From luke-tierney at uiowa.edu  Fri Mar 16 16:37:03 2018
From: luke-tierney at uiowa.edu (luke-tierney at uiowa.edu)
Date: Fri, 16 Mar 2018 10:37:03 -0500 (CDT)
Subject: [Rd] Discrepancy: R sum() VS C or Fortran sum
In-Reply-To: <1885e1e6-374f-21c0-4ea3-5aa2e8587ac2@mailservices.uwaterloo.ca>
References: <1885e1e6-374f-21c0-4ea3-5aa2e8587ac2@mailservices.uwaterloo.ca>
Message-ID: <alpine.DEB.2.20.1803161026280.4709@luke-Latitude>

Install the gmp package, run your code, and then try this:

bu <- gmp::as.bigq(u)
bs4 <- bu[1] + bu[2] + bu[3] + bu[4] + bu[5]
s4 <- as.double(bs4)
s1 - s4
##  [1] 0
s2[[2]] - s4
##  [1] 7.105427e-15
s3 - s4
##  [1] 7.105427e-15
identical(s1, s4)
##  [1] TRUE

`bs4` is the exact sum of the binary rationals in your `u` vector;
`s4` is the closest double precision to this exact sum.

Looking at the C source code for sum() will show you that it makes
some extra efforts to get a more accurate sum than your simple
version.

Best,

luke

On Fri, 16 Mar 2018, Pierre Chausse wrote:

> Hi all,
>
> I found a discrepancy between the sum() in R and either a sum done in C or 
> Fortran for vector of just 5 elements. The difference is very small, but this 
> is a very small part of a much larger numerical problem in which first and 
> second derivatives are computed numerically. This is part of a numerical 
> method course I am teaching in which I want to compare speeds of R versus 
> Fortran (We solve a general equilibrium problem all numerically, if you want 
> to know). Because of this discrepancy, the Jacobian and Hessian in R versus 
> in Fortran are quite different, which results in the Newton method producing 
> a different solution (for a given stopping rule). Since the solution computed 
> in Fortran is almost identical to the analytical solution, I suspect that the 
> sum in Fortran may be more accurate (That's just a guess).  Most of the time 
> the sum produces identical results, but for some numbers, it is different. 
> The following example, shows what happens:
>
> set.seed(12233)
> n <- 5
> a <- runif(n,1,5)
> e <- runif(n, 5*(1:n),10*(1:n))
> s <- runif(1, 1.2, 4)
> p <- runif(5, 3, 10)
> x <- c(e[-5], (sum(e*p)-sum(e[-5]*p[-5]))/p[5])
> u <- a^(1/s)*x^((s-1)/s)
> dyn.load("sumF.so")
>
> u[1] <- u[1]+.0001 ### If we do not add .0001, all differences are 0
> s1 <- sum(u)
> s2 <- .Fortran("sumf", as.double(u), as.integer(n), sf1=double(1),
>               sf2=double(1))[3:4]
> s3 <- .C("sumc", as.double(u), as.integer(n), sC=double(1))[[3]]
>
> s1-s2[[1]] ## R versus compiler sum() (Fortran)
>
> [1] -7.105427e-15
>
> s1-s2[[2]] ## R versus manual sum (Fortran
>
> [1] -7.105427e-15
>
> s1-s3 ## R Versus manual sum in C
>
> [1] -7.105427e-15
>
> s2[[2]]-s2[[1]] ## manual sum versus compiler sum() (Fortran)
>
> [1] 0
>
> s3-s2[[2]] ## Fortran versus C
>
> [1] 0
>
> My sumf and sumc are
>
>      subroutine sumf(x, n, sx1, sx2)
>      integer i, n
>      double precision x(n), sx1, sx2
>      sx1 = sum(x)
>      sx2 = 0.0d0
>      do i=1,n
>         sx2 = sx2+x(i)
>      end do
>      end
>
> void sumc(double *x, int *n, double *sum)
> {
>  int i;
>  double sum1 = 0.0;
>  for (i=0; i< *n; i++) {
>    sum1 += x[i];
>  }
>  *sum = sum1;
> }
>
> Can that be a bug?  Thanks.
>
>

-- 
Luke Tierney
Ralph E. Wareham Professor of Mathematical Sciences
University of Iowa                  Phone:             319-335-3386
Department of Statistics and        Fax:               319-335-3017
    Actuarial Science
241 Schaeffer Hall                  email:   luke-tierney at uiowa.edu
Iowa City, IA 52242                 WWW:  http://www.stat.uiowa.edu



[Rd] Discrepancy: R sum() VS C or Fortran sum
luke-tierney at uiowa.edu