Problem 98: Anagramic squares

(see projecteuler.net/problem=98)

By replacing each of the letters in the word CARE with 1, 2, 9, and 6 respectively, we form a square number: 1296 = 362.
What is remarkable is that, by using the same digital substitutions, the anagram, RACE, also forms a square number: 9216 = 962.
We shall call CARE (and RACE) a square anagram word pair and specify further that leading zeroes are not permitted, neither may a different letter have the same digital value as another letter.

Using words.txt (right click and 'Save Link/Target As...'), a 16K text file containing nearly two-thousand common English words,
find all the square anagram word pairs (a palindromic word is NOT considered to be an anagram of itself).

What is the largest square number formed by any member of such a pair?

NOTE: All anagrams formed must be contained in the given text file.

My Algorithm

Two numbers are permutations of each other if they share the same digits.
Two words are permutations of each other if they share the same letters.

My fingerprint from problem 49 (and used in many more problems) returns the same result if two numbers share the same digits.

Two words share the same letters if sorting their letters produces the same string: sort(KEEP)=EEKP and sort(PEEK)=EEKP.

After all input has been read, my problem looks for the longest word with at least one "anagram sibling".
Hint: it's 9 for the anagram siblings ("INTRODUCE", "REDUCTION")

Then all square numbers 1^2, 2^2, 3^2, ... with at most 9 digits are computed.
Their fingerprints are stored in permutations along with the base number (where square = base^2).
A second container fingerprintLength keeps track of the number of digits of square.
C++'s log10 produces minor rounding artefacts - that's why I had to subtract 1.

All words with same anagram are then matched against each other.

Modifications by HackerRank

The Hackerrank problem is such different that I decided to use completely different code:
I have to find the largest square with n digits that is an anagram of another square.
My solution is extremely short compared to the original problem.

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.

Input data (separated by spaces or newlines):

This is equivalent to
echo 4 | ./98

Output:

(please click 'Go !')

(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, as well as the input data, too.

The code contains #ifdefs 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 <cmath>
 
#include <vector>
#include <set>
#include <map>
 
#include <string>
#include <algorithm>
 
#include <iostream>
 
// count digits, two numbers have the same fingerprint if they are permutations of each other
unsigned long long fingerprint(unsigned long long x)
{
unsigned long long result = 0;
while (x > 0)
{
auto digit = x % 10;
x /= 10;
 
result += 1LL << (4 * digit);
}
return result;
}
 
 
//#define ORIGINAL
#ifdef ORIGINAL // the original problem differs significantly from Hackerrank
 
// read a single word from STDIN, syntax: "abc","def","xyz"
std::string readWord()
{
std::string result;
while (true)
{
// read one character
char c = std::cin.get();
// no more input ?
if (!std::cin)
break;
 
// ignore quotes
if (c == '"')
continue;
// finish when a comma appears
if (c == ',')
break;
 
// nope, just an ordinary letter (no further checks whether c in 'A'..'Z')
result += c;
}
return result;
}
 
// return biggest square if both a and b can be mapped to anagram squares, else return 0
unsigned long long match(const std::string& a, const std::string& b, const std::vector<unsigned long long>& squares)
{
unsigned long long result = 0;
// try all combinations
for (auto i : squares)
for (auto j : squares)
{
// don't match a word with itself
if (i == j)
continue;
 
// convert squares to strings
auto replaceA = std::to_string(i);
auto replaceB = std::to_string(j);
 
// 1. replace all digits of squareA by the letters of a
// 2. at the same time, whenever such a digit can be found in squareB replace it by the same letter
 
// [digit] => letter
std::map<char, char> replaceTable;
bool valid = true;
for (size_t k = 0; k < replaceA.size(); k++)
{
char original = replaceA[k];
// no replacement rule found ? => abort
if (replaceTable.count(original) != 0 &&
replaceTable[original] != a[k])
valid = false;
 
// replacement successful
replaceTable[original] = a[k];
}
 
// two digits must not map to the same letter, though
std::set<char> used;
for (auto x : replaceTable)
{
// already used ?
if (used.count(x.second) != 0)
valid = false;
// mark as used
used.insert(x.second);
}
 
// any constraint violation ?
if (!valid)
continue;
 
// using that mapping, can "a" be constructed ?
std::string aa;
for (auto x : replaceA)
aa += replaceTable[x];
if (aa != a)
continue;
 
// using that mapping, can "b" be constructed ?
std::string bb;
for (auto x : replaceB)
bb += replaceTable[x];
if (bb != b)
continue;
 
// new bigger square ?
if (result < i)
result = i;
if (result < j)
result = j;
}
 
return result;
}
 
int main()
{
// find word anagrams: sort letters of each word
// [sorted letters] => [list of words]
std::map<std::string, std::vector<std::string>> anagrams;
 
// read all words from STDIN and fill "anagram" container
while (true)
{
// read a single word, abort when empty
auto word = readWord();
if (word.empty())
break;
 
auto sorted = word;
std::sort(sorted.begin(), sorted.end());
// add to word anagrams
anagrams[sorted].push_back(word);
}
 
// find longest anagram
size_t maxDigits = 0;
for (auto i : anagrams)
if (i.second.size() > 1) // at least two words share the same letters ?
if (maxDigits < i.second.front().size())
maxDigits = i.second.front().size();
// maxDigits will be 9 for the given input ("INTRODUCE", "REDUCTION")
 
unsigned long long maxNumber = 1;
for (size_t i = 0; i < maxDigits; i++)
maxNumber *= 10;
 
// generate all squares
// for each square, compute its fingerprint
std::map<unsigned long long, std::vector<unsigned long long>> permutations;
std::map<unsigned int, std::set <unsigned long long>> fingerprintLength;
// walk through all square numbers (base^2)
unsigned long long base = 1;
while (base*base <= maxNumber)
{
auto square = base*base;
auto id = fingerprint(square);
permutations[id].push_back(square);
 
auto numDigits = log10(square - 1) + 1;
fingerprintLength[numDigits].insert(id);
 
base++;
}
 
// only process non-unique words (size > 1)
unsigned long long result = 0;
for (auto i : anagrams)
{
auto pairs = i.second;
// no other word with the same letters ?
if (pairs.size() == 1)
continue;
 
// there is a chance that not all words of a permutation group are squares,
// there only need to be at least two matching words
auto length = pairs.front().size();
// compare each word with each other
for (size_t i = 0; i < pairs.size(); i++)
for (size_t j = i + 1; j < pairs.size(); j++)
{
// extract all relevant squares
for (auto id : fingerprintLength[length])
{
// and perform the matching process ...
auto best = match(pairs[i], pairs[j], permutations[id]);
// bigger square found ?
if (result < best)
result = best;
}
}
}
 
std::cout << result << std::endl;
}
 
 
#else // ---------- completely different algorithms for Hackerrank problem ----------
 
int main()
{
unsigned int digits;
std::cin >> digits;
 
// find smallest and largest number with the number of digits
// e.g. digits = 5 ==> minNumber = 10000 and maxNumber = 99999
unsigned long long minNumber = 1;
for (unsigned int i = 1; i < digits; i++)
minNumber *= 10;
unsigned long long maxNumber = minNumber * 10 - 1;
 
// generate all squares between minNumber and maxNumber
// for each square, compute its fingerprint
unsigned long long base = sqrt(minNumber);
if (base*base < minNumber)
base++;
std::map<unsigned long long, std::vector<unsigned long long>> permutations;
while (base*base <= maxNumber)
{
auto square = base*base;
permutations[fingerprint(square)].push_back(square);
base++;
}
 
// find most common fingerprint and its highest value
size_t bestCount = 0;
unsigned long long highestSquare = 0;
for (auto p : permutations)
{
auto size = p.second.size();
auto high = p.second.back();
 
// same number of elements but highest square exceeds the previous record ?
if (bestCount == size && highestSquare < high)
highestSquare = high;
 
// new record number of elements
if (bestCount < size)
{
bestCount = size;
highestSquare = high;
}
}
 
std::cout << highestSquare << std::endl;
}
 
#endif

This solution contains 42 empty lines, 49 comments and 10 preprocessor commands.

Benchmark

The correct solution to the original Project Euler problem was found in 0.24 seconds on an Intel® Core™ i7-2600K CPU @ 3.40GHz.
Peak memory usage was about 5 MByte.

(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 14, 2017 submitted solution
May 9, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler098

My code solves 11 out of 11 test cases (score: 100%)

Difficulty

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

Hackerrank describes this problem as hard.

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.

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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The 289 solved problems (level 11) had an average difficulty of 32.1% at Project Euler and
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.

more about me can be found on my homepage, especially in my coding blog.
some names mentioned on this site may be trademarks of their respective owners.
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