<< problem 103 - Special subset sums: optimum | Special subset sums: testing - problem 105 >> |

# Problem 104: Pandigital Fibonacci ends

(see projecteuler.net/problem=104)

The Fibonacci sequence is defined by the recurrence relation:

F_n = F_{n-1} + F_{n-2}, where F_1 = 1 and F_2 = 1.

It turns out that F_541, which contains 113 digits, is the first Fibonacci number for which the last nine digits are 1-9 pandigital (contain all the digits 1 to 9, but not necessarily in order).

And F_2749, which contains 575 digits, is the first Fibonacci number for which the first nine digits are 1-9 pandigital.

Given that F_k is the first Fibonacci number for which the first nine digits AND the last nine digits are 1-9 pandigital, find k.

# Algorithm

My solution is based on a stripped down version of my `BigNum`

class and stores 9 digits per cell, that's why it's called `BillionNum`

(because `MaxDigit = 1000000000`

→ 10^9).

It only supports `operator+=`

.

`isPandigital(x, digits)`

returns true if `x`

is `1..digits`

-pandigital, e.g. `isPandigital(312, 3) == true`

because 312 is 3-pandigital.

The original problem assumes `digits = 9`

.

`x`

's digits are chopped off step-by-step and a bitmask tracks which digits we have already seen.

Zero is not part of any pandigital number, not even implicit leading zeros.

The `main`

routines analyzes the lower digits first.

Only when they are pandigital then the highest digits are checked, too, because that's bit slower:

I take all digits from the highest cell of my `BillionNum`

. If there are too few digits, all digits in its neighboring cell are included.

We might have too many digits now, therefore I remove the lowest digit until the number of digits is correct.

If these digits are pandigital, then we are done.

Finding the next Fibonacci number involves `operator+=`

of `BillionNum`

. The default algorithm is:

F_{next} = F_a + F_b

F_a = F_b

F_b = F_{next}

→ quite a few memory allocations, and many object constructor/destruction will take place behind the curtain.

A simple trick to improve performance is to use these equations instead, which get rid of these "memory/object bookkeeping" effects:

F_a += F_b (add in-place)

F_a <=> F_b (swap contents of both object, which is technically just swapping two pointers)

The main performance boost comes from my next trick:

- we check the lowest digit whether they are pandigital

- we check the highest digit whether they are pandigital

- *but we don't care what the other digits are*

→ whenever a number exceeds a certain size, I delete digits from the middle

Currently I delete if there are more than 10 cells, which represent 9 x 10 = 90 digits.

There is no special reason why I chose 10, it was the first number that I tried and it worked immediately.

The solution is found 100x faster now ...

## Modifications by HackerRank

The user can define F_1 and F_2 as well as how many digits should be pandigital.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

The code contains `#ifdef`

s to switch between the original problem and the Hackerrank version.

Enable `#ifdef ORIGINAL`

to produce the result for the original problem (default setting for most problems).

#include <iostream>
#include <vector>
//#define ORIGINAL

// store single digits with lowest digits first
// e.g. 1024 is stored as { 4,2,0,1 }

struct BillionNum : public std::vector<unsigned int>
{
// must store exactly 9 digits per cell
static const unsigned int MaxDigit = 1000000000;
// fill one cell
BillionNum(unsigned int x)
{
push_back(x);
}
// add a big number
void operator+=(const BillionNum& other)
{
// this code is a 95% copy of my BigNum class
// add in-place, make sure it's big enough
if (size() < other.size())
resize(other.size(), 0);
unsigned int carry = 0;
for (size_t i = 0; i < size(); i++)
{
// perform standard addition algorithm
carry += operator[](i);
if (i < other.size())
carry += other[i];
else
if (carry == 0)
return;
if (carry < MaxDigit)
{
// no overflow
operator[](i) = carry;
carry = 0;
}
else
{
// yes, we have an overflow
operator[](i) = carry - MaxDigit;
carry = 1;
}
}
if (carry > 0)
push_back(carry);
}
};
// return true if x is pandigital (1..digits)

bool isPandigital(unsigned int x, unsigned int digits = 9)
{
unsigned int mask = 0; // zero is not allowed, only 1 to 9
for (unsigned int i = 0; i < digits; i++)
{
// get next digit
auto current = x % 10;
if (current == 0 || current > digits)
return false;
auto bit = 1 << current;
// already seen ?
if ((mask & bit) != 0)
return false;
// mark that digit as "used"
mask |= bit;
x /= 10;
}
return true;
}
int main()
{
unsigned int first = 1; // F_1 = 1
unsigned int second = 1; // F_2 = 1
unsigned int digits = 9; // 9-pandigital (1..9)
#ifndef ORIGINAL
std::cin >> first >> second >> digits;
#endif
if (first == 1 && digits == 1)
{
std::cout << "1" << std::endl;
return 0;
}
unsigned long long Modulo = 1; // 10^digits
for (unsigned i = 1; i <= digits; i++)
Modulo *= 10;
// convert to a simplified BigNum
BillionNum a = first;
BillionNum b = second;
for (unsigned int i = 2; i <= 2000000; i++)
{
// analyze the lowest digits
auto lowest = b.front() % Modulo;
bool isPanLow = isPandigital(lowest, digits);
// look at the highest digits
if (isPanLow)
{
unsigned long long highest = b.back();
// maybe too few digits: use next cells, too
if (highest < Modulo && b.size() > 1)
highest = highest * BillionNum::MaxDigit + b[b.size() - 2];
// too many digits ? shrink until we have the right number of digits
while (highest >= Modulo)
highest /= 10;
// check if pandigital
if (isPandigital(highest, digits))
{
// yes, pandigital on both ends !
std::cout << i << std::endl;
return 0;
}
}
// next Fibonacci number
a += b;
std::swap(a, b);
// don't compute all digits:
// - look at the lowest digits
// - and look at the highest (incl. some of its neighbors because of carry)
// => remove those in the middle
// => my "magic numbers" 10 and 2 were chosen pretty much at random ...
if (a.size() > 10)
{
a.erase(a.begin() + 2);
b.erase(b.begin() + 2);
}
}
// failed
std::cout << "no solution" << std::endl;
return 0;
}

This solution contains 24 empty lines, 29 comments and 4 preprocessor commands.

# Interactive test

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

This live test is based on the Hackerrank problem.

This is equivalent to`echo "1 1 2" | ./104`

Output:

*Note:* the original problem's input `1 1 9`

__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 0.05 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

May 12, 2017 submitted solution

May 12, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler104

My code solves **15** out of **15** test cases (score: **100%**)

# Difficulty

Project Euler ranks this problem at **25%** (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=104 - **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-104-fibonacci-pandigital/ (written by Kristian Edlund)

Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p104.java (written by Nayuki)

Scala: github.com/samskivert/euler-scala/blob/master/Euler104.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 103 - Special subset sums: optimum | Special subset sums: testing - problem 105 >> |