<< problem 358 - Cyclic numbers Bézier Curves - problem 363 >>

# Problem 359: Hilbert's New Hotel

An infinite number of people (numbered 1, 2, 3, etc.) are lined up to get a room at Hilbert's newest infinite hotel.
The hotel contains an infinite number of floors (numbered 1, 2, 3, etc.), and each floor contains an infinite number of rooms (numbered 1, 2, 3, etc.).

Initially the hotel is empty. Hilbert declares a rule on how the nth person is assigned a room:
person n gets the first vacant room in the lowest numbered floor satisfying either of the following:

• the floor is empty
• the floor is not empty, and if the latest person taking a room in that floor is person m, then m + n is a perfect square
Person 1 gets room 1 in floor 1 since floor 1 is empty.
Person 2 does not get room 2 in floor 1 since 1 + 2 = 3 is not a perfect square.
Person 2 instead gets room 1 in floor 2 since floor 2 is empty.
Person 3 gets room 2 in floor 1 since 1 + 3 = 4 is a perfect square.

Eventually, every person in the line gets a room in the hotel.

Define P(f, r) to be n if person n occupies room r in floor f, and 0 if no person occupies the room. Here are a few examples:
P(1, 1) = 1
P(1, 2) = 3
P(2, 1) = 2
P(10, 20) = 440
P(25, 75) = 4863
P(99, 100) = 19454

Find the sum of all P(f, r) for all positive f and r such that f * r = 71328803586048 and give the last 8 digits as your answer.

# My Algorithm

My standard procedure is to make sure I can solve the problem for small input values:

• fillHotel() "puts" guests into their rooms
• Pslow() performs a simple lookup in rooms
That's pretty fast for a few thousands guests but there's no chance of solving the real problem.

Printing the guest IDs of the first floors / rooms showed a certain pattern:
floor \ room123456789
1136101521283645
2279162029354654
34511142227374456
481719303447536876
5121323263843576480
61831334852697594102
72425394258638188108
8324951707495101124132
9404159628287109116140

• the first floor is the triangular series n(n+1)/2
• the difference between the first and the second guest is 1 on odd floors and 2f+1 on even floors
• the difference between the second and the third guest is 2 on odd floors and 2f on even floors
• differences between odd rooms on the second floor is 9-2=7, 20-9=11, 35-20=15, 54-35=19, ... → difference increases by 4 with each step
• differences between even rooms on the second floor is 16-7=9, 29-16=13, 46-29=17, ... → difference increases by 4 with each step
• the third, fourth, fifth, ... floor show a pattern similar to the pattern I found in the second floor
1. be able to compute the first floor P(1, r)
2. find the first guest of a floor P(f, 1)
3. find the increment between first and second room P(f, 2) - P(f, 1)
4. find the increment between second and third room P(f, 3) - P(f, 2)
5. Based on 3. and 4. find the initial differences between odd rooms P(f, 3) - P(f, 1) and even rooms P(f, 4) - P(f, 2)

With lots of trial-and-error I found that the first guest of floor f is
(1) even-numbered floors: P(f, 1) = \lfloor dfrac{floor^2}{2} \rfloor

(2) odd-numbered floors: P(f, 1) = dfrac{(floor + 1) * (floor - 1)}{2} except for P(1, 1) = 1

C++ automatically truncates quotients so (floor + 1) / 2 * floor is good enough for (1) while also helpful for (2).

incrementOdd and incrementEven are set according to the rules I defined above.
Unfortunately, a simple loop would take forever for large room numbers.
However, the constant increase of 4 per step lets me use the triangle formula dfrac{x(x+1)}{2} with some modifications.

The input value 71328803586048 has only two prime factors: 2^27 * 3^12.
Iterating over each combination 2^i * 3^j where 0 <= i <= 27 and 0 <= j <= 12 returns all 364 divisors.

## Note

Well, this is one of those problems where I ask myself: whyyyyyyy ?
I couldn't find a scientific approach and everything I did was playing around with weird formulas until P() produced the same output as Pslow().
I didn't learn anything new nor did I have fun.
It's kind of strange that so many people solved this problem (with ease ?!) because I'm pretty sure I saw a few patterns just by pure luck.

Please keep in mind that I had to use G++'s 128 bit integer extension and therefore the code doesn't compile with Visual C++.

# 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):
Note: Enter x and y to find all solutions for 2^x * 3^y

This is equivalent to
echo "1 1" | ./359

Output:

Note: the original problem's input 27 12 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)

# My code

… was written in C++11 and can be compiled with G++, Clang++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>
#include <vector>
#include <cmath>

// ---------- slow algorithm ----------

std::vector<std::vector<unsigned int>> rooms;

// look up result in precomputed 2D vector "rooms"
unsigned int Pslow(unsigned int floor, unsigned int room)
{
// from one-indexed to zero-based
floor--;
room--;

if (floor >= rooms.size())
return 0;
if (room >= rooms[floor].size())
return 0;

return rooms[floor][room];
}

// return true if x is a perfect square
bool isSquare(unsigned int x)
{
unsigned int root = sqrt(x);
return root * root == x;
}

// precompute Hilbert's New Hotel, too slow for the problem, but solves the examples
void fillHotel(unsigned int numPersons)
{
// one iteration per person
for (unsigned int person = 1; person < numPersons; person++)
{
bool needNewFloor = true;
// try to place person in an existing floor
for (size_t floor = 0; floor < rooms.size(); floor++)
{
// last + person should be a perfect square
if (isSquare(rooms[floor].back() + person))
{
rooms[floor].push_back(person);
needNewFloor = false;
break;
}
}

// no suitable floor found, create a new floor
if (needNewFloor)
{
std::vector<unsigned int> newFloor = { person };
rooms.push_back(newFloor);
}
}
}

// ---------- fast algorithm ----------

// compute P(floor, room)
unsigned int P(unsigned long long floor, unsigned long long room, unsigned int modulo = 100000000)
{
// compute number in first room of that floor
__int128 result = (__int128)(floor + 1) / 2 * floor;
if (floor % 2 == 1 && floor > 1)
result -= (floor + 1) / 2;

// separate increments for odd and even rooms
__int128 incrementEven = 1;
if (floor % 2 == 0)
incrementEven = 2 * floor + 1;

__int128 incrementOdd = 2;
if (floor % 2 == 1)
incrementOdd = 2 * floor;

// and they are a bit different on the first floor
if (floor == 1)
{
incrementOdd  = 3;
incrementEven = 2;
}

// my original code:
//for (unsigned int i = 2; i <= room; i += 2)
//{
//  result += incrementEven;
//  incrementEven += 2;
//}
//for (unsigned int i = 3; i <= room; i += 2)
//{
//  result += incrementOdd;
//  incrementOdd += 2;
//}
// and converted to a closed form:

// number of rooms with even room numbers
__int128 numEven = room / 2;
// sum of 1+2+3+...+numEven
auto triangleEven = numEven * (numEven + 1) / 2;
// sum of 2+4+6+...+2*numEven
triangleEven *= 2;

// number of rooms with even odd numbers
__int128 numOdd = (room - 1) / 2;
// sum of 1+2+3+...+numOdd
auto triangleOdd = numOdd * (numOdd + 1) / 2;
// sum of 2+4+6+...+2*numOdd
triangleOdd *= 2;

result += numEven * (incrementEven - 2) + triangleEven;
result += numOdd  * (incrementOdd  - 2) + triangleOdd;

return result % modulo;
}

int main()
{
auto maxExponentTwo   = 27;
auto maxExponentThree = 12;
auto number = 71328803586048ULL;

std::cin >> maxExponentTwo >> maxExponentThree;
number = pow(2ULL, maxExponentTwo) * pow(3ULL, maxExponentThree);

const unsigned int Modulo = 100000000;

// solve examples with slow algorithm
//fillHotel(20000);
//std::cout << Pslow( 1,  1) << " " << Pslow( 1,  2) << " " << Pslow( 2,   1) << " "
//          << Pslow(10, 20) << " " << Pslow(25, 75) << " " << Pslow(99, 100) << std::endl;

unsigned int sum = 0;
// 71328803586048 = 2^27 * 3^12
auto two = 1ULL;
// iterate over all exponents of 2 and 3
for (auto expTwo = 0; expTwo <= maxExponentTwo; expTwo++, two *= 2)
{
auto three = 1ULL;
for (auto expThree = 0; expThree <= maxExponentThree; expThree++, three *= 3)
{
auto floor = two * three;    //         2^expTwo      * 3^expThree
auto room  = number / floor; // will be 2^(27-expTwo) * 3^(12-expThree)
sum += P(floor, room, Modulo);
sum %= Modulo;
}
}

// finally ...
std::cout << sum << std::endl;
return 0;
}


This solution contains 24 empty lines, 39 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in less than 0.01 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

September 6, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 25% (out of 100%).

# Heatmap

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

 green solutions solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too yellow solutions score less than 100% at Hackerrank (but still solve the original problem easily) gray problems are already solved but I haven't published my solution yet blue solutions are relevant for Project Euler only: there wasn't a Hackerrank version of it (at the time I solved it) or it differed too much orange problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte red problems are not solved yet but I wrote a simulation to approximate the result or verified at least the given example - usually I sketched a few ideas, too black problems are solved but access to the solution is blocked for a few days until the next problem is published [new] the flashing problem is the one I solved most recently

I stopped working on Project Euler problems around the time they released 617.
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The 310 solved problems (that's level 12) had an average difficulty of 32.6% at Project Euler and
I scored 13526 points (out of 15700 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

My username at Project Euler is stephanbrumme while it's stbrumme at Hackerrank.

Look at my progress and performance pages to get more details.

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