A new Optimize me challenge.

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Summary:

I've optimized a code. Can you optimize it further? (Test the byte count. The bytes contained in the program name also count.)

A new Optimize me challenge.

RandomGuest (guest) 28 Feb 2015 16:48

PROGRAM:TOHSOLVE
Repeat A>=1 and A<=16
Input "NO. OF RINGS? ",A
End
List>matr(seq(R,R,1,A),[C]
{A+1,3->dim([C]
[[1,2][1,3][2,3->[E]
A+1
{1,Ans,Ans->L1
Delvar R Clrhome
"ABCDEFGHIJKLMNOP->Str0
Disp sub(Ans,1,A
For(M,2,2^A
R+1-R(R=3->R
[E](Ans,2
[C](L1(Ans),Ans->J
[E](R,1
[C](L1(Ans),Ans
J and J<Ans or not(Ans ->J
[E](R,Ans+1->B
L1(Ans->I
Output(B,Ans," \\One space
[C](Ans,B->C
I+1->L1(B
[E](R,2-J ->B
L1(Ans ->I
C->[C](Ans-1,B
I-1->L1(B
Output(B,Ans,sub(Str0,C,1
End

310 bytes.

Re: A new Optimize me challenge.

Timtech 01 Mar 2015 01:18

Nice work. Just a readability note, that A+1:{1,Ans,Ans->L1 looks better as {1,A+1,A+1->L1 but you did a great job! Without algorithm changes this is pretty much as small as it will get.

Last edited on 01 Mar 2015 01:18 by Timtech Show more

RandomGuest (guest) 05 Mar 2015 15:35

I did do some changes, and got it down to 281 bytes.
(Yes, I did change that Input into a Prompt.)
Here's basically what I did:
Try using L2 instead of [E].
{1,2,1,3,2,3 -> L2
Also, increment R by two instead of by one, and loop when it equals six.
(About the readability note - I'm looking at byte size, not readability.)
I tried using a For( loop when checking the columns (first 5 lines inside main loop). That got it down to 280, but a major speed cost. I'll post the code if you're interested.

I did not find the original code. I optimized a code I found on Rosetta Code.

Timtech 08 Mar 2015 01:37

Okay, I understand 100% but since they are the same byte size I suggested it. I guess for speed your way would probably be faster.

RandomGuest (guest) 08 Mar 2015 19:35

Oh. Whoops. Miscounted the bytes, or attributed the byte size to the wrong change.

RandomGuest (guest) 11 Mar 2015 14:21

Repeat A≥1 and A≤16
Prompt A
End
List>matr(seq(R,R,1,A),[C]
{A+1,3 -> dim([C]
{1,A+1,A+1 -> L1
{1,2,1,3,2,3 -> L2
Delvar R ClrHome \\The space between is for readability purposes only.
"ABCDEFGHIJKLMNOP ->Str0
Disp sub(Ans,1,A
For(M,2,2^A
R+2-6(R=6 -> R
L2(R
[C](L1(Ans),Ans -> J
L2(R-1
[C](L1(Ans),Ans
J and J<Ans or not(Ans -> J
L2(R-1+J -> B
L1(Ans -> I
Output(B,Ans," \\ One space
[C](I,B -> C
I+1 -> L1(B
L2(R-J -> B
L1(B)-1 -> L1(B
Output(B,Ans,sub(Str0,C,1
C -> [C](Ans,B
End

281 bytes

lirtosiast 12 Apr 2015 05:11

I see a couple of extra bytes.

{1,A+1,A+1
can be 1+{0,A,A

R+2-6(R=6 -> R
can be 2+R(R≠6

RandomGuest (guest) 12 Apr 2015 05:13

Thank you!!!!
That brought it down to 277.
I never thought of those before!