Detailed mathematical approach

Problem Summary

We have \(N\) identical balls moving in a one-dimensional frictionless tube of length \(L\). Their empty gaps and initial directions are generated deterministically from the recurrence

$r_1=6563116,\qquad r_{n+1}\equiv r_n^2 \pmod{32745673}.$

From each state we obtain a gap

$d_n=(r_n\bmod 1000)+1,$

and the direction is to the right when \(r_n\le 10^7\), otherwise to the left. The goal is not the whole collision history, but only the exit time of the ball that starts in position \(j\) from the left. The implementations avoid explicit collision simulation by turning the problem into a selection problem on a derived list of times.

Mathematical Approach

The key observation is that equal-mass elastic collisions in one dimension only exchange velocities. That lets us replace the real system by non-colliding ghost trajectories, then recover the required labeled ball through an order statistic.

Step 1: Build the Reduced Starting Coordinates

The recurrence gives one positive gap for each ball. Define the cumulative reduced position

$s_n=\sum_{m=1}^{n} d_m.$

This quantity is exactly what the implementations maintain while scanning the pseudo-random sequence. It is the natural coordinate after collapsing away the fixed diameter contribution of the balls already placed to the left, so

$s_1<s_2<\cdots<s_N.$

All geometry-dependent formulas in the algorithm depend only on these cumulative values, not on a step-by-step physical reconstruction of every collision.

Step 2: Replace Collisions by Ghost Trajectories

In one dimension, two identical balls undergoing an elastic collision simply exchange velocities. If we ignore labels, this is indistinguishable from the balls passing through each other. Therefore the multiset of unlabeled trajectories is the same as if no collisions happened at all.

So each ball can be treated as a ghost particle that moves in a straight line until it leaves the tube. The real system is recovered afterward by restoring labels according to left-to-right order. This is the standard label-exchange trick and is the reason the problem can be reduced to pure arithmetic.

Step 3: Derive the Candidate Exit Time for Each Ghost

For the queried ball index \(j\), the implementations use the fixed geometry term

$B_j=L-10-20(j-1).$

The subtraction of \(20(j-1)\) accounts for the diameters of the \(j-1\) balls initially to its left, and the remaining \(10\) is the edge-radius correction appearing in the tube geometry. Once the reduced coordinate \(s_n\) is known, the \(n\)-th ghost contributes the candidate time

$\tau_n=\begin{cases} B_j-s_n, & r_n\le 10^7,\\ B_j+s_n, & r_n>10^7. \end{cases}$

The sign is determined entirely by the initial direction. A right-moving ghost has less distance left when its reduced coordinate is larger, while a left-moving ghost contributes the symmetric expression with a plus sign. The C++, Python, and Java implementations compute exactly these values.

Step 4: Convert Label Tracking into an Order Statistic

Let the candidate times be sorted in nondecreasing order:

$\tau_{(1)}\le \tau_{(2)}\le \cdots \le \tau_{(N)}.$

A larger candidate means that the corresponding ghost trajectory remains inside the tube longer. Because real collisions only swap labels, the ball that starts in position \(j\) from the left follows the \(j\)-th longest surviving ghost schedule. Equivalently, its exit time is the \((N-j+1)\)-st smallest candidate:

$T_j=\tau_{(N-j+1)}.$

In zero-based array indexing this is exactly the element at position

$N-j.$

Step 5: Final Formula

The whole problem is therefore reduced to a deterministic recurrence plus one order statistic:

$\boxed{T_j=\operatorname{OS}_{N-j+1}\!\left(\left\{\tau_1,\tau_2,\dots,\tau_N\right\}\right),\qquad \tau_n=\begin{cases} L-10-20(j-1)-s_n, & r_n\le 10^7,\\ L-10-20(j-1)+s_n, & r_n>10^7. \end{cases}}$

Here \(\operatorname{OS}_{k}\) denotes the \(k\)-th smallest element. No collision tree, event queue, or pairwise interaction simulation is needed.

Worked Example: \((L,N,j)=(5000,3,2)\)

For this sample we have

$B_2=5000-10-20(2-1)=4970.$

The first three recurrence values generate the following data:

$\begin{aligned} r_1&=6563116,& d_1&=117,& s_1&=117,& r_1\le 10^7,& \tau_1&=4970-117=4853,\\ r_2&=14723431,& d_2&=432,& s_2&=549,& r_2>10^7,& \tau_2&=4970+549=5519,\\ r_3&=19804172,& d_3&=173,& s_3&=722,& r_3>10^7,& \tau_3&=4970+722=5692. \end{aligned}$

After sorting,

$4853<5519<5692.$

Because \(N-j=1\), we read the element at zero-based index \(1\), namely

$T_2=5519,$

which matches the sample check used by the implementation.

How the Code Works

The implementations iterate once through the recurrence, maintaining the running cumulative gap \(s_n\). For each ball they immediately convert the current state into one candidate value \(\tau_n\) using the sign rule above and store that value in an array or list.

After all \(N\) candidates are produced, the C++ implementation performs direct order-statistic selection, while the Python and Java implementations sort the full collection and read the element at index \(N-j\). In every language the mathematical object being computed is the same: the \((N-j+1)\)-st smallest candidate exit time.

Complexity Analysis

Generating the recurrence, cumulative gaps, and candidate times costs \(O(N)\) time. The C++ implementation then performs expected \(O(N)\) selection, so its overall expected running time is \(O(N)\). The Python and Java implementations sort all candidates, giving \(O(N\log N)\) total time. All three implementations use \(O(N)\) memory to store the candidate times.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=653
2. Elastic collision in one dimension: Wikipedia — Elastic collision
3. Order statistic: Wikipedia — Order statistic
4. Selection algorithms: Wikipedia — Selection algorithm
5. Modular arithmetic: Wikipedia — Modular arithmetic