This is my first stab at machine learning, and I can implement the code anyway that I want. I have Matlab access, which I think will be simpler than Python, and I have pseudo code for implementing a PLA. The last part of the code, however, absolutely baffles me, though it is simpler than the code I have seen on here thus far. It seems to be calling for the use of variables not declared. Here's what I have. I'll point out the number line at which I get stuck.
1) w <- (n + 1) X m (matrix of small random nums)
2) I <- I augmented with col. of 1s
3) for 1 = 1 to 1000
4) delta_W = (N + 1) X m (matrix of zeros) // weight changes
5) for each pattern 1 <= j <= p
6) Oj = (Ij * w) > 0 // j's are subscript/vector matrix product w/ threshold
7) Dj = = Tj - Oj // diff. between target and actual
8) w = w + Ij(transpose)*Dj // the learning rule
Lines 1 thru 4 are coded.
My questions are on line 5: What does "for each pattern mean" (i.e., how does one say it in code). Also, which j are they interested in, I have a j in the observation matrix and a j in the target matrix. Also, where did "p" come from (I have i's, j's, m's and n's but no p's)? Any thoughts would be appreciated.
"for each pattern" refers to the inputs. All they are saying is to run that loop where Ij is the input to the perceptron.
To write this in MATLAB, it really depends on how your data is oriented. I would store your inputs as a mXn matrix, where m is the number of inputs and n is the size of each input.
Say our inputs look like :
input = [1 5 -1;
2 3 2;
4 5 6;
... ]
First 'augment' this with a column of ones for the bias input:
[r c] = size(input);
input = [input ones(r,1)];
Then, your for loop will simply be:
for inputNumber = 1:r
pattern = input(inputNumber,:);
and you can continue from there.
Related
Looking within the xcorr function, most of it is pretty straightforward, except for one function within xcorr called "findTransformLength".
function m = findTransformLength(m)
m = 2*m;
while true
r = m;
for p = [2 3 5 7]
while (r > 1) && (mod(r, p) == 0)
r = r / p;
end
end
if r == 1
break;
end
m = m + 1;
end
With no comments, i fail to understand what this function is meant to acheive and what is the significance of p = [2 3 5 7]. Why those numbers specifically? Why not take a fixed FFT size instead? Is there a disadvantage(cause errors) to taking a fixed FFT size?
This part is used to get the integer closest to 2*m that can be written in the form:
Either:
m is already of this form, then the loop
for p = [2 3 5 7]
while (r > 1) && (mod(r, p) == 0)
r = r / p;
end
end
Will decrease r down to 1 and the break will be reached.
Or m has at least one other prime factor, and r will not reach 1. You go back to the look with m+1 and so on until you reach a number of the right form.
As per why they do this, you can see on the fft doc, in the Input arguments section:
n — Transform length [] (default) | nonnegative integer scalar
Transform length, specified as [] or a nonnegative integer scalar.
Specifying a positive integer scalar for the transform length can
increase the performance of fft. The length is typically specified as
a power of 2 or a value that can be factored into a product of small
prime numbers. If n is less than the length of the signal, then fft
ignores the remaining signal values past the nth entry and returns the
truncated result. If n is 0, then fft returns an empty matrix.
Example: n = 2^nextpow2(size(X,1))
I am working on MATLAB on my own, and was doing problem 9 on Project Euler
It states
" A Pythagorean triplet is a set of three natural numbers, a < b < c, for which,
a2 + b2 = c2
For example, 32 + 42 = 9 + 16 = 25 = 52.
There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc."
Below is the code I wrote; however, it compiles, but does not produce and output. I was hoping to get some feedback on what's wrong, so I can fix it.
Thanks,
syms a;
syms b;
syms c;
d= 1000;
d= a + b + c ;
ab= a.^2 + b.^2;
ab= c.^2;
c
I propose a vectorized way (that is, without using loops) to solve the problem. It may seem relatively complicated, especially if you come from other programming languages; but for Matlab you should get used to this way of approaching problems.
Ingredients:
Vectorization;
Indexing;
Transpose;
Implicit singleton expansion;
hypot;
find.
Read up on these concepts if you are not familiar with them, and then try to solve the problem yourself (which of course is the whole point of Project Euler). As a hint, the code below proceeds along these lines:
Generate a 1×1000 vector containing all possible values for a and b.
Compute a 1000×1000 matrix with the values of c corresponding to each pair a, b
From that compute a new matrix such that each entry contains a+b+c
Find the row and column indices where that matrix equals 1000. Those indices are the desired a and b (why?).
You'll get more than one solution (why?). Pick one.
Compute the product of the obtained a and b and the corresponding c.
Once you have tried yourself, you may want to check the code (move the mouse over it):
ab = 1:1000; % step 1
cc = hypot(ab,ab.'); % step 2
sum_abc = ab+ab.'+cc; % step 3
[a, b] = find(sum_abc==1000); % step 4
a = a(1); b = b(1); % step 5
prod_abc = a*b*cc(a,b); % step 6
So I am very new to Matlab and I have been tasked with implementing LU factorization. I have to do it recursively because we are not allowed to use for loops in our code, and recursion will give us optimal marks. I have written this code. The code works for the first step, and does what it is supposed to, but for the next two steps, the matrix is not modified at all. I have this code:
function[L, U] = myLU(A,B, pos)
%A = Mtrix that becomes U
%B = Matrix that becomes L
tmp_L = B;
[x,y] = size(A);
if pos > x
L = B;
U = A;
return
else
pos %<-- just to see if it iterates through everything
[tmp_U,tmp_L] = elimMat(A,pos);
myLU(tmp_U,tmp_L, pos+1);
end
L = tmp_L;
U = tmp_U;
end
I where elimMat(A, pos) returns the elimination matrix for column pos. as well as another matrix, which will end up being the matrix of multipliers. What i tried doing is then finding the LU factorization of this matrix A. since elimMat returns L and U(this works, if i do it manually it works), i had to make a function that allows me to do it automatically without using a for loop. I thought i would do it recursively. What i ended up doing is adding another variable B to the function so that i can store intermediate values of my matrix obtained in each step and put it all together later.
So here is my question. Am i implementing the recursion wrong? and if i am how can i fix it? The other thing i wanted to ask is how can i implement this so i do not need variable B as an additional imput, and only use the existing variables, or variables previously defined, to find the solution? I would really like only two inputs in my function: The matrix name and the starting index.
here is elimMat if if helps:
function [M,L] = elimMat(A,k)
%find the size of the matrix
[x,y] = size(A);
tmp_mat = zeros(x,y);
%M = The current matrix we are working on for Elimination -> going to
%become U.
%L = The L part of the matrix we are working on. Contains all the
%multipliers. This is going to be our L matrix.
for i = 1:x
mult = A(i,k)/A(k,k);
if i > k
tmp_mat(i,k) = mult;
P = A(k,:)*mult;
A(i,:) = A(i,:)-P;
elseif i == k
tmp_mat(k,k) = 1;
end
end
M = A;
L = tmp_mat;
end
thanks for any feedback you can provide.
Here is the output: WHAT I GET VS what it should be:
[U = VS [U =
1 2 2 1 2 2
0 -4 -6 0 -4 -6
0 -2 -4] 0 0 2
L = VS [L=
1 0 0 1 0 0
4 0 0 4 1 0
4 0 0] 4 0.5 1
As you can see only the first column is changed
You forgot to catch the output of your recursive call:
[tmp_L, tmp_U] = myLU(tmp_U,tmp_L, pos+1);
Matlab passes variables by value, so a function cannot change its input variable itself (ok, it can, but it's tricky and unsafe).
Your original version didn't return the updated matrices, so the outermost function call encountered the myLU() call, let the recursion unfold and finish, and then went on to use tmp_L and tmp_U as returned from the very first call to elimMAT(A,1).
Note that you might want to standardize your functions such that they return U and L in the same order to avoid confusion.
Problem : How do I use a continuous map - The Link1: Bernoulli Shift Map to model binary sequence?
Concept :
The Dyadic map also called as the Bernoulli Shift map is expressed as x(k+1) = 2x(k) mod 1. In Link2: Symbolic Dynamics, explains that the Bernoulli Map is a continuous map and is used as the Shift Map. This is explained further below.
A numeric trajectory can be symbolized by partitioning into appropriate regions and assigning it with a symbol. A symbolic orbit is obtained by writing down the sequence of symbols corresponding to the successive partition elements visited by the point in its orbit. One can learn much about the dynamics of the system by studying its symbolic orbits. This link also says that the Bernoulli Shift Map is used to represent symbolic dynamics.
Question :
How is the Bernoulli Shift Map used to generate the binary sequence? I tried like this, but this is not what the document in Link2 explains. So, I took the numeric output of the Map and converted to symbols by thresholding in the following way:
x = rand();
y = mod(2* x,1) % generate the next value after one iteration
y =
0.3295
if y >= 0.5 then s = 1
else s = 0
where 0.5 is the threshold value, called the critical value of the Bernoulli Map.
I need to represent the real number as fractions as explained here on Page 2 of Link2.
Can somebody please show how I can apply the Bernoulli Shift Map to generate symbolized trajectory (also called time series) ?
Please correct me if my understanding is wrong.
How do I convert a real valued numeric time series into symbolized i.e., how do I use the Bernoulli Map to model binary orbit /time series?
You can certainly compute this in real number space, but you risk hitting precision problems (depending on starting point). If you're interested in studying orbits, you may prefer to work in a rational fraction representation. There are more efficient ways to do this, but the following code illustrates one way to compute a series derived from that map. You'll see the period-n definition on page 2 of your Link 2. You should be able to see from this code how you could easily work in real number space as an alternative (in that case, the matlab function rat will recover a rational approximation from your real number).
[EDIT] Now with binary sequence made explicit!
% start at some point on period-n orbit
period = 6;
num = 3;
den = 2^period-1;
% compute for this many steps of the sequence
num_steps = 20;
% for each step
for n = 1:num_steps
% * 2
num = num * 2;
% mod 1
if num >= den
num = num - den;
end
% simplify rational fraction
g = gcd(num, den);
if g > 1
num = num / g;
den = den / g;
end
% recover 8-bit binary representation
bits = 8;
q = 2^bits;
x = num / den * q;
b = dec2bin(x, bits);
% display
fprintf('%4i / %4i == 0.%s\n', num, den, b);
end
Ach... for completeness, here's the real-valued version. Pure mathematicians should look away now.
% start at some point on period-n orbit
period = 6;
num = 3;
den = 2^period-1;
% use floating point approximation
x = num / den;
% compute for this many steps of the sequence
num_steps = 20;
% for each step
for n = 1:num_steps
% apply map
x = mod(x*2, 1);
% display
[num, den] = rat(x);
fprintf('%i / %i\n', num, den);
end
And, for extra credit, why is this implementation fast but daft? (HINT: try setting num_steps to 50)...
% matlab vectorised version
period = 6;
num = 3;
den = 2^period-1;
x = zeros(1, num_steps);
x(1) = num / den;
y = filter(1, [1 -2], x);
[a, b] = rat(mod(y, 1));
disp([a' b']);
OK, this is supposed to be an answer, not a question, so let's answer my own questions...
It's fast because it uses Matlab's built-in (and highly optimised) filter function to handle the iteration (that is, in practice, the iteration is done in C rather than in M-script). It's always worth remembering filter in Matlab, I'm constantly surprised by how it can be turned to good use for applications that don't look like filtering problems. filter cannot do conditional processing, however, and does not support modulo arithmetic, so how do we get away with it? Simply because this map has the property that whole periods at the input map to whole periods at the output (because the map operation is multiply by an integer).
It's daft because it very quickly hits the aforementioned precision problems. Set num_steps to 50 and watch it start to get wrong answers. What's happening is the number inside the filter operation is getting to be so large (order 10^14) that the bit we actually care about (the fractional part) is no longer representable in the same double-precision variable.
This last bit is something of a diversion, which has more to do with computation than maths - stick to the first implementation if your interest lies in symbol sequences.
If you only want to deal with rational type of output, you'll first have to convert the starting term of your series into a rational number if it is not. You can do that with:
[N,D] = rat(x0) ;
Once you have a numerator N and a denominator D, it is very easy to calculate the series x(k+1)=mod(2*x(k), 1) , and you don't even need a loop.
for the part 2*x(k), it means all the Numerator(k) will be multiplied by successive power of 2, which can be done by matrix multiplication (or bsxfun for the lover of the function):
so 2*x(k) => in Matlab N.*(2.^(0:n-1)) (N is a scalar, the numerator of x0, n is the number of terms you want to calculate).
The Mod1 operation is also easy to translate to rational number: mod(x,1)=mod(Nx,Dx)/Dx (Nx and Dx being the numerator and denominator of x.
If you do not need to simplify the denominator, you could get all the numerators of the series in one single line:
xn = mod( N.*(2.^(0:n-1).'),D) ;
but for visual comfort, it is sometimes better to simplify, so consider the following function:
function y = dyadic_rat(x0,n)
[N,D] = rat(x0) ; %// get Numerator and Denominator of first element
xn = mod( N.*(2.^(0:n-1).'),D) ; %'// calculate all Numerators
G = gcd( xn , D ) ; %// list all "Greatest common divisor"
y = [xn./G D./G].' ; %'// output simplified Numerators and Denominators
If I start with the example given in your wiki link (x0=11/24), I get:
>> y = dyadic_rat(11/24,8)
y =
11 11 5 2 1 2 1 2
24 12 6 3 3 3 3 3
If I start with the example given by Rattus Ex Machina (x0=3/(2^6-1)), I also get the same result:
>> y = dyadic_rat(3/63,8)
y =
1 2 4 8 16 11 1 2
21 21 21 21 21 21 21 21
In Matlab, suppose there is a vector whose elements can be complex or real. I was wondering how to remove the nonreal elements, and consequently reduce the size of the vector? Thanks and regards!
Use the REAL and IMAG functions:
>> x = [1+i; 4+3i; 5+6i]
x =
1 + 1i
4 + 3i
5 + 6i
>> real(x)
ans =
1
4
5
>> imag(x)
ans =
1
3
6
EDIT
The above doesn't answer the poster's question. This does.
Use the FIND and REAL functions:
>> v = [1+i; 2; 3]
v =
1 + 1i
2
3
>> v(v == real(v))
ans =
2
3
You can also avoid testing in a loop using Matlab's vector syntax:
x = [1, 2, 3+1i, 4+2i, 5+3i, 6, 7, 8+4i, 9+0.000001i]
y = x(imag(x) == 0);
z = real(x(abs(imag(x)) < 0.00001));
y should be [1,2,6,7] and z should be [1,2,6,7,9]
imag(x) == 0 yields a logical vector with true values whenever the equality condition holds. x(<logical vector>) yields a new vector with only the elements of x where <logical vector> was true.
That's a very unusual thing to ask. Since the imaginary part is a floating point number, you can't really tell which number is a real number and which number is very close to a real number. Such function doesn't exist in Matlab since it's not very clear how it would be useful for anything (it doesn't make much sense to remove those numbers). Specifying your actual purpose here might help you get better answers.
If you want to ensure that only real numbers are left in the vector, you can use the following (it doesn't work with matrices and vertical rows, but you've got the idea):
x = [1, 2, 3+1i, 4+2i, 5+3i, 6, 7, 8+4i, 9+0i]
z = []
for k = [1:length(x)]
if imag(x(k)) == 0
z = [z, real(x(k))]
endif
endfor
If you want to keep all numbers that are close to a real number, but could have some small non-zero imaginary part, you can use the following:
x = [1, 2, 3+1i, 4+2i, 5+3i, 6, 7, 8+4i, 9+0.000001i]
z = []
for k = [1:length(x)]
if abs(imag(x(k))) < 0.00001
z = [z, real(x(k))]
endif
endfor
Of course, if you tell us what your actual criterion is, it would be much easier to give you a better idea. Are you looking for the real solutions to some sort of equation or system of equations, real roots of a polynomial? In this case, the first one might miss a real solution because of the approximation error, and the second one can give you things that aren't solutions.