fibbing is like lying... or just not telling the whole truth
Because the same calculations are done over and over again. Fib(n) - the nth. number in the sequence - is equal to fib(n-1) + fib(n-2). For example, fib(10) - the 10th. number in the sequence - is equal to fib(9) + fib(8). If you expand this, you get fib(8) + fib(7) + fib(7) + fib(6). If you expand again, you get fib(7) + fib(6) + fib(6) + fib(5) + fib(6) + fib(5) + fib(5) + fib(4). You can already see that some of the numbers have to be evaluated several times. In fact, the amount of calculations increases exponentially; whereas with a simple loop, to add numbers up to fib(n), this is not the case.
Fib as a noun "he told a fib about eating his spinach" fib as verb "Fibbing is not acceptable, even if you don't call it lying"
No, it is not. The noun "fib" means a subjectively minor lie. The verb "to fib" means to tell a lie.
A recursive formula is one that references itself. The famous example is the Fibonacci function: fib(n) := fib(n-1) + fib(n-2), with the terminating proviso that fib(0) = 0 and fib(1) = 1.
"fib" is an English slang word for a 'small' lie
public static int fib(int n) {return fib(n-1) + fib(n-2);}
FIB
A fib about ecosystems.... Ecosystems are made of cheese......
Maryoku Yummy - 2010 Fij Fij and the Fib Fib Now You're Cooking was released on: USA: 15 October 2010
Fib is basically an insignificant or a harmless lie , or u can say fibbing is milder form of lying . . . sentence : she fibbed about going early to bed last night when she actually did not she told a fib to everyone about being the topper in her school
You don't actually need a program to prove this. Fib[n] is the sum of Fib[n-1] and Fib[n-2]. it therefore follows that Fib[n-2] must be the sum of Fib[n-3] and Fib[n-4]. That being the case, it stands to reason that Fib[n] must be the sum of Fib[n-1], Fib[n-3] and Fib[n-4]. The "not necessarily different" part of the problem is only required to cater for the first part of the sequence where either Fib[n-1], Fib[n-3] or Fib[n-4] do not exist. E.g.,: 0 = 0+0+0 1 = 1+0+0 2 = 1+1+0 or 2+0+0 3 = 1+1+1 or 2+1+0 or 3+0+0 5 = 2+2+1 or 3+1+1 or 3+2+0 or 5+0+0 From there onwards, there is always at least one solution where all three numbers are different and greater than zero. Note that every Fibonacci is the sum of itself, zero and zero, so there is always at least one solution for every Fibonacci number. int main (void) { /* Fact: there are only 44 Fibonacci numbers in the range 0:1,000,000,000 */ const int max=44; int fib[max]; int index; /* Note: the sequence begins 0, 1, 1, 2, 3, 5... The value 1 appears twice so we'll begin the sequence there and then replace the first 1 with a 0. */ fib[0]=1; fib[1]=1; index=2; while (index<max) { fib[index]=fib[index-1]+fib[index-2]; ++index; } fib[0]=0; /* Test each Fibonacci... */ for (index=0; index<max; ++index) { bool found = false; /* Toggle this when we find a solution */ int a, b, c; for (a=0; a<max; ++a) for (b=0; b<max; ++b) for (c=0; c<max; ++c) if (fib[index]==fib[a]+fib[b]+fib[c]) { found = true; printf ("%d = %d + %d + %d\n", fib[index], fib[a], fib[b], fib[c]); } if (!found) { printf ("I'm too stupid to solve this problem!\n"); return -1; } } return 0; }
a fib