<< problem 147 - Rectangles in cross-hatched grids Searching for a maximum-sum subsequence - problem 149 >>

# Problem 148: Exploring Pascal's triangle

We can easily verify that none of the entries in the first seven rows of Pascal's triangle are divisible by 7:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

However, if we check the first one hundred rows, we will find that only 2361 of the 5050 entries are not divisible by 7.

Find the number of entries which are not divisible by 7 in the first one billion (109) rows of Pascal's triangle.

# My Algorithm

I needed a few attempts to finally solve this problem.
My first idea was to iteratively create the triangle's rows, one at a time.
To find the binomial coefficient C(n+1,k+1) you only need the previous row:
C(n+1,k+1) = C(n,k) + C(n,k+1)
To avoid overflows, each value should be the original value modulo 7.
That works because (a+b) mod 7 = ((a mod 7) + (b mod 7)) mod 7 (that's no special property of 7, you can pick any integer)

Unfortunately I knew right from the start that this approach would be slow ... and memory-consuming (roughly 2 GByte).
Nevertheless I used the code (see nextRow) to show me the results for the first 1000 rows. And then a pattern appeared (my rows start at index 0):
rowfound
01
12
23
34
45
56
67
72
84
96
108
1314
143
156

If I convert the row from decimal system to a number in base 7 then
found(row) = (digit_1(row) + 1) * (digit_2(row) + 1) * ... (digit_n(row) + 1)
For example: found(15_10) = found(111_7) = (1+1) * (1+1) * (1+1) = 6

My function countNonDivisible does exactly that: it converts its parameter to base 7 and multiplies all digits plus one.
It takes 34 seconds until the correct result is displayed on my computer.

The inner-most loop of countNonDivisible consists of modulo and division operations - they are extremely slow in comparision to addition, subtractio, multiplication.
And actually there is no need to perform these divisions:
I process all numbers consecutively in ascending order.

Therefore my final algorithm stores all digits of row in base 7. To speed up the program, not the true digits but the digits plus one are stored
because when multiplying all digits I have to add one.
That means that my array base7 has up to 12 digits (7^12 > 10^9), each from 1 to 7.
Incrementing by one can cause some digits to become 7+1=8 → reset them to 1 and carry over 1.
That algorithm needs about 3.5 seconds.

## Alternative Approaches

My final algorithm is still a brute-force algorithm. You can find a closed formula, too:
The sum of the first 7^1 = 7 rows is 28. The sum of the first 7^2 = 49 rows is 28^2 = 784 ... and so on.

## Note

A substantial reason for the performance gain of my third algorithm is that base7 fits into the CPU cache and/or CPU registers (it's just 12 bytes).
After changing the data type from std::vector<unsigned char> to std::vector<unsigned int> the execution time explodes from 3.5 to 28 seconds
(and unsigned short → 3.9 seconds).

# 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):

This is equivalent to
echo 100 | ./148

Output:

Note: the original problem's input 1000000000 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++, Visual C++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>
#include <vector>

typedef std::vector<unsigned char> Row;
const unsigned int Modulo = 7;

// generate next row of Pascal's triangle modulo a number (> 1)
// return count of elements that are not a multiple of modulo (in C++ speak: x % modulo != 0)
unsigned long long nextRow(Row& row)
{
// last value is always 1
row.push_back(1);
if (row.size() == 1)
return 1;

// first and last value are never a multiple of 7
unsigned long long result = 2;

// C(n+1,k+1) = C(n,k) + C(n,k+1)
for (size_t k = row.size() - 2; k > 0; k--)
{
// note: I'm processing the row back-to-front
// therefore minus 1 instead of plus 1
unsigned char current = row[k] + row[k - 1];

// subtraction is faster than modulo: current %= modulo
// all values must be 0 ... 2*(modulo-1)
if (current >= Modulo)
current -= Modulo;

// not divisible ?
if (current != 0)
result++;

row[k] = current;
}

return result;
}

// convert to base 7 and multiply all digits plus 1
unsigned long long countNonDivisible(unsigned int row)
{
unsigned long long result = 1;
while (row > 0)
{
// one more digit ...
result *= (row % Modulo) + 1;
row    /= Modulo;
}
return result;
}

int main()
{
unsigned int numRows = 1000000000;
std::cin >> numRows;

// for simple algorithm based on nextRow()
Row current = { 1 };

// for my fastest pseudo brute-force algorithm
std::vector<unsigned char> base7(12, 1); // 7^12 > 10^9

unsigned long long count = 1;
for (unsigned int row = 1; row < numRows; row++)
{
// simple algorithm (basically takes forever and needs tons of memory)
//auto found = nextRow(current);
//std::cout << row << " " << found << std::endl;

//auto found = countNonDivisible(row);

// and my fastest (still pseudo brute-force) algorithm:
// keep all digits of row in base 7 in an array base7 with a twist:
// each digit is one higher than it should be

// next number
base7[0]++;
// carry over to next digits
auto carryPos = 0;
while (base7[carryPos] == Modulo + 1)
{
base7[carryPos] = 1; // remember: start at 1 instead of 0
base7[carryPos + 1]++;
carryPos++;
}

// multiply all digits
unsigned long long found = 1;
for (auto& x : base7)
found *= x;

// keep track of the sum of all rows
count += found;
}

std::cout << count << std::endl;
return 0;
}


This solution contains 19 empty lines, 27 comments and 2 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 3.5 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

July 12, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 50% (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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