notes.html

problem 94 · 12 lines

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<h3>Problem Analysis</h3>
<p>An almost equilateral triangle has two equal sides of length <code>a</code> and a third side that differs from them by exactly one unit, so the base is either <code>a-1</code> or <code>a+1</code>. The challenge is to require that both the side lengths <em>and</em> the area be integers. The natural tool is <a class="reference-source" href="https://en.wikipedia.org/wiki/Heron%27s_formula" rel="noopener noreferrer" target="_blank">Heron's formula</a>. For an isosceles triangle with equal sides <code>a</code> and base <code>b</code>, dropping a perpendicular from the apex to the midpoint of the base splits the figure into two congruent right triangles, and the area becomes <code>A = (b/2) * sqrt(a^2 - (b/2)^2)</code>. For <code>A</code> to be a positive integer, the quantity <code>a^2 - (b/2)^2</code> under the square root must itself be a perfect square. That is the integrality condition that drives everything.</p>
<p>Substituting <code>b = a+1</code> (and noting <code>a</code> must be odd so that <code>b/2</code> stays rational), then clearing denominators, the perfect-square requirement collapses into a <a class="reference-source" href="https://en.wikipedia.org/wiki/Pell%27s_equation" rel="noopener noreferrer" target="_blank">Pell-like equation</a> - a Diophantine equation of the form <code>x^2 - D*y^2 = N</code> for small integer constants <code>D</code> and <code>N</code>, here built around <code>D = 3</code>. The case <code>b = a-1</code> yields a sibling equation. The decisive fact about <a class="reference-source" href="https://en.wikipedia.org/wiki/Pell%27s_equation" rel="noopener noreferrer" target="_blank">Pell equations</a> is that their solutions, though infinite in number, are not scattered at random: they obey a <a class="reference-source" href="https://en.wikipedia.org/wiki/Recurrence_relation" rel="noopener noreferrer" target="_blank">linear recurrence relation</a>. Each valid triangle can therefore be generated from the previous one by a fixed arithmetic update, instead of testing every integer side length up to the bound. Because the valid side lengths grow geometrically (roughly by a factor of 4 per step), fewer than twenty almost-equilateral triangles exist below a perimeter of one billion. The algorithm runs in <code>O(log N)</code> time in the perimeter bound <code>N</code> - effectively instantaneous.</p>
<h3>Solution Approach</h3>
<h4>Recurrence-Based Pell Equation Enumeration (Python and C)</h4>
<p>The Python and C solutions implement the same algorithm and differ only in language syntax. The state is five variables: <code>s</code> (the running sum of perimeters), <code>s1</code> and <code>s2</code> (two consecutive terms of the <a class="reference-source" href="https://en.wikipedia.org/wiki/Recurrence_relation" rel="noopener noreferrer" target="_blank">recurrence</a> for the equal side length, both initialised to 1), <code>m</code> (a sign alternator initialised to +1), and <code>p</code> (the current candidate perimeter, initialised to 0).</p>
<p>The loop runs while <code>p</code> does not exceed the bound. Each iteration first folds the current perimeter into the accumulator (<code>s + p</code>), then advances the recurrence. The next side length is computed as <code>4*s2 - s1 + 2*m</code>. The factor of 4 is the multiplier coming from the fundamental solution of the underlying <a class="reference-source" href="https://en.wikipedia.org/wiki/Pell%27s_equation" rel="noopener noreferrer" target="_blank">Pell equation</a> <code>x^2 - 3*y^2 = 1</code>; the <code>+2*m</code> term is a small correction whose sign alternates between the two families of triangles (base <code>a+1</code> versus <code>a-1</code>). The sign alternator then flips with <code>m = -m</code>, and the new perimeter follows directly as <code>p = 3*s2 - m</code>. When the base is <code>a+1</code> we have <code>m = -1</code> so <code>p = 3a+1</code>; when it is <code>a-1</code> we have <code>m = +1</code> so <code>p = 3a-1</code>. A single expression <code>3*s2 - m</code> captures both cases.</p>
<p>The two families are interleaved into one monotonically increasing sequence. The loop therefore terminates the instant a generated perimeter exceeds the bound, with no further searching. Because each new term is roughly four times the previous, the loop executes only a logarithmic number of times.</p>
<p>The Python version updates all four state variables in one <a class="reference-source" href="https://docs.python.org/3/reference/simple_stmts.html#assignment-statements" rel="noopener noreferrer" target="_blank">parallel-assignment</a> statement, exploiting the fact that the entire right-hand side is evaluated before any name is rebound. The C version computes the two dependent values (<code>new_s2</code> and <code>new_m</code>) into temporaries before writing back, which is the idiomatic way to express the same simultaneous update in a language without parallel assignment. The implementations are logically identical, and both complete in unmeasurable time (recorded as 0.0000 s).</p>
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