Splitting Area under curve into equal portions

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Hi Milton,

Not quite, that would be an equal number of data points in each colour group.
What I want is an unequal number of points in each group such that:
sum(work[group.members]) is approximately the same for each group of data points.

In the mean time, I came up with the following, and took a leaf out of your book
with the colouring for example:

<code>
n <- 2002
work <- vector()
for(x in 1:(n-2)) {
  work[x] <- ((n-1-x)*(n-x))/2
}
plot(work)

tasks <- vector('list')
tasks_per_slave <- 1
work_per_task <- sum(work) / (n_slaves * tasks_per_slave)

# Now define ranges of x of equal "work"
block_start <- 1
for(x in (1:(length(work)))) {
  if(x == length(work)) {
    # this will be the last block
    tasks[[length(tasks)+1]] <- list(x=block_start:length(work))
    break
  }
  work_in_block_to_x <- sum(work[block_start:(x)])

  if(work_in_block_to_x > work_per_task) {
    # use this value of x as the chunk end
    tasks[[length(tasks)+1]] <- list(x=block_start:x)

    # move the block_start position
    block_start <- x+1
  }
}

colours <- vector()
for(i in 1:length(tasks)) {
  colours <- append(colours,rep(i,length(tasks[[i]]$x)))
}

plot(work, col=colours)
</code>

Essentially, the area under the line for each of the coloured groups (i.e. the
total work associated with those values of x) should be approximately equal and
I believe the above code achieves this. Just found the cumsum() function. You
could look at it this way:

<code>
plot(cumsum(work), col=colours)
</code>

The coloured groupings coincide with splitting the cumulative total (y-axis)
into 4 approximately equal bits.

There must be a nicer way to do this!
Nathan

Splitting Area under curve into equal portions

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