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.
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
There is a puzzle which I am writing code to solve that goes as follows.
Consider a binary vector of length n that is initially all zeros. You choose a bit of the vector and set it to 1. Now a process starts that sets the bit that is the greatest distance from any 1 bit to $1$ (or an arbitrary choice of furthest bit if there is more than one). This happens repeatedly with the rule that no two 1 bits can be next to each other. It terminates when there is no more space to place a 1 bit. The goal is to place the initial 1 bit so that as many bits as possible are set to 1 on termination.
Say n = 2. Then wherever we set the bit we end up with exactly one bit set.
For n = 3, if we set the first bit we get 101 in the end. But if we set the middle bit we get 010 which is not optimal.
For n = 4, whichever bit we set we end up with two set.
For n = 5, setting the first gives us 10101 with three bits set in the end.
For n = 7, we need to set the third bit to get 1010101 it seems.
I have written code to find the optimal value but it does not scale well to large n. My code starts to get slow around n = 1000 but I would like to solve the problem for n around 1 million.
#!/usr/bin/python
from __future__ import division
from math import *
def findloc(v):
count = 0
maxcount = 0
id = -1
for i in xrange(n):
if (v[i] == 0):
count += 1
if (v[i] == 1):
if (count > maxcount):
maxcount = count
id = i
count = 0
#Deal with vector ending in 0s
if (2*count >= maxcount and count >= v.index(1) and count >1):
return n-1
#Deal with vector starting in 0s
if (2*v.index(1) >= maxcount and v.index(1) > 1):
return 0
if (maxcount <=2):
return -1
return id-int(ceil(maxcount/2))
def addbits(v):
id = findloc(v)
if (id == -1):
return v
v[id] = 1
return addbits(v)
#Set vector length
n=21
max = 0
for i in xrange(n):
v = [0]*n
v[i] = 1
v = addbits(v)
score = sum([1 for j in xrange(n) if v[j] ==1])
# print i, sum([1 for j in xrange(n) if v[j] ==1]), v
if (score > max):
max = score
print max
Latest answer (O(log n) complexity)
If we believe the conjecture by templatetypedef and Aleksi Torhamo (update: proof at the end of this post), there is a closed form solution count(n) calculable in O(log n) (or O(1) if we assume logarithm and bit shifting is O(1)):
Python:
from math import log
def count(n): # The count, using position k conjectured by templatetypedef
k = p(n-1)+1
count_left = k/2
count_right = f(n-k+1)
return count_left + count_right
def f(n): # The f function calculated using Aleksi Torhamo conjecture
return max(p(n-1)/2 + 1, n-p(n-1))
def p(n): # The largest power of 2 not exceeding n
return 1 << int(log(n,2)) if n > 0 else 0
C++:
int log(int n){ // Integer logarithm, by counting the number of leading 0
return 31-__builtin_clz(n);
}
int p(int n){ // The largest power of 2 not exceeding n
if(n==0) return 0;
return 1<<log(n);
}
int f(int n){ // The f function calculated using Aleksi Torhamo conjecture
int val0 = p(n-1);
int val1 = val0/2+1;
int val2 = n-val0;
return val1>val2 ? val1 : val2;
}
int count(int n){ // The count, using position k conjectured by templatetypedef
int k = p(n-1)+1;
int count_left = k/2;
int count_right = f(n-k+1);
return count_left + count_right;
}
This code can calculate the result for n=100,000,000 (and even n=1e24 in Python!) correctly in no time1.
I have tested the codes with various values for n (using my O(n) solution as the standard, see Old Answer section below), and they still seem correct.
This code relies on the two conjectures by templatetypedef and Aleksi Torhamo2. Anyone wants to proof those? =D (Update 2: PROVEN)
1By no time, I meant almost instantly
2The conjecture by Aleksi Torhamo on f function has been empirically proven for n<=100,000,000
Old answer (O(n) complexity)
I can return the count of n=1,000,000 (the result is 475712) in 1.358s (in my iMac) using Python 2.7. Update: It's 0.198s for n=10,000,000 in C++. =)
Here is my idea, which achieves O(n) time complexity.
The Algorithm
Definition of f(n)
Define f(n) as the number of bits that will be set on bitvector of length n, assuming that the first and last bit are set (except for n=2, where only the first or last bit is set). So we know some values of f(n) as follows:
f(1) = 1
f(2) = 1
f(3) = 2
f(4) = 2
f(5) = 3
Note that this is different from the value that we are looking for, since the initial bit might not be at the first or last, as calculated by f(n). For example, we have f(7)=3 instead of 4.
Note that this can be calculated rather efficiently (amortized O(n) to calculate all values of f up to n) using the recurrence relation:
f(2n) = f(n)+f(n+1)-1
f(2n+1) = 2*f(n+1)-1
for n>=5, since the next bit set following the rule will be the middle bit, except for n=1,2,3,4. Then we can split the bitvector into two parts, each independent of each other, and so we can calculate the number of bits set using f( floor(n/2) ) + f( ceil(n/2) ) - 1, as illustrated below:
n=11 n=13
10000100001 1000001000001
<----> <----->
f(6)<----> f(7) <----->
f(6) f(7)
n=12 n=14
100001000001 10000010000001
<----> <----->
f(6)<-----> f(7) <------>
f(7) f(8)
we have the -1 in the formula to exclude the double count of the middle bit.
Now we are ready to count the solution of original problem.
Definition of g(n,i)
Define g(n,i) as the number of bits that will be set on bitvector of length n, following the rules in the problem, where the initial bit is at the i-th bit (1-based). Note that by symmetry the initial bit can be anywhere from the first bit up to the ceil(n/2)-th bit. And for those cases, note that the first bit will be set before any bit in between the first and the initial, and so is the case for the last bit. Therefore the number of bit set in the first partition and the second partition is f(i) and f(n+1-i) respectively.
So the value of g(n,i) can be calculated as:
g(n,i) = f(i) + f(n+1-i) - 1
following the idea when calculating f(n).
Now, to calculate the final result is trivial.
Definition of g(n)
Define g(n) as the count being looked for in the original problem. We can then take the maximum of all possible i, the position of initial bit:
g(n) = maxi=1..ceil(n/2)(f(i) + f(n+1-i) - 1)
Python code:
import time
mem_f = [0,1,1,2,2]
mem_f.extend([-1]*(10**7)) # This will take around 40MB of memory
def f(n):
global mem_f
if mem_f[n]>-1:
return mem_f[n]
if n%2==1:
mem_f[n] = 2*f((n+1)/2)-1
return mem_f[n]
else:
half = n/2
mem_f[n] = f(half)+f(half+1)-1
return mem_f[n]
def g(n):
return max(f(i)+f(n+1-i)-1 for i in range(1,(n+1)/2 + 1))
def main():
while True:
n = input('Enter n (1 <= n <= 10,000,000; 0 to stop): ')
if n==0: break
start_time = time.time()
print 'g(%d) = %d, in %.3fs' % (n, g(n), time.time()-start_time)
if __name__=='__main__':
main()
Complexity Analysis
Now, the interesting thing is, what is the complexity of calculating g(n) with the method described above?
We should first note that we iterate over n/2 values of i, the position of initial bit. And in each iteration we call f(i) and f(n+1-i). Naive analysis will lead to O(n * O(f(n))), but actually we used memoization on f, so it's much faster than that, since each value of f(i) is calculated only once, at most. So the complexity is actually added by the time required to calculate all values of f(n), which would be O(n + f(n)) instead.
So what's the complexity of initializing f(n)?
We can assume that we precompute every value of f(n) first before calculating g(n). Note that due to the recurrence relation and the memoization, generating the whole values of f(n) takes O(n) time. And the next call to f(n) will take O(1) time.
So, the overall complexity is O(n+n) = O(n), as evidenced by this running time in my iMac for n=1,000,000 and n=10,000,000:
> python max_vec_bit.py
Enter n (1 <= n <= 10,000,000; 0 to stop): 1000000
g(1000000) = 475712, in 1.358s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
>
> <restarted the program to remove the effect of memoization>
>
> python max_vec_bit.py
Enter n (1 <= n <= 10,000,000; 0 to stop): 10000000
g(10000000) = 4757120, in 13.484s
Enter n (1 <= n <= 10,000,000; 0 to stop): 6745231
g(6745231) = 3145729, in 3.072s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
And as a by-product of memoization, the calculation of lesser value of n will be much faster after the first call to large n, as you can also see in the sample run. And with language better suited for number crunching such as C++, you might get significantly faster running time
I hope this helps. =)
The code using C++, for performance improvement
The result in C++ is about 68x faster (measured by clock()):
> ./a.out
Enter n (1 <= n <= 10,000,000; 0 to stop): 1000000
g(1000000) = 475712, in 0.020s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
>
> <restarted the program to remove the effect of memoization>
>
> ./a.out
Enter n (1 <= n <= 10,000,000; 0 to stop): 10000000
g(10000000) = 4757120, in 0.198s
Enter n (1 <= n <= 10,000,000; 0 to stop): 6745231
g(6745231) = 3145729, in 0.047s
Enter n (1 <= n <= 10,000,000; 0 to stop): 0
Code in C++:
#include <cstdio>
#include <cstring>
#include <ctime>
int mem_f[10000001];
int f(int n){
if(mem_f[n]>-1)
return mem_f[n];
if(n%2==1){
mem_f[n] = 2*f((n+1)/2)-1;
return mem_f[n];
} else {
int half = n/2;
mem_f[n] = f(half)+f(half+1)-1;
return mem_f[n];
}
}
int g(int n){
int result = 0;
for(int i=1; i<=(n+1)/2; i++){
int cnt = f(i)+f(n+1-i)-1;
result = (cnt > result ? cnt : result);
}
return result;
}
int main(){
memset(mem_f,-1,sizeof(mem_f));
mem_f[0] = 0;
mem_f[1] = mem_f[2] = 1;
mem_f[3] = mem_f[4] = 2;
clock_t start, end;
while(true){
int n;
printf("Enter n (1 <= n <= 10,000,000; 0 to stop): ");
scanf("%d",&n);
if(n==0) break;
start = clock();
int result = g(n);
end = clock();
printf("g(%d) = %d, in %.3fs\n",n,result,((double)(end-start))/CLOCKS_PER_SEC);
}
}
Proof
note that for the sake of keeping this answer (which is already very long) simple, I've skipped some steps in the proof
Conjecture of Aleksi Torhamo on the value of f
For `n>=1`, prove that:
f(2n+k) = 2n-1+1 for k=1,2,…,2n-1 ...(1)
f(2n+k) = k for k=2n-1+1,…,2n ...(2)
given f(0)=f(1)=f(2)=1
The result above can be easily proven using induction on the recurrence relation, by considering the four cases:
Case 1: (1) for even k
Case 2: (1) for odd k
Case 3: (2) for even k
Case 4: (2) for odd k
Suppose we have the four cases proven for n. Now consider n+1.
Case 1:
f(2n+1+2i) = f(2n+i) + f(2n+i+1) - 1, for i=1,…,2n-1
= 2n-1+1 + 2n-1+1 - 1
= 2n+1
Case 2:
f(2n+1+2i+1) = 2*f(2n+i+1) - 1, for i=0,…,2n-1-1
= 2*(2n-1+1) - 1
= 2n+1
Case 3:
f(2n+1+2i) = f(2n+i) + f(2n+i+1) - 1, for i=2n-1+1,…,2n
= i + (i+1) - 1
= 2i
Case 4:
f(2n+1+2i+1) = 2*f(2n+i+1) - 1, for i=2n-1+1,…,2n-1
= 2*(i+1) - 1
= 2i+1
So by induction the conjecture is proven.
Conjecture of templatetypedef on the best position
For n>=1 and k=1,…,2n, prove that g(2n+k) = g(2n+k, 2n+1)
That is, prove that placing the first bit on the 2n+1-th position gives maximum number of bits set.
The proof:
First, we have
g(2n+k,2n+1) = f(2n+1) + f(k-1) - 1
Next, by the formula of f, we have the following equalities:
f(2n+1-i) = f(2n+1), for i=-2n-1,…,-1
f(2n+1-i) = f(2n+1)-i, for i=1,…,2n-2-1
f(2n+1-i) = f(2n+1)-2n-2, for i=2n-2,…,2n-1
and also the following inequality:
f(k-1+i) <= f(k-1), for i=-2n-1,…,-1
f(k-1+i) <= f(k-1)+i , for i=1,…,2n-2-1
f(k-1+i) <= f(k-1)+2n-2, for i=2n-2,…,2n-1
and so we have:
f(2n+1-i)+f(k-1+i) <= f(2n+1)+f(k-1), for i=-2n-1,…,2n-1
Now, note that we have:
g(2n+k) = maxi=1..ceil(2n-1+1-k/2)(f(i) + f(2n+k+1-i) - 1)
<= f(2n+1) + f(k-1) - 1
= g(2n+k,2n+1)
And so the conjecture is proven.
So in a break with my normal tradition of not posting algorithms I don't have a proof for, I think I should mention that there's an algorithm that appears to be correct for numbers up to 50,000+ and runs in O(log n) time. This is due to Sophia Westwood, who I worked on this problem with for about three hours today. All credit for this is due to her. Empirically it seems to work beautifully, and it's much, much faster than the O(n) solutions.
One observation about the structure of this problem is that if n is sufficiently large (n ≥ 5), then if you put a 1 anywhere, the problem splits into two subproblems, one to the left of the 1 and one to the right. Although the 1s might be placed in the different halves at different times, the eventual placement is the same as if you solved each half separately and combined them back together.
The next observation is this: suppose you have an array of size 2k + 1 for some k. In that case, suppose that you put a 1 on either side of the array. Then:
The next 1 is placed on the other side of the array.
The next 1 is placed in the middle.
You now have two smaller subproblems of size 2k-1 + 1.
The important part about this is that the resulting bit pattern is an alternating series of 1s and 0s. For example:
For 5 = 4 + 1, we get 10101
For 9 = 8 + 1, we get 101010101
For 17 = 16 + 1, we get 10101010101010101
The reason this matters is the following: suppose you have n total elements in the array and let k be the largest possible value for which 2k + 1 ≤ n. If you place the 1 at position 2k + 1, then the left part of the array up to that position will end up getting tiled with alternating 1s and 0s, which puts a lot of 1s into the array.
What's not obvious is that placing the 1 bit there, for all numbers up to 50,000, appears to yield an optimal solution! I've written a Python script that checks this (using a recurrence relation similar to the one #justhalf) and it seems to work well. The reason that this fact is so useful is that it's really easy to compute this index. In particular, if 2k + 1 ≤ n, then 2k ≤ n - 1, so k ≤ lg (n - 1). Choosing the value ⌊lg (n - 1) ⌋ as your choice of k then lets you compute the bit index by computing 2k + 1. This value of k can be computed in O(log n) time and the exponentiation can be done in O(log n) time as well, so the total runtime is Θ(log n).
The only issue is that I haven't formally proven that this works. All I know is that it's right for the first 50,000 values we've tried. :-)
Hope this helps!
I'll attach what I have. Same as yours, alas, time is basically O(n**3). But at least it avoids recursion (etc), so won't blow up when you get near a million ;-) Note that this returns the best vector found, not the count; e.g.,
>>> solve(23)
[6, 0, 11, 0, 1, 0, 0, 10, 0, 5, 0, 9, 0, 3, 0, 0, 8, 0, 4, 0, 7, 0, 2]
So it also shows the order in which the 1 bits were chosen. The easiest way to get the count is to pass the result to max().
>>> max(solve(23))
11
Or change the function to return maxsofar instead of best.
If you want to run numbers on the order of a million, you'll need something radically different. You can't even afford quadratic time for that (let alone this approach's cubic time). Unlikely to get such a huge O() improvement from fancier data structures - I expect it would require deeper insight into the mathematics of the problem.
def solve(n):
maxsofar, best = 1, [1] + [0] * (n-1)
# by symmetry, no use trying starting points in last half
# (would be a mirror image).
for i in xrange((n + 1)//2):
v = [0] * n
v[i] = count = 1
# d21[i] = distance to closest 1 from index i
d21 = range(i, 0, -1) + range(n-i)
while 1:
d, j = max((d, j) for j, d in enumerate(d21))
if d >= 2:
count += 1
v[j] = count
d21[j] = 0
k = 1
while j-k >= 0 and d21[j-k] > k:
d21[j-k] = k
k += 1
k = 1
while j+k < n and d21[j+k] > k:
d21[j+k] = k
k += 1
else:
if count > maxsofar:
maxsofar = count
best = v[:]
break
return best