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Assume I have a multilevel structure of data. With a global distribution, from which I draw a highlevel distribution from which i draw a lowlevel distribution from which I draw my response variable. How would I implement such a thing in a stan model.
Below is a minimal example which I hope illustrates the problem. In the stan code there is
one commented "model" section which is working, but ignores the mutlilevel aspect and treats every lower level equal, irrespective of the highlevel origin and provides therefor not shrinkage by the highlevel order (see pic).
A "model"section with a forloop, which I though would do what I want, but takes forever to finish, and with a lot of warnings (Rhat, treedepth, Bayesion Fraction, low ESS)
I am quite inexperienced with modeling and all tutorials on ML-Modeling do not have the Loop-Approach I though would make sense here, so I suspect I am completely heading in the wrong direction with that. So any help will be highly appreciated.
R-Code to generate and run the model
```{r}
library(ggplot2)
library(rstan)
library(bayesplot)
library(loo)
set.seed(0)
numberOFTest <- 50
resposeList <- c()
highLevelIndexList <- c()
lowLevelIndexList <- c()
for (highLevelIndex in seq(3)) {
highLevelMu <- rnorm(1,0,5) #High Level avarage drawn drom normal dist
for (lowLevelIndex in seq(6)) {
lowLevelMu <- rnorm(1,highLevelMu,3) #low Level avarage drawn drom normal dist
resposeList <- c(resposeList, rnorm(numberOFTest,lowLevelMu,1))
highLevelIndexList <- c(highLevelIndexList, rep(highLevelIndex,numberOFTest) )
lowLevelIndexList <- c(lowLevelIndexList, rep(paste0(highLevelIndex,lowLevelIndex),numberOFTest) )
}
}
lowLevelIndexList <- as.integer( as.factor(lowLevelIndexList))
dat <- list(Nr=length(resposeList),
Nh=length(unique(highLevelIndexList)),
Nl=length(unique(lowLevelIndexList)),
#############
r=seq(length(resposeList)),
response=resposeList,
highLevelIndex=(highLevelIndexList),
lowLevelIndex=(lowLevelIndexList)
)
model <- stan(data=dat, file="modle.stan",chains = 4, cores = 8, control=list(adapt_delta=0.95),iter = 2000)
The stan code
data {
int<lower=1> Nr; // number of rows (entries)
int<lower=1> Nh; // number of highlevels
int<lower=1> Nl; // number of low levels
int<lower=1> r[Nr]; // rownumber
real response[Nr]; //
int<lower=1> highLevelIndex[Nr]; //
int<lower=1> lowLevelIndex[Nr]; //
}
parameters {
real<lower=0> sigma; //
real<lower=0> sigma_global; //
real<lower=0> sigma_hL; //
real<lower=0> sigma_lL; //
real a_global; //
real a_hL[Nh]; //
real a_lL[Nl]; //
}
transformed parameters {
vector[Nr] mu;
for (rowIndex in 1:Nr) { //
mu[rowIndex] = a_lL[lowLevelIndex[rowIndex]];
}
}
//Idea of what I want
model {
sigma_global ~ exponential(0.1); //hyper
sigma_hL ~ exponential(0.1); //hyper
sigma_lL ~ exponential(0.1); //hyper
a_global ~ normal(0,sigma_global);
//-----------
for (rowIndex in 1:Nr) { //
a_hL[highLevelIndex[rowIndex]] ~ normal(a_global, sigma_hL); //hyper
a_lL[lowLevelIndex[rowIndex]] ~ normal(a_hL[highLevelIndex[rowIndex]], sigma_lL);
}
////
sigma ~ exponential(0.1);
response ~ normal(mu, sigma);
}
/*
model {///// Working but not what I want
a_lL ~ normal(0, 10);
////
sigma ~ exponential(0.1);
response ~ normal(mu, sigma);
}
*/
generated quantities { // for the PPC and loo
real n_rep[Nr];
vector[Nr] log_lik;
for (i in 1:Nr){
n_rep[i] = normal_rng(mu[i], sigma);
log_lik[i] = normal_lpdf(response[i] | mu[i], sigma);
}
}
mcmc_areas of the working modle with no shrinkage within each block (1-6),(7-12) and (12-18)
found the mistake: I needed to map the lowlevel values to the highlevel ones, with a look up table. Below is now a working version, which also just takes a second to finish.
data {
int<lower=1> Nr; // number of rows (entries)
int<lower=1> Nh; // number of highlevels
int<lower=1> Nl; // number of low levels
real response[Nr]; //
int<lower=1> highLevelIndex[Nr]; //
int<lower=1> lowLevelIndex[Nr]; //
int<lower=1> lookUp[Nl];//lookuptable which highlevel value a lowlevel value fits to
}
parameters {
real<lower=0> sigma; //
real<lower=0> sigma_hL; //
real<lower=0> sigma_lL; //
real a_hL[Nh]; //
real a_lL[Nl]; //
}
transformed parameters {
vector[Nr] mu;
for (rowIndex in 1:Nr) { //
mu[rowIndex] = a_lL[lowLevelIndex[rowIndex]];
}
}
model {
sigma_hL ~ exponential(0.01);
a_hL ~ normal(0, sigma_hL);
sigma_lL ~ exponential(0.01);
for (lLIdx in 1:Nl) { //
a_lL[lLIdx] ~ normal(a_hL[lookUp[lLIdx]],sigma_lL);
}
////
sigma ~ exponential(0.1);
response ~ normal(mu, sigma);
}
generated quantities { // for the PPC and loo
real n_rep[Nr];
vector[Nr] log_lik;
for (i in 1:Nr){
n_rep[i] = normal_rng(mu[i], sigma);
log_lik[i] = normal_lpdf(response[i] | mu[i], sigma);
}
}
I have a 6-DOF robot arm model:
robot arm structure
I want to calculate forward kinematics, so I uses the D-H matrix. the D-H parameters are:
static const std::vector<float> theta = {
0,0,90.0f,0,-90.0f,0};
// d
static const std::vector<float> d = {
380.948f,0,0,-560.18f,0,0};
// a
static const std::vector<float> a = {
-220.0f,522.331f,80.0f,0,0,94.77f};
// alpha
static const std::vector<float> alpha = {
90.0f,0,90.0f,-90.0f,-90.0f,0};
and the calculation :
glm::mat4 Robothand::armForKinematics() noexcept
{
glm::mat4 pose(1.0f);
float cos_theta, sin_theta, cos_alpha, sin_alpha;
for (auto i = 0; i < 6;i++)
{
cos_theta = cosf(glm::radians(theta[i]));
sin_theta = sinf(glm::radians(theta[i]));
cos_alpha = cosf(glm::radians(alpha[i]));
sin_alpha = sinf(glm::radians(alpha[i]));
glm::mat4 Ai = {
cos_theta, -sin_theta * cos_alpha,sin_theta * sin_alpha, a[i] * cos_theta,
sin_theta, cos_theta * cos_alpha, -cos_theta * sin_alpha,a[i] * sin_theta,
0, sin_alpha, cos_alpha, d[i],
0, 0, 0, 1 };
pose = pose * Ai;
}
return pose;
}
the problem I have is that, I can't get the correct result, for example, I want to calculate the transformation matrix from first joint to the 4th joint, I will change the for loop i < 3,then I can get the pose matrix, and I can the origin coordinate in 4th coordinate system by pose * (0,0,0,1).but the result (380.948,382.331,0) seems not correct because it should be move along x-axis not y-axis. I have read many books and materials about D-H matrix, but I can't figure out what's wrong with it.
I have figured it out by myself, the real problem behind is glm::mat, glm::mat is col-type which means columns will be initialized before rows,I changed the code and get the correct result:
for (int i = 0; i < joint_num; ++i)
{
pose = glm::rotate(pose, glm::radians(degrees[i]), glm::vec3(0, 0, 1));
pose = glm::translate(pose,glm::vec3(0,0,d[i]));
pose = glm::translate(pose, glm::vec3(a[i], 0, 0));
pose = glm::rotate(pose,glm::radians(alpha[i]),glm::vec3(1,0,0));
}
then I can get the position by:
auto pos = pose * glm::vec4(x,y,z,1);
I'm making a bidirectional path tracer and I have some troubles.
To be clear :
1) One point light
2) All objects are diffuse
3) All objects are spheres, even walls (they are very large)
4) NO MIS WEIGHTING
The light emission is a 3D vector. The BRDF of a sphere is a 3D vector. Hard coded.
In the main function below I generate EyePath and LightPath then I connect them. At least I try.
In this post I will talking about the main function then EyePath then LightPath. The talking about connecting function will appear once EyePath and Light are good.
First questions :
Does the generation of the first light point is good ?
Do I need to compute this point according to the emission of the light source? or is it just the emission ? The line is commented where i'm filling the Vertices structure.
Do I need to translate fromlight ? In order to put it on the sphere
The code below is sampled in the main function. Above it there is two for loops going through all pixels. Camera.o is the eye. CameraRayDir is the direction to the current pixel.
//The path light starting point is at the same position as the light
Ray fromLight(Vec(0, 24.3, 0), Vec());
Sphere light = spheres[7];
#define PDF 0.15915494309 // 1 / (2 * PI)
for(int i = 0; i < samps; ++i)
{
std::vector<Vertices> PathEye;
std::vector<Vertices> PathLight;
Vec cameraRayDir = cx * (double(x) / w - .5) + cy * (double(y) / h - .5) + camera.d;
Ray rayEye(camera.o, cameraRayDir.norm());
// Hemisphere oriented towards the top
fromLight.d = generateRayInHemisphere(fromLight.o,Vec(0,1,0)).d;
double f = clamp(n.dot(fromLight.d.norm()));
Vertices vert;
vert.d = fromLight.d;
vert.x = fromLight.o;
vert.id = 7;
vert.cos = f;
vert.n = Vec(0,1,0).norm();
// this one ?
//vert.couleur = spheres[7].e * f / PDF;
// Or this one ?
vert.couleur = spheres[7].e;
PathLight.push_back(vert);
int sizeEye = generateEyePath(PathEye, rayEye, maxDepth);
int sizeLight = generateLightPath(PathLight, fromLight, maxDepth);
for (int s = 0; s < sizeLight; ++s)
{
for (int t = 1; t < sizeEye; ++t)
{
int depth = t + s - 1;
if ((s == 0 && t == 0) || depth < 0 || depth > maxDepth)
continue;
pixelValue = pixelValue + connectPaths(PathEye, PathLight, s, t);
}
}
}
For the EyePath I intersect the geometry then I compute the illumination according to the distance with the light. The colour is black if the point is in the shadow.
Second question : For the eye path and the direct illumination, is the computation good ? I've seen in many code, people use the pdf even in direct illumination. But I'm only using point light and spheres.
int generateEyePath(std::vector<Vertices>& v, Ray eye, int maxDepth)
{
double t;
int id = 0;
Vertices vert;
int RussianRoulette;
while(v.size() <= maxDepth)
{
if(distribRREye(generatorRREye) < 10)
break;
// Intersect all the geometry
// id is the id of the intersected geometry in an array
intersect(eye, t, id);
const Sphere& obj = spheres[id];
// Intersection point
Vec x = eye.o + eye.d * t;
// normal
Vec n = (x - obj.p).norm();
Vec direction = light.p - x;
// Shadow ray
Ray RaytoLight = Ray(x, direction.norm());
const float distance = direction.length();
// shadow
const bool visibility = intersect(RaytoLight, t, id);
const Sphere &lumiere = spheres[id];
float degree = clamp(n.dot((lumiere.p - x).norm()));
// If the intersected geometry is not a light, then in shadow
if(lumiere.e.x == 0)
{
vert.couleur = Vec();
}
else // else we compute the colour
// obj.c is the brdf, lumiere.e is the emission
vert.couleur = (obj.c).mult(lumiere.e / (distance * distance)) * degree;
vert.x = x;
vert.id = id;
vert.n = n;
vert.d = eye.d.normn();
vert.cos = degree;
v.push_back(vert);
eye = generateRayInHemisphere(x,n);
}
return v.size();
}
For the LightPath, for a given point, I compute it according to the previous one and the values at this point. Like in a common path tracing.\n
Third question: Is the colour computation good ?
int generateLightPath(std::vector<Vertices>& v, Ray fromLight, int maxDepth)
{
double t;
int id = 0;
Vertices vert;
Vec previous;
while(v.size() <= maxDepth)
{
if(distribRRLight(generatorRRLight) < 10)
break;
previous = v.back().couleur;
intersect(fromLight, t, id);
// intersected geometry
const Sphere& obj = spheres[id];
// Intersection point
Vec x = fromLight.o + fromLight.d * t;
// normal
Vec n = (x - obj.p).norm();
double f = clamp(n.dot(fromLight.d.norm()));
// obj.c is the brdf
vert.couleur = previous.mult(((obj.c / M_PI) * f) / PDF);
vert.x = x;
vert.id = id;
vert.n = n;
vert.d = fromLight.d.norm();
vert.cos = f;
v.push_back(vert);
fromLight = generateRayInHemisphere(x,n);
}
return v.size();
}
For the moment I get this result.
enter image description here
The connecting function will come once EyePath and LightPath are good.
Thank you all
Try the spherical reference scene mentioned in this paper. I think then you can work out most of your questions by yourself since it has an analytical solution.
https://www.researchgate.net/publication/221546261_Testing_Monte-Carlo_Global_Illumination_Methods_with_Analytically_Computable_Scenes
It would save your time to implement and verify your understanding with path tracing and light tracing first, then try to combine them with weights.
Can someone give me example of an algorithm that has square root(n) time complexity. What does square root time complexity even mean?
Square root time complexity means that the algorithm requires O(N^(1/2)) evaluations where the size of input is N.
As an example for an algorithm which takes O(sqrt(n)) time, Grover's algorithm is one which takes that much time. Grover's algorithm is a quantum algorithm for searching an unsorted database of n entries in O(sqrt(n)) time.
Let us take an example to understand how can we arrive at O(sqrt(N)) runtime complexity, given a problem. This is going to be elaborate, but is interesting to understand. (The following example, in the context for answering this question, is taken from Coding Contest Byte: The Square Root Trick , very interesting problem and interesting trick to arrive at O(sqrt(n)) complexity)
Given A, containing an n elements array, implement a data structure for point updates and range sum queries.
update(i, x)-> A[i] := x (Point Updates Query)
query(lo, hi)-> returns A[lo] + A[lo+1] + .. + A[hi]. (Range Sum Query)
The naive solution uses an array. It takes O(1) time for an update (array-index access) and O(hi - lo) = O(n) for the range sum (iterating from start index to end index and adding up).
A more efficient solution splits the array into length k slices and stores the slice sums in an array S.
The update takes constant time, because we have to update the value for A and the value for the corresponding S. In update(6, 5) we have to change A[6] to 5 which results in changing the value of S1 to keep S up to date.
The range-sum query is interesting. The elements of the first and last slice (partially contained in the queried range) have to be traversed one by one, but for slices completely contained in our range we can use the values in S directly and get a performance boost.
In query(2, 14) we get,
query(2, 14) = A[2] + A[3]+ (A[4] + A[5] + A[6] + A[7]) + (A[8] + A[9] + A[10] + A[11]) + A[12] + A[13] + A[14] ;
query(2, 14) = A[2] + A[3] + S[1] + S[2] + A[12] + A[13] + A[14] ;
query(2, 14) = 0 + 7 + 11 + 9 + 5 + 2 + 0;
query(2, 14) = 34;
The code for update and query is:
def update(S, A, i, k, x):
S[i/k] = S[i/k] - A[i] + x
A[i] = x
def query(S, A, lo, hi, k):
s = 0
i = lo
//Section 1 (Getting sum from Array A itself, starting part)
while (i + 1) % k != 0 and i <= hi:
s += A[i]
i += 1
//Section 2 (Getting sum from Slices directly, intermediary part)
while i + k <= hi:
s += S[i/k]
i += k
//Section 3 (Getting sum from Array A itself, ending part)
while i <= hi:
s += A[i]
i += 1
return s
Let us now determine the complexity.
Each query takes on average
Section 1 takes k/2 time on average. (you might iterate atmost k/2)
Section 2 takes n/k time on average, basically number of slices
Section 3 takes k/2 time on average. (you might iterate atmost k/2)
So, totally, we get k/2 + n/k + k/2 = k + n/k time.
And, this is minimized for k = sqrt(n). sqrt(n) + n/sqrt(n) = 2*sqrt(n)
So we get a O(sqrt(n)) time complexity query.
Prime numbers
As mentioned in some other answers, some basic things related to prime numbers take O(sqrt(n)) time:
Find number of divisors
Find sum of divisors
Find Euler's totient
Below I mention two advanced algorithms which also bear sqrt(n) term in their complexity.
MO's Algorithm
try this problem: Powerful array
My solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1E6 + 10, k = 500;
struct node {
int l, r, id;
bool operator<(const node &a) {
if(l / k == a.l / k) return r < a.r;
else return l < a.l;
}
} q[N];
long long a[N], cnt[N], ans[N], cur_count;
void add(int pos) {
cur_count += a[pos] * cnt[a[pos]];
++cnt[a[pos]];
cur_count += a[pos] * cnt[a[pos]];
}
void rm(int pos) {
cur_count -= a[pos] * cnt[a[pos]];
--cnt[a[pos]];
cur_count -= a[pos] * cnt[a[pos]];
}
int main() {
int n, t;
cin >> n >> t;
for(int i = 1; i <= n; i++) {
cin >> a[i];
}
for(int i = 0; i < t; i++) {
cin >> q[i].l >> q[i].r;
q[i].id = i;
}
sort(q, q + t);
memset(cnt, 0, sizeof(cnt));
memset(ans, 0, sizeof(ans));
int curl(0), curr(0), l, r;
for(int i = 0; i < t; i++) {
l = q[i].l;
r = q[i].r;
/* This part takes O(n * sqrt(n)) time */
while(curl < l)
rm(curl++);
while(curl > l)
add(--curl);
while(curr > r)
rm(curr--);
while(curr < r)
add(++curr);
ans[q[i].id] = cur_count;
}
for(int i = 0; i < t; i++) {
cout << ans[i] << '\n';
}
return 0;
}
Query Buffering
try this problem: Queries on a Tree
My solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 10, k = 333;
vector<int> t[N], ht;
int tm_, h[N], st[N], nd[N];
inline int hei(int v, int p) {
for(int ch: t[v]) {
if(ch != p) {
h[ch] = h[v] + 1;
hei(ch, v);
}
}
}
inline void tour(int v, int p) {
st[v] = tm_++;
ht.push_back(h[v]);
for(int ch: t[v]) {
if(ch != p) {
tour(ch, v);
}
}
ht.push_back(h[v]);
nd[v] = tm_++;
}
int n, tc[N];
vector<int> loc[N];
long long balance[N];
vector<pair<long long,long long>> buf;
inline long long cbal(int v, int p) {
long long ans = balance[h[v]];
for(int ch: t[v]) {
if(ch != p) {
ans += cbal(ch, v);
}
}
tc[v] += ans;
return ans;
}
inline void bal() {
memset(balance, 0, sizeof(balance));
for(auto arg: buf) {
balance[arg.first] += arg.second;
}
buf.clear();
cbal(1,1);
}
int main() {
int q;
cin >> n >> q;
for(int i = 1; i < n; i++) {
int x, y; cin >> x >> y;
t[x].push_back(y); t[y].push_back(x);
}
hei(1,1);
tour(1,1);
for(int i = 0; i < ht.size(); i++) {
loc[ht[i]].push_back(i);
}
vector<int>::iterator lo, hi;
int x, y, type;
for(int i = 0; i < q; i++) {
cin >> type;
if(type == 1) {
cin >> x >> y;
buf.push_back(make_pair(x,y));
}
else if(type == 2) {
cin >> x;
long long ans(0);
for(auto arg: buf) {
hi = upper_bound(loc[arg.first].begin(), loc[arg.first].end(), nd[x]);
lo = lower_bound(loc[arg.first].begin(), loc[arg.first].end(), st[x]);
ans += arg.second * (hi - lo);
}
cout << tc[x] + ans/2 << '\n';
}
else assert(0);
if(i % k == 0) bal();
}
}
There are many cases.
These are the few problems which can be solved in root(n) complexity [better may be possible also].
Find if a number is prime or not.
Grover's Algorithm: allows search (in quantum context) on unsorted input in time proportional to the square root of the size of the input.link
Factorization of the number.
There are many problems that you will face which will demand use of sqrt(n) complexity algorithm.
As an answer to second part:
sqrt(n) complexity means if the input size to your algorithm is n then there approximately sqrt(n) basic operations ( like **comparison** in case of sorting). Then we can say that the algorithm has sqrt(n) time complexity.
Let's analyze the 3rd problem and it will be clear.
let's n= positive integer. Now there exists 2 positive integer x and y such that
x*y=n;
Now we know that whatever be the value of x and y one of them will be less than sqrt(n). As if both are greater than sqrt(n)
x>sqrt(n) y>sqrt(n) then x*y>sqrt(n)*sqrt(n) => n>n--->contradiction.
So if we check 2 to sqrt(n) then we will have all the factors considered ( 1 and n are trivial factors).
Code snippet:
int n;
cin>>n;
print 1,n;
for(int i=2;i<=sqrt(n);i++) // or for(int i=2;i*i<=n;i++)
if((n%i)==0)
cout<<i<<" ";
Note: You might think that not considering the duplicate we can also achieve the above behaviour by looping from 1 to n. Yes that's possible but who wants to run a program which can run in O(sqrt(n)) in O(n).. We always look for the best one.
Go through the book of Cormen Introduction to Algorithms.
I will also request you to read following stackoverflow question and answers they will clear all the doubts for sure :)
Are there any O(1/n) algorithms?
Plain english explanation Big-O
Which one is better?
How do you calculte big-O complexity?
This link provides a very basic beginner understanding of O() i.e., O(sqrt n) time complexity. It is the last example in the video, but I would suggest that you watch the whole video.
https://www.youtube.com/watch?v=9TlHvipP5yA&list=PLDN4rrl48XKpZkf03iYFl-O29szjTrs_O&index=6
The simplest example of an O() i.e., O(sqrt n) time complexity algorithm in the video is:
p = 0;
for(i = 1; p <= n; i++) {
p = p + i;
}
Mr. Abdul Bari is reknowned for his simple explanations of data structures and algorithms.
Primality test
Solution in JavaScript
const isPrime = n => {
for(let i = 2; i <= Math.sqrt(n); i++) {
if(n % i === 0) return false;
}
return true;
};
Complexity
O(N^1/2) Because, for a given value of n, you only need to find if its divisible by numbers from 2 to its root.
JS Primality Test
O(sqrt(n))
A slightly more performant version, thanks to Samme Bae, for enlightening me with this. 😉
function isPrime(n) {
if (n <= 1)
return false;
if (n <= 3)
return true;
// Skip 4, 6, 8, 9, and 10
if (n % 2 === 0 || n % 3 === 0)
return false;
for (let i = 5; i * i <= n; i += 6) {
if (n % i === 0 || n % (i + 2) === 0)
return false;
}
return true;
}
isPrime(677);
Is there any logarithm function implemented in the GMP library?
I know you didn't ask how to implement it, but...
You can implement a rough one using the properties of logarithms: http://gnumbers.blogspot.com.au/2011/10/logarithm-of-large-number-it-is-not.html
And the internals of the GMP library: https://gmplib.org/manual/Integer-Internals.html
(Edit: Basically you just use the most significant "digit" of the GMP representation since the base of the representation is huge B^N is much larger than B^{N-1})
Here is my implementation for Rationals.
double LogE(mpq_t m_op)
{
// log(a/b) = log(a) - log(b)
// And if a is represented in base B as:
// a = a_N B^N + a_{N-1} B^{N-1} + ... + a_0
// => log(a) \approx log(a_N B^N)
// = log(a_N) + N log(B)
// where B is the base; ie: ULONG_MAX
static double logB = log(ULONG_MAX);
// Undefined logs (should probably return NAN in second case?)
if (mpz_get_ui(mpq_numref(m_op)) == 0 || mpz_sgn(mpq_numref(m_op)) < 0)
return -INFINITY;
// Log of numerator
double lognum = log(mpq_numref(m_op)->_mp_d[abs(mpq_numref(m_op)->_mp_size) - 1]);
lognum += (abs(mpq_numref(m_op)->_mp_size)-1) * logB;
// Subtract log of denominator, if it exists
if (abs(mpq_denref(m_op)->_mp_size) > 0)
{
lognum -= log(mpq_denref(m_op)->_mp_d[abs(mpq_denref(m_op)->_mp_size)-1]);
lognum -= (abs(mpq_denref(m_op)->_mp_size)-1) * logB;
}
return lognum;
}
(Much later edit)
Coming back to this 5 years later, I just think it's cool that the core concept of log(a) = N log(B) + log(a_N) shows up even in native floating point implementations, here is the glibc one for ia64
And I used it again after encountering this question
No there is no such function in GMP.
Only in MPFR.
The method below makes use of mpz_get_d_2exp and was obtained from the gmp R package. It can be found under the function biginteger_log in the file bigintegerR.cc (You first have to download the source (i.e. the tar file)). You can also see it here: biginteger_log.
// Adapted for general use from the original biginteger_log
// xi = di * 2 ^ ex ==> log(xi) = log(di) + ex * log(2)
double biginteger_log_modified(mpz_t x) {
signed long int ex;
const double di = mpz_get_d_2exp(&ex, x);
return log(di) + log(2) * (double) ex;
}
Of course, the above method could be modified to return the log with any base using the properties of logarithm (e.g. the change of base formula).
Here it is:
https://github.com/linas/anant
Provides gnu mp real and complex logarithm, exp, sine, cosine, gamma, arctan, sqrt, polylogarithm Riemann and Hurwitz zeta, confluent hypergeometric, topologists sine, and more.
As other answers said, there is no logarithmic function in GMP. Part of answers provided implementations of logarithmic function, but with double precision only, not infinite precision.
I implemented full (arbitrary) precision logarithmic function below, even up to thousands bits of precision if you wish. Using mpf, generic floating point type of GMP.
My code uses Taylor serie for ln(1 + x) plus mpf_sqrt() (for boosting computation).
Code is in C++, and is quite large due to two facts. First is that it does precise time measurements to figure out best combinations of internal computational parameters for your machine. Second is that it uses extra speed improvements like extra usage of mpf_sqrt() for preparing initial value.
Algorithm of my code is following:
Factor out exponent of 2 from input x, i.e. rewrite x = d * 2^exp, with usage of mpf_get_d_2exp().
Make d (from step above) such that 2/3 <= d <= 4/3, this is achieved by possibly multiplying d by 2 and doing --exp. This ensures that d always differs from 1 by at most 1/3, in other words d extends from 1 in both directions (negative and positive) in equal distance.
Divide x by 2^exp, with usage of mpf_div_2exp() and mpf_mul_2exp().
Take square root of x several times (num_sqrt times) so that x becomes closer to 1. This ensures that Taylor Serie converges more rapidly. Because computation of square root several times is faster than contributing much more time in extra iterations of Taylor Serie.
Compute Taylor Serie for ln(1 + x) up to desired precision (even thousands of bit of precision if needed).
Because in Step 4. we took square root several times, now we need to multiply y (result of Taylor Serie) by 2^num_sqrt.
Finally because in Step 1. we factored out 2^exp, now we need to add ln(2) * exp to y. Here ln(2) is computed by just one recursive call to same function that implements whole algorithm.
Steps above come from sequence of formulas ln(x) = ln(d * 2^exp) = ln(d) + exp * ln(2) = ln(sqrt(...sqrt(d))) * num_sqrt + exp * ln(2).
My implementation automatically does timings (just once per program run) to figure out how many square roots is needed to balance out Taylor Serie computation. If you need to avoid timings then pass 3rd parameter sqrt_range to mpf_ln() equal to 0.001 instead of zero.
main() function contains examples of usage, testing of correctness (by comparing to lower precision std::log()), timings and output of different verbose information. Function is tested on first 1024 bits of Pi number.
Before call to my function mpf_ln() don't forget to setup needed precision of computation by calling mpf_set_default_prec(bits) with desired precision in bits.
Computational time of my mpf_ln() is about 40-90 micro-seconds for 1024 bit precision. Bigger precision will take more time, that is approximately linearly proportional to the amount of precision bits.
Very first run of a function takes considerably longer time becuse it does pre-computation of timings table and value of ln(2). So it is suggested to do first single computation at program start to avoid longer computation inside time critical region later in code.
To compile for example on Linux, you have to install GMP library and issue command:
clang++-14 -std=c++20 -O3 -lgmp -lgmpxx -o main main.cpp && ./main
Try it online!
#include <cstdint>
#include <iomanip>
#include <iostream>
#include <cmath>
#include <chrono>
#include <mutex>
#include <vector>
#include <unordered_map>
#include <gmpxx.h>
double Time() {
static auto const gtb = std::chrono::high_resolution_clock::now();
return std::chrono::duration_cast<std::chrono::duration<double>>(
std::chrono::high_resolution_clock::now() - gtb).count();
}
mpf_class mpf_ln(mpf_class x, bool verbose = false, double sqrt_range = 0) {
auto total_time = verbose ? Time() : 0.0;
int const prec = mpf_get_prec(x.get_mpf_t());
if (sqrt_range == 0) {
static std::mutex mux;
std::lock_guard<std::mutex> lock(mux);
static std::vector<std::pair<size_t, double>> ranges;
if (ranges.empty())
mpf_ln(3.14, false, 0.01);
while (ranges.empty() || ranges.back().first < prec) {
size_t const bits = ranges.empty() ? 64 : ranges.back().first * 3 / 2;
mpf_class x = 3.14;
mpf_set_prec(x.get_mpf_t(), bits);
double sr = 0.35, sr_best = 1, time_best = 1000;
size_t constexpr ntests = 5;
while (true) {
auto tim = Time();
for (size_t i = 0; i < ntests; ++i)
mpf_ln(x, false, sr);
tim = (Time() - tim) / ntests;
bool updated = false;
if (tim < time_best) {
sr_best = sr;
time_best = tim;
updated = true;
}
sr /= 1.5;
if (sr <= 1e-8) {
ranges.push_back(std::make_pair(bits, sr_best));
break;
}
}
}
sqrt_range = std::lower_bound(ranges.begin(), ranges.end(), size_t(prec),
[](auto const & a, auto const & b){
return a.first < b;
})->second;
}
signed long int exp = 0;
// https://gmplib.org/manual/Converting-Floats
double d = mpf_get_d_2exp(&exp, x.get_mpf_t());
if (d < 2.0 / 3) {
d *= 2;
--exp;
}
mpf_class t;
// https://gmplib.org/manual/Float-Arithmetic
if (exp >= 0)
mpf_div_2exp(x.get_mpf_t(), x.get_mpf_t(), exp);
else
mpf_mul_2exp(x.get_mpf_t(), x.get_mpf_t(), -exp);
auto sqrt_time = verbose ? Time() : 0.0;
// Multiple Sqrt of x
int num_sqrt = 0;
if (x >= 1)
while (x >= 1.0 + sqrt_range) {
// https://gmplib.org/manual/Float-Arithmetic
mpf_sqrt(x.get_mpf_t(), x.get_mpf_t());
++num_sqrt;
}
else
while (x <= 1.0 - sqrt_range) {
mpf_sqrt(x.get_mpf_t(), x.get_mpf_t());
++num_sqrt;
}
if (verbose)
sqrt_time = Time() - sqrt_time;
static mpf_class const eps = [&]{
mpf_class eps = 1;
mpf_div_2exp(eps.get_mpf_t(), eps.get_mpf_t(), prec + 8);
return eps;
}(), meps = -eps;
// Taylor Serie for ln(1 + x)
// https://math.stackexchange.com/a/878376/826258
x -= 1;
mpf_class k = x, y = x, mx = -x;
size_t num_iters = 0;
for (int32_t i = 2;; ++i) {
k *= mx;
y += k / i;
// Check if error is small enough
if (meps <= k && k <= eps) {
num_iters = i;
break;
}
}
auto VerboseInfo = [&]{
if (!verbose)
return;
total_time = Time() - total_time;
std::cout << std::fixed << "Sqrt range " << sqrt_range << ", num sqrts "
<< num_sqrt << ", sqrt time " << sqrt_time << " sec" << std::endl;
std::cout << "Ln number of iterations " << num_iters << ", ln time "
<< total_time << " sec" << std::endl;
};
// Correction due to multiple sqrt of x
y *= 1 << num_sqrt;
if (exp == 0) {
VerboseInfo();
return y;
}
mpf_class ln2;
{
static std::mutex mutex;
std::lock_guard<std::mutex> lock(mutex);
static std::unordered_map<size_t, mpf_class> ln2s;
auto it = ln2s.find(size_t(prec));
if (it == ln2s.end()) {
mpf_class sqrt_sqrt_2 = 2;
mpf_sqrt(sqrt_sqrt_2.get_mpf_t(), sqrt_sqrt_2.get_mpf_t());
mpf_sqrt(sqrt_sqrt_2.get_mpf_t(), sqrt_sqrt_2.get_mpf_t());
it = ln2s.insert(std::make_pair(size_t(prec), mpf_class(mpf_ln(sqrt_sqrt_2, false, sqrt_range) * 4))).first;
}
ln2 = it->second;
}
y += ln2 * exp;
VerboseInfo();
return y;
}
std::string mpf_str(mpf_class const & x) {
mp_exp_t exp;
auto s = x.get_str(exp);
return s.substr(0, exp) + "." + s.substr(exp);
}
int main() {
// https://gmplib.org/manual/Initializing-Floats
mpf_set_default_prec(1024); // bit-precision
// http://www.math.com/tables/constants/pi.htm
mpf_class x(
"3."
"1415926535 8979323846 2643383279 5028841971 6939937510 "
"5820974944 5923078164 0628620899 8628034825 3421170679 "
"8214808651 3282306647 0938446095 5058223172 5359408128 "
"4811174502 8410270193 8521105559 6446229489 5493038196 "
"4428810975 6659334461 2847564823 3786783165 2712019091 "
"4564856692 3460348610 4543266482 1339360726 0249141273 "
"7245870066 0631558817 4881520920 9628292540 9171536436 "
);
std::cout << std::boolalpha << std::fixed << std::setprecision(14);
std::cout << "x:" << std::endl << mpf_str(x) << std::endl;
auto cmath_val = std::log(mpf_get_d(x.get_mpf_t()));
std::cout << "cmath ln(x): " << std::endl << cmath_val << std::endl;
auto volatile tmp = mpf_ln(x); // Pre-Compute to heat-up timings table.
auto time_start = Time();
size_t constexpr ntests = 20;
for (size_t i = 0; i < ntests; ++i) {
auto volatile tmp = mpf_ln(x);
}
std::cout << "mpf ln(x) time " << (Time() - time_start) / ntests << " sec" << std::endl;
auto mpf_val = mpf_ln(x, true);
std::cout << "mpf ln(x):" << std::endl << mpf_str(mpf_val) << std::endl;
std::cout << "equal to cmath: " << (std::abs(mpf_get_d(mpf_val.get_mpf_t()) - cmath_val) <= 1e-14) << std::endl;
return 0;
}
Output:
x:
3.141592653589793238462643383279502884197169399375105820974944592307816406286208998628034825342117067982148086513282306647093844609550582231725359408128481117450284102701938521105559644622948954930381964428810975665933446128475648233786783165271201909145648566923460348610454326648213393607260249141273724587007
cmath ln(x):
1.14472988584940
mpf ln(x) time 0.00004426845000 sec
Sqrt range 0.00000004747981, num sqrts 23, sqrt time 0.00001440000000 sec
Ln number of iterations 42, ln time 0.00003873100000 sec
mpf ln(x):
1.144729885849400174143427351353058711647294812915311571513623071472137769884826079783623270275489707702009812228697989159048205527923456587279081078810286825276393914266345902902484773358869937789203119630824756794011916028217227379888126563178049823697313310695003600064405487263880223270096433504959511813198
equal to cmath: true