samcarter
> This is part of the Summer of Code 2022 series, see https://topanswers.xyz/tex?q=2059 for more information
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 ≤ 1000, is the number of solutions maximised?
(this programming puzzle is taken from https://projecteuler.net/problem=39, licensed under CC BY-NC-SA 4.0)
Top Answer
samcarter
*No spoiler*
Expanding the solution to SoC Day 1 a bit gives an inefficient but working solution:
```
% !TeX TS-program = pdflatex
\documentclass{article}
\begin{document}
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 $p \leq 1000$, is the number of solutions maximised?
\ExplSyntaxOn
\int_new:N \l_sam_a_int
\int_new:N \l_sam_b_int
\int_new:N \l_sam_p_int
\int_new:N \l_sam_count_int
\int_new:N \l_sam_tot_int
\int_new:N \l_sam_tot_p_int
\int_step_inline:nn { 1000 }
{
\int_set:Nn \l_sam_p_int { #1 }
\int_zero:N \l_sam_count_int
% loop over b (next biggest integer)
\int_step_inline:nn {\l_sam_p_int}
{
\int_set:Nn \l_sam_b_int { ##1 }
% loop over a (smalles integer)
\int_step_inline:nn {\l_sam_b_int-1}
{
\int_set:Nn \l_sam_a_int { ####1 }
% check if a^2 + b^2 = (p - a - b)^2
\int_compare:nNnT
{
\l_sam_b_int*\l_sam_b_int + \l_sam_a_int*\l_sam_a_int
} = {
(\l_sam_p_int-\l_sam_a_int-\l_sam_b_int)*(\l_sam_p_int-\l_sam_a_int-\l_sam_b_int)
}{
\int_incr:N \l_sam_count_int
}% if a^2 + b^2 = (p - a - b)^2
}% loop a
}% loop b
\par \int_use:N \l_sam_p_int :~ \int_use:N \l_sam_count_int
\int_compare:nNnT { \l_sam_count_int } > { \l_sam_tot_int }
{
\int_set:Nn \l_sam_tot_int { \l_sam_count_int }
\int_set:Nn \l_sam_tot_p_int { \l_sam_p_int }
}
}
\par Maximum~ solution~ for~ $p = \int_use:N \l_sam_tot_p_int$:~
\int_use:N \l_sam_tot_int \nobreakspace solutions.
\ExplSyntaxOff
\end{document}
```
Answer #2
CrazyHorse
without sorting the values and using `lualatex`
```
\documentclass{article}
\usepackage{luacode}
\begin{luacode}
function triangle(p)
tex.print("Triangle Numbers for "..tostring(p).."\\par")
local found = false
for b = 1, math.floor(p/3)-1 do
for a = 1, p-b do
if (a*a+b*b) == (p-a-b)^2 then
c = p - a - b
tex.print("("..tostring(a)..","..tostring(b)..","..tostring(c).."), ")
found = true
end
end
end
if not found then tex.print("---") end
end
\end{luacode}
\def\triangle#1{\directlua{triangle(#1)}}
\begin{document}
\triangle{120}\par
\triangle{200}\par
\triangle{500}\par
\triangle{1000}
\end{document}
```
