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Alphametics exercise updated #1119
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Update example.py
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Merge branch 'master' into patch-3
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Original file line number | Diff line number | Diff line change |
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@@ -1,98 +1,176 @@ | ||
from itertools import permutations | ||
from string import ascii_uppercase as acu | ||
acuset = set(acu) | ||
dset = set(range(10)) | ||
nzdset = dset.copy() | ||
nzdset.remove(0) | ||
from itertools import permutations, chain, product | ||
""" | ||
This solution will first parse the alphametic expression | ||
grouping and counting letters buy digit ranks | ||
then trace recursively all possible permutations starting from | ||
the lowest rank and genrating additional permutations for new digits | ||
at higer ranks as necessary. | ||
This will allow to avoid unnesessarily large permutations to scan. | ||
Also leading letters in words will be treated as non-zero digits only | ||
to reduce the number of permutations | ||
""" | ||
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def solve(an): | ||
# Break down to words | ||
an = an.upper() | ||
alphaexp = [tuple(map(str.strip, s.split("+"))) | ||
for s in an.split("==")] | ||
# Sum powers of 10 for letters ready for computation | ||
expdict = dict() | ||
loexpdict = dict() | ||
for si, s in enumerate(alphaexp): | ||
esign = 1 - (si << 1) | ||
for t in s: | ||
lowletter = t[-1] | ||
if lowletter not in loexpdict: | ||
loexpdict[lowletter] = 0 | ||
loexpdict[lowletter] += esign | ||
for p, letter in enumerate(reversed(t)): | ||
if letter not in expdict: | ||
expdict[letter] = 0 | ||
expdict[letter] += esign * (10 ** p) | ||
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# Extract all letters and check if they are really letters | ||
alldigits = set(expdict.keys()) | ||
if not alldigits <= acuset: | ||
raise ValueError | ||
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# extract high and low digigts | ||
hidigits = set([w[0] for s in alphaexp for w in s]) | ||
lodigits = set([w[-1] for s in alphaexp for w in s]) | ||
def digPerms(digset, nzcharset, okzcharset): | ||
"""This function creates permutations given the set of digits, | ||
letters not alllowed to be 0, and letters allowed to be 0 | ||
""" | ||
nzcnt = len(nzcharset) # How many letters are non-0 | ||
okzcnt = len(okzcharset) # How many letters are allowed 0 | ||
totcnt = nzcnt + okzcnt # Total number of letters | ||
if totcnt < 1: # if total numbers of letters is 0 | ||
return [()] # return a singe empty permutation | ||
nzdigset = digset - set((0,)) # generate a non-zero digit set | ||
nzdigsetcnt = len(nzdigset) # how many non-zero digits are available | ||
digsetcnt = len(digset) # how many ok zero digits are available | ||
# if either fewer digits than letters at all or fewer non-0 digits | ||
# than letters that need to be non-zero | ||
if digsetcnt < totcnt or nzdigsetcnt < nzcnt: | ||
return [] # Return no permutations possible | ||
# Simple case when zeros are allowed everwhere | ||
# or no zero is containted within the given digits | ||
elif nzcnt == 0 or digsetcnt == nzdigsetcnt: | ||
return permutations(digset, totcnt) | ||
# Another simple case all letters are non-0 | ||
elif okzcnt == 0: | ||
return permutations(nzdigset, totcnt) | ||
else: | ||
# General case | ||
# Generate a list of possible 0 positions | ||
poslst = list(range(nzcnt, totcnt)) | ||
# Chain two iterators | ||
# first iterator with all non-0 permutations | ||
# second iterator with all permulations without 1 letter | ||
# insert 0 in all possible positions of that permutation | ||
return chain(permutations(nzdigset, totcnt), | ||
map(lambda x: x[0][:x[1]] + (0,) + x[0][x[1]:], | ||
product(permutations(nzdigset, totcnt - 1), | ||
poslst))) | ||
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# Break down low digits to nonzeros (also high digits) and possible zeros | ||
lonzdigits = lodigits & hidigits | ||
lorestdigits = lodigits - lonzdigits | ||
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# Main digits, all but not low | ||
maindigits = alldigits - lodigits | ||
def check_rec(eqparams, tracecombo=(dict(), 0, set(range(10))), p=0): | ||
"""This function recursively traces a parsed expression from lowest | ||
digits to highest, generating additional digits when necessary | ||
checking the digit sum is divisible by 10, carrying the multiple of 10 | ||
up to the next level | ||
""" | ||
# Basic parameters of the equation, | ||
# maximal digit rank | ||
# characters with multipliers by rank | ||
# unique non-zero characters by rank | ||
# unique zero-allowed characters by rank | ||
# all unique characters by rank | ||
maxp, tchars, unzchars, uokzchars, uchars = eqparams | ||
# recursion cumulative parameters | ||
# established characters with digits | ||
# carry-over from the previous level | ||
# remaining unassigned digits | ||
prevdict, cover, remdigs = tracecombo | ||
# the maximal 10-power (beyond the maximal rank) | ||
# is reached | ||
if p == maxp: | ||
# Carry-over is zero, meaning solution is found | ||
if cover == 0: | ||
return prevdict | ||
else: | ||
# Otherwise the solution in this branch is not found | ||
# return empty | ||
return dict() | ||
diglets = uchars[p] # all new unique letters from the current level | ||
partsum = cover # Carry over from lower level | ||
remexp = [] # TBD letters | ||
# Break down the current level letter into what can be | ||
# calculated in the partial sum and remaining TBD letter-digits | ||
for c, v in tchars[p]: | ||
if c in prevdict: | ||
partsum += v * prevdict[c] | ||
else: | ||
remexp.append((c, v)) | ||
# Generate permutations for the remaining digits and currecnt level | ||
# non-zero letters and zero-allowed letters | ||
for newdigs in digPerms(remdigs, unzchars[p], uokzchars[p]): | ||
# build the dictionary for the new letters and this level | ||
newdict = dict(zip(diglets, newdigs)) | ||
# complete the partial sum into test sum using the current permutation | ||
testsum = partsum + sum([newdict[c] * v | ||
for c, v in remexp]) | ||
# check if the sum is divisible by 10 | ||
d, r = divmod(testsum, 10) | ||
if r == 0: | ||
# if divisible, update the dictionary to all established | ||
newdict.update(prevdict) | ||
# proceed to the next level of recursion with | ||
# the same eqparams, but updated digit dictionary, | ||
# new carry over and remaining digits to assign | ||
rectest = check_rec(eqparams, | ||
(newdict, d, remdigs - set(newdigs)), | ||
p + 1) | ||
# if the recursive call returned a non-empty dictionary | ||
# this means the recursion has found a solution | ||
# otherwise, proceed to the new permutation | ||
if len(rectest) > 0: | ||
return rectest | ||
# if no permutations are avaialble or no | ||
# permutation gave the result return the empty dictionary | ||
return dict() | ||
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# Break down main digit list into nonzeroees and possible zeroes | ||
mainnzdigits = maindigits & hidigits | ||
mainrestdigits = maindigits - mainnzdigits | ||
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# change sets to tuples to guarantee the stable order | ||
t_lorestdigits = tuple(lorestdigits) | ||
t_lonzdigits = tuple(lonzdigits) | ||
t_lowdigs = t_lorestdigits + t_lonzdigits | ||
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t_mainrestdigits = tuple(mainrestdigits) | ||
t_mainnzdigits = tuple(mainnzdigits) | ||
t_maindigs = t_mainrestdigits + t_mainnzdigits | ||
t_alldigs = t_lowdigs + t_maindigs | ||
def solve(an): | ||
"""A function to solve the alphametics problem | ||
""" | ||
# First, split the expresion into left and right parts by == | ||
# split each part into words by + | ||
# strip spaces fro, each word, reverse each work to | ||
# enumerate the digit rank from lower to higer | ||
fullexp = [list(map(lambda x: list(reversed(x.strip())), s.split("+"))) | ||
for s in an.strip().upper().split("==")] | ||
# Find the maximal lenght of the work, maximal possive digit rank or | ||
# the power of 10, should the < maxp | ||
maxp = max([len(w) for s in fullexp for w in s]) | ||
# Extract the leading letters for each (reversed) word | ||
# those cannot be zeros as the number cannot start with 0 | ||
nzchars = set([w[-1] for s in fullexp for w in s]) | ||
# initialize the lists for digit ranks | ||
unzchars = [] # non-zero letters unique at level | ||
uokzchars = [] # zero-allowed letters unique at level | ||
uchars = [] # all letters unique at level | ||
tchars = [] # all letter with multipliers per level | ||
for i in range(maxp): | ||
tchars.append(dict()) | ||
unzchars.append(set()) | ||
uokzchars.append(set()) | ||
# Now lets scan the expression and accumulate the letter counts | ||
for si, s in enumerate(fullexp): | ||
sgn = 1 - (si << 1) # left side (0) is +1, right right (1) is -1 | ||
for w in s: # for each word in the side (already reversed) | ||
for p, c in enumerate(w): # enumerate with ranks | ||
if c not in tchars[p]: # check if the letter was alread there | ||
tchars[p][c] = 0 | ||
tchars[p][c] += sgn # append to the rank dictionary | ||
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# Check all possible digit permunations with zeros | ||
for lorest in permutations(dset, len(lorestdigits)): | ||
remnzdigs = nzdset - set(lorest) | ||
# Generate addtional non-zero digit permutations | ||
for lonz in permutations(remnzdigs, len(lonzdigits)): | ||
# Build a dictionary for to test the expression | ||
t_digvals = lorest + lonz | ||
# Evaluate the expression sides | ||
testsum = sum([dig * loexpdict[let] | ||
for let, dig in zip(t_lowdigs, t_digvals)]) | ||
if testsum % 10 == 0: | ||
# Low digit test passed, check the main digits | ||
# if there are no other digits that low digits, | ||
# test the whole expression and return if OK | ||
if len(maindigits) == 0: | ||
testsum = sum([dig * expdict[let] | ||
for let, dig in zip(t_lowdigs, t_digvals)]) | ||
if testsum == 0: | ||
return dict(zip(t_lowdigs, t_digvals)) | ||
totchars = set() # Keep track of letters already seen at lower ranks | ||
# go through the accumulated rank dictionaries | ||
for p, chardict in enumerate(tchars): | ||
for c, cnt in tuple(chardict.items()): | ||
if cnt == 0: # if the cumulative is 0 | ||
del chardict[c] # remove the letter from check dictionry | ||
# it does not impact the sum with 0-multiplier | ||
# if the letter contributes to the sum | ||
# and was not yet seen at lower ranks | ||
elif c not in totchars: | ||
# add the letter to either non-zero set | ||
# or allowed-zero set | ||
if c in nzchars: | ||
unzchars[p].add(c) | ||
else: | ||
# non-assigned digits | ||
remdigs = dset - set(t_digvals) | ||
# non-assigned without 0 | ||
remnzdigs = remdigs - set((0,)) | ||
# permutations for the rest of the digits | ||
for mainrest in permutations(remdigs, | ||
len(t_mainrestdigits)): | ||
lastnzdigs = remnzdigs - set(mainrest) | ||
# permutations for the non-zero rest of the digits | ||
for mainnz in permutations(lastnzdigs, | ||
len(t_mainnzdigits)): | ||
# Evaluate | ||
t_alldigvals = lorest + lonz + mainrest + mainnz | ||
testsum = sum([dig * expdict[let] | ||
for let, dig in zip(t_alldigs, | ||
t_alldigvals)]) | ||
if testsum == 0: | ||
return dict(zip(t_alldigs, t_alldigvals)) | ||
|
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return {} | ||
uokzchars[p].add(c) | ||
# add to the list as seen letter to ignore at the next | ||
# ranks | ||
totchars.add(c) | ||
# pre-build the combo list of letters for the rank | ||
# non-zero first, followed by zero-allowed | ||
uchars.append(tuple(unzchars[p]) + tuple(uokzchars[p])) | ||
# pre-convert check dictionaries to tuples | ||
tchars[p] = tuple(chardict.items()) | ||
# go for the recursion | ||
return check_rec([maxp, tchars, unzchars, uokzchars, uchars]) |
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This should run fine without wrappingchardict.items()
intuple()
.This is fine.