Lets say I have a simple equation as follows:
x = 50 * (20 * item)
Is there a system or language to reverse this automatically? So lets say I have x as an input and want to find out which item it is.
item = (x - 50) / 20
A very trivial and probably misleading example, but just to illustrate the point. I'm not talking about solving equations with least squares fitting or such, but instead a system that can reverse equations effectively generating what you would have to code by hand. I'm not a mathematician so forgive the incorrect terminology (if any).
A colleage of mine said Matlab can do this natively. Anyone has any idea how? Can I actually specify which inputs I have and the output I'm looking for? Or do I simply have to use the inbuilt math functions to for a handwritten reversed equation?
You're probably looking for http://www.mathworks.com/help/symbolic/solve.html
You should first declare a symbolic variable: syms item
and then: solve(x == 50 * (20 * item), item) should give You what You're looking for.
Related
There are many mathematical programs out there out of which some are able to solve calculus-based problems, GeoGebra, Qalculate! to name a few.
How are those programs able to solve calculus-based problems which humans need to evaluate using a long procedure?
For example, the problem:
It takes a lot of steps for humans to solve this problem as shown here on Quora.
How can those mathematical programs solve them with such a good accuracy?
The Church-Turing thesis implies that anything a human being can calculate can be calculated by any Turing-equivalent system of computation - including programs running on computers. That is to say, if we can solve the problem (or calculate an approximate answer that meets some criteria) then a computer program can be made to do the same thing. Let's consider a simpler example:
f(x) = x
a = Integral(f, 0, 1)
A human being presented with this problem has two options:
try to compute the antiderivative using some procedure, then use procedures to evaluate the definite integral over the supplied range
use some numerical method to calculate an approximate value for the definite integral which meets some criteria for closeness to the true value
In either case, human beings have a set of tools that allow them to do this:
recognize that f(x) is a polynomial in x. There are rules for constructing the antiderivatives of polynomials. Specifically, each term ax^b in the polynomial can be converted to a/(b+1)x^(b+1) and then an arbitrary constant c added to the end. We then say Sf(x)dx = (1/2)x^2 + c. Now that we have the antiderivative, we have a procedure for computing the antiderivative over a range: calculate Sf(x)dx for the high value, then subtract from that the result of calculating Sf(x)dx for the low value. This gives ((1/2)1^2) - ((1/2)0^2) = 1/2 - 0 = 1/2.
decide that for our purposes a Riemann sum with dx=1/10 is sufficient and that we'll take the midpoint value. We get 10 rectangles with base 1/10 and heights 1/20, 3/20, 5/20, 7/20, 9/20, 11/20, 13/20, 15/20, 17/20 and 19/20, respectively. The areas are 1/200, 3/200, 5/200, 7/200, 9/200, 11/200, 13/200, 15/200, 17/200 and 19/200. The sum of these is (1+3+5+7+9+11+13+15+17+19)/200 = 100/200 = 1/2. We happened to get the exact answer since we used the midpoint value and evaluated the definite integral of a linear function; in general, we'd have been close but not exact.
The only difficulty is in adequately specifying the procedure human beings use to solve these problems in various ways. Once specified, computers are perfectly capable of doing them. And make no mistake, human beings have a procedure - conscious or subconscious - for doing these problems reliably.
I was solving a RSA problem and facing difficulty to compute d
plz help me with this
given p-971, q-52
Ø(n) - 506340
gcd(Ø(n),e) = 1 1< e < Ø(n)
therefore gcd(506340, 83) = 1
e= 83 .
e * d mod Ø(n) = 1
i want to compute d , i have all the info
can u help me how to computer d from this.
(83 * d) mod 506340 = 1
i am a little wean in maths so i am having difficulties finding d from the above equation.
Your value for q is not prime 52=2^2 * 13. Therefore you cannot find d because the maths for calculating this relies upon the fact the both p and q are prime.
I suggest working your way through the examples given here http://en.wikipedia.org/wiki/RSA_%28cryptosystem%29
Normally, I would hesitate to suggest a wikipedia link such as that, but I found it very useful as a preliminary source when doing a project on RSA as part of my degree.
You will need to be quite competent at modular arithmetic to get to grips with how RSA works. If you want to understand how to find d you will need to learn to find the Modular multiplicative inverse - just google this, I didn't come across anything incorrect when doing so myself.
Good luck.
A worked example
Let's take p=11, q=5. In reality you would use very large primes but we are going to be doing this by hand to we want smaller numbers. Keep both of these private.
Now we need n, which is given as n=pq and so in our case n=55. This needs to be made public.
The next item we need is the totient of n. This is simply phi(n)=(p-1)(q-1) so for our example phi(n)=40. Keep this private.
Now you calculate the encryption exponent, e. Defined such that 1<e<phi(n) and gcd(e,phi(n))=1. There are nearly always many possible different values of e - just pick one (in a real application your choice would be determined by additional factors - different choices of e make the algorithm easier/harder to crack). In this example we will choose e=7. This needs to be made public.
Finally, the last item to be calculated is d, the decryption exponent. To calculate d we must solve the equation ed mod phi(n) = 1. This is most commonly calculated using the Extended Euclidean Algorithm. This algorithm solves the equation phi(n)x+ed=1 subject to 1<d<phi(n), where x is an unknown multiplicative factor - which is identical to writing the previous equation without using mod. In our particular example, solving this leads to d=23. This should be kept private.
Then your public key is: n=55, e=7
and your private key is: n=55, d=23
To see the workthrough of the Extended Euclidean Algorithm check out this youtube video https://www.youtube.com/watch?v=kYasb426Yjk. The values used in that video are the same as the ones used here.
RSA is complicated and the mathematics gets very involved. Try solving a couple of examples with small values of p and q until you are comfortable with the method before attempting a problem with large values.
I would like to know how can I get both the positive and the negative solution from a sqrt in Matlab.
For example if I have:
sin(a) = sqrt(1-cos(a)^2);
The docs don't say anything specific about always only providing the positive square root but it does seem like a fair assumption in which case you can get the negative square pretty easily like this:
p = sqrt(1-cos(a)^2);
n = -sqrt(1-cos(a)^2);
btw assigning to sin(a) like that is going to create a variable called sin which will hide the sin function leading to many possible errors, so I would highly recommend choosing a different variable name.
MATLAB (and every other programming language that I know of) only returns the principal square root of x when calling sqrt(x) or equivalent.
How you'd write the square root of x mathematically, is
s = ±√x
which is just a shorthand for writing the whole solution set
s = {+√x -√x}
In MATLAB, you'd write it the same as this last case, but with slightly different syntax,
s = [+sqrt(x) -sqrt(x)]
which can be computed more efficiently if you "factor out" the sqrt:
s = sqrt(x) * [1 -1]
So, for your case,
s = sqrt(1-cos(a)^2) * [1 -1]
or, if you so desire,
s = sin(acos(a)) * [1 -1]
which is a tad slower, but perhaps more readable (and actually a bit more accurate as well).
Now of course, if you can somehow find the components whose quotient results in the value of your cosine, then you wouldn't have to deal with all this messy business of course....
sqrt does not solve equations, only gives numerical output. You will need to formulate your equation as you need it, and then you can use sqrt(...) -1*sqrt(...) to give your positive and negative outputs.
I have three values X,Y and Z. These values have a range of values between 0 and 1 (0 and 1 included).
When I call a function f(X,Y,Z) it returns a value V (value between 0 and 1). My Goal is to choose X,Y,Z so that the returned value V is as close as possible to 1.
The selection Process should be automated and the right values for X,Y,Z are unknown.
Due to my Use Case it is possible to set Y and Z to 1 (the value 1 hasn't any influence on the output) and search for the best value of X.
After that I can replace X by that value and do the same for Y. Same procedure for Z.
How can I find the "maximum of the function"? Is there somekind of "gradient descend" or hill climbing algorithm or something like that?
The whole modul is written in perl so maybe there is an package for perl that can solve that problem?
You can use Simulated Annealing. Its a multi-variable optimization technique. It is also used to get a partial solution for the Travelling Salesperson problem. Its one of the search algorithms mentioned in Peter Norvig's Intro to AI book as well.
Its a hill climbing algorithm which depends on random variables. Also it won't necessarily give you the 'optimal' answer. You can also vary the iterations required by it as per your computational/time needs.
http://en.wikipedia.org/wiki/Simulated_annealing
http://www1bpt.bridgeport.edu/sed/projects/449/Fall_2000/fangmin/chapter2.htm
I suggest you take a look at Math::Amoeba which implements the Nelder–Mead method for finding stationary points on functions.
Let's assume that we have a vector like
x = -1:0.05:1;
ids = randperm(length(x));
x = x(ids(1:20));
I would like to calculate the maximum distance between the elements of x in some idiomatic way. It would be easy to just iterate over all possible combinations of x's elements but I feel like there could be a way to do it with MATLAB's built-in functions in some crazy but idiomatic way.
What about
max_dist = max(x) - min(x)
?
Do you mean the difference between the largest and smallest elements in your vector ? If you do, then something like this will work:
max(x) - min(x)
If you don't, then I've misunderstood the question.
This is an interpoint distance computation, although a simple one, since you are working in one dimension. Really that point which falls at a maximum distance in one dimension is always one of two possible points. So all you need do is grab the minimum value and the maximum value from the list, and see which is farther away from the point in question. So assuming that the numbers in x are real numbers, this will work:
xmin = min(x);
xmax = max(x);
maxdistance = max(x - xmin,xmax - x);
As an alternative, some time ago I put a general interpoint distance computation tool up on the file exchange (IPDM). It is smart enough to special case simple problems like the 1-d farthest point problem. This call would do it for you:
D = ipdm(x,'subset','farthest','result','struct');
Of course, it will not be as efficient as the simple code I wrote above, since it is a fully general tool.
Uhh... would love to have a MATLAB at my hands and its still early in the morning, but what about something like:
max_dist = max(x(2:end) - x(1:end-1));
I don't know if this is what You are looking for.