Constraints in Dynamic Time Warping for Speech

After some thought, I see that $Q_{max}$ and $Q_{min}$ are to be interpreted as slopes. Here is the diagram that shows the area bounded by 4 lines (when the inequalities become equal at extremes).

$Q_{max}$ and $Q_{min}$ are $1/tan$ in nature. This is because the increments $p_i$ and $q_i$ are along x-axis and y-axis respectively. Hence $Q_{max}$ is capturing average slope value along that path, but in inverse.

Remembered that any line takes the form

$$ y = m \cdot x + c $$

Why not reformulate it this way? (We can always adjust c' accordingly)

$$ y = m \cdot (x - c1) + c' $$

This should be useful when I already know for fact that the start point is $(1,1)$ and end is $(T_x,T_y)$.

Considering the first global path constraint.

$$ 1+\frac{[\phi _x(k) - 1]}{Q_{max}} \leq \phi _y(k) \leq 1 + Q_{max}\cdot [\phi _x(k) - 1] $$

At its extremes, we have two equations ; both sharing $(1,1)$

$$ 1+\frac{[\phi _x(k) - 1]}{Q_{max}} = \phi _y(k) $$

$$ \phi _y(k) = 1 + Q_{max}\cdot [\phi _x(k) - 1] $$

Given its slope and just one coordinate, the y-intercept is findable. For $(1,1)$ we verify its correctness by substituting $\phi _x(k) = 1$ and we indeed get $\phi _y(k)$ as 1.

I expected the second equation to be $$ \phi _y(k) = 1 + \frac {1}{Q_{min}} \cdot [\phi _x(k) - 1] $$

But it makes sense that $$ Q_{min} = \frac {1}{Q_{max}} $$

Similarly, the second global path constraint gives us another pair of lines; both sharing $(T_{x},T_{y})$

$$ T_y + Q{max}\cdot [\phi _x(k) - Tx] = \phi y(k) $$

$$ \phi y(k) = T_y+\frac{[\phi x(k) - T_x]}{Q{max}} $$

Finally, when we look at the path that is outside this region, we see that it starts with a high slope but then has to flatten out quick so as to meet $(T_{x},T_{y})$ in time. Its average slope could not be higher than $Q_{max}$. Such a path shape is also unreasonable.