Problem 458: Permutations of Project

(see projecteuler.net/problem=458)

Consider the alphabet A made out of the letters of the word "project": A= \{c,e,j,o,p,r,t\}.
Let T(n) be the number of strings of length n consisting of letters from A that do not have a substring that is one of the 5040 permutations of "project".
T(7)=7^7 - 7! = 818503.

Find T(10^12). Give the last 9 digits of your answer.

My Algorithm

First let's state the most important property of the word "project": all its letters are distinct.

I wrote three (!) algorithms but only the last actually solves T(10^12) in a reasonable amount of time (function fast):
bruteForce() was a quick hack to ensure I can solve T(n) for small parameters (n < 10 is okay)
It iterates over all possible 7^n strings and checks every subset of 7 consecutive letters whether they can produce the word "project".
I assign each letter P,R,O,J,E,C,T a number between 0 and 6 which in turn is used as the position in a bitmask.
Only if seven consecutive letters are distinct then the ORed bitmask of those seven letters will be 1111111 in binary (= 2^7 - 1 = 127 in decimal).

The next algorithm's implementation is the function slow. It works quite differently because it treats the whole thing as a state machine:

Initially, the variable state[8] = { 1,0,0,0,0,0,0,0 } because there is only one empty string and it's in state 0.
No empty string can be in state 1,2,...,7.

When the first letter is processed then there are 7 different letters. Each causes a transition from state 0 to state 1.
When the second letter is processed then there are 7 differenz letters as well but I have to be careful:
if the second letter is equal to the first letter, then I remain in state 1 else I jump to state 2.
Remember: the current state's ID indicates how many of the last letters were distinct.

When the third letter is processed then there are 3 different state transitions:
The same concept repeats with state 3,4,5,6.

State 7 stands out because once I reach state 7 then I stay there, no matter what future letters will arrive.
That's because I'm interesting in all strings containing the word "project" at least once - and it doesn't matter if it appears 1x, 10x, 100x, ...
I wrote a simple loop that repeatedly adds/multiplies each state by the number of different state transitions.
That O(n) algorithm can solve T(10^12) in a few hours - still too slow !

So I wrote the fast() function: the state transitions look a lot like a matrix - why not rewrite it as a Matrix ?
The matrix's number at position x,y contains the number of transitions from state x to state y.
For example, M_{4,5} = M[4,5] = 3 because 3 (out of 7) letters cause a transition from state 4 to 5.
You'll find it in the fifth column and sixth line (the upper-right corner has index (0,0) because I used zero-based indices):

M = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 7 & 1 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 6 & 1 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 5 & 1 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 4 & 1 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 3 & 1 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 2 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 7 \end{pmatrix}

The state variable'' looks like a vector such that the computation per step is:
(1) state_{i+1} = M \bullet state_i

And the next step:
(2) state_{i+2} = M \bullet state_{i+1}
(3) state_{i+2} = M \bullet (M \bullet state_i)
(4) state_{i+2} = (M \bullet M) \bullet state_i
(5) state_{i+2} = M^2 \bullet state_i

The last equation can be generalized to:
(6) state_{final} = M^n \bullet state_0

When solving problem 137 I wrote a fast exponentiation algorithm for a 2x2 matrix.
The Matrix class written for the current problem generalizes that algorithm to larger quadratic matrices, see its powmod member function.

I was too lazy to write a Vector class to represent the state variable - and I can get away with that lazy attitude ;-)
The meaning of M^n_{x,y} was that the matrix M contains the total number of state transitions from state x to y after n steps.
Hence M^n_{0,7} is the total number of strings with the word "project" after those n = 10^12 steps.

Now I need 7^n mod 10^9 - and that number is part of the matrix, too !
The number M_{7,7} is initially 7 and multiplied by itself n times, that's exactly what I need.
However, I saw that M^n_{7,7} = 1 after n = 10^12 steps - which I found strange and I was sure I had a bug in my code.
But Wolfram Alpha confirmed that it's correct: www.wolframalpha.com/input/?i=PowerMod[7,10^12,10^9]
So the result is M^n_{7,7} - M^n_{0,7} (mod 1000000000).

Fast exponentiation is truly fast and the correct result is displayed after a few milliseconds.

Alternative Approaches

After submitting my result I looked at the official forum and noticed that pretty much everyone solved it the same way.
But some postings differed completely: they found the generating function and solved it.

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 <array>
 
// "project" has 7 distinct letters
const auto WordSize = 7;
enum Letters { P = 0, R = 1, O = 2, J = 3, E = 4, C = 5, T = 6 };
 
// only the last nine digits
const unsigned int Modulo = 1000000000;
 
// ---------- first algorithm ----------
 
// super-slow brute-force algorithm, becomes unusable for numLetters > 10
unsigned long long bruteForce(unsigned int numLetters)
{
// bruteForce( 7) = 818503
// bruteForce( 8) = 5699280
// bruteForce( 9) = 39688327
// bruteForce(10) = 276386929
unsigned long long result = 0;
 
// 7^numLetters
unsigned long long maxId = 1;
for (unsigned long long exponent = 1; exponent <= numLetters; exponent++)
maxId *= WordSize;
 
// iterate over all 7^numLetters combinations
for (unsigned long long i = 0; i < maxId; i++)
{
// convert the current number to letters (it's like converting from decimal to base 7)
auto id = i;
Letters letters[20];
for (unsigned int letter = 0; letter < numLetters; letter++)
{
// I use the data type "Letters" to simplify debugging
letters[letter] = (Letters)(id % WordSize);
id /= WordSize;
}
 
// look at each group of 7 consecutive letters
bool isProject = false;
for (unsigned int from = 0; from + WordSize <= numLetters; from++)
{
// build a bitmask:
unsigned int mask = 0;
for (unsigned int current = 0; current < WordSize; current++)
mask |= 1 << letters[from + current];
 
// all seven letters used ?
if (mask == 127) // 2^7 - 1 = 127
{
isProject = true;
break;
}
}
 
// no seven consecutive letters can be re-order to the word "PROJECT"
if (!isProject)
result++;
}
 
return result;
}
 
// ---------- second algorithm ----------
 
// O(n) algorithm, needs about 2 seconds for numLetters = 10^8
unsigned long long slow(unsigned long long numLetters)
{
// state x means that currently x distinct letters were observed in the last x steps
typedef std::array<unsigned long long, 7 + 1> State;
// once we reach state 7 we have the word P,R,O,J,E,C,T (in any order) and remain in that state
// no matter what follows
State state = { 0 };
state[0] = 1; // seed
 
unsigned long long all = 1;
for (unsigned long long i = 1; i <= numLetters; i++)
{
// total number of strings (regardless whether they contain PROJECT or not)
all *= 7;
all %= Modulo;
 
State next = { 0 };
 
// compute all state transitions
next[1] = 7 * state[0] + 1 * state[1] + 1 * state[2] + 1 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[2] = 6 * state[1] + 1 * state[2] + 1 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[3] = 5 * state[2] + 1 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[4] = 4 * state[3] + 1 * state[4] + 1 * state[5] + 1 * state[6];
next[5] = 3 * state[4] + 1 * state[5] + 1 * state[6];
next[6] = 2 * state[5] + 1 * state[6];
next[7] = + 1 * state[6];
 
// once I'm in state 7 I stay there
next[7] += 7 * state[7];
 
// keep only the last 9 digits
for (auto& x : next)
x %= Modulo;
 
state = std::move(next);
}
 
// state[7] represents the number of strings WITH the word "project"
auto withProject = state[7];
// without = all - with, but avoid negative results (due to modulo)
if (all < withProject)
all += Modulo;
return all - withProject;
}
 
// ---------- third and final algorithm ----------
 
// quadratic 2D matrix
template <typename Number, unsigned int Size>
class Matrix
{
// store all elements
std::array<std::array<Number, Size>, Size> data; // same as Number data[Size][Size];
 
public:
// set all elements to zero
Matrix()
: data()
{
for (unsigned int i = 0; i < Size; i++)
data[i].fill(0);
}
 
// access a field (read/write), indices are zero-based
Number& operator()(unsigned int column, unsigned int row)
{
return data[row][column];
}
// access a field (read-only), indices are zero-based
Number get(unsigned int column, unsigned int row) const
{
return data[row][column];
}
 
// multiply two matrices
Matrix operator*(const Matrix& other) const
{
Matrix result; // initially all fields are zero
for (unsigned int i = 0; i < Size; i++)
for (unsigned int j = 0; j < Size; j++)
for (unsigned int k = 0; k < Size; k++)
result(i,k) += get(j,k) * other.get(i,j);
return result;
}
 
// fast exponentiation with modulo
Matrix powmod(unsigned long long exponent, unsigned int modulo) const
{
// more or less the same concept as powmod from my toolbox (which works on integers instead of matrices)
 
// start with identity matrix
Matrix result;
for (unsigned int i = 0; i < Size; i++)
result(i,i) = 1;
Matrix base = *this;
 
while (exponent > 0)
{
// fast exponentation:
// odd exponent ? a^b = a*a^(b-1)
if (exponent & 1)
{
result = result * base;
 
// modulo
for (unsigned int i = 0; i < Size; i++)
for (unsigned int k = 0; k < Size; k++)
result(i,k) = result(i,k) % modulo;
}
 
// even exponent ? a^b = (a*a)^(b/2)
base = base * base;
 
// modulo
for (unsigned int i = 0; i < Size; i++)
for (unsigned int k = 0; k < Size; k++)
base(i,k) = base(i,k) % modulo;
 
exponent >>= 1;
}
return result;
}
};
 
// solve "almost instantly", too fast to measure execution time ...
unsigned long long fast(unsigned long long numLetters)
{
// same concept as slow() but rewritten with Matrix
 
// state transitions
Matrix<unsigned long long, 8> mat;
mat(0,0) = 0; mat(1,0) = 0; mat(2,0) = 0; mat(3,0) = 0; mat(4,0) = 0; mat(5,0) = 0; mat(6,0) = 0; mat(7,0) = 0;
mat(0,1) = 7; mat(1,1) = 1; mat(2,1) = 1; mat(3,1) = 1; mat(4,1) = 1; mat(5,1) = 1; mat(6,1) = 1; mat(7,1) = 0;
mat(0,2) = 0; mat(1,2) = 6; mat(2,2) = 1; mat(3,2) = 1; mat(4,2) = 1; mat(5,2) = 1; mat(6,2) = 1; mat(7,2) = 0;
mat(0,3) = 0; mat(1,3) = 0; mat(2,3) = 5; mat(3,3) = 1; mat(4,3) = 1; mat(5,3) = 1; mat(6,3) = 1; mat(7,3) = 0;
mat(0,4) = 0; mat(1,4) = 0; mat(2,4) = 0; mat(3,4) = 4; mat(4,4) = 1; mat(5,4) = 1; mat(6,4) = 1; mat(7,4) = 0;
mat(0,5) = 0; mat(1,5) = 0; mat(2,5) = 0; mat(3,5) = 0; mat(4,5) = 3; mat(5,5) = 1; mat(6,5) = 1; mat(7,5) = 0;
mat(0,6) = 0; mat(1,6) = 0; mat(2,6) = 0; mat(3,6) = 0; mat(4,6) = 0; mat(5,6) = 2; mat(6,6) = 1; mat(7,6) = 0;
mat(0,7) = 0; mat(1,7) = 0; mat(2,7) = 0; mat(3,7) = 0; mat(4,7) = 0; mat(5,7) = 0; mat(6,7) = 1; mat(7,7) = 7;
// note: I could have skipped all those mat(x,y) = 0 because all cell are initialized with zero anyway
 
// exponentiate matrix
auto superMatrix = mat.powmod(numLetters, Modulo);
 
// the number of strings WITH the word "project" (number of ways to transition from state 0 to 7)
auto withProject = superMatrix.get(0,7);
// total number of combinations is 7^numLetters
// => I could use powmod(7, 10^12, Modulo) from my toolbox, but there's a 7 in the lower-right corner of the matrix at 7,7
// and since it's exponated as well, it becomes (7^numLetters) % Modulo and that's just what I need
auto all = superMatrix.get(7,7); // and surprisingly 7^(10^12) % 10^9 = 1 !
 
// without = all - with, but avoid negative results (due to modulo)
if (all < withProject)
all += Modulo;
return all - withProject;
}
 
int main()
{
unsigned long long limit = 1000000000000ULL; // 10^12
std::cin >> limit;
//std::cout << bruteForce(limit) << std::endl;
//std::cout << slow(limit) << std::endl;
std::cout << fast(limit) << std::endl;
return 0;
}

This solution contains 37 empty lines, 54 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

December 18, 2017 submitted solution
December 18, 2017 added comments

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