How can I get equation numbers added to asciimath:[] expression?
That is, how can I get the '(5)' in the formula such as
E = m c^2 (5)
Is there a method to refer to equation numbers, similar to LaTeX?
Related
I am looking into a octave/matlab code and find the following:
deltaT = 1; % sampling period for data
......
R = rcValues(2:2:end)/1000; % convert these also
C = rcValues(3:2:end)*1000; % convert kF to F
RCfact = exp(-deltaT./(R.*C));
What does the point (.) mean in -deltaT. and R. in this mathematical expression?
Thanks
The dot operator is used to execute an operation on each element of a matrix. In your case, if deltaT and R are single elements, using the dot operator doesn't do anything. HOWEVER, if they were a matrix, then the operation would've been executed in each element of the matrix.
The operator is used with multiplication, division, and exponentiation.
For more info visit https://www.mathworks.com/matlabcentral/answers/506078-please-help-me-understand-the-use-of-dot-operator#accepted_answer_416043
Effectively what I'm looking for is a function f(x) that outputs into a range that is pre-defined. Calling f(f(x)) should be valid as well. The function should be cyclical, so calling f(f(...(x))) where the number of calls is equal to the size of the range should give you the original number, and f(x) should not be time dependent and will always give the same output.
While I can see that taking a list of all possible values and shuffling it would give me something close to what I want, I'd much prefer it if I could simply plug values into the function one at a time so that I do not have to compute the entire range all at once.
I've looked into Minimal Perfect Hash Functions but haven't been able to find one that doesn't use external libraries. I'm okay with using them, but would prefer to not do so.
If an actual range is necessary to help answer my question, I don't think it would need to be bigger than [0, 2^24-1], but the starting and ending values don't matter too much.
You might want to take a look at Linear Congruential Generator. You shall be looking at full period generator (say, m=224), which means parameters shall satisfy Hull-Dobell Theorem.
Calling f(f(x)) should be valid as well.
should work
the number of calls is equal to the size of the range should give you the original number
yes, for LCG with parameters satisfying Hull-Dobell Theorem you'll get full period covered once, and 'm+1' call shall put you back at where you started.
Period of such LCG is exactly equal to m
should not be time dependent and will always give the same output
LCG is O(1) algorithm and it is 100% reproducible
LCG is reversible as well, via extended Euclid algorithm, check Reversible pseudo-random sequence generator for details
Minimal perfect hash functions are overkill, all you've asked for is a function f that is,
bijective, and
"cyclical" (ie fN=f)
For a permutation to be cyclical in that way, its order must divide N (or be N but in a way that's just a special case of dividing N). Which in turn means the LCM of the orders of the sub-cycles must divide N. One way to do that is to just have one "sub"-cycle of order N. For power of two N, it's also really easy to have lots of small cycles of some other power-of-two order. General permutations do not necessarily satisfy the cycle-requirement, of course they are bijective but the LCM of the orders of the sub-cycles may exceed N.
In the following I will leave all reduction modulo N implicit. Without loss of generality I will assume the range starts at 0 and goes up to N-1, where N is the size of the range.
The only thing I can immediately think of for general N is f(x) = x + c where gcd(c, N) == 1. The GCD condition ensures there is only one cycle, which necessarily has order N.
For power-of-two N I have more inspiration:
f(x) = cx where c is odd. Bijective because gcd(c, N) == 1 so c has a modular multiplicative inverse. Also cN=1, because φ(N)=N/2 (since N is a power of two) so cφ(N)=1 (Euler's theorem).
f(x) = x XOR c where c < N. Trivially bijective and trivially cycles with a period of 2, which divides N.
f(x) = clmul(x, c) where c is odd and clmul is carry-less multiplication. Bijective because any odd c has a carry-less multiplicative inverse. Has some power-of-two cycle length (less than N) so it divides N. I don't know why though. This is a weird one, but it has decent special cases such as x ^ (x << k). By symmetry, the "mirrored" version also works.
Eg x ^ (x >> k).
f(x) = x >>> k where >>> is bit-rotation. Obviously bijective, and fN(x) = x >>> Nk, where Nk mod N = 0 so it rotates all the way back to the unrotated position regardless of what k is.
I want to calculate the integration of a matrix over a path. This matrix is in fact dependent on two variables. the answer of this integral would be a vector. it is:
Fn=integral(-(q ) Wn dГ)
q is a constant. Wn is a 2D matrix, N*n, which N is the number of the points (x,y) and n is the number of source points which create element of function and refers to different columns of this matrix. for example W2(1,2) is the matrix function value at point (x1,y1) for the source n=2.
I cannot use "trapz" for calculation of this integral, because in trapz(X,Y) the X should be a vector but in my case the function Wn is dependent on two variable (x,y), So the X in trapz would be a matrix instead of a vector.
how can I calculate this integral?
also, how should I implement the path in the calculation of my integral. My current path for integral calculation is a vertical line at x=0, 0
so many thanks in advance.
I found the answer. I should devide the boundary to strait lines between each 2 node,then calculate the integral using gauss-lojander method.
I have been trying to reverse a quite simple looking function.
the function is presented in assembly:
(Argument is loaded into AX)
AND AX, 0xFFFE (round down to even number)
MUL AX (Multiply AX by AX ; the result is represented as DX:AX)
XOR AX,DX
The function can be described as: H(X) = F(X & 0xFFFE); F(X) = ((X * X) mod 2^16) xor ((X * X) div 2^16)
Calculated all of the values from 1 to 2^16 and plotted on matlab in order to "see" some function.
Can anyone help me find an answer to this? (when given y what is the argument x).
It might be that for some values there is more than one answer, so narrowing it down is my goal.
Thanks,
Or.
It's a hash function.
You can't reverse a hash function, because the whole point of it is that it's a one way function.
The multiply is clearly reversible, it's the xor that's not. By combining the low and high part of the multiplication you lose information.
As you can see in the plot there are some white spaces, because there are 2^16 spaces in that plot that means there are also different input values that hash to the same value.
This is common in a hash function.
The only way to 'reverse' it is to build a lookup table that translates output values into possible input values. However you will find that for every output values that be 1 or more input values.
An even number x an even number is always a multiple of 4.
So the low 2 bits are always 0, ergo the low 2 bits of the result are bits 16+17 of the multiplication.
Bits 2..15 are a mix of bits 2..15 xor bits 18..31.
A quick simulation shows 24350 unique outputs ergo on average 1.34 0.34 duplicates for every input value, not bad.
The maximum number of collisions is 6, but most numbers don't collide.
For all those numbers that don't collide you can uniquely lookup your input value in the lookup table (all this disregarding odd input values obviously).
I am unsure how to use the Distributive property on the following function:
F = B'D + A'D + BD
I understand that F = xy + x'z would become (xy + x')(xy + z) but I'm not sure how to do this with three terms with two variables.
Also another small question:
I was wondering how to know what number a minterm is without having to consult (or memorise) the table of minterms.
For example how can I tell that xy'z' is m4?
When you're trying to use the distributive property there, what you're doing is converting minterms to maxterms. This is actually very related to your second question.
To tell that xy'z' is m4, think of function as binary where false is 0 and true is 1. xy'z' then becomes 100, binary for the decimal 4. That's really what a k-map/minterm table is doing for you to give a number.
Now an important extension of this: the number of possible combinations is 2^number of different variables. If you have 3 variables, there are 2^3 or 8 different combinations. That means you have min/maxterm possible numbers from 0-7. Here's the cool part: anything that isn't a minterm is a maxterm, and vice versa.
So, if you have variables x and y, and you have the expression xy', you can see that as 10, or m2. Because the numbers go from 0-3 with 2 variables, m2 implies M0, M1, and M3. Therefore, xy'=(x+y)(x+y')(x'+y').
In other words, the easiest way to do the distributive property in either direction is to note what minterm or maxterm you're dealing with, and just switch it to the other.
For more info/different wording.