# Custom constraints¶

Constraining Order is designed to make it easy to add custom constraints. This tutorial will show this for the example of one of the most prominent constraint satisfaction problems, the 8 queens problem.

## 8 queens problem¶

The task is to place 8 queens on a chessboard in such a way that no queen can attack any other queen, i.e. no two queens occupy the same row, column or same diagonals.

One way to model this is by using 8 variables, one for a queen in each column. In this way, the requirement that there has to be one queen in every column is already built in, which reduces the number of constraints and the domain of the variables and improves performance.

As values for the variables we choose tuples with the actual coordinates on the board, this makes it easier to formulate the diagonal constraints. We name the variables with lowercase letter, the traditional naming schema of the columns of a chess board. Coordinates will be zero indexed, as this is more convenient in python.

from constrainingorder.sets import DiscreteSet
from constrainingorder.variables import DiscreteVariable

variables = {}
for i,col in enumerate('abcdefgh'):
domain = DiscreteSet([(j,i) for j in range(8)])
variables[col] = DiscreteVariable(col,domain=domain)


As we already built in the column constraint, it remains to express the row and diagonal constraints. We will take care of all of them by creating a new constraint type, a QueensConstraint:

from constrainingorder.constraints import Constraint
from itertools import product

class QueensConstraint(Constraint):
"""
Constraint that ensures that a number of queens on a chessboard can
not attack each other
"""
def __init__(self,queens):
"""
Create a new Queens constraint.

:param variable: Variables representing the position of queens on a chess board
:type variable: list of DiscreteVariables
"""
Constraint.__init__(self,dict((var,var.domain) for var in queens))

def _conflict(self,val1,val2):
#check for row conflict
if val1 == val2:
return True
#check for rising diagonal conflict
if val1 - val1 == val2 - val2:
return True
#check for falling diagonal conflict
if val1 + val1 == val2 + val2:
return True
def satisfied(self,lab):
for v1,v2 in product(self.vnames,repeat=2):
if v1 == v2:
continue
if v1 not in lab or v2 not in lab:
return False
if self._conflict(lab[v1],lab[v2]):
return False
return True
def consistent(self,lab):
for v1,v2 in product(self.vnames,repeat=2):
if v1 not in lab or v2 not in lab or v1 == v2:
continue
if self._conflict(lab[v1],lab[v2]):
return False
return True


A constraint needs to derive from the Constraint class and implement the two methods satisfied and consistent.

In the constructor we pass a dictionary of variables and the values for them which are consistent with this constraint. In this case, there is no field on the board excluded a priori, so we use the full domain of the variable.

As both of the methods we have to implement check for conflicts between the queens, it makes sense to write a small utility method that does this check to avoid code duplication. It compares the first component of the tuples to check for a row conflict. To check whether the two queens are on the same diagonal, we compare the sum and difference of the row and column components. It might not be obvious, but it is easy to check that fields with the same sum or difference of rows and columns are on the same diagonal. Not that we don’t check for column conflicts, as they are taken care of by the setup of our variables.

The satisfied method checks that the labelling (a dictionary with variable names and values) assigns values to all variables affected by this constraint, and that there are no conflicts. The parameter names of the affected variables are accessible in the attribute vnames, that the Constraint class sets up for us.

The consistent method is a bit weaker, as it just checks for conflicts, but doesn’t care about missing values. It allows the solution and reduction algorithms to detect inconsistencies even if not all queens are placed yet.

And thats it. We can now use this constraint just like the in-built ones:

from constrainingorder import Space
from constrainingorder.solver import solve

constraint = QueensConstraint(variables.values())
space = Space(variables.values(),[constraint])

for solution in solve(space,method='backtrack'):
for i in range(8):
for j in range(8):
if (i,j) in solution.values():
print 'X',
else:
print '.',
print
break


In contrast to the sudoku solver discussed in the Quickstart, the 8 queens problem space can not be reduced, as no fields can be eliminated a priori, for every field there exist solutions where a queen occupies this field.

We also don’t specify a variable ordering, as in this case the total number of variables is rather low, and solution is quick in any case.

X . . . . . . .
. . . . . X . .
. . . . . . . X
. . X . . . . .
. . . . . . X .
. . . X . . . .
. X . . . . . .
. . . . X . . .


## Custom Binary relations¶

A riddle from this weeks newspaper:

Professor Knooster is visiting Shroombia. The people of Shroombia is divided into two groups, the shrimpfs that always lie and the wex that always tell the truth. For his research the professor asked 10 Shroombians about their groups. The answers:

• Zeroo: Onesy is a shrimpf
• Onesy: Threesy is a shrimpf
• Twoo: Foursy is a shrimpf
• Threesy: Sixee is a shrimpf
• Foursy: Seveen is a shrimpf
• Fivsy: Ninee is a shrimpf
• Sixee: Twoo is a shrimpf
• Seveen: Eightsy is a shrimpf
• Eightsy: Fivsy is a shrimpf

The professor sighed: “I will never find out who is in which group if you continue like this.” Then the last Shroombian answered

• Ninee: Zeroo and Sixee belong to different groups

This riddle can be modelled as a CSP, and it gives the opportunity to discuss a special kind of constraint, namely binary relations.

First set up the variables

from constrainingorder.variables import DiscreteVariable
from constrainingorder.sets import DiscreteSet

domain = DiscreteSet(['Shrimpf','Wex'])
variables = []
for i in range(10):
variables.append(DiscreteVariable(str(i),domain=domain))


So every variable represents one Shroombian, who can be either a shrimpf or a wex.

Almost all hints are of the same structure: one Shroombian accuses another Shroombian of being a shrimpf. The hint is fulfilled either if the accusing shroombian is a Shrimpf (who is always lying) and the accused shroombian is not actually a Shrimpf, or if the accusor is a Wex (who is always telling the truth) and the accused is a in fact a Shrimpf.

We can represent this in form of a custom constraint. As each hint affects two shroombians, such constraints are binary relations. The implementation of binary relations is much simpler than for general constraints.

from constrainingorder.constraints import BinaryRelation

class Accusation(BinaryRelation):
def relation(self,val1,val2):
return (val1 == 'Shrimpf' and val2 == 'Wex') or\
(val1 == 'Wex' and val2 == 'Shrimpf')


For classes derived from BinaryRelations it suffices to implement a single method that returns True if the two values fulfill the relation and False otherwise. Specific constraints are obtained by instantiating this class with two variables.

For DiscreteVariables with small domains one can represent binary relations also by listing all tuples of values that fulfill the relation. An equivalent implementation would be derived from DiscreteBinaryRelation.

from constrainingorder.constraints import DiscreteBinaryRelation

class Accusation(DiscreteBinaryRelation):
def __init__(self,var1,var2):
DiscreteBinaryRelation.__init__(self,var1,var2,[
('Shrimpf','Wex'), ('Wex','Shrimpf')
])


In addition we need to implement a new constraint for the last hint. As it affects three shroombians, this is not a binary relation.

from constrainingorder.constraints import Constraint
class AllegedNonEqual(Constraint):
def __init__(self,var1,var2,var3):
Constraint.__init__(self,{
var1 : var1.domain,
var2 : var2.domain,
var3 : var3.domain}
)
self.v1 = var1.name
self.v2 = var2.name
self.v3 = var3.name

def satisfied(self,lab):
if not (self.v1 in lab and self.v2 in lab and self.v3 in lab):
return False
elif lab[self.v1] == 'Shrimpf':
return lab[self.v2] == lab[self.v3]
elif lab[self.v1] == 'Wex':
return lab[self.v2] != lab[self.v3]

def consistent(self,lab):
if not (self.v1 in lab and self.v2 in lab and self.v3 in lab):
return True
elif lab[self.v1] == 'Shrimpf':
return lab[self.v2] == lab[self.v3]
elif lab[self.v1] == 'Wex':
return lab[self.v2] != lab[self.v3]


Now we can specify the constraints

cons = []
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))
cons.append(Accusation(variables,variables))

cons.append(AllegedNonEqual(variables,variables,variables))


And solve the problem

from constrainingorder import Space
from constrainingorder.solver import solve

space = Space(variables,cons)

for solution in solve(space,method='backtrack'):
for name, group in sorted(solution.items()):
print name, group

0 Shrimpf
1 Wex
2 Shrimpf
3 Shrimpf
4 Wex
5 Shrimpf
6 Wex
7 Shrimpf
8 Wex
9 Wex