aoc2023day24.md

Let $p,v,t$ be the position, velocity and time unknowns such that for all $i$ we have a collision

$$p-p_i - t\times(v-v_i)=0.$$

To eliminate $t$ we wedge (exterior) product with $v-v_i$ to get

$$0=(p-p_i)\wedge (v-v_i)= p\wedge v - p \wedge v_i - p_i \wedge v + p_i \wedge v_i.$$

The nonlinear term $p\wedge v$ is constant for all $i$, so we subtract equations to get a linear equation

$$p \wedge (v_i-v_j) + (p_i - p_j) \wedge v - p_i \wedge v_i + p_j \wedge v_j=0.$$

Then wedge with $p_i - p_j$ to eliminate the $v$ unknowns

$$p \wedge (v_i-v_j) \wedge (p_i - p_j) + p_i \wedge v_i \wedge p_j + p_j \wedge v_j \wedge p_i =0.$$

This is an equation with three unknowns so pick three combinations for $i,j$ and solve the system

@abroy77 copy the second formula putting j instead of i, then subtract the original one with the i from it.

This is done to get rid of the nonlinear term $p\wedge v$ and land on a linear equation (with respect to $p$ that is what we are looking for).

Hi! sorry for another question:
Should the first formula be

p - p_i - t x (v_i - v) = 0

Or is the original order correct because of the cross product with 't'. I'm also confused why there's a cross product with 't' since 't' is a scalar?

@abroy77 I just mean multiplication, it's given the same symbol as the cross product.

Ambivalence like this is an antipattern since you have to know the arguments to deduce the operator.