Stack Overflow Asked by Jacky Chan on December 9, 2020

I am dealing with the following question:

Create a function that accepts two arguments, the number of dice rolled, and the outcome of the roll. The function returns the number of possible combinations that could produce that outcome. The number of dice can vary from 1 to 6.

And below is my code for 4 dice (x=4). But I am not able to extend this to any number of dices (x).

```
def dice_roll(x, y):
all_comb = []
for i in range(1, 7):
for j in range(1, 7):
for q in range(1, 7):
for r in range(1, 7):
total = i+j+q+r
all_comb.append(total)
return all_comb.count(y)
```

Is there a way to use recursion to deal with variable numbers of nested loops? And are there other more elegant ways apart from recursion for this question?

```
def combin_list(n):
# n is number of dice
if n == 1:
return [1,2,3,4,5,6]
else:
temp = combin_list(n-1)
resualt = []
for i in range(1,7):
for item in temp:
resualt.append(i+item)
return resualt
def dice_roll(x, y):
list = combin_list(x)
return list.count(y)
```

Correct answer by hr.mousavi on December 9, 2020

Here's a version that calculates the number of rolls for each total in O(n*s) time where n is the number of dice, and s the number of sides. It uses O(n) storage space.

If R[k, t] is the number of rolls of k dice that total t (keeping the number of sides fixed), then the recurrence relations are:

```
R[0, t] = 1 if t=0, 0 otherwise
R[k, t] = 0 if t < 0
R[k, t] = R[k-1, t-1] + R[k-1, t-2] + ... + R[k-1, t-s]
```

Then we solving this with dynamic programming.

```
def dice_roll(n, sides):
A = [1] + [0] * (sides * n)
for k in range(n):
T = sum(A[k] for k in range(max(0, sides*k), sides*(k+1)))
for i in range(sides*(k+1), -1, -1):
A[i] = T
if i > 0:
T -= A[i-1]
if i - sides - 1 >= 0:
T += A[i - sides - 1]
return A
print(dice_roll(9, 4))
```

The program returns an array `A`

with `A[i]`

storing the number of ways of rolling `n`

dice with `s`

sides that sum to `i`

.

Answered by Paul Hankin on December 9, 2020

a more mathematical approach using `sympy`

.

for larger numbers of dice this will scale way better than iterating over all possibilities; for small numbers of dice the start-up time of `sympy`

will probably not be worth the trouble.

```
from sympy.utilities.iterables import partitions, multiset_permutations
def dice_roll(n_dice, total):
ret = 0
for item in partitions(total, k=6, m=n_dice):
if sum(item.values()) != n_dice:
continue
print(f"{item}")
# this for loop is only needed if you want the combinations explicitly
for comb in multiset_permutations(item):
print(f" {comb}")
ret += sum(1 for _ in multiset_permutations(item))
return ret
```

i get the number of partitions of `total`

first (limiting the maximum value with `k=6`

as needed for a dice) and then count the number of possible multiset partitions.

the example:

```
r = dice_roll(n_dice=3, total=5)
print(r)
```

outputs:

```
{3: 1, 1: 2}
[1, 1, 3]
[1, 3, 1]
[3, 1, 1]
{2: 2, 1: 1}
[1, 2, 2]
[2, 1, 2]
[2, 2, 1]
6
```

meaning there are 6 ways to get to 5 with 3 dice. the combinations are shown.

in order to speed up things you could calculate the number of multiset combinations directly (you loose the explicit representation of the possibilities, though):

```
from sympy.utilities.iterables import partitions, multiset_permutations
from math import comb
def number_multiset_comb(dct):
ret = 1
n = sum(dct.values()) # number of slots
for v in dct.values():
ret *= comb(n, v)
n -= v
return ret
def dice_roll(n_dice, total):
ret = 0
for item in partitions(total, k=6, m=n_dice):
if sum(item.values()) != n_dice:
continue
ret += number_multiset_comb(dct=item)
return ret
```

Answered by hiro protagonist on December 9, 2020

As mentioned in the comments you should use `itertools.product`

like this:

```
import itertools
def dice_roll(dices: int, result: int) -> int:
combos = [x for x in itertools.product(range(1, 7), repeat=dices) if sum(x) == result]
return len(combos)
# combos = [(1, 3), (2, 2), (3, 1)]
if __name__ == '__main__':
print(dice_roll(2, 4))
# returns 3
```

With `itertools.product`

you get all possible combinations of the given amount of dices. with the list comprehension we filter the values by the correct sum.

Answered by D-E-N on December 9, 2020

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