I've been trying to implement the algorithm which does raising to a power every previous digit to current digit, which is also raised to. Then I find the last digit of this number. Here is the formula of this algorithm:
(x0 ** (x1 ** (x2 ** (x3 ** (...) **(Xn))))))
Then I find the last digit like that:
return find_last_digit % 10
If the list is empty, programm must return 1.
I have the Solution of this problem:
def last_digit(lst):
if len(lst) > 0:
temp = lst[-1]
for i in range(len(lst) - 2, -1, -1):
temp = lst[i] ** temp
return temp % 10
else:
return 1
But as you can see, this code takes a lot of time to be implemented if any value of the input list is large. Could you answer me, how can I make this code more effecient? Thx a lot
Here are some observations that can make the calculations more efficient:
As we need the last digit, and we are essentially doing multiplications, we can use the rules of modular arithmetic. If 𝑎⋅𝑏 = 𝑐, then 𝑎(mod 𝑚)⋅𝑏(mod 𝑚) = 𝑐(mod 𝑚). So a first idea could be to take 𝑚 as 10, and perform the multiplications. But we don't want to split up exponentiation in individual mutliplications, so then see the next point:
For all unsigned integers 𝑏 it holds that 𝑏2 = 𝑏6 modulo 20. You can verify this by doing this for all values of 𝑏 in the range {0,...,19}. By consequence, 𝑏𝑛 = 𝑏𝑛+4 for 𝑛 > 1. We choose 20 as modulus as that is both a multiple of 10 and 4. A multiple of 10, because we need to maintain the last digit in the process, and 4 as we will reduce the exponent by a multiple of 4. Both are necessary conditions at the same time, so not to lose out on the final digit. In the end we have that 𝑎(mod 20)(mod 10) = 𝑎(mod 10)
With these simplification rules, you can keep the involved exponents limited to at most 5, the base to at most 21, and the resulting power to at most 215 = 4084101.
The code could become:
def last_digit(lst):
power = 1
for base in reversed(lst):
power = (base if base < 2 else (base - 2) % 20 + 2) ** (
power if power < 2 else (power - 2) % 4 + 2)
return power % 10
In practice you can skip the reduction of base to (base - 2) % 20 + 2 if these input numbers are not very large.
I have created a recursive function to calculate the total numbers divisible by n within a range of (start,end). The function works for smaller numbers except when they start getting larger for ex. start=1 and end=10**12 - 1 gives me an error saying that the maximum recursion depth has been exceeded. How do I fix my code to stop this error:
def count(start, end, n, tot=0):
if start > end:
return tot
else:
if start % n == 0:
tot += 1
return count(start + 1, end, n, tot)
start = 1
end = 10**12 - 1
n = 5
print(count(start, end, n))
You can solve this problem with simple arithmetic. Thinking about a range such as (3, 18) (inclusive) and an n of 5, the actual range of interest is from 5 to 15 (since there are no multiples of 5 below 5 or above 15. You can round the start and end values to the new endpoints like this:
start = (start + 4) // n * n
end = end // n * n
What you need to do then is just count the values from 5 to 15 (3) which you can do subtracting the new start from the new end, dividing by n and adding 1 i.e.
(end // n * n - (start + 4) // n * n) // n + 1
which can be simplified to
end // n - (start + 4) // n + 1
So in total your function becomes:
def count(start, end, n):
return end // n - (start + 4) // n + 1
Two points.
(directly answering your question) You can increase the recursion depth by using the sys module.
import sys
sys.setrecursionlimit(10**12)
(towards making you a better programmer) You don't need to use recursion for this problem. In fact, it's a really bad idea as others have noted. Recursion like what you would be looking to do is O(n^2) which is, also known as 'really bad'. A linear approach would be O(n) which is pretty good. Better yet is to take the mathematical approach as Samwise and btilly suggested because you arent tasked to return the numbers that meet the criteria.
I understand Big-O notation, but I don't know how to calculate it for many functions. In particular, I've been trying to figure out the computational complexity of the naive version of the Fibonacci sequence:
int Fibonacci(int n)
{
if (n <= 1)
return n;
else
return Fibonacci(n - 1) + Fibonacci(n - 2);
}
What is the computational complexity of the Fibonacci sequence and how is it calculated?
You model the time function to calculate Fib(n) as sum of time to calculate Fib(n-1) plus the time to calculate Fib(n-2) plus the time to add them together (O(1)). This is assuming that repeated evaluations of the same Fib(n) take the same time - i.e. no memoization is used.
T(n<=1) = O(1)
T(n) = T(n-1) + T(n-2) + O(1)
You solve this recurrence relation (using generating functions, for instance) and you'll end up with the answer.
Alternatively, you can draw the recursion tree, which will have depth n and intuitively figure out that this function is asymptotically O(2n). You can then prove your conjecture by induction.
Base: n = 1 is obvious
Assume T(n-1) = O(2n-1), therefore
T(n) = T(n-1) + T(n-2) + O(1) which is equal to
T(n) = O(2n-1) + O(2n-2) + O(1) = O(2n)
However, as noted in a comment, this is not the tight bound. An interesting fact about this function is that the T(n) is asymptotically the same as the value of Fib(n) since both are defined as
f(n) = f(n-1) + f(n-2).
The leaves of the recursion tree will always return 1. The value of Fib(n) is sum of all values returned by the leaves in the recursion tree which is equal to the count of leaves. Since each leaf will take O(1) to compute, T(n) is equal to Fib(n) x O(1). Consequently, the tight bound for this function is the Fibonacci sequence itself (~θ(1.6n)). You can find out this tight bound by using generating functions as I'd mentioned above.
Just ask yourself how many statements need to execute for F(n) to complete.
For F(1), the answer is 1 (the first part of the conditional).
For F(n), the answer is F(n-1) + F(n-2).
So what function satisfies these rules? Try an (a > 1):
an == a(n-1) + a(n-2)
Divide through by a(n-2):
a2 == a + 1
Solve for a and you get (1+sqrt(5))/2 = 1.6180339887, otherwise known as the golden ratio.
So it takes exponential time.
I agree with pgaur and rickerbh, recursive-fibonacci's complexity is O(2^n).
I came to the same conclusion by a rather simplistic but I believe still valid reasoning.
First, it's all about figuring out how many times recursive fibonacci function ( F() from now on ) gets called when calculating the Nth fibonacci number. If it gets called once per number in the sequence 0 to n, then we have O(n), if it gets called n times for each number, then we get O(n*n), or O(n^2), and so on.
So, when F() is called for a number n, the number of times F() is called for a given number between 0 and n-1 grows as we approach 0.
As a first impression, it seems to me that if we put it in a visual way, drawing a unit per time F() is called for a given number, wet get a sort of pyramid shape (that is, if we center units horizontally). Something like this:
n *
n-1 **
n-2 ****
...
2 ***********
1 ******************
0 ***************************
Now, the question is, how fast is the base of this pyramid enlarging as n grows?
Let's take a real case, for instance F(6)
F(6) * <-- only once
F(5) * <-- only once too
F(4) **
F(3) ****
F(2) ********
F(1) **************** <-- 16
F(0) ******************************** <-- 32
We see F(0) gets called 32 times, which is 2^5, which for this sample case is 2^(n-1).
Now, we want to know how many times F(x) gets called at all, and we can see the number of times F(0) is called is only a part of that.
If we mentally move all the *'s from F(6) to F(2) lines into F(1) line, we see that F(1) and F(0) lines are now equal in length. Which means, total times F() gets called when n=6 is 2x32=64=2^6.
Now, in terms of complexity:
O( F(6) ) = O(2^6)
O( F(n) ) = O(2^n)
There's a very nice discussion of this specific problem over at MIT. On page 5, they make the point that, if you assume that an addition takes one computational unit, the time required to compute Fib(N) is very closely related to the result of Fib(N).
As a result, you can skip directly to the very close approximation of the Fibonacci series:
Fib(N) = (1/sqrt(5)) * 1.618^(N+1) (approximately)
and say, therefore, that the worst case performance of the naive algorithm is
O((1/sqrt(5)) * 1.618^(N+1)) = O(1.618^(N+1))
PS: There is a discussion of the closed form expression of the Nth Fibonacci number over at Wikipedia if you'd like more information.
You can expand it and have a visulization
T(n) = T(n-1) + T(n-2) <
T(n-1) + T(n-1)
= 2*T(n-1)
= 2*2*T(n-2)
= 2*2*2*T(n-3)
....
= 2^i*T(n-i)
...
==> O(2^n)
Recursive algorithm's time complexity can be better estimated by drawing recursion tree, In this case the recurrence relation for drawing recursion tree would be T(n)=T(n-1)+T(n-2)+O(1)
note that each step takes O(1) meaning constant time,since it does only one comparison to check value of n in if block.Recursion tree would look like
n
(n-1) (n-2)
(n-2)(n-3) (n-3)(n-4) ...so on
Here lets say each level of above tree is denoted by i
hence,
i
0 n
1 (n-1) (n-2)
2 (n-2) (n-3) (n-3) (n-4)
3 (n-3)(n-4) (n-4)(n-5) (n-4)(n-5) (n-5)(n-6)
lets say at particular value of i, the tree ends, that case would be when n-i=1, hence i=n-1, meaning that the height of the tree is n-1.
Now lets see how much work is done for each of n layers in tree.Note that each step takes O(1) time as stated in recurrence relation.
2^0=1 n
2^1=2 (n-1) (n-2)
2^2=4 (n-2) (n-3) (n-3) (n-4)
2^3=8 (n-3)(n-4) (n-4)(n-5) (n-4)(n-5) (n-5)(n-6) ..so on
2^i for ith level
since i=n-1 is height of the tree work done at each level will be
i work
1 2^1
2 2^2
3 2^3..so on
Hence total work done will sum of work done at each level, hence it will be 2^0+2^1+2^2+2^3...+2^(n-1) since i=n-1.
By geometric series this sum is 2^n, Hence total time complexity here is O(2^n)
The proof answers are good, but I always have to do a few iterations by hand to really convince myself. So I drew out a small calling tree on my whiteboard, and started counting the nodes. I split my counts out into total nodes, leaf nodes, and interior nodes. Here's what I got:
IN | OUT | TOT | LEAF | INT
1 | 1 | 1 | 1 | 0
2 | 1 | 1 | 1 | 0
3 | 2 | 3 | 2 | 1
4 | 3 | 5 | 3 | 2
5 | 5 | 9 | 5 | 4
6 | 8 | 15 | 8 | 7
7 | 13 | 25 | 13 | 12
8 | 21 | 41 | 21 | 20
9 | 34 | 67 | 34 | 33
10 | 55 | 109 | 55 | 54
What immediately leaps out is that the number of leaf nodes is fib(n). What took a few more iterations to notice is that the number of interior nodes is fib(n) - 1. Therefore the total number of nodes is 2 * fib(n) - 1.
Since you drop the coefficients when classifying computational complexity, the final answer is θ(fib(n)).
It is bounded on the lower end by 2^(n/2) and on the upper end by 2^n (as noted in other comments). And an interesting fact of that recursive implementation is that it has a tight asymptotic bound of Fib(n) itself. These facts can be summarized:
T(n) = Ω(2^(n/2)) (lower bound)
T(n) = O(2^n) (upper bound)
T(n) = Θ(Fib(n)) (tight bound)
The tight bound can be reduced further using its closed form if you like.
It is simple to calculate by diagramming function calls. Simply add the function calls for each value of n and look at how the number grows.
The Big O is O(Z^n) where Z is the golden ratio or about 1.62.
Both the Leonardo numbers and the Fibonacci numbers approach this ratio as we increase n.
Unlike other Big O questions there is no variability in the input and both the algorithm and implementation of the algorithm are clearly defined.
There is no need for a bunch of complex math. Simply diagram out the function calls below and fit a function to the numbers.
Or if you are familiar with the golden ratio you will recognize it as such.
This answer is more correct than the accepted answer which claims that it will approach f(n) = 2^n. It never will. It will approach f(n) = golden_ratio^n.
2 (2 -> 1, 0)
4 (3 -> 2, 1) (2 -> 1, 0)
8 (4 -> 3, 2) (3 -> 2, 1) (2 -> 1, 0)
(2 -> 1, 0)
14 (5 -> 4, 3) (4 -> 3, 2) (3 -> 2, 1) (2 -> 1, 0)
(2 -> 1, 0)
(3 -> 2, 1) (2 -> 1, 0)
22 (6 -> 5, 4)
(5 -> 4, 3) (4 -> 3, 2) (3 -> 2, 1) (2 -> 1, 0)
(2 -> 1, 0)
(3 -> 2, 1) (2 -> 1, 0)
(4 -> 3, 2) (3 -> 2, 1) (2 -> 1, 0)
(2 -> 1, 0)
The naive recursion version of Fibonacci is exponential by design due to repetition in the computation:
At the root you are computing:
F(n) depends on F(n-1) and F(n-2)
F(n-1) depends on F(n-2) again and F(n-3)
F(n-2) depends on F(n-3) again and F(n-4)
then you are having at each level 2 recursive calls that are wasting a lot of data in the calculation, the time function will look like this:
T(n) = T(n-1) + T(n-2) + C, with C constant
T(n-1) = T(n-2) + T(n-3) > T(n-2) then
T(n) > 2*T(n-2)
...
T(n) > 2^(n/2) * T(1) = O(2^(n/2))
This is just a lower bound that for the purpose of your analysis should be enough but the real time function is a factor of a constant by the same Fibonacci formula and the closed form is known to be exponential of the golden ratio.
In addition, you can find optimized versions of Fibonacci using dynamic programming like this:
static int fib(int n)
{
/* memory */
int f[] = new int[n+1];
int i;
/* Init */
f[0] = 0;
f[1] = 1;
/* Fill */
for (i = 2; i <= n; i++)
{
f[i] = f[i-1] + f[i-2];
}
return f[n];
}
That is optimized and do only n steps but is also exponential.
Cost functions are defined from Input size to the number of steps to solve the problem. When you see the dynamic version of Fibonacci (n steps to compute the table) or the easiest algorithm to know if a number is prime (sqrt(n) to analyze the valid divisors of the number). you may think that these algorithms are O(n) or O(sqrt(n)) but this is simply not true for the following reason:
The input to your algorithm is a number: n, using the binary notation the input size for an integer n is log2(n) then doing a variable change of
m = log2(n) // your real input size
let find out the number of steps as a function of the input size
m = log2(n)
2^m = 2^log2(n) = n
then the cost of your algorithm as a function of the input size is:
T(m) = n steps = 2^m steps
and this is why the cost is an exponential.
Well, according to me to it is O(2^n) as in this function only recursion is taking the considerable time (divide and conquer). We see that, the above function will continue in a tree until the leaves are approaches when we reach to the level F(n-(n-1)) i.e. F(1). So, here when we jot down the time complexity encountered at each depth of tree, the summation series is:
1+2+4+.......(n-1)
= 1((2^n)-1)/(2-1)
=2^n -1
that is order of 2^n [ O(2^n) ].
No answer emphasizes probably the fastest and most memory efficient way to calculate the sequence. There is a closed form exact expression for the Fibonacci sequence. It can be found by using generating functions or by using linear algebra as I will now do.
Let f_1,f_2, ... be the Fibonacci sequence with f_1 = f_2 = 1. Now consider a sequence of two dimensional vectors
f_1 , f_2 , f_3 , ...
f_2 , f_3 , f_4 , ...
Observe that the next element v_{n+1} in the vector sequence is M.v_{n} where M is a 2x2 matrix given by
M = [0 1]
[1 1]
due to f_{n+1} = f_{n+1} and f_{n+2} = f_{n} + f_{n+1}
M is diagonalizable over complex numbers (in fact diagonalizable over the reals as well, but this is not usually the case). There are two distinct eigenvectors of M given by
1 1
x_1 x_2
where x_1 = (1+sqrt(5))/2 and x_2 = (1-sqrt(5))/2 are the distinct solutions to the polynomial equation x*x-x-1 = 0. The corresponding eigenvalues are x_1 and x_2. Think of M as a linear transformation and change your basis to see that it is equivalent to
D = [x_1 0]
[0 x_2]
In order to find f_n find v_n and look at the first coordinate. To find v_n apply M n-1 times to v_1. But applying M n-1 times is easy, just think of it as D. Then using linearity one can find
f_n = 1/sqrt(5)*(x_1^n-x_2^n)
Since the norm of x_2 is smaller than 1, the corresponding term vanishes as n tends to infinity; therefore, obtaining the greatest integer smaller than (x_1^n)/sqrt(5) is enough to find the answer exactly. By making use of the trick of repeatedly squaring, this can be done using only O(log_2(n)) multiplication (and addition) operations. Memory complexity is even more impressive because it can be implemented in a way that you always need to hold at most 1 number in memory whose value is smaller than the answer. However, since this number is not a natural number, memory complexity here changes depending on whether if you use fixed bits to represent each number (hence do calculations with error)(O(1) memory complexity this case) or use a better model like Turing machines, in which case some more analysis is needed.
Let a,b,c be the first digits of a number (e.g. 523 has a=5, b=2, c=3). I am trying to check if abc == sqrt(a^b^c) for many values of a,b,c. (Note: abc = 523 stands for the number itself.)
I have tried this with Python, but for a>7 it already took a significant amount of time to check just one digit combination. I have tried rewriting the equality as multiple logs, like log_c[log_b[log_a[ (abc)^2 ]]] == 1, however, I encountered Math Domain Errors.
Is there a fast / better way to check this equality (preferably in Python)?
Note: Three digits are an example for StackOverflow. The goal is to test much higher powers with seven to ten digits (or more).
Here is the very basic piece of code I have used so far:
for a in range(1,10):
for b in range(1,10):
for c in range(1,10):
N = a*10**2 + b*10 + c
X = a**(b**c)
if N == X:
print a,b,c
The problem is that you are uselessly calculating very large integers, which can take much time as Python has unlimited size for them.
You should limit the values of c you test.
If your largest possible number is 1000, you want a**b**c < 1000**2, so b**c < log(1000**2, a) = 2*log(1000, a)), so c < log(2*log(1000, a), b)
Note that you should exclude a = 1, as any power of it is 1, and b = 1, as b^c would then be 1, and the whole expression is just a.
To test if the square root of a^b^c is abc, it's better to test if a^b^c is equal to the square of abc, in order to avoid using floats.
So, the code, that (as expected) doesn't find any solution under 1000, but runs very fast:
from math import log
for a in range(2,10):
for b in range(2,10):
for c in range(1,int(log(2*log(1000, a), b))):
N2 = (a*100 + b*10 + c)**2
X = a**(b**c)
if N2 == X:
print(a,b,c)
You are looking for numbers whose square root is equal to a three-digit integer. That means your X has to have at most 6 digits, or more precisely log10(X) < 6. Once your a gets larger, the potential solutions you're generating are much larger than that, so we can eliminate large swathes of them without needing to check them (or needing to calculate a ** b ** c, which can get very large: 9 ** 9 ** 9 has 369_693_100 DIGITS!).
log10(X) < 6 gives us log10(a ** b ** c) < 6 which is the same as b ** c * log10(a) < 6. Bringing it to the other side: log10(a) < 6 / b ** c, and then a < 10 ** (6 / b ** c). That means I know I don't need to check for any a that exceeds that. Correcting for an off-by-one error gives the solution:
for b in range(1, 10):
for c in range(1, 10):
t = b ** c
for a in range(1, 1 + min(9, int(10 ** (6 / t)))):
N = a * 100 + b * 10 + c
X = a ** t
if N * N == X:
print(a, b, c)
Running this shows that there aren't any valid solutions to your equation, sadly!
a**(b**c) will grow quite fast and most of the time it will far exceed three digit number. Most of the calculations you are doing will be useless. To optimize your solution do the following:
Iterate over all 3 digit numbers
For each of these numbers square it and is a power of the first digit of the number
For those that are, check if this power is in turn a power of the second digit
And last check if this power is the third digit
How do I go about problems like this one: T(n) = T(n/3) + T(n/4) + O(n^2)
I am able to use two for loops which will give me the O(n^2), right?
To interpret the equation, read it in English as: "the running time for an input of size n equals the running time for an input of size n/3, plus the running time for an input of size n/4, plus a time proportional to n^2".
Writing code that runs in time proportional to n^2 is possible using nested loops, for example, though it's simpler to write a single loop like for i in range(n ** 2): ....
Writing code that runs in time equal to the time it takes for the algorithm with an input of size n/3 or n/4 is even easier - just call the algorithm recursively with an input of that size. (Don't forget a base case for the recursion to terminate on.)
Putting it together, the code could look something like this:
def nonsense_algorithm(n):
if n <= 0:
print('base case')
else:
nonsense_algorithm(n // 3)
nonsense_algorithm(n // 4)
for i in range(n ** 2):
print('constant amount of text')
I've used integer division (rounding down) because I don't think it makes sense otherwise.
Here, we would use for and range(start, stop, steps), maybe with a simple to understand algorithm, similar to:
def algorithm_n2(n: int) -> int:
"""Returns an integer for a O(N2) addition algorithm
:n: must be a positive integer
"""
if n < 1:
return False
output = 0
for i in range(n):
for j in range(n):
output += n * 2
for i in range(0, n, 3):
output += n / 3
for i in range(0, n, 4):
output += n / 4
return int(output)
# Test
if __name__ == "__main__":
print(algorithm_n2(24))
Using print() inside a method is not a best practice.
Output
27748