<< problem 208 - Robot Walks Divisor Square Sum - problem 211 >>

# Problem 209: Circular Logic

A k-input binary truth table is a map from k input bits (binary digits, 0 [false] or 1 [true]) to 1 output bit.
For example, the 2-input binary truth tables for the logical AND and XOR functions are:

xyx AND y
000
010
100
111

xyx XOR y
000
011
101
110

How many 6-input binary truth tables, tau, satisfy the formula
tau(a, b, c, d, e, f) \space \text{AND} \space tau(b, c, d, e, f, a \space \text{XOR} \space (b \space \text{AND} \space c)) = 0
for all 6-bit inputs (a, b, c, d, e, f)?

# My Algorithm

The most crucial step is to recognize that there is a 1:1 mapping (bijective mapping) between (a, b, c, d, e, f) and (b, c, d, e, f, a \space \text{XOR} \space (b \space \text{AND} \space c)) = 0.
Therefore my program maps each possible input (a, b, c, d, e, f) to (b, c, d, e, f, a \space \text{XOR} \space (b \space \text{AND} \space c)) = 0.
These six variables are equivalent to a set of all binary numbers from 0 to 2^6 - 1 (=63).
I store that mapping in connections.

Then I have to find all cycles: how to long it take that a number is mapped to itself: connections[connections[connections[...connections[x]]]] = x.
Each number is part of exactly one cycle - most are very short but the longest has 46 elements.
Note that if x maps to itself after s steps, then each other number of the same cycle maps to itself after s steps, too.
Therefore all numbers in the same cycle can be treated equally.

I computed the number of possibilities for small cycle lengths (essentially they can't can have consecutive ones) and looked up my values in OEIS.
They are called Lucas numbers (see en.wikipedia.org/wiki/Lucas_number) and extremely similar to the Fibonacci sequence.

Each cycle is independent of each other therefore the Lucas numbers of all cycles have to be multiplied.

## Note

There is no live test available for my solution.

# 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. Or just jump to my GitHub repository.

       #include <iostream>
#include <bitset> // I often prefer std::vector<bool> but let's try some of the rarely included C++ containers !

int main()
{
// 2^6 = 64
const auto SixtyFour = 64;

// go through all 2^6 = 64 possible inputs of tau(a,b,c,d,e,f) and find the corresponding state tau(b,c,d,e,f,a XOR (b AND c))
unsigned char connections[SixtyFour];
for (auto from = 0; from < SixtyFour; from++) // state of first tau
{
std::bitset<6> leftSide = from;
std::bitset<6> rightSide;

rightSide[0] = leftSide[1]; // b => a
rightSide[1] = leftSide[2]; // c => b
rightSide[2] = leftSide[3]; // d => c
rightSide[3] = leftSide[4]; // e => d
rightSide[4] = leftSide[5]; // f => e
rightSide[5] = leftSide[0] ^ (leftSide[1] & leftSide[2]); // a ^ (b & c) => f

// connections[from] = to
connections[from] = rightSide.to_ulong();
}

// precompute Lucas numbers
unsigned long long lucas[SixtyFour + 1] = { 2, 1 }; // seeds
for (auto i = 2; i <= SixtyFour; i++)     // actually I don't need all of them, longest cycle is < 64
lucas[i] = lucas[i - 2] + lucas[i - 1]; // computation is similar to Fibonacci, but different seeds

// multiply Lucas numbers of each cycle length
unsigned long long result = 1;

// find cycles
std::bitset<SixtyFour> used = 0; // set used[x] to true if state x was processed
while (!used.all())
{
// pick a randomly chosen available state
// I always take the smallest / lowest but it doesn't really matter which one I pick
auto start = 0;
while (used[start])
start++;

// walk through the states until the initial state is reached again
auto current = start;
auto cycleLength = 0;
do
{
// "use" this state
used[current] = true;
cycleLength++;

// continue with next state in this cycle
current = connections[current];
} while (current != start);

// include all those combinations
result *= lucas[cycleLength];
}

// hooray ...
std::cout << result << std::endl;
return 0;
}


This solution contains 11 empty lines, 13 comments and 2 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 12, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 60% (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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