This function takes a float then spits out the two integers for the decimal value. At least that was the intention
let flr (x:float) = float(int(x))
let f x =
let r y = let x = x * y in x = flr(x)
let rec f y =
if r(y)
then x*y,y
else f(y+1.0)
f 1.0
f 0.2;;
val it: float * float = (1.0, 5.0)
f 3.14;;
val it: float * float = (157.0, 50.0)
Here is an example where the integers, er will be integers eventually rather, have not been "simplified"
f 0.14;;
val it: float * float = (35.0, 250.0)
Checking the fractional part to be less than .01, as opposed to equaling exactly zero, got around this issue but I don't really like that solution. So I set it back to what you see in the code above. I am using the function below for some of the values that do not simplify though:
let g (x,y) =
let rec f k =
if x/k = flr(x/k)
then g(k)
else f(k-1.0)
and g k =
if y/k = flr(y/k)
then x/k,y/k
else f(k-1.0)
if x < y then f x else f y
Anyway, the main issue is this value:
3.142857143
Homeboy just keeps grinding without stack errors and I'm not sure what I've ran into here. Any clarity would be awesome! Thanks y'all.
Your algorithm is trying to find a rational number to represent a decimal number (represented as a floating point number).
For any input x, you are looking for a number represented as p/q such that x=p/q and you do this by incrementing q, starting from 1 and checking if you can find an integer p to make this work.
This works fine for numbers that have a nice rational representation like 0.2, but it does not work great for numbers like 3.142857 that do not have a simpler rational representation. For 3.142857, you will just keep iterating until you reach 3142857/1000000 (which is technically correct, but not very helpful).
As mentioned in the comments, there are issues caused by the fact that floating-point numbers cannot be precisely compared, but also, iterating like this for 3.142857143 might just take too long.
You can look up better algorithms for finding a rational number for a given decimal. You could also see if you can accept some margin of error. If you do not need a completely precise solution, you could for example change your r test function to something like:
let r y =
let x = x * y
x < flr(x) + 0.0001 && x > flr(x) + 0.0001
This will not give you exactly the same number, but it will likely find a solution that is good enough.
Related
Say you have two column vectors vv and ww, each with 7 elements (i.e., they have dimensions 7x1). Consider the following code:
z = 0;
for i = 1:7
z = z + v(i) * w(i)
end
A) z = sum (v .* w);
B) z = w' * v;
C) z = v * w;
D) z = w * v;
According to the solutions, answers (A) AND (B) are the right answers, can someone please help me understand why?
Why is z = v * w' which is similar to answer (B) but only the order of the operation changes, is false? Since we want a vector that by definition only has one column, wouldn't we need a matrix of this size: 1x7 * 7x1 = 1x1 ? So why is z = v' * w false ? It gives the same dimension as answer (B)?
z = v'*w is true and is equal to w'*v.
They both makes 1*1 matrix, which is a number value in octave.
See this:
octave:5> v = rand(7, 1);
octave:6> w = rand(7, 1);
octave:7> v'*w
ans = 1.3110
octave:8> w'*v
ans = 1.3110
octave:9> sum(v.*w)
ans = 1.3110
Answers A and B both perform a dot product of the two vectors, which yields the same result as the code provided. Answer A first performs the element-wise product (.*) of the two column vectors, then sums those intermediate values. Answer B performs the same mathematical operation but does so via a dot product (i.e., matrix multiplication).
Answer C is incorrect because it would be performing a matrix multiplication on misaligned matrices (7x1 and 7x1). The same is true for D.
z = v * w', which was not one of the options, is incorrect because it would yield a 7x7 matrix (instead of the 1x1 scalar value desired). The point is that order matters when performing matrix multiplication. (1xN)X(Nx1) -> (1x1), whereas (Nx1)X(1xN) -> (NxN).
z = v' * w is actually a correct solution but was simply not provided as one of the options.
Suppose I have an image A, I applied Gaussian Blur on it with Sigam=3 So I got another Image B. Is there a way to know the applied sigma if A,B is given?
Further clarification:
Image A:
Image B:
I want to write a function that take A,B and return Sigma:
double get_sigma(cv::Mat const& A,cv::Mat const& B);
Any suggestions?
EDIT1: The suggested approach doesn't work in practice in its original form(i.e. using only 9 equations for a 3 x 3 kernel), and I realized this later. See EDIT1 below for an explanation and EDIT2 for a method that works.
EDIT2: As suggested by Humam, I used the Least Squares Estimate (LSE) to find the coefficients.
I think you can estimate the filter kernel by solving a linear system of equations in this case. A linear filter weighs the pixels in a window by its coefficients, then take their sum and assign this value to the center pixel of the window in the result image. So, for a 3 x 3 filter like
the resulting pixel value in the filtered image
result_pix_value = h11 * a(y, x) + h12 * a(y, x+1) + h13 * a(y, x+2) +
h21 * a(y+1, x) + h22 * a(y+1, x+1) + h23 * a(y+1, x+2) +
h31 * a(y+2, x) + h32 * a(y+2, x+1) + h33 * a(y+2, x+2)
where a's are the pixel values within the window in the original image. Here, for the 3 x 3 filter you have 9 unknowns, so you need 9 equations. You can obtain those 9 equations using 9 pixels in the resulting image. Then you can form an Ax = b system and solve for x to obtain the filter coefficients. With the coefficients available, I think you can find the sigma.
In the following example I'm using non-overlapping windows as shown to obtain the equations.
You don't have to know the size of the filter. If you use a larger size, the coefficients that are not relevant will be close to zero.
Your result image size is different than the input image, so i didn't use that image for following calculation. I use your input image and apply my own filter.
I tested this in Octave. You can quickly run it if you have Octave/Matlab. For Octave, you need to load the image package.
I'm using the following kernel to blur the image:
h =
0.10963 0.11184 0.10963
0.11184 0.11410 0.11184
0.10963 0.11184 0.10963
When I estimate it using a window size 5, I get the following. As I said, the coefficients that are not relevant are close to zero.
g =
9.5787e-015 -3.1508e-014 1.2974e-015 -3.4897e-015 1.2739e-014
-3.7248e-014 1.0963e-001 1.1184e-001 1.0963e-001 1.8418e-015
4.1825e-014 1.1184e-001 1.1410e-001 1.1184e-001 -7.3554e-014
-2.4861e-014 1.0963e-001 1.1184e-001 1.0963e-001 9.7664e-014
1.3692e-014 4.6182e-016 -2.9215e-014 3.1305e-014 -4.4875e-014
EDIT1:
First of all, my apologies.
This approach doesn't really work in the practice. I've used the filt = conv2(a, h, 'same'); in the code. The resulting image data type in this case is double, whereas in the actual image the data type is usually uint8, so there's loss of information, which we can think of as noise. I simulated this with the minor modification filt = floor(conv2(a, h, 'same'));, and then I don't get the expected results.
The sampling approach is not ideal, because it's possible that it results in a degenerated system. Better approach is to use random sampling, avoiding the borders and making sure the entries in the b vector are unique. In the ideal case, as in my code, we are making sure the system Ax = b has a unique solution this way.
One approach would be to reformulate this as Mv = 0 system and try to minimize the squared norm of Mv under the constraint squared-norm v = 1, which we can solve using SVD. I could be wrong here, and I haven't tried this.
Another approach is to use the symmetry of the Gaussian kernel. Then a 3x3 kernel will have only 3 unknowns instead of 9. I think, this way we impose additional constraints on v of the above paragraph.
I'll try these out and post the results, even if I don't get the expected results.
EDIT2:
Using the LSE, we can find the filter coefficients as pinv(A'A)A'b. For completion, I'm adding a simple (and slow) LSE code.
Initial Octave Code:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = conv2(a, h, 'same');
% use non-overlapping windows to for the Ax = b syatem
% NOTE: boundry error checking isn't performed in the code below
s = floor(size(a)/2);
y = s(1);
x = s(2);
w = k*k;
y1 = s(1)-floor(w/2) + r;
y2 = s(1)+floor(w/2);
x1 = s(2)-floor(w/2) + r;
x2 = s(2)+floor(w/2);
b = [];
A = [];
for y = y1:k:y2
for x = x1:k:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
A = [A; f(:)'];
end
end
% estimated filter kernel
g = reshape(A\b, k, k)
LSE method:
clear all
im = double(imread('I2vxD.png'));
k = 5;
r = floor(k/2);
a = im(:, :, 1); % take the red channel
h = fspecial('gaussian', [3 3], 5); % filter with a 3x3 gaussian
filt = floor(conv2(a, h, 'same'));
s = size(a);
y1 = r+2; y2 = s(1)-r-2;
x1 = r+2; x2 = s(2)-r-2;
b = [];
A = [];
for y = y1:2:y2
for x = x1:2:x2
b = [b; filt(y, x)];
f = a(y-r:y+r, x-r:x+r);
f = f(:)';
A = [A; f];
end
end
g = reshape(A\b, k, k) % A\b returns the least squares solution
%g = reshape(pinv(A'*A)*A'*b, k, k)
I searched around and it turns out the answer to this is surprising hard to find. Theres algorithm out there that can generate a random orientation in quaternion form but they involve sqrt and trig functions. I dont really need a uniformly distributed orientation. I just need to generate (many) quaternions such that their randomness in orientation is "good enough." I cant specify what is "good enough" except that I need to be able to do the generation quickly.
Quoted from http://planning.cs.uiuc.edu/node198.html:
Choose three points u, v, w ∈ [0,1] uniformly at random. A uniform, random quaternion is given by the simple expression:
h = ( sqrt(1-u) sin(2πv), sqrt(1-u) cos(2πv), sqrt(u) sin(2πw), sqrt(u) cos(2πw))
From Choosing a Point from the Surface of a Sphere by George Marsaglia:
Generate independent x, y uniformly in (-1..1) until z = x²+y² < 1.
Generate independent u, v uniformly in (-1..1) until w = u²+v² < 1.
Compute s = √((1-z) / w).
Return the quaternion (x, y, su, sv). It's already normalized.
This will generate a uniform random rotation because 4D spheres, unit quaternions and 3D rotations have equivalent measures.
The algorithm uses one square root, one division, and 16/π ≈ 5.09 random numbers on average. C++ code:
Quaternion random_quaternion() {
double x,y,z, u,v,w, s;
do { x = random(-1,1); y = random(-1,1); z = x*x + y*y; } while (z > 1);
do { u = random(-1,1); v = random(-1,1); w = u*u + v*v; } while (w > 1);
s = sqrt((1-z) / w);
return Quaternion(x, y, s*u, s*v);
}
Simplest way to generate it, just generate 4 random float and normalize it if required. If you want to produce rotation matrices later , than normalization can be skipped and convertion procedure should note nonunit quaternions.
According to the TryF#.org site this function below returns quadruple of the number entered.
let quadruple x =
let double x = x * 2
double(double(x))
Can anyone explain why as I interpret it as like follows? Quadruple doesn't perform any mutation or multiple calls.
function quadruple(x)
return function double(x)
return x * 2
or C#
int a(int x) { return b(x); }
int b(int x) { return x * 2; }
I think this is just a confused indentation. The function should probably look like this:
let quadruple x =
let double x = x * 2
double(double(x))
This should hopefully make more sense - the quadruple function defines a function double and then calls it on the input x (multiplying it by 2) and then applies double on the result, multiplying it by 2 again, so the result is (x * 2) * 2.
Using the indentation in your sample, the code would not compile, because it is not syntactically valid (a function body cannot end with a let line - it needs to end with an expression representing some result to be returned).
I am developing a program that solves a system of equations. When it gives me the results, it is like: "x1= 1,36842". I'd like to get the fraction of that "1,36842", so I wrote this code.
procedure TForm1.Button1Click(Sender: TObject);
var numero,s:string;
a,intpart,fracpart,frazfatta:double;
y,i,mcd,x,nume,denomin,R:integer;
begin
a:=StrToFloat(Edit1.Text); //get the value of a
IntPart := Trunc(a); // here I get the numerator and the denominator
FracPart := a-Trunc(a);
Edit2.Text:=FloatToStr(FracPart);
numero:='1';
for i:= 1 to (length(Edit2.Text)-2) do
begin
numero:=numero+'0';
end; //in this loop it creates a string that has many 0 as the length of the denominator
Edit3.text:=FloatToStr(IntPart);
y:=StrToInt(numero);
x:=StrToInt(Edit3.Text);
while y <> 0 do
begin
R:= x mod y;
x:=y;
y:=R;
end;
mcd:=x; //at the end of this loop I have the greatest common divisor
nume:= StrToInt(Edit3.Text) div mcd;
denomin:= StrToInt(numero) div mcd;
Memo1.Lines.Add('fraction: '+IntToStr(nume)+'/'+IntToStr(denomin));
end;
It doesn't work correctly because the fraction that it gives to me is wrong. Could anyone help me please?
Your code cannot work because you are using binary floating point. And binary floating point types cannot represent the decimal numbers that you are trying to represent. Representable binary floating point numbers are of the form s2e where s is the significand and e is the exponent. So, for example, you cannot represent 0.1 as a binary floating point value.
The most obvious solution is to perform the calculation using integer arithmetic. Don't call StrToFloat at all. Don't touch floating point arithmetic. Parse the input string yourself. Locate the decimal point. Use the number of digits that follow to work out the decimal scale. Strip off any leading or trailing zeros. And do the rest using integer arithmetic.
As an example, suppose the input is '2.79'. Convert that, by processing the text, into numerator and denominator variables
Numerator := 279;
Denominator := 100;
Obviously you'd have to code string parsing routines rather than use integer literals, but that is routine.
Finally, complete the problem by finding the gcd of these two integers.
The bottom line is that to represent and operate on decimal data you need a decimal algorithm. And that excludes binary floating point.
I recommend defining a function GreaterCommonDivisor function first (wiki reference)
This is going to be Java/C like code since I'm not familiar with Delphi
let
float x = inputnum // where inputnum is a float
// eg. x = 123.56
Then, multiplying
int n = 1;
while(decimalpart != 0){// or cast int and check if equal-> (int)x == x
x = x * 10;
decimalpart = x % 1;
// or a function getting the decimal part if the cast does work
n *= 10;
}
// running eg. x = 123.56 now x = 12356
// n = 100
Then you should have (float)x/n == inputnum at this point eg. (12356/100 == 123.56)
This mean you have a fraction that may not be simpified at this point. All you do now is implement and use the GCD function
int gcd = GreaterCommonDivisor(x, n);
// GreaterCommonDivisor(12356, 100) returns 4
// therefore for correct implementation gcd = 4
x /= gcd; // 12356 / 4 = 3089
n /= gcd; // 100 / 4 = 25
This should be quick and simple to implement, but:
Major Pitfalls:
Float must be terminating. For example expected value for 0.333333333333333333 won't be rounded to 1/3
Float * n <= max_int_value, otherwise there will be a overflow, there are work around this, but there may be another solutions more fitting to these larger numbers
Continued fractions can be used to find good rational approximations to real numbers. Here's an implementation in JavaScript, I'm sure it's trivial to port to Delphi:
function float2rat(x) {
var tolerance = 1.0E-6;
var h1=1; var h2=0;
var k1=0; var k2=1;
var b = x;
do {
var a = Math.floor(b);
var aux = h1; h1 = a*h1+h2; h2 = aux;
aux = k1; k1 = a*k1+k2; k2 = aux;
b = 1/(b-a);
} while (Math.abs(x-h1/k1) > x*tolerance);
return h1+"/"+k1;
}
For example, 1.36842 is converted into 26/19.
You can find a live demo and more information about this algorithm on my blog.
#Joni
I tried 1/2 and the result was a "division by zero" error;
I correct the loop adding:
if b - a = 0 then BREAK;
To avoid
b:= 1 / (b - a);