<< 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?

# 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 never 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.

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

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My username at Project Euler is

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

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