Datalog w Pythonie

Ostatnio na studiach miałem coś o języku Datalog. Okazuje się, że jest i Pythonowa biblioteka, która realizuje ten paradygmat. https://sites.google.com/site/pydatalog/Online-datalog-tutorial

Poniżej cytuję jeden z przykładów z dokumentacji, rozwiązanie problemu ośmiu hetmanów:

from pyDatalog import pyDatalog
import time

@pyDatalog.program()
def _():
 # when is it ok to have a queen in row X1 and another in row X2, separated by N columns
 # this is memoized !
 ok(X1, N, X2) <= (X1 != X2) & (X1 != X2+N) & (X1 != X2-N)

 queens(X0)                      <= (X0._in(range(8)))
 queens(X0,X1)                   <= queens(X0)                   & next_queen(X0,X1)
 queens(X0,X1,X2)                <= queens(X0,X1)                & next_queen(X0,X1,X2)
 queens(X0,X1,X2,X3)             <= queens(X0,X1,X2)             & next_queen(X0,X1,X2,X3)
 queens(X0,X1,X2,X3,X4)          <= queens(X0,X1,X2,X3)          & next_queen(X0,X1,X2,X3,X4)
 queens(X0,X1,X2,X3,X4,X5)       <= queens(X0,X1,X2,X3,X4)       & next_queen(X0,X1,X2,X3,X4,X5)
 queens(X0,X1,X2,X3,X4,X5,X6)    <= queens(X0,X1,X2,X3,X4,X5)    & next_queen(X0,X1,X2,X3,X4,X5,X6)
 queens(X0,X1,X2,X3,X4,X5,X6,X7) <= queens(X0,X1,X2,X3,X4,X5,X6) & next_queen(X0,X1,X2,X3,X4,X5,X6,X7)

 next_queen(X0,X1)                   <= queens(X1)                       & ok(X0,1,X1)
 next_queen(X0,X1,X2)                <= next_queen(X1,X2)                & ok(X0,2,X2)
 next_queen(X0,X1,X2,X3)             <= next_queen(X1,X2,X3)             & ok(X0,3,X3)
 next_queen(X0,X1,X2,X3,X4)          <= next_queen(X1,X2,X3,X4)          & ok(X0,4,X4)
 next_queen(X0,X1,X2,X3,X4,X5)       <= next_queen(X1,X2,X3,X4,X5)       & ok(X0,5,X5)
 next_queen(X0,X1,X2,X3,X4,X5,X6)    <= next_queen(X1,X2,X3,X4,X5,X6)    & ok(X0,6,X6)
 next_queen(X0,X1,X2,X3,X4,X5,X6,X7) <= next_queen(X1,X2,X3,X4,X5,X6,X7) & ok(X0,7,X7)

 # counting is 0-based, so this is actually the 8-queens solution
 start_time = time.time()
 print(queens(X0,X1,X2,X3,X4,X5,X6,X7))
 print("First datalog run in %f seconds" % (time.time() - start_time))

 start = time.time()
 for i in range(20):  
  # there is a warm-up period for the JIT --> let's compute it again
  start_time = time.time()
  datalog_count = len(queens(X0,X1,X2,X3,X4,X5,X6,X7))
  datalog_time = (time.time() - start_time)
  print(datalog_time)
 print("Average : %s" % ((time.time() - start)/20))
 print("%i solutions by datalog in %f seconds" % (datalog_count, datalog_time))

# pure python solution found on http://rosettacode.org/wiki/N-Queens#Python, for comparison purposes
from itertools import permutations

n = 8
cols = range(n)
def queens():
 for vec in permutations(cols):
  if n == len(set(vec[i]+i for i in cols)) \
    == len(set(vec[i]-i for i in cols)):
   #print ( vec )
   pass

start_time = time.time()
queens() 
python_time = time.time() - start_time

print("python : %f seconds" % python_time)

# results with pypy 2.6 on Intel Core i7-4810 QM CPU @ 2.8 GHz (run from Command prompt):
# 0.07 sec for Datalog in average
# 0.03 sec for python