12 random numbers, 1-36, no repeats

Stephen · **Posted:** January 28th, 2014, 2:52 am

Hi,

Was looking for a random number no repeat script.

Checking this forum and elsewhere online, I came up with a script that works, but is very long (my scripting skills weren't terribly efficient).

I suspect there is a more efficient way to populate the arrays and set up checking loops.

Rather than put the entire lengthy script below, I can include a smaller version.

I would appreciate any suggestions on how to write this more efficiently (and shorter).

Code:

function chooser100(){
var myArray = new Array(); 
 myArray[0] ="F1";  
 myArray[1] ="F2";
 myArray[2] ="F3";
 myArray[3] ="F4";
 myArray[4] ="F5";
 myArray[5] ="F6";
 myArray[6] ="F7";
 myArray[7] ="F8";
 myArray[8] ="F9";
 myArray[9] ="F10";
 myArray[10] ="F11";
 myArray[11] ="F12";
 myArray[12] ="F13";
 myArray[13] ="F14";
 myArray[14] ="F15";
 myArray[15] ="F16";
 myArray[16] ="F17";  
 myArray[17] ="F18";
 myArray[18] ="F19";
 myArray[19] ="F20";
 myArray[20] ="F21";
 myArray[21] ="F22";
 myArray[22] ="F23";
 myArray[23] ="F24";
 myArray[24] ="F25";
 myArray[25] ="F26";
 myArray[26] ="F27";
 myArray[27] ="F28";
 myArray[28] ="F29";
 myArray[29] ="F30";
 myArray[30] ="F31";
 myArray[31] ="F32";
 myArray[32] ="F33";
 myArray[33] ="F34";
 myArray[34] ="F35";
 myArray[35] ="F36";

var len = myArray.length;
var choice = new Array();

for (var i = 0; i<=11; i++) {
var rand1 = Math.floor(Math.random() * len);  
choice[i] = myArray[rand1];

for (k=0; k<=i-1; k++) {
if (choice[k] == choice[i]) {
i--;
 
}
}
}
if (choice[0] == "F1")
{
F1.Rotate(90)
}
if (choice[0] == "F2")
{
F2.Rotate(90)
}
if (choice[0] == "F3")
{
F3.Rotate(90)
}
if (choice[0] == "F4")
{
F4.Rotate(90)
}
if (choice[0] == "F5")
{
F5.Rotate(90)
}
if (choice[0] == "F6")
{
F6.Rotate(90)
}
if (choice[0] == "F7")
{
F7.Rotate(90)
}
if (choice[0] == "F8")
{
F8.Rotate(90)
}
if (choice[0] == "F9")
{
F9.Rotate(90)
}
if (choice[0] == "F10")
{
F10.Rotate(90)
}

//etc.-up to F36, then the same for array choices [1], [2], [3], [4] and [5], then on to more for Rotate.(270)-much, much too long, but works.

One item that I tried, without success:

Code:

for (var m = 0;m<=5;m++){
if ((choice[m] == "F1")||(choice[m] == "F2")||(choice[m] == "F3")||(choice[m] == "F4")||(choice[m] == "F5")||(choice[m] == "F6")||(choice[m] == "F7")||
(choice[m] == "F8")||(choice[m] == "F9")||(choice[m] == "F10")||(choice[m] == "F11")||(choice[m] == "F12")||(choice[m] == "F13")||(choice[m] == "F14")||
(choice[m] == "F15")||(choice[m] == "F16")||(choice[m] == "F17")||(choice[m] == "F18")||(choice[m] == "F19")||(choice[m] == "F20")||(choice[m] == "F21")||
(choice[m] == "F22")||(choice[m] == "F23")||(choice[m] == "F24")||(choice[m] == "F25")||(choice[m] == "F26")||(choice[m] == "F27")||(choice[m] == "F28")||
(choice[m] == "F29")||(choice[m] == "F30")||(choice[m] == "F31")||(choice[m] == "F32")||(choice[m] == "F33")||(choice[m] == "F34")||(choice[m] == "F35")||
(choice[m] == "F36"))
{
choice[m].Rotate(90)
}
}

Must be a simpler way to write this? :oops:

mackavi · **Posted:** February 3rd, 2014, 9:19 am

You don't specify what type of publication you're building

The following works for Opus Pro Types:

Code:

/*USAGE
len = number range - IE 36 is 1,2,3...34,35,36
sub = number sub-range - IE 4 is first four random numbers
*/

doRandomRotate(10,4) //generate numbers 1 to 10 and pick the first four random numbers

function doRandomRotate(len, sub)
{
   var a  =new Array;

   for (var i=0;i<len;i++) //generate numbers
      a[i] = i+1

      while ( --len ) //shuffle numbers
      {
        var j = Math.floor( Math.random() * ( len + 1 ) )
        var tempi = a[len]
        var tempj = a[j]
        a[len] = tempj
        a[j] = tempi
      }

   for (var i=0;i<sub;i++) //pick 'sub' and perform action
      eval("F"+a[i] +".Rotate(90)")
}

For example: If I have 35 frames on screen named F1 to F35 and want to rotate a random selection of 11 frames, I call

Code:

doRandomRotate(35,11);

Mack
https://twitter.com/imsmackavi

Stephen · **Posted:** February 3rd, 2014, 1:12 pm

Hi Mack,

Much thanks!

I'll try it out (36 puzzle pieces). (Apologies: is a "standard" Opus Pro 9 pub for Windows.)

mackavi · **Posted:** February 3rd, 2014, 4:55 pm

That should work fine then.

Just create your frames as F1 to F36 and run the function as described to randomly rotate X pieces of the set.

Mack

Stephen · **Posted:** February 3rd, 2014, 5:26 pm

Hi Mack,

Yes, this works and reduces greatly the size of the script. Very nice!

A little more complexity (and puzzle challenge):

What if I would like to rotate 6 @ 90° and 6 more @ 270 :total of 12 from 36, comprised of 6 (rotate 90°) and 6 (rotate 270°).

I tried to substitute the following, some success, but not randomly chosen (instead uses F1-18 or F19-36. Must be a better, random method?):

Code:

for (var i=0;i<sub;i++){     
   if (a[i]<=18){
   eval("F"+a[i] +".Rotate(90)")
   }
   if (a[i]>18){
   eval("F"+a[i] +".Rotate(270)")
   }
   }

So, not quite sure how to change the script for this, especially without repeats?

mackavi · **Posted:** February 3rd, 2014, 9:47 pm

That depends on whether set two has to be distinct - IE it cannot have already occurred in set one.

If this is the case, you simply need to repeat the whole loop, adjusting for the next range in the array:

Code:

   for (var i=0;i<sub;i++) //pick 'sub' and perform action
      eval("F"+a[i] +".Rotate(270)")

Change the var i=0 and i<sub sections to accommodate the new range with with fixed values or by passing a new parameter to the function. EG:

Code:

   for (var i=10;i<20;i++) //pick 'sub' and perform action
      eval("F"+a[i] +".Rotate(180)")

Will select those elements from the original shuffle

Alternatively, if it doesn't matter whether the sets are distinct - IE a piece can be in the 90 and 270 set, then I'd just convert the Rotate value to a variable, pass it as a parameter and call the function each time specifying the angle.

Mack
https://twitter.com/imsmackavi

Stephen · **Posted:** February 3rd, 2014, 10:12 pm

Thank you again, Mack.

Used the "distinct" sets, and it works great!

Digital Workshop

12 random numbers, 1-36, no repeats

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