csp.py

"""CSP (Constraint Satisfaction Problems) problems and solvers. (Chapter 5)."""

from __future__ import generators
from utils import *
import search
import types

class CSP(search.Problem):
    """This class describes finite-domain Constraint Satisfaction Problems.
    A CSP is specified by the following inputs:
        vars        A list of variables; each is atomic (e.g. int or string).
        domains     A dict of {var:[possible_value, ...]} entries.
        neighbors   A dict of {var:[var,...]} that for each variable lists
                    the other variables that participate in constraints.
        constraints A function f(A, a, B, b) that returns true if neighbors
                    A, B satisfy the constraint when they have values A=a, B=b
    In the textbook and in most mathematical definitions, the
    constraints are specified as explicit pairs of allowable values,
    but the formulation here is easier to express and more compact for
    most cases. (For example, the n-Queens problem can be represented
    in O(n) space using this notation, instead of O(N^4) for the
    explicit representation.) In terms of describing the CSP as a
    problem, that's all there is.

    However, the class also supports data structures and methods that help you
    solve CSPs by calling a search function on the CSP.  Methods and slots are
    as follows, where the argument 'a' represents an assignment, which is a
    dict of {var:val} entries:
        assign(var, val, a)     Assign a[var] = val; do other bookkeeping
        unassign(var, a)        Do del a[var], plus other bookkeeping
        nconflicts(var, val, a) Return the number of other variables that
                                conflict with var=val
        curr_domains[var]       Slot: remaining consistent values for var
                                Used by constraint propagation routines.
    The following methods are used only by graph_search and tree_search:
        successor(state)        Return a list of (action, state) pairs
        goal_test(state)        Return true if all constraints satisfied
    The following are just for debugging purposes:
        nassigns                Slot: tracks the number of assignments made
        display(a)              Print a human-readable representation

    >>> search.depth_first_graph_search(australia)
    <Node (('WA', 'B'), ('Q', 'B'), ('T', 'B'), ('V', 'B'), ('SA', 'G'), ('NT', 'R'), ('NSW', 'R'))>
    """

    def __init__(self, vars, domains, neighbors, constraints):
        "Construct a CSP problem. If vars is empty, it becomes domains.keys()."
        vars = vars or domains.keys()
        update(self, vars=vars, domains=domains,
               neighbors=neighbors, constraints=constraints,
               initial=(), curr_domains=None, nassigns=0)
        
    def assign(self, var, val, assignment):
        "Add {var: val} to assignment; Discard the old value if any."
        assignment[var] = val
        self.nassigns += 1

    def unassign(self, var, assignment):
        """Remove {var: val} from assignment.
        DO NOT call this if you are changing a variable to a new value;
        just call assign for that."""
        if var in assignment:
            del assignment[var]

    def nconflicts(self, var, val, assignment):
        "Return the number of conflicts var=val has with other variables."
        # Subclasses may implement this more efficiently
        def conflict(var2):
            val2 = assignment.get(var2, None)
            return val2 != None and not self.constraints(var, val, var2, val2)
        return count_if(conflict, self.neighbors[var])

    def display(self, assignment):
        "Show a human-readable representation of the CSP."
        # Subclasses can print in a prettier way, or display with a GUI
        print 'CSP:', self, 'with assignment:', assignment

    ## These methods are for the tree and graph search interface:

    def successor(self, state):
        "Return a list of (action, state) pairs."
        if len(state) == len(self.vars):
            return []
        else:
            assignment = dict(state)
            var = find_if(lambda v: v not in assignment, self.vars)
            return [((var, val), state + ((var, val),))
                    for val in self.domains[var]
                    if self.nconflicts(var, val, assignment) == 0]

    def goal_test(self, state):
        "The goal is to assign all vars, with all constraints satisfied."
        assignment = dict(state)
        return (len(assignment) == len(self.vars) and
                every(lambda var: self.nconflicts(var, assignment[var],
                                                  assignment) == 0,
                      self.vars))

    ## These are for constraint propagation

    def support_pruning(self):
        """Make sure we can prune values from domains. (We want to pay
        for this only if we use it.)"""
        if self.curr_domains is None:
            self.curr_domains = dict((v, self.domains[v][:]) for v in self.vars)

    def suppose(self, var, value):
        "Start accumulating inferences from assuming var=value."
        self.support_pruning()
        removals = [(var, a) for a in self.curr_domains[var] if a != value]
        self.curr_domains[var] = [value]
        return removals

    def prune(self, var, value, removals):
        "Rule out var=value."
        self.curr_domains[var].remove(value)
        if removals is not None: removals.append((var, value))

    def choices(self, var):
        "Return all values for var that aren't currently ruled out."
        return (self.curr_domains or self.domains)[var]

    def restore(self, removals):
        "Undo all inferences from a supposition."
        for B, b in removals:
            self.curr_domains[B].append(b)

    def infer_assignment(self):
        "Return the partial assignment implied by the current inferences."
        self.support_pruning()
        return dict((v, self.curr_domains[v][0]) 
                    for v in self.vars if 1 == len(self.curr_domains[v]))

    ## This is for min_conflicts search

    def conflicted_vars(self, current):
        "Return a list of variables in current assignment that are in conflict"
        return [var for var in self.vars
                if self.nconflicts(var, current[var], current) > 0]

#______________________________________________________________________________
# CSP Backtracking Search

## Variable ordering

def first_unassigned_variable(assignment, csp):
    "The default variable order."
    return find_if(lambda var: var not in assignment, csp.vars)

def mrv(assignment, csp):
    "Minimum-remaining-values heuristic."
    return argmin_random_tie(
        [v for v in csp.vars if v not in assignment],
        lambda var: num_legal_values(csp, var, assignment))

def num_legal_values(csp, var, assignment):
    if csp.curr_domains:
        return len(csp.curr_domains[var])
    else:
        return count_if(lambda val: csp.nconflicts(var, val, assignment) == 0,
                        csp.domains[var])

## Value ordering

def unordered_domain_values(var, assignment, csp):
    "The default value order."
    return csp.choices(var)

def lcv(var, assignment, csp):
    "Least-constraining-values heuristic."
    return sorted(csp.choices(var),
                  key=lambda val: csp.nconflicts(var, val, assignment))

## Inference

def no_inference(assignment, csp, var, value):
    return []

def forward_checking(assignment, csp, var, value):
    "Prune neighbor values inconsistent with var=value."
    removals = csp.suppose(var, value)
    for B in csp.neighbors[var]:
        if B not in assignment:
            for b in csp.curr_domains[B][:]:
                if not csp.constraints(var, value, B, b):
                    csp.prune(B, b, removals)
    return removals

def mac(assignment, csp, var, value):
    "Maintain arc consistency."
    removals = csp.suppose(var, value)
    AC3(csp, [(X, var) for X in csp.neighbors[var]], removals)
    return removals

## The search, proper

def backtracking_search(csp,
                        select_unassigned_variable = first_unassigned_variable,
                        order_domain_values = unordered_domain_values,
                        inference = no_inference):
    """Fig. 6.5 (3rd edition)
    >>> backtracking_search(australia) is not None
    True
    >>> backtracking_search(australia, select_unassigned_variable=mrv) is not None
    True
    >>> backtracking_search(australia, order_domain_values=lcv) is not None
    True
    >>> backtracking_search(australia, select_unassigned_variable=mrv, order_domain_values=lcv) is not None
    True
    >>> backtracking_search(australia, inference=forward_checking) is not None
    True
    >>> backtracking_search(australia, inference=mac) is not None
    True
    >>> backtracking_search(usa, select_unassigned_variable=mrv, order_domain_values=lcv, inference=mac) is not None
    True
    """

    if inference is not no_inference:
        csp.curr_domains = dict((v, csp.domains[v][:])
                                for v in csp.vars)

    def backtrack(assignment):
        if len(assignment) == len(csp.vars):
            return assignment
        var = select_unassigned_variable(assignment, csp)
        for value in order_domain_values(var, assignment, csp):
            if 0 == csp.nconflicts(var, value, assignment):
                csp.assign(var, value, assignment)
                removals = inference(assignment, csp, var, value)
                result = backtrack(assignment)
                if result is not None:
                    return result
                csp.restore(removals)
                csp.unassign(var, assignment)
        return None

    result = backtrack({})
    assert result is None or csp.goal_test(result)
    return result
                
#______________________________________________________________________________
# Constraint Propagation with AC-3

def AC3(csp, queue=None, removals=None):
    """[Fig. 5.7]"""
    if queue is None:
        queue = [(Xi, Xk) for Xi in csp.vars for Xk in csp.neighbors[Xi]]
    csp.support_pruning()
    while queue:
        (Xi, Xj) = queue.pop()
        if remove_inconsistent_values(csp, Xi, Xj, removals):
            for Xk in csp.neighbors[Xi]:
                queue.append((Xk, Xi))

def remove_inconsistent_values(csp, Xi, Xj, removals):
    "Return true if we remove a value."
    removed = False
    for x in csp.curr_domains[Xi][:]:
        # If Xi=x conflicts with Xj=y for every possible y, eliminate Xi=x
        if every(lambda y: not csp.constraints(Xi, x, Xj, y),
                 csp.curr_domains[Xj]):
            csp.prune(Xi, x, removals)
            removed = True
    return removed

#______________________________________________________________________________
# Min-conflicts hillclimbing search for CSPs

def min_conflicts(csp, max_steps=100000):
    """Solve a CSP by stochastic hillclimbing on the number of conflicts."""
    # Generate a complete assignment for all vars (probably with conflicts)
    csp.current = current = {}
    for var in csp.vars:
        val = min_conflicts_value(csp, var, current)
        csp.assign(var, val, current)
    # Now repeatedly choose a random conflicted variable and change it
    for i in range(max_steps):
        conflicted = csp.conflicted_vars(current)
        if not conflicted:
            return current
        var = random.choice(conflicted)
        val = min_conflicts_value(csp, var, current)
        csp.assign(var, val, current)
    return None

def min_conflicts_value(csp, var, current):
    """Return the value that will give var the least number of conflicts.
    If there is a tie, choose at random."""
    return argmin_random_tie(csp.domains[var],
                             lambda val: csp.nconflicts(var, val, current)) 

#______________________________________________________________________________
# Map-Coloring Problems

class UniversalDict:
    """A universal dict maps any key to the same value. We use it here
    as the domains dict for CSPs in which all vars have the same domain.
    >>> d = UniversalDict(42)
    >>> d['life']
    42
    """
    def __init__(self, value): self.value = value
    def __getitem__(self, key): return self.value
    def __repr__(self): return '{Any: %r}' % self.value

def different_values_constraint(A, a, B, b):
    "A constraint saying two neighboring variables must differ in value."
    return a != b

def MapColoringCSP(colors, neighbors):
    """Make a CSP for the problem of coloring a map with different colors
    for any two adjacent regions.  Arguments are a list of colors, and a
    dict of {region: [neighbor,...]} entries.  This dict may also be
    specified as a string of the form defined by parse_neighbors"""

    if isinstance(neighbors, str):
        neighbors = parse_neighbors(neighbors)     
    return CSP(neighbors.keys(), UniversalDict(colors), neighbors,
               different_values_constraint)

def parse_neighbors(neighbors, vars=[]):
    """Convert a string of the form 'X: Y Z; Y: Z' into a dict mapping
    regions to neighbors.  The syntax is a region name followed by a ':'
    followed by zero or more region names, followed by ';', repeated for
    each region name.  If you say 'X: Y' you don't need 'Y: X'.
    >>> parse_neighbors('X: Y Z; Y: Z')
    {'Y': ['X', 'Z'], 'X': ['Y', 'Z'], 'Z': ['X', 'Y']}
    """
    dict = DefaultDict([])
    for var in vars:
        dict[var] = []
    specs = [spec.split(':') for spec in neighbors.split(';')]
    for (A, Aneighbors) in specs:
        A = A.strip()
        dict.setdefault(A, [])
        for B in Aneighbors.split():
            dict[A].append(B)
            dict[B].append(A)
    return dict

australia = MapColoringCSP(list('RGB'),
                           'SA: WA NT Q NSW V; NT: WA Q; NSW: Q V; T: ')
    
usa = MapColoringCSP(list('RGBY'),
        """WA: OR ID; OR: ID NV CA; CA: NV AZ; NV: ID UT AZ; ID: MT WY UT;
        UT: WY CO AZ; MT: ND SD WY; WY: SD NE CO; CO: NE KA OK NM; NM: OK TX;
        ND: MN SD; SD: MN IA NE; NE: IA MO KA; KA: MO OK; OK: MO AR TX;
        TX: AR LA; MN: WI IA; IA: WI IL MO; MO: IL KY TN AR; AR: MS TN LA;
        LA: MS; WI: MI IL; IL: IN; IN: KY; MS: TN AL; AL: TN GA FL; MI: OH;
        OH: PA WV KY; KY: WV VA TN; TN: VA NC GA; GA: NC SC FL;
        PA: NY NJ DE MD WV; WV: MD VA; VA: MD DC NC; NC: SC; NY: VT MA CT NJ;
        NJ: DE; DE: MD; MD: DC; VT: NH MA; MA: NH RI CT; CT: RI; ME: NH;
        HI: ; AK: """)
#______________________________________________________________________________
# n-Queens Problem

def queen_constraint(A, a, B, b):
    """Constraint is satisfied (true) if A, B are really the same variable,
    or if they are not in the same row, down diagonal, or up diagonal."""
    return A == B or (a != b and A + a != B + b and A - a != B - b)

class NQueensCSP(CSP):
    """Make a CSP for the nQueens problem for search with min_conflicts.
    Suitable for large n, it uses only data structures of size O(n).
    Think of placing queens one per column, from left to right.
    That means position (x, y) represents (var, val) in the CSP.
    The main structures are three arrays to count queens that could conflict:
        rows[i]      Number of queens in the ith row (i.e val == i)
        downs[i]     Number of queens in the \ diagonal
                     such that their (x, y) coordinates sum to i
        ups[i]       Number of queens in the / diagonal
                     such that their (x, y) coordinates have x-y+n-1 = i
    We increment/decrement these counts each time a queen is placed/moved from
    a row/diagonal. So moving is O(1), as is nconflicts.  But choosing
    a variable, and a best value for the variable, are each O(n).
    If you want, you can keep track of conflicted vars, then variable
    selection will also be O(1).
    >>> len(backtracking_search(NQueensCSP(8)))
    8
    """
    def __init__(self, n):
        """Initialize data structures for n Queens."""
        CSP.__init__(self, range(n), UniversalDict(range(n)),
                     UniversalDict(range(n)), queen_constraint)
        update(self, rows=[0]*n, ups=[0]*(2*n - 1), downs=[0]*(2*n - 1))

    def nconflicts(self, var, val, assignment): 
        """The number of conflicts, as recorded with each assignment.
        Count conflicts in row and in up, down diagonals. If there
        is a queen there, it can't conflict with itself, so subtract 3."""
        n = len(self.vars)
        c = self.rows[val] + self.downs[var+val] + self.ups[var-val+n-1]
        if assignment.get(var, None) == val:
            c -= 3
        return c

    def assign(self, var, val, assignment):
        "Assign var, and keep track of conflicts."
        oldval = assignment.get(var, None)
        if val != oldval:
            if oldval is not None: # Remove old val if there was one
                self.record_conflict(assignment, var, oldval, -1)
            self.record_conflict(assignment, var, val, +1)
            CSP.assign(self, var, val, assignment)

    def unassign(self, var, assignment):
        "Remove var from assignment (if it is there) and track conflicts."
        if var in assignment:
            self.record_conflict(assignment, var, assignment[var], -1)
        CSP.unassign(self, var, assignment)
        
    def record_conflict(self, assignment, var, val, delta):
        "Record conflicts caused by addition or deletion of a Queen."
        n = len(self.vars)
        self.rows[val] += delta
        self.downs[var + val] += delta
        self.ups[var - val + n - 1] += delta

    def display(self, assignment):
        "Print the queens and the nconflicts values (for debugging)."
        n = len(self.vars)
        for val in range(n):
            for var in range(n):
                if assignment.get(var,'') == val: ch ='Q'
                elif (var+val) % 2 == 0: ch = '.'
                else: ch = '-'
                print ch,
            print '    ',
            for var in range(n):
                if assignment.get(var,'') == val: ch ='*'
                else: ch = ' '
                print str(self.nconflicts(var, val, assignment))+ch, 
            print        

#______________________________________________________________________________
# Sudoku

import itertools, re

def flatten(seqs): return sum(seqs, [])

easy1   = '..3.2.6..9..3.5..1..18.64....81.29..7.......8..67.82....26.95..8..2.3..9..5.1.3..'
harder1 = '4173698.5.3..........7......2.....6.....8.4......1.......6.3.7.5..2.....1.4......'

class Sudoku(CSP):
    """A Sudoku problem.
    The box grid is a 3x3 array of boxes, each a 3x3 array of cells.
    Each cell holds a digit in 1..9. In each box, all digits are
    different; the same for each row and column as a 9x9 grid.
    >>> e = Sudoku(easy1)
    >>> e.display(e.infer_assignment())
    . . 3 | . 2 . | 6 . .
    9 . . | 3 . 5 | . . 1
    . . 1 | 8 . 6 | 4 . .
    ------+-------+------
    . . 8 | 1 . 2 | 9 . .
    7 . . | . . . | . . 8
    . . 6 | 7 . 8 | 2 . .
    ------+-------+------
    . . 2 | 6 . 9 | 5 . .
    8 . . | 2 . 3 | . . 9
    . . 5 | . 1 . | 3 . .
    >>> AC3(e); e.display(e.infer_assignment())
    4 8 3 | 9 2 1 | 6 5 7
    9 6 7 | 3 4 5 | 8 2 1
    2 5 1 | 8 7 6 | 4 9 3
    ------+-------+------
    5 4 8 | 1 3 2 | 9 7 6
    7 2 9 | 5 6 4 | 1 3 8
    1 3 6 | 7 9 8 | 2 4 5
    ------+-------+------
    3 7 2 | 6 8 9 | 5 1 4
    8 1 4 | 2 5 3 | 7 6 9
    6 9 5 | 4 1 7 | 3 8 2
    >>> h = Sudoku(harder1)
    >>> None != backtracking_search(h, select_unassigned_variable=mrv, inference=forward_checking)
    True
    """
    R3 = range(3)
    Cell = itertools.count().next
    bgrid = [[[[Cell() for x in R3] for y in R3] for bx in R3] for by in R3]
    boxes = flatten([map(flatten, brow)       for brow in bgrid])
    rows  = flatten([map(flatten, zip(*brow)) for brow in bgrid])
    units = map(set, boxes + rows + zip(*rows))
    neighbors = dict([(v, set.union(*[u for u in units if v in u]) - set([v]))
                      for v in flatten(rows)])

    def __init__(self, grid):
        squares = re.findall(r'\d|\.', grid)
        domains = dict((var, [int(ch)] if ch.isdigit() else range(1, 10))
                       for var, ch in zip(flatten(self.rows), squares))
        CSP.__init__(self, None, domains, self.neighbors,
                     different_values_constraint)

    def display(self, assignment):
        def show_box(box): return [' '.join(map(show_cell, row)) for row in box]
        def show_cell(cell): return str(assignment.get(cell, '.'))
        def abut(lines1, lines2): return map(' | '.join, zip(lines1, lines2))
        print '\n------+-------+------\n'.join(
            '\n'.join(reduce(abut, map(show_box, brow))) for brow in self.bgrid)

#______________________________________________________________________________
# The Zebra Puzzle

def Zebra():
    "Return an instance of the Zebra Puzzle."
    Colors = 'Red Yellow Blue Green Ivory'.split()
    Pets = 'Dog Fox Snails Horse Zebra'.split()
    Drinks = 'OJ Tea Coffee Milk Water'.split()
    Countries = 'Englishman Spaniard Norwegian Ukranian Japanese'.split()
    Smokes = 'Kools Chesterfields Winston LuckyStrike Parliaments'.split()
    vars = Colors + Pets + Drinks + Countries + Smokes
    domains = {}
    for var in vars:
        domains[var] = range(1, 6)
    domains['Norwegian'] = [1]
    domains['Milk'] = [3]
    neighbors = parse_neighbors("""Englishman: Red;
                Spaniard: Dog; Kools: Yellow; Chesterfields: Fox;
                Norwegian: Blue; Winston: Snails; LuckyStrike: OJ;
                Ukranian: Tea; Japanese: Parliaments; Kools: Horse;
                Coffee: Green; Green: Ivory""", vars)
    for type in [Colors, Pets, Drinks, Countries, Smokes]:
        for A in type:
            for B in type:
                if A != B:
                    if B not in neighbors[A]: neighbors[A].append(B)
                    if A not in neighbors[B]: neighbors[B].append(A)
    def zebra_constraint(A, a, B, b, recurse=0):
        same = (a == b)
        next_to = abs(a - b) == 1
        if A == 'Englishman' and B == 'Red': return same
        if A == 'Spaniard' and B == 'Dog': return same
        if A == 'Chesterfields' and B == 'Fox': return next_to
        if A == 'Norwegian' and B == 'Blue': return next_to
        if A == 'Kools' and B == 'Yellow': return same
        if A == 'Winston' and B == 'Snails': return same
        if A == 'LuckyStrike' and B == 'OJ': return same
        if A == 'Ukranian' and B == 'Tea': return same
        if A == 'Japanese' and B == 'Parliaments': return same
        if A == 'Kools' and B == 'Horse': return next_to
        if A == 'Coffee' and B == 'Green': return same
        if A == 'Green' and B == 'Ivory': return (a - 1) == b
        if recurse == 0: return zebra_constraint(B, b, A, a, 1)
        if ((A in Colors and B in Colors) or
            (A in Pets and B in Pets) or
            (A in Drinks and B in Drinks) or
            (A in Countries and B in Countries) or
            (A in Smokes and B in Smokes)): return not same        
        raise 'error'
    return CSP(vars, domains, neighbors, zebra_constraint)

def solve_zebra(algorithm=min_conflicts, **args):
    z = Zebra()
    ans = algorithm(z, **args)
    for h in range(1, 6):
        print 'House', h,
        for (var, val) in ans.items():
            if val == h: print var,
        print
    return ans['Zebra'], ans['Water'], z.nassigns, ans


__doc__ += random_tests("""
>>> min_conflicts(australia)
{'WA': 'B', 'Q': 'B', 'T': 'G', 'V': 'B', 'SA': 'R', 'NT': 'G', 'NSW': 'G'}
>>> min_conflicts(NQueensCSP(8), max_steps=10000)
{0: 5, 1: 0, 2: 4, 3: 1, 4: 7, 5: 2, 6: 6, 7: 3}
""")