Detailed mathematical approach

Problem Summary

Stone Game IV studies a one-heap take-away game controlled by a real parameter \(r\). On the opening turn, a player may remove any positive number of stones except the whole heap. After a move of size \(x\), the next player may remove at most \(r x\) stones. For each fixed \(r\), some starting heap sizes are losing positions, meaning the player to move has no winning strategy. These losing positions form an increasing sequence, and that sequence changes only when \(r\) crosses certain rational thresholds. The problem asks for the \(123456\)th value in the increasing threshold sequence obtained by starting from \(r_1=1\) and repeatedly jumping to the next larger transition.

Mathematical Approach

The implementation does not search whole game trees. Instead it tracks how the losing-position sequence depends on \(r\), then extracts the next threshold where that sequence must change.

Step 1: Define the Smallest Winning Opening Move

For a fixed \(r\), let \(w_r(n)\) be the smallest opening move that wins from a heap of size \(n\). If no winning opening move exists, define \(w_r(n)=n\). With this convention, \(n\) is a losing position exactly when

$w_r(n)=n.$

The recursion comes directly from the move rule. If the current player removes \(k\) stones, the opponent faces \(n-k\) stones and may remove at most \(r k\). That move is winning exactly when the opponent has no winning reply of size at most \(r k\). Since the opponent's smallest winning reply from \(n-k\) stones is \(w_r(n-k)\), we obtain

$w_r(1)=1,\qquad w_r(n)=\min\left\{1\le k\le n:\; r k<w_r(n-k)\right\}\quad (n\ge 2).$

The value \(k=n\) is a sentinel: it means every legal opening move \(1\le k\le n-1\) fails, so \(n\) is losing.

Step 2: Extract the Losing Positions

Let

$L_1=1<L_2=2<L_3<L_4<\cdots$

be the increasing sequence of losing starting heaps. If a move of size \(k\) leaves \(L_j\) stones, then that move is winning precisely when

$r k<L_j,$

because \(L_j\) itself has no winning reply smaller than \(L_j\). Therefore the profitable gaps after \(L_j\) are exactly the earlier losing positions whose size is still strictly below \(L_j/r\).

Step 3: Recurrence for the Losing Sequence

Suppose \(L_1,\dots,L_{i-1}\) are already known. Define \(d_i\) as the largest lag such that

$\frac{L_{i-1}}{L_{i-d_i}}\le r.$

Equivalently, \(L_{i-d_i}\) is the smallest earlier losing position that is at least \(L_{i-1}/r\). Then the next losing position is

$L_i=L_{i-1}+L_{i-d_i}.$

Why does this work? Every smaller earlier losing position \(L_t\) with \(L_t<L_{i-1}/r\) gives a winning move from \(L_{i-1}+L_t\): remove \(L_t\) stones and leave the losing heap \(L_{i-1}\), while the reply bound \(rL_t\) is still too small to unlock a winning response. The first earlier losing position that fails this inequality is exactly the gap where the winning block ends, so adding that gap produces the next losing heap.

Step 4: Critical Transition Fractions

For a fixed \(r\), the lag \(d_i\) stays unchanged until one more earlier losing position becomes admissible. The critical value for step \(i\) is therefore

$c_i=\frac{L_{i-1}}{L_{i-(d_i+1)}},$

whenever \(d_i<i-1\). Crossing \(c_i\) makes the next smaller denominator available, so the recurrence changes. Hence the next global transition above \(r\) is

$T(r)=\min\left\{c_i:\; d_i<i-1,\ c_i>r\right\}.$

This is the key observation behind the solver: build the losing sequence for the current threshold, record every candidate fraction just above \(r\), and keep the smallest one.

Worked Example: \(r=2\)

Start with \(L_1=1\) and \(L_2=2\).

For \(L_3\), the largest admissible lag is \(d_3=2\) because

$\frac{L_2}{L_1}=\frac{2}{1}=2\le 2.$

So

$L_3=L_2+L_1=3.$

For \(L_4\), we have

$\frac{L_3}{L_2}=\frac{3}{2}\le 2,\qquad \frac{L_3}{L_1}=3>2,$

so \(d_4=2\) and

$L_4=L_3+L_2=5.$

Next,

$\frac{L_4}{L_3}=\frac{5}{3}\le 2,\qquad \frac{L_4}{L_2}=\frac{5}{2}>2,$

which gives

$L_5=L_4+L_3=8.$

Continuing in the same way yields

$1,2,3,5,8,13,\dots,$

so the losing positions are the Fibonacci numbers. The transition candidates visible in these first steps are

$c_4=3,\qquad c_5=\frac{5}{2},\qquad c_6=\frac{8}{3},\dots$

The smallest candidate strictly above \(2\) is \(5/2\), therefore

$T(2)=\frac{5}{2}.$

Likewise, when \(r=1\), the recurrence doubles every term and the losing positions become \(1,2,4,8,\dots\), so the first transition is \(T(1)=2\).

How the Code Works

The implementation stores the current threshold as an exact reduced fraction. For one transition evaluation, it generates the losing-position sequence only up to a finite horizon \(M\). A monotone lag pointer is advanced while the inequality \(L_{i-1}/L_{i-d}\le r\) remains true, so each new term is produced in amortized constant time. At the same moment, the implementation also forms the critical fraction \(L_{i-1}/L_{i-(d+1)}\) and keeps the smallest candidate that is strictly larger than the current threshold.

All fraction comparisons are done by exact cross-multiplication, avoiding floating-point error. Sequence growth is protected by overflow checks before every addition. Because the decisive transition might occur beyond the initial horizon, the implementation repeats the computation with larger and larger horizons until the same best candidate reappears far enough before the end of the computed region. That stabilization test makes the returned fraction reliable without pretending that a short finite prefix is automatically sufficient.

Finally, the transition map is iterated from

$r_1=1,\qquad r_{m+1}=T(r_m),$

until \(m=123456\). The C++, Python, and Java solutions all expose this same exact recurrence-based computation, and built-in checkpoints verify early values such as \(r_2=2\) and \(r_{22}=145/23\).

Complexity Analysis

For a fixed horizon \(M\), one transition evaluation performs \(O(M)\) arithmetic operations and stores \(O(M)\) sequence values. The lag pointer only moves forward, so the inner search is amortized linear rather than quadratic. If \(M_{\mathrm{eff}}(r)\) is the final stabilized horizon needed for threshold \(r\), then a single transition costs \(O(M_{\mathrm{eff}}(r))\) time and \(O(M_{\mathrm{eff}}(r))\) memory. Reaching the target index therefore costs

$O\left(\sum_{m=1}^{123455} M_{\mathrm{eff}}(r_m)\right)$

time and

$O\left(\max_m M_{\mathrm{eff}}(r_m)\right)$

memory. In practice, the method is efficient because it follows only the recurrence of losing positions and the exact transition fractions, not the full game tree.

Footnotes and References

1. Problem page: https://projecteuler.net/problem=870
2. Combinatorial game theory: Wikipedia - Combinatorial game theory
3. Fibonacci Nim: Wikipedia - Fibonacci Nim
4. Rational number: Wikipedia - Rational number
5. Recurrence relation: Wikipedia - Recurrence relation