exponential

I'm looking for an intuitive, real-world example of a problem that takes (worst case) exponential time complexity to solve for a talk I am giving. Here are examples for other time complexities I have come up with (many of them taken from this SO question ): O(1) - determining if a number is odd or even O(log N) - finding a word in the dictionary (using binary search) O(N) - reading a book O(N log N) - sorting a deck of playing cards (using merge sort) O(N^2) - checking if you have everything on your shopping list in your trolley O(infinity) - tossing a coin until it lands on heads Any ideas? O

Which function grows faster, exponential (like 2^n, n^n, e^n etc) or factorial (n!)? Ps: I just read somewhere, n! grows faster than 2^n. n! eventually grows faster than an exponential with a constant base (2^n and e^n), but n^n grows faster than n! since the base grows as n increases. n! = n * (n-1) * (n-2) * ... n^n = n * n * n * ... Every term after the first one in n^n is larger, so n^n will grow faster. n^n grows larger than n! -- for an excellent explanation, see the answer by @AlexQueue. For the other cases, read on: Factorial functions do asymptotically grow larger than exponential

Currently I need to calculate 2^N , however N can be as large as 1929238932899 and I'm stuck using a long data type which can't hold a number that large. I've currently tried converting to 'BigInt' however I'm still stuck with the long data type restriction from what I've seen as well. I have a function which calculates the power. However, with the long data type, it just returns 0 when the number gets too big. Note that this is just a generic recursive power function. For example, with 2^6 its meant to return 64 and with 2^47 to return 140737488355328. However, when it becomes 2^8489289 , it

I've been trying to fit an exponential curve to my data using ggplot and geom_smooth. I'm trying to replicate the answer to a similar problem ( geom_smooth and exponential fits ) but keep getting following error message: > exp.model <-lm(y ~ exp(x), df) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : NA/NaN/Inf in 'x' I don't understand the error, as there is not NA/NaN/Inf values in the dataset: >df x y 1 1981 3.262897 2 1990 2.570096 3 2000 7.098903 4 2001 5.428424 5 2002 6.056302 6 2003 5.593942 7 2004 10.869635 8 2005 12.425793 9 2006 5.601889 10 2007 6.498187 11

Is it possible to fix parameters while fitting distributions in SciPy? For example, this code: import scipy.stats as st xx = st.expon.rvs(size=100) print st.expon.fit(xx, loc=0) results in non-zero location ( loc ). When some parameter is provided to the fit function it is considered as an initial guess. And if it is provided to the constructor ( st.expon(loc=0) ) the distribution becomes "frozen" and can not be used for fitting. Warren Weckesser To fix loc , use the argument floc : print st.expon.fit(xx, floc=0) E.g. In [33]: import scipy.stats as st In [34]: xx = st.expon.rvs(size=100) In

I'm working on Project Euler to brush up on my C++ coding skills in preparation for the programming challenge(s) we'll be having this next semester (since they don't let us use Python, boo!). I'm on #16, and I'm trying to find a way to keep real precision for 2¹°°° For instance: int main(){ double num = pow(2, 1000); printf("%.0f", num): return 0; } prints

So, I was just playing around with manually calculating the value of e in R and I noticed something that was a bit disturbing to me. The value of e using R's exp() command... exp(1) #[1] 2.718282 Now, I'll try to manually calculate it using x = 10000 x <- 10000 y <- (1 + (1 / x)) ^ x y #[1] 2.718146 Not quite but we'll try to get closer using x = 100000 x <- 100000 y <- (1 + (1 / x)) ^ x y #[1] 2.718268 Warmer but still a bit off... x <- 1000000 y <- (1 + (1 / x)) ^ x y #[1] 2.71828 Now, let's try it with a huge one x <- 5000000000000000 y <- (1 + (1 / x)) ^ x y #[1] 3.035035 Well, that's not