Is there anyway to calculate the sum of 1 to n in Theta(log n)?
Of course, the obvious way to do it is sum = n*(n+1)/2.
However, for practicing, I want to calculate in Theta(log n).
For example,
sum=0; for(int i=1; i<=n; i++) { sum += i}
this code will calculate in Theta(n).
Fair way (without using math formulas) assumes direct summing all n values, so there is no way to avoid O(n) behavior.
If you want to make some artificial approach to provide exactly O(log(N)) time, consider, for example, using powers of two (knowing that Sum(1..2^k = 2^(k-1) + 2^(2*k-1) - for example, Sum(8) = 4 + 32). Pseudocode:
function Sum(n)
if n < 2
return n
p = 1 //2^(k-1)
p2 = 2 //2^(2*k-1)
while p * 4 < n:
p = p * 2;
p2 = p2 * 4;
return p + p2 + ///sum of 1..2^k
2 * p * (n - 2 * p) + ///(n - 2 * p) summands over 2^k include 2^k
Sum(n - 2 * p) ///sum of the rest over 2^k
Here 2*p = 2^k is the largest power of two not exceeding N. Example:
Sum(7) = Sum(4) + 5 + 6 + 7 =
Sum(4) + (4 + 1) + (4 + 2) + (4 + 3) =
Sum(4) + 3 * 4 + Sum(3) =
Sum(4) + 3 * 4 + Sum(2) + 1 * 2 + Sum(1) =
Sum(4) + 3 * 4 + Sum(2) + 1 * 2 + Sum(1) =
2 + 8 + 12 + 1 + 2 + 2 + 1 = 28
Related
I would like to know how I can calculate the number of compares for an optimized merge sort described below, on an array already sorted.
It is like an ordinary merge-sort but does not 'merge' if two subarrays are already sorted.
mergesort(arr, auxArr, lo, hi)
if (hi <= lo) return;
mid = lo + (hi + lo) / 2
sort(arr, auxArr, lo, mid)
sort(arr, auxArr, mid+1, hi)
if (less(arr[mid], arr[mid+1]) return; // optimization
merge(arr, auxArr, lo, mid ,hi)
Here is my attempt to find out the number of compares
Let C(N) be the number of compares
C(N) = 2 * C(N/2) + 1 (number of compares to sort two subarrays + 1 compare for our optimization)
Using recurrence relation,
C(N) = = 2 * C(N/2) + 1
= 2 ( 2C(N/4) + 1 ) + 1
= 4 C(N/4) + 1 + y
...
= N C(N/N) + 1 + 2 + 4 + ... + 2^(log2 (N/2))
= 1 + 2 + 4 + ... + 2^(log2 (N/2))
But the answer says it is ~ log2(N), and I am stuck. Any ideas?
f(N) = 0^0 + 1^1 + 2^2 + 3^3 + 4^4 + ... + N^N.
I want to calculate (f(N) mod M).
These are the constraints.
1 ≤ N ≤ 10^9
1 ≤ M ≤ 10^3
Here is my code
test=int(input())
ans = 0
for cases in range(test):
arr=[int(x) for x in input().split()]
N=arr[0]
mod=arr[1]
#ret=sum([int(y**y) for y in range(N+1)])
#ans=ret
for i in range(1,N+1):
ans = (ans + pow(i,i,mod))%mod
print (ans)
I tried another approach but in vain.
Here is code for that
from functools import reduce
test=int(input())
answer=0
for cases in range(test):
arr=[int(x) for x in input().split()]
N=arr[0]
mod=arr[1]
answer = reduce(lambda K,N: x+pow(N,N), range(1,N+1)) % M
print(answer)
Two suggestions:
Let 0^0 = 1 be what you use. This seems like the best guidance I have for how to handle that.
Compute k^k by multiplying and taking the modulus as you go.
Do an initial pass where all k (not exponents) are changed to k mod M before doing anything else.
While computing (k mod M)^k, if an intermediate result is one you've already visited, you can cut back on the number of iterations to continue by all but up to one additional cycle.
Example: let N = 5 and M = 3. We want to calculate 0^0 + 1^1 + 2^2 + 3^3 + 4^4 + 5^5 (mod 3).
First, we apply suggestion 3. Now we want to calculate 0^0 + 1^1 + 2^2 + 0^3 + 1^4 + 2^5 (mod 3).
Next, we begin evaluating and use suggestion 1 immediately to get 1 + 1 + 2^2 + 0^3 + 1^4 + 2^5 (mod 3). 2^2 is 4 = 1 (mod 3) of which we make a note (2^2 = 1 (mod 3)). Next, we find 0^1 = 0, 0^2 = 0 so we have a cycle of size 1 meaning no further multiplication is needed to tell 0^3 = 0 (mod 3). Note taken. Similar process for 1^4 (we tell on the second iteration that we have a cycle of size 1, so 1^4 is 1, which we note). Finally, we get 2^1 = 2 (mod 3), 2^2 = 1(mod 3), 2^3 = 2(mod 3), a cycle of length 2, so we can skip ahead an even number which exhausts 2^5 and without checking again we know that 2^5 = 2 (mod 3).
Our sum is now 1 + 1 + 1 + 0 + 1 + 2 (mod 3) = 2 + 1 + 0 + 1 + 2 (mod 3) = 0 + 0 + 1 + 2 (mod 3) = 0 + 1 + 2 (mod 3) = 1 + 2 (mod 3) = 0 (mod 3).
These rules will be helpful to you since your cases see N much larger than M. If this were reversed - if N were much smaller than M - you'd get no benefit from my method (and taking the modulus w.r.t. M would affect the outcome less).
Pseudocode:
Compute(N, M)
1. sum = 0
2. for i = 0 to N do
3. term = SelfPower(i, M)
4. sum = (sum + term) % M
5. return sum
SelfPower(k, M)
1. selfPower = 1
2. iterations = new HashTable
3. for i = 1 to k do
4. selfPower = (selfPower * (k % M)) % M
5. if iterations[selfPower] is defined
6. i = k - (k - i) % (i - iterations[selfPower])
7. clear out iterations
8. else iterations[selfPower] = i
9. return selfPower
Example execution:
resul = Compute(5, 3)
sum = 0
i = 0
term = SelfPower(0, 3)
selfPower = 1
iterations = []
// does not enter loop
return 1
sum = (0 + 1) % 3 = 1
i = 1
term = SelfPower(1, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 1 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 1
return 1
sum = (1 + 1) % 3 = 2
i = 2
term = SelfPower(2, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 2 % 3) % 3 = 2
iterations[2] is not defined
iterations[2] = 1
i = 2
selfPower = (2 * 2 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 2
return 1
sum = (2 + 1) % 3 = 0
i = 3
term = SelfPower(3, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 3 % 0) % 3 = 0
iterations[0] is not defined
iterations[0] = 1
i = 2
selfPower = (0 * 3 % 0) % 3 = 0
iterations[0] is defined as 1
i = 3 - (3 - 2) % (2 - 1) = 3
iterations is blank
return 0
sum = (0 + 0) % 3 = 0
i = 4
term = SelfPower(4, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 4 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 1
i = 2
selfPower = (1 * 4 % 3) % 3 = 1
iterations[1] is defined as 1
i = 4 - (4 - 2) % (2 - 1) = 4
iterations is blank
return 1
sum = (0 + 1) % 3 = 1
i = 5
term = SelfPower(5, 3)
selfPower = 1
iterations = []
i = 1
selfPower = (1 * 5 % 3) % 3 = 2
iterations[2] is not defined
iterations[2] = 1
i = 2
selfPower = (2 * 5 % 3) % 3 = 1
iterations[1] is not defined
iterations[1] = 2
i = 3
selfPower = (1 * 5 % 3) % 3 = 2
iterations[2] is defined as 1
i = 5 - (5 - 3) % (3 - 1) = 5
iterations is blank
return 2
sum = (1 + 2) % 3 = 0
return 0
why not just use simple recursion to find the recursive sum of the powers
def find_powersum(s):
if s == 1 or s== 0:
return 1
else:
return s*s + find_powersum(s-1)
def find_mod (s, m):
print(find_powersum(s) % m)
find_mod(4, 4)
2
Given a number N, print in how many ways it can be represented as
N = a + b + c + d
with
1 <= a <= b <= c <= d; 1 <= N <= M
My observation:
For N = 4: Only 1 way - 1 + 1 + 1 + 1
For N = 5: Only 1 way - 1 + 1 + 1 + 2
For N = 6: 2 ways - 1 + 1 + 1 + 3
1 + 1 + 2 + 2
For N = 7: 3 ways - 1 + 1 + 1 + 4
1 + 1 + 2 + 3
1 + 2 + 2 + 2
For N = 8: 5 ways - 1 + 1 + 1 + 5
1 + 1 + 2 + 4
1 + 1 + 3 + 3
1 + 2 + 2 + 3
2 + 2 + 2 + 2
So I have reduced it to a DP solution as follows:
DP[4] = 1, DP[5] = 1;
for(int i = 6; i <= M; i++)
DP[i] = DP[i-1] + DP[i-2];
Is my observation correct or am I missing any thing. I don't have any test cases to run on. So please let me know if the approach is correct or wrong.
It's not correct. Here is the correct one:
Lets DP[n,k] be the number of ways to represent n as sum of k numbers.
Then you are looking for DP[n,4].
DP[n,1] = 1
DP[n,2] = DP[n-2, 2] + DP[n-1,1] = n / 2
DP[n,3] = DP[n-3, 3] + DP[n-1,2]
DP[n,4] = DP[n-4, 4] + DP[n-1,3]
I will only explain the last line and you can see right away, why others are true.
Let's take one case of n=a+b+c+d.
If a > 1, then n-4 = (a-1)+(b-1)+(c-1)+(d-1) is a valid sum for DP[n-4,4].
If a = 1, then n-1 = b+c+d is a valid sum for DP[n-1,3].
Also in reverse:
For each valid n-4 = x+y+z+t we have a valid n=(x+1)+(y+1)+(z+1)+(t+1).
For each valid n-1 = x+y+z we have a valid n=1+x+y+z.
Unfortunately, your recurrence is wrong, because for n = 9, the solution is 6, not 8.
If p(n,k) is the number of ways to partition n into k non-zero integer parts, then we have
p(0,0) = 1
p(n,k) = 0 if k > n or (n > 0 and k = 0)
p(n,k) = p(n-k, k) + p(n-1, k-1)
Because there is either a partition of value 1 (in which case taking this part away yields a partition of n-1 into k-1 parts) or you can subtract 1 from each partition, yielding a partition of n - k. It's easy to show that this process is a bijection, hence the recurrence.
UPDATE:
For the specific case k = 4, OEIS tells us that there is another linear recurrence that depends only on n:
a(n) = 1 + a(n-2) + a(n-3) + a(n-4) - a(n-5) - a(n-6) - a(n-7) + a(n-9)
This recurrence can be solved via standard methods to get an explicit formula. I wrote a small SAGE script to solve it and got the following formula:
a(n) = 1/144*n^3 + 1/32*(-1)^n*n + 1/48*n^2 - 1/54*(1/2*I*sqrt(3) - 1/2)^n*(I*sqrt(3) + 3) - 1/54*(-1/2*I*sqrt(3) - 1/2)^n*(-I*sqrt(3) + 3) + 1/16*I^n + 1/16*(-I)^n + 1/32*(-1)^n - 1/32*n - 13/288
OEIS also gives the following simplification:
a(n) = round((n^3 + 3*n^2 -9*n*(n % 2))/144)
Which I have not verified.
#include <iostream>
using namespace std;
int func_count( int n, int m )
{
if(n==m)
return 1;
if(n<m)
return 0;
if ( m == 1 )
return 1;
if ( m==2 )
return (func_count(n-2,2) + func_count(n - 1, 1));
if ( m==3 )
return (func_count(n-3,3) + func_count(n - 1, 2));
return (func_count(n-1, 3) + func_count(n - 4, 4));
}
int main()
{
int t;
cin>>t;
cout<<func_count(t,4);
return 0;
}
I think that the definition of a function f(N,m,n) where N is the sum we want to produce, m is the maximum value for each term in the sum and n is the number of terms in the sum should work.
f(N,m,n) is defined for n=1 to be 0 if N > m, or N otherwise.
for n > 1, f(N,m,n) = the sum, for all t from 1 to N of f(S-t, t, n-1)
This represents setting each term, right to left.
You can then solve the problem using this relationship, probably using memoization.
For maximum n=4, and N=5000, (and implementing cleverly to quickly work out when there are 0 possibilities), I think that this is probably computable quickly enough for most purposes.
for (int i=0; i < n; i++)
for (j=0;j<i*i;j++)
x++
Would the big O be O(n^3)? I'm just confused about how the i's relate to the n.
The required math (summation) is:
0 + 1 + 4 + 9 + ... + n * n = n**3 / 3 + n**2 / 2 + n / 6 = O(n**3)
So, you're right: it's O(n**3); moreover
0 + 1 + ... + n**k = O(n**(k + 1))
Methodically, to proceed using Sigma notation like below, would get you where you need to go:
Determine the positive number c & n0 for the following recurrences (Using Substitution Method):
T(n) = T(ceiling(n/2)) + 1 ... Guess is Big-Oh(log base 2 of n)
T(n) = 3T(floor(n/3)) + n ... Guess is Big-Omega (n * log base 3 of n)
T(n) = 2T(floor(n/2) + 17) + n ... Guess is Big-Oh(n * log base 2 of n).
I am giving my Solution for Problem 1:
Our Guess is: T(n) = O (log_2(n)).
By Induction Hypothesis assume T(k) <= c * log_2(k) for all k < n,here c is a const & c > 0
T(n) = T(ceiling(n/2)) + 1
<=> T(n) <= c*log_2(ceiling(n/2)) + 1
<=> " <= c*{log_2(n/2) + 1} + 1
<=> " = c*log_2(n/2) + c + 1
<=> " = c*{log_2(n) - log_2(2)} + c + 1
<=> " = c*log_2(n) - c + c + 1
<=> " = c*log_2(n) + 1
<=> T(n) not_<= c*log_2(n) because c*log_2(n) + 1 not_<= c*log_2(n).
To solve this remedy used a trick a follows:
T(n) = T(ceiling(n/2)) + 1
<=> " <= c*log(ceiling(n/2)) + 1
<=> " <= c*{log_2 (n/2) + b} + 1 where 0 <= b < 1
<=> " <= c*{log_2 (n) - log_2(2) + b) + 1
<=> " = c*{log_2(n) - 1 + b} + 1
<=> " = c*log_2(n) - c + bc + 1
<=> " = c*log_2(n) - (c - bc - 1) if c - bc -1 >= 0
c >= 1 / (1 - b)
<=> T(n) <= c*log_2(n) for c >= {1 / (1 - b)}
so T(n) = O(log_2(n)).
This solution is seems to be correct to me ... My Ques is: Is it the proper approach to do?
Thanks to all of U.
For the first exercise:
We want to show by induction that T(n) <= ceiling(log(n)) + 1.
Let's assume that T(1) = 1, than T(1) = 1 <= ceiling(log(1)) + 1 = 1 and the base of the induction is proved.
Now, we assume that for every 1 <= i < nhold that T(i) <= ceiling(log(i)) + 1.
For the inductive step we have to distinguish the cases when n is even and when is odd.
If n is even: T(n) = T(ceiling(n/2)) + 1 = T(n/2) + 1 <= ceiling(log(n/2)) + 1 + 1 = ceiling(log(n) - 1) + 1 + 1 = ceiling(log(n)) + 1.
If n is odd: T(n) = T(ceiling(n/2)) + 1 = T((n+1)/2) + 1 <= ceiling(log((n+1)/2)) + 1 + 1 = ceiling(log(n+1) - 1) + 1 + 1 = ceiling(log(n+1)) + 1 = ceiling(log(n)) + 1
The last passage is tricky, but is possibile because n is odd and then it cannot be a power of 2.
Problem #1:
T(1) = t0
T(2) = T(1) + 1 = t0 + 1
T(4) = T(2) + 1 = t0 + 2
T(8) = T(4) + 1 = t0 + 3
...
T(2^(m+1)) = T(2^m) + 1 = t0 + (m + 1)
Letting n = 2^(m+1), we get that T(n) = t0 + log_2(n) = O(log_2(n))
Problem #2:
T(1) = t0
T(3) = 3T(1) + 3 = 3t0 + 3
T(9) = 3T(3) + 9 = 3(3t0 + 3) + 9 = 9t0 + 18
T(27) = 3T(9) + 27 = 3(9t0 + 18) + 27 = 27t0 + 81
...
T(3^(m+1)) = 3T(3^m) + 3^(m+1) = ((3^(m+1))t0 + (3^(m+1))(m+1)
Letting n = 3^(m+1), we get that T(n) = nt0 + nlog_3(n) = O(nlog_3(n)).
Problem #3:
Consider n = 34. T(34) = 2T(17+17) + 34 = 2T(34) + 34. We can solve this to find that T(34) = -34. We can also see that for odd n, T(n) = 1 + T(n - 1). We continue to find what values are fixed:
T(0) = 2T(17) + 0 = 2T(17)
T(17) = 1 + T(16)
T(16) = 2T(25) + 16
T(25) = T(24) + 1
T(24) = 2T(29) + 24
T(29) = T(28) + 1
T(28) = 2T(31) + 28
T(31) = T(30) + 1
T(30) = 2T(32) + 30
T(32) = 2T(33) + 32
T(33) = T(32) + 1
We get T(32) = 2T(33) + 32 = 2T(32) + 34, meaning that T(32) = -34. Working backword, we get
T(32) = -34
T(33) = -33
T(30) = -38
T(31) = -37
T(28) = -46
T(29) = -45
T(24) = -96
T(25) = -95
T(16) = -174
T(17) = -173
T(0) = -346
As you can see, this recurrence is a little more complicated than the others, and as such, you should probably take a hard look at this one. If I get any other ideas, I'll come back; otherwise, you're on your own.
EDIT:
After looking at #3 some more, it looks like you're right in your assessment that it's O(nlog_2(n)). So you can try listing a bunch of numbers - I did it from n=0 to n=45. You notice a pattern: it goes from negative numbers to positive numbers around n=43,44. To get the next even-index element of the sequence, you add powers of two, in the following order: 4, 8, 4, 16, 4, 8, 4, 32, 4, 8, 4, 16, 4, 8, 4, 64, 4, 8, 4, 16, 4, 8, 4, 32, ...
These numbers are essentially where you'd mark an arbitary-length ruler... quarters, halves, eights, sixteenths, etc. As such, we can solve the equivalent problem of finding the order of the sum 1 + 2 + 1 + 4 + 1 + 2 + 1 + 8 + ... (same as ours, divided by 4, and ours is shifted, but the order will still work). By observing that the sum of the first k numbers (where k is a power of 2) is equal to sum((n/(2^(k+1))2^k) = (1/2)sum(n) for k = 0 to log_2(n), we get that the simple recurrence is given by (n/2)log_2(n). Multiply by 4 to get ours, and shift x to the right by 34 and perhaps add a constant value to the result. So we're playing around with y = 2nlog_n(x) + k' for some constant k'.
Phew. That was a tricky one. Note that this recurrence does not admit of any arbitary "initial condiditons"; in other words, the recurrence does not describe a family of sequences, but one specific one, with no parameterization.