Can anyone explain to me how do I use a recursion, if I don't know the limit. For example, I need the remainder r of the Euclidean algorithm for gcd(a,b) which equals 0. I figured out that the recursive formula I need is
r[n]=r[n-2]-Floor[r[n-2]/r[n-1] * r[n-1]
r[1] = a - Floor[a/b]* b;
r[2] = b - Floor[b/r1] r1;
Related
Hi i wanted to know how can i solve the tine complexity of this algorithm
I solved with f(n/4) but not f(n/i)
void f(int n){
if (n<4) return;
for (int i=0;i*i<n;i++)
printf("-");
for (int i=2;i<4;i++)
f(n/i); // solved the case f(n/4) but stuck f(n/i)
}
Note that the loop condition is i<4, so i never reaches 4. i.e. the only recursive terms are f(n/2) and f(n/3).
Recurrence relation:
T(n) = T(n/2) + T(n/3) + Θ(sqrt(n))
There are two ways to approach this problem:
Find upper and lower bounds by replacing one of the recursive terms with the other:
R(n) = 2T(n/3) + Θ(sqrt(n))
S(n) = 2T(n/2) + Θ(sqrt(n))
R(n) ≤ T(n) ≤ S(n)
You can easily solve for both bounds by substitution or applying the Master Theorem:
R(n) = O(n^[log3(2)]) = O(n^0.63...)
S(n) = O(n)
If you need an exact answer, use the Akra-Bazzi method:
a1 = a2 = 1
h1(x) = h2(x) = 0
g(x) = sqrt(x)
b1 = 1/2
b2 = 1/3
You need to solve for a power p such that [1/2]^p + [1/3]^p = 1. Do this numerically with e.g. Newton-Raphson, to obtain p = 0.78788.... Perform the integral:
‒ to obtain T(n) = O(n^0.78...), which is consistent with the bounds found before.
I think this is about O(sqrt(9/2) * sqrt(n)) time, but I'd go with O(sqrt(n)) to be safe. It's admittedly been a while since I worked with time complexity.
If n < 4, the function returns immediately, at constant time O(1)
If n >= 4, the function's for loop, for (int i=0; i*i<n; i++) performs the constant-time function printf("-"); a total number of sqrt(n) times. So far we're at O(sqrt(n)) time.
The next for loop performs two recursive calls: one for f(n/2) and one for f(n/3)
The first runs in O(sqrt(n/2)) time, the second in O(sqrt(n/4)) time, and so on - this series converges to O(sqrt(2n))
Likewise, the function f(n/3) converges to O(sqrt(3/2 n))
This doesn't factor in the fact that each recursive call also invokes a little extra time by calling both of these functions when it runs, but I believe this converges to about O(sqrt(n)) + O(sqrt(2n)) + O(sqrt(3/2 n)), which itself converges to O(sqrt(9/2) * sqrt(n))
This is likely a little low bit low for an exact constant value, but I believe you can safely say this runs at O(sqrt(n)) time, with some small-ish constant out front.
I want to learn some prolog and found the exercise to calculate pi recursively for a given predicat pi(10, Result). I don't want it to be tail recursive because I find tail recursion to be easier. I've been trying to do this for hours now but it seems like I can't come to a solution, this is how far I've come:
(I'm using Leibniz' pi formula as reference)
pi(0, 0).
pi(Next, Result) :-
Num is -1**(Next + 1),
Part is Num / (2 * Next - 1),
N1 is Next -1,
pi(N1, R),
Result is Part + R.
Now, I'm aware that the addition at the end is wrong. Also I need to multiply the end result by 4 and I don't know how to do that. Would be glad if anyone could help out. And no, this is not a homework or anything. :)
Here's a slightly different twist that terminates based upon reaching a given precision. It also is tail recursive. Because Leibniz converges very slowly, the formula is a stack hog when done using simple recursion. it's not an algorithm well-suited for a recursive solution in any language. However, a smart Prolog interpreter can take advantage of the tail recursion and avoid that. Just by way of example, it only allows precision within a specific range.
pi(Precision, Pi) :-
Precision > 0.0000001,
Precision < 0.1,
pi_over_4(1, 1, Precision/4, 1, Pi_over_4), % Compensate for *4 later
Pi is Pi_over_4 * 4.
pi_over_4(AbsDenominator, Numerator, Precision, Sum, Result) :-
NewAbsDenominator is AbsDenominator + 2,
NewNumerator is -Numerator,
NewSum is Sum + NewNumerator/NewAbsDenominator,
( abs(NewSum - Sum) < Precision
-> Result = NewSum
; pi_over_4(NewAbsDenominator, NewNumerator, Precision, NewSum, Result)
).
2 ?- pi(0.0001, P).
P = 3.1416426510898874.
3 ?- pi(0.00001, P).
P = 3.141597653564762.
4 ?- pi(0.000005, P).
P = 3.141595153583494.
This is strictly an imperative use of Prolog, which isn't what Prolog is strong for.
Ok,
so this is a application of existing mathematical practices, but I can't really apply them to my case.
So, I have x of a currency to increase the level of a game-object y for cost z.
z is calculated in cost(y.lvl) = c_1 * c_2^y.lvl / c_3, where the c's are constants.
I am seeking an efficient way to calculate, how often I can increase the level of y, given x. Currently I'm using a loop that does something like this:
double tempX = x;
int counter = 0;
while(tempX >= cost(y.lvl+counter)){
tempX-=cost(y.lvl)+counter;
counter++;
}
The problem is, that in some cases, this loop has to iterate too many times to stay performant.
What I am looking for is essentially a function
int howManyCanBeBought(x,y.lvl), which calculates it's result in a single go, instead of looping a lot of times.
I've read something about transforming recursions to generating functions and transforming them to closed formulas, but I didn't get the math behind it. Is there an easy way to it?
If I understand correctly, you're looking for the largest n such that:
Σi=0..n c1/c3 c2lvl+i ≤ x
Dividing by the constant factor:
Σi=0..n c2i ≤ c3 / (c1 c2lvl) x
Using the formula for the sum of a geometric series:
(c2n+1 - 1) / (c2 - 1) ≤ c3 / (c1 c2lvl) x
And solving for the maximum integer:
n = floor(logc2(c3 (c2 - 1) / (c1 c2lvl) x + 1) - 1)
I'm trying to find time complexity (big O) of a recursive formula.
I tried to find a solution, you may see the formula and my solution below:
Like Brenner said, your last assumption is false. Here is why: Let's take the definition of O(n) from the Wikipedia page (using n instead of x):
f(n) = O(n) if and only if there exist constants c, n0 s.t. |f(n)| <= c |g(n)|, for alln >= n0.
We want to check if O(2^n^2) = O(2^n). Clearly, 2^n^2 is in O(2^n^2), so let's pick f(n) = 2^n^2 and check if this is in O(2^n). Put this into the above formula:
exists c, n0: 2^n^2 <= c * 2^n for all n >= n0
Let's see if we can find suitable constant values n0 and c for which the above is true, or if we can derive a contradiction to proof that it is not true:
Take the log on both sides:
log(2^n^2) <= log(c * 2 ^ n)
Simplify:
2 ^n log(2) <= log(c) + n * log(2)
Divide by log(2):
n^2 <= log(c)/log(2) * n
It's easy to see know that there is no c, n0 for which the above is true for all n >= n0, thus O(2^n^2) = O(n^2) is not a valid assumption.
The last assumption you've specified with the question mark is false! Do not make such assumptions.
The rest of the manipulations you've supplied seem to be correct. But they actually bring you nowhere.
You should have finished this exercise in the middle of your draft:
T(n) = O(T(1)^(3^log2(n)))
And that's it. That's the solution!
You could actually claim that
3^log2(n) == n^log2(3) ==~ n^1.585
and then you get:
T(n) = O(T(1)^(n^1.585))
which is somewhat similar to the manipulations you've made in the second part of the draft.
So you can also leave it like this. But you cannot mess with the exponent. Changing the value of the exponent changes the big-O classification.
For an ocean shader, I need a fast function that computes a very approximate value for sin(x). The only requirements are that it is periodic, and roughly resembles a sine wave.
The taylor series of sin is too slow, since I'd need to compute up to the 9th power of x just to get a full period.
Any suggestions?
EDIT: Sorry I didn't mention, I can't use a lookup table since this is on the vertex shader. A lookup table would involve a texture sample, which on the vertex shader is slower than the built in sin function.
It doesn't have to be in any way accurate, it just has to look nice.
Use a Chebyshev approximation for as many terms as you need. This is particularly easy if your input angles are constrained to be well behaved (-π .. +π or 0 .. 2π) so you do not have to reduce the argument to a sensible value first. You might use 2 or 3 terms instead of 9.
You can make a look-up table with sin values for some values and use linear interpolation between that values.
A rational algebraic function approximation to sin(x), valid from zero to π/2 is:
f = (C1 * x) / (C2 * x^2 + 1.)
with the constants:
c1 = 1.043406062
c2 = .2508691922
These constants were found by least-squares curve fitting. (Using subroutine DHFTI, by Lawson & Hanson).
If the input is outside [0, 2π], you'll need to take x mod 2 π.
To handle negative numbers, you'll need to write something like:
t = MOD(t, twopi)
IF (t < 0.) t = t + twopi
Then, to extend the range to 0 to 2π, reduce the input with something like:
IF (t < pi) THEN
IF (t < pi/2) THEN
x = t
ELSE
x = pi - t
END IF
ELSE
IF (t < 1.5 * pi) THEN
x = t - pi
ELSE
x = twopi - t
END IF
END IF
Then calculate:
f = (C1 * x) / (C2 * x*x + 1.0)
IF (t > pi) f = -f
The results should be within about 5% of the real sine.
Well, you don't say how accurate you need it to be. The sine can be approximated by straight lines of slopes 2/pi and -2/pi on intervals [0, pi/2], [pi/2, 3*pi/2], [3*pi/2, 2*pi]. This approximation can be had for the cost of a multiplication and an addition after reducing the angle mod 2*pi.
Using a lookup table is probably the best way to control the tradeoff between speed and accuracy.