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 am trying to implement RSA prime generation for P and Q based on FIP186-4 specification. The specification describes two different implementations: Section 3.2 Provable Prime Construction vs. Section 3.3 Probable Prime Construction. Initially, I tried implementing the probable prime approach because it is easier to understand and implement, but I discovered it is very slow because of the number of iterations needed to find P and Q primes (worst case it takes 15 minutes). Next, I decided to try the provable prime approach but I found out the algorithm is much more complex and might be slow as well. Below are my two issues:
In Section C.10, Step 12, how to eliminate the sqrt(2) to the expression x = floor(sqrt(2))(2^(L−1))) + (x mod (2^L − floor((sqrt(2)(2^(L−1))))) so that I can represent it as whole numbers using BigNum representation?
In Section C.10, Step 14, is there a fast way to compute y in the interval [1, p2] such that 0 = ( y p0 p1–1) mod p2? The specification doesn't specify a method to implement this. My initial thought was to perform a linear search staring from integer 1 and up but that can be very slow because p2 can be a very large number.
I tried searching online for help on this issue, but I discovered a lot of examples don't even comply with FIPS186-4. I assume it is because these two methods are too slow.
This is my first thread here and I would like to ask you a couple of questions for universal hashing of integers.
A universal hashing algorithm is supposed to use this:
equation =
((a*x+b)mod p) mod m
a=random number from 1 to p-1
b=random number from 0 to p-1
x= the Key
p= a prime number >=m
m=the size of the array
I know the numbers I am going to hash are on the range of 1-2969.
But I cannot understand how to use this equation in order to make as low collisions as possible.
At the time a and b are random I cannot do anything about it.
My question is how I am supposed to pick the prime if I have more than one choice, the range of primes I can use are from 2 to 4999.
I tried to pick the first available that corresponds the requirements for the function but sometimes it can return negative numbers. I have searched on Google and Stackoverflow but I could not figure out what I am not doing wrong.
I am coding in C. Also, I can use only universal hashing.
Thank your for your time.
I hope you can help me with this one.
I am using cointegration to discover potential pairs trading opportunities within stocks and more precisely I am utilizing the Johansen trace test for only two stocks at a time.
I have several securities, but for each test I only test two at a time.
If two stocks are found to be cointegrated using the Johansen test, the idea is to define the spread as
beta' * p(t-1) - c
where beta'=[1 beta2] and p(t-1) is the (2x1) vector of the previous stock prices. Notice that I seek a normalized first coefficient of the cointegration vector. c is a constant which is allowed within the cointegration relationship.
I am using Matlab to run the tests (jcitest), but have also tried utilizing Eviews for comparison of results. The two programs yields the same.
When I run the test and find two stocks to be cointegrated, I usually get output like
beta_1 = 12.7290
beta_2 = -35.9655
c = 121.3422
Since I want a normalized first beta coefficient, I set beta1 = 1 and obtain
beta_2 = -35.9655/12.7290 = -2.8255
c =121.3422/12.7290 = 9.5327
I can then generate the spread as beta' * p(t-1) - c. When the spread gets sufficiently low, I buy 1 share of stock 1 and short beta_2 shares of stock 2 and vice versa when the spread gets high.
~~~~~~~~~~~~~~~~ The problem ~~~~~~~~~~~~~~~~~~~~~~~
Since I am testing an awful lot of stock pairs, I obtain a lot of output. Quite often, however, I receive output where the estimated beta_1 and beta_2 are of the same sign, e.g.
beta_1= -1.4
beta_2= -3.9
When I normalize these according to beta_1, I get:
beta_1 = 1
beta_2 = 2.728
The current pairs trading literature doesn't mention any cases where the betas are of the same sign - how should it be interpreted? Since this is pairs trading, I am supposed to long one stock and short the other when the spread deviates from its long run mean. However, when the betas are of the same sign, to me it seems that I should always go long/short in both at the same time? Is this the correct interpretation? Or should I modify the way in which I normalize the coefficients?
I could really use some help...
EXTRA QUESTION:
Under some of my tests, I reject both the hypothesis of r=0 cointegration relationships and r<=1 cointegration relationships. I find this very mysterious, as I am only considering two variables at a time, and there can, at maximum, only be r=1 cointegration relationship. Can anyone tell me what this means?
I would like to partition a number into an almost equal number of values in each partition. The only criteria is that each partition must be in between 60 to 80.
For example, if I have a value = 300, this means that 75 * 4 = 300.
I would like to know a method to get this 4 and 75 in the above example. In some cases, all partitions don't need to be of equal value, but they should be in between 60 and 80. Any constraints can be used (addition, subtraction, etc..). However, the outputs must not be floating point.
Also it's not that the total must be exactly 300 as in this case, but they can be up to a maximum of +40 of the total, and so for the case of 300, the numbers can sum up to 340 if required.
Assuming only addition, you can formulate this problem into a linear programming problem. You would choose an objective function that would maximize the sum of all of the factors chosen to generate that number for you. Therefore, your objective function would be:
(source: codecogs.com)
.
In this case, n would be the number of factors you are using to try and decompose your number into. Each x_i is a particular factor in the overall sum of the value you want to decompose. I'm also going to assume that none of the factors can be floating point, and can only be integer. As such, you need to use a special case of linear programming called integer programming where the constraints and the actual solution to your problem are all in integers. In general, the integer programming problem is formulated thusly:
You are actually trying to minimize this objective function, such that you produce a parameter vector of x that are subject to all of these constraints. In our case, x would be a vector of numbers where each element forms part of the sum to the value you are trying to decompose (300 in your case).
You have inequalities, equalities and also boundaries of x that each parameter in your solution must respect. You also need to make sure that each parameter of x is an integer. As such, MATLAB has a function called intlinprog that will perform this for you. However, this function assumes that you are minimizing the objective function, and so if you want to maximize, simply minimize on the negative. f is a vector of weights to be applied to each value in your parameter vector, and with our objective function, you just need to set all of these to -1.
Therefore, to formulate your problem in an integer programming framework, you are actually doing:
(source: codecogs.com)
V would be the value you are trying to decompose (so 300 in your example).
The standard way to call intlinprog is in the following way:
x = intlinprog(f,intcon,A,b,Aeq,beq,lb,ub);
f is the vector that weights each parameter of the solution you want to solve, intcon denotes which of your parameters need to be integer. In this case, you want all of them to be integer so you would have to supply an increasing vector from 1 to n, where n is the number of factors you want to decompose the number V into (same as before). A and b are matrices and vectors that define your inequality constraints. Because you want equality, you'd set this to empty ([]). Aeq and beq are the same as A and b, but for equality. Because you only have one constraint here, you would simply create a matrix of 1 row, where each value is set to 1. beq would be a single value which denotes the number you are trying to factorize. lb and ub are the lower and upper bounds for each value in the parameter set that you are bounding with, so this would be 60 and 80 respectively, and you'd have to specify a vector to ensure that each value of the parameters are bounded between these two ranges.
Now, because you don't know how many factors will evenly decompose your value, you'll have to loop over a given set of factors (like between 1 to 10, or 1 to 20, etc.), place your results in a cell array, then you have to manually examine yourself whether or not an integer decomposition was successful.
num_factors = 20; %// Number of factors to try and decompose your value
V = 300;
results = cell(1, num_factors);
%// Try to solve the problem for a number of different factors
for n = 1 : num_factors
x = intlinprog(-ones(n,1),1:n,[],[],ones(1,n),V,60*ones(n,1),80*ones(n,1));
results{n} = x;
end
You can then go through results and see which value of n was successful in decomposing your number into that said number of factors.
One small problem here is that we also don't know how many factors we should check up to. That unfortunately I don't have an answer to, and so you'll have to play with this value until you get good results. This is also an unconstrained parameter, and I'll talk about this more later in this post.
However, intlinprog was only released in recent versions of MATLAB. If you want to do the same thing without it, you can use linprog, which is the floating point version of integer programming... actually, it's just the core linear programming framework itself. You would call linprog this way:
x = linprog(f,A,b,Aeq,beq,lb,ub);
All of the variables are the same, except that intcon is not used here... which makes sense as linprog may generate floating point numbers as part of its solution. Due to the fact that linprog can generate floating point solutions, what you can do is if you want to ensure that for a given value of n, you could loop over your results, take the floor of the result and subtract with the final result, and sum over the result. If you get a value of 0, this means that you had a completely integer result. Therefore, you'd have to do something like:
num_factors = 20; %// Number of factors to try and decompose your value
V = 300;
results = cell(1, num_factors);
%// Try to solve the problem for a number of different factors
for n = 1 : num_factors
x = linprog(-ones(n,1),[],[],ones(1,n),V,60*ones(n,1),80*ones(n,1));
results{n} = x;
end
%// Loop through and determine which decompositions were successful integer ones
out = cellfun(#(x) sum(abs(floor(x) - x)), results);
%// Determine which values of n were successful in the integer composition.
final_factors = find(~out);
final_factors will contain which number of factors you specified that was successful in an integer decomposition. Now, if final_factors is empty, this means that it wasn't successful in finding anything that would be able to decompose the value into integer factors. Noting your problem description, you said you can allow for tolerances, so perhaps scan through results and determine which overall sum best matches the value, then choose whatever number of factors that gave you that result as the final answer.
Now, noting from my comments, you'll see that this problem is very unconstrained. You don't know how many factors are required to get an integer decomposition of your value, which is why we had to semi-brute-force it. In fact, this is a more general case of the subset sum problem. This problem is NP-complete. Basically, what this means is that it is not known whether there is a polynomial-time algorithm that can be used to solve this kind of problem and that the only way to get a valid solution is to brute-force each possible solution and check if it works with the specified problem. Usually, brute-forcing solutions requires exponential time, which is very intractable for large problems. Another interesting fact is that modern cryptography algorithms use NP-Complete intractability as part of their ciphertext and encrypting. Basically, they're banking on the fact that the only way for you to determine the right key that was used to encrypt your plain text is to check all possible keys, which is an intractable problem... especially if you use 128-bit encryption! This means you would have to check 2^128 possibilities, and assuming a moderately fast computer, the worst-case time to find the right key will take more than the current age of the universe. Check out this cool Wikipedia post for more details in intractability with regards to key breaking in cryptography.
In fact, NP-complete problems are very popular and there have been many attempts to determine whether there is or there isn't a polynomial-time algorithm to solve such problems. An interesting property is that if you can find a polynomial-time algorithm that will solve one problem, you will have found an algorithm to solve them all.
The Clay Mathematics Institute has what are known as Millennium Problems where if you solve any problem listed on their website, you get a million dollars.
Also, that's for each problem, so one problem solved == 1 million dollars!
(source: quickmeme.com)
The NP problem is amongst one of the seven problems up for solving. If I recall correctly, only one problem has been solved so far, and these problems were first released to the public in the year 2000 (hence millennium...). So... it has been about 14 years and only one problem has been solved. Don't let that discourage you though! If you want to invest some time and try to solve one of the problems, please do!
Hopefully this will be enough to get you started. Good luck!