Derive the Gibbs sampler for this bivariate distribution I understand the theory of Gibbs sampling. It is an iterative sampling algorithm that defines, sequence of random variables with the property of a Markov chain. Specifically, I choose any starting value, $(x_0, y_0)$ and discard the initial samples if my guess is poor (known as burn-in), then sample $x_1 \sim p(x|y_0)$ and $y_1 \sim p(y|x_1)$ for the first iteration. 
How do I derive the Gibbs sampler for the bivariate distribution with pdf:

$p(x,y) \propto \mathbb{1}(|x - y| < c) \mathbb{1}(x,y \in (0,1))$

In the examples I've seen online, like on page 3 of Casella's 1992 paper (http://biostat.jhsph.edu/~mmccall/articles/casella_1992.pdf), the conditional distributions can be written as a known distribution with varying parameters. In the bivariate pdf above, I don't see how the indicator functions can be interpreted as a known distribution.
Specifically, how do I solve for $p(x|y)$ and $p(y|x)$? 
My work thus far:
\begin{equation}
\begin{array}{lcl}
p(x,y) &\propto & p(x|y) \\[2ex]
&\propto & \mathbb{1}(-c < |x-y| < c) \mathbb{1}(x,y \in (0,1)) \\[2ex]
&\propto & \mathbb{1}(-c +y < x < c + y) \mathbb{1}(x \in (0,1)) \\[2ex]
p(x,y) &\propto & p(y|x) \\[2ex]
&\propto & \mathbb{1}(-c < |x-y| < c) \mathbb{1}(x,y \in (0,1)) \\[2ex]
&\propto & \mathbb{1}(-c - x < -y < c - x) \mathbb{1}(x,y \in (0,1)) \\[2ex]
&\propto & \mathbb{1}(c + x < y < -c + x) \mathbb{1}(y \in (0,1)) \\[2ex]
\end{array}
\end{equation}
 A: First, it is useful to understand what $p(x,y)$ is. The superposition of an area which is $1\times\text{constant}$ (the proportionality constant that makes everything integrate to 1) in every point where $x$ and $y$ are at distance $d<c$ (if you go through the $x$ axis, you draw a band from $x+c$ to $x-c$) and a square from $(0,0)$ to $(1,1)$.
If your imagination fails, there is Wolfram Alpha :)

To get $p(x | y)$ you must get the slice of the figure taken at the given $y$. You see it is not the same slice at every $y$?. That means that, of course, $p(x | y)$ should include $y$ somewhere. Otherwise you are saying that $x$ is independent of $y$, that is $p(x | y) = p(x)$, and $p(x,y) = p(x)p(y)$, which would be a super boring joint distribution and you wouldn't need Gibbs sampling anymore.
Note: on "translating" the indicator function to a known distribution. What Gibbs sampling does is to attack every dimension separately. You don't know how to deal with the ugly bi-dimensional distribution of the figure. But you know how to deal with its uni-dimensional slices. 
You just have to take the red pill and stop getting frightened by the bi-dimensional figure. We are doing Gibbs sampling, we live in 1-dimensional slices! ;)
