[R] Season's Greetings (and great news ... )!
Suzen, Mehmet
msuzen at gmail.com
Sun Dec 22 17:54:50 CET 2013
I wouldn't blame R for floating-point arithmetic and our personal
feeling of what 'zero' should be.
> options(digits=20)
> pi
[1] 3.141592653589793116
> sqrt(pi)^2
[1] 3.1415926535897926719
> (pi - sqrt(pi)^2) < 1e-15
[1] TRUE
There was a similar post before, for example see:
http://r.789695.n4.nabble.com/Why-does-sin-pi-not-return-0-td4676963.html
There is an example by Martin Maechler (author of Rmpfr) on how to use
arbitrary precision
with your arithmetic.
On 22 December 2013 10:59, Ted Harding <Ted.Harding at wlandres.net> wrote:
> Greetings All!
> With the Festive Season fast approaching, I bring you joy
> with the news (which you will surely wish to celebrate)
> that R cannot do arithmetic!
>
> Usually, this is manifest in a trivial way when users report
> puzzlement that, for instance,
>
> sqrt(pi)^2 == pi
> # [1] FALSE
>
> which is the result of a (generally trivial) rounding or
> truncation error:
>
> sqrt(pi)^2 - pi
> [1] -4.440892e-16
>
> But for some very simple calculations R goes off its head.
>
> I had originally posted this example some years ago, but I
> have since generalised it, and the generalisation is even
> more entertaining than the original.
>
> The Original:
> Consider a sequence generated by the recurrence relation
>
> x[n+1] = 2*x[n] if 0 <= x[n] <= 1/2
> x[n+1] = 2*(1 - x[n]) if 1/2 < x[n] <= 1
>
> (for 0 <= x[n] <= 1).
>
> This has equilibrium points (x[n+1] = x[n]) at x[n] = 0
> and at x[n] = 2/3:
>
> 2/3 -> 2*(1 - 2/3) = 2/3
>
> It also has periodic points, e.g.
>
> 2/5 -> 4/5 -> 2/5 (period 2)
> 2/9 -> 4/9 -> 8/9 -> 2/9 (period 3)
>
> The recurrence relation can be implemented as the R function
>
> nextx <- function(x){
> if( (0<=x)&(x<=1/2) ) {x <- 2*x} else {x <- 2*(1 - x)}
> }
>
> Now have a look at what happens when we start at the equilibrium
> point x = 2/3:
>
> N <- 1 ; x <- 2/3
> while(x > 0){
> cat(sprintf("%i: %.9f\n",N,x))
> x <- nextx(x) ; N <- N+1
> }
> cat(sprintf("%i: %.9f\n",N,x))
>
> Run that, and you will see that successive values of x collapse
> towards zero. Things look fine to start with:
>
> 1: 0.666666667
> 2: 0.666666667
> 3: 0.666666667
> 4: 0.666666667
> 5: 0.666666667
> ...
>
> but, later on,
>
> 24: 0.666666667
> 25: 0.666666666
> 26: 0.666666668
> 27: 0.666666664
> 28: 0.666666672
> ...
>
> 46: 0.667968750
> 47: 0.664062500
> 48: 0.671875000
> 49: 0.656250000
> 50: 0.687500000
> 51: 0.625000000
> 52: 0.750000000
> 53: 0.500000000
> 54: 1.000000000
> 55: 0.000000000
>
> What is happening is that, each time R multiplies by 2, the binary
> representation is shifted up by one and a zero bit is introduced
> at the bottom end. To illustrate this, do the calculation in
> 7-bit arithmetic where 2/3 = 0.1010101, so:
>
> 0.1010101 x[1], >1/2 so subtract from 1 = 1.0000000 -> 0.0101011,
> and then multiply by 2 to get x[2] = 0.1010110. Hence
>
> 0.1010101 x[1] -> 2*(1 - 0.1010101) = 2*0.0101011 ->
> 0.1010110 x[2] -> 2*(1 - 0.1010110) = 2*0.0101010 ->
> 0.1010100 x[3] -> 2*(1 - 0.1010100) = 2*0.0101100 ->
> 0.1011000 x[4] -> 2*(1 - 0.1011000) = 2*0.0101000 ->
> 0.1010000 x[5] -> 2*(1 - 0.1010000) = 2*0.0110000 ->
> 0.1100000 x[6] -> 2*(1 - 0.1100000) = 2*0.0100000 ->
> 0.1000000 x[7] -> 2*0.1000000 = 1.0000000 ->
> 1.0000000 x[8] -> 2*(1 - 1.0000000) = 2*0 ->
> 0.0000000 x[9] and the end of the line.
>
> The final index of x[i] is i=9, 2 more than the number of binary
> places (7) in this arithmetic, since 8 successive zeros have to
> be introduced. It is the same with the real R calculation since
> this is working to .Machine$double.digits = 53 binary places;
> it just takes longer (we reach 0 at x[55])! The above collapse
> to 0 occurs for any starting value in this simple example (except
> for multiples of 1/(2^k), when it works properly).
>
> Generalisation:
> This is basically the same, except that everything is multiplied
> by a scale factor S, so instead of being on the interval [0,1].
> it is on [0,S], and
>
> x[n+1] = 2*x[n] if 0 <= x[n] <= S/2
> x[n+1] = 2*(S - x[n]) if S/2 < x[n] <= S
> (for 0 <= x[n] <= S).
>
> Again, x[n] = 2*S/3 is an equilibrium point. 2*S/3 > S/2, so
>
> x[n] -> 2*(S - 2*S/3) = 2*(S/3) = 2*S/3
>
> Functions to implement this:
>
> nxtS <- function(x,S){
> if((x >= 0)&(x <= S/2)){ x<- 2*x } else {x <- 2*(S-x)}
> }
>
> S <- 6 ## Or some other value of S
> Nits <- 100
> x <- 2*S/3
> N <- 1 ; print(c(N,x))
> while(x>0){
> if(N > Nits) break ### to stop infinite looping
> N <- (N+1) ; x <- nxtS(x,S)
> print(c(N,x))
> }
>
> The behaviour of the sequence now depends on the value of S.
>
> If S is a multiple of 3, then with x[1] = 2*S/3 the equilibrium
> is immediately attained and x[n] = 2*S/3 forever after, since
> R is now calculating with integers. E.g. try the above with S<-6
> That is what arithmetic ought to be like! But for S not a multiple
> of 3 one can get the impression that R is on some sort of drug!
>
> For other values of S (but not all) we observe the same collapse
> to x=0 as before, and again it takes 54 steps (ending with x[55]).
> Try e.g. S <- 16
>
> For some values of S, however, the iteration ends up in a periodic loop.
>
> For example, with S<-7, at x[52] we get x[52]=4, x[53]=6, x[54]=2,
> and then 4 6 2 4 6 2 4 6 2 ... forever (or until Nits cuts in),
> so period = 3.
>
> For S<-11, x[52]=8 then 6 then 10 then 2 then 4 then 8 6 10 2 4 ...
> so period = 5.
>
> For S<-13, x[51]=4 then 8 10 6 12 2 4 8 10 6 12 2 4 8 ...
> so period = 6.
>
> For S<-19, x[51]=12 then 14 10 18 2 4 8 16 6 12 ...
> so period = 9.
>
> And so on ...
>
> So, one sniff of something like S<-19, and R is off its head!
>
> All it has to do is multiply by 2 -- and it gets it cumulatively wrong!
> R just doesn't add up ...
>
> Season's Greetings to all -- and may your calculations always
> be accurate -- to within machine precision ...
>
> Ted.
>
> -------------------------------------------------
> E-Mail: (Ted Harding) <Ted.Harding at wlandres.net>
> Date: 22-Dec-2013 Time: 09:59:00
> This message was sent by XFMail
>
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