Perfect shuffle: Difference between revisions
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→Sequence solution: Improved syntax.
ReeceGoding (talk | contribs) m (→{{header|R}}: Syntax highlighting.) |
ReeceGoding (talk | contribs) m (→Sequence solution: Improved syntax.) |
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The previous solution exploits R's matrix construction; This solution exploits its array indexing.
<lang rsplus>#A strict reading of the task description says that we need a function that does exactly one shuffle.
pShuffle <- function(deck)
{
n <- length(deck)#Assumed even (as in task description).
new <- array(n)#Maybe not as general as it could be, but the task said to use whatever was convenient.
new[seq(from = 1, to = n, by = 2)] <- deck[seq(from = 1, to = n/2, by = 1)]
new[seq(from = 2, to = n, by = 2)] <- deck[seq(from = 1 + n/2, to = n, by = 1)]
new
}
task2 <- function(deck)
{
new <- deck
count <- 0
repeat
{
new <- pShuffle(new)
count <- count + 1
if(all(new == deck))
}
cat("It takes", count, "shuffles of a deck of size", length(deck), "to return to the original deck.","\n")
invisible(count)#For the unit tests. The task wanted this printed so we only return it invisibly.
}
#Tests - All done in one line.
mapply(function(x, y) task2(1:x) == y, c(8, 24, 52, 100, 1020, 1024, 10000), c(3, 11, 8, 30, 1018, 10, 300))</lang>
{{out}}
<pre>> mapply(function(x,y) task2(1:x)==y,c(8,24,52,100,1020,1024,10000),c(3,11,8,30,1018,10,300))
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