<< 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]
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:
This is equivalent toecho 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
// 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.
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
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:
C# 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)
Those links are just an unordered selection of source code I found with a semi-automatic search script on Google/Bing/GitHub/whatever.
You will probably stumble upon better solutions when searching on your own. Maybe not all linked resources produce the correct result and/or exceed time/memory limits.
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 | |
the flashing problem is the one I solved most recently |
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I scored 13,486 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.
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 >> |