Of Balloons and Bumps; or, some very angry Balloonists - A quest of the 14th Anniversary

[Aia del Mana]

For the Anniversary celebrations, Aelis the Sky Scouter did create of two thousand and nineteen Hot-Air Balloons, through those resources which he did scrounge from those new upon the Island of the Gateway. As a seasoned Sky-Scouter, he did plot of their course in a perfect, straight path toward Mur in the East, as a surprise to Mur, with much fanfare; and did launch of all of them from the peak of the mountain upon the Gateway, each following of the previous in perfect synchrony of speed.

In coincidence, and mayhap also in response to the many newlings that did complain of the broken balloon upon the Island of the Gateway, Mur did also create of two thousand and nineteen Hot-Air Balloons, and did plot their course toward the Island, that each should also travel in a perfect, straight path, at constant speed, from Mur toward the Island, such that they may be of service for those who did wish to leave the Island.

In further coincidence, this path were exactly that of Aelis, in reverse; Mur, and Aelis, did release these balloons in perfect synchrony - such were the workings of the Soldiers of the Inner Sun.

A curious incident, then, did occur within the skies - when any balloon did meet another balloon which were travelling in a direction opposite, they did collide, sending each directly away from each other, still traversing along this perfect path, at the same speed, yet in the opposite direction. A surprise, then, for Aelis, and Mur, who did each find each created balloon returned to themselves.

The balloonists did indeed experience of many collisions, in many bumpy rides, and did curse of their ill luck.

The first to message to myself upon the Forum a solution for the total number of collisions, and an explanation thereof, shall be awarded an anniversary creature. As this were a race - only the first correct answer shall win. Do assume that Chewett hath movelocked of Mur while any balloon were airborne.

Edited April 15, 2019 by Aia del Mana

[Aia del Mana]

Samon has, in utter expedience, solved of this quest, and earned of the anniversary creature.

I shall leave this quest open for those who wish to attempt it; a solution shall be posted upon the Anniversary close.

I shall be quite happy to verify any answer sent to myself upon the Forum, if one were so inclined to do so.

Edited April 15, 2019 by Aia del Mana

[Aia del Mana]

Samon has offered a joker to the next to solve of this quest; so it is open again to all, and I shall await further entries of those that wish to partake.

Of course, participants will also receive plushies, as with any other quest.

Edited April 15, 2019 by Aia del Mana

[Kaya]

I had fun puzzling this out, so would love to donate a Joker for the next person to solve it

[Aia del Mana]

Aelis is the second to answer correctly.

 

Plushies do remain for all who attempt this puzzle in good faith.

[Aia del Mana]

@fallen god (Demonic God within the realm) hath also submitted of a correct solution to this puzzle, and hath earned of plushies, which shall be granted upon the Anniversary close.

[Aia del Mana]

I hereby close of all entries to this quest.

Samon, Demonic God, and Aelis hath answered it correctly; Lazarus also hath submitted an entry, although it were an incorrect solution; all shall receive of five plushies.

Herewith do find Aelis's solution:
 

Quote

Let's consider the last balloon sent by either person. In order to return, this balloon only needs a single collision with the next one. However, the next one must be already heading in the opposite direction from a previous collision, so that will count as its second collision. Then again, the second balloon will now be last in line and will require an additional collision in order to return to its sender, that makes it three. The same reasoning folllows and we can observe that each following balloon will require 2 additional collisions.

Thus, let us we assign an index i to the balloon as we send them, starting with 2018 (2019 - 1) so that the last balloon has index i = 0, we know that each balloon will collide 2*i + 1 times (e.g. balloon 0 collides once, balloon 1 collides 3 times, etc). We should consider both my balloon's and Mur's, but since each collision involves two balloons we end up multiplying by 2 and dividing by 2, so, the total collision count is given by:

2*i + 1 ; for 0 <= i <= 2018, which yields 4076361

However, one should note that one may simply assume the balloons instead pass through each other, as the end result were still two balloons travelling in opposite directions. One may then simply multiply 2019 by 2019 for the desired answer of 4076361 collisions.