Why am I getting the word Return in the output?

Questions
1

I wrote the following code:

ptable = Map[(3 #^2 - #)/2 &, Range@4500];

For[k = 2250, k > 1, k--, atk = ptable[[k]];
  For[j = k - 1, j >= 1, j--, atj = ptable[[j]];
   If[MemberQ[ptable, atk - atj] && MemberQ[ptable, atk + atj], Return[atk - atj]]]] // Timing

I’m getting the following:

{34.491821, Return[5482660]}

Instead of what I expected:

{34.491821, 5482660}

Why?

thanks,
-Joe

2

Return is only use for functions, not for For-loops, you simply remove ‘Return[]’ for the case of a For-loop, it will return atk - atj.

3 
boo := Module[{},
  ptable = Map[(3 #^2 - #)/2 &, Range@4500];
  For[k = 2250, k > 1, k--,
   atk = ptable[[k]];
   For[j = k - 1, j >= 1, j--,
    atj = ptable[[j]];
    If[MemberQ[ptable, atk - atj] && MemberQ[ptable, atk + atj],
     Return[atk - atj, Module]]
    ]
   ]
  ]

Now

boo // Timing
{22.401744, 5482660}
4

Thanks for the help… I’m still learning some of the basics it seems!

5

The following code is faster, and in your PC much faster than mine…

In my Pc, i get: boo // Timing, = {33.633816, 5482660}.

But with:

JesusCA =
Compile[{{kk, _Integer}},
Module[{}, ptable = Map[(3 #^2 - #)/2 &, Range@4500];
    k = kk; out = 0; While[And[k > 1, out == 0],
     atk = ptable[];
     Do[atj = ptable[];
      If[MemberQ[ptable, atk - atj] && MemberQ[ptable, atk + atj],
       out = atk - atj;]
      , {j, k - 1, 1, -1}]; k--]; out
    ], RuntimeAttributes → {Listable}, Parallelization → True];

JesusCA[2250] // Timing

{24.866559, 5482660}

My pc is slower than yours, but my code in faster. Bye.