Add many more solutions and notes

14 files changed, 417 insertions, 95 deletions

diff --git a/sem6/com/m3/channel_test.py b/sem6/com/m3/channel_test.py
deleted file mode 100644
index c8f9e85..0000000
--- a/sem6/com/m3/channel_test.py
+++ /dev/null
@@ -1,90 +0,0 @@
-import numpy as np
-
-class MatrixEncoder:
-    def ident(dim, value=1):
-        return np.identity(dim) * value
-
-class Vector:
-    def __init__(self, arr):
-        self.arr = arr
-        self.len = len(arr)
-
-    def add(self, other):
-        return self.__class__(self.arr + other.arr)
-
-    def __str__(self):
-        return str(self.arr)
-
-    def print(self, f="{}"):
-        print(f.format(self))
-        return self
-
-    def gaussian(size, mu, sigma):
-        return Vector(sigma * np.random.randn(size) + mu)
-
-
-class BitVector(Vector):
-    def gen_uniform(size):
-        arr = np.random.randint(0, 2, size)
-        return BitVector(arr)
-
-    def to_int(self):
-        reverse = np.flip(self.arr)
-
-        res = 0
-        for (i, b) in enumerate(reverse):
-            res += b * 2**i
-        return res
-
-    def encode(self, matrix):
-        return SymbolVector(matrix[self.to_int()])
-
-    def from_int(index, size=None):
-        bits = []
-        while index > 0:
-            remainder = index % 2
-            index = index // 2
-            bits.insert(0, remainder)
-
-        # Padding if they want
-        if size is not None:
-            missing = max(0, size - len(bits))
-            bits = [0] * missing + bits
-
-        return BitVector(bits)
-
-    def check_if_error(self, other):
-        return np.array_equal(self.arr, other.arr)
-
-class SymbolVector(Vector):
-    def with_noise(self, c):
-        noise = Vector.gaussian(self.len, 0, c)
-        return self.add(noise)
-
-    def decide_and_decode(self, matrix):
-        dim = len(matrix)
-        # Collect all distances in vector
-        dists = np.arange(dim)
-        for (i, vec) in enumerate(matrix):
-            dists[i] = np.dot(self.arr, vec)
-
-        # Find index with smallest
-        index = np.argmax(dists)
-
-        # Convert integer back to bit vector
-        return BitVector.from_int(index, size=int(np.log2(dim)))
-
-encoder = MatrixEncoder.ident(4, np.sqrt(1))
-
-faults = 0
-total = 10000
-for i in range(total):
-    original = BitVector.gen_uniform(2)
-    after = original.encode(encoder) \
-        .with_noise(1) \
-        .decide_and_decode(encoder)
-    if original.check_if_error(after):
-        faults += 1
-
-print(f"Fault percentage {(faults / total) * 100}%")
-
diff --git a/sem6/dig/m2/Makefile b/sem6/dig/m2/Makefile
index 4f83294..2a4d25a 100644
--- a/sem6/dig/m2/Makefile
+++ b/sem6/dig/m2/Makefile
@@ -1,4 +1,4 @@
 
-INPUTFILES = nor_gate ex2 ex5 ex6
+INPUTFILES = nor_gate ex1 ex2 ex5 ex6 ex3
 
 include ../common.mk
diff --git a/sem6/dig/m2/ex1.vhdl b/sem6/dig/m2/ex1.vhdl
new file mode 100644
index 0000000..33b187c
--- /dev/null
+++ b/sem6/dig/m2/ex1.vhdl
@@ -0,0 +1,23 @@
+-- TEST_START{"inputs": ["sw1", "sw2", "sw3", "sw4"], "outputs": ["l1", "l2", "l3", "l4"], "testin": ["0000", "0001", "0101", "1101", "1010"]}TEST_STOP
+LIBRARY IEEE;
+USE IEEE.STD_LOGIC_1164.all;
+
+ENTITY ex1 IS
+    PORT(
+        sw1: IN STD_LOGIC;
+        sw2: IN STD_LOGIC;
+        sw3: IN STD_LOGIC;
+        sw4: IN STD_LOGIC;
+        l1: OUT STD_LOGIC;
+        l2: OUT STD_LOGIC;
+        l3: OUT STD_LOGIC;
+        l4: OUT STD_LOGIC);
+END ex1;
+
+ARCHITECTURE impl OF ex1 IS
+BEGIN
+    l1 <= sw1;
+    l2 <= sw2;
+    l3 <= sw3;
+    l4 <= sw4;
+END impl;
diff --git a/sem6/dig/m2/ex3.vhdl b/sem6/dig/m2/ex3.vhdl
new file mode 100644
index 0000000..8b3603a
--- /dev/null
+++ b/sem6/dig/m2/ex3.vhdl
@@ -0,0 +1,15 @@
+-- TEST_START{"inputs": ["a", "b"], "outputs": ["o"], "testin": ["00", "01", "10", "11"]}TEST_STOP
+LIBRARY IEEE;
+USE IEEE.STD_LOGIC_1164.all;
+
+ENTITY ex3 IS
+    PORT(
+        a: IN STD_LOGIC;
+        b: IN STD_LOGIC;
+        o: OUT STD_LOGIC);
+END ex3;
+
+ARCHITECTURE impl OF ex3 IS
+BEGIN
+    o <= a and b;
+END impl;
diff --git a/sem6/dig/m2/noter.tex b/sem6/dig/m2/noter.tex
new file mode 100644
index 0000000..7a5c531
--- /dev/null
+++ b/sem6/dig/m2/noter.tex
@@ -0,0 +1,35 @@
+\title{Mininoter}
+
+Der findes har 3 forskellige typer:
+\begin{itemize}
+    \item Island style eller array type som brugt af Xilinx
+    \item Hierachical brugt af Altera
+    \item Logarithmic hvilket er mere eksotisk
+\end{itemize}
+
+Island style indeholder CE (Configurable Elements), IM (Interconnection Matrix) og CB (Connection Block).
+
+Hierachical har mere ting i hierachier i stedet for et grid.
+
+\section{VHDL}
+
+Architecture er en implementation af et design.
+Et design kan have flere implementation hvilket er grunden til at man skal give den et unikt navn.
+
+\texttt{inout} er også en mulighed for pin type, og giver en bidirectional port.
+
+Forskellige værdier til \texttt{STD_LOGIC} og \texttt{STD_LOGIC_VECTOR}.
+
+\begin{itemize}
+    \item \texttt{1} -- forced high
+    \item \texttt{0} -- forced low
+    \item \texttt{Z} -- high impedance
+    \item \texttt{U} -- Uninitialized
+    \item \texttt{X} -- unknown, low impedance
+    \item \texttt{W} -- unknown (weak), low impedance
+    \item \texttt{L} -- weak 0
+    \item \texttt{H} -- weak 1
+    \item \texttt{-} -- wild card, don't care
+\end{itemize}
+
+Port maps are used to compose modules at a higher level.
diff --git a/sem6/dig/m4/Makefile b/sem6/dig/m4/Makefile
index 1c04cdd..7c53246 100644
--- a/sem6/dig/m4/Makefile
+++ b/sem6/dig/m4/Makefile
@@ -1,4 +1,4 @@
 
-INPUTFILES=ex1 ex2 ex3
+INPUTFILES=ex1 ex2 ex3 ex4
 
 include ../common.mk
diff --git a/sem6/dig/m4/ex2.vhdl b/sem6/dig/m4/ex2.vhdl
index 76883ad..cb7e952 100644
--- a/sem6/dig/m4/ex2.vhdl
+++ b/sem6/dig/m4/ex2.vhdl
@@ -1,4 +1,4 @@
--- TEST_START{"inputs": ["sw0"], "outputs": ["o,3,0"], "clk": "bt0", "testin": "000111001001"}TEST_STOP
+-- TEST_START{"inputs": ["sw0"], "outputs": ["o,3,0"], "clk": "bt0", "testin": "010111110000001100"}TEST_STOP
 library ieee;
 use ieee.std_logic_1164.all;
 
@@ -17,12 +17,19 @@ begin
     -- Implement shifting
     next_state(3 downto 1) <= state(2 downto 0);
     next_state(0) <= sw0;
-    o <= state;
 
     process (bt0)
+        variable bit_count : integer range 0 to 3;
     begin
         if (bt0'event and bt0 = '1') then
             state <= next_state;
+
+            if (bit_count = 3) then
+                o <= next_state;
+                bit_count := 0;
+            else
+                bit_count := bit_count + 1;
+            end if;
         end if;
     end process;
 end impl;
diff --git a/sem6/dig/m4/ex3.vhdl b/sem6/dig/m4/ex3.vhdl
index 650d303..0fc1062 100644
--- a/sem6/dig/m4/ex3.vhdl
+++ b/sem6/dig/m4/ex3.vhdl
@@ -1,4 +1,4 @@
--- TEST_START{"inputs": ["input,3,0", "write", "read"], "outputs": ["output,3,0"], "testin": [["0000", 0, 0], ["0000", 0, 1], ["1010", 1, 0], ["0000", 0, 1], ["1100", 1, 1], ["0011", 1, 1]]}TEST_STOP
+-- TEST_START{"inputs": ["input,3,0", "write", "read"], "outputs": ["output,3,0"], "testin": [["0000", 0, 0], ["0000", 0, 1], ["1010", 1, 0], ["0000", 0, 1], ["0000", 0, 0], ["0000", 0, 1], ["1100", 1, 1], ["0011", 1, 1]]}TEST_STOP
 library ieee;
 use ieee.std_logic_1164.all;
 
diff --git a/sem6/dig/m4/ex4.vhdl b/sem6/dig/m4/ex4.vhdl
new file mode 100644
index 0000000..57537bd
--- /dev/null
+++ b/sem6/dig/m4/ex4.vhdl
@@ -0,0 +1,32 @@
+-- TEST_START{"inputs": ["input,7,0", "write", "read"], "outputs": ["output,7,0"], "testin": [["00000000", 0, 0], ["00000000", 0, 1], ["10101010", 1, 0], ["00000000", 0, 1], ["00000000", 0, 0], ["00000000", 0, 1], ["11001010", 1, 1], ["00111111", 1, 1]]}TEST_STOP
+library ieee;
+use ieee.std_logic_1164.all;
+
+entity ex4 is
+    port (
+        input: in std_logic_vector(7 downto 0);
+        write: in std_logic;
+        read: in std_logic;
+        output: out std_logic_vector(7 downto 0)
+    );
+end ex4;
+
+architecture impl of ex4 is
+begin
+
+    mem_low : ENTITY work.ex3
+    PORT MAP (
+                input => input(3 downto 0),
+                write => write,
+                read => read,
+                output => output(3 downto 0)
+             );
+
+    mem_high : ENTITY work.ex3
+    PORT MAP (
+                input => input(7 downto 4),
+                write => write,
+                read => read,
+                output => output(7 downto 4)
+             );
+end impl;
diff --git a/sem6/dig/m4/noter.tex b/sem6/dig/m4/noter.tex
new file mode 100644
index 0000000..de217fd
--- /dev/null
+++ b/sem6/dig/m4/noter.tex
@@ -0,0 +1,75 @@
+\title{Noter til Lektion}
+
+Dette handlede meget om VHDL men vil mest forklare lidt om logic og nogle forskellige ting man kan lave.
+
+\paragraph{Combinatorial Cirquits} er når man har et kredsløb som er en direkt function af input, som for eksempel en and gate.
+
+Herfra kan man implementere memory så output fra kreds afhænger af alt tidligere input.
+Dette bliver kaldt \textbf{Sequential Cirquits}.
+For det meste bliver denne afhængighed implementeret med en slags \emph{state variabel}.
+
+De forskellige kan ses i slides og der er nok en god ide at finde dem frem til eksamnen.
+
+\paragraph{SR latch} er en meget simpel hvor man en set og en reset input, og et output.
+Den har følgende sandhedstabel.
+
+\begin{tabular}{ll|ll} \toprule
+    $S$ & $R$ & $Q$ & $\neg Q$ \\ \midrule
+    0 & 0 & Last $Q$ & Last $\neg Q$ \\
+    0 & 1 & 0 & 1 \\
+    1 & 0 & 1 & 0 \\
+    1 & 1 & 0 & 0 \\ \bottomrule
+\end{tabular}
+
+Her er det lidt et problem at have en state hvor både $Q$ og $\neg Q$ er 0.
+Dette giver jo ikke så meget mening.
+
+Det også tit en fordel at have en \emph{clock} så man kan bestemme hvornår værdier skal gemmes.
+Dette kan let implementeres med nogle NOR gates.
+
+\paragraph{D latch} udskifter $R$ og $S$ med en et enkelt input og en clock.
+Dette giver mening da man gemmer en værdi på $D$ ved at clock $C$, også er den gemt til man giver den en ny værdi.
+
+\begin{tabular}{ll|ll} \toprule
+    $C$ & $D$ & $Q$ & $\neg Q$ \\ \midrule
+    0 & x & Last $Q$ & Last $\neg Q$ \\
+    1 & 0 & 0 & 1 \\
+    1 & 1 & 1 & 0 \\ \bottomrule
+\end{tabular}
+
+Her tilføjer man tit en \emph{edge triggering} så den kun gemmer når $C$ går fra høj til lav, eller omvendt.
+Dette giver en \textbf{D Flip-Flop} og er implementeret med to D latches.
+
+\paragraph{JK Flip-Flop} fungere som en SR latch men trigger på clock rise eller fall.
+Den har heller ikke problemet med at have en mærkelig state da output bliver flipped når $J=1$ og $K=1$.
+
+\section{Characteristic Equation}
+
+Her beskriver man sin Sequential kreds med en formel.
+Man kan beskrive en D Flip-Flop med:
+\begin{equation*}
+    \begin{split}
+        Q = D \\
+        Q_{n+1} = D_n
+    \end{split}
+\end{equation*}
+
+Eller en JK Flip-Flop med:
+\[
+    Q_{n+1} = J \cdot \neg Q_n + \neg K \cdot Q_n
+\]
+
+\section{State Automaton}
+
+Her har man en enhed hvor output, $O_n$, kan beskrives som følgende.
+\begin{equation*}
+    \begin{split}
+        Q_{n+1} = F(Q_n, I_n) \\
+        O_{n} = G(Q_n, I_n)
+    \end{split}
+\end{equation*}
+
+Her har man altså noget kombinatorisk logic før og efter en block memory, hvilket man kan bruge til at implementere ret avancerede ting.
+Her er det vigtigt at man tager højde for timing mellem disse blokke.
+
+Man kan implementere memory med \texttt{process} functionen i VHDL.
diff --git a/sem6/dig/mm1/opgaver.md b/sem6/dig/mm1/opgaver.md
index 509c1c5..9f80ad4 100644
--- a/sem6/dig/mm1/opgaver.md
+++ b/sem6/dig/mm1/opgaver.md
@@ -1,3 +1,5 @@
+
+
 # Opgave 1
 
 Nope
diff --git a/sem6/dig/mm1/opgaver.tex b/sem6/dig/mm1/opgaver.tex
new file mode 100644
index 0000000..4b8dc91
--- /dev/null
+++ b/sem6/dig/mm1/opgaver.tex
@@ -0,0 +1,169 @@
+\title{Opgaver til Lektion 1}
+
+\let\inv\overline
+
+\section{Opgave 1}
+
+Perfect induction af $(A + B) \cdot (A + C) = A + (B \cdot C)$:
+
+\begin{tabular}{lll|ll} \toprule
+    $A$ & $B$ & $C$ & $(A + B) \cdot (A + C)$ & $A + (B \cdot C)$ \\ \midrule
+    0 & 0 & 0 & 0 & 0 \\
+    0 & 0 & 1 & 0 & 0 \\
+    0 & 1 & 0 & 0 & 0 \\
+    0 & 1 & 1 & 1 & 1 \\
+    1 & 0 & 0 & 1 & 1 \\
+    1 & 0 & 1 & 1 & 1 \\
+    1 & 1 & 0 & 1 & 1 \\
+    1 & 1 & 1 & 1 & 1 \\ \bottomrule
+\end{tabular}
+
+Perfect induction af $A \cdot (A + B) = A$:
+
+\begin{tabular}{ll|ll} \toprule
+    $A$ & $B$ & $A + B$ & $A$ \\ \midrule
+    0 & 0 & 0 & 0 \\
+    0 & 1 & 0 & 0 \\
+    1 & 0 & 1 & 1 \\
+    1 & 1 & 1 & 1 \\ \bottomrule
+\end{tabular}
+
+Perfect induction of $A + \inv{A} = 1$:
+
+\begin{tabular}{l|ll} \toprule
+    $A$ & $A + \inv{A} $ & $1$ \\ \midrule
+    0 & 1 & 1 \\
+    1 & 1 & 1 \\ \bottomrule
+\end{tabular}
+
+Perfect induction of $\inv{A + B + C} = \inv{A} \cdot \inv{B} \cdot \inv{C}$:
+
+\begin{tabular}{lll|ll} \toprule
+    $A$ & $B$ & $C$ & $\inv{A + B + C}$ & $\inv{A} \cdot \inv{B} \cdot \inv{C}$ \\ \midrule
+    0 & 0 & 0 & 1 & 1 \\
+    0 & 0 & 1 & 0 & 0 \\
+    0 & 1 & 0 & 0 & 0 \\
+    0 & 1 & 1 & 0 & 0 \\
+    1 & 0 & 0 & 0 & 0 \\
+    1 & 0 & 1 & 0 & 0 \\
+    1 & 1 & 0 & 0 & 0 \\
+    1 & 1 & 1 & 0 & 0 \\ \bottomrule
+\end{tabular}
+
+\section{Opgave 2}
+
+\begin{opg}
+    Show that the following
+
+    \begin{equation*}
+        R = \inv{\inv{A \cdot \inv{B}} \cdot \inv{\inv{A} \cdot B}}
+    \end{equation*}
+
+    is XOR.
+\end{opg}
+
+Man kan skrive XOR operator som:
+\begin{equation*}
+    A \oplus B = (A + B) \cdot \inv{(A \cdot B)}
+\end{equation*}
+
+\begin{tabular}{ll|ll} \toprule
+    $A$ & $B$ & $A \oplus B$ & $(A + B) \cdot \inv{(A \cdot B)}$ \\ \midrule
+    0 & 0 & 0 & 0 \\
+    0 & 1 & 1 & 1 \\
+    1 & 0 & 1 & 1 \\
+    1 & 1 & 0 & 0 \\ \bottomrule
+\end{tabular}
+
+Kan herefter vise at dette er $\inv{\inv{A \cdot \inv{B}} \cdot \inv{\inv{A} \cdot B}}$.
+\begin{equation*}
+    \inv{\inv{A \cdot \inv{B}} \cdot \inv{\inv{A} \cdot B}} = (A + B) \cdot \inv{(A \cdot B)}\\
+\end{equation*}
+Vi kan starte med at bruge DeMorgans lov.
+\begin{equation*}
+    = (A + B) \cdot (\inv{A} + \inv{B}) \\
+\end{equation*}
+Man kan skrive det ud.
+\begin{equation*}
+    = A \cdot \inv{A} + A \cdot \inv{B} + B \cdot \inv{A} + B \cdot \inv{B}
+\end{equation*}
+Vi kan bruge axiomen $X \cdot \inv{X} = 0$.
+\begin{equation*}
+    = A \cdot \inv{B} + B \cdot \inv{A}
+\end{equation*}
+Og nu kan vi bruge DeMorgans lov på venstre side.
+\begin{equation*}
+    \inv{ \inv{A \cdot \inv{B}}} + \inv{ \inv{\inv{A} \cdot B}} = A \cdot \inv{B} + B \cdot \inv{A}
+\end{equation*}
+Og dette passer hvis man lader inverserne gå ud med hinnanden.
+
+\section{Opgave 3}
+
+\begin{opg}
+    Reduce the following expressions:
+    \begin{align*}
+        A \cdot \inv{B} \cdot \inv{C} + A \cdot B \cdot \inv{C} + \inv{A} \cdot \inv{C} \\
+        M \cdot \inv N \cdot P + \inv L \cdot M \cdot P + \inv L \cdot M \cdot \inv N + \inv L \cdot M \cdot N \cdot \inv P + \inv L \cdot \inv N \cdot \inv P
+    \end{align*}
+\end{opg}
+
+Vi starter med den første opgave.
+Her tager vi og bruger den distributive teorem et par gange.
+\begin{equation*}
+    \begin{split}
+        \inv{C} \cdot (A \cdot \inv B + A \cdot B + \inv A) \\
+        \inv{C} \cdot (A \cdot (\inv B + B) + \inv A)
+    \end{split}
+\end{equation*}
+Nu kan vi bruge axiomen \(X + \inv X = 1\) og \(X \cdot 1 = X\).
+\[
+    \inv C \cdot (A + \inv A) = \inv C
+\]
+
+Lad og tage den anden.
+Vi starter med at bruge den distributive teorem et par gange.
+\begin{equation*}
+    \begin{split}
+        M \cdot (\inv N \cdot P + \inv L \cdot P + \inv L \cdot \inv N + \inv L \cdot N \cdot \inv P) + \inv L \cdot \inv N \cdot \inv P \\
+        M \cdot (\inv N \cdot P + \inv L \cdot (P + \inv N + N \cdot \inv P)) + \inv L \cdot \inv N \cdot \inv P
+    \end{split}
+\end{equation*}
+Den inerste parantes er på formen $A + B + \inv A \cdot \inv B$.
+Dette kan vise altid er lig med $1$, ved at starte med DeMorgans lov.
+\[
+    A + B + \inv{A + B}
+\]
+Ud fra axiomen $X + \inv X = 1$ kan vi se at dette altid er 1.
+Nu kan vi sætte dette ind og bruge den distributive lov igen omvendt.
+\[
+    M \cdot (\inv N \cdot P + \inv L) + \inv L \cdot \inv N \cdot \inv P = M \cdot \inv N \cdot P + M \cdot \inv L + \inv L \cdot \inv N \cdot \inv P
+\]
+Mere kan jeg desværre ikke reducere den.
+
+\section{Opgave 4}
+
+\begin{opg}
+    Find expression
+\end{opg}
+
+\begin{verbatim}
+X = ~(A * B)
+Y = ~(A * X)
+Z = ~(B * X)
+C = ~(Y * Z)
+D = ~X
+
+C = ~(~(A * ~(A * B)) * ~(B * ~(A * B)))
+D = ~~(A * B) = A * B
+\end{verbatim}
+
+
+\section{Gate Input Cost}
+
+Skriver lige hurtigt formlen op for GIC.
+\begin{verbatim}
+    GIC = LC + number of terms in boolean expr + number of unique inversions
+\end{verbatim}
+Her er LC litteral cost, hvilket er antallet af ikke unikke variabler.
+
+GIC kan også findes ved at tælle antal inputs i et logic diagram.
diff --git a/sem6/dig/shell.nix b/sem6/dig/shell.nix
new file mode 100644
index 0000000..28b6ba9
--- /dev/null
+++ b/sem6/dig/shell.nix
@@ -0,0 +1,7 @@
+{ pkgs ? import <nixpkgs> {}}:
+
+pkgs.mkShell {
+    buildInputs = with pkgs; [
+        ghdl-llvm zlib
+    ];
+}
diff --git a/sem6/prob/eksamnen/notes.tex b/sem6/prob/eksamnen/notes.tex
new file mode 100644
index 0000000..4dfee30
--- /dev/null
+++ b/sem6/prob/eksamnen/notes.tex
@@ -0,0 +1,47 @@
+\title{Eksamnens Noter}
+
+
+The universal set or sample space is the set everything, and is denoted $S$.
+Therefore the probability of hitting $S$ is $P(S) = 1$.
+
+This is the first of 3 axioms repeated below.
+
+\begin{enumerate}
+    \item For any event $A$, $P(A) \geq 0$.
+    \item The probability of hitting sample space is always 1, $P(S) = 1$.
+    \item If events $A_1, A_2, ...$ are \textbf{disjoint} event, then
+        \begin{equation}
+            P(A_1 \cup A_2 ...) = P(A_1) + P(A_2)\,.
+        \end{equation}
+\end{enumerate}
+
+The last axiom requires that the events $A_n$ are disjoint.
+If they aren't one should subtract the part they have in common.
+This is called the \emph{Inclusion-Exclusion Principle}.
+
+\begin{principle}
+    The \emph{Inclusion-Exclusion Principle} is defined as
+    \begin{equation}
+        P(A \cup B) = P(A) + P(B) - P(A \cap B)\,.
+    \end{equation}
+    Definition with 3 events can be found in the in the book.
+\end{principle}
+
+\section{Counting}
+
+The probability of a event $A$ can be found by
+\begin{equation}
+    P(A) = \frac {|A|} {|S|}\,.
+\end{equation}
+It is therefore required to count how many elements are in $S$ and $A$.
+The most simple method is the \emph{multiplication principle}.
+
+\begin{principle}[Multiplication principle]
+    Let there be $r$ random experiments, where the $k$'th experiment has $n_k$ outcomes.
+    Then there are
+    \begin{equation}
+        n_1 \cdot n_2 \cdot ... \cdot n_r
+    \end{equation}
+    possible outcomes over all $r$ experiments.
+\end{principle}
+