Sunday Times Teaser 2504 – No Title

by Bob Walker

Published September 19 2010 (link)

Here are 18 black cards and there is a cross on the back of some of them.

The clue in any gap indicates how many crosses there are on cards that are adjacent to that gap.

How many black cards have a cross on them?

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Brian Gladman permalink


from operator import eq, ne, gt, lt

grid = '''
+---+---+---+---+---+---+
| B |   | B |   | B |<>0|
+---+---+---+---+---+---+
|==1| B | >2| B | <2| B |
+---+---+---+---+---+---+
| B |   | B |<>1| B |   |
+---+---+---+---+---+---+
|   | B |   | B | >1| B |
+---+---+---+---+---+---+
| B | >2| B |<>3| B |   |
+---+---+---+---+---+---+
|<>2| B |   | B |==1| B |
+---+---+---+---+---+---+'''

# black cell n (n in 0..17), when present in this dictionary,
# indicates the conditions that can be computed when the cell
# has just been filled
chk = { 5:(0,),  6:(1,),    7:(2,),  8: (3,), 10:(4,),
       14:(5,), 15:(6, 8), 16:(7,), 17: (9,)} 

# for 'condition' cell n (n in 0..9), item[n] in this tuple gives the numbers
# of the black cells that contribute to its value
adj = ((2, 5),   (0, 3, 6),  (1, 3, 4, 7), (2, 4, 5, 8), (4, 7, 8, 10), 
       (8, 10, 11, 14), (9, 12, 13, 15), (10, 13, 14, 16), (12, 15), (14, 16, 17))

# the operators and the values for the 10 condition cells 
op = ((ne, 0), (eq, 1), (gt, 2), (lt, 2), (ne, 1),
      (gt, 1), (gt, 2), (ne, 3), (ne, 2), (eq, 1))

def set_cards(g, adj, chk):
  ln = len(g)
  # can we check any condition cells?
  if ln - 1 in chk:
    # for each cell that we can check
    for cc in chk[ln - 1]:
      # sum its adjacent (black card) cells 
      v = sum(g[i] for i in adj[cc])
      # exit if the condition is not met
      if not op[cc][0](v, op[cc][1]):
        return
  if ln == 18:
    yield g
  else:
    # try the next card without and with a cross
    for x in (0, 1):
      # continue to the next black card
      yield from set_cards(g + (x,), adj, chk)

for g in set_cards(tuple(), adj, chk):
  # output the grid
  ss, cnt = '', 0
  for c in grid:
    if c == 'B':
      ss += str(g[cnt])
      cnt += 1
    else:
      ss += c
  print(ss, ' count =', sum(g))

from operator import eq, ne, gt, lt

grid = '''

+---+---+---+---+---+---+

| B | | B | | B |<>0|

+---+---+---+---+---+---+

|==1| B | >2| B | <2| B |

+---+---+---+---+---+---+

| B | | B |<>1| B | |

+---+---+---+---+---+---+

| | B | | B | >1| B |

+---+---+---+---+---+---+

| B | >2| B |<>3| B | |

+---+---+---+---+---+---+

|<>2| B | | B |==1| B |

+---+---+---+---+---+---+'''

# black cell n (n in 0..17), when present in this dictionary,

# indicates the conditions that can be computed when the cell

# has just been filled

chk = { 5:(0,), 6:(1,), 7:(2,), 8: (3,), 10:(4,),

14:(5,), 15:(6, 8), 16:(7,), 17: (9,)}

# for 'condition' cell n (n in 0..9), item[n] in this tuple gives the numbers

# of the black cells that contribute to its value

adj = ((2, 5), (0, 3, 6), (1, 3, 4, 7), (2, 4, 5, 8), (4, 7, 8, 10),

(8, 10, 11, 14), (9, 12, 13, 15), (10, 13, 14, 16), (12, 15), (14, 16, 17))

# the operators and the values for the 10 condition cells

op = ((ne, 0), (eq, 1), (gt, 2), (lt, 2), (ne, 1),

(gt, 1), (gt, 2), (ne, 3), (ne, 2), (eq, 1))

def set_cards(g, adj, chk):

ln = len(g)

# can we check any condition cells?

if ln - 1 in chk:

# for each cell that we can check

for cc in chk[ln - 1]:

# sum its adjacent (black card) cells

v = sum(g[i] for i in adj[cc])

# exit if the condition is not met

if not op[cc][0](v, op[cc][1]):

return

if ln == 18:

yield g

else:

# try the next card without and with a cross

for x in (0, 1):

# continue to the next black card

yield from set_cards(g + (x,), adj, chk)

for g in set_cards(tuple(), adj, chk):

# output the grid

ss, cnt = '', 0

for c in grid:

if c == 'B':

ss += str(g[cnt])

cnt += 1

else:

ss += c

print(ss, ' count =', sum(g))

Sunday Times Teaser 2504 – No Title

by Bob Walker

Published September 19 2010 (link)

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