My python one-liner scramble generator

[Aditya Sriram]

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'

[qqwref]

Your code seems to make every other move an R or L turn, and the rest are never R or L turns. This does avoid canceling moves but it also provides far less possibilities than a normal scramble. The other scramblers in the thread were made to use all of the moves equally but also avoid cancellations - it's a little trickier, but see if you can make one that does that.

[qq280833822]

1429-characters random-state 2x2 scrambler, and 1798-characters random-state pyraminx scrambler

var get2x2optscramble=function(){function o(a,c,b,d,e){if(0!=b){if(j[a]>b||k[c]>b)return!1;var h,g,f,i;for(i=0;3>i;i++)if(i!=d){h=a;g=c;for(f=0;3>f;f++)if(h=m[h][i],g=n[g][i],o(h,g,b-1,i,e))return e.push("URF".charAt(i)+" 2'".charAt(f)),!0}}else return 0==a&&0==c}function p(){p=function(){};var a,c,b,d,e;for(a=0;5040>a;a++){j[a]=-1;m[a]=[];for(c=0;3>c;c++)m[a][c]=q(a,c)}for(a=j[0]=0;7>a;a++)for(c=0;5040>c;c++)if(j[c]==a)for(b=0;3>b;b++){d=c;for(e=0;3>e;e++)d=m[d][b],0>j[d]&&(j[d]=a+1)}for(a=0;729>a;a++){k[a]=-1;n[a]=[];for(c=0;3>c;c++)n[a][c]=r(a,c)}for(a=k[0]=0;6>a;a++)for(c=0;729>c;c++)if(k[c]==a)for(b=0;3>b;b++){d=c;for(e=0;3>e;e++)d=n[d][b],0>k[d]&&(k[d]=a+1)}}function q(a,c){var b,d,e,h=a,g=[];for(b=1;7>=b;b++){d=h%b;h=(h-d)/b;for(e=b-1;e>=d;e--)g[e+1]=g[e];g[d]=7-b}0==c?l(g,1,3,2):1==c?l(g,4,5,1):2==c&&l(g,2,6,4);for(b=h=0;7>b;b++){for(e=d=0;7>e&&!(g[e]==b);e++)g[e]>b&&d++;h=h*(7-b)+d}return h}function r(a,c){var b,d,e,h=0,g=a,f=[];for(b=0;5>=b;b++)e=~~(g/3),d=g-3*e,g=e,f[b]=d,h-=d,0>h&&(h+=3);f[6]=h;0==c?l(f,1,3,2):1==c?(l(f,4,5,1),f[0]+=2,f[1]++,f[5]+=2,f[4]++):2==c&&(l(f,2,6,4),f[2]+=2,f[0]++,f[4]+=2,f[6]++);g=0;for(b=5;0<=b;b--)g=3*g+f[b]%3;return g}function l(a,c,b,d){var e=a[0];a[0]=a[c];a[c]=a[b];a[b]=a[d];a[d]=e}var j=[],k=[],m=[],n=[];return function(a){p();var c=~~(5040*Math.random()),b=~~(729*Math.random()),d=[];if(0!=c||0!=b)for(;99>a&&!o(c,b,a,-1,d);a++);return d.reverse().join(" ")}}();

//1429 chars


var getpyraoptscramble=(function(){function r(){function l(a,c){for(var e=[],j=5517840,f=0,b=0;5>b;b++){var h=k[5-b],d=~~(a/h),a=a-d*h,f=f^d,d=d<<2;e[b]=j>>d&15;h=(1<<d)-1;j=(j&h)+(j>>4&~h)}0==(f&1)?e[5]=j:(e[5]=e[4],e[4]=j);0==c&&g(e,0,3,1);1==c&&g(e,1,5,2);2==c&&g(e,0,2,4);3==c&&g(e,3,4,5);a=0;j=5517840;for(b=0;4>b;b++)d=e[b]<<2,a*=6-b,a+=j>>d&15,j-=1118480<<d;return a}function i(a,c){var e,d,f;d=0;var b=[],h=a;for(e=0;4>=e;e++)b[e]=h&1,h>>=1,d^=b[e];b[5]=d;for(e=6;9>=e;e++)f=~~(h/3),d=h-3*f,h=f,b[e]=d;0==c&&(b[6]++,3==b[6]&&(b[6]=0),g(b,0,3,1),b[1]^=1,b[3]^=1);1==c&&(b[7]++,3==b[7]&&(b[7]=0),g(b,1,5,2),b[2]^=1,b[5]^=1);2==c&&(b[8]++,3==b[8]&&(b[8]=0),g(b,0,2,4),b[0]^=1,b[2]^=1);3==c&&(b[9]++,3==b[9]&&(b[9]=0),g(b,3,4,5),b[3]^=1,b[4]^=1);h=0;for(e=9;6<=e;e--)h=3*h+b[e];for(e=4;0<=e;e--)h=2*h+b[e];return h}function g(a,c,d,f){var g=a[c];a[c]=a[d];a[d]=a[f];a[f]=g}r=function(){};var k=[1,1,1,3,12,60,360],m,a,f,d,c;for(a=0;360>a;a++){n[a]=-1;p[a]=[];for(c=0;4>c;c++)p[a][c]=l(a,c)}for(d=n[0]=0;5>d;d++)for(a=0;360>a;a++)if(n[a]==d)for(c=0;4>c;c++){f=a;for(m=0;2>m;m++)f=p[f][c],-1==n[f]&&(n[f]=d+1)}for(a=0;2592>a;a++){o[a]=-1;q[a]=[];for(c=0;4>c;c++)q[a][c]=i(a,c)}for(d=o[0]=0;5>=d;d++)for(a=0;2592>a;a++)if(o[a]==d)for(c=0;4>c;c++){f=a;for(m=0;2>m;m++)f=q[f][c],-1==o[f]&&(o[f]=d+1)}}function s(l,i,g,k,m){if(0==g)return 0==l&&0==i;if(n[l]>g||o[i]>g)return!1;for(var a=0;4>a;a++)if(a!=k)for(var f=l,d=i,c=0;2>c;c++)if(f=p[f][a],d=q[d][a],s(f,d,g-1,a,m))return m.push("ULRB".charAt(a)+["","'"][c]),!0;return!1}var n=[],o=[],p=[],q=[];return function(l){r();var i=~~(360*Math.random()),g=~~(2592*Math.random()),k=[];if(0!=i||0!=g)for(;99>l&&!s(i,g,l,-1,k);l++);k=k.reverse().join(" ")+" ";for(g=0;4>g;g++)i=~~(3*Math.random()),2>i&&(k+="lrbu".charAt(g)+["","'"][i]+" ");return k}})();

1798 chars

[onionhoney]

for(int i=25,a,b=6,c;i;){do c=rand()%6;while(!(c-b&&(c+b!=5||c-a)));cout<<"UFLRBD"[i--,a=b,b=c]<<"'2 "[rand()%3];}

My first try with C++. 114 chars.

EDIT: tried in vain to modify the 102-char solution by saving a rand()....end up getting one more..

for(int c=0,b=6,j=25,m;j;m%6-c|c+b-5&&b-m%6&&cout<<"URFBLD"[j--,c=b,b=m%6]<<" 2'"[m/6%3]<<" ")m=rand();

[qqwref]

Hey all, here's a scrambler in K Yeah, a bit silly, but under 100 characters. Can probably be optimized a ton.

,/("FRUDLB"@25#({,/1_'(0,&((x=q)|(:':t)&t:5=x+q: :':x))_x:0,x}/99?6)),'25?(" ";"' ";"2 ")

Edit: Typical output:
"U R F2 B' R2 U' D2 L2 U' L B' L2 U2 B' F U' L2 R D' B' D2 F U' R2 L' "
And you can get K from kx.com, they have a Q console and from there you can press \ to switch between Q and K. K is a lot less verbose but is missing several useful functions.

[Baian Liu]

Even shorter for TI-83/84.

:ClrHome
:0→E
:7→D
:For(A,1,6
:For(B,2,14,3
:E→F
:D→E
:While D=E or D+E=7 and D=F
:randInt(1,6→D
:End
:Output(A,B,sub("URFBLD",D,1)+sub(" '2",randInt(1,3),1
:End
:End

[qqwref]

Much better K scrambler:

,/("FRUDLB"@{#&(=':x),2_&':5=+':x}{25?6}/2#2),'25?" ","'2",'" "

Only 63 characters, by my count. I think that's pretty solid.
(I had one that was 61 before until I realized it didn't allow sequences starting with 5.)

Explanation of what it does:

Spoiler:

Let's first dissect this weird-looking function: {#&(=':x),2_&':5=+':x}. The curly braces signify a function, and it's implied that the argument is x, so we don't need to declare it. The expression f':x for a function f basically returns the list you get from applying f to each adjacent pair of elements of x. So +':x gives us a list of sums of two adjacent elements. 5=+':x gives us a list of bits, with a 1 set in every position where the sum of two adjacent elements is equal to 5. &':5=+':x gives us another list of bits, but this time there is a 1 only when there were two sums-to-5 in a row (which in this case represents sequences like LRL or UDU). The 2_ removes the first two elements from that so we don't get problems with sequences that start with 5 or 5 0. So that part gives us a 1 for each LRL-type sequence. The part to the left of the comma, =':x, gives us another list of bits, with a 1 for each repeated number (so, a sequence like RR). The comma joins those two sequences together; and then the #& at the left returns the number of 1s we have. So this whole function ends up giving us the number of bad sequences in our potential scramble.

In the middle we have {stuff}{25?6}/2#2. 2#2 makes a list of two 2's so we have (2 2) there. So what does this thing do? Well, f g/ x (with f and g functions) applies g to x repeatedly until f of the result is zero. Now 25?6 generates a list of 25 random integers from 0 to 5, so we are generating lists of 25 random integers until they pass the first function (i.e. the result is 0). We needed that 2#2 so the function would start off giving a 1; otherwise the expression would immediately terminate. Notice that a list passes the first function if it has zero mistakes - that is, the list of faces to move is valid. So great - now we have generated a random list of 25 faces until it was valid for making a scramble out of. The list will look something like this: 0 3 2 5 3 0 3 5 ...

The rest is more straightforward. Let's call the part we looked at so far X; then we have ,/("FRUDLB"@X),'25?" ","'2",'" ". The "FRUDLB"@X just indexes our list into the string "FRUDLB" so now we have a list of 25 letters, such as "FDUBDFDB...". The " ","'2",'" " at the far right generates a list of the 3 possible suffixes (" ", "' ", and "2 ") and the 25? chooses 25 random suffixes, just like the previous use of it. Finally, the ,' joins each move letter up with one of the suffixes, and the ,/ at the beginning takes the whole list of strings and joins it together into one. Pretty tricky, but it makes sense if you analyze it.

[Escher]

Explanation of what it does:

Spoiler:

Let's first dissect this weird-looking function: {#&(=':x),2_&':5=+':x}. The curly braces signify a function, and it's implied that the argument is x, so we don't need to declare it. The expression f':x for a function f basically returns the list you get from applying f to each adjacent pair of elements of x. So +':x gives us a list of sums of two adjacent elements. 5=+':x gives us a list of bits, with a 1 set in every position where the sum of two adjacent elements is equal to 5. &':5=+':x gives us another list of bits, but this time there is a 1 only when there were two sums-to-5 in a row (which in this case represents sequences like LRL or UDU). The 2_ removes the first two elements from that so we don't get problems with sequences that start with 5 or 5 0. So that part gives us a 1 for each LRL-type sequence. The part to the left of the comma, =':x, gives us another list of bits, with a 1 for each repeated number (so, a sequence like RR). The comma joins those two sequences together; and then the #& at the left returns the number of 1s we have. So this whole function ends up giving us the number of bad sequences in our potential scramble.

In the middle we have {stuff}{25?6}/2#2. 2#2 makes a list of two 2's so we have (2 2) there. So what does this thing do? Well, f g/ x (with f and g functions) applies g to x repeatedly until f of the result is zero. Now 25?6 generates a list of 25 random integers from 0 to 5, so we are generating lists of 25 random integers until they pass the first function (i.e. the result is 0). We needed that 2#2 so the function would start off giving a 1; otherwise the expression would immediately terminate. Notice that a list passes the first function if it has zero mistakes - that is, the list of faces to move is valid. So great - now we have generated a random list of 25 faces until it was valid for making a scramble out of. The list will look something like this: 0 3 2 5 3 0 3 5 ...

The rest is more straightforward. Let's call the part we looked at so far X; then we have ,/("FRUDLB"@X),'25?" ","'2",'" ". The "FRUDLB"@X just indexes our list into the string "FRUDLB" so now we have a list of 25 letters, such as "FDUBDFDB...". The " ","'2",'" " at the far right generates a list of the 3 possible suffixes (" ", "' ", and "2 ") and the 25? chooses 25 random suffixes, just like the previous use of it. Finally, the ,' joins each move letter up with one of the suffixes, and the ,/ at the beginning takes the whole list of strings and joins it together into one. Pretty tricky, but it makes sense if you analyze it.

I think this applies to you, too.

o_________________________o

[Stefan]

Your explanation is about 41 times as long as the program. Can't decide whether that's good or bad.

[qqwref]

Here's a 2x2 random-state scrambler in Q (a more verbose version of K, as above), in 593 characters:

system"S ",string 6h$.z.T
e:{m:enlist();x:enlist x;do[y;t:asc each raze each(3*(til 3)except/:(first each m)div 3)+'/:\:til 3;x:raze ({y x/z}./:({x:x@\:0 5 1 3 4 6 2;x[1]:(x[1]+0 2 1 0 0 1 2)mod 3;x};{x@\:1 2 3 0 4 5 6};{x:x@\:4 0 2 3 5 1 6;x[1]:(x[1]+2 1 0 0 1 2 0)mod 3;x})cross 1+til 3)[t]@\:'x;m:raze t,\:'m];(m;x)}
w:{a:e[(til 7;7#0);ceiling y%2];b:e[x;floor y%2];$[(s:(raze a[1]~/:\:b 1)?1b)=(c:count b 0)*count a 0;0N;" "sv ("RUF"cross("";"2";"'"))(reverse b[0;s mod c]),2 1 0 5 4 3 8 7 6@a[0;s div c]]}
scramble:{i:0;s:0N;h:(-7?7;k,(neg sum k:6?3)mod 3);while[0N~s;i+:1;s:w[h;i]];s}

I'm obviously not gonna explain this in full detail, but the general gist is that we get a random position and then try 1, 2, ... moves until we find a solution. We look for solutions of n moves by generating all positions ceiling(n/2) moves away from solved and all positions floor(n/2) moves away from our position, and then matching every pair of elements and seeing if we get any successes. This isn't the fastest possible way to do this but it seems fast enough.

Sample output:
q)scramble[]
"U2 F2 U F2 U R2 U2 F R'"

In K: 513 characters :3

."\\S ",$6h$.z.T
e:{m:,();x:,x;do[y;t:.q.asc',/'(3*(!3).q.except/:(*:'m)div 3)+'/:\:!3;x:,/({y x/z}./:,/({x:x@\:.:'"0513462";x[1]:(x[1]+.:'"0210012").q.mod'3;x};{x@\:.:'"1230456"};{x:x@\:.:'"4023516";x[1]:(x[1]+.:'"2100120").q.mod'3;x}),/:\:1+!3)[t]@\:'x;m:,/t,\:'m];(m;x)}
w:{a:e[(!7;7#0);-_-y%2];b:e[x;_y%2];$[(s:(,/a[1]~/:\:b 1)?1b)=(c:#b 0)*#a 0;0N;" "/:(,/"RUF",/:\:("";"2";"'"))(|b[0;s .q.mod'c]),.:'"210543876"@a[0;s div c]]}
scramble:{i:0;s:0N;h:(-7?7;k,(-+/k:6?3).q.mod'3);while[0N~s;i+:1;s:w[h;i]];s}