Problem 310: Nim Square

(see projecteuler.net/problem=310)

Alice and Bob play the game Nim Square.
Nim Square is just like ordinary three-heap normal play Nim, but the players may only remove a square number of stones from a heap.
The number of stones in the three heaps is represented by the ordered triple (a,b,c).
If 0 <= a <= b <= c <= 29 then the number of losing positions for the next player is 1160.

Find the number of losing positions for the next player if 0 <= a <= b <= c <= 100 000.

My Algorithm

Problem 301 introduced the Nim game and problem 306 re-used some of the knowledge gained and added the concept of a mex function.
After reading the blog posting mathstrek.blog/2012/08/05/combinatorial-game-theory-iii/ I immediately realized that I already knew enough to solve the current problem -
but was way to stupid "to open my eyes".

Anyway, writing a quick'n'dirty bruteForce solver is good idea because it takes just a few minutes and helps verifying that the "smart" solver is correct (for small values).
See the bruteForce function and #define BRUTE_FORCE.

The basic idea is to look at a single heap.
The mex-value of an empty heap is zero because then the game is lost.
Whenever I want to determine the mex-value of a heap with size elements then I analyze all heaps size - i*i.
For example, to compute the mex-value of 13 I need the mex-values of mex(13 - 1^2) = mex(12), mex(13 - 2^2) = mex(9) and mex(13 - 3^2) = mex(4).
The mex-value of 13 is the smallest number >= 0 which is not among these values.

Maybe it becomes a bit clearer when looking at the mex-values between 0 and 13:

heap size012345678910111213
mex01012010120101

Since mex(12) = 0, mex(9) = 2 and mex(4) = 2 the set of its "children mex-values" is \{ 0, 2 \}.
The smallest number >= 0 which is not part of that set is 1.

The first part of search implements this straight-forward algorithm.
Looking at my logging output, the maximum mex-value for heaps smaller than 100000 is surprisingly small: just 74.
A simple std::bitset instead of std::set makes this process pretty fast.

However, the original problem was about three heaps - and not a single heap.
I could write three nested loops and check whether mex(a) ^ mex(b) ^ mex(c) is zero (= lost, same XOR idea like in problem 301).
But that's not very efficient: it would require approx 100000^3 iterations (actually a bit less, but still a huge number).

The following observations speed up the process (\oplus is the XOR-operation):
(1) x \oplus y is only zero if x = y
(2) x \oplus y \oplus z = x \oplus (y \oplus z)

To count the number of lost positions in a three heap Nim Square game:

Alternative Approaches

Clever use of combinatorics can eliminate the last step. I opted for the "programming" solution instead of the hairy "mathematical approach".

Interactive test

This feature is not available for the current problem.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <vector>
#include <bitset>
#include <unordered_set>
#include <algorithm>
 
// needed to compute a unique ID for each position (bruteForce only)
const unsigned long long MaxValue = 100000;
 
// determine whether a single position is won (true) or lost (false)
// reasonably fast for small values but eats quite a good amount of memory (about 800 MByte to fully analyze a,b,c <= 500)
bool bruteForce(unsigned int a, unsigned int b, unsigned int c)
{
// game over ?
if (a == 0 && b == 0 && c == 0)
return false;
 
// sort them in ascending order: a <= b <= c
if (a > b)
std::swap(a, b);
if (b > c)
std::swap(b, c);
if (a > b)
std::swap(a, b);
 
// memoize
auto id = a * MaxValue * MaxValue + b * MaxValue + c;
 
// two separate caches
static std::unordered_set<unsigned long long> cacheWon;
if (cacheWon .count(id) != 0)
return true;
static std::unordered_set<unsigned long long> cacheLost;
if (cacheLost.count(id) != 0)
return false;
 
// try every possible move:
// take a square number of stones
for (unsigned int i = 1; i*i <= c; i++)
{
// take a square number of stones from the smallest stack
if (i*i <= a && !bruteForce(a - i*i, b, c))
{
cacheWon.insert(id);
return true;
}
 
// take a square number of stones from the stack in the middle
if (i*i <= b && !bruteForce(a, b - i*i, c))
{
cacheWon.insert(id);
return true;
}
 
// take a square number of stones from the largest stack
if (!bruteForce(a, b, c - i*i))
{
cacheWon.insert(id);
return true;
}
}
 
// no winning move ... meaning that the current player loses
cacheLost.insert(id);
return false;
}
 
// count all lost positions, just under 5 seconds for limit = 100000
unsigned long long search(unsigned int limit)
{
// based on this great posting: https://mathstrek.blog/2012/08/05/combinatorial-game-theory-iii/
 
// compute Nim values for a single pile
std::vector<unsigned int> mex(limit + 1, 0); // fill with zeros
for (unsigned int size = 0; size <= limit; size++)
{
// I saw in my experiments that Nim values are pretty small for limit <= 100000 (<= 74)
const size_t NimLimit = 80;
 
// collect all Nim values after each possible move
std::bitset<NimLimit> moves; // initialized with zeros
for (unsigned int i = 1; i*i <= size; i++)
// take i^2 stones from the heap
moves[mex[size - i*i]] = true;
 
// find the smallest non-negative number which is NOT part of the "moves" container
unsigned int available = 0;
while (moves[available])
available++;
mex[size] = available;
}
 
// the largest Nim value
auto maxNimValue = *std::max_element(mex.begin(), mex.end());
// find largest possible number when XORing two Nim values
unsigned int nextPowerOfTwo = 1; // it's 128
while (nextPowerOfTwo < maxNimValue)
nextPowerOfTwo *= 2;
 
// count how often value[b] ^ value[c] occurs where a <= b <= c <= limit
std::vector<unsigned int> frequency(nextPowerOfTwo, 0);
 
unsigned long long numLost = 0;
for (int a = limit; a >= 0; a--)
{
// modify the "frequency" container to cover new pairs (b,c), too
// it already contains all pairs (b+1,c), now a is one unit smaller so b can be one smaller, too
for (auto b = a; b == a; b++) // I know that it looks silly -
// but I want to make clear it's two nested loops
// where the outer needs only one iteration
for (auto c = b; c <= (int)limit; c++)
frequency[mex[b] ^ mex[c]]++;
 
// look at Nim value of a
// how many pairs (b,c) have to same Nim value such that value[a] ^ (value[b] ^ value[c]) = 0 => lost
numLost += frequency[mex[a]];
}
 
return numLost;
}
 
int main()
{
unsigned int limit = 100000;
std::cin >> limit;
 
//#define BRUTE_FORCE
#ifdef BRUTE_FORCE
// if limit is 500: => 21084251 positions, 2018811 lost => 800 MByte caches in bruteForce(), finish in about two minutes
// brute-force
unsigned long long numLost = 0;
// generate every possible position and count how many are lost
for (unsigned int a = 0; a <= limit; a++)
for (auto b = a; b <= limit; b++)
for (auto c = b; c <= limit; c++)
// lost ?
if (!bruteForce(a,b,c))
numLost++;
 
std::cout << numLost << std::endl;
#endif
 
std::cout << search(limit) << std::endl;
return 0;
}

This solution contains 22 empty lines, 34 comments and 7 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 4.6 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=gnu++11 -DORIGINAL)

See here for a comparison of all solutions.

Note: interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without -DORIGINAL.

Changelog

December 12, 2017 submitted solution
December 12, 2017 added comments

Difficulty

40% Project Euler ranks this problem at 40% (out of 100%).

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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
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