My python one-liner scramble generator

[qqwref]

If I do that on C++, does the count have to include the includes and the "void main" or only the line?

The line is as follows:
for(int i=25;i;i--)printf("%c%c ","UDFBRL"[rand()%6]," 2'"[rand()%3]);

(70 chars)

output :
B' R2 B2 D2 L2 U' F' U R2 F D2 F R' F2 D' F2 L U' L2 U2 F' B U D2 L

This wouldn't quite work. There is no provision in your code to prevent sequences such as R R or R L R.

Someone should start an official scrambler programming contest.

I don't know about official, but this topic was a pretty decent contest.

 

[Carrot]

Someone should start an official scrambler programming contest.

Here's my pathetic attempt in Ruby:


m= ["R", "U", "D", "B", "F", "L"]
x = [" ", "2 ", "' "]
o = 0
d = 0
while d < 26
n = m.at(rand(6))
if n != o
print n
o = n
print x.at(rand(3))
d = d + 1
end
end


148 characters.

Doesn't work..
U' D' F' L2 F2 D2 R' D2 U2 D' F D L B D' R B' D' R L B R2 L2 B' F B'

And it doesn't allow R as the first move if I'm right

 

[Owen]

Doesn't work..
U' D' F' L2 F2 D2 R' D2 U2 D' F D L B D' R B' D' R L B R2 L2 B' F B'

And it doesn't allow R as the first move if I'm right

Blah. I'm bad at programming. I shortened all the variables, so now I can't even understand how it works.

But it can start with R, I don't know where you got that idea.


EDIT:

http://codepad.org/7EHN0e5M

This appears to solve the problem, but it seems too good to be true...

 

[Carrot]

Blah. I'm bad at programming. I shortened all the variables, so now I can't even understand how it works.

But it can start with R, I don't know where you got that idea.


EDIT:

http://codepad.org/7EHN0e5M

This appears to solve the problem, but it seems too good to be true...

True, it didn't solve any problems

Because of this line:
if n != o
you claim o=0 in the beginning, so n cannot be 0 for the first run, therefore you'll never get position 0 (R move) for the first move

This is my attempt to save your code:

m,x=["R","U","B","F","D","L"],[" ","2 ","' "]
a=b=c=d=6
while d<31
a=rand(6)
if a!=b&&!(a==c&&b+a=5)
print m.at(a)+x.at(rand(3))
c,b=b,a
d+=1
end
end

 

[joey]

b=c=d=6
while d<32
if (a=rand 6)!=b&&!(a==c&&b+a==5)
$><<%w[R U B F D L][a]+[" ","2 ","' "][rand 3]
c,b=b,a
d+=1
end
end

Just a few mins on it.

 

[Owen]

Nice job!

 

[sz35]

This wouldn't quite work. There is no provision in your code to prevent sequences such as R R or R L R.

Right.

Fixed solution:
for(int i=25,t=6,k;i;(k=rand()%6)-t&&t%2-1|k-t+1&&printf("%c%c ","UDFBRL"[i--,t=k]," 2'"[rand()%3]));
101 chars.

I wanted to write on pure C, but cout makes it shorter:
for(int i=25,t=6,k;i;(k=rand()%6)-t&&t%2-1|k-t+1&&cout<<"UDFBRL"[i--,t=k]<<" 2'"[rand()%3]<<" ");
97 chars

L after R is possible but R after L isn't. Each L->R move can be shown as R->L, and we avoid R->L->R.

 

[Carrot]

b=c=d=6
while d<32
if (a=rand 6)!=b&&!(a==c&&b+a==5)
$><<%w[R U B F D L][a]+[" ","2 ","' "][rand 3]
c,b=b,a
d+=1
end
end

Just a few mins on it.

a=b=d=6
while d<32
c,b=b,a
while c+b==5&&a==c||a==b:a=rand 6;end
$><<%w[R U B F D L][a]+[" ","2 ","' "][rand 3]
d+=1
end

(120 chars) damn... same number of chars

 

[qq280833822]

b=d=6
while d<32
if (a=rand 6)!=b&&a!=b+3
$><<%w[U R F D L B][b=a]+[" ","2 ","' "][rand 3]
d+=1
end
end

(103 chars)

 

[joey]

Odder.. edit your post =D

 

[Stefan]

Because of this line:
if n != o
you claim o=0 in the beginning, so n cannot be 0 for the first run, therefore you'll never get position 0 (R move) for the first move

Is that some weird Ruby stuff I don't know? I'd say n=m.at(rand(6)) never gets you n=0, just n="R". So that shouldn't be a problem.

 

[Owen]

I think so, because my original code generated scrambles beginning with R on multiple occasions.

 

[Carrot]

Is that some weird Ruby stuff I don't know? I'd say n=m.at(rand(6)) never gets you n=0, just n="R". So that shouldn't be a problem.

I think so, because my original code generated scrambles beginning with R on multiple occasions.

I'm no good at Ruby Ruby is retarded xD I'was god damn sure that rand(6) generated 0-5 -.-' (I generated like 30 scrambles without a leading R move :O )


EDIT:

Chen Shuang:
Your distribution is wrong..

U->L=25%
D->L=20%

 

[Stefan]

I'was god damn sure that rand(6) generated 0-5

And it probably does. But the code says n=m.at(rand(6)), not n=rand(6).

 

[Carrot]

And it probably does. But the code says n=m.at(rand(6)), not n=rand(6).

Ohh ****... Can you tell me the best place to hide? (I had already changed his code inside my head when I commented his code... stupid Odder)

 

[qq280833822]

Its algorithm can be describe as:
generate random moves
and then check it, let RR->R, LL->L, RL->LR, and UD, FB as well
is it really wrong?

 

[Carrot]

Its algorithm can be describe as:
generate random moves
and then check it, let RR->R, LL->L, RL->LR, and UD, FB as well
is it really wrong?

I'm pretty confident that I'm right. But I'm already looking for a place to hide when you prove me wrong.

(a=rand 6)!=b&&a!=b+3

same as:
a!=b and a!=b+3

well, a!=b is obvious..

a!=b+3 can be described like this:
setA=[U,R,F], setB=[D,L,B]
then we have: setA[x]+3=setB[x], where x is the position.
but setB[x]+3 != setA[x].... (as 5+3 != 2)
this means that for any moves in setA, there is 4 valid next-moves, but for setB we have 5.

The goal for the scrambles is to have:
5 valid moves, unless we have just had two opposite moves in a row, then it should be 4

 

[qq280833822]

why should valid moves of setA be the same as of setB
if we only allow "LR" but disallow "RL", of course setA is difference to setB...

 

[Carrot]

Imagine you have the move R

R doesn't allow the next move to be L, therefore you can't get RL.. (RL is not converted to LR) (which leaves 4 valid moves only)

now take L

L does allow the next move to be R... (5 valid moves, YAY!! )

 

[Aditya Sriram]

Another Python code(prevents R R' or R L)

I know the code doesn't suit the title and it is slightly long, but my purpose was to avoid Cancellation moves.
Here is my code (Python 2.7)

import random                                                                   

no=input("Input the number of scrambles you require.")                          
length=input("Input the length of the scramble.")                               
scramble=''

set_1=[" R"," R'"," R2"," L"," L'"," L2"]                                       
set_2=[" F"," F'"," F2"," B"," B'"," B2"," U"," U'"," U2"," D"," D'"," D2"]     

for i in range(no):                                                             
    scramble= scramble + '('+str(i+1)+')'
    for i in range(length):                                                     
        if i%2 == 0:                                                            
            scramble=scramble + random.choice(set_1)
        else:
            scramble=scramble + random.choice(set_2)
    print scramble                                                              
                                                                                
    print "\n"                                                                  
    scramble=''

Output:

(1) L2 D' L2 B' L U2 L' U L2 U2 R' F2 R2 D2 R' U2 R U2 L' D' R' F2 R' F2 L'
(2) L2 U' R2 U2 R' B' R2 F2 L' F L' F' L' B L2 U2 L2 B L' F R2 D2 L2 D R2
(3) L F2 L' U' R' B2 L F2 L B2 L U L' D2 R' D' R' U2 R F2 R U R2 F2 R'
(4) R2 F2 R2 U' R U2 L2 B R2 B R B R F2 R' F R2 D' L B' R F2 L B2 R
(5) L F R D R D L D' R' B2 R2 F R2 U2 R D2 L' U' R' U R2 B' R2 U' L'