach \i in {0,2,...,26}
    {
        \draw (\i,1.25) -- (\i,1.75);
    }
    \footnotesize
    \node at (0,2.5) {0};
    \node at (10,2.5) {5};
    \node at (20,2.5) {10};

  \end{scope}
\end{tikzpicture}
\end{center}
In this solution, $C$ yields 5 points,
$B$ yields 0 points, $A$ yields $-7$ points
and $D$ yields $-8$ points,
so the total score is $-10$.

Surprisingly, the optimal solution for the problem
doesn't depend on the dedalines at all,
but a correct greedy strategy is to simply
perform the tasks \emph{sorted by their durations}
in increasing order.
The reason for this is that if we ever perform
two successive tasks such that the first task
takes longer than the second task,
we can obtain a better solution if we swap the tasks.
For example, if the successive tasks are
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw (0, 0) rectangle (8, -1);
    \draw (8, 0) rectangle (12, -1);
    \node at (0.5,-0.5) {$X$};
    \node at (8.5,-0.5) {$Y$};

\draw [decoration={brace}, decorate, line width=0.3mm] (7.75,-1.5) -- (0.25,-1.5);
\draw [decoration={brace}, decorate, line width=0.3mm] (11.75,-1.5) -- (8.25,-1.5);

\footnotesize
\node at (4,-2.5) {$a$};
\node at (10,-2.5) {$b$};

  \end{scope}
\end{tikzpicture}
\end{center}
and $a>b$, the swapped order of the tasks
\begin{center}
\begin{tikzpicture}[scale=.4]
  \begin{scope}
    \draw (0, 0) rectangle (4, -1);
    \draw (4, 0) rectangle (12, -1);
    \node at (0.5,-0.5) {$Y$};
    \node at (4.5,-0.5) {$X$};

\draw [decoration={brace}, decorate, line width=0.3mm] (3.75,-1.5) -- (0.25,-1.5);
\draw [decoration={brace}, decorate, line width=0.3mm] (11.75,-1.5) -- (4.25,-1.5);

\footnotesize
\node at (2,-2.5) {$b$};
\node at (8,-2.5) {$a$};

  \end{scope}
\end{tikzpicture}
\end{center}
gives $b$ points less to $X$ and $a$ points more to $Y$,
so the total score increases by $a-b > 0$.
In an optimal solution,
for each two successive tasks,
it must hold that the shorter task comes
before the longer task.
Thus, the tasks must be performed
sorted by their durations.

\section{Minimizing sums}

We will next consider a problem where
we are given $n$ numbers $a_1,a_2,\ldots,a_n$
and our task is to find a value $x$
such that the sum
\[|a_1-x|^c+|a_2-x|^c+\cdots+|a_n-x|^c\]
becomes as small as possible.
We will focus on the cases $c=1$ and $c=2$.

\subsubsection{Case $c=1$}

In this case, we should minimize the sum
\[|a_1-x|+|a_2-x|+\cdots+|a_n-x|.\]
For example, if the numbers are $[1,2,9,2,6]$,
the best solution is to select $x=2$
which produces the sum
\[
|1-2|+|2-2|+|9-2|+|2-2|+|6-2|=12.
\]
In the general case, the best choice for $x$
is the \textit{median} of the numbers,
i.e., the middle number after sorting.
For example, the list $[1,2,9,2,6]$
becomes $[1,2,2,6,9]$ after sorting,
so the median is 2.

The median is the optimal choice,
because if $x$ is smaller than the median,
the sum becomes smaller by increasing $x$,
and if $x$ is larger then the median,
the sum becomes smaller by decreasing $x$
Thus, we should move $x$ as near the median
as possible, so the optimal solution that $x$
is the median.
If $n$ is even and there are two medians,
both medians and all values between them
are optimal solutions.

\subsubsection{Case $c=2$}

In this case, we should minimize the sum
\[(a_1-x)^2+(a_2-x)^2+\cdots+(a_n-x)^2.\]
For example, if the numbers are $[1,2,9,2,6]$,
the best solution is to select $x=4$
which produces the sum
\[
(1-4)^2+(2-4)^2+(9-4)^2+(2-4)^2+(6-4)^2=46.
\]
In the general case, the best choice for $x$
is the \emph{average} of the numbers.
In the example the average is $(1+2+9+2+6)/5=4$.
This result can be derived by presenting
the sum as follows:
\[
nx^2 - 2x(a_1+a_2+\cdots+a_n) + (a_1^2+a_2^2+\cdots+a_n^2).
\]
The last part doesn't depend on $x$,
so we can ignore it.
The remaining parts form a function
$nx^2-2xs$ where $s=a_1+a_2+\cdots+a_n$.
This is a parabola opening upwards
with roots $x=0$ and $x=2s/n$,
and the minimum value is the average
of the roots $x=s/n$, i.e.,
the average of the numbers $a_1,a_2,\ldots,a_n$.

\section{Data compression}

\index{data compression}
\index{binary code}
\index{codeword}

We are given a string, and our task is to
\emph{compress} it so that it requires less space.
We will do this using a \key{binary code}
that determines for each character
a \key{codeword} that consists of bits.
After this, we can compress the string
by replacing each character by the
corresponding codeword.
For example, the following binary code
determines codewords for characters
\texttt{A}–\texttt{D}:
\begin{center}
\begin{tabular}{rr}
character & codeword \\
\hline
\texttt{A} & 00 \\
\texttt{B} & 01 \\
\texttt{C} & 10 \\
\texttt{D} & 11 \\
\end{tabular}
\end{center}
This is a \key{constant-length} code
which means that the length of each
codeword is the same.
For example, the compressed form of the string
\texttt{AABACDACA} is
\[000001001011001000,\]
so 18 bits are needed.
However, we can compress the string better
by using a \key{variable-length} code
where codewords may have different lengths.
Then we can give short codewords for
characters that appear often,
and long codewords for characters
that appear rarely.
It turns out that the \key{optimal} code
for the aforementioned string is as follows:
\begin{center}
\begin{tabular}{rr}
character & codeword \\
\hline
\texttt{A} & 0 \\
\texttt{B} & 110 \\
\texttt{C} & 10 \\
\texttt{D} & 111 \\
\end{tabular}
\end{center}
The optimal code produces a compressed string
that is as short as possible.
In this case, the compressed form using
the optimal code is
\[001100101110100,\]
so only 15 bits are needed.
Thus, thanks to a better code it was possible to
save 3 bits in the compressed string.

Note that it is required that no codeword
is a prefix of another codeword.
For example, it is not allowed that a code
would contain both codewords 10
and 1011.
The reason for this is that we also want
to be able to generate the original string
from the compressed string.
If a codeword could be a prefix of another codeword,
this would not always be possible.
For example, the following code is \emph{not} valid:
\begin{center}
\begin{tabular}{rr}
merkki & koodisana \\
\hline
\texttt{A} & 10 \\
\texttt{B} & 11 \\
\texttt{C} & 1011 \\
\texttt{D} & 111 \\
\end{tabular}
\end{center}
Using this code, it would not be possible to know
if the compressed string 1011 means
the string \texttt{AB} or the string \texttt{C}.

\index{Huffman coding}

\subsubsection{Huffman coding}

\key{Huffman coding} is a greedy algorithm
that constructs an optimal code for
compressing a string.
The algorithm builds a binary tree
based on the frequencies of the characters
in the string,
and a codeword for each characters can be read
by following a path from the root to
the corresponding node.
A move to the left correspons to bit 0,
and a move to the right corresponds to bit 1.

Initially, each character of the string is
represented by a node whose weight is the
number of times the character appears in the string.
Then at each step two nodes with minimum weights
are selected and they are combined by creating
a new node whose weight is the sum of the weights
of the original nodes.
The process continues until all nodes have been
combined and the code is ready.

Next we will see how Huffman coding creates
the optimal code for the string
\texttt{AABACDACA}.
Initially, there are four nodes that correspond
to the characters in the string:

\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,0) {$5$};
\node[draw, circle] (2) at (2,0) {$1$};
\node[draw, circle] (3) at (4,0) {$2$};
\node[draw, circle] (4) at (6,0) {$1$};

\node[color=blue] at (0,-0.75) {\texttt{A}};
\node[color=blue] at (2,-0.75) {\texttt{B}};
\node[color=blue] at (4,-0.75) {\texttt{C}};
\node[color=blue] at (6,-0.75) {\texttt{D}};

%\path[draw,thick,-] (4) -- (5);
\end{tikzpicture}
\end{center}
The node that represents character \texttt{A}
has weight 5 because character \texttt{A}
appears 5 times in the string.
The other weights have been calculated
in the same way.

The first step is to combine the nodes that
correspond to characters \texttt{B} and \texttt{D},
both with weight 1.
The result is:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (0,0) {$5$};
\node[draw, circle] (3) at (2,0) {$2$};
\node[draw, circle] (2) at (4,0) {$1$};
\node[draw, circle] (4) at (6,0) {$1$};
\node[draw, circle] (5) at (5,1) {$2$};

\node[color=blue] at (0,-0.75) {\texttt{A}};
\node[color=blue] at (2,-0.75) {\texttt{C}};
\node[color=blue] at (4,-0.75) {\texttt{B}};
\node[color=blue] at (6,-0.75) {\texttt{D}};

\node at (4.3,0.7) {0};
\node at (5.7,0.7) {1};

\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (4) -- (5);
\end{tikzpicture}
\end{center}
After this, the nodes with weight 2 are combined:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (1,0) {$5$};
\node[draw, circle] (3) at (3,1) {$2$};
\node[draw, circle] (2) at (4,0) {$1$};
\node[draw, circle] (4) at (6,0) {$1$};
\node[draw, circle] (5) at (5,1) {$2$};
\node[draw, circle] (6) at (4,2) {$4$};

\node[color=blue] at (1,-0.75) {\texttt{A}};
\node[color=blue] at (3,1-0.75) {\texttt{C}};
\node[color=blue] at (4,-0.75) {\texttt{B}};
\node[color=blue] at (6,-0.75) {\texttt{D}};

\node at (4.3,0.7) {0};
\node at (5.7,0.7) {1};
\node at (3.3,1.7) {0};
\node at (4.7,1.7) {1};

\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (4) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (5) -- (6);
\end{tikzpicture}
\end{center}
Finally, the two remaining nodes are combined:
\begin{center}
\begin{tikzpicture}[scale=0.9]
\node[draw, circle] (1) at (2,2) {$5$};
\node[draw, circle] (3) at (3,1) {$2$};
\node[draw, circle] (2) at (4,0) {$1$};
\node[draw, circle] (4) at (6,0) {$1$};
\node[draw, circle] (5) at (5,1) {$2$};
\node[draw, circle] (6) at (4,2) {$4$};
\node[draw, circle] (7) at (3,3) {$9$};

\node[color=blue] at (2,2-0.75) {\texttt{A}};
\node[color=blue] at (3,1-0.75) {\texttt{C}};
\node[color=blue] at (4,-0.75) {\texttt{B}};
\node[color=blue] at (6,-0.75) {\texttt{D}};

\node at (4.3,0.7) {0};
\node at (5.7,0.7) {1};
\node at (3.3,1.7) {0};
\node at (4.7,1.7) {1};
\node at (2.3,2.7) {0};
\node at (3.7,2.7) {1};

\path[draw,thick,-] (2) -- (5);
\path[draw,thick,-] (4) -- (5);
\path[draw,thick,-] (3) -- (6);
\path[draw,thick,-] (5) -- (6);
\path[draw,thick,-] (1) -- (7);
\path[draw,thick,-] (6) -- (7);
\end{tikzpicture}
\end{center}

Now all nodes are in the tree, so the code is ready.
The following codewords can be read from the tree:
\begin{center}
\begin{tabular}{rr}
character & codeword \\
\hline
\texttt{A} & 0 \\
\texttt{B} & 110 \\
\texttt{C} & 10 \\
\texttt{D} & 111 \\
\end{tabular}
\end{center}

% \subsubsection{Miksi algoritmi toimii?}
% 
% Huffmanin koodaus on ahne algoritmi, koska se
% yhdistää aina kaksi solmua, joiden painot ovat
% pienimmät.
% Miksi on varmaa, että tämä menetelmä tuottaa
% aina optimaalisen koodin?
% 
% Merkitään $c(x)$ merkin $x$ esiintymiskertojen
% määrää merkkijonossa sekä $s(x)$
% merkkiä $x$ vastaavan koodisanan pituutta.
% Näitä merkintöjä käyttäen merkkijonon
% bittiesityksen pituus on
% \[\sum_x c(x) \cdot s(x),\]
% missä summa käy läpi kaikki merkkijonon merkit.
% Esimerkiksi äskeisessä esimerkissä
% bittiesityksen pituus on
% \[5 \cdot 1 + 1 \cdot 3 + 2 \cdot 2 + 1 \cdot 3 = 15.\]
% Hyödyllinen havainto on, että $s(x)$ on yhtä suuri kuin
% merkkiä $x$ vastaavan solmun \emph{syvyys} puussa
% eli matka puun huipulta solmuun.
% 
% Perustellaan ensin, miksi optimaalista koodia vastaa
% aina binääripuu, jossa jokaisesta solmusta lähtee
% alaspäin joko kaksi haaraa tai ei yhtään haaraa.
% Tehdään vastaoletus, että jostain solmusta lähtisi
% alaspäin vain yksi haara.
% Esimerkiksi seuraavassa puussa tällainen tilanne on solmussa $a$:
% \begin{center}
% \begin{tikzpicture}[scale=0.9]
% \node[draw, circle, minimum size=20pt] (3) at (3,1) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (2) at (4,0) {$b$};
% \node[draw, circle, minimum size=20pt] (5) at (5,1) {$a$};
% \node[draw, circle, minimum size=20pt] (6) at (4,2) {\phantom{$a$}};
% 
% \path[draw,thick,-] (2) -- (5);
% \path[draw,thick,-] (3) -- (6);
% \path[draw,thick,-] (5) -- (6);
% \end{tikzpicture}
% \end{center}
% Tällainen solmu $a$ on kuitenkin aina turha, koska se
% tuo vain yhden bitin lisää polkuihin, jotka kulkevat
% solmun kautta, eikä sen avulla voi erottaa kahta
% koodisanaa toisistaan. Niinpä kyseisen solmun voi poistaa
% puusta, minkä seurauksena syntyy parempi koodi,
% eli optimaalista koodia vastaavassa puussa ei voi olla
% solmua, josta lähtee vain yksi haara.
% 
% Perustellaan sitten, miksi on joka vaiheessa optimaalista
% yhdistää kaksi solmua, joiden painot ovat pienimmät.
% Tehdään vastaoletus, että solmun $a$ paino on pienin,
% mutta sitä ei saisi yhdistää aluksi toiseen solmuun,
% vaan sen sijasta tulisi yhdistää solmu $b$
% ja jokin toinen solmu:
% \begin{center}
% \begin{tikzpicture}[scale=0.9]
% \node[draw, circle, minimum size=20pt] (1) at (0,0) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (2) at (-2,-1) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (3) at (2,-1) {$a$};
% \node[draw, circle, minimum size=20pt] (4) at (-3,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (5) at (-1,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (8) at (-2,-3) {$b$};
% \node[draw, circle, minimum size=20pt] (9) at (0,-3) {\phantom{$a$}};
% 
% \path[draw,thick,-] (1) -- (2);
% \path[draw,thick,-] (1) -- (3);
% \path[draw,thick,-] (2) -- (4);
% \path[draw,thick,-] (2) -- (5);
% \path[draw,thick,-] (5) -- (8);
% \path[draw,thick,-] (5) -- (9);
% \end{tikzpicture}
% \end{center}
% Solmuille $a$ ja $b$ pätee
% $c(a) \le c(b)$ ja $s(a) \le s(b)$.
% Solmut aiheuttavat bittiesityksen pituuteen lisäyksen
% \[c(a) \cdot s(a) + c(b) \cdot s(b).\]
% Tarkastellaan sitten toista tilannetta,
% joka on muuten samanlainen kuin ennen,
% mutta solmut $a$ ja $b$ on vaihdettu keskenään:
% \begin{center}
% \begin{tikzpicture}[scale=0.9]
% \node[draw, circle, minimum size=20pt] (1) at (0,0) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (2) at (-2,-1) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (3) at (2,-1) {$b$};
% \node[draw, circle, minimum size=20pt] (4) at (-3,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (5) at (-1,-2) {\phantom{$a$}};
% \node[draw, circle, minimum size=20pt] (8) at (-2,-3) {$a$};
% \node[draw, circle, minimum size=20pt] (9) at (0,-3) {\phantom{$a$}};
% 
% \path[draw,thick,-] (1) -- (2);
% \path[draw,thick,-] (1) -- (3);
% \path[draw,thick,-] (2) -- (4);
% \path[draw,thick,-] (2) -- (5);
% \path[draw,thick,-] (5) -- (8);
% \path[draw,thick,-] (5) -- (9);
% \end{tikzpicture}
% \end{center}
% Osoittautuu, että tätä puuta vastaava koodi on
% \emph{yhtä hyvä tai parempi} kuin alkuperäinen koodi, joten vastaoletus
% on väärin ja Huffmanin koodaus
% toimiikin oikein, jos se yhdistää aluksi solmun $a$
% jonkin solmun kanssa.
% Tämän perustelee seuraava epäyhtälöketju:
% \[\begin{array}{rcl}
% c(b) & \ge & c(a) \\
% c(b)\cdot(s(b)-s(a)) & \ge & c(a)\cdot (s(b)-s(a)) \\
% c(b)\cdot s(b)-c(b)\cdot s(a) & \ge & c(a)\cdot s(b)-c(a)\cdot s(a) \\
% c(a)\cdot s(a)+c(b)\cdot s(b) & \ge & c(a)\cdot s(b)+c(b)\cdot s(a) \\
% \end{array}\]