I have a Ax =b type linear system - where A is an upper-triangular matrix. The structure of A is defined as follows:
comp.Amat <- function(i,j,prob) ifelse(i > j, 0, dbinom(x=i, size=j, prob=prob))
prob <- 1/4
A <- outer(1:50, 1:50 , FUN=function(r,c) comp.Amat(r,c,prob) )
The entries in A are binomial probabilities - and the issue is the diagonal entries fastly approach to 0 when the size of A grows.
If we define the vector b as follows as well:
b <- seq(1,50,1);
Then solve(a=A,b=b) - gives an error:
" system is computationally singular: reciprocal condition number = 1.07584e-64"
That makes sense, since the diagonal entries are almost 0, so the matrix becomes non-invertible.
As a work-around, I have written the following recursive function - which starts to compute the value of last diagonal entry, then replaces that value in the previous rows. Since each entry in matrix is dbinom(j,i, prob) for j=>i :I can get a solution via this way.
solve.for.x.custom <- function(A, b, prob)
{
n =length(A[1,])
m =length(A[,1])
x = seq(1,n, 1);
x[x> 0] = -1000;
calc.inv.Aii <- function(i,j, prob)
{
res = (1 / (prob*(1-prob)))^i;
return(res);
}
for (i in m:1 )
{
if(i ==m)
{
rhs =0;
}else
{
rhs=0;
for(j in m:(i+1))
{
rhs = dbinom(x=i,size=j,prob=prob)*x[j] + rhs;
}
}
x[i] = (b[i] - rhs)*calc.inv.Aii(i,i);
}
print(x)
return(x)
}
My problem is - when I multiply this solution x' by matrix A, the errors (Ax'- b) are huge. Since I have an analytical solution (each entry in x_i can be described as a in terms of binomial probabilities multiplies by previous values) - the error I should get is 0- in each row.
I see that (1 / (1/a)) may not be equal to a because of these issues. However, the current errors are really big( -1.13817489781529e+168).
x_prime=solve.for.x.custom(A, b, prob)
A%*%x_prime - b
#output
[,1]
[1,] -1.13817489781529e+168
[2,] 2.11872209742428e+167
[3,] -1.58403954589004e+166
[4,] 6.52328959209082e+164
[5,] -1.69562573261261e+163
[6,] 3.00614551450976e+161
***
[49,] -7.58010305220250e+08
[50,] 9.65162608741321e+03
I would really appreciate it you'd recommend any suggestions or efficient methods. I gave the size of A and b as 50 -but I intend to grow them as well thus in that case this the error will increase also.
If your matrix A is upper triangular you probably want to use backsolve(A, b) rather than solve(A, b).
You can do arbitrary precision in R with Rmpfr, which will require writing a compatible version of backsolve. With the code below the break we can get
> print(max(abs(b - .b)), digits=5)
1 'mpfr' number of precision 1024 bits
[1] 2.9686e-267
There is one important caveat though: the values in A may not be accurate enough since they come from dbinom rather than using mpfr objeccts. Depending on your end goal, you may need to write your own version of dbinom using Rmpfr.
library(Rmpfr)
logcomp.Amat <- function(i,j,prob) ifelse(i > j, -Inf, dbinom(x=i, size=j, prob=prob, log=TRUE))
nbits <- 1024
.backsolve <- function(A, b) {
n <- length(b)
x <- mpfr(numeric(n), nbits)
for(i in rev(seq_len(n))) {
known <- i + seq_len(n - i)
z <- if(length(known) > 0) sum(A[i,known] * x[known]) else 0
x[i] <- (b[i] - z) / A[i,i]
}
return(x)
}
logA <- outer(1:50, 1:50, logcomp.Amat, prob=1/4)
b <- 1:50
A <- exp(mpfr(logA, nbits))
b <- mpfr(b, nbits)
x <- .backsolve(A, b)
.b <- as.vector(A %*% x)
Related
I'm trying to set a function in R to compute the Greatest Common RATIONAL Divisor of a vector. So I'm not working with a vector of integers, but of numerics. And from this vector I would like to automatically determine the highest numeric that can divide all the values in the vector and result in a integer. Which is very difficult with floating-point arithmetic used in R.
To give an example, lets say that I would like to find the highest common rational divisor of 5, 0.37 and 0.02. It's 0.01, but how can I automate this, taking into account that with floating-point arithmetic 0.37 will for instance be considered like 0.3700000000000000000000000005271 in R (something like that) ? With that problem I can't easily compute the lowest order of decimal (10^-2 in my example), or if you prefer the lowest one that has significance for me as user.
The fact that the result in itself will have floating-point-like error (e.g. 0.0100000000000000000000008465 in place of 0.01) is not a problem. However ideally the solution should be the most general possible (capable of working with vectors having extremely different values (10^20 and 10^-20 for instance).
I got a solution. So the basic idea is to divide everything by the smallest value, to multiply by integers, until every value is made of integers. This is controlled via the floor() function, which allows to have explicit control over the level of tolerance. I added some control to the amount of possibilities it tests, to make it kind of efficient, but I'm not sure this is the best method. Anyway I'll put this in the StratigrapheR package
divisor <- function(x, tolerance = 8, relative = T, tries = 4, speak = T)
{
if(!(isFALSE(relative) | isTRUE(relative))) {
stop("The 'relative' parameter should be TRUE or FALSE'")
}
if(!(isFALSE(speak) | isTRUE(speak))) {
stop("The 'speak' parameter should be TRUE or FALSE'")
}
x <- unique(x)
x <- x[x != 0]
# Divide by smallest
mx <- min(x)
d <- x/mx
if(!relative){
# Test if tolerance is of lower order than the smallest
if(-log10(mx) > tolerance) {
stop(paste("If 'relative' is FALSE, the smallest value (zero excepted)",
"should of higher order than the order",
"defined by the 'tolerance' parameter"))
}
}
# Test the dispersion of values
general_tolerance <- 15 # Order of digits affected by floating-point
if(log10(max(x)) >= (general_tolerance - tolerance)){
stop(paste("The range of 'x' values is too large to find a meaningful",
"greatest common rational divisor.",
"To solve this problem you can change the values in x or",
" lower the 'tolerance' parameter (i.e. the",
"tolerance for floating-point aritmetics):in the later case be",
"critical of the result."))
}
# Test and remove values that are multiples of the smallest value
remain1 <- (d - floor(d + 10^-(tolerance - 1)))
if(!relative) remain1 <- remain1 * mx
rzero1 <- abs(remain1) < 10^-tolerance
d <- d[!rzero1]
if(length(d) == 0) {
if(!relative){
res <- round(mx, tolerance)
} else {
res <- signif(mx, tolerance)
}
} else {
# Multiply d [x/min(x)] by integers, and test if
# this returns only integers within tolerance
ld <- length(d)
try_order_OLD <- 0
try_order_i <- 6 - ceiling(log10(ld))
for(i in seq_len(as.integer(tries))){
if(speak) {
print(paste("Try ",i,": 10^",try_order_i,
" possibilities tested", sep = ""))
}
t <- 1:(10^try_order_i)
t <- t[-(1:(10^try_order_OLD))]
lt <- length(t)
tmat <- matrix(rep(t,ld), ncol = ld)
dmat <- matrix(rep(d, lt), ncol = ld, byrow = T)
test <- dmat * tmat
remain2 <- (test - floor(test + 10^-(tolerance - 1)))
remain2 <- remain2/tmat
if(!relative) remain2 <- remain2 * mx
rzero2 <- abs(remain2) < 10^-tolerance
rzero2 <- matrix(as.integer(rzero2), ncol = ld)
test[which(rowSums(rzero2) == ld),]
res <- mx/t[which(rowSums(rzero2) == ld)[1]]
if(!is.na(res)) break
try_order_OLD <- try_order_i
try_order_i <- try_order_i + 1
}
}
return(res)
}
Given a function f(x,c,d) of x that also depends on some parameters c and d. I would like to find the zeroes for a cartesian product of certain values c_1,...,c_n and d_1,...,d_m of the parameters, i.e. an x_ij such that f(x_ij,c_i,d_j)=0 for i=1,...,n and j=1,...,m. Although not that crucial I am applying a Newton-Raphson algorithm for the root finding:
newton.raphson <- function(f, a, b, tol = 1e-5, n = 1000){
require(numDeriv) # Package for computing f'(x)
x0 <- a # Set start value to supplied lower bound
k <- n # Initialize for iteration results
# Check the upper and lower bounds to see if approximations result in 0
fa <- f(a)
if (fa == 0.0){
return(a)
}
fb <- f(b)
if (fb == 0.0) {
return(b)
}
for (i in 1:n) {
dx <- genD(func = f, x = x0)$D[1] # First-order derivative f'(x0)
x1 <- x0 - (f(x0) / dx) # Calculate next value x1
k[i] <- x1 # Store x1
# Once the difference between x0 and x1 becomes sufficiently small, output the results.
if (abs(x1 - x0) < tol) {
root.approx <- tail(k, n=1)
res <- list('root approximation' = root.approx, 'iterations' = k)
return(res)
}
# If Newton-Raphson has not yet reached convergence set x1 as x0 and continue
x0 <- x1
}
print('Too many iterations in method')
}
The actual function that I am interest is more complicated, but the following example illustrates my problem.
test.function <- function(x=1,c=1,d=1){
return(c*d-x)
}
Then for any given c_i and d_j I can easily calculate the zero by
newton.raphson(function(x) test.function(x,c=c_i,d=d_j),0,1)[1]
which here is obviously just the product c_i*d_j.
Now I tried to define a function that finds for two given vectors (c_1,...,c_n) and (d_1,...,d_m) the zeroes for all combinations. For this, I tried to define
zeroes <- function(ci=1,dj=1){
x<-newton.raphson(function(x) test.function(x,c=ci,d=dj),0,1)[1]
return(as.numeric(x))
}
and then use the outer-function, e.g.
outer(c(1,2),c(1,2,3),FUN=zeroes)
Unfortunately, this did not work. I got an error message
Error during wrapup: dims [product 6] do not match the length of object [1]
There might be also a much better solution to my problem. I am happy for any input.
I am trying to create a function that computes the sum of digits of large numbers, of the order of 100^100. The approach described in this question does not work, as shown below. I tried to come up with a function that does the job, but have not been able to get very far.
The inputs would be of the form a^b, where 1 < a, b < 100 and a and b are integers. So, in that sense, I am open to making digitSumLarge a function that accepts two arguments.
digitSumLarge <- function(x) {
pow <- floor(log10(x)) + 1L
rem <- x
i <- 1L
num <- integer(length = pow)
# Individually isolate each digit starting from the largest and store it in num
while(rem > 0) {
num[i] <- rem%/%(10^(pow - i))
rem <- rem%%(10^(pow - i))
i <- i + 1L
}
return(num)
}
# Function in the highest voted answer of the linked question.
digitsum <- function(x) sum(floor(x / 10^(0:(nchar(x) - 1))) %% 10)
Consider the following tests:
x <- c(1,2,3,4,5,6,7,8,9,1,2,3,4,5,6,7,8,9)
as.numeric(paste(x, collapse = ''))
# [1] 1.234568e+17
sum(x)
# 90
digitSumLarge(as.numeric(paste(x, collapse = '')))
# 85
digitsum(as.numeric(paste(x, collapse = '')))
# 81, with warning message about loss of accuracy
Is there any way I can write such a function in R?
You need arbitrary precision numbers. a^b with R's numerics (double precision floats) can be only represented with limited precision and not exactly for sufficiently large input.
library(gmp)
a <- as.bigz(13)
b <- as.bigz(67)
sum(as.numeric(strsplit(as.character(a^b), split = "")[[1]]))
#[1] 328
I am re-writting an algorithm I did in C++ in R for practice called the Finite Difference Method. I am pretty new with R so I don't know all the rules regarding vector/matrix multiplication. For some reason I am getting a non-conformable arguments error when I do this:
ST_u <- matrix(0,M,1)
ST_l <- matrix(0,M,1)
for(i in 1:M){
Z <- matrix(gaussian_box_muller(i),M,1)
ST_u[i] <- (S0 + delta_S)*exp((r - (sigma*sigma)/(2.0))*T + sigma*sqrt(T)%*%Z)
ST_l[i] <- (S0 - delta_S)*exp((r - (sigma*sigma)/(2.0))*T + sigma*sqrt(T)%*%Z)
}
I get this error:
Error in sqrt(T) %*% Z : non-conformable arguments
Here is my whole code:
gaussian_box_muller <- function(n){
theta <- runif(n, 0, 2 * pi)
rsq <- rexp(n, 0.5)
x <- sqrt(rsq) * cos(theta)
return(x)
}
d_j <- function(j, S, K, r, v,T) {
return ((log(S/K) + (r + (-1^(j-1))*0.5*v*v)*T)/(v*(T^0.5)))
}
call_delta <- function(S,K,r,v,T){
return (S * dnorm(d_j(1, S, K, r, v, T))-K*exp(-r*T) * dnorm(d_j(2, S, K, r, v, T)))
}
Finite_Difference <- function(S0,K,r,sigma,T,M,delta_S){
ST_u <- matrix(0,M,1)
ST_l <- matrix(0,M,1)
for(i in 1:M){
Z <- matrix(gaussian_box_muller(i),M,1)
ST_u[i] <- (S0 + delta_S)*exp((r - (sigma*sigma)/(2.0))*T + sigma*sqrt(T)%*%Z)
ST_l[i] <- (S0 - delta_S)*exp((r - (sigma*sigma)/(2.0))*T + sigma*sqrt(T)%*%Z)
}
Delta <- matrix(0,M,1)
totDelta <- 0
for(i in 1:M){
if(ST_u[i] - K > 0 && ST_l[i] - K > 0){
Delta[i] <- ((ST_u[i] - K) - (ST_l[i] - K))/(2*delta_S)
}else{
Delta <- 0
}
totDelta = totDelta + exp(-r*T)*Delta[i]
}
totDelta <- totDelta * 1/M
Var <- 0
for(i in 1:M){
Var = Var + (Delta[i] - totDelta)^2
}
Var = Var*1/M
cat("The Finite Difference Delta is : ", totDelta)
call_Delta_a <- call_delta(S,K,r,sigma,T)
bias <- abs(call_Delta_a - totDelta)
cat("The bias is: ", bias)
cat("The Variance of the Finite Difference method is: ", Var)
MSE <- bias*bias + Var
cat("The marginal squared error is thus: ", MSE)
}
S0 <- 100.0
delta_S <- 0.001
K <- 100.0
r <- 0.05
sigma <- 0.2
T <- 1.0
M <- 10
result1 <- Finite_Difference(S0,K,r,sigma,T,M,delta_S)
I can't seem to figure out the problem, any suggestions would be greatly appreciated.
In R, the %*% operator is reserved for multiplying two conformable matrices. As one special case, you can also use it to multiply a vector by a matrix (or vice versa), if the vector can be treated as a row or column vector that conforms to the matrix; as a second special case, it can be used to multiply two vectors to calculate their inner product.
However, one thing it cannot do is perform scalar multipliciation. Scalar multiplication of vectors or matrices always uses the plain * operator. Specifically, in the expression sqrt(T) %*% Z, the first term sqrt(T) is a scalar, and the second Z is a matrix. If what you intend to do here is multiply the matrix Z by the scalar sqrt(T), then this should just be written sqrt(T) * Z.
When I made this change, your program still didn't work because of another bug -- S is used but never defined -- but I don't understand your algorithm well enough to attempt a fix.
A few other comments on the program not directly related to your original question:
The first loop in Finite_Difference looks suspicious: guassian_box_muller(i) generates a vector of length i as i varies in the loop from 1 up to M, and forcing these vectors into a column matrix of length M to generate Z is probably not doing what you want. It will "reuse" the values in a cycle to populate the matrix. Try these to see what I mean:
matrix(gaussian_box_muller(1),10,1) # all one value
matrix(gaussian_box_muller(3),10,1) # cycle of three values
You also use loops in many places where R's vector operations would be easier to read and (typically) faster to execute. For example, your definition of Var is equivalent to:
Var <- sum((Delta - totDelta)^2)/M
and the definitions of Delta and totDelta could also be written in this simplified fashion.
I'd suggest Googling for "vector and matrix operations in r" or something similar and reading some tutorials. Vector arithmetic in particular is idiomatic R, and you'll want to learn it early and use it often.
You might find it helpful to consider the rnorm function to generate random Gaussians.
Happy R-ing!
This question already has answers here:
How would you program Pascal's triangle in R?
(2 answers)
How to work with large numbers in R?
(1 answer)
Closed 6 years ago.
For a class assignment, I need to create a function that calculates n Choose k. I did just that, and it works fine with small numbers (e.g. 6 choose 2), but I'm supposed to get it work with 200 choose 50, where it naturally doesn't.
The answer is too large and R outputs NaN or Inf, saying:
> q5(200, 50)
[1] "NaN"
Warning message:
In factorial(n) : value out of range in 'gammafn'
I tried using logs and exponents, but it doesn't cut it.
q5 <- function (n, k) {
answer <- log(exp( factorial(n) / ( (factorial(k)) * (factorial(n - k)) )))
paste0(answer)
}
The answer to the actual question is that R cannot show numbers it cannot represent, and some of the terms in your equation are too big to represent. So it fails. However there are approximations to factorial that can be used - they work with logarithms which get big a lot slower.
The most famous one, Sterling's approximation, was not accurate enough, but the Ramanujan's approximation came to the rescue :)
ramanujan <- function(n){
n*log(n) - n + log(n*(1 + 4*n*(1+2*n)))/6 + log(pi)/2
}
nchoosek <- function(n,k){
factorial(n)/(factorial(k)*factorial(n-k))
}
bignchoosek <- function(n,k){
exp(ramanujan(n) - ramanujan(k) - ramanujan(n-k))
}
nchoosek(20,5)
# [1] 15504
bignchoosek(20,5)
# [1] 15504.06
bignchoosek(200,50)
# [1] 4.538584e+47
You can try this too:
q5 <- function (n, k) {
# nchoosek = (n-k+1)(n-k+2)...n / (1.2...k)
return(prod(sapply(1:k, function(i)(n-k+i)/(i))))
}
q5(200, 50)
#[1] 4.538584e+47
or in log domain
q5 <- function (n, k) {
# ln (nchoosek) = ln(n-k+1) + ln(n-k+2) + ...+ ln(n) - ln(1) -ln(2) - ...- ln(k)
return(exp(sum(sapply(1:k, function(i)(log(n-k+i) - log(i))))))
}
q5(200, 50)
#[1] 4.538584e+47
The packages for large numbers:
Brobdingnag package for "Very large numbers in R":
https://cran.r-project.org/web/packages/Brobdingnag/index.html
Paper: https://www.researchgate.net/publication/251996764_Very_large_numbers_in_R_Introducing_package_Brobdingnag
library(Brobdingnag)
googol <- as.brob(10)^100 # googol:=10^100
googol
# [1] +exp(230.26) # exponential notation is convenient for very large numbers
gmp package for multiple Precision Arithmetic (big integers and rationals, prime number tests, matrix computation):
https://cran.r-project.org/web/packages/gmp/index.html
This solution calculates the complete row of the Pascal triangle:
x <- 1
print(x)
for (i in 1:200) { x <- c(0, x) + c(x, 0); print(x) }
x[51] ### 200 choose 50
## > x[51]
## [1] 4.538584e+47
(as I proposed for How would you program Pascal's triangle in R? )
If you want to speed up the code then do not the print(x) (output is a relative slow operation).
To put the code in a function we can do
nchoosek <- function(n,k) {
x <- 1
for (i in 1:n) x <- c(0, x) + c(x, 0)
x[k+1] ### n choose k
}
nchoosek(200, 50) ### testing the function
## [1] 4.538584e+47
Here is a more refined version of my function:
nchoosek <- function(n, k) {
if (k==0) return(1)
if (k+k > n) k <- n-k
if (k==0) return(1)
x <- 1
for (i in 1:k) x <- c(0, x) + c(x, 0)
for (i in 1:(n-k)) x <- x + c(0, head(x, -1))
tail(x, 1)
}
nchoosek(200, 50) ### testing the function
## [1] 4.538584e+47