Rand.Int Replication

[Fashoomp]

how can i use randint, so it gives my a random pattern of numbers, so it does not give me the same value twice. I want all the values between 1 and 10, given to me, in a random order.

[Clayton]

you'll have to loop until you get all of the numbers. Now, you'll need to have some sort of array to hold all of the numbers that you've gotten. Just compare that array of numbers to the number you get from Rand.Int(), and if that number is not in the array, display it, and then add that number to the array.

[Saad]

This is the fastest way i could come up with.

code:

var number : array 1 .. 10 of int var repeted : boolean := false var minnum : int := 1  var maxnum : int := 10  % Then random number range must be greater or equal then the upper if number or else it will never stop for i : 1 .. upper (number)     number (i) := Rand.Int (minnum, maxnum)     if (i > 1) then         loop             repeted := false             for b : 1 .. i - 1                 if (number (i) = number (b)) then                     number (i) := Rand.Int (minnum, maxnum)                     repeted := true                 end if             end for             exit when not repeted         end loop     end if end for for i : 1 .. upper (number)     put number (i) end for

Edit: Fixed some spelling mistakes

[Carey]

Here is a shorter and more efficiant method.

Turing:

var numbers : array 1 .. 10 of int %holds numbers var i:= lower (numbers)-1 %-1 is to counteract the i += 1 at the start loop     i +=1     numbers (i) := Rand.Int (lower (numbers), upper (numbers)) %choses a random #     for j : lower (numbers) .. i - 1 %for every # chosen already not including itself         if numbers (i) = numbers (j) then %if the # has already been picked             i -= 1  %decrease count to stay at the same array spot             exit    %exit to the lop that pickes another random #         end if     end for     exit when i = upper(numbers) %exit when done end loop %TEST for k : 1 .. upper (numbers)     put numbers (k) end for

[Saad]

Is it me or does Rand.Int use system time to generate a random number, since if you ran the prog 10 times in .5 sec all the random values would be the same

[DIIST]

A recursive solution:

Turing:

procedure random (l, h : int)     if not (l > h) then         var m : int := Rand.Int (l, h)         put m         random (l, m - 1)         random (m + 1, h)     end if end random randomize random (1, 10)

Not the best but it works for the most part .

a100 @ Tue May 22, 2007 2:17 pm wrote:

Is it me or does Rand.Int use system time to generate a random number, since if you ran the prog 10 times in .5 sec all the random values would be the same

Yes i do believe system time is used as the random seed generator. But you can easily reset it by doing randomize (I think). Its all machine dependant. Some machines, regardless of what you do may still spit out the same set of value again and again at each runtime.

[Cervantes]

I'm pretty sure randomize is obsolete in the newer versions of Turing. Yes, the system time is used. Theoretically, I think it uses milliseconds, so you probably shouldn't be seeing any patterns. However, practically, I've seen them too.

Clayton wrote:

you'll have to loop until you get all of the numbers. Now, you'll need to have some sort of array to hold all of the numbers that you've gotten. Just compare that array of numbers to the number you get from Rand.Int(), and if that number is not in the array, display it, and then add that number to the array.

I can't believe you just said this. This scheme can take an unbounded amount of time. This really sucks, when you compare it to the better solution, which takes exactly 10 iterations, where each iteration requires a little search (could easily be done in log n time, if you do a binary search).

[Clayton]

I'm not sure if you're thinking about this the same way I was thinking about it when I wrote it. Perhaps I just didn't convey my message clearly. For that, I apologize.