Ehrenfest's urn

[zahi48g]

Ehrenfest's urn
Hi Everyone (from France )

I'm trying this algorythm on my prime without success , of course i've tried to debug .

EXPORT EHRENFEST1()
BEGIN
LOCAL n,p,K;
INPUT(p);
L1:={1,2};
L2:={};
FOR K FROM 1 TO p DO
n:=RANDINT(1,2);
IF L1(1)==n OR L1(2)==n THEN
L1:=SUPPRESS(L1,n);
L2:=INSERT(L2,1,n);
ELSE
L2:=SUPPRESS(L2,n);
L1:=INSERT(L1,1,n);
END;
END;
PRINT(L1);
PRINT(L2);
END;

Could you help me ?
It seems that the problem occurs when i write " n:=RANDINT(1,2); " .

Thank you .

[toml_12953]

(12-29-2021 10:16 PM)zahi48g Wrote:  Ehrenfest's urn
Hi Everyone (from France )

I'm trying this algorythm on my prime without success , of course i've tried to debug .

EXPORT EHRENFEST1()
BEGIN
LOCAL n,p,K;
INPUT(p);
L1:={1,2};
L2:={};
FOR K FROM 1 TO p DO
n:=RANDINT(1,2);
IF L1(1)==n OR L1(2)==n THEN
L1:=SUPPRESS(L1,n);
L2:=INSERT(L2,1,n);
ELSE
L2:=SUPPRESS(L2,n);
L1:=INSERT(L1,1,n);
END;
END;
PRINT(L1);
PRINT(L2);
END;

Could you help me ?
It seems that the problem occurs when i write " n:=RANDINT(1,2); " .

Thank you .

L2 only has one element. If n=2 you get an error when you try to SUPPRESS L2(n) since L2(2) doesn't exist.

[zahi48g]

Hi

Thank you for replying.

First , the algorithme is from a book, I've just translated it into the prime language .

Second , it does not try to remove L2(2) from L2 but n from L2 . According to the Ehrenfest'urns model an urn can be empty (this is anticipated in the IF...THEN...ELSE),
so L2(2) may not exist . I should have the following outcomes :

{1,2}
{}
or
{1}
{2}
or
{}
{1,2}
or
{2}
{1}

If you get these results , could copy and paste you code ?

Thank you .

[Didier Lachieze]

(12-29-2021 11:53 PM)zahi48g Wrote:  it does not try to remove L2(2) from L2 but n from L2 .

With SUPPRESS(L2,n) you are removing from L2 the element at the nth position, so if L2={2} then SUPRESS(L2,2) will fail because there is only one element in the list. The correct instruction to remove 2 from the list is SUPPRESS(L2,POS(L2,2))) which will remove 2 from the list whatever it’s position in the list.

(12-29-2021 11:53 PM)zahi48g Wrote:  If you get these results , could copy and paste you code ?

Here is a working code, I’ve added the number of elements in the urns as a parameter of the program so you can test with more than two:

Code:

#pragma mode( separator(.,;) integer(h32) )

EXPORT EHRENFEST1(N)
BEGIN
 LOCAL n,p,K;
 INPUT(p);
 L1:=MAKELIST(I,I,1,N);
 L2:={};
 FOR K FROM 1 TO p DO
  n:=RANDINT(1,N);
  IF POS(L1,n) THEN
   L1:=SUPPRESS(L1,POS(L1,n));
   L2(0):=n;
  ELSE
   L2:=SUPPRESS(L2,POS(L2,n));
   L1(0):=n;
  END;
 END;
 PRINT(SORT(L1));
 PRINT(SORT(L2));
END;

[John Keith]

Thanks for the interesting post, this problem is new to me.

Section 2.1 of this article suggests an alternate representation of the urns that may lead to a simpler program. Instead of two lists, one initially empty, one has a single list of (initially) all zeros. At each iteration, the state of a randomly selected element is flipped using NOT. The number of balls in urn "A" is the number of zeros in the list; the number of balls in urn "B" is similarly the number of ones.

My PPL is a bit rusty but it should be fairly simple.

[zahi48g]

Thank you D Lachieze and John Keith.

I've learned two things : How to remove properly an object from a list

and that L1(0) refers to the last element of list L1 (am I right ?) .

But i've got one question for D Lachieze . In order to know if n belongs to L1, shouldn't you write

" IF POS(L1,n)>0 THEN ..." instead of " IF POS(L1,n) THEN ..." ?

Thank you

[toml_12953]

(01-03-2022 11:10 PM)zahi48g Wrote:  Thank you D Lachieze and John Keith.

I've learned two things : How to remove properly an object from a list

and that L1(0) refers to the last element of list L1 (am I right ?) .

But i've got one question for D Lachieze . In order to know if n belongs to L1, shouldn't you write

" IF POS(L1,n)>0 THEN ..." instead of " IF POS(L1,n) THEN ..." ?

Thank you

Since false is 0 and the true condition is anything other than 0, either test works.