Project Euler Problem 39 Solution

Project Euler Problem 39 Statement

If $p$ is the perimeter of a right angle triangle with integral length sides, $\{a, b, c\}$, there are exactly three solutions for $p = 120$.

$\{20,48,52\}$, $\{24,45,51\}$, $\{30,40,50\}$

For which value of $p \le 1000$, is the number of solutions maximised?

Solution

HackerRank Project Euler Problem 39 overview:

Given an upper limit \( L \) on the perimeter of right‐angled triangles with integer sides, determine the perimeter \( p \leq L \) that maximizes the number of distinct right‐angled triangles that can be formed.

Solution Summary

The solution precomputes the number of right‐angled triangles for each perimeter up to \(N = 5\,000\,000\) by generating primitive Pythagorean triples using a modified version of Euclid’s formula and updating all their multiples, then builds a prefix maximum array that records the best perimeter (with the most solutions) up to each value, enabling each query for a given \(L\) to be answered in constant time.

Solution Details

1. Initial Settings

The program begins by importing the required modules and setting up an array c of length \( N \):

import itertools, math
N = 5_000_000
c = [0] * N

Here:

• \( N = 5\,000\,000 \) is the limit up to which perimeters will be considered.
• c is an array where c[p] will eventually store the number of different right‐angled triangles that have perimeter \( p \).

2. Generating Pythagorean Triples

Next, the code generates Pythagorean triples using a variation of Euclid’s formula. Recall that the primitive Pythagorean triple can be generated by:

\( a = m^2 - n^2,\quad b = 2mn,\quad c = m^2 + n^2, \) with \( m > n \), \( m \) and \( n \) coprime and of opposite parity.

Their perimeter is:

\( p = a + b + c = 2m(m+n). \)

The program uses a modified parameterization:

for t, s in itertools.combinations_with_replacement(range(1, math.isqrt(N), 2), 2):
    if math.gcd(s, t) == 1:
        x = s * (s + t)
        c[x::x] = [v+1 for v in c[x::x]]

Let's break down this loop:

• Iteration Over \( s \) and \( t \):
  The loop iterates over pairs \( (t, s) \) where both \( t \) and \( s \) are odd numbers from \( 1 \) up to \( \texttt{math.isqrt(N)} \). Using itertools.combinations_with_replacement ensures that \( t \leq s \). This smartly reduces the parameter space.
• Ensuring Primitiveness:
  The condition if math.gcd(s, t) == 1: guarantees that \( s \) and \( t \) are coprime, ensuring the resulting triple is primitive.
• Calculating \( x \):
  The value \( x \) is computed as:
  \( x = s(s+t). \)
  Note that in Euclid’s formula the full perimeter of a primitive triple is \( 2m(m+n) \). Here, by working with odd numbers for \( s \) and \( t \), we treat \( x \) as half of the full perimeter.
• Incrementing Counts Using Slicing:
  The slicing assignment:
  c[x::x] = [v+1 for v in c[x::x]]
  is equivalent to:
  for j in range(x, N, x): c[j] += 1
  This loop increments the count for every multiple of \( x \) (i.e. \( x, 2x, 3x, \dots \) up to \( N \)). Since each right-angled triangle (primitive or its multiples) contributes a solution for its perimeter, this effectively counts all solutions.

3. Building the “Best-So-Far” Array

After the counts are in place, the program constructs an array \( r \) such that for each perimeter \( i \), r[i] holds the perimeter \(\leq i\) which has the maximum number of solutions. The code is:

m, a, r = 0, 0, []
for i, v in enumerate(c):
    if v > m:
        m, a = v, i
    r.append(a)

Here:

• m holds the maximum count found so far.
• a stores the corresponding perimeter.
• The loop goes through all perimeters \( 0 \) to \( N-1 \) and updates \( m \) and \( a \) whenever a new maximum is found. Then, it appends the current best perimeter \( a \) to the array \( r \).

4. Answering the Queries

The final part of the code quickly answers the input queries using the precomputed array \( r \):

for _ in range(int(input())):
    print(r[int(input())])

For each test case, the program reads an integer \( L \) (the maximum allowed perimeter) and immediately outputs r[L], which is the perimeter \( \leq L \) having the most right-angled triangle solutions.

HackerRank version

Solves all test cases for Hackerrank Project Euler Problem 39

HackerRank Project Euler 39 includes either 1—9 or 1—8 pandigital numbers and has you find the multipliers 100 ≤ N ≤ 10⁵ in ascending order.