Skip to content

Latest commit

 

History

History
265 lines (211 loc) · 13.8 KB

12b.md

File metadata and controls

265 lines (211 loc) · 13.8 KB
Raw

Day 12: N-Body Problem - Code

The space near Jupiter is not a very safe place; you need to be careful of a big distracting red spot, extreme radiation, and a whole lot of moons swirling around. You decide to start by tracking the four largest moons: Io, Europa, Ganymede, and Callisto.

After a brief scan, you calculate the position of each moon (your puzzle input). You just need to simulate their motion so you can avoid them.

Each moon has a 3-dimensional position (x, y, and z) and a 3-dimensional velocity. The position of each moon is given in your scan; the x, y, and z velocity of each moon starts at 0.

Simulate the motion of the moons in time steps. Within each time step,

  1. update the velocity of every moon by applying gravity. Then, once all moons' velocities have been updated,
  2. update the position of every moon by applying velocity.
  3. Time progresses by one step once all of the positions are updated.

To apply gravity, consider every pair of moons. On each axis (x, y, and z), the velocity of each moon changes by exactly +1 or -1 to pull the moons together. For example, if Ganymede has an x position of 3, and Callisto has a x position of 5, then Ganymede's x velocity changes by +1 (because 5 > 3) and Callisto's x velocity changes by -1 (because 3 < 5). However, if the positions on a given axis are the same, the velocity on that axis does not change for that pair of moons.

Once all gravity has been applied, apply velocity: simply add the velocity of each moon to its own position. For example, if Europa has a position of x=1, y=2, z=3 and a velocity of x=-2, y=0,z=3, then its new position would be x=-1, y=2, z=6. This process does not modify the velocity of any moon.

For example, suppose your scan reveals the following positions:

<x=-1, y=0, z=2>
<x=2, y=-10, z=-7>
<x=4, y=-8, z=8>
<x=3, y=5, z=-1>

Simulating the motion of these moons would produce the following:

After 0 steps:
pos=<x=-1, y=  0, z= 2>, vel=<x= 0, y= 0, z= 0>
pos=<x= 2, y=-10, z=-7>, vel=<x= 0, y= 0, z= 0>
pos=<x= 4, y= -8, z= 8>, vel=<x= 0, y= 0, z= 0>
pos=<x= 3, y=  5, z=-1>, vel=<x= 0, y= 0, z= 0>

After 1 step:
pos=<x= 2, y=-1, z= 1>, vel=<x= 3, y=-1, z=-1>
pos=<x= 3, y=-7, z=-4>, vel=<x= 1, y= 3, z= 3>
pos=<x= 1, y=-7, z= 5>, vel=<x=-3, y= 1, z=-3>
pos=<x= 2, y= 2, z= 0>, vel=<x=-1, y=-3, z= 1>

After 2 steps:
pos=<x= 5, y=-3, z=-1>, vel=<x= 3, y=-2, z=-2>
pos=<x= 1, y=-2, z= 2>, vel=<x=-2, y= 5, z= 6>
pos=<x= 1, y=-4, z=-1>, vel=<x= 0, y= 3, z=-6>
pos=<x= 1, y=-4, z= 2>, vel=<x=-1, y=-6, z= 2>

After 3 steps:
pos=<x= 5, y=-6, z=-1>, vel=<x= 0, y=-3, z= 0>
pos=<x= 0, y= 0, z= 6>, vel=<x=-1, y= 2, z= 4>
pos=<x= 2, y= 1, z=-5>, vel=<x= 1, y= 5, z=-4>
pos=<x= 1, y=-8, z= 2>, vel=<x= 0, y=-4, z= 0>

After 4 steps:
pos=<x= 2, y=-8, z= 0>, vel=<x=-3, y=-2, z= 1>
pos=<x= 2, y= 1, z= 7>, vel=<x= 2, y= 1, z= 1>
pos=<x= 2, y= 3, z=-6>, vel=<x= 0, y= 2, z=-1>
pos=<x= 2, y=-9, z= 1>, vel=<x= 1, y=-1, z=-1>

After 5 steps:
pos=<x=-1, y=-9, z= 2>, vel=<x=-3, y=-1, z= 2>
pos=<x= 4, y= 1, z= 5>, vel=<x= 2, y= 0, z=-2>
pos=<x= 2, y= 2, z=-4>, vel=<x= 0, y=-1, z= 2>
pos=<x= 3, y=-7, z=-1>, vel=<x= 1, y= 2, z=-2>

After 6 steps:
pos=<x=-1, y=-7, z= 3>, vel=<x= 0, y= 2, z= 1>
pos=<x= 3, y= 0, z= 0>, vel=<x=-1, y=-1, z=-5>
pos=<x= 3, y=-2, z= 1>, vel=<x= 1, y=-4, z= 5>
pos=<x= 3, y=-4, z=-2>, vel=<x= 0, y= 3, z=-1>

After 7 steps:
pos=<x= 2, y=-2, z= 1>, vel=<x= 3, y= 5, z=-2>
pos=<x= 1, y=-4, z=-4>, vel=<x=-2, y=-4, z=-4>
pos=<x= 3, y=-7, z= 5>, vel=<x= 0, y=-5, z= 4>
pos=<x= 2, y= 0, z= 0>, vel=<x=-1, y= 4, z= 2>

After 8 steps:
pos=<x= 5, y= 2, z=-2>, vel=<x= 3, y= 4, z=-3>
pos=<x= 2, y=-7, z=-5>, vel=<x= 1, y=-3, z=-1>
pos=<x= 0, y=-9, z= 6>, vel=<x=-3, y=-2, z= 1>
pos=<x= 1, y= 1, z= 3>, vel=<x=-1, y= 1, z= 3>

After 9 steps:
pos=<x= 5, y= 3, z=-4>, vel=<x= 0, y= 1, z=-2>
pos=<x= 2, y=-9, z=-3>, vel=<x= 0, y=-2, z= 2>
pos=<x= 0, y=-8, z= 4>, vel=<x= 0, y= 1, z=-2>
pos=<x= 1, y= 1, z= 5>, vel=<x= 0, y= 0, z= 2>

After 10 steps:
pos=<x= 2, y= 1, z=-3>, vel=<x=-3, y=-2, z= 1>
pos=<x= 1, y=-8, z= 0>, vel=<x=-1, y= 1, z= 3>
pos=<x= 3, y=-6, z= 1>, vel=<x= 3, y= 2, z=-3>
pos=<x= 2, y= 0, z= 4>, vel=<x= 1, y=-1, z=-1>

Then, it might help to calculate the total energy in the system. The total energy for a single moon is its potential energy multiplied by its kinetic energy. A moon's potential energy is the sum of the absolute values of its x, y, and z position coordinates. A moon's kinetic energy is the sum of the absolute values of its velocity coordinates. Below, each line shows the calculations for a moon's potential energy (pot), kinetic energy (kin), and total energy:

Energy after 10 steps:
pot: 2 + 1 + 3 =  6;   kin: 3 + 2 + 1 = 6;   total:  6 * 6 = 36
pot: 1 + 8 + 0 =  9;   kin: 1 + 1 + 3 = 5;   total:  9 * 5 = 45
pot: 3 + 6 + 1 = 10;   kin: 3 + 2 + 3 = 8;   total: 10 * 8 = 80
pot: 2 + 0 + 4 =  6;   kin: 1 + 1 + 1 = 3;   total:  6 * 3 = 18
Sum of total energy: 36 + 45 + 80 + 18 = 179

In the above example, adding together the total energy for all moons after 10 steps produces the total energy in the system, 179.

What is the total energy in the system after simulating the moons given in your scan for 1000 steps?

Your puzzle answer was 6678.

Part 2 - Code

All this drifting around in space makes you wonder about the nature of the universe. Does history really repeat itself? You're curious whether the moons will ever return to a previous state.

Determine the number of steps that must occur before all of the moons' positions and velocities exactly match a previous point in time.

For example, the first example above takes 2772 steps before they exactly match a previous point in time; it eventually returns to the initial state:

After 0 steps:
pos=<x= -1, y=  0, z=  2>, vel=<x=  0, y=  0, z=  0>
pos=<x=  2, y=-10, z= -7>, vel=<x=  0, y=  0, z=  0>
pos=<x=  4, y= -8, z=  8>, vel=<x=  0, y=  0, z=  0>
pos=<x=  3, y=  5, z= -1>, vel=<x=  0, y=  0, z=  0>

After 2770 steps:
pos=<x=  2, y= -1, z=  1>, vel=<x= -3, y=  2, z=  2>
pos=<x=  3, y= -7, z= -4>, vel=<x=  2, y= -5, z= -6>
pos=<x=  1, y= -7, z=  5>, vel=<x=  0, y= -3, z=  6>
pos=<x=  2, y=  2, z=  0>, vel=<x=  1, y=  6, z= -2>

After 2771 steps:
pos=<x= -1, y=  0, z=  2>, vel=<x= -3, y=  1, z=  1>
pos=<x=  2, y=-10, z= -7>, vel=<x= -1, y= -3, z= -3>
pos=<x=  4, y= -8, z=  8>, vel=<x=  3, y= -1, z=  3>
pos=<x=  3, y=  5, z= -1>, vel=<x=  1, y=  3, z= -1>

After 2772 steps:
pos=<x= -1, y=  0, z=  2>, vel=<x=  0, y=  0, z=  0>
pos=<x=  2, y=-10, z= -7>, vel=<x=  0, y=  0, z=  0>
pos=<x=  4, y= -8, z=  8>, vel=<x=  0, y=  0, z=  0>
pos=<x=  3, y=  5, z= -1>, vel=<x=  0, y=  0, z=  0>

Of course, the universe might last for a very long time before repeating. Here's a copy of the second example from above:

<x=-8, y=-10, z=0>
<x=5, y=5, z=10>
<x=2, y=-7, z=3>
<x=9, y=-8, z=-3>

This set of initial positions takes 4686774924 steps before it repeats a previous state! Clearly, you might need to find a more efficient way to simulate the universe.

How many steps does it take to reach the first state that exactly matches a previous state?

Your puzzle answer was 496734501382552.

Solution

Two insights:

  1. Each directional component moves independently. x component has no influence on the y component and y has no influence on z.
  2. It's not possible for step=N @ v.x=5, p.x=10 to repeat but not step=0 @ v.x=0, p.x=-1. The entire cycle must repeat. Thus, all we have to do is figure out when the state @ step=0 appears again. Why? Check out this proof by contradiction:

This, it must be the case that step=0 @ v.x=0, p.x=-1 repeats.

We can now solve this problem like so:

  1. Figure out what steps and happens again. Call this .
  2. Figure out what steps and happens again. Call this .
  3. Figure out what steps and happens again. Call this .

Then figure out the total step by doing , right? No, but super close!

Imagine it's a 2D plane, and . If we just multiply them together, we'd get . But there's a smaller that satisfies both , and that's . Both and divide into . We get that by getting the Least Common Multiple (LCM) of :

export interface IPoint3D {
    x: number;
    y: number;
    z: number;
}

export const stepsForComponent = (moons: IPoint3D[], c: (moon: IPoint3D) => number) => {
    const ms = moons.map(c);
    const v = Array.from({length: moons.length}, () => 0);
    let steps = 0;
    while (steps++ < Number.MAX_SAFE_INTEGER) {
        // velocities
        for (let i = 0; i < moons.length; i++) {
            for (let j = i + 1; j < moons.length; j++) {
                const delta = cDelta(ms[i], ms[j]);
                v[i] += delta;
                v[j] -= delta;
            }
        }

        // distance
        for (let i = 0; i < moons.length; i++)
            ms[i] += v[i];

        if (v.every(vi => vi === 0) && ms.every((m, i) => c(moons[i]) === m))
            return steps;
    }
    return Infinity;
};

export const getStepsForCircle = (moons: IPoint3D[]) => lcm(
    stepsForComponent(moons, m => m.x),
    stepsForComponent(moons, m => m.y),
    stepsForComponent(moons, m => m.z)
);

export const run = () => {
    const sims = getSimsFromFile();
    for (const s of sims) {
        console.log(timer.start(`12b - ${s.name}`));
        console.log(getStepsForCircle(s.moons));
        console.log(timer.stop());
    }
};

Appendix

Calculating LCM

LCM of two numbers a and b can be defined as:

E.g.,

LCM(6, 10) = 6 * 10 / gcd(6, 10) = 60 / 2 = 30

For three numbers,

E.g.,

LCM(6, 10, 12)
    = LCM(6, 10) * 12 / gcd(LCM(6, 10), 12)
    = 30 * 12 / gcd(30, 12)
    = 30 * 12 / 6
    = 60

export const getgcd = (a: number, b: number): number => {
    if (a === 0)
        return b;
    if (b === 0)
        return a;
    const numerator = Math.max(a, b);
    const denominator = Math.min(a, b);
    const remainder = numerator % denominator;

    return remainder === 0 ? denominator : getgcd(denominator, remainder);
};

export const lcm = (...n: number[]) => {
    return n.reduce((a, c) => a * c / getgcd(a, c), 1);
};

console.log(lcm(6, 6, 6)); // = 6