<< problem 91 - Right triangles with integer coordinates Arithmetic expressions - problem 93 >>

# Problem 92: Square digit chains

A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.

For example,

44 → 32 → 13 → 10 → 1 → 1
85 → 89 → 145 → 42 → 20 → 4 → 16 → 37 → 58 → 89

Therefore any chain that arrives at 1 or 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 or 89.

How many starting numbers below ten million will arrive at 89?

# My Algorithm

becomes89(x) checks whether repeated application of the "digit-square-adding" terminates with 89 (true) or 1 (false).

The real magic is hidden in main - it's an iterative approach similar to Dynamic Programming and was inspired by rosettacode.org/wiki/Iterated_digits_squaring

I count how many numbers sums[x] share the same squared digit sum x:

• initially sums[] contains a bunch of zeros
• after that, all single digit numbers 0..9 are processed:
sums[0*0] = 1; sums[1*1] = 1; sums[2*2] = 1; sums[3*3] = 1; ...

Then all two-digit numbers 10..99 are processed:
• they can have squared digit sums between 1 and 2*9*9
• the highest digit will be between 1 and 9 (I call it high)
• a squared sum x should become the sum of all one-digit numbers with that squared digit sum and all two-digit number with that squared digit sum
• the count of one-digit numbers is already stored in sum[x], it's either 0 or 1
• the count of two-digit numbers is the count of sum[x - high*high]
→ e.g. when high is 1, then sum[5] += sum[5 - 1*1]1
→ when high becomes 2, then sum[5] += sum[5 - 2*2]2
→ there are 2 one-digit or two-digit numbers with a squared digit sum of 5

Please note that in-place updates of sum[x] are in descending order because otherwise we encounter collisions:
sum[x - high*high] should not have been updated in the current pass. That means, x must decrease while high increases in their respective loops.

That algorithm is repeated for all digits

When I know how many numbers share the same squared digit sum, I can simply count all squared digit sums which lead to 89.

## Modifications by HackerRank

They want you to compute the result for values up to 10^{200}. The final number must be modulo 1000000007, too (otherwise it gets huge).

## Note

You have to enter the number of digits of the upper limit. It's 7 for the original problem because 10^7=10000000.

# 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):
Note: Enter the exponent k of 10^k => "enter the number of digits"

This is equivalent to
echo 1 | ./92

Output:

Note: the original problem's input 7 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.

       #include <iostream>

// return true, if x will arrive at 89
bool becomes89(unsigned int x)
{
do
{
// sum of squares of all digits
unsigned int squareDigitSum = 0;
auto reduce = x;
while (reduce > 0)
{
auto digit = reduce % 10;
reduce /= 10;
squareDigitSum += digit * digit;
}

// terminated ?
if (squareDigitSum == 89)
return true;
if (squareDigitSum ==  1)
return false;

// not done yet ... next iteration
x = squareDigitSum;
} while (true);
}

int main()
{
unsigned int digits = 7; // 10^7 = 10,000,000
std::cin >> digits;

// Hackerrank's result will become quite big, they only want the final number modulo 1000000007
// (doesn't affect the original problem)
const unsigned int Modulo = 1000000007;

// count how many numbers with the same digit sum exist
// inspired by https://rosettacode.org/wiki/Iterated_digits_squaring

// initialized with zeros
unsigned int sums[200*9*9 + 1] = { 0 };

// single-digit numbers
for (unsigned int first = 0; first <= 9; first++)
sums[first * first]++;
// alternative approach: set sums[0] = 1; and start the new loop with length = 1

for (unsigned int length = 2; length <= digits; length++)
// go through all potential sums (including the most recently added digit)
for (unsigned int sum = length*9*9; sum > 0; sum--)
// what sum is it without that most recently added digit ?
for (unsigned int high = 1; high <= 9; high++)
{
// square of the just added digit
auto square = high * high;
// this digit can't be part of the current digit sum because it's too big
if (square > sum)
break;

// add count of all numbers without the new digit
sums[sum] += sums[sum - square];
// avoid overflows (Hackerrank only)
sums[sum] %= Modulo;
}

// now we know how many numbers sums[x] exist with digit sum x
// let's check which digit sums will be converted to 89
unsigned int count89 = 0;
// check all sums
for (unsigned int i = 1; i <= digits*9*9; i++)
if (becomes89(i))
{
count89 += sums[i]; // yes, all these numbers turn to 89
count89 %= Modulo;  // Hackerrank only
}

std::cout << count89 << std::endl;
return 0;
}


This solution contains 12 empty lines, 21 comments and 1 preprocessor command.

# 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

March 9, 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 easy.

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 91 - Right triangles with integer coordinates Arithmetic expressions - problem 93 >>
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