Barnsley fern: Difference between revisions
m
→'Obvious' solution: Improved syntax.
ReeceGoding (talk | contribs) m (→{{header|R}}: Syntax highlighting.) |
ReeceGoding (talk | contribs) m (→'Obvious' solution: Improved syntax.) |
||
Line 2,416:
==='Obvious' solution===
The matrix solution above is a clever approach, but the following solution is more readable if you're unfamiliar with linear algebra. This is very much a blind "just do what the task says" solution. It's so simple that it probably runs unadapted in S. I suspect that there is room for an interesting use of R's ifelse function somewhere, but I couldn't find a clean way.
<lang rsplus>fernOfNPoints <- function(n)
{
currentX <- currentY <- newX <- newY <- 0
plot(0, 0, xlim = c(-2, 3), ylim = c(0, 10), xlab = "", ylab = "", pch = 20, col = "darkgreen", cex = 0.1)
f1 <- function()#ran 1% of the time
{
newX <<- 0
newY <<- 0.16 * currentY
}
f2 <- function()#ran 85% of the time
{
newX <<- 0.85 * newX + 0.04 * newY
newY <<- -0.04 * newX + 0.85 * newY + 1.6
}
f3 <- function()#ran 7% of the time
{
newX <<- 0.2 * newX - 0.26 * newY
newY <<- 0.23 * newX + 0.22 * newY + 1.6
}
f4 <- function()#ran 7% of the time
{
newX <<- -0.15 * newX + 0.28 * newY
newY <<- 0.26 * newX + 0.24 * newY + 0.44
}
{
case <- runif(1)▼
if(case <= 0.01) f1()▼
else if(case <= 0.86) f2()▼
else if(case <= 0.93) f3()▼
else f4()▼
points(newX, newY, pch = 20, col = "darkgreen", cex = 0.1)▼
}
▲ for(i in 2:n)#We've already plotted (0,0), so we can skip one run.
▲ case<-runif(1)
▲ if(case<=0.01)f1()
▲ else if(case<=0.86)f2()
▲ else if(case<=0.93)f3()
▲ else f4()
▲ points(newX,newY,pch=20,col="darkgreen",cex=0.1)
return(invisible())
}
| |||