neural network deciding the next output - matlab

I'd like to make a piece of music and there are 16 sounds possible, but not every sound can follow the other (for example sound 'A' can be followed by sound 'B', 'D' and 'F' but not by 'c' and 'E' etc). I have substituted this as a vector with 16 state spaces (values between 0 and 9) and I'd like to use them as inputs for neural network. Let's say I get new values as random (from 0 to 9) for the 16 elements of my vector. Now what I'd like my network to decide is which is the best suited value from the 16 based on the value of the last output, because as I already mentioned, each element has a certain number of possible elements that can come afterwards. DO YOU THINK IT COULD BE SOLVED BY NEURAL NETWORK and WHAT KIND OF NN WOULD BE BEST SUITED. Does anyone have similar examples. THANKS!

Rather than using a neural network, you may just want to create a table of probabilities that mark transitions between your notes. Imagining you could only play A, C, and E:
A C E
A 0.2 0.1 0.7
C 0.4 0.2 0.4
E 0.2 0.5 0.3
Using that table as an example, you could start on a random note (let's say a C). Then, using the table and starting from the row for 'C', we see that with a 0.4 probability we go to A, a 0.2 probability we go to C again, and a 0.4 probability we go to E. Then you just keep transitioning between your notes. A lot simpler than a neural network, and it may be closer to what you're looking for?

Related

How to define Traingular membership function for fuzzy controller design?

I am designing a fuzzy controller and for that, I have to define 3 triangular function sets. They are:
1 large
2 medium
3 small
But my problem is I have following data only:
Maximum input = 3 Minimum input= 0.1
Maximum output = 5.5 Minimum output= 0.8
How to define 3 triangular set range based on only this given information?
Here is the formula for a triangular membership function
f=0 if x<=a
f=(x-a)/(b-a) if a<=x<=b
f=(c-x)/(c-b) if b<=x<=c
f=0 if x>c
where a is the min, c is the max and b is the midpoint.
In your case, take the top situation where the max is 3 and the min is 0.1. The midpoint is (3+0.1)/2=1.55, so you have
f=0 if x<=0.1
f=(x-0)/(1.55-1) if 0.1<=x<=1.55
f=(3-x)/(3-1.55) if 1.55<=x<=3
f=0 if x>3
You should be able to take the 2nd example from here, but if not let me know. Something worth pointing out is that the midpoint may not be the ideal b in your situation. Any point between a and c could serve as your b, just know that it is the point where the membership function equals 1.
It is difficult to tell, but it looks like maybe you just have given parameters for two of the functions, perhaps for small and large or medium and large. You may need to use some judgement for the 3rd membership function.

Training simple artifical neural network to solve XOR sometimes doesn't work

I use NN with:
1 input layer (2 neurons),
1 hidden layer (2 neurons),
1 output layer (1 neuron).
I train it using feed forward and backpropagation algorithm. I also initialize weights randomly from range [-1, 1] (also tried [0, 1], but it doesn't change anything actually). And often (like 4/5 times) everything is trained properly (for inputs
[00, 01, 10, 11]
it outputs
[~0.1, ~0.9, ~0.9, ~0.1]
, respectively), but the remaining 1/5 times it outputs like
[~0.5, ~0.6, ~0.4, ~0.1]
(by ~number I mean a value around the number e.g. ~0.1 may be either 0.098 or 0.132 or similiar)
It doesn't matter if I train it for, let's say, 20 seconds or 10 minutes, it's still the same.
I'm pretty sure it's because of the randomly intialized weights, but I'm not sure how to fix that.
How shall I initialize the weights for this problem as well as for others (if it causes the problem)?
How do you do that? Have you any idea what causes this problem or you need some code? Thanks in advance.

Dot Product: * Command vs. Loop gives different results

I have two vectors in Matlab, z and beta. Vector z is a 1x17:
1 0.430742139435890 0.257372971229541 0.0965909090909091 0.694329541928697 0 0.394960106863064 0 0.100000000000000 1 0.264704325268675 0.387774594078319 0.269207605609567 0.472226643323253 0.750000000000000 0.513121013402805 0.697062571025173
... and beta is a 17x1:
6.55269487769363e+26
0
0
-56.3867588816768
-2.21310778926413
0
57.0726052009847
0
3.47223691057151e+27
-1.00249317882651e+27
3.38202232046686
1.16425987969027
0.229504956512063
-0.314243264212449
-0.257394312588330
0.498644243389556
-0.852510642195370
I'm dealing with some singularity issues, and I noticed that if I want to compute the dot product of z*beta, I potentially get 2 different solutions. If I use the * command, z*beta = 18.5045. If I write a loop to compute the dot product (below), I get a solution of 0.7287.
summation=0;
for i=1:17
addition=z(1,i)*beta(i);
summation=summation+addition;
end
Any idea what's going on here?
Here's a link to the data: https://dl.dropboxusercontent.com/u/16594701/data.zip
The problem here is that addition of floating point numbers is not associative. When summing a sequence of numbers of comparable magnitude, this is not usually a problem. However, in your sequence, most numbers are around 1 or 10, while several entries have magnitude 10^26 or 10^27. Numerical problems are almost unavoidable in this situation.
The wikipedia page http://en.wikipedia.org/wiki/Floating_point#Accuracy_problems shows a worked example where (a + b) + c is not equal to a + (b + c), i.e. demonstrating that the order in which you add up floating point numbers does matter.
I would guess that this is a homework assignment designed to illustrate these exact issues. If not, I'd ask what the data represents to suss out the appropriate approach. It would probably be much more productive to find out why such large numbers are being produced in the first place than trying to make sense of the dot product that includes them.

Removing extreme values in a vector in Matlab?

So say, I have a = [2 7 4 9 2 4 999]
And I'd like to remove 999 from the matrix (which is an obvious outlier).
Is there a general way to remove values like this? I have a set of vectors and not all of them have extreme values like that. prctile(a,99.5) is going to output the largest number in the vector no matter how extreme (or non-extreme) it is.
There are several way to do that, but first you must define what is "extreme'? Is it above some threshold? above some number of standard deviations?
Or, if you know you have exactly n of these extreme events and that their values are larger than the rest, you can use sort and the delete the last n elements. etc...
For example a(a>threshold)=[] will take care of a threshold like definition, while a(a>mean(a)+n*std(a))=[] will take care of discarding values that are n standard deviation above the mean of a.
A completely different approach is to use the median of a, if the vector is as short as you mention, you want to look on a median value and then you can either threshold anything above some factor of that value a(a>n*median(a))=[] .
Last, a way to assess an approach to treat these spikes would be to take a histogram of the data, and work from there...
I can think of two:
Sort your matrix and remove n-elements from top and bottom.
Compute the mean and the standard deviation and discard all values that fall outside:
mean +/- (n * standard deviation)
In both cases n must be chosen by the user.
Filter your signal.
%choose the value
N = 10;
filtered = filter(ones(1,N)/N, 1, signal);
Find the noise
noise = signal - filtered;
Remove noisy elements
THRESH = 50;
signal = signal(abs(noise) < THRESH);
It is better than mean+-n*stddev approach because it looks for local changes so it won't fail on a slowly changing signal like [1 2 3 ... 998 998].

explanation of roulette wheel selection example

I have a lisp program on roulette wheel selection,I am trying to understand the theory behind it but I cannot understand anything.
How to calculate the fitness of the selected strng?
For example,if I have a string 01101,how did they get the fitness value as 169?
Is it that the binary coding of 01101 evaluates to 13,so i square the value and get the answer as 169?
That sounds lame but somehow I am getting the right answers by doing that.
The fitness function you have is therefore F=X^2.
The roulette wheel calculates the proportion (according to its fitness) of the whole that that individual (string) takes, this is then used to randomly select a set of strings for the next generation.
Suggest you read this a few times.
The "fitness function" for a given problem is chosen (often) arbitrarily keeping in mind that as the "fitness" metric rises, the solution should approach optimality. For example for a problem in which the objective is to minimize a positive value, the natural choice for F(x) would be 1/x.
For the problem at hand, it seems that the fitness function has been given as F(x) = val(x)*val(x) though one cannot be certain from just a single value pair of (x,F(x)).
Roulette-wheel selection is just a commonly employed method of fitness-based pseudo-random selection. This is easy to understand if you've ever played roulette or watched 'Wheel of Fortune'.
Let us consider the simplest case, where F(x) = val(x),
Suppose we have four values, 1,2,3 and 4.
This implies that these "individuals" have fitnesses 1,2,3 and 4 respectively. Now the probability of selection of an individual 'x1' is calculated as F(x1)/(sum of all F(x)). That is to say here, since the sum of the fitnesses would be 10, the probabilities of selection would be, respectively, 0.1,0.2,0.3 and 0.4.
Now if we consider these probabilities from a cumulative perspective the values of x would be mapped to the following ranges of "probability:
1 ---> (0.0, 0.1]
2 ---> (0.1, (0.1 + 0.2)] ---> (0.1, 0.3]
3 ---> (0.3, (0.1 + 0.2 + 0.3)] ---> (0.3, 0.6]
4 ---> (0.6, (0.1 + 0.2 + 0.3 + 0.4)] ---> (0.6, 1.0]
That is, an instance of a uniformly distributed random variable generated, say R lying in the normalised interval, (0, 1], is four times as likely to be in the interval corresponding to 4 as to that corresponding to 1.
To put it another way, suppose you were to spin a roulette-wheel-type structure with each x assigned a sector with the areas of the sectors being in proportion to their respective values of F(x), then the probability that the indicator will stop in any given sector is directly propotional to the value of F(x) for that x.