Unsigned long long outputs 0 - c

Question

I have this issue where I need to square line in Pascal's triangle and when it comes to big numbers it only outputs 0
unsigned long sum_squared(const int line){
unsigned long long n = 2*line;
unsigned long long x1 = factorial(line);
unsigned long long k = x1*x1;
unsigned long long x = factorial(n)/(k);
return x;
}
unsigned long long factorial(unsigned long long n) {
if (n == 0){
return 1;
}
return n * factorial(n - 1);
}
I test it with:
printf("%lu\n", sum_squared(14));

Answer 1

The key word here is "outputs". You need the correct format specifier to output unsigned long long: %llu.

Answer 2

The maximum number should be 7219428434016265740 and it equals sum_squared(33)
The code tries to evaluate the sum using (directly) the formula
⎛ 2n ⎞ (2n)!
⎜ ⎟ = ―――――
⎝ n ⎠ n!·n!
As noted in the other answers and comments, this leads to overflow of the intermediate results.
Luckily, this expression can be simplified and evaluated safely (without overflowing 64-bit wide unsigned integers) up to the required value.
We can start dividing both the numerator and the denominator by n!
(2n)! (n+1)·(n+2)· ... ·(n+n) (n+1)·(n+2)· ... ·(2n-2)·(2n-1)·(2n)
――――― = ――――――――――――――――――――――― = ――――――――――――――――――――――――――――――――――――
n!·n! n! 1·2· ... ·(n-1)·n
Also note that the higher half terms in the denominator have the corresponding doubles in the numerator. Let's see a couple of examples
⎛ 8 ⎞ 8! 5·6·7·8 5·2·7·2
⎜ ⎟ = ――――― = ――――――― = ――――――― = 70
⎝ 4 ⎠ 4!·4! 1·2·3·4 1·2
⎛ 10 ⎞ 10! 6·7·8·9·10 2·7·2·9·2 70
⎜ ⎟ = ――――― = ―――――――――― = ――――――――― = 252 = ―― · 9 · 2
⎝ 5 ⎠ 5!·5! 1·2·3·4·5 1·2 5
⎛ 12 ⎞ 12! 7·8·9·10·11·12 7·2·9·2·11·2 252
⎜ ⎟ = ――――― = ―――――――――――――― = ―――――――――――― = 924 = ――― · 11
⎝ 6 ⎠ 6!·6! 1·2·3·4·5·6 1·2·3 3
Now we can derive a couple of recurrence relations that let us evaluate the sum at a given n from the sum at n - 1
sum(1) = 2
sum(n-1)
sum(n) = ―――――――― · (2·n - 1) when n is even
n/2
sum(n-1)
sum(n) = ―――――――― · (2·n - 1) · 2 when n is odd
n
Note that the divisions (which are always exact, but I won't demonstrate that) are performed before the multiplications, avoiding unnecessary overflows.
Those formulas produce the following sequence
1 2
2 6
3 20
4 70
5 252
6 924
7 3432
8 12870
9 48620
10 184756
11 705432
12 2704156
13 10400600
14 40116600
15 155117520
16 601080390
17 2333606220
18 9075135300
19 35345263800
20 137846528820
21 538257874440
22 2104098963720
23 8233430727600
24 32247603683100
25 126410606437752
26 495918532948104
27 1946939425648112
28 7648690600760440
29 30067266499541040
30 118264581564861424
31 465428353255261088
32 1832624140942590534
33 7219428434016265740
The next value, 34, can't be calculated this way (using a 64-bit type).

Answer 3

28! = 304888344611713860501504000000 which is larger than maximum unsigned long long 18446744073709551615. It wraps over (modulo ULLONG_MAX)and the result is smaller than k. The division will be 0
You use the wrong printf format.
unsigned long long factorial(unsigned long long n)
{
if (n == 0) return 1;
return n * factorial(n - 1);
}
unsigned long sum_squared(const int line)
{
unsigned long long n = 2*line;
unsigned long long x1 = factorial(line);
unsigned long long k = x1*x1;
unsigned long long x = factorial(n)/(k);
printf("%llu %llu %llu %llu %llu\n", n, x1, k, x , factorial(n));
return x;
}
int main(void)
{
printf("%llu\n", sum_squared(14));
}
output:
28 87178291200 18442642257371725824 0 12478583540742619136
18417982275456073728