Ask HN: algorithm to slice a number into parts

6 pointsby admvor 18 Jahren14 comments

I am trying to write a algorithm to slice a number into parts; such that those slices add upto original number.

  e.g. n = (n1 + n2+ .. + nm)
  where n is number and m no of slices.

  n will be positive integer and m won't exceed 50.

Do you know any algorithm for this?

Edit: slices should be of distinct values.

14 comments

cpercivavor 18 Jahren

I think your problem description is missing something; otherwise n1 = n2 = n3 ... = n[m-1] = 1, nm = n - m + 1 is a trivial solution.

cpercivavor 18 Jahren

Edit: slices should be of distinct values.

In that case, [1, 2, 3, ... m - 1, n - m*(m-1)/2] is a trivial solution.

inklesspenvor 18 Jahren

Is this your homework?

RiderOfGiraffesvor 18 Jahren

Edit: slices should be of distinct values.

You can't do it with positive integers if n<m(m+1)/2, or with non-negative integers if n<m(m-1)/2. If there is no restriction then there are, again, trivial solutions.

Why are you doing this? As someone else has said, this does smell of homework. If it is homework, then tell us what you've tried. If it isn't, then tell us what it's for.

You continue to underspecify your problem. What are you trying to achieve?

RiderOfGiraffesvor 18 Jahren

As others have said, your problem as stated is trivial. Here's a degenerate solution:

    n1 = n2 = ... = n(m-1) = 0
    nm = n

I'm sure that's not what you want.

Here's another solution:

    n1 = n2 = ... = nm = n/m

I'm sure that's not what you want either.

Others have asked relevant questions. Assuming you want the numbers to be integers, and as equal as possible, then compute:

    k_min = floor(n/m)
    excess = n-k_min*m

All will be at least k_min. If they are all k_min, then you will have a total of k_min*m. You need an additional "excess", so assign them, one each, to the first bunch.

    n1 = k_min + 1
    ...
    n(excess) = k_min + 1
    n(excess+1) = k_min
    ...
    nm = k_min

If you want you can reverse this so that the larger ones come after. That's left as an exercise for the interested reader.

You could also put all the excess in one place, so you have

  n1 = n2 = ... = n(m-1) = k_min
  nm = k_min + excess

So really, it all depends on what you want.

VarunGuptavor 18 Jahren

There can be many ways to do so, one of them is already described by cperciva.

Another can be,

a) All Integers n1 = n2 = n3 =.....= n[m-1] = int(n / m) and nm = n - (n1 + n2 + n3 +.....+ n[m-1])

b) Floats n1 = n2 = n3 =.....= n[m] = n / m

I feel the problem can be more serious if you want the parts to be in certain fashion or distribution.

gjm11vor 18 Jahren

Sure (even with the distinctness requirement that wasn't originally there). n1=1, n2=2, ..., and nm = whatever's left. This will only work if n is at least 1+2+...+m = m(m+1)/2, but if that isn't true then there's no solution.

kurtosisvor 18 Jahren

if interested in integer partitions this python should do the job

def intpart(n):

    if n == 0:

        yield []

    for k in range(1,n+1):

        for p in intpart(n-k):

            yield [k] + p

i believe that this isn't truly integer partitions because it will count 4 = 1,1,2,1 and 1,1,1,2 as different partitions but that's all I can give you between breakfast and work.

see wikipedia for integer partiton or young tableau for more on these interesting objects

wlievensvor 18 Jahren

Do n1, n2, ... have to have different values? Increasing values?

mooism2vor 18 Jahren

n[i] = i for 1 <= i < m

n[m] = n - m.(m-1)/2

This is still fairly trivial.

I sense there are yet more requirements you're not telling us.

kraematevor 18 Jahren

protip: Use Generating functions. Piece of cake if you use them for this problem.

arstotelvor 17 Jahren

write an algorithm that compute the factorial of a given whole number

arstotelvor 17 Jahren

how to write an algorithm

alexkayvor 18 Jahren

http://stackoverflow.com/ is a good place for this kind of questions.