Re: [pymprog] Constraints and Iteration

[Nico G. <nic...@gm...>]

Hi Yingjie,

not, it's not too late at all - this project is a persistent one. ^_-
Thank you very much - I see now that I must get more accustomed to the generator syntax.

Generally, is there a way to initialize an "empty" pymprog variable?

I stumbled upon this problem when I tried to change the constraint from the zero-sum-game example for arbitrarily large games.
In:

st(p[1]+p[2]+p[3] == 1)

the number of ps is already fixed to three. So, my idea was:

------------------------------
p = var([1,2,3]) 
sum = var()

for i in range (len(p)):
    sum = sum + p[i+1]

st(sum == 1)

--------------------------------

The problem is that sum = var() leads sum being to X1=0.000000.
And with that, my final condition ends up being:
X0[1]+ X0[2]+ X0[3]+ X1

To get around this, I used a rather clumsy trick:
-------------------------------
p = var([1,2,3])
sum = p[1]
        for i in range (len(p)-1):
            sum = sum + p[i+2]
        st(sum == 1)
--------------------------------

I can't remember why I didn't just initialize sum as an int with value zero, but there must have been a reason for this. -_-°
Anyway: Is there a way to initialize an empty pymprog variable? If not, is there a smart way to do this with generators?

Many Thanks

Bye
Nico 


-----Ursprüngliche Nachricht-----
Von: Yingjie Lan [mailto:la...@ya...] 
Gesendet: Montag, 8. März 2010 06:50
An: pym...@li...; Nico Grupp
Betreff: Re: [pymprog] Constraints and Iteration

> 
> m = 3 # 3
> strategies 
> 
> M =
> range(m) # payoffs of player A 
> 
> n = 3 # 3
> strategies 
> 
> N = range(n) # payoffs
> of player B 
> 
>    
> 
> (…) 
> 
>    
> 
>    
> 
> class Test: 
> 
>    
>         def
> __init__(self): 
> 
>        
>     beginModel('game') 
> 
>      
> self.x =
> var(A, 'x') 
> 
>    
> 
>    
> 
> So far, so good.
> But when I try to
> implement point 3, it doesn’t work. The idea was to
> ensure that: 
> 
>    
> 
> payoff of
> self.x[i,j] = - payoff of self.x[j,i] 
> 
>    
> 
> and it’d best
> be done with
> iterations. But when trying something like this:
> 
> 
>    
> 
>        
> for i in
> M:       
> 
>            
> for j in N:  
> 
>                
> st((self.x[i,j]) ==
> -(self.x[i,j])) 
> 
>    
> 
>    
> 
> it leads
> to: 
> 
Hi, Nico:


It is probably too late, but I tested with the code below:

#--------------------Cut-----------------------
from pymprog import *
m = 3 # 3 strategies
M = range(m) # payoffs of player A
n = 3 # 3 strategies
N = range(n) # payoffs of player B
A = iprod(M,N)

class Test:
   def __init__(me):
      beginModel('game')
      me.x = var(A,'x')
      #st((me.x[i,j]) == -(me.x[j,i]) for i,j in A)
      for i in M:
         for j in N:
            print st((me.x[i,j]) == - (me.x[j,i]))

t=Test()
#--------------------Cut-----------------------



And this is the output (note: you might want to check if you are using the latest version of pymprog):

#--------------------Cut-----------------------
yaellan:pymprog xiaoliz$ python testreg.py 
s.t. R0: 2 x[(0, 0)] == 0
s.t. R1: x[(0, 1)]+ x[(1, 0)] == 0
s.t. R2: x[(2, 0)]+ x[(0, 2)] == 0
s.t. R3: x[(0, 1)]+ x[(1, 0)] == 0
s.t. R4: 2 x[(1, 1)] == 0
s.t. R5: x[(1, 2)]+ x[(2, 1)] == 0
s.t. R6: x[(2, 0)]+ x[(0, 2)] == 0
s.t. R7: x[(1, 2)]+ x[(2, 1)] == 0
s.t. R8: 2 x[(2, 2)] == 0
#--------------------Cut-----------------------


Cheers,

Yingjie