Enigmatic Code
Programming Enigma Puzzles
Enigma 1430: Doubt and fury
Posted by on 15 October 2012
From New Scientist #2591, 17th February 2007
The latest creation acquired by our local art gallery consists of six separate squares with labels such as “Doubt” and “Fury”, each of a different colour, including yellow and purple. The six squares have to be arranged at the gallery into a rectangle consisting of two rows with three squares in each row.
The curator remembered that “Charm” is above the red square (which is not “Angst”), that the blue square is between “Belief” and the orange square, and that the green square has no edge in common with “Envy”. He also remembered that the three labels in the top row are in alphabetical order, but in the bottom row neither the labels nor the colours are in alphabetical order.
So he arranged the squares satisfying all these conditions, but unfortunately he still got it wrong, thus infuriating the artist.
What (in alphabetical order of the labels) were the colours of the squares?
[enigma1430]
Enigmatic Code 
This Python program runs in 206ms. It’s a direct interpretation of the problem statement and checks the conditions over every possible permutation of the labels and colours, then looks for arrangements that have ambiguous orders.
from itertools import permutations, product from collections import defaultdict labels = ( 'A', 'B', 'C', 'D', 'E', 'F' ) colours = ( 'b', 'g', 'o', 'p', 'r', 'y' ) # adjacency matrix (squares that share edges) adj = ( (1, 3), (0, 2, 4), (1, 5), (0, 4), (1, 3, 5), (2, 4) ) # count the results r = defaultdict(list) # arrange the labels and colours for (l, c) in product(permutations(labels), permutations(colours)): # "Charm is above the Red square (which is not Angst)" i = c.index('r') # index of Red if i < 3: continue # Red must be in the bottom row if l[i] == 'A': continue # Red is not Angst if l[i - 3] != 'C': continue # Charm must be above Red # "the Blue square is between Belief and the Orange square" i = c.index('b') # index of Blue if i not in (1, 4): continue # Blue must be in the central column # and Blue is between Belief and Orange if not((l[i - 1] == 'B' and c[i + 1] == 'o') or (l[i + 1] == 'B' and c[i - 1] == 'o')): continue # "the Green square has no edge in common with Envy" i = c.index('g') # index of Green j = l.index('E') # index of Envy if i == j or j in adj[i]: continue # "labels in the top row are in alphabetical order" if not(l[0] < l[1] < l[2]): continue # "in the bottom row neither the labels nor the colours are in alphabetical order" if l[3] < l[4] < l[5]: continue if c[3] < c[4] < c[5]: continue # colours of the squares, in order of the labels s = tuple(c[l.index(i)] for i in labels) r[s].append((l, c)) for (k, v) in r.items(): if len(v) < 2: continue print(k, v)Solution: The squares, in alphabetical order of the labels are: Green, Yellow, Blue, Orange, Purple and Red.
If you rearrange so the tests are in a different order to that stated in the puzzle text you can get the run time down to 77ms.
from itertools import permutations from collections import defaultdict labels = ( 'A', 'B', 'C', 'D', 'E', 'F' ) colours = ( 'b', 'g', 'o', 'p', 'r', 'y' ) # adjacency matrix (squares that share edges) adj = ( (1, 3), (0, 2, 4), (1, 5), (0, 4), (1, 3, 5), (2, 4) ) # count the results r = defaultdict(list) # arrange the labels for l in permutations(labels): # "labels in the top row are in alphabetical order" if not(l[0] < l[1] < l[2]): continue # "in the bottom row neither the labels nor the colours are in alphabetical order" if l[3] < l[4] < l[5]: continue # "the Blue square is between Belief and the Orange square" # so Belief cannot be in the central column if 'B' in (l[1], l[4]): continue # and now arrange the colours for c in permutations(colours): # "in the bottom row neither the labels nor the colours are in alphabetical order" if c[3] < c[4] < c[5]: continue # "Charm is above the Red square (which is not Angst)" i = c.index('r') # index of Red if i < 3: continue # Red must be in the bottom row if l[i] == 'A': continue # Red is not Angst if l[i - 3] != 'C': continue # Charm must be above Red # "the Blue square is between Belief and the Orange square" i = c.index('b') # index of Blue if i not in (1, 4): continue # Blue must be in the central column # and Blue is between Belief and Orange if not((l[i - 1] == 'B' and c[i + 1] == 'o') or (l[i + 1] == 'B' and c[i - 1] == 'o')): continue # "the Green square has no edge in common with Envy" i = c.index('g') # index of Green j = l.index('E') # index of Envy if i == j or j in adj[i]: continue # colours of the squares, in order of the labels s = tuple(c[l.index(i)] for i in labels) r[s].append((l, c)) for (k, v) in r.items(): if len(v) < 2: continue print(k, v)Here is a version adapted from my Python code for the classic Zebra/Water puzzle. http://arthurvause.blogspot.co.uk/2012/10/who-owns-zebra.html
The adaptation isn’t entirely satisfactory as there are several “reverse lookups” required.
# arrangement of squares is: 1 2 3 # 4 5 6 from itertools import permutations as perm Labels = ['Angst','Belief','Charm','Doubt','Envy','Fury'] Colours = ['Red','Blue','Orange','Green','Yellow','Purple'] def lab_pos( ): for p in perm(Labels): yield {a:b for (a,b) in zip(p,range(1,7))} def col_lab( ): for p in perm(Labels): yield {a:b for (a,b) in zip(Colours,p)} def key(d, v): # find a key in dictionary d with value v return [name for name, val in d.items() if val == v][0] label_positions = [ x for x in lab_pos() if x['Charm'] < 4 and key(x,1) < key(x,2) < key(x,3) and not (key(x,4) < key(x,5) and key(x,5) < key(x,6)) ] for cl in col_lab(): solutions = [ lp for lp in label_positions if lp[cl['Blue']] in [2,5] and abs(lp[cl['Blue']]-lp[cl['Orange']])==1 and lp['Charm']+3 == lp[cl['Red']] and lp['Angst'] != lp[cl['Red']] and abs(lp[cl['Blue']]-lp['Belief'])==1 and lp['Belief'] != lp[cl['Orange']] and not ( sorted( (lp['Envy'], lp[cl['Green']])) in [[1,2],[2,3],[1,4],[2,5],[3,6],[4,5],[5,6]]) and lp['Envy'] != lp[cl['Green']] and not ( key(cl,key(lp,4)) < key(cl,key(lp,5)) and key(cl,key(lp,5)) < key(cl,key(lp,6)) ) ] if len(solutions) > 1: for lab in Labels: print lab, key(cl,lab)