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I was using this formula to calculate last m digits of 2^n.
pow=2+(n-m)%(4*5^(m-1))
ans =(2^pow)%(10^m)**
But this is not working for n=2009 and m=3.
Suggest any error in my calculation or a better formula if there is.
I don't understand what your formula is doing, but the simplest way is to calculate (2^2009)%(10^m) . Here is a pseudo code to find (x^y)%mod in O(log y). Put x=2, y=2009 and mod=10^m
power(x,y)
{
if( y == 0)
return 1
temp = power(x, y/2)
if (y%2 == 0)
return (temp*temp)%mod
else
return ((x*temp%mod)*temp)%mod
}
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What type of algorithm would this be called? Is it any sorting algorithm or bruteforce algorithm?
boolean isPrime = true;
for(int i=2; i<=number/2; i++){
if(number%i == 0){
isPrime = false;
break;
}
}
if(isPrime){
System.out.println("Prime");
}
System.out.println("Not Prime");
The algorithm apparently tests the variable number for primality by using all smaller numbers as potential factors; the approach could be considered brute-force.
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How can I generate a matrix in Matlab that has 5 rows and some specific number of columns and the elements may only be binary numbers and the column sum has to be less or equal to 3?
Some possibilites without loops:
Using strings:
D = 5;
S = 3;
numbers = str2mat(dec2bin(0:2^D-1))-'0';
numbers = numbers(sum(numbers,2)<=S,:);
Using combinatorial numbers, one line:
numbers = [zeros(1,D); cell2mat(arrayfun(#(s) fliplr(full(sparse((1:nchoosek(D,s)).'*ones(1,s), nchoosek(1:D,s), 1))), 0:S, 'uni', 0).')];
How about this: The maximum binary number, that you can represent by 5bit is 2^5-1 = 31 and skip through these to find the ones with sum of digits <= 3.
Something like
n = 1:1:31;
for ii = 1:length(ii)
bin = dec2bin(ii)
digitSum = 0
for d = 1:length(bin)
digitSum = digitSum + str2num(bin(d))
end
if (digitSum <= 3)
%store results
end
end
Here is a vecotorized solution to provide all occurences efficiently:
Bstr =dec2bin(1:31);
Bstr(sum(dec2bin(0:31),2)<=sum('00111'),:)=='1'
Inspired by the solution of #pyStarter
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How to calculate the sum of (1+a%m+a^2%m……+a^n%m) where
m=k!, 1<=k<=12, n<=10^18. How to Calculate this sum.
Using computer and the time limit is 3 sec.
Sorry about my mistake
1+a+a^2+...+a^n = (1+a+a^2+...+a^n)*(1-a)/(1-a) =
= (1 - a^(n+1))/(1-a)
In other words, your expression can be computed as:
(1 - a^(n+1))/(1-a) % m
Or, in programmatic form,
fmod((1-pow(a,n+1))/(1-a), m)
sum = 0;
i = 0;
while(i <= n){
sum = sum + math.pow(a,i);
i++;
}
result = sum % m;
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i am wondering if it is possible to simplify:
T*V + V*T // V = V^(t) symmetric
where both operands are matrixes
I don't think this is possible due to the following considerations:
If we multiply two matrices A and T, where A is symmetric (i.e. A(i,j) = A(j,i)), we have the following:
For A*T we have that the item in row z and column s is computed as:
__n__
\
/ A(z,i)*T(i,s)
-----
i=1
For the other way around, T*A, we get for row z, column s:
__n__ __n__
\ \
/ T(z,i)*A(i,s) = / T(z,i)*A(s,i)
----- -----
i=1 i=1
So, as long as we do not know anything about the entries T(i,j) in T, I think we can not say how these sums relate to each other.
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Is there some way to calculate the inverse factorials of real numbers?
For example - 1.5 ! = 1.32934039
Is there some way to obtain 1.5 back if I have the value 1.32934039?
I am trying
http://www.wolframalpha.com/input/?i=Gamma^(-1)[1.32934039]
but that is a fail.
Using wolframalpha.com, you can ask for
Solve[Gamma[x+1]==1.32934039,x]
As mentioned in the comments, Gamma does not have a unique inverse. True even when you are solving for a conventional factorial, e.g.
Solve[Gamma[x+1]==6,x]
yields several answers, of which one is 3.
Instead of using Gamma[] in WolframAlpha, you can also use Factorial[]:
Solve[Factorial[x]==6,x]
Solve[Factorial[x]==1.32934039,x]
David Cantrell gives a good approximation of Γ-1(n) on this page:
k = the positive zero of the digamma function, approximately 1.461632
c = Sqrt(2*pi)/e - Γ(k), approximately 0.036534
L(x) = ln((x+c)/Sqrt(2*pi))
W(x) = Lambert W function
ApproxInvGamma(x) = L(x) / W(L(x) / e) + 1/2
For integers you can do:
i = 2
n = someNum
while (n != 1):
n /= i
i += 1
return (i==1 ? i : None)
The factorial for real numbers has no inverse. You say that "each function must have an inverse". That is incorrect. Consider the constant function f(x)=0. What is f^-1(42)? For a function to be inverse it must be both an injection and a surjection.