def problem460():
    """


    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 (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 (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205.
       Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
       From P[1](1, 3) to P[2](3, 3) the ant travels at velocity 3 so the
       required time is 2 / 3 ≈ 0.6667. From P[2](3, 3) to B(4, 1) the ant
       travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required
       time is 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) ≈ 2.960516287.
       We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.

       Find F(10000). Give your answer rounded to nine decimal places.
    """

    pass

def problem460():
    """


    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 (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 (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205.
       Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.
       From P[1](1, 3) to P[2](3, 3) the ant travels at velocity 3 so the
       required time is 2 / 3 ≈ 0.6667. From P[2](3, 3) to B(4, 1) the ant
       travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required
       time is 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) ≈ 2.960516287.
       We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.

       Find F(10000). Give your answer rounded to nine decimal places.
    """

    pass

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 (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 (3 - 1) / (ln(3) - ln(1)) ≈ 1.8205.        Then the required time is sqrt(5) / 1.8205 ≈ 1.2283.        From P[1](1, 3) to P[2](3, 3) the ant travels at velocity 3 so the        required time is 2 / 3 ≈ 0.6667. From P[2](3, 3) to B(4, 1) the ant        travels at velocity (1 - 3) / (ln(1) - ln(3)) ≈ 1.8205 so the required        time is 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) ≈ 2.960516287.        We can verify that F(10) ≈ 4.668187834 and F(100) ≈ 9.217221972.         Find F(10000). Give your answer rounded to nine decimal places.