Latest revision as of 14:26, 22 March 2014

다른 언어라면 이렇게 쉽게 안되었을 문제인데, 역시 Python입니다 ^ 오 ^

EightQueenProblem

Start Time : 2014. 03. 16. 1:08
End Time : 2014. 03. 16. 1:24

~Python
from itertools import permutations
n = 8
cols = range(n)
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)):
    for i in cols:
      for j in cols:
        if vec[i] == j: print ('Q'),
        else: print '#',
      print 
    print 
    break

EightQueenProblem2

~Python
from itertools import permutations
n = 8
cols = range(n)
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

EightQueenProblem, EightQueenProblem2, 안정원

@@ Line 1: / Line 1: @@
-[[최다인]] 학우가 단톡에 Javascript가 어쩌고 Eight Queen이 어쩌고 하길래 짜보았습니다.
+다른 언어라면 이렇게 쉽게 안되었을 문제인데, 역시 Python입니다 ^ 오 ^
-다른 언어라면 이렇게 쉽게 안되었을 문제인데, 역시 Python입니다^ 오 ^
+= EightQueenProblem =
 # Start Time : 2014. 03. 16. 1:08
 # End Time : 2014. 03. 16. 1:24
+ ~Python
   from itertools import permutations
   n = 8
@@ Line 16: / Line 17: @@
         print ''
       print ''
+     break
+= EightQueenProblem2 =
+ ~Python
+ from itertools import permutations
+ n = 8
+ cols = range(n)
+ 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
+----
+[[EightQueenProblem]], [[EightQueenProblem2]], [[안정원]]