<< problem 461 - Almost Pi Maximum number of divisors - problem 485 >>

# Problem 473: Phigital number base

Let phi be the golden ratio: phi = 1 + 5 sqrt{2}.
Remarkably it is possible to write every positive integer as a sum of powers of phi
even if we require that every power of phi is used at most once in this sum.
Even then this representation is not unique.
We can make it unique by requiring that no powers with consecutive exponents are used and that the representation is finite.
E.g: 2 = phi + phi^-2 and 3 = phi^2 + phi^-2

To represent this sum of powers of phi we use a string of 0's and 1's with a point to indicate where the negative exponents start.
We call this the representation in the phigital numberbase.
So 1=1_{phi}, 2=10.01_{phi}, 3=100.01_{phi} and 14=100100.001001_{phi}

The strings representing 1, 2 and 14 in the phigital number base are palindromic, while the string representing 3 is not.
(the phigital point is not the middle character).

The sum of the positive integers not exceeding 1000 whose phigital representation is palindromic is 4345.

Find the sum of the positive integers not exceeding 10^10 whose phigital representation is palindromic.

# My Algorithm

According to the problem statement, two 1s must be separated by at least one zero.
When I looked at the found numbers, there were always at least TWO zeros between two 1s.

In search() I generate only one half of the palindrome: the right-hand side. Obviously the left-hand side must be the mirrored version.
If current is the sum of a few phi^n then I add phi^{n+2}, then phi^{n+3}, phi^{n+4}, ... until the sum gets too large.

Unfortunately, I can't work around precision issues and need GCC's __float128 extension.
Computing phi^n on-the-fly caused even more precision problems - so I precomputed more accurate values with Wolfram Alpha (see large table in phipow().

## Alternative Approaches

I didn't realize that the golden ratio can be found in Fibonacci numbers, too (see en.wikipedia.org/wiki/Fibonacci_number):
F_n = dfrac{phi^n - (-phi)^{-n}}{2 phi - 1}
This way all my precision problem can be avoided - and even more important, a solution can be found in a few milliseconds.

## Note

Working with GCC's __float128 extension requires the std=gnu++11 compiler switch.

# 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):

This is equivalent to
echo 1000 | ./473

Output:

Note: the original problem's input 10000000000 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++. You can download it, too. Or just jump to my GitHub repository.

       #include <iostream>
#include <vector>
#include <cmath>

typedef __float128 Number; // G++ only !
// acceptable error
const Number Epsilon = 1e-15Q;

// ln(10^10)/ln((1+sqrt(5))/2) is about 47.85, so at most 48 exponents
unsigned int maxExponent = 48;

// return phi^exponent
Number phipow(int exponent)
{
// golden ratio is (1 + sqrt(5)) / 2
//const Number Phi = 1.618033988749894848204586834365638117720309179805762862135Q;

// precompute values Phi^0 ... Phi^48
static const Number precomputed[] =
{
1,
1.618033988749894848204586834365638117720309179805762862135Q,
2.618033988749894848204586834365638117720309179805762862135Q,
4.236067977499789696409173668731276235440618359611525724270Q,
6.854101966249684544613760503096914353160927539417288586406Q,
11.09016994374947424102293417182819058860154589902881431067Q,
17.94427190999915878563669467492510494176247343844610289708Q,
29.03444185374863302665962884675329553036401933747491720776Q,
46.97871376374779181229632352167840047212649277592102010484Q,
76.01315561749642483895595236843169600249051211339593731260Q,
122.9918693812442166512522758901100964746170048893169574174Q,
199.0050249987406414902082282585417924771075170027128947300Q,
321.9968943799848581414605041486518889517245218920298521475Q,
521.0019193787254996316687324071936814288320388947427468775Q,
842.9988137587103577731292365558455703805565607867725990250Q,
1364.000733137435857404797968963039251809388599681515345902Q,
2206.999546896146215177927205518884822189945160468287944927Q,
3571.000280033582072582725174481924073999333760149803290830Q,
5777.999826929728287760652380000808896189278920618091235757Q,
9349.000106963310360343377554482732970188612680767894526588Q,
15126.99993389303864810402993448354186637789160138598576234Q,
24476.00004085634900844740748896627483656650428215388028893Q,
39602.99997474938765655143742344981670294439588353986605128Q,
64079.00001560573666499884491241609153951090016569374634021Q,
103681.9999903551243215502823358659082424552960492336123914Q,
167761.0000059608609865491272482819997819661962149273587317Q,
271442.9999963159853080994095841479080244214922641609711232Q,
439204.0000022768462946485368324299078063876884790883298549Q,
710646.9999985928316027479464165778158308091807432493009781Q,
1149851.00000086967789739648324900772363719686922233763Q,
1860497.99999946250950014442966558553946800604996558693Q,
3010349.00000033218739754091291459326310520291918792456Q,
4870846.99999979469689768534258017880257320896915351149Q,
7881196.00000012688429522625549477206567841188834143605Q,
12752042.9999999215811929115980749508682516208574949475Q,
20633239.0000000484654881378535697229339300327458363836Q,
33385281.9999999700466810494516446738021816536033313311Q,
54018521.0000000185121691873052143967361116863491677147Q,
87403802.9999999885588502367568590705382933399524990459Q,
141422324.000000007071019424062073467274405026301666760Q,
228826126.999999995629869660818932537812698366254165806Q,
370248451.000000002700889084881006005087103392555832567Q,
599074577.999999998330758745699938542899801758809998373Q,
969323029.000000001031647830580944547986905151365830941Q,
1568397606.99999999936240657628088309088670691017582931Q,
2537720636.00000000039405440686182763887361206154166025Q,
4106118242.99999999975646098314271072976031897171748957Q,
6643838879.00000000015051539000453836863393103325914982Q,
10749957121.999999999906976373147249098394250004976639Q // already too big: > 10^10
};

// lookup cached values
if (exponent >= 0)
return        precomputed[+exponent];
else
return 1.0Q / precomputed[-exponent];
}

// return phi^pos + phi^-(pos+1)
Number phipowBoth(unsigned int pos)
{
// precompute phi^i + phi^-(i+1)
// => those are the pairs in the palindromic representation
static std::vector<Number> cache;
if (cache.empty())
{
cache.push_back(phipow(0));
for (int i = 1; i < 49; i++)
cache.push_back(phipow(i) + phipow(-(i+1)));
}

return cache[pos];
}

// generate all strings where there are at least two zeros between two ones
unsigned long long search(Number limit, unsigned int exponent = 0, Number current = 0)
{
// "out of bounds" ?
if (current > limit)
return 0;

unsigned long long result = 0;

// an integer ?
auto rounded = round(current); // avoid rounding issues, add a small "buffer"
auto diff = current - rounded;
if (diff > -Epsilon && diff < +Epsilon)
result += rounded;

// it turns out at least two zeros between a pair of ones
// go deeper by appending "001", "0001", "00001", etc.
// => at least two zeros and a single one
// except next to the dot: a zero on one side generates a zero on the other side, too (palindrome)
if (exponent == 0)
exponent++;
else
exponent += 2;

// until phi^48 > 10^10
while (exponent <= maxExponent)
{
// add phi^exponent (and its mirrored "twin" phi^-(exponent+1)
result += search(limit, exponent+1, current + phipowBoth(exponent));
exponent++;
}

return result;
}

int main()
{
// 10^10
unsigned long long limit = 10000000000ULL;
std::cin >> limit;

// find maximum length of one side of the palindrome
maxExponent = 0;
while (phipow(maxExponent) <= limit)
maxExponent++;

// only one phigital number can have a 1 next to the dot => that's the degenerated string "1"
// all other strings must have zeros next to the dot bceause "no powers with consecutive exponents"
std::cout << 1 + search(limit) << std::endl;
return 0;
}


This solution contains 16 empty lines, 23 comments and 3 preprocessor commands.

# Benchmark

The correct solution to the original Project Euler problem was found in 19.3 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

September 30, 2017 submitted solution

# Difficulty

Project Euler ranks this problem at 35% (out of 100%).

# 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 461 - Almost Pi Maximum number of divisors - problem 485 >>
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