# Re: [Rd] When 1+2 != 3 (PR#9895)

From: Petr Savicky <savicky_at_cs.cas.cz>
Date: Mon, 03 Sep 2007 16:39:57 +0200

On Mon, Sep 03, 2007 at 08:59:22AM +0200, marco.vicentini_at_gmail.com wrote:
> Full_Name: Marco Vicentini, University of Verona
> Version: 2.4.1 & 2.5.1
> OS: OsX & WinXP
> Submission from: (NULL) (157.27.253.46)
>
>
> When I proceed to test the following equation 1 + 2 == 3, I obviously obtain the
> value TRUE. But when I tryed to do the same using real number (i.e. 0.1 + 0.2 ==
> 0.3) I obtained an unusual FALSE.
> In the online help there are some tricks for this problem. It suggests to use
> identical(...) which again answer FALSE. Only using isTRUE(all.equal(0.3, 0.1 +
> 0.2)) I can obtain the true value TRUE.

A rational number has a finite binary expansion iff its denominator is a power of 2. Numbers 0.1 and 0.2 are 1/10 and 1/5, so they have 5 in their denominator. Their binary expansion is
0.1 = .0001100110011001100110011001100110...  0.2 = .0011001100110011001100110011001100... A double variable stores the numbers rounded to 53 significant binary digits. Hence, they are not exactly 0.1 and 0.2, as may be seen in

```  formatC(0.1,digits=30) # [1] "0.100000000000000005551115123126"
```
formatC(0.2,digits=30) # [1] "0.200000000000000011102230246252"

In order to compare numbers with some tolerance, the function all.equal may be used, which you also mention below. See its help page, which specifies the tolerance to be .Machine\$double.eps ^ 0.5.

> But the problem does not concern only the operator ==. Many other functions,
> among over: sort, order, unique, duplicate, identical are not able to deal with
> this problem. This is very dangerous because no advice are provide by the online
> help, and anybody can use these functions no think to unusual results.
>
> I think that the problem is due to how double number are store by the C
> compiler.

Not C compiler, but the hardware.

Petr Savicky.

> If it may be usefull, I have written to small function (Unique and isEqual)
> which can deal with this problem of the double numbers.
>
> I also add some other conditions for the same problem.
>
> 0.3 == 0.15 + 0.15
> 0.3 == 0.1 + 0.2
> 1 - 0.7 == 0.3
> 0.1 == 1 - 0.9
>
> 0.2 == 1 - 0.2 - 0.2 - 0.2 - 0.2
> -0.2 == 1 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2
>
> identical (0.3, 0.1 + 0.2)
> all.equal (0.3, 0.1 + 0.2)
>
> identical (-0.2 , 1 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2)
> all.equal (-0.2 , 1 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2)
>
> isTRUE( all.equal (-0.2 , 1 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2) )
>
>
> -0.2 == 1 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2
>
> a= -0.2
> b= 1 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2 - 0.2
>
> x<-c(a,b)
> sprintf("%.15f",x)
> sprintf("%.50f",x)
>
>
>
> Unique <- function(x, digits = 8, fast = TRUE) {
>
> if (fast) {
> unique (round(x * 10^digits)) / 10^digits
> } else {
> x = sort(x)
> for (i in 1:(length(x)-1))
> if (isTRUE(all.equal(x[i],x[i+1]))) x[i] = NaN
> x [ which (!is.nan(x)) ]
> }}
>
> isEqual <- function (object, x, tol = 1e-9) {
> if (!is.vector(object)) stop("Object must be a vector")
> if (is.character(object)) stop("Object can not be a character")
> if (!is.real(x)) stop("x must be a real number")
> if (any(is.na(c(object,x)))) stop("NA is not supported")
> if (length(x) != 1) stop("length x must equal to 1")
>
> ifelse (abs(object - x) < tol, TRUE,FALSE)
> # .Call("isEqual",as.real(object),as.real(x),as.real(tol), PACKAGE="mvUtils")
> }
>
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