Problem 90: Cube digit pairs

(see projecteuler.net/problem=90)

Each of the six faces on a cube has a different digit (0 to 9) written on it; the same is done to a second cube.
By placing the two cubes side-by-side in different positions we can form a variety of 2-digit numbers.

For example, the square number 64 could be formed:
cubes

In fact, by carefully choosing the digits on both cubes it is possible to display all of the square numbers below one-hundred:
01, 04, 09, 16, 25, 36, 49, 64, and 81.

For example, one way this can be achieved is by placing {0, 5, 6, 7, 8, 9} on one cube and {1, 2, 3, 4, 8, 9} on the other cube.

However, for this problem we shall allow the 6 or 9 to be turned upside-down so that an arrangement like {0, 5, 6, 7, 8, 9} and {1, 2, 3, 4, 6, 7}
allows for all nine square numbers to be displayed; otherwise it would be impossible to obtain 09.

In determining a distinct arrangement we are interested in the digits on each cube, not the order.

{1, 2, 3, 4, 5, 6} is equivalent to {3, 6, 4, 1, 2, 5}
{1, 2, 3, 4, 5, 6} is distinct from {1, 2, 3, 4, 5, 9}

But because we are allowing 6 and 9 to be reversed, the two distinct sets in the last example both represent the extended set {1, 2, 3, 4, 5, 6, 9}
for the purpose of forming 2-digit numbers.

How many distinct arrangements of the two cubes allow for all of the square numbers to be displayed?

Algorithm

[TODO] write down a few words

My code

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

#include <algorithm>
#include <vector>
#include <iostream>
 
// all sides of a dice
typedef std::vector<unsigned int> Dice;
 
// each cube has 6 different sides, we can choose any 6 out of 10
const Dice Sides = { 0,1,2,3,4,5,6,7,8,9 };
 
const unsigned int Skip = 0;
const unsigned int Take = 1;
// if container[x] is Take, then Sides[x] is part of the dice
const std::vector<unsigned int> Initial = { Skip,Skip,Skip,Skip, Take,Take,Take,Take,Take,Take };
const std::vector<unsigned int> Unused = { Take };
 
int main()
{
unsigned int dices = 2;
unsigned int limit = 9; // up to the 9th square (9^2 = 81)
std::cin >> limit >> dices;
 
// Hackerrank extended the problem to three dices
const unsigned int AllDices = 3;
unsigned int maxSquare = 0;
 
// generate all square numbers
std::vector<unsigned short> squares;
for (unsigned int i = 1; i <= limit; i++)
{
auto reduce = i*i;
maxSquare = reduce;
 
std::vector<unsigned int> digits;
// no matter what, always generate a three-digit square (maybe with some leading zeros)
for (unsigned int j = 0; j < AllDices; j++)
{
auto digit = reduce % 10;
reduce /= 10;
// convert all 9s to 6s
if (digit == 9)
digit = 6;
 
digits.push_back(digit);
}
 
// digits in ascending order
std::sort(digits.begin(), digits.end());
// convert to a fingerprint
// e.g. 9^2 = 081 (with leading zero)
// => 018 (sorted)
// => 18 (sorted, converted to an integer)
auto sortedSquare = digits[0] * 100 + digits[1] * 10 + digits[2];
if (std::find(squares.begin(), squares.end(), sortedSquare) == squares.end())
squares.push_back(sortedSquare);
}
 
// will contain all solutions
unsigned int valid = 0;
// all possible label combinations for first dice
Dice dice1, dice2, dice3;
auto open = squares;
 
auto permutationDice1 = Initial;
auto permutationDice2 = Initial;
auto permutationDice3 = Initial;
 
do
{
dice1.clear();
for (size_t i = 0; i < permutationDice1.size(); i++)
if (permutationDice1[i] == Take)
dice1.push_back(Sides[i]);
 
// if both 6 and 9 are contained, then they behave identical
//if (permutationDice1[6] == Take && permutationDice1[9] == Take)
//dice1.pop_back();
 
// second dice is "lexicographically" bigger than or equal to first dice
permutationDice2 = (dices >= 2 ? permutationDice1 : Unused);
do
{
dice2.clear();
for (size_t i = 0; i < permutationDice2.size(); i++)
if (permutationDice2[i] == Take)
dice2.push_back(Sides[i]);
 
// some digits need to occur at least twice
if (maxSquare >= 100)
{
// 100 requires two zeros, so we must already have at least one
if (std::count(dice1.begin(), dice1.end(), 0) +
std::count(dice2.begin(), dice2.end(), 0) < 1)
continue;
}
if (maxSquare >= 144)
{
// 144 requires two 4s, so we must have at least one by now
if (std::count(dice1.begin(), dice1.end(), 4) +
std::count(dice2.begin(), dice2.end(), 4) < 1)
continue;
}
 
// if less than three dices are requested then add a dummy dice with one side
permutationDice3 = (dices == 3 ? permutationDice2 : Unused);
do
{
dice3.clear();
for (size_t i = 0; i < permutationDice3.size(); i++)
if (permutationDice3[i] == Take)
dice3.push_back(Sides[i]);
 
// simple pre-check
unsigned int frequency[10] = { 0,0,0,0,0, 0,0,0,0,0 };
for (auto x : dice1)
frequency[x]++;
for (auto x : dice2)
frequency[x]++;
for (auto x : dice3)
frequency[x]++;
 
// for performance optimization only: reject impossible combinations
if (frequency[1] < 1)
continue;
if (maxSquare >= 4 && frequency[4] < 1)
continue;
if (maxSquare >= 25 && frequency[2] < 1)
continue;
if (maxSquare >= 25 && frequency[5] < 1)
continue;
if (maxSquare >= 36 && frequency[3] < 1)
continue;
if (maxSquare >= 81 && frequency[8] < 1)
continue;
if (maxSquare >= 100 && frequency[0] < 2)
continue;
if (maxSquare >= 144 && frequency[4] < 2)
continue;
 
std::vector<unsigned int> matches;
// build all combinations and remove any squares we encounter along the way
for (auto one : dice1)
{
// 6 is 9 (upside down)
if (one == 9)
one = 6;
for (auto two : dice2)
{
// 6 is 9 (upside down)
if (two == 9)
two = 6;
 
for (auto three : dice3)
{
// 6 is 9 (upside down)
if (three == 9)
three = 6;
 
unsigned int current[AllDices] = { one, two, three };
 
// std::sort is much slower for such small containers
if (current[0] > current[1])
std::swap(current[0], current[1]);
if (current[1] > current[2])
std::swap(current[1], current[2]);
if (current[0] > current[1])
std::swap(current[0], current[1]);
 
auto sortedSquare = 100 * current[0] + 10 * current[1] + current[2];
// if successful then another square number was matched
auto match = std::find(squares.begin(), squares.end(), sortedSquare);
// remove it from the list
if (match != squares.end())
matches.push_back(sortedSquare);
}
}
}
 
if (matches.size() < squares.size())
continue;
 
std::sort(matches.begin(), matches.end());
auto last = std::unique(matches.begin(), matches.end());
 
open = squares;
for (auto m = matches.begin(); m != last; m++)
{
auto match = std::find(open.begin(), open.end(), *m);
open.erase(match);
}
 
// all squares matched ?
if (open.empty())
valid++;
} while (std::next_permutation(permutationDice3.begin(), permutationDice3.end()));
} while (std::next_permutation(permutationDice2.begin(), permutationDice2.end()));
} while (std::next_permutation(permutationDice1.begin(), permutationDice1.end()));
 
std::cout << valid;
return 0;
}

This solution contains 28 empty lines, 32 comments and 3 preprocessor commands.

Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Input data (separated by spaces or newlines):

This is equivalent to
echo "3 1" | ./90

Output:

(please click 'Go !')

Note: the original problem's input 9 2 cannot be entered
because just copying results is a soft skill reserved for idiots.

(this interactive test is still under development, computations will be aborted after one second)

Benchmark

The correct solution to the original Project Euler problem was found in 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.
(compiled for x86_64 / Linux, GCC flags: -O3 -march=native -fno-exceptions -fno-rtti -std=c++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

March 19, 2017 submitted solution
May 9, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler090

My code solves 34 out of 43 test cases (score: 77.5%)

I failed 0 test cases due to wrong answers and 9 because of timeouts

Difficulty

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

Hackerrank describes this problem as easy.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is never an option.

Links

projecteuler.net/thread=90 - the best forum on the subject (note: you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-90-cubes-make-square/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p090.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler090.scala (written by Michael Bayne)

Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.
yellow problems score less than 100% at Hackerrank (but still solve the original problem).
gray problems are already solved but I haven't published my solution yet.
blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

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The 163 solved problems had an average difficulty of 22.2% at Project Euler and I scored 11,907 points (out of 13200) at Hackerrank's Project Euler+.
My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.
more about me can be found on my homepage, especially in my coding blog.
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