<< problem 89 - Roman numerals Right triangles with integer coordinates - problem 91 >>

# Problem 90: Cube digit pairs

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:

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:

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

# Hackerrank

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.

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.

Please click on a problem's number to open my solution to that problem:

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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.
 << problem 89 - Roman numerals Right triangles with integer coordinates - problem 91 >>
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