I´m trying to make very fast MOD operations.
I saw in several pages we can do alternatively calculate the MOD using the AND operator with (Divisor-1). Eg.:
result = (100 mod 8) is the same as
result = (100 and 7)
It functions perfectly if the divisor is less than 8 bits, but if we calculate (1245 mod 67) we can see the result is different of (1245 and 66).
So, how can I calculate this faster than using the MOD operator provided by VB.NET language?
Thanks!
Using a bitwise AND only works for modulus powers of 2 (and only positive ones, at that). It does not work for other numbers. See this link
I think that the modulus operator built into the framework is fast and you probably won't be able to improve on it.
Well, 100 mod 8 = 100 and 7 works, because 7 is binary 111b.
Whatever number you have (100 decimal is 1100100b), the last 3 binary digits (bits) are kept due to the 111b. For 100, that would be 100b = 4.
Now consider what you did attempt with 1245 mod 67.
1245 is binary 10011011101b.
67 is binary 1000011b.
Can you see, why it's not working?
I am super confused what the percentage sign does in Objective C. Can someone explain to me in language that an average idiot like myself can understand?! Thanks.
% is the modulo operator, so for example 10 % 3 would result in 1.
If you have some numbers a and b, a % b gives you just the remainder of a divided by b.
So in the example 10 % 3, 10 divided by 3 is 3 with remainder 1, so the answer is 1.
If there is no remainder to a divided by b, the answer is zero, so for example, 4 % 2 = 0.
Here's a relevant SO question about modular arithmetic.
Same as what it does in C, it's "modulo" (also known as integer remainder).
% is the modulo operator. It returns the remainder of <number> / <number>. For example:
5 % 2
means 5 / 2, which equals 2 with a remainder of 1, so, 1 is the value that is returned. Here's some more examples:
3 % 3 == 0 //remainder of 3/3 is 0
6 % 3 == 0 //remainder of 6/3 is 0
5 % 3 == 2 //remainder of 5/3 is 2
15 % 4 == 3 //remainder of 15/4 is 3
99 % 30 == 9 //remainder of 99/30 is 9
The definition of modulo is:
mod·u·lo
(in number theory) with respect to or using a modulus of a specified number. Two numbers are congruent modulo a given number if they give the same remainder when divided by that number.
In Mathematics, The Percentage Sign %, Called Modulo (Or Sometimes The Remainder Operator) Is A Operator Which Will Find The Remainder Of Two Numbers x And y. Mathematically Speaking, If x/y = {(z, r) : y * z + r = x}, Where All x, y, and z Are All Integers, Then
x % y = {r: ∃z: x/y = (z, r)}. So, For Example, 10 % 3 = 1.
Some Theorems And Properties About Modulo
If x < y, Then x % y = x
x % 1 = 0
x % x = 0
If n < x, Then (x + n) % x = n
x Is Even If And Only If x % 2 = 0
x Is Odd If And Only If x % 2 = 1
And Much More!
Now, One Might Ask: How Do We Find x % y? Well, Here's A Fairly Simple Way:
Do Long Division. I Could Explain How To Do It, But Instead, Here's A Link To A Page Which Explains Long Division: https://www.mathsisfun.com/numbers/long-division-index.html
Stop At Fractions. Once We Reach The Part Where We Would Normally Write The Answer As A Fraction, We Should Stop. So, For Example, 101/2 Would Be 50.5, But, As We Said, We Would Stop At The Fractions, So Our Answer Ends Up Being 50.
Output What's Left As The Answer. Here's An Example: 103/3. First, Do Long Division. 103 - 90 = 13. 13 - 12 = 1. Now, As We Said, We Stop At The Fractions. So Instead Of Continuing The Process And Getting The Answer 34.3333333..., We Get 34. And Finally, We Output The Remainder, In This Case, 1.
NOTE: Some Mathematicians Write x mod y Instead Of x % y, But Most Programming Languages Only Understand %.
I'm new to the topic here :/
Could anyone please tell me how to solve the following?
Show that 36^2004 + 17^768 x 27^412 is divisible by 19.
Thanks!
Simple identities can be used to solve the above, the important of them being:
(a + b) mod c = a mod c + b mod c
Also,
ab mod c = (a mod c)*(b mod c)
This can be used to solve very big exponents also, for example, if you are to solve:
24^3100 mod 19
You could probably break it up as:
24^(310*100) mod 19
which can be further written as:
24^310 mod 19 x 24^100 mod 19
You can further break it down to values you could actually compute and solve. For example, if you keep on breaking down 100, you could end up solving
(24^4 mod 19)^25
and so on and so forth. Since this is a homework question, I can only provide hints and not the complete solution.
You can also do it with the fast exponentiation method where the exponent is expressed in powers of two.
How do you compute the following using Fermat's Little Theorem?
2^1,000,006 mod 101
2^-1,000,005 mod 11
You know that a^(p-1) === 1 mod p, so...
2^10 === 1 mod 11
2^(-1,000,005) = 2^(-1,000,000)*2^(-5) = 1 * 2^(-5) = 2^(-5)*2^(10) = 32 mod 11 = -1 = 10
From this, can you see how to work the larger number? The process is the same.
It's FLT all the way. I messed up.
Since 101 and 11 are prime, then (respectively) 2^100 and 2^10 are congruent to 1 mod 101 and 11.
Try to express 2^1000006 in terms of 2^100 and 2^-1000005 in terms of 2^10. You should be able to reduce each problem to something easy to calculate.
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.