In JAGS I'd like to define a Poisson distribution for parameter w[i] which is also truncated (greater than or equal to 2) if another parameter, e[i], is greater than 0.
Essentially I want it to represent:
w[i] ~ ifelse( e[i] > 0, dpois(mu) T(2,) , dpois(mu) )
I've tried using the step function by adapting the code that was given in response to someone else's post which was requesting something similar: Choosing Different Distributions based on if - else condition in WinBugs/JAGS
But this doesn't seem to work?
Thank you
Maybe something like this?
pois1 ~ dpois(mu) T(2,)
pois2 ~ dpois(mu)
for(i in 1:N){
indicator1[i] <- ifelse(e[i] > 0, 1, 0)
indicator2[i] <- ifelse(e[i] <= 0, 1, 0)
w[i] <- (pois1 * indicator1[i]) + (pois2 * indicator2[i])
}
When e[i] is greater than 1 w[i] takes the value from pois1. If it is not w[i] takes the value from pois2.
EDIT: Or, you could define only one indicator variable and do it like this.
pois1 ~ dpois(mu) T(2,)
pois2 ~ dpois(mu)
for(i in 1:N){
indicator[i] <- ifelse(e[i] > 0, 1, 0)
w[i] <- (pois1 * indicator[i]) + (pois2 * (1 - indicator[i]))
}
You could try this
w[i] ~ dpois(mu) T(ifelse(e[i] > 0), 2, 0), )
A lower bound of 0 on the Poisson distribution is equivalent to not having a lower bound.
Related
Consider there are N houses on a single road. I have M lightpoles. Given that M < N. Distance between all adjacent houses are different. Lightpole can be placed at the house only. And I have to place all lightpoles at house so that sum of distances from each house to its nearest lightpole is smallest. How can I code this problem?
After a little research I came to know that I have to use dynamic programming for this problem. But I don't know how to approach it to this problem.
Here's a naive dynamic program with search space O(n^2 * m). Perhaps others know of another speedup? The recurrence should be clear from the function f in the code.
JavaScript code:
// We can calculate these in O(1)
// by using our prefixes (ps) and
// the formula for a subarray, (j, i),
// reaching for a pole at i:
//
// ps[i] - ps[j-1] - (A[i] - A[j-1]) * j
//
// Examples:
// A: [1,2,5,10]
// ps: [0,1,7,22]
// (2, 3) =>
// 22 - 1 - (10 - 2) * 2
// = 5
// = 10-5
// (1, 3) =>
// 22 - 0 - (10 - 1) * 1
// = 13
// = 10-5 + 10-2
function sumParts(A, j, i, isAssigned){
let result = 0
for (let k=j; k<=i; k++){
if (isAssigned)
result += Math.min(A[k] - A[j], A[i] - A[k])
else
result += A[k] - A[j]
}
return result
}
function f(A, ps, i, m, isAssigned){
if (m == 1 && isAssigned)
return ps[i]
const start = m - (isAssigned ? 2 : 1)
const _m = m - (isAssigned ? 1 : 0)
let result = Infinity
for (let j=start; j<i; j++)
result = Math.min(
result,
sumParts(A, j, i, isAssigned)
+ f(A, ps, j, _m, true)
)
return result
}
var A = [1, 2, 5, 10]
var m = 2
var ps = [0]
for (let i=1; i<A.length; i++)
ps[i] = ps[i-1] + (A[i] - A[i-1]) * i
var result = Math.min(
f(A, ps, A.length - 1, m, true),
f(A, ps, A.length - 1, m, false))
console.log(`A: ${ JSON.stringify(A) }`)
console.log(`ps: ${ JSON.stringify(ps) }`)
console.log(`m: ${ m }`)
console.log(`Result: ${ result }`)
I got you covered bud. I will write to explain the dynamic programming algorithm first and if you are not able to code it, let me know.
A-> array containing points so that A[i]-A[i-1] will be the distance between A[i] and A[i-1]. A[0] is the first point. When you are doing memoization top-down, you will have to handle cases when you would want to place a light pole at the current house or you would want to place it at a lower index. If you place it now, you recurse with one less light pole available and calculate the sum of distances with previous houses. You handle the base case when you are not left with any ligh pole or you are done with all the houses.
double k = 0;
int l = 1;
double digits = pow(0.1, 5);
do
{
k += (pow(-1, l - 1)/l);
l++;
} while((log(2)-k)>=digits);
I'm trying to write a little program based on an example I seen using a series of Σ_(l=1) (pow(-1, l - 1)/l) to estimate log(2);
It's supposed to be a guess refinement thing where time it gets closer and closer to the right value until so many digits match.
The above is what I tried but but it's not coming out right. After messing with it for quite a while I can't figure out where I'm messing up.
I assume that you are trying to extimate the natural logarithm of 2 by its Taylor series expansion:
∞ (-1)n + 1
ln(x) = ∑ ――――――――(x - 1)n
n=1 n
One of the problems of your code is the condition choosen to stop the iterations at a specified precision:
do { ... } while((log(2)-k)>=digits);
Besides using log(2) directly (aren't you supposed to find it out instead of using a library function?), at the second iteration (and for every other even iteration) log(2) - k gets negative (-0.3068...) ending the loop.
A possible (but not optimal) fix could be to use std::abs(log(2) - k) instead, or to end the loop when the absolute value of 1.0 / l (which is the difference between two consecutive iterations) is small enough.
Also, using pow(-1, l - 1) to calculate the sequence 1, -1, 1, -1, ... Is really a waste, especially in a series with such a slow convergence rate.
A more efficient series (see here) is:
∞ 1
ln(x) = 2 ∑ ――――――― ((x - 1) / (x + 1))2n + 1
n=0 2n + 1
You can extimate it without using pow:
double x = 2.0; // I want to calculate ln(2)
int n = 1;
double eps = 0.00001,
kpow = (x - 1.0) / (x + 1.0),
kpow2 = kpow * kpow,
dk,
k = 2 * kpow;
do {
n += 2;
kpow *= kpow2;
dk = 2 * kpow / n;
k += dk;
} while ( std::abs(dk) >= eps );
I am trying to write my own codes for Gaussian pyramid using c++.
I tried both reduce and expand equations as stated in http://persci.mit.edu/pub_pdfs/pyramid83.pdf, the equation (1) and (2). However, my array index is out of bounds when I am trying to access
[2i + m][2j + n] and [(i - m) / 2][(j - n) / 2], respectively.
My Gaussian kernel: the 5x5 matrix; g1Image: the original image reduced by 1 level, both row and column are half of the dimensions of the original image's.
My m and n are set to -2 < m/n <= 2, thus when i access my Gaussian kernel, i add 2 to the index, becoming
w[m + 2][n + 2] * original_image[2i + m][2j + n]
I did try to set my m and n to 0 < m/n <=4 as well, equation becomes
w[m][n] * original_image[2i + m][2j + n] or w[m][n] * original_image[2i + m - 2][2j + n - 2]
Any of the mentioned equations are out of bounds.
w[m][n] * original_image[2i][2j] for reduce equation and
w[m][n] * g1Image[i / 2][j / 2] for expand equation are working though.
However, the displayed image seems like there is no smoothing effect.
Can anyone explain to me how should I set my image dimension for each Gaussian Pyramid Reduction, Gaussian Pyramid Expansion and the m and n boundaries?
I have solved the problem by including this line
index1 = (2 * h) + m; index2 = (2 * w) + n;
if(index1 >= 0 && index1 < Height && index2 >= 0 && index2 < Width)
temp = w[m + 2][n + 2] * original_image[index1][index2];
More information at :
http://www.songho.ca/dsp/convolution/convolution.html
For example, I have a three variables: "x", "y" and "z". They are all from certain range {min, max}.
Now, I want to compute a new variable, let's say p = x + F*(y-z), where F is some constant between 0 and 1. This new computed variable "p" needs to be mapped into above {min, max} range. How do I do that?
EDIT 1
Generating numbers into array:
population[D*id]=0;
population[D*id+N]=0;
population[D*id +2*N]=0;
population[D*id+1]=rndFloat(globalState,threadIdx.x,4);
population[D*id+N+1]=0;
population[D*id +2*N+1]=0;
for(int i=2; i<N; i++){
float min= -4 - 1/4*abs((int)((i-4)/3));
float max= 4 + 1/4*abs((int)((i-4)/3));
if(i==2)
{
population[D*id+2]=rndFloat(globalState,threadIdx.x,3.14159265359);
population[D*id+N+2]=rndFloat(globalState,threadIdx.x,min,max);
population[D*id +2*N+2]=0;
}
else
{
population[D*id +i]=rndFloat(globalState,threadIdx.x,min,max);
population[D*id+N+i]=rndFloat(globalState,threadIdx.x,min,max);
population[D*id +2*N+i]=rndFloat(globalState,threadIdx.x,min,max);
}
}
Computing a new variable:
for(int i=0; i<D-1; i++)
{
pop[D*id+i]= population[D*a +i] + F*(population[D*b +i]-population[D*c +i]);
}
Indices a, b and c are picked randomly.
The important thing to notice is min and max range and its dependance by given indices:
float min= -4 - 1/4*abs((int)((i-4)/3));
float max= 4 + 1/4*abs((int)((i-4)/3));
where i is replaced by an a, b and c for each range.
EDIT 2
to simplify, let's just say that there are 3 variables x, y and z which are in the certain range. Each variable has its own range. I want to compute new variable p = x + F*(y-z) and it needs to be mapped appropriately into its own range. How do I do that?
It depends what your min and max values are.
For example, to map p between 0 and 1:
if (p < 10)
p /= 9.0f;
else if (p < 100)
p /= 99.0f;
else if (p < 1000)
p /= 999.0f;
And so on...
If you just want to take some variable x which is in the range [A, B] and scale it to a different range [C, D], then you should note that:
x - A is in the range [0, B - A].
Scaling a number in the range [0, B - A] is effectively the same problem being solved in this question: How do I scale down numbers from rand()?.
Note that if you use some really naieve technique like the modulus operator, you will have a non-uniform distribution in your mappings. The linked question contains some reasonable techniques for achieving a mostly-uniform mapping.
In your case, p is in the range [min - max + min, max + max - min].
I'm running a simple for loop with some if statements. In this for loop, 3 variables are to be given a value depending on the index value in the for loop. It seems fairly simple, however, when I run the code, the values always come out as zero and I have no idea why this is happening. My for loop is provided below. I appreciate any suggestions.
double A [N+1];
double r;
double s;
double v;
for(int i = 2; i < N+1; i++)
{
if(i == 2)
{
r = 1/2/i/(i-1);
s = -1/2/(i*i - 1);
v = 1/4/i/(i+1);
}
else if(i <= N-2 && i > 2)
{
r = 1/4/i/(i-1);
s = -1/2/(i*i - 1);
v = 1/4/i/(i+1);
}
else if(i <= N-4 && i > N-2)
{
r = 1/4/i/(i-1);
s = 0;
v = 1/4/i/(i+1);
}
else
{
r = 1/4/i/(i-1);
s = 0;
v = 0;
}
A[i] = r*F[i-2] + s*F[i] + v*F[i+2];
cout << r << s << v << endl;
}
It’s happening because you’re using integer division. An example:
r = 1/2/i/(i-1);
This is the same as:
r = ((1 / 2) / i) / (i - 1);
Which is the same as:
r = (0 / i) / (i - 1);
… which is the same as:
r = 0 / (i - 1);
… which is 0.
Because 1 / 2 is 0 in integer arithmetic. To fix this, use floating point values.
Three things:
else if(i <= N-4 && i > N-2) makes no sense, that condition cannot hold
all your divisions are integer divisions - to fix, convert one of the numbers to a double.
as a result of 1, when i = N-1, and i = N, then the last branch is taken where you force two variables to 0 anyway!
1, 2 and 4 are integers. In integerland 1/2 = 0 and 1/4 = 0
With integers, 1/2 is zero. I would suggest (for a start) changing constants like 2 into 2.0 to ensure they're treated as doubles.
You may also want to (though it may not be necessary) cast all your i variables to floating point values as well, just for completeness, such as:
r = 1.0 / 2.0 / (double)i / ((double)i - 1.0);
The fact that r is a double in no way affects the calculations done on the right of the =. It only affects the final bit (the actual assignment).
1/2, 1/4 and -1/2 will always be zero because of the integer division.So try with 1.0/2.0, 1.0/4.0 and -1.0/2.0 to get it sorted out quickly. But follow the basics and do not use many magic numbers inside a code. Consider creating constants for them and use .