Problem 327: Rooms of Doom

(see projecteuler.net/problem=327)

A series of three rooms are connected to each other by automatic doors.

maze

Each door is operated by a security card. Once you enter a room the door automatically closes and that security card cannot be used again.
A machine at the start will dispense an unlimited number of cards, but each room (including the starting room) contains scanners
and if they detect that you are holding more than three security cards or if they detect an unattended security card on the floor,
then all the doors will become permanently locked.
However, each room contains a box where you may safely store any number of security cards for use at a later stage.

If you simply tried to travel through the rooms one at a time then as you entered room 3 you would have used all three cards and would be trapped in that room forever!

However, if you make use of the storage boxes, then escape is possible. For example, you could enter room 1 using your first card,
place one card in the storage box, and use your third card to exit the room back to the start.
Then after collecting three more cards from the dispensing machine you could use one to enter room 1 and collect the card you placed in the box a moment ago.
You now have three cards again and will be able to travel through the remaining three doors.
This method allows you to travel through all three rooms using six security cards in total.

It is possible to travel through six rooms using a total of 123 security cards while carrying a maximum of 3 cards.

Let C be the maximum number of cards which can be carried at any time.

Let R be the number of rooms to travel through.

Let M(C,R) be the minimum number of cards required from the dispensing machine to travel through R rooms carrying up to a maximum of C cards at any time.

For example, M(3,6)=123 and M(4,6)=23.
And, sum{M(C,6)}=146 for 3 <= C <= 4.

You are given that sum{M(C,10)}=10382 for 3 <= C <= 10.

Find sum{M(C,30)} for 3 <= C <= 40.

My Algorithm

I process all rooms from the right to the left side. In each room I calculate the number of cards needed to escape.
Only one card is required in the right-most room: just insert it into the scanner - done.
I need two cards in the room left to the right-most room: one card to enter the last room and another card to escape.
Similarly, I need three cards in the room before that room.

Let's assume there are only three rooms - the problem statement gives away that a total of six cards are needed.
And further assume that in front of the first room is room "zero".
To exit the maze I need three cards when I'm inside the first room.
However, I cannot hold more than three cards at the same time and already needed one card to enter the first room.
So I need a few extra moves: go back to room zero, takes as many cards as possible, enter room one, deposit as many as possible and go back to room room.
These extra moves "transport" cards: I can transport C - (1+1) cards from the previous to the current room with one extra move:
one card is needed to enter the current room and another to leave the current room.

In order to have three cards available in the first room, the following sequence needs to be followed:

→ a total of six cards were taken from the dispenser

So my algorithm is as follows:
Surprisingly, a simple while loop finds the correct result if C >= 4 almost instantly (disable #define FAST).
However, the results for C = 3 are enormous and that loop would take a long time to finish.
Each iteration of the simple while loop does the same job: adding as subtracting constants.
So my #define FAST just calculates the number of iterations and finishes in less than 0.01 seconds.

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>
 
unsigned long long search(unsigned int cards, unsigned int rooms)
{
// need one card to exit the right-most room
unsigned long long need = 1;
 
// I need one card to enter the next room and one to come back, thus I can transport "cards - 2" to the next room in one step
const auto Transport = cards - (1 + 1); // these 2 cards are "wasted"
 
// compute in each room how many cards I need to exit, beginning with the right-most room
for (auto currentRoom = rooms; currentRoom > 0; currentRoom--)
{
// how many cards are required to satisfy "need", including cards wasted moving forth and back
auto consumed = 0ULL;
 
#define FAST
#ifdef FAST
// all iterations of the following "while"-loop are basically the same, let's process them quickly
if (need >= cards)
{
// determine number of iterations
auto moves = (need - cards) / Transport;
 
// do I need one more to ensure need < cards at the end ?
if (need - moves * Transport >= cards)
moves++;
 
// compute all iterations at once
need -= moves * Transport;
consumed += moves * cards;
}
#else
// the basic algorithm becomes slow if needing many cards and Transport is small
while (need >= cards)
{
need -= Transport;
consumed += cards;
}
#endif
 
// carry the remaining cards in the last move, including a card to enter the next room
auto previous = need + consumed + 1;
// ... and continue with the previous room (the one on the left side of the current room)
need = previous;
}
 
return need;
}
 
int main()
{
unsigned int rooms = 30;
unsigned int cards = 40;
std::cin >> rooms >> cards;
 
unsigned long long result = 0;
for (unsigned int i = 3; i <= cards; i++)
result += search(i, rooms);
 
std::cout << result << std::endl;
return 0;
}

This solution contains 11 empty lines, 11 comments and 5 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

October 27, 2017 submitted solution
October 27, 2017 added comments

Difficulty

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

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