Merge pull request #803 from jeqcho/master

Fix typo on tuple encoding

Author: boazbk

lec_02_representation.md

@@ -682,7 +682,7 @@ The following theorem shows that if $E$ is a prefix-free encoding of $\mathcal{O
 
 ::: {.theorem title="Prefix-free implies tuple encoding" #prefixfreethm}
 Suppose that $E:\mathcal{O} \rightarrow \{0,1\}^*$ is prefix-free.
-Then the following map $\overline{E}:\mathcal{O}^* \rightarrow \{0,1\}^*$ is one to one, for every $o_0,\ldots,o_{k-1} \in \mathcal{O}^*$, we define
+Then the following map $\overline{E}:\mathcal{O}^* \rightarrow \{0,1\}^*$ is one to one, for every $(o_0,\ldots,o_{k-1}) \in \mathcal{O}^*$, we define
 $$
 \overline{E}(o_0,\ldots,o_{k-1}) = E(o_0)E(o_1) \cdots E(o_{k-1}) \;.
 $$

lec_04_code_and_data.md

@@ -229,7 +229,7 @@ f_i(x) = \begin{cases} b & x=x^* \\ f_{i-1}(x) & x \neq x^*
 $$
 or in other words
 $$
-f_i(x) = f_{i-1}(x) \wedge \neg EQUAL(x^*,x) \; \vee \;  b \wedge EQUAL(x^*,x)
+f_i(x) = IF(EQUAL(x^*,x),b,f_{i-1}(x))
 $$
 where $EQUAL:\{0,1\}^{2n} \rightarrow \{0,1\}$ is the function that maps $x,x' \in \{0,1\}^n$ to $1$ if they are equal and to $0$ otherwise.
 Since (by our choice of $i$), $f_{i-1}$ can be computed using at most $s$ gates and (as can be easily verified) that $EQUAL \in SIZE_n(9n)$,
@@ -539,7 +539,7 @@ Please make sure that you understand why `GET` and $LOOKUP_\ell$ are the same fu
 We saw that we can compute $LOOKUP_\ell$  in time $O(2^\ell) =  O(s)$ for our choice of $\ell$.
 
 For every $\ell$, let $UPDATE_\ell:\{0,1\}^{2^\ell + \ell +1} \rightarrow \{0,1\}^{2^\ell}$ correspond to the  `UPDATE` function for arrays of length $2^\ell$.
-That is,  on input $V\in \{0,1\}^{2^\ell}$ , $i\in \{0,1\}^\ell$, $b\in \{0,1\}$, $UPDATE_\ell(V,b,i)$ is equal to $V' \in \{0,1\}^{2^\ell}$ such that
+That is,  on input $V\in \{0,1\}^{2^\ell}$ , $i\in \{0,1\}^\ell$, $b\in \{0,1\}$, $UPDATE_\ell(V,i,b)$ is equal to $V' \in \{0,1\}^{2^\ell}$ such that
 $$
 V'_j = \begin{cases} V_j & j \neq i \\ b & j = i \end{cases}
 $$

lec_06_loops.md

@@ -146,7 +146,7 @@ We describe the operation of our Turing machine $M$ in words:
 
 * $M$ starts in state `START` and goes right, looking for the first symbol that is $0$ or $1$. If it finds $\varnothing$ before it hits such a symbol then it moves to the `OUTPUT_1` state described below.
 
-* Once $M$ finds such a symbol $b \in \{0,1\}$, $M$ deletes $b$ from the tape by writing the $\times$ symbol, it enters either the `RIGHT_0` or `RIGHT_1` mode according to the value of $b$ and starts moving rightwards until it hits the first $\varnothing$ or $\times$ symbol.
+* Once $M$ finds such a symbol $b \in \{0,1\}$, $M$ deletes $b$ from the tape by writing the $\times$ symbol, it enters either the `RIGHT_`$b$ mode and starts moving rightwards until it hits the first $\varnothing$ or $\times$ symbol.
 
 * Once $M$ finds this symbol, it goes into the state `LOOK_FOR_0` or `LOOK_FOR_1` depending on whether it was in the state `RIGHT_0` or `RIGHT_1` and makes one left move.
 
@@ -169,7 +169,7 @@ The above description can be turned into a table describing for each one of the
 The formal definition of Turing machines is as follows:
 
 :::  {.definition title="Turing Machine" #TM-def}
-A (one tape) _Turing machine_ with $k$ states and alphabet $\Sigma \supseteq \{0,1, \triangleright, \varnothing \}$ is represented by a _transition function_
+A (one tape) _Turing machine_ $M$ with $k$ states and alphabet $\Sigma \supseteq \{0,1, \triangleright, \varnothing \}$ is represented by a _transition function_
 $\delta_M:[k]\times \Sigma \rightarrow [k] \times \Sigma  \times \{\mathsf{L},\mathsf{R}, \mathsf{S}, \mathsf{H} \}$.
 
 For every $x\in \{0,1\}^*$, the _output_ of $M$ on input $x$, denoted by $M(x)$, is the result of the following process: