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I am trying to solve T(n) = 2T(n/2) + log n
substituted n = 2^k
T(2^k) = 2T(2^(k-1)) + k
T(2^k) = 2^2 T(2^(k-1)) + 2(k-1) + k
after k steps
T(2^k) = 2^k T(1) + 2^(k-1) + 2 * (2^(k-2)) +....+k
So basically I need to sum a term of i*2^i where i = 1 to log n - 1.
And I could not find an easy way to sum these terms. Am I doing something wrong ? Is there any other way to solve this recursion ? would master theorem work her ? if yes than how ?
Thanks.
first you should define a recursive export,say T(1)
then:
since T(2^k) = 2T(2^(k-1)) + k; *
we define g(k) = T(2^k)/2^k;
then * come into:
g(k) = g(k-1) + k/2^k = g(1) + sum(i/2^i); i=2,3,4...k
where g(1) = T(1)/2 = c;
where you could then unfold the sum expression and define it = y;
then unfold the expression of y/2;
y-y/2 is a geometric progression, so youcan solve it
as I worked out, sum = 3/2 - (k+2)/2^k;
so T(n) = 2^k * g(k) = (3/2+c)*n - (2+logn)
Wolfram|Alpha gives a closed form solution:
for a constant c_1 that is fixed by the initial condition.
By the way, log(n)/log(2) = lg(n), where lg is the base two logarithm.
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Given N. Find number of all the integer pairs x, y such that
1<=x, y<=N
and x^2 - y is a perfect square
the N is large but O(sqrt(N)) will be fine to solve this problem.
I tried to solve this problem like, letting z^2 be the square number
x^2 - z^2 = y = (x+z)(x-z)
then let x + z = p and x - z = q;
then x = (p+q)/2 and z = (p-q)/2;
and (p+q)/2<=N;
and p and q should have same parity (both even or odd as (p+q)/2 is integer)
also pq<=N
Now I don't know how to proceed from here
or tell me some other method to solve this problem efficiently.
This solution solves the problem in O(sqrt N).
Rephrasing the problem
Let z^2 = x^2 - y, z ≥ 0, or equivalently 0 < y = x^2 - z^2 ≤ N
We need pairs of perfect squares under N^2 whose differences are less than or equal to N. By arithmetic series,
1 + 3 + 5 + 7 + ... + (2k - 1) = k^2
That means x^2 - z^2 is a sum of some n consecutive odd integers.
Counting odd integers
z^2 + (2z + 1) + (2z + 3) + ... + (2x - 1) = x^2. Apply arithmetic series formula
z^2 + n/2 * (4z + 2 + 2(n - 1)) = x^2
z^2 + n * (2z + n) = x^2
n(2z + n) ≤ N
z ≤ floor((N/n - n)/2)
We are thus able to find the last values of z for which at least n+1 odd consecutive integers are needed for their sum to exceed N.
For each z, the x can be z+1, z+2 ... z+n, for a total of n pairs.
#include <cmath>
#include <iostream>
int N = 99;
int main(void){
int z = -1;
// z = 0 is valid for x^2 < N, so -1 is largest invalid z.
int count = 0;
for (int n = std::sqrt(N); n > 0; n--){
int zNew = (N/n - n)/2;
// zNew is max z that has n perfect squares from z + 1 to z + n
count += (zNew - z) * n;
z = zNew;
}
std::cout << count << '\n';
}
A version in Java passed these unit tests.
(N, count) = (1, 1), (3, 2), (5, 4), (8, 6), (60, 68), (99, 124), (500, 808)
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Let X be a binomial random variable with mean Np and variance Np(1−p).
Show that the ratio X/N also has a binomial distribution with mean p and
variance p(1 − p)/N.
Let r = X/N. Since X has a binomial distribution, r also has the
same distribution. The mean and variance for r can be computed as follows:
Mean , E[r] = E[X/N] = E[X]/N = (N p)/N = p .
Variance , E[(r − E[r])2] = E[(X/N − E[X/N])2] = E[(X − E[X])2]/N2 = N p(1 − p)/N2 = p(1 − p)/N
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I know that it may seem a duplicate question, but I could not find my answer in previous questions.
I mean how to write a log base 10 function by simple loops and not using built in log function in c++.
The easiest way is to calculate the natural logarithm (ln) with a Taylor series. Once you have found the natural logarithm, just divide it by ln(10) and you get the base-10 log.
The Taylor series is quite simple to implement in C. If z is the number for which you are seeking the log, you just have to loop a few iterations multiplying an accumulator by (z-1) each time. To a limit, the more iterations you run, the more accurate your result will be. Check it a few times against the libC log10() version until you are happy with the precision.
This is a "numeric approach". There are other numeric solutions to finding the logarithm of a number which can give more accurate results. Some of them can be found in that Wikipedia link I gave you.
Assuming by "log base 10" you mean "the number of times n can be divided by 10 before resulting in a value < 10":
log = 0;
// Assume n has initial value N
while ( n >= 10 ) {
// Invariant: N = n * 10^log
n /= 10;
log += 1;
}
You'll get faster convergence with Newton's Method. Use something like this (hand written not compiled or tested uses f(r) = 2**r - x to compute log2(x) ):
double next(double r, double x) {
static double one_over_ln2 = 1.4426950408889634;
return r - one_over_ln2 * (1 - x / (1 << static_cast<int>(r)));
double log2(double x) {
static double epsilon = 0.000000001; // change this to change accuracy
double r = x / 2;. // better first guesses converge faster
double r2 = next(r, x);
double delta = r - r2;
while (delta * delta > epsilon) {
r = r2;
r2 = next(r, x);
delta = r - r2
}
return r2;
}
double log10(double x) {
static double log2_10 = log2(10);
return log2(x) / log2_10;
}
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Is it possible to simplify (a+b)xor(c+b)? What is the contribution of b to the final result?
Note that I'm mixing boolean algebra with arithmetic, xor is a bitwise exclusive or on corresponding bits and + is a standard addition on 8 bits, that wraps around when overflown.
a, b, c are unsigned char;
We can use an SMT solver to test our hypothesis that your formula can be simplified. You can head over to http://rise4fun.com:
x = BitVec('x', 8)
y = BitVec('y', 8)
z = BitVec('z', 8)
print simplify((x + z) ^ (y + z))
and the result, anticlimactically, is:
x + z ^ y + z
Which means your formula cannot be further simplified.
(a+b)xor(c+b)
--------------
=((not(a+b))*(c+b))+((a+b)*(not(c+b)))
-----------------------
=((not a)*(not b)*(c+b))+((a+b)*(not c)*(not b))
----
=((not a)(not b)*c) + (a*(not c)(not b))
----
=(not b)((not a)c + a(not c))
----
=(not b)(a xor c)
----
Hi I need to calculate (2^n + (-1)^n) % 10000007
where 1 < n < 10^9
How should I go about writing a program for it in c++?
I know this mod property
(a + b)%n = (a%n + b%n)%n but this wont help me.
Given
(a + b)%m = (a%m + b%m)%m
Then, replace both a and b with the same power of 2, and you get the recurrence:
2k+1%m = (2k%m + 2k%m)%m
You probably already figured your formula allows you to break down your problem into:
(2n + (-1)n)%P = (2n%P + (-1)n%P)%P
Then, note that (-1)k is either 1 or -1, and you should be able to calculate your problem in O(n) time.