Project Euler Problem 94 Solution

Project Euler Problem 94 Statement

It is easily proved that no equilateral triangle exists with integral length sides and integral area. However, the almost equilateral triangle $5$-$5$-$6$ has an area of $12$ square units.

We shall define an almost equilateral triangle to be a triangle for which two sides are equal and the third differs by no more than one unit.

Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed one billion ($1\,000\,000\,000$).

Solution

HackerRank Project Euler Problem 79 overview:

Below is a step-by-step explanation of how (and why) this code solves Project Euler #94. The problem (in short) is to find the sum of the perimeters of all “almost equilateral” triangles with integer sides and area, where those perimeters do not exceed a given limit $N$. “Almost equilateral” here means that two sides are equal and the third side differs by 1 (i.e., sides are $(s,s,s+1)$ or $(s,s,s−1)$)

Solution Summary

This script solves Project Euler #94 by directly generating the valid side lengths $ s $ (for almost-equilateral triangles) via a Pell-type recurrence instead of brute force. It stores two consecutive $ s $-values (s0 and s1), alternates the sign (m) to handle both $ s+1 $ and $ s-1 $ triangles, and calculates each valid perimeter $ p = 3s \pm 1 $. As long as $ p $ is within the limit $ N $, it adds that perimeter to the running total (s). Once the newly computed perimeter exceeds $ N $, it stops and prints the sum of all such perimeters.

1. Understanding the Mathematical Background

1.1 The Geometry / Number Theory Setup

We’re looking for integer-sided triangles $ (s, s, s\pm1) $ that have an integer area.
For a triangle with sides $ (s, s, s+1) $ or $ (s, s, s-1) $ to have integer area, $ s $ can only take certain values. These values come from solving a Pell-type equation that emerges when enforcing the area to be an integer (specifically using Heron’s formula).

1.2 Pell Equations and Recurrences

Solutions to certain Pell equations or Pell-like equations often appear in sequences that satisfy linear recurrences.
In particular, once you find the first few valid $ s $ values by direct calculation, you can generate the next ones via a recurrence of the form:
$s_{k+1} = a \cdot s_k - s_{k-1} + (\text{some constant terms}),$ and so forth, where $ a $ is often related to the fundamental solution of the Pell equation.

1.3 Relation to the Perimeter

For an “almost equilateral” triangle with sides $ (s, s, s\pm1) $, its perimeter $ p $ is $ p = s + s + (s\pm1) = 3s \pm 1 $. Hence, when the code computes $ s $ (the “base” side length) and decides which sign ($ +1 $ or $ -1 $) is relevant, we get $ p = 3s \pm 1 $.

2. Variables in the Code

Let’s introduce the code’s variables with short explanations:

s0, s1:
- These hold two consecutive terms in the sequence of valid side lengths $ s $.
- The sequence of $ s $ values is generated by a linear recurrence to ensure integer area.
m:
- This toggles between $ +1 $ and $ -1 $, effectively switching between triangles $ (s, s, s+1) $ and $ (s, s, s-1) $.
- Notice the assignment m = -m in each iteration.
p:
- This represents the perimeter of the current triangle under consideration.
- p is set to $ 3 \times s1 - m $, which is $ 3s \pm 1 $.
s:
- This accumulates (sums up) all valid perimeters that do not exceed $ N $.
- In other words, s will be the final answer each time we process a limit N.
N:
- The upper limit for the perimeter.

3. Stepping Through the Code

3.1 Initialization

s0, s1, s, p, m = 1, 1, 0, 0, 1

s0 = 1, s1 = 1: We start with the first two terms in the special sequence of side lengths both as 1. These seed values come from the minimal valid solutions for near-equilateral triangles with integer area.
s = 0: Will accumulate the sum of perimeters.
p = 0: Perimeter will be computed in the loop before we check its value.
m = 1: We start assuming the triangle is of the form $ (s, s, s+1) $. Each iteration will toggle this to $ -1 $ for the next one, capturing $ (s, s, s-1) $.

3.2 The `while` Loop

while p <= N:
    s0, s1, m = s1, 4 * s1 - s0 + 2 * m, -m
    s += p
    p = 3 * s1 - m

Check condition: while p <= N:
We will keep generating new solutions (new perimeters) as long as the current perimeter p does not exceed N.
Recurrence update:
```
s0, s1, m = s1, 4*s1 - s0 + 2*m, -m
```
- s0, s1 become the next pair of consecutive terms in the side-length sequence.
- The “new” side length is computed by $ 4 \times s1 - s0 + 2 \times m $. This is a Pell-type recurrence that generates the next valid side $ s $.
- We then flip m = -m to alternate between the $ (s, s, s+1) $ form and the $ (s, s, s-1) $ form.
Accumulate the previous perimeter:
```
s += p
```
- We add the previously computed (valid) perimeter to our total sum.
- Note that on the first iteration, p was 0 (initial), so it just adds nothing, but ensures the loop logic remains consistent as we iterate.
Compute the new perimeter:
```
p = 3 * s1 - m
```
- Given the current s1 and the sign m, the new perimeter is $ 3s1 - m $.
- If $ m = +1 $, perimeter = $ 3s - 1 $; if $ m = -1 $, perimeter = $ 3s + 1 $.
The loop will stop once this new perimeter p becomes larger than N.

3.3 Printing the Result

After the loop ends, we print s, which holds the sum of all valid triangle perimeters that are $ \leq N $.

4. Key Takeaways

Generating Valid Triangles: The code uses a known linear recurrence relation to jump directly from one valid near-equilateral triangle (with integer area) to the next, rather than checking all $ s $ one by one.
Efficient Summation: Because of the recurrence, each valid perimeter is found in constant time per iteration, making it very efficient for large $ N $.
Toggling m: The value m toggles between $ +1 $ and $ -1 $, which accounts for the two possibilities $ s+1 $ and $ s-1 $ in the triangle’s third side.
Why Pell-like Recurrence? Heron’s formula demands the area $ \sqrt{s(s)(s\pm1)(\dots)} $ to be an integer. Solving that diophantine condition leads to a Pell-type equation, whose integer solutions appear in recursive sequences.

HackerRank version

Solves all test cases for Hackerrank Project Euler Problem 94

HackerRank Project Euler 94: Find the sum of the perimeters of all almost equilateral triangles with integral side lengths and area and whose perimeters do not exceed $N$, (15 ≤ N ≤ 10^{18}).

Python Source Code

for _ in range(int(input())):
    N = int(input())
    s0, s1, s, p, m = 1, 1, 0, 0, 1
    while p <= N:
        s0, s1, m = s1, 4 * s1 - s0 + 2 * m, -m
        s+= p
        p = 3 * s1 - m
    print(s)