fix remaining refs

exams/least-common-ancestor/index.md

@@ -20,7 +20,7 @@ To find the least common ancestor of two nodes $x$ and $y$ in a tree $t$, we fol
 3. consider the common prefix of $p_x$ and $p_y$, the last node in the common prefix is the least
    common ancestor. 
 
-Consider, for example, the binary tree depicted in Figure~\ref{fig:tree}. The least common ancestor 
+Consider, for example, the binary tree depicted below. The least common ancestor 
 of $3$ and $5$ is $2$. Indeed, the path from the root $1$ to $3$ is $1,2,3$. The path from $1$
 to $5$ is $1,2,4,5$. Their common prefix is $1,2$ whose last element is $2$.
 

exams/unit-propagation/index.md

@@ -17,7 +17,7 @@ $$
 $$
 is represented as 
 $$
-  \varphi=\{\{a,b,\neg c,\neg f\}, \{b,c\}, \{\neg b, e\}, \{\neg b\}\}.
+  \varphi=\{\{a,b,\neg c,\neg f\}, \{b,c\}, \{\neg b, e\}, \{\neg b\}\}. \qquad (1)
 $$
 
 One of the subroutines of the DPLL algorithm is the unit propagation simplifying the input formula.
@@ -29,7 +29,7 @@ $c_k=\{u\}$ for some $k$ and literal $u$, then $\varphi$ can be simplified by th
 1. if $u\in c_i$, then $c_i$ can be removed from $\varphi$,
 2. if $\neg u\in c_i$, then $\neg u$ can be removed from $c_i$.
 
-For example, the formula $\varphi$ in~(\ref{ex1}) has a unit $\{\neg b\}$, so we can simplify to
+For example, the formula $\varphi$ in $(1)$ has a unit $\{\neg b\}$, so we can simplify to
 $\{\{a,\neg c,\neg f\}, \{c\}\}$. Note that by propagating the unit, a new unit was created. Thus we
 can continue and propagate the unit $\{c\}$ obtaining $\{\{a,\neg f\}\}$. The resulting set of
 clauses has no unit.

homework/hw01.md

@@ -202,9 +202,9 @@ $\lfloor{i}\rfloor$ is that $i$ might be slightly larger than $255$ due to round
 
 The function returned by `ascii-art` should split the matrix of intensities into blocks. The size of
 blocks is given by the arguments `width` and `height` The separation of the matrix into blocks is
-depicted in Figure~\ref{fig:separation}. In case the width (resp. height) of the matrix is not
+depicted below. In case the width (resp. height) of the matrix is not
 divisible by `width` (resp. `height`) the incomplete blocks have to be removed from the matrix as is
-illustrated by the red area in the figure below:
+illustrated by the red area:
 
 <img src="/img/blocks.svg" style="width: 50%; margin-left: auto; margin-right: auto;" />