Hofstadter Figure-Figure sequences: Difference between revisions
Added R. Rather short!
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ReeceGoding (talk | contribs) (Added R. Rather short!) |
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=={{header|R}}==
Global variables aren't idiomatic R, but this is otherwise an ideal task for the language. Comments aside, this is easily one of the shortest solutions on this page. This is mostly due to how R treats most things as a vector. For example, rValues starts out as the number 1, but repeatedly has new values appended to it without much ceremony.
<lang R>rValues <- 1
sValues <- 2
ffr <- function(n)
{
if(!is.na(rValues[n])) rValues[n] else (rValues[n] <<- ffr(n-1) + ffs(n-1))
}
#In theory, generating S requires computing ALL values not in R.
#That would be infinitely many values.
#However, to generate S(n) we only need to observe that its value cannot exceed R(n)+1.
ffs <- function(n)
{
if(!is.na(sValues[n])) sValues[n] else (sValues[n] <<- setdiff(seq_len(1 + ffr(n)), rValues)[n])
}
#Task 1
invisible(ffr(10))
print(rValues)
#Task 2
#If we try to call ffs(960) directly, R will complain about the stack being too big.
#Calling ffs(500) first solves this problem.
invisible(ffs(500))
invisible(ffs(960))
#In R, "the first 40 values of ffr plus the first 960 values of ffs" can easily be misread.
#rValues[1:40]+sValues[1:960] is valid R code. It will duplicate the first 40 rValues 24
#times, append them, and add that vector to the first 960 sValues. This gives an output of
#length 960, which clearly cannot contain 1000 different values.
#Presumably, the task wants us to append rValues[1:40] and sValues[1:960].
counts <- table(c(rValues[1:40], sValues[1:960]))
print(all(counts == 1))</lang>
{{out}}
<pre>> print(rValues)
[1] 1 3 7 12 18 26 35 45 56 69
> print(all(counts == 1))
[1] TRUE</pre>
=={{header|Racket}}==
{{trans|Java}}
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