CODEWITHSUNDEEP
pythoncombinationspermutationprobability
geek101
geek101

Reputation: 194

A Confusing task

My approach to the problem was as follows-

a=[ ]
for i in range(7):
    a.append([0]*7)

c=dict()
for i in range(7):
    for j in range(7):
        a[i][j]=(i,j)

for i in range(7):
    for j in range(7):
        c[i+j]=tuple((i*j+j+c))

print c

But this produces:

{0: (0, 0), 1: (1, 0), 2: (2, 0), 3: (3, 0), 4: (4, 0), 5: (5, 0), 6: (6, 0), 7: (6, 1), 8: (6, 2), 9: (6, 3), 10: (6, 4), 11: (6, 5), 12: (6, 6)}

Upvotes: 1

Views: 78

Answers (3)

emvee
emvee

Reputation: 4449

Using some standard Python utilities:

from collections import defaultdict
from itertools import product

All numbers on a single dice:

numbers = range(1,7)

And now, for each combination, sum the combination and append the combination to the list for that combination:

reduce(lambda acc, comb: acc[sum(comb)].append(comb) or acc, product(numbers, numbers), defaultdict(list))

Resulting in:

defaultdict(<type 'list'>, {2: [(1, 1)], 3: [(1, 2), (2, 1)], 4: [(1, 3), (2, 2), (3, 1)], 5: [(1, 4), (2, 3), (3, 2), (4, 1)], 6: [(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)], 7: [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)], 8: [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)], 9: [(3, 6), (4, 5), (5, 4), (6, 3)], 10: [(4, 6), (5, 5), (6, 4)], 11: [(5, 6), (6, 5)], 12: [(6, 6)]})

Upvotes: 0

Cory Kramer
Cory Kramer

Reputation: 117981

A step by step way to do this would be

pairs = {}
for first in range(1,7):
    for second in range(1,7):
        total = first + second
        if total in pairs:
            # If sum exists, add this tuple to the list for this key.
            pairs[total] += [(first,second)]
        else:
            # If sum doesn't exist, start a new list for this key
            pairs[total] = [(first,second)]

Result

>>> pairs
{2: [(1, 1)],
 3: [(1, 2), (2, 1)],
 4: [(1, 3), (2, 2), (3, 1)],
 5: [(1, 4), (2, 3), (3, 2), (4, 1)],
 6: [(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)],
 7: [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)],
 8: [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)],
 9: [(3, 6), (4, 5), (5, 4), (6, 3)],
 10: [(4, 6), (5, 5), (6, 4)],
 11: [(5, 6), (6, 5)],
 12: [(6, 6)]}

Since this sounds like an academic exercise I assume you cannot use some of the pre-existing Python modules. Otherwise, you may want to check out collections.defaultdict and itertools.product. The former can handle the "does this key exist yet or not?" and the latter can handle combinations to remove your nested for loops.

Upvotes: 2

Jamie Cockburn
Jamie Cockburn

Reputation: 7555

This will work:

combinations = {}
for a in range(1, 7):
    for b in range(1, 7):
        combinations.setdefault(a+b, []).append((a, b))

Result:

{2: [(1, 1)],
 3: [(1, 2), (2, 1)],
 4: [(1, 3), (2, 2), (3, 1)],
 5: [(1, 4), (2, 3), (3, 2), (4, 1)],
 6: [(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)],
 7: [(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)],
 8: [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)],
 9: [(3, 6), (4, 5), (5, 4), (6, 3)],
 10: [(4, 6), (5, 5), (6, 4)],
 11: [(5, 6), (6, 5)],
 12: [(6, 6)],
 }

Upvotes: 1

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