<< problem 39 - Integer right triangles Pandigital prime - problem 41 >>

# Problem 40: Champernowne's constant

An irrational decimal fraction is created by concatenating the positive integers:

0.12345678910\red{1}112131415161718192021...

It can be seen that the 12th digit of the fractional part is 1.
If d_n represents the nth digit of the fractional part, find the value of the following expression.

d_1 * d_10 * d_100 * d_1000 * d_10000 * d_100000 * d_1000000

# My Algorithm

The original problem can be solved in a trivial way:

• a for-loop appends numbers to a long string until that string contains enough digits
• read relevant digits, convert them from ASCII to integers and multiply them
The Hackerrank problem asks for digits at positions up to 2^18 which cannot be done the brute force way
because we would be running out of memory (and CPU time).

My function getDigit finds a digit without building such a long string.
It is based on the following observation:
• there are 9 numbers with one digit (1 to 9)
• there are 90 numbers with one digit (10 to 99)
• there are 900 numbers with one digit (100 to 999)
• ... and so on
The first part of getDigit figures out how many digits the number has which is pointed to by the parameter pos.
• the first 9 numbers are represented by 1*9 digits in Champernowne's constant
• the next 90 numbers are represented by 2*90=180 digits
• the next 900 numbers are represented by 3*900=2700 digits
• ... and so on: range will be 9, 90, 900, ... and digits will be 1, 2, 3, ...
The variable skip will contain 9, 9+2*90 = 189, 9+2*90+3*900 = 2890 until the next step would exceed pos.
Now that the function knows how many digit the number (pointed to by pos) has, getDigit process its digits.

To do so, it moves first closer to pos by repeated adding range.
Whenever range becomes too large, the next (smaller) digit has to be adjusted until we have the final value of first.
That number is converted to a string and the desired digit returned.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 1 2 3 4 5 6 7" | ./40

Output:

Note: the original problem's input 1 10 100 1000 10000 100000 1000000 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 <string>
#include <iostream>

// return the digit at position "pos"
// first digit after then decimal dot has pos = 1 (not zero !)
unsigned int getDigit(unsigned long long pos)
{
// assume pos has one digit
unsigned int       digits = 1;
// then there are 9 other numbers
unsigned long long range  = 9;
// the smallest of them is 1
unsigned long long first  = 1;

// there are    9 numbers with 1 digit
// there are   90 numbers with 2 digits
// there are  900 numbers with 3 digits
// there are 9000 numbers with 4 digits
// ...
// let's figure out the number of digits

// skip numbers with too few digits
unsigned long long skip = 0;
while (skip + digits*range < pos)
{
skip += digits*range;
// digits = 2 => range = 90 and
// digits = 3 => range = 900
// digits = 4 => range = 9000, etc.
digits++;
range *= 10;
first *= 10;
}

// now that we know the number of digits
// adjust "first" and "skip" such that the left-most/highest digit of pos and skip are identical
// then continue with the next digit
while (range > 9)
{
// could be replace by some modular arithmetic, but I'm too lazy for tough thinking ;-)
while (skip + digits*range < pos)
{
skip  += digits*range;
first += range;
}

// next lower digit
range /= 10;
}

// right-most digit (basically same inner loop as above when range == 1)
while (skip + digits < pos)
{
first++;
skip += digits;
}

// skip all "skippable" digits
pos -= skip;
// strings are zero-based whereas input is one-based
pos--;

// create a string version of our number
auto s = std::to_string(first);
// extract digit and convert from ASCII to an integer
return s[pos] - '0';
}

int main()
{
unsigned int tests;
std::cin >> tests;
while (tests--)
{
unsigned int product = 1;
for (unsigned int i = 0; i < 7; i++)
{
unsigned long long pos;
std::cin >> pos;

// multiply all digits
product *= getDigit(pos);
}

// print result
std::cout << product << std::endl;
}

return 0;
}


This solution contains 12 empty lines, 28 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

February 25, 2017 submitted solution

# Hackerrank

My code solves 9 out of 9 test cases (score: 100%)

# Difficulty

Project Euler ranks this problem at 5% (out of 100%).

Hackerrank describes this problem as medium.

Note:
Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.
In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# 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.

 << problem 39 - Integer right triangles Pandigital prime - problem 41 >>
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