Problem 93: Arithmetic expressions

(see projecteuler.net/problem=93)

By using each of the digits from the set, {1, 2, 3, 4}, exactly once, and making use of the four arithmetic operations (+, , *, /) and brackets/parentheses,
it is possible to form different positive integer targets.

For example,

8 = (4 * (1 + 3)) / 2
14 = 4 * (3 + 1 / 2)
19 = 4 * (2 + 3) − 1
36 = 3 * 4 * (2 + 1)

Note that concatenations of the digits, like 12 + 34, are not allowed.

Using the set, {1, 2, 3, 4}, it is possible to obtain thirty-one different target numbers of which 36 is the maximum,
and each of the numbers 1 to 28 can be obtained before encountering the first non-expressible number.

Find the set of four distinct digits, a < b < c < d, for which the longest set of consecutive positive integers, 1 to n,
can be obtained, giving your answer as a string: abcd.

Algorithm

No matter what brackets or operators are used, the basic pattern is always the same:
1. an operation on two out of the four numbers is executed
2. the result will become part of the equation which consists now of three numbers
3. an operation on two out of the three numbers is executed
4. the result will become part of the equation which consists now of two numbers
5. an operation on the final two numbers is executed

I generate all sets of digits {a, b, c, d} where a < b < c < d. There are 126 such sets and four nested loops produce these sets in main.
My eval function accepts an arbitrary set of numbers and does the following:
- pick any two numbers and execute all operations on these numbers, thus reducing numbers by one element
- call itself recursively
- when only one element is left in numbers then mark the result as used

There a a few pitfalls:
- values may temporarily be non-integer, e.g. (1/3)*6+7 = 9, but may produce a valid integer result.
- values may temporarily be negative, e.g. (1-2)+3+4 = 6, but may produce a valid positive integer result.
- division by zero is not allowed, not even temporarily
- any order of digits must be checked: a op b and b op a

Especially the first pitfall (rational numbers) is tough to handle without floating-point numbers.
However, double (or float) may have some rounding issues, that's why I added an Epsilon to make sure that results like 3.9999 are treated as 4.

getSequenceLength is the glue between main and eval:
It finds the first gap, that means the first number not represented by any combination of arithmetic operations.

My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, 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 <vector>
#include <iostream>
#include <cmath>
 
const double Epsilon = 0.00001;
 
// try all arithmetic operations of any two elements of "numbers", set their result in "used" to true
void eval(const std::vector<double>& numbers, std::vector<bool>& used)
{
// 1. if array holds just one element, add it to the "used" list and we are done
// 2. pick any two numbers
// 3. loop through all operators
// 4. add result to the array and call eval() recursively
 
// step 1
if (numbers.size() == 1)
{
auto result = numbers.front() + Epsilon;
// reject non-integer result (caused by division)
if (fmod(result, 1) > 10*Epsilon)
return;
 
int index = int(result + Epsilon);
// reject negative and very large results
if (index >= 0 && index < (int)used.size())
used[index] = true;
 
return;
}
 
// step 2
auto next = numbers;
for (size_t i = 0; i < numbers.size(); i++)
for (size_t j = i + 1; j < numbers.size(); j++)
{
// fetch two numbers
double a = numbers[i];
double b = numbers[j];
 
// prepare for next recursive step
next = numbers;
next.erase(next.begin() + j); // delete the higher index first
next.erase(next.begin() + i);
 
// steps 3 and 4 (unrolled)
next.push_back(a + b); // add
eval(next, used);
next.back() = a - b; // subtract (I)
eval(next, used);
next.back() = b - a; // subtract (II)
eval(next, used);
next.back() = a * b; // multiply
eval(next, used);
if (b != 0)
{
next.back() = a / b; // divide (I)
eval(next, used);
}
if (a != 0)
{
next.back() = b / a; // divide (II)
eval(next, used);
}
 
// note: I overwrite the last element because that's simpler than
// pop_back(); push_back(...);
}
}
 
// evaluate all expressions and count how many numbers from 1 to x can be expressed without any gaps
unsigned int getSequenceLength(const std::vector<double>& numbers)
{
// find all results
std::vector<bool> used(1000, false);
eval(numbers, used);
 
// longest sequence, beginning from 1
unsigned int result = 0;
while (used[result + 1])
result++;
return result;
}
 
int main()
{
#define ORIGINAL
#ifdef ORIGINAL
unsigned int longestSequence = 0;
unsigned int longestDigits = 0;
 
// four different digits
for (unsigned int a = 1; a <= 6; a++)
for (unsigned int b = a+1; b <= 7; b++)
for (unsigned int c = b+1; c <= 8; c++)
for (unsigned int d = c+1; d <= 9; d++)
{
// evaluate all combinations
auto sequenceLength = getSequenceLength({ double(a), double(b), double(c), double(d) });
 
// new record ?
if (longestSequence < sequenceLength)
{
longestSequence = sequenceLength;
// build one number out of the four digits
longestDigits = a * 1000 + b * 100 + c * 10 + d;
}
}
 
// print result
std::cout << longestDigits << std::endl;
 
#else
 
// 1..5 digits
unsigned int numDigits;
std::cin >> numDigits;
// read those digits
std::vector<double> numbers(numDigits);
for (auto& x : numbers)
std::cin >> x;
 
// print length of longest sequence
std::cout << getSequenceLength(numbers);
 
#endif
 
return 0;
}

This solution contains 20 empty lines, 25 comments and 7 preprocessor commands.

Interactive test

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

Number of test cases (1-5):

Input data (separated by spaces or newlines):

This is equivalent to
echo "1 " | ./93

Output:

(please click 'Go !')

(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 0.02 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 12, 2017 submitted solution
May 8, 2017 added comments

Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler093

My code solved 25 out of 25 test cases (score: 100%)

Difficulty

Project Euler ranks this problem at 15% (out of 100%).

Hackerrank describes this problem as medium.

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=93 - 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-93-longest-sequence/ (written by Kristian Edlund)
Java: github.com/nayuki/Project-Euler-solutions/blob/master/java/p093.java (written by Nayuki)
Scala: github.com/samskivert/euler-scala/blob/master/Euler093.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 already 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.

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The 133 solved problems had an average difficulty of 16.9% at Project Euler and I scored 11,174 points (out of 12300) at Hackerrank's Project Euler+.
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