I thought that this was a fun problem, so I decided to demonstrate how to do it efficiently. On my system it takes about half a second to find all 4193 solutions.
The trick is to use dynamic programming to build a data structure from which the expressions can all be found efficiently.
from fractions import Fraction
def bit_subset(index, things):
i = 0
bit = 1
answer = []
while bit <= index:
if bit & index:
answer.append(things[i])
i += 1
bit += bit
return answer
def indexes_to_bit_subset(indexes):
answer = 0
for index in indexes:
answer += 2**index
return answer
def calculation_tree (digits):
# Here is the tree structure:
#
# By bit_integer of subset used
# By value reached
# [
# count_of_ways,
# [
# [
# count,
# first subset bit_integer,
# first value,
# op ('+', '-', '*', '/', None),
# second subset bit_integer (may be 0),
# second value (may be None)
# ]
# ]
# ]
#
tree = {}
# Populate with all of the raw numnbers.
for i in range(len(digits)):
# By bit_integer for 1 thing.
tree[2**i] = {
# By value reached (use fractions to avoid roundoff
Fraction(digits[i], 1): [
# Just 1 way to do it.
1,
# What are the ways? Let me tell you!
[[1, 2**i, digits[i], None, 0, None]]
]
}
# This loops over all subsets with something in the first.
for subset_index in range(1, 2**len(digits)):
# The indexes into the chosen subset.
index_subset = bit_subset(subset_index, list(range(len(digits))))
if subset_index not in tree:
subtree = {}
tree[subset_index] = subtree
# Look at ways to split it with something in both sides
for sub_subset_index in range(1, 2**len(index_subset)-1):
co_subset_index = 2**(len(index_subset)) - 1 - sub_subset_index
# We have a split by indexes into index_subset.
# We need to turn that into a split by indexes into digits
# Which we represent by integers representing bits.
index_sub_subset = bit_subset(sub_subset_index, index_subset)
sub_subset_bit_index = indexes_to_bit_subset(index_sub_subset)
index_co_subset = bit_subset(co_subset_index, index_subset)
co_subset_bit_index = indexes_to_bit_subset(index_co_subset)
# Let's pull out the already known calculation results.
subtree1 = tree[sub_subset_bit_index]
subtree2 = tree[co_subset_bit_index]
for value1 in subtree1.keys():
count1 = subtree1[value1][0]
for value2 in subtree2.keys():
count2 = subtree2[value2][0]
# We now have to add each possible operation result to subtree
options = {
'+': value1 + value2,
'-': value1 - value2,
'*': value1 * value2,
}
if value2 != 0:
options['/'] = value1 / value2
for op, value in options.items():
if value in subtree:
subtree[value][0] += count1 * count2
subtree[value][1].append([
count1 * count2,
sub_subset_bit_index,
value1,
op,
co_subset_bit_index,
value2
])
else:
subtree[value] = [
count1 * count2,
[[
count1 * count2,
sub_subset_bit_index,
value1,
op,
co_subset_bit_index,
value2
]]
]
return tree
# Yields the expressions that result in value
def expression_iter (tree, bit_integer, value):
if bit_integer in tree:
subtree = tree[bit_integer]
if value in subtree:
ways = subtree[value][1]
for (
count,
bit_integer1, value1,
op,
bit_integer2, value2
) in ways:
if op is None:
yield str(value1)
else:
for expr1 in expression_iter(tree, bit_integer1, value1):
for expr2 in expression_iter(tree, bit_integer2, value2):
yield f'({expr1} {op} {expr2})'
def all_expressions(digits, target):
tree = calculation_tree(digits)
frac_target = Fraction(target, 1)
for bit_integer, subtree in tree.items():
if frac_target in subtree:
for expr in expression_iter(tree, bit_integer, frac_target):
yield expr
for expr in all_expressions([2, 1, 4, 6, 7], 3):
print(expr)