Histogram of the distribution of dice rolls - c++

Question

I saw a question on careercup, but I do not get the answer I want there. I wrote an answer myself and want your comment on my analysis of time complexity and comment on the algorithm and code. Or you could provide a better algorithm in terms of time. Thanks.
You are given d > 0 fair dice with n > 0 "sides", write an function that returns a histogram of the frequency of the result of dice rolls.
For example, for 2 dice, each with 3 sides, the results are:
(1, 1) -> 2
(1, 2) -> 3
(1, 3) -> 4
(2, 1) -> 3
(2, 2) -> 4
(2, 3) -> 5
(3, 1) -> 4
(3, 2) -> 5
(3, 3) -> 6
And the function should return:
2: 1
3: 2
4: 3
5: 2
6: 1
(my sol). The time complexity if you use a brute force depth first search is O(n^d). However, you can use the DP idea to solve this problem. For example, d=3 and n=3. You can use the result of d==1 when computing d==2:
d==1
num #
1 1
2 1
3 1
d==2
first roll second roll is 1
num # num #
1 1 2 1
2 1 -> 3 1
3 1 4 1
first roll second roll is 2
num # num #
1 1 3 1
2 1 -> 4 1
3 1 5 1
first roll second roll is 3
num # num #
1 1 4 1
2 1 -> 5 1
3 1 6 1
Therefore,
second roll
num #
2 1
3 2
4 3
5 2
6 1
The time complexity of this DP algorithm is
SUM_i(1:d) {n*[n(d-1)-(d-1)+1]} ~ O(n^2*d^2)
~~~~~~~~~~~~~~~ <--eg. d=2, n=3, range from 2~6
The code is written in C++ as follows
vector<pair<int,long long>> diceHisto(int numSide, int numDice) {
int n = numSide*numDice;
vector<long long> cur(n+1,0), nxt(n+1,0);
for(int i=1; i<=numSide; i++) cur[i]=1;
for(int i=2; i<=numDice; i++) {
int start = i-1, end = (i-1)*numSide; // range of previous sum of rolls
//cout<<"start="<<start<<" end="<<end<<endl;
for(int j=1; j<=numSide; j++) {
for(int k=start; k<=end; k++)
nxt[k+j] += cur[k];
}
swap(cur,nxt);
for(int j=start; j<=end; j++) nxt[j]=0;
}
vector<pair<int,long long>> result;
for(int i=numDice; i<=numSide*numDice; i++)
result.push_back({i,cur[i]});
return result;
}

Answer 1

You can do it in O(n*d^2). First, note that the generating function for an n-sided dice is p(n) = x+x^2+x^3+...+x^n, and that the distribution for d throws has generating function p(n)^d. Representing the polynomials as arrays, you need O(nd) coefficients, and multiplying by p(n) can be done in a single pass in O(nd) time by keeping a rolling sum.
Here's some python code that implements this. It has one non-obvious optimisation: it throws out a factor x from each p(n) (or equivalently, it treats the dice as having faces 0,1,2,...,n-1 rather than 1,2,3,...,n) which is why d is added back in when showing the distribution.
def dice(n, d):
r = [1] + [0] * (n-1) * d
nr = [0] * len(r)
for k in xrange(d):
t = 0
for i in xrange(len(r)):
t += r[i]
if i >= n:
t -= r[i-n]
nr[i] = t
r, nr = nr, r
return r
def show_dist(n, d):
for i, k in enumerate(dice(n, d)):
if k: print i + d, k
show_dist(6, 3)
The time and space complexity are easy to see: there's nested loops with d and (n-1)*d iterations so the time complexity is O(n.d^2), and there's two arrays of size O(nd) and no other allocation, so the space complexity is O(nd).

Answer 2

Just in case, here a simple example in Python using the OpenTurns platform.
import openturns as ot
d = 2 # number of dice
n = 6 # number of sides per die
# possible values
dice_distribution = ot.UserDefined([[i] for i in range(1, n + 1)])
# create the sum distribution d times the sum
sum_distribution = sum([dice_distribution] * d)
That's it!
print(sum_distribution)
will show you all the possible values and their corresponding probabilities:
>>> UserDefined(
{x = [2], p = 0.0277778},
{x = [3], p = 0.0555556},
{x = [4], p = 0.0833333},
{x = [5], p = 0.111111},
{x = [6], p = 0.138889},
{x = [7], p = 0.166667},
{x = [8], p = 0.138889},
{x = [9], p = 0.111111},
{x = [10], p = 0.0833333},
{x = [11], p = 0.0555556},
{x = [12], p = 0.0277778}
)
You can also draw the probability distribution function:
sum_distribution.drawPDF()