Problem 313: Sliding Game
View on Project EulerProject Euler Problem 313 Solution
EulerSolve provides an optimized solution for Project Euler Problem 313, Sliding Game, with C++, Python, Java, and a step-by-step mathematical explanation.
Problem Summary Let \(S(m,n)\) be the minimum number of moves needed to move the red counter from the top-left corner of an \(m\times n\) board to the bottom-right corner, with the empty cell initially at the bottom-right corner. The problem asks for the number of grids such that $$S(m,n)=p^2,\qquad p<10^6,\ p\text{ prime}.$$ The key fact is that the sliding game has an affine move formula, so the problem becomes a pure lattice-point count for each prime square. Mathematical Approach 1) Symmetry and small data. Because rotating the board swaps \(m\) and \(n\) without changing the puzzle, $$S(m,n)=S(n,m).$$ A brute-force BFS on small boards gives $$S(2,2)=5,\quad S(3,2)=9,\quad S(4,2)=15,\quad S(5,2)=21,$$ $$S(4,3)=17,\quad S(5,4)=25,\quad S(6,5)=33.$$ The pattern is clear: $$S(m,2)=6m-9\qquad (m\ge 3),$$ and every extra row beyond \(2\) adds exactly two more moves. 2) Closed form for non-square boards. Assume \(m>n\ge 2\). Starting from the \(m\times 2\) strip, each extra row gives two extra blank-cell shuffles, so $$S(m,n)=S(m,2)+2(n-2)=6m+2n-13.$$ This formula matches the sample from the statement: $$S(5,4)=6\cdot 5+2\cdot 4-13=25.$$ 3) What about square boards?...
Detailed mathematical approach
Problem Summary
Let \(S(m,n)\) be the minimum number of moves needed to move the red counter from the top-left corner of an \(m\times n\) board to the bottom-right corner, with the empty cell initially at the bottom-right corner.
The problem asks for the number of grids such that
$$S(m,n)=p^2,\qquad p<10^6,\ p\text{ prime}.$$
The key fact is that the sliding game has an affine move formula, so the problem becomes a pure lattice-point count for each prime square.
Mathematical Approach
1) Symmetry and small data.
Because rotating the board swaps \(m\) and \(n\) without changing the puzzle,
$$S(m,n)=S(n,m).$$
A brute-force BFS on small boards gives
$$S(2,2)=5,\quad S(3,2)=9,\quad S(4,2)=15,\quad S(5,2)=21,$$
$$S(4,3)=17,\quad S(5,4)=25,\quad S(6,5)=33.$$
The pattern is clear:
$$S(m,2)=6m-9\qquad (m\ge 3),$$
and every extra row beyond \(2\) adds exactly two more moves.
2) Closed form for non-square boards.
Assume \(m>n\ge 2\). Starting from the \(m\times 2\) strip, each extra row gives two extra blank-cell shuffles, so
$$S(m,n)=S(m,2)+2(n-2)=6m+2n-13.$$
This formula matches the sample from the statement:
$$S(5,4)=6\cdot 5+2\cdot 4-13=25.$$
3) What about square boards?
For \(m=n\ge 3\), brute force gives
$$S(3,3)=13,\quad S(4,4)=21,\quad S(5,5)=29,$$
so the square case follows
$$S(n,n)=8n-11.$$
But for an odd prime \(p\),
$$p^2\equiv 1\pmod 8,$$
whereas
$$8n-11\equiv 5\pmod 8.$$
So no square board can satisfy \(S(n,n)=p^2\). The case \(p=2\) also fails. Therefore only the non-square formula matters for prime squares.
4) Convert the move equation into an integer interval.
Set
$$q=p^2.$$
We must count integer pairs \((m,n)\) with \(m>n\ge2\) such that
$$6m+2n-13=q.$$
Divide by \(2\):
$$3m+n=\frac{q+13}{2}=:h.$$
Now solve for \(n\):
$$n=h-3m.$$
The conditions become:
From \(n\ge 2\),
$$m\le \left\lfloor\frac{h-2}{3}\right\rfloor.$$
From \(m>n\),
$$m>h-3m\iff 4m>h\iff m\ge \left\lfloor\frac{h}{4}\right\rfloor+1.$$
So the valid values of \(m\) lie in the interval
$$m_{\min}=\left\lfloor\frac{h}{4}\right\rfloor+1,\qquad m_{\max}=\left\lfloor\frac{h-2}{3}\right\rfloor.$$
5) Why the final count is multiplied by \(2\).
The interval above counts only boards with \(m>n\). But each such board has a transposed partner \(n\times m\), and both are distinct grids in the problem statement. Therefore the contribution of \(q=p^2\) is
$$f(q)=2\cdot \max\bigl(0,\ m_{\max}-m_{\min}+1\bigr).$$
This is exactly the formula implemented in the C++ solver.
6) Simplify the formula for odd primes \(p>3\).
Every odd prime larger than \(3\) satisfies
$$p^2\equiv 1\pmod{24},$$
so write
$$p^2=24k+1.$$
Then
$$h=\frac{p^2+13}{2}=12k+7,$$
and therefore
$$m_{\min}=\left\lfloor\frac{12k+7}{4}\right\rfloor+1=3k+2,$$
$$m_{\max}=\left\lfloor\frac{12k+5}{3}\right\rfloor=4k+1.$$
The number of admissible \(m\) values is
$$m_{\max}-m_{\min}+1=(4k+1)-(3k+2)+1=k,$$
so
$$f(p^2)=2k=\frac{p^2-1}{12}\qquad (p>3).$$
The exceptional prime \(p=3\) gives
$$q=9,\quad h=11,\quad m_{\min}=3,\quad m_{\max}=3,\quad f(9)=2.$$
And \(p=2\) contributes \(0\).
7) Final counting formula.
Hence
$$\#\{(m,n):S(m,n)=p^2,\ p<P\} = 2+\sum_{\substack{3<p<P\\ p\text{ prime}}}\frac{p^2-1}{12},$$
with the leading \(2\) coming from \(p=3\).
Worked Examples
The sample \(p<10\). The primes are \(2,3,5,7\).
\(p=2\) contributes \(0\), \(p=3\) contributes \(2\), \(p=5\) contributes \(2\), and \(p=7\) contributes \(4\). So the total is
$$0+2+2+4=8,$$
which matches the checkpoint in the code.
The sample \(p<100\). Summing the same formula over all primes below \(100\) gives
$$5482,$$
exactly as stated in the problem.
Algorithm
1) Generate all primes \(p<10^6\) with the sieve of Eratosthenes.
2) For each prime, compute \(q=p^2\).
3) Evaluate
$$h=\frac{q+13}{2},\qquad m_{\min}=\left\lfloor\frac{h}{4}\right\rfloor+1,\qquad m_{\max}=\left\lfloor\frac{h-2}{3}\right\rfloor.$$
4) Add
$$2\cdot \max(0,m_{\max}-m_{\min}+1).$$
This is \(O(1)\) work per prime.
Complexity Analysis
The sieve costs \(O(P\log\log P)\) time and \(O(P)\) memory. After that, the summation is linear in the number of primes, with constant-time arithmetic for each one.
Checks And Final Result
The implementation verifies
$$\text{solve}(10)=8,\qquad \text{solve}(100)=5482.$$
The full answer is
$$2057774861813004.$$
Further Reading
- Problem page: https://projecteuler.net/problem=313
- Sieve of Eratosthenes: https://en.wikipedia.org/wiki/Sieve_of_Eratosthenes
- Diophantine equations: https://en.wikipedia.org/wiki/Diophantine_equation
Problem 313 source code
C++
#include <cstdint>
#include <iostream>
#include <string>
#include <vector>
namespace {
using u64 = std::uint64_t;
struct Options {
int prime_limit = 1000000;
bool run_checkpoints = true;
};
bool parse_int_after_prefix(const std::string& arg, const std::string& prefix, int& value) {
if (arg.rfind(prefix, 0U) != 0U) {
return false;
}
const std::string tail = arg.substr(prefix.size());
if (tail.empty()) {
return false;
}
int parsed = 0;
for (char c : tail) {
if (c < '0' || c > '9') {
return false;
}
parsed = parsed * 10 + static_cast<int>(c - '0');
}
value = parsed;
return true;
}
bool parse_arguments(int argc, char** argv, Options& options) {
for (int i = 1; i < argc; ++i) {
const std::string arg(argv[i]);
if (arg == "--skip-checkpoints") {
options.run_checkpoints = false;
continue;
}
if (parse_int_after_prefix(arg, "--prime-limit=", options.prime_limit)) {
continue;
}
std::cerr << "Unknown argument: " << arg << '\n';
return false;
}
return options.prime_limit >= 2;
}
std::vector<int> sieve_primes(const int limit) {
std::vector<std::uint8_t> is_prime(static_cast<std::size_t>(limit + 1), 1U);
is_prime[0] = 0U;
is_prime[1] = 0U;
for (int p = 2; static_cast<u64>(p) * p <= static_cast<u64>(limit); ++p) {
if (is_prime[static_cast<std::size_t>(p)] == 0U) {
continue;
}
for (int q = p * p; q <= limit; q += p) {
is_prime[static_cast<std::size_t>(q)] = 0U;
}
}
std::vector<int> primes;
for (int p = 2; p <= limit; ++p) {
if (is_prime[static_cast<std::size_t>(p)] != 0U) {
primes.push_back(p);
}
}
return primes;
}
u64 count_grids_for_prime_square(const u64 p2) {
if ((p2 & 1ULL) == 0ULL) {
return 0ULL;
}
const u64 h = (p2 + 13ULL) / 2ULL;
const u64 low = h / 4ULL + 1ULL;
const u64 high = (h >= 2ULL) ? ((h - 2ULL) / 3ULL) : 0ULL;
if (high < low) {
return 0ULL;
}
return 2ULL * (high - low + 1ULL);
}
u64 solve(const int prime_limit) {
const std::vector<int> primes = sieve_primes(prime_limit - 1);
u64 total = 0ULL;
for (int p : primes) {
const u64 p2 = static_cast<u64>(p) * static_cast<u64>(p);
total += count_grids_for_prime_square(p2);
}
return total;
}
bool run_checkpoints() {
if (solve(100) != 5482ULL) {
std::cerr << "Checkpoint failed for prime limit 100 sample" << '\n';
return false;
}
if (solve(10) != 8ULL) {
std::cerr << "Checkpoint failed for prime limit 10" << '\n';
return false;
}
return true;
}
} // namespace
int main(int argc, char** argv) {
Options options;
if (!parse_arguments(argc, argv, options)) {
return 1;
}
if (options.run_checkpoints && !run_checkpoints()) {
return 2;
}
std::cout << solve(options.prime_limit) << '\n';
return 0;
}
Python
def sieve_primes(limit):
is_prime = bytearray([1] * (limit + 1))
is_prime[0] = 0
is_prime[1] = 0
for p in range(2, int(limit ** 0.5) + 1):
if is_prime[p]:
for q in range(p * p, limit + 1, p):
is_prime[q] = 0
return [p for p in range(2, limit + 1) if is_prime[p]]
def count_grids_for_prime_square(p2):
if (p2 & 1) == 0:
return 0
h = (p2 + 13) // 2
low = h // 4 + 1
high = (h - 2) // 3 if h >= 2 else 0
if high < low:
return 0
return 2 * (high - low + 1)
def solve(prime_limit=1000000):
primes = sieve_primes(prime_limit - 1)
total = 0
for p in primes:
p2 = p * p
total += count_grids_for_prime_square(p2)
return str(total)
if __name__ == '__main__':
print(solve())
Java
import java.util.ArrayList;
import java.util.List;
public class Euler313 {
static List<Integer> sievePrimes(int limit) {
byte[] isPrime = new byte[limit + 1];
for (int i = 2; i <= limit; i++) {
isPrime[i] = 1;
}
for (int p = 2; (long) p * p <= limit; ++p) {
if (isPrime[p] == 0)
continue;
for (int q = p * p; q <= limit; q += p) {
isPrime[q] = 0;
}
}
List<Integer> primes = new ArrayList<>();
for (int p = 2; p <= limit; ++p) {
if (isPrime[p] == 1) {
primes.add(p);
}
}
return primes;
}
static long countGridsForPrimeSquare(long p2) {
if ((p2 & 1) == 0)
return 0;
long h = (p2 + 13) / 2;
long low = h / 4 + 1;
long high = (h >= 2) ? ((h - 2) / 3) : 0;
if (high < low)
return 0;
return 2 * (high - low + 1);
}
public static String solve() {
int primeLimit = 1000000;
List<Integer> primes = sievePrimes(primeLimit - 1);
long total = 0;
for (int p : primes) {
long p2 = (long) p * p;
total += countGridsForPrimeSquare(p2);
}
return String.valueOf(total);
}
public static void main(String[] args) {
System.out.println(solve());
}
}