freeCodeCamp/curriculum/challenges/english/10-coding-interview-prep/project-euler/problem-460-an-ant-on-the-move.md

id, title, challengeType, forumTopicId, dashedName

id	title	challengeType	forumTopicId	dashedName
5900f5381000cf542c51004b	Problem 460: An ant on the move	5	302135	problem-460-an-ant-on-the-move

--description--

On the Euclidean plane, an ant travels from point A(0, 1) to point B(d, 1) for an integer d.

In each step, the ant at point (x_0, y_0) chooses one of the lattice points (x_1, y_1) which satisfy x_1 ≥ 0 and y_1 ≥ 1 and goes straight to (x_1, y_1) at a constant velocity v. The value of v depends on y_0 and y_1 as follows:

• If y_0 = y_1, the value of v equals y_0.
• If y_0 ≠ y_1, the value of v equals \frac{y_1 - y_0}{\ln y_1 - \ln y_0}.

The left image is one of the possible paths for d = 4. First the ant goes from A(0, 1) to P_1(1, 3) at velocity \frac{3 - 1}{\ln 3 - \ln 1} ≈ 1.8205. Then the required time is \frac{\sqrt{5}}{1.820} ≈ 1.2283.

From P_1(1, 3) to P_2(3, 3) the ant travels at velocity 3 so the required time is \frac{2}{3} ≈ 0.6667. From P_2(3, 3) to B(4, 1) the ant travels at velocity \frac{1 - 3}{\ln 1 - \ln 3} ≈ 1.8205 so the required time is \frac{\sqrt{5}}{1.8205} ≈ 1.2283.

Thus the total required time is 1.2283 + 0.6667 + 1.2283 = 3.1233.

The right image is another path. The total required time is calculated as 0.98026 + 1 + 0.98026 = 2.96052. It can be shown that this is the quickest path for d = 4.

Let F(d) be the total required time if the ant chooses the quickest path. For example, F(4) ≈ 2.960\\,516\\,287. We can verify that F(10) ≈ 4.668\\,187\\,834 and F(100) ≈ 9.217\\,221\\,972.

Find F(10\\,000). Give your answer rounded to nine decimal places.

--hints--

antOnTheMove() should return 18.420738199.

assert.strictEqual(antOnTheMove(), 18.420738199);

--seed--

--seed-contents--

function antOnTheMove() {

  return true;
}

antOnTheMove();

--solutions--

// solution required