Detailed mathematical approach

Problem Summary

There are \(m=c+1\) coins on a strip of \(n\) squares, and exactly one of them is the silver dollar. A legal move is either a regular left move of one coin or the special move of pocketing the leftmost coin. The first player wins iff they eventually pocket the silver dollar.

So we must count all configurations from which the first player can force that outcome. If we first choose the occupied squares and then choose which of the \(m\) coins is silver, the total number of raw configurations is

$T=m\binom{n}{m}.$

Mathematical Approach

Gap Model

Write the coin positions as

$0\le x_0 \lt x_1 \lt \cdots \lt x_{m-1}\le n-1.$

Instead of working with positions directly, the code switches to gaps:

$g_0=x_0,\qquad g_i=x_i-x_{i-1}-1\ (1\le i\le m-1),\qquad g_m=n-1-x_{m-1}.$

These are the empty squares before the first coin, between consecutive coins, and after the last coin. Every legal placement corresponds to exactly one nonnegative gap vector \((g_0,\dots,g_m)\), and the total number of empty squares is

$g_0+g_1+\cdots+g_m=n-m=:N.$

This is why the implementation begins with the parameter reduction

$m=c+1,\qquad N=n-m.$

Ordinary Silver Dollar Positions as Nim Heaps

The crucial combinatorial fact is that the ordinary silver dollar game can be described by alternating gaps. Let

$k=\left\lceil\frac{m}{2}\right\rceil,\qquad r=m-k+1.$

The active gaps are the ones that contribute to the Nim-style xor condition:

If \(m\) is even, the active gaps are

$g_1,g_3,\dots,g_{m-1}.$

If \(m\) is odd, the active gaps are

$g_0,g_2,\dots,g_{m-1}.$

All remaining gaps are passive: they affect the total length, but not the xor criterion.

For the ordinary silver dollar game, a position is losing precisely when the xor of the active gaps is zero:

$a_1\oplus a_2\oplus\cdots\oplus a_k=0,$

where \(\oplus\) denotes bitwise xor, i.e. the Nim-sum. This reduces the geometric game on coin positions to a counting problem on nonnegative integers with a fixed total sum and an xor constraint.

How the Pocket Rule Changes the Losing Positions

Now restore the fact that one coin is distinguished as the silver dollar, and let \(j\in\{0,\dots,m-1\}\) be its rank from the left.

1) The leftmost silver dollar is never losing. If \(j=0\), the current player pockets it immediately and wins.

2) Even number of coins. If \(m\) is even and \(j\ge1\), the losing configurations are exactly the ordinary xor-zero gap configurations. The silver dollar may be any of the \(m-1\) non-leftmost coins, so

$L=(m-1)L_0,\qquad L_0=\text{Count}(N;\text{ways}_k).$

3) Odd number of coins. If \(m\) is odd, two different families appear:

When the silver dollar is the second coin (\(j=1\)), we again obtain the ordinary xor-zero count

$L_0=\text{Count}(N;\text{ways}_k).$

When the silver dollar is any coin from the third onward (\(j\ge2\)), the pocket move shifts the effective left boundary by one square. In gap language this becomes the same xor-zero counting problem, but with total sum \(N+1\) and the extra requirement \(g_0>0\). The code denotes this count by \(L_1\):

$L_1=\text{Count}(N+1;\text{ways}_k)-\text{Count}(N+1;\text{ways}_{k-1}).$

The subtraction removes the cases with \(g_0=0\): if the first active gap is zero, it contributes nothing to the xor and the problem collapses to the same count with only \(k-1\) active gaps.

Therefore, for odd \(m\),

$L=L_0+(m-2)L_1.$

Finally the number of winning configurations is always

$W=T-L.$

Counting the Xor-Zero Gap Tuples

The function \(\text{Count}(S;\text{ways})\) counts nonnegative gap tuples whose total sum is \(S\) and whose active gaps xor to zero. Direct enumeration is impossible for \(n=10^6\), so the code counts them bit by bit.

Fix one binary column. Suppose exactly \(x\) of the \(k\) active gaps have bit 1. To keep xor zero in that column, \(x\) must be even. If exactly \(y\) of the \(r\) passive gaps also have bit 1, then the total number of ones in that column is \(s=x+y\), and the number of assignments for that column is

$\binom{k}{x}\binom{r}{y}.$

Summing over all even \(x\) gives the coefficient array used by the DP:

$\text{ways}(s)=\sum_{\substack{0\le x\le k\\x\text{ even}}}\binom{k}{x}\binom{r}{s-x},\qquad 0\le s\le k+r.$

This explains the code in build_ways_even_*: it is not an arbitrary precomputation, but the exact count of all valid bit-column patterns.

Binary Carry Dynamic Programming

After the bit-column counts are known, the remaining task is to ensure that the total arithmetic sum of all gaps is exactly \(S\). That is what count_with_ways does.

Process the binary expansion of \(S\) from least significant bit to most significant bit. Let \(c\) be the incoming carry at bit \(b\), and let \(\beta_b\in\{0,1\}\) be the \(b\)-th bit of \(S\). If a chosen column has \(s\) ones in total, then it is valid exactly when

$s+c\equiv \beta_b\pmod2.$

The outgoing carry is then

$c'=\frac{s+c-\beta_b}{2}.$

Hence the transition is

$dp_{b+1}(c')\mathrel{+}=dp_b(c)\cdot \text{ways}(s).$

This is a standard digit-DP with carries, but here the column multiplicities already encode the xor-zero condition on the active gaps.

Worked Example: \(W(10,2)=324\)

Here \(m=c+1=3\) and \(N=n-m=7\). The total number of configurations is

$T=3\binom{10}{3}=3\cdot120=360.$

Because \(m\) is odd, the losing count splits into two parts. The silver dollar on the second coin contributes

$L_0=\text{Count}(7;\text{ways}_2)=20.$

The silver dollar on the third coin contributes

$L_1=\text{Count}(8;\text{ways}_2)-\text{Count}(8;\text{ways}_1)=16.$

Therefore

$L=20+(3-2)\cdot16=36,$

and

$W=360-36=324,$

which matches the published checkpoint.

How the Code Works

The C++, Python, and Java implementations follow the same pipeline. First, they build small binomial tables for the gap-layer coefficients. Next, build_ways_even_* constructs the array \(\text{ways}(s)\). Then count_with_ways_* performs the carry DP for an exact target sum. Finally, compute_W_* applies the even/odd formulas for the losing count and subtracts it from \(T\).

For the final Euler input, the modulus is composite:

$M=1000036000099=1000003\cdot1000033.$

So the code computes binomial coefficients modulo each prime via factorial and inverse-factorial tables, then merges the two residues with the Chinese Remainder Theorem. The C++ version additionally computes the small checkpoints exactly before switching to modular arithmetic for \(W(1\,000\,000,100)\).

Complexity Analysis

For fixed \(m\), a single Count call scans \(O(\log N)\) bit levels. Each level considers \(O(m)\) carry states and \(O(m)\) possible column sums, so the DP cost is

$O(m^2\log N)$

time and \(O(m)\) memory. The factorial / inverse-factorial preprocessing for modular binomial coefficients is \(O(n)\) for each prime modulus.

Further Reading

1. Problem page: https://projecteuler.net/problem=344
2. Silver Dollar Game overview: Cut-the-Knot - The Silver Dollar Game
3. Nim and xor / nim-sum: Wikipedia - Nim
4. Dynamic programming: Wikipedia - Dynamic programming
5. Chinese Remainder Theorem: Wikipedia - Chinese remainder theorem