M src/index.tex → src/index.tex

@@ -60,12 +60,12 @@
 \end{definition}
 \begin{example}
 Suppose $n=5$.
-\begin{equation}
+\begin{equation*}
     |(1, 0, 0, 1, 1)| = 3 \quad\quad |(1, 0, 0, 0, 0)| = 1 \quad\quad |(0, 0, 1, 1, 1)| = 4
-\end{equation}
+\end{equation*}
 \end{example}
 
-\begin{definition}[bit-wise xor]
+\begin{definition}[sum modulo 2 (bit-wise xor)]
 Given two words $\textbf{w}, \textbf{w}' \in \{0,1\}^n$, we can define the following operation:
 \begin{equation}
     \textbf{w} \oplus \textbf{w}' = ((w_1 + w_1')\text{mod}2, (w_2 + w_2')\text{mod}2, \ldots, (w_n + w_n')\text{mod}2) \in \{0,1\}^n
@@ -79,7 +79,7 @@ \end{itemize}
 
 \begin{example}
 Suppose $n=6, \textbf{w} = (1, 1, 0, 0, 1, 0), \textbf{w}' = (0, 1, 1, 0, 0, 1)$.
-\begin{equation}
+\begin{equation*}
 \begin{array}{l}
 (1, 1, 0, 0, 1, 0) \,+ \\
 (0, 1, 1, 0, 0, 1) = \\  \hline
@@ -90,7 +90,7 @@ \begin{array}{l}
 (1, 2, 1, 0, 1, 1)\text{mod} 2 = \\ \hline
 (1, 0, 1, 0, 1, 1) = \textbf{w} \oplus \textbf{w}'
 \end{array}.
-\end{equation}
+\end{equation*}
 \end{example}
 
 At this point, it's clear how this identity holds true:
@@ -109,7 +109,7 @@ \end{proposition}
 \begin{example}
     Suppose we want to count the words which have distance $i=2$ from the word $001$ ($n=3$).
     
-    \begin{equation}
+    \begin{equation*}
         \begin{aligned}
             \mathbf{w} = 001 \\
             n = 3 \\
@@ -137,13 +137,13 @@ 01011\\
             01101
         \end{aligned}
         \right\} 5
-    \end{equation}
+    \end{equation*}
 
-    \begin{equation}
+    \begin{equation*}
         {3 \choose 2} = \frac{3!}{2!(3-2)!}=\frac{3!}{2!}=3
         \quad\quad
         {5 \choose 4} = \frac{5!}{4!(5-4)!}=\frac{5!}{4!}=5
-    \end{equation}
+    \end{equation*}
 
 \end{example}
 
@@ -162,7 +162,7 @@
 \begin{example}
 For $n=7$, a valid code of distance $d=3$ could be the set:
 
-\begin{equation}
+\begin{equation*}
 \label{sample-code}
 \mathcal{C} =
 \left.
@@ -172,7 +172,7 @@ & 0100110, 0101101, 0110011, 0111000, \\
 & 1000111, 1001100, 1010010, 1011001, \\
 & 1100001, 1101010, 1110100, 1111111 
 \end{cases}\right\}.
-\end{equation}
+\end{equation*}
 
 Notice how, in this code, every two distinct words have an Hamming distance of at least $3$.
 \end{example}
@@ -182,16 +182,16 @@ A code $\mathcal{C}$ can correct at most $r$ errors if and only if it has distance $d \geq 2r+1$.
 \end{proposition}
 \begin{example}
     Suppose $\mathcal{C}$ contains two distinct words $\textbf{w}', \textbf{w}''$ that differ in at most $2r$ bits. For example, with $n=7$ and $r=1$, suppose we receive $\hat{\textbf{w}}$, obtained by flipping $r=1$ bits in one of them:
-\begin{equation}
+\begin{equation*}
 \textbf{w}' = 0000011 \quad \textbf{w}'' =0000101 \quad \hat{\textbf{w}} = 0000001
-\end{equation}
+\end{equation*}
 At this point, it's impossible to know whether the originally transmitted word was $\textbf{w}'$ or $\textbf{w}''$. This test can be repeated by choosing any code with distance $d=2r$ and flipping $r$ of them.
 
 On the other hand, if $\mathcal{C}$ had a distance of $2r+1$, we would be sure it could correct up to $r$ errors:
 
-\begin{equation}
+\begin{equation*}
 \textbf{w}' = 0000011 \quad \textbf{w}'' =0010101 \quad \hat{\textbf{w}} = 0000001
-\end{equation}
+\end{equation*}
 
 In this case, we can be sure that $\textbf{w}''$ was transmitted.
 \end{example}
@@ -210,8 +210,8 @@ \end{lemma}
 \subsubsection*{Proof}
 We need to prove two things:
 \begin{enumerate}
-    \item $\exists (\mathbf{w},\mathbf{w}') : d_H(\mathbf{w},\mathbf{w}')=2\quad\forall \mathbf{w},\mathbf{w}' \in \mathcal{C}$
-    \item $\nexists (\mathbf{w},\mathbf{w}') : d_H(\mathbf{w},\mathbf{w}')=1\quad\forall \mathbf{w},\mathbf{w}' \in \mathcal{C}$
+    \item $\exists (\mathbf{w},\mathbf{w}') : d_H(\mathbf{w},\mathbf{w}')=2\quad\forall\; \mathbf{w},\mathbf{w}' \in \mathcal{C}$
+    \item $\nexists (\mathbf{w},\mathbf{w}') : d_H(\mathbf{w},\mathbf{w}')=1\quad\forall\; \mathbf{w},\mathbf{w}' \in \mathcal{C}$
 \end{enumerate}
 
 To prove point 1, we can just take any word $\mathbf{w}\in \mathcal{C}:|\mathbf{w}|\geq 2$. Then, consider the word $\mathbf{w}'$, obtained by replacing two of the 1's in the word with two 0's.
@@ -224,7 +224,7 @@ This means that if $\mathbf{w}$ has even weight, $\mathbf{w}'$ has to have odd weight, and viceversa. This is a contradiction to the hypothesis of all words in $\mathcal{C}$ having even weight, so $\mathbf{w}$ and $\mathbf{w}'$ cannot have Hamming distance 1.
 
 Finally, let's find an upper bound for this code.
 \begin{lemma}
-    Fixed $n$, if $ \mathcal{C} \subseteq \{0,1\}^n : |w| \text{ is even } \forall \mathbf{w}\in \mathcal{C}$, then $|\mathcal{C}| \geq 2^{n-1}$.%TODO: should be equal?
+    Fixed $n$, if $ \mathcal{C} \subseteq \{0,1\}^n : |w| \text{ is even }\forall\; \mathbf{w}\in \mathcal{C}$, then $|\mathcal{C}| \geq 2^{n-1}$.%TODO: should be equal?
 \end{lemma}
 \subsubsection*{Proof}
 We know that the total number of words in a $n$-bit code is $2^n$. We start with a code made up of $(n-1)$-bit words, and consider the code $\mathcal{C} \subseteq \{0,1\}^n$ obtained by applying the following transformation to each word of the code:
@@ -265,9 +265,9 @@ So we have $|\mathcal{C}|$ balls of radius $r$ and distance at least $2r+1$ from one another. This means the balls are disjoint, they're not overlapped. The total number of balls (words in $\mathcal{C}$) cannot be larger than the total number of words in $\{0,1\}^n$  divided by the number of words in a single ball.
 
 The number of words at Hamming distance exactly $i$ from any word $w$ is ${n \choose i}$, which implies:
 
-\begin{equation}
+\begin{equation*}
     |B(\mathbf{w}, r)| = 1 + {n \choose 1} + {n \choose 2} + \dots + {n \choose r} = \sum_{i=0}^{r} {n \choose i},
-\end{equation}
+\end{equation*}
 where each iteration of the sum adds the amount of elements of weight $i$.
 
 Finally, as we said before, the total number of elements in $\{0,1\}^n$ is $2^n$. So, we get the following upper bound:
@@ -282,16 +282,16 @@ \end{lemma}
 
 \begin{example}
     For $n=7$ and $d=3$, we have $r=1$ so:
-    \begin{equation}
+    \begin{equation*}
         A(7,3) \leq \floor*{\frac{2^7}{\sum\limits_{i=0}^{1}{7 \choose i}}} = \floor*{\frac{128}{{7 \choose 0} + {7 \choose 1}}} = \floor*{\frac{128}{1 + 7}} = 16.
-    \end{equation}
+    \end{equation*}
 \end{example}
 
 \begin{example}
     For $n=17$ and $d=3$:
-    \begin{equation}
+    \begin{equation*}
         A(17,3) \leq \floor*{\frac{2^{17}}{\sum\limits_{i=0}^{1}{17 \choose i}}} = \floor*{\frac{131072}{{17 \choose 0} + {17 \choose 1}}} = \floor*{\frac{131072}{1 + 17}} = 7281.
-    \end{equation}
+    \end{equation*}
 \end{example}
 
 \section{The Delsarte bound}
@@ -317,7 +317,7 @@ \end{equation}
 \end{theorem}
 
 \begin{example}
-    % TODO: insert examplea
+    % TODO: insert examples
 \end{example}
 \subsection{Making sense of these constraints}
 For each code $\mathcal{C} \subseteq \{0,1\}^n$ we associate $\mathbf{\tilde{x}}=(\tilde{x}_1, \tilde{x}_2, \ldots, \tilde{x}_n)$ such that each $\tilde{x}_i$ is a nonnegative real number and $\tilde{x}_1+\tilde{x}_2+\ldots+\tilde{x}_n=|\mathcal{C}|$.
@@ -346,12 +346,12 @@ \tilde{x}_0 + \tilde{x}_1 + \tilde{x}_2 + \tilde{x}_3 = 1 + \frac{3}{2} + 1 + \frac{1}{2} = 4 = |\mathcal{C}|
         \end{array}
     \end{equation*}
 \end{example}
-\subsection{One last constraint}
+\subsection{The final constraint}
 Let $\mathcal{C} \subseteq \{0,1\}^n$ be an arbitrary set, and let $\tilde{x}_i$ be defined as above (\ref{xtildei}) for $i=0,1,\ldots,n$ and $t=1,2,\ldots,n$.
 Then,
-\begin{equation}
+\begin{equation*}
     \sum_{i=0}^{n}K_t(n,i) \tilde{x}_i \geq 0.
-\end{equation}
+\end{equation*}
 To prove this, let $I \subseteq \{1,2,\ldots n\}$ be a set of indices and let's define the Hamming distance $d_H^I(\mathbf{w}, \mathbf{w}')$ as the largest number of indices $i \in I$ with $w_i \neq w_i'$:
 \begin{equation}
     d_H^I(\mathbf{w}, \mathbf{w}') = |\{i \in I : w_i \neq w_i'\}| \quad\forall\; \mathbf{w}, \mathbf{w}' \in \mathcal{C}.
@@ -377,15 +377,15 @@ \begin{proof}
         First of all, let's prove that any couple of words $\mathbf{w}, \mathbf{w}' \in \mathcal{C}$ has even $d_H$ if $|\mathbf{w}|$ and $|\mathbf{w}'|$ have the same parity.
         We previously defined $d_H=|\textbf{w} \oplus \textbf{w}'|$. To obtain the weight of the result of the $\oplus$ operation we can also sum the number of 1's in each word and subtract twice the number of positions where both words have 1, $D$ for brevity:
         
-        \begin{equation}
+        \begin{equation*}
             d_H(\mathbf{w}, \mathbf{w}') = |\mathbf{w} \oplus \mathbf{w}'| = |\textbf{w}| + |\textbf{w}'| + 2D,
-        \end{equation}
+        \end{equation*}
         where $D=|\{i \in \{1,2,\ldots,n\}:w_i=w_i'=1\}|$.
         First of all, $2D$ is even, so it doesn't change the parity of result. That only depends on the weight of $\mathbf{w}$ and $\mathbf{w}'$.
         Let's call $\mathcal{E}$ the set of all words in $\mathcal{C}$ with even weight and $\mathcal{O}$ the set of all words in $\mathcal{C}$ with odd weight. Of course, $\mathcal{E} \cup \mathcal{O} = \mathcal{C}$.
-        \begin{equation}
+        \begin{equation*}
             \mathcal{E} = \{\mathbf{w} \in \mathcal{C} : |\mathbf{w}|\text{ is even}\}, \quad \mathcal{O} = \{\mathbf{w} \in \mathcal{C} : |\mathbf{w}|\text{ is odd}\}.
-        \end{equation}
+        \end{equation*}
         Then:
         \begin{itemize}
             \item for $(\mathbf{w}, \mathbf{w}'): \mathbf{w} \in \mathcal{E} \land \mathbf{w}' \in \mathcal{E}$, $d_H(\mathbf{w}, \mathbf{w}')$ is even;
@@ -396,14 +396,14 @@ \end{itemize}
         So we can conclude that any couple of words $\mathbf{w}, \mathbf{w}' \in \mathcal{C}$ has even $d_H$ if $|\mathbf{w}|$ and $|\mathbf{w}'|$ have the same parity.
         That said, the couples of words with even distance are the ones obtained by the cartesian products $\mathcal{E} \times \mathcal{E}=\mathcal{E}^2$ and $\mathcal{O} \times \mathcal{O}=\mathcal{O}^2$; furthermore, the couples of words with odd distance are the ones obtained by the cartesian products $\mathcal{E} \times \mathcal{O}$ and $\mathcal{O} \times \mathcal{E}$.
         If we want to write our original thesis in these terms, we get the following form, which is clearly always true:
-        \begin{equation}
+        \begin{equation*}
             \begin{array}{l}
                 |\mathcal{E}|^2+|\mathcal{O}|^2\geq |\mathcal{E}|\cdot|\mathcal{O}| + |\mathcal{O}|\cdot|\mathcal{E}|;\\\\
                 |\mathcal{E}|^2+|\mathcal{O}|^2\geq 2\cdot|\mathcal{E}|\cdot|\mathcal{O}|;\\\\
                 |\mathcal{E}|^2+|\mathcal{O}|^2 - 2\cdot|\mathcal{E}|\cdot|\mathcal{O}|\geq 0; \\\\
                 (|\mathcal{E}| - |\mathcal{O}|)^2 \geq 0.
             \end{array}
-        \end{equation}
+        \end{equation*}
     \end{proof}
 \end{lemma}
 \begin{lemma}
@@ -419,37 +419,43 @@ \label{evens-odd-geq-zero}
             \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}(-1)^{d_H^I(\mathbf{w}, \mathbf{w}')} \geq 0.
         \end{equation}
         If we set $I=\{i:v_i=1\}$, then $d_H^I(\mathbf{w}, \mathbf{w}') = (\textbf{w} \oplus \textbf{w}')^T\mathbf{v}$: transposing and doing a scalar product with the vector $\mathbf{v}$ gives the same result as computing the Hamming distance on all indices $i\in I$.
-        \begin{example}
-            Suppose $\mathbf{w}=(10010)$ and $\mathbf{w}'=(00111)$ with $\mathbf{v}=(10011)$.
-            Then, $I = \{1,4,5\}$. Let's show that $(\textbf{w} \oplus \textbf{w}')^T\mathbf{v} = d_H^I(\mathbf{w}, \mathbf{w}') = 2$:
-            \begin{equation*}
-                [(10010) \oplus (00111)]^T \cdot (10011) = (10101)^T \cdot (10011) = 1+0+0+0+1 = 2.
-            \end{equation*}
-        \end{example}
-        We can now go back to (\ref{evens-odd-geq-zero}) and replace the distance:
-        \begin{equation}
+
+        This means we can replace the distance with this product:
+        \begin{equation*}
             \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}(-1)^{d_H^I(\mathbf{w}, \mathbf{w}')} \geq 0
             =
             \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}(-1)^{(\textbf{w} \oplus \textbf{w}')^T\mathbf{v}} \geq 0.
-        \end{equation}
+        \end{equation*}
     \end{proof}
+    \begin{example}
+        Suppose $\mathbf{w}=(10010)$ and $\mathbf{w}'=(00111)$ with $\mathbf{v}=(10011)$.
+        Then, $I = \{1,4,5\}$. Let's show that $(\textbf{w} \oplus \textbf{w}')^T\mathbf{v} = d_H^I(\mathbf{w}, \mathbf{w}') = 2$:
+        \begin{equation*}
+            [(10010) \oplus (00111)]^T \cdot (10011) = (10101)^T \cdot (10011) = 1+0+0+0+1 = 2.
+        \end{equation*}
+    \end{example}
+        
+    
     \begin{proof}[Proof (algebraic)]
         %TODO: fill this
     \end{proof}
 \end{lemma}
 
 \begin{lemma}
-    Now, the only thing left to prove is $\sum_{i=0}^{n}K_t(n,i)\tilde{x}_i\geq 0 \forall t\in\{1,2,\ldots,n\}$.
+    Now, the only thing left to prove is:
+    \begin{equation*}
+        \sum\limits_{i=0}^{n}K_t(n,i)\tilde{x}_i\geq 0 \quad\forall\; t\in\{1,2,\ldots,n\}.
+    \end{equation*}
 
     \begin{proof}
         We start by summing the inequality in (\ref{sum-almost-here}) over all $\mathbf{v}\in \{0,1\}^n$ of weight $t$:
-        \begin{equation}
+        \begin{equation*}
             \sum_{\mathbf{v}\in \{0,1\}^n:|\mathbf{v}|=t}\;\sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}(-1)^{(\textbf{w} \oplus \textbf{w}')^T\mathbf{v}} \geq 0;
-        \end{equation}
-        then, we interchange the summation order:
-        \begin{equation}
+        \end{equation*}
+        then, we change the summation order:
+        \begin{equation*}
             \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}\;\sum_{\mathbf{v}\in \{0,1\}^n:|\mathbf{v}|=t}(-1)^{(\textbf{w} \oplus \textbf{w}')^T\mathbf{v}} \geq 0.
-        \end{equation}
+        \end{equation*}
 
         At this point, let us fix $\mathbf{u} = \textbf{w} \oplus \textbf{w}'$ and re-write our first inequality as:
         \begin{equation}
@@ -457,30 +463,30 @@ \label{s-u-inequality}
             \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}S(\mathbf{u}) \geq 0,
         \end{equation}
         where:
-        \begin{equation}
+        \begin{equation*}
             S(\mathbf{u})=\sum_{\mathbf{v}\in \{0,1\}^n:|\mathbf{v}|=t}(-1)^{\mathbf{u}^T\mathbf{v}}.
-        \end{equation}
+        \end{equation*}
         
 
-        In the sum Of $S(\mathbf{u})$, the $\mathbf{v}$ with $\mathbf{u}^T\mathbf{v}=j$ are counted with sign $(-1)^j$. We can count the number of vectors $\mathbf{v}$ of weight $t$ and with $\mathbf{u}^T\mathbf{v}=j$.
+        In the sum of $S(\mathbf{u})$, the $\mathbf{v}$ vectors with $\mathbf{u}^T\mathbf{v}=j$ are counted with sign $(-1)^j$. We can count the number of vectors $\mathbf{v}$ of weight $t$ and with $\mathbf{u}^T\mathbf{v}=j$.
 
-        By definition, $|\mathbf{u}|=d_H(\mathbf{w}, \mathbf{w}')$ is the number of 1's in $\mathbf{u}$. To obtain a valid vector $\mathbf{v}$, we need to put $j$ 1's in positions where $\mathbf{u}$ has 0's.
+        By definition, $|\mathbf{u}|=d_H(\mathbf{w}, \mathbf{w}') = i$ is the number of 1's in $\mathbf{u}$. To obtain a valid vector $\mathbf{v}$, we need to put $j$ 1's in positions where $\mathbf{u}$ has 0's.
         Hence, the number of these $\mathbf{v}$ is ${i \choose j}{n-i \choose t-j}$. Then,
 
-        \begin{equation}
+        \begin{equation*}
             S(\mathbf{u})=\sum_{j=0}^{\min(i,t)}(-1)^j{i \choose j}{n-i \choose t-j},
-        \end{equation}
+        \end{equation*}
         which is the Krawtchouk polynomial of degree $t$: $K_t(n,i)$.
         
         We can now go back to (\ref{s-u-inequality}), which becomes:
-        \begin{equation}
+        \begin{equation*}
             \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}S(\mathbf{u}) = \sum_{(\mathbf{w}, \mathbf{w}')\in\mathcal{C}^2}K_t(n,d_H(\mathbf{w}, \mathbf{w}')) \geq 0.
-        \end{equation}
+        \end{equation*}
         For each $i$, $K_t(n,i)$ appears in this sum $|\mathcal{C}|\cdot \tilde{x}_i$ times, because each $\tilde{x}_i$ is the number of couples $(\mathbf{w}, \mathbf{w}')$ with distance $i$. 
 
-        We obtain the thesis by changing the summation to reflect our $\tilde{x}_i$ variables:
-        \begin{equation}
+        We obtain the thesis by changing the sum to include our $\tilde{x}_i$ variables:
+        \begin{equation*}
             \sum_{i=0}^{n}K_t(n,i)\cdot\tilde{x}_i \geq 0.
-        \end{equation}
+        \end{equation*}
     \end{proof}
 \end{lemma}