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

# Problem 92: Square digit chains

(see projecteuler.net/problem=92)

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

Let's start with the simple stuff:

`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.

# 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
// start with one digit and iteratively add digits
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.

# Interactive test

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

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)*

# 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

March 9, 2017 submitted solution

May 6, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler092

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.

# Links

projecteuler.net/thread=92 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-92-square-digits-number-chain/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p092.java (written by Nayuki)

Scala: github.com/samskivert/euler-scala/blob/master/Euler092.scala (written by Michael Bayne)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

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

blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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

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I scored 12,983 points (out of 15100 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

Look at my progress and performance pages to get more details.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

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