<< problem 207 - Integer partition equations | Divisor Square Sum - problem 211 >> |

# Problem 209: Circular Logic

(see projecteuler.net/problem=209)

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.

#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 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

September 12, 2017 submitted solution

September 12, 2017 added comments

# Difficulty

Project Euler ranks this problem at **60%** (out of 100%).

# Links

projecteuler.net/thread=209 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

# Heatmap

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

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gray | problems are already solved but I haven't published my solution yet | |

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orange | problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte | |

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I scored 13,386 points (out of 15600 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.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort !!!

<< problem 207 - Integer partition equations | Divisor Square Sum - problem 211 >> |