scoptrial 0.0.5

SCOP Trial Version

SCOP Trial

SCOP (SOlver for Constraint Programming) trial

Install

pip install scoptrial

How to use

See https://www.logopt.com/scop2/

from scoptrial.scop import *

'''
Example 1 (Assignment Problem):
Three jobs (0,1,2) must be assigned to three workers (A,B,C)
so that each job is assigned to exactly one worker.
The cost matrix is represented by the list of lists
Cost=[[15, 20, 30],
      [7, 15, 12],
      [25,10,13]],
where rows of the matrix are workers, and columns are jobs.
Find the minimum cost assignment of workers to jobs.
'''

workers=['A','B','C']
Jobs   =[0,1,2]
Cost={ ('A',0):15, ('A',1):20, ('A',2):30,
       ('B',0): 7, ('B',1):15, ('B',2):12,
       ('C',0):25, ('C',1):10, ('C',2):13 }

m=Model()
x={}
for i in workers:
    x[i]=m.addVariable(name=i,domain=Jobs)

xlist=[]
for i in x:
    xlist.append(x[i])

con1=Alldiff('AD',xlist,weight='inf')

con2=Linear('linear_constraint',weight=1,rhs=0,direction='<=')
for i in workers:
    for j in Jobs:
        con2.addTerms(Cost[i,j],x[i],j)

m.addConstraint(con1)
m.addConstraint(con2)

print(m)

m.Params.TimeLimit=1
sol,violated=m.optimize()

if m.Status==0:
    print('solution')
    for x in sol:
        print (x,sol[x])
    print ('violated constraint(s)')
    for v in violated:
        print (v,violated[v])

Model: 
number of variables = 3  
number of constraints= 2  
variable A:['0', '1', '2'] = None 
variable B:['0', '1', '2'] = None 
variable C:['0', '1', '2'] = None 
AD: weight= inf type=alldiff  C A B ;  :LHS =0  
linear_constraint: weight= 1 type=linear 15(A,0) 20(A,1) 30(A,2) 7(B,0) 15(B,1) 12(B,2) 25(C,0) 10(C,1) 13(C,2) <=0 :LHS =0 

 ================ Now solving the problem ================ 

Status= 127
Output=