Law of cosines - triples: Difference between revisions

=={{header|Phix}}==

Using a simple flat sequence of 100 million elements (well within the language limits) proved significantly faster than a dictionary (5x or so).

<lang Phix>sequence squares = repeat(0,10000*10000)

for c=1 to 10000 do

squares[c*c] = c

end for

function solve(integer angle, maxlen, bool samelen=true)

sequence res = {}

for a=1 to maxlen do

integer a2 = a*a

for b=a to maxlen do

integer c2 = a2+b*b

if angle!=90 then

if angle=60 then c2 -= a*b

elsif angle=120 then c2 += a*b

else crash("angle must be 60/90/120")

end if

end if

integer c = iff(c2>length(squares)?0:squares[c2])

if c!=0 and c<=maxlen then

if samelen or a!=b or b!=c then

res = append(res,{a,b,c})

end if

end if

end for

end for

return res

end function

procedure show(string fmt,sequence res, bool full=true)

printf(1,fmt,{length(res),iff(full?sprint(res):"")})

end procedure

puts(1,"Integer triangular triples for sides 1..13:\n")

show("Angle 60 has %2d solutions: %s\n",solve( 60,13))

show("Angle 90 has %2d solutions: %s\n",solve( 90,13))

show("Angle 120 has %2d solutions: %s\n",solve(120,13))

show("Non-equilateral angle 60 triangles for sides 1..10000: %d%s\n",solve(60,10000,false),false)</lang>

{{out}}

<pre style="font-size: 11px">

Integer triangular triples for sides 1..13:

Angle 60 has 15 solutions: {{1,1,1},{2,2,2},{3,3,3},{3,8,7},{4,4,4},{5,5,5},{5,8,7},{6,6,6},{7,7,7},{8,8,8},{9,9,9},{10,10,10},{11,11,11},{12,12,12},{13,13,13}}

Angle 90 has 3 solutions: {{3,4,5},{5,12,13},{6,8,10}}

Angle 120 has 2 solutions: {{3,5,7},{7,8,13}}

Non-equilateral angle 60 triangles for sides 1..10000: 18394

</pre>