2.1 KiB
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 ofv
equalsy_0
. - If
y_0 ≠ y_1
, the value ofv
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