Construct homotopy from $(\alpha \cdot \beta) \cdot \gamma$ to $\alpha \cdot (\beta \cdot \gamma)$ explicitly I understand the general idea of a homotopy, but I'm a little lost on how to create them myself. For example if I wanted to show 
$$ \text{Let} \: \alpha, \beta, \text{and} \: \gamma \: \text{be paths} \: I \to X, \: \text{from} \: x_{0} \: \text{to} \: y_{0}, y_{0} \: \text{to} \: z_{0}, \: \text{and}  \: z_{0} \: \text{to} \: u_{0}. \: \text{Then} \: \\
(\alpha \cdot \beta) \cdot \gamma \sim \alpha \cdot (\beta \cdot \gamma) $$
A possible homotopy is $F: I \times I \to X$, given by 
$$\\ F(t,s) = 
\begin{cases}
\alpha(\frac{4t}{1+s}) & 0 \leq t \leq \frac{s+1}{4} \\
\beta(4t-1-s) & \frac{s+1}{4} \leq t \leq \frac{s+2}{4} \\
\gamma(\frac{4t - 2 - s}{2-s}) & \frac{s+2}{4} \leq t \leq 1 \\
\end{cases}$$
What I don't understand is where this comes from. What is the intuition here and how can I form explicit homotopies like this?

See the picture below.

The idea is that, for any choice of $s$, the loop $t \mapsto F(t, s)$ consists of walking along $\alpha$, $\beta$ and $\gamma$, in that order. The difference between the various choices of $s$ is in the "time schedule": each choice of $s$ allocates different amounts of time to $\alpha$, $\beta$ and $\gamma$. 


*

*If $s = 0$, you walk path $\alpha$ in time $[0, \tfrac 1 4]$, then walk $\beta$ in time $[\tfrac 1 4, \tfrac 1 2 ]$, then walk $\gamma$ in time $[\tfrac 1 2 , 1]$.

*If $s = 1$, you walk path $\alpha$ in time $[0, \tfrac 1 2]$, then walk $\beta$ in time $[\tfrac 1 2, \tfrac 3 4]$, then walk $\gamma$ in time $[\tfrac 3 4 , 1]$.
For intermediate choices of $s$, the schedule is given by interpolation.
So for example, if $s = \tfrac 1 2$, you walk $\alpha$ in time $[0, \tfrac 3 8]$, then walk $\beta$ in time $[\tfrac 3 8, \tfrac 5 8]$, then walk $\gamma$ in time $[\tfrac 5 8, 1]$. And so on.
