Velocity Reviews > set partitioning

# set partitioning

hymort@hotmail.com
Guest
Posts: n/a

 05-01-2006
Can someone tell me of a good algorithm to partition a set of n
elements into m non-empty, disjoint subsets, where each subset has size
k?

hymort@hotmail.com
Guest
Posts: n/a

 05-01-2006
Also, if I do not care about the number of subsets, what is a good
algorithm to partition a set of n elements into non-empty, disjoint
subsets of size k?

aaronwmail-usenet@yahoo.com
Guest
Posts: n/a

 05-01-2006
Something like this, or am I missing something?

def partition(List, n, m, k):
if n!=m*k:
raise "sorry, too many or too few elts"
D = {}
for x in List:
D[x] = 1
if len(D)!=n:
raise "sorry (2) you lied about the number"
List2 = D.keys()
result = []
for i in range(m):
result.append( List2[i*k: i*k+k] )
return result
?

If this was a take home exam problem,
you should be ashamed of yourself!
-- Aaron Watters

===

It's easy. All you have to do is hit
the right keys at the right time and
the organ plays itself. -- J.S. Bach

hymort@hotmail.com
Guest
Posts: n/a

 05-01-2006
Hello,

Not quite what I'm looking for. I would like a list of all partitions
with each partition having k or less elements, not just one instance.

(E-Mail Removed) wrote:
> Something like this, or am I missing something?
>
> def partition(List, n, m, k):
> if n!=m*k:
> raise "sorry, too many or too few elts"
> D = {}
> for x in List:
> D[x] = 1
> if len(D)!=n:
> raise "sorry (2) you lied about the number"
> List2 = D.keys()
> result = []
> for i in range(m):
> result.append( List2[i*k: i*k+k] )
> return result
> ?
>
> If this was a take home exam problem,
> you should be ashamed of yourself!
> -- Aaron Watters
>
> ===
>
> It's easy. All you have to do is hit
> the right keys at the right time and
> the organ plays itself. -- J.S. Bach

Michael Ekstrand
Guest
Posts: n/a

 05-01-2006
On Mon, May 01, 2006 at 03:42:53PM -0700, http://www.velocityreviews.com/forums/(E-Mail Removed) wrote:
> Not quite what I'm looking for. I would like a list of all partitions
> with each partition having k or less elements, not just one instance.

def partition(S, k):
parts = []
ct = 0
cp = []
for elem in S:
if ct > k:
parts.append(cp)
cp = []
ct = 0
cp.append(elem)
ct += 1
parts.append(cp)
return parts

> > If this was a take home exam problem,
> > you should be ashamed of yourself!
> > -- Aaron Watters

- Michael

--
mouse, n: a device for pointing at the xterm in which you want to type.
-- Fortune
Visit me on the Web: http://www.elehack.net

hymort@hotmail.com
Guest
Posts: n/a

 05-02-2006
For the list [1,2,3,4], I'm looking for the following for k = 2:

[[1,2], [3,4]]
[[1,3], [2,4]]
[[1,4], [2,3]]

for k = 3:

[[1,2,3], [4]]
[[1,2,4], [3]]
[[1,3,4], [2]]
[[2,3,4], [1]]

James Waldby
Guest
Posts: n/a

 05-02-2006
"(E-Mail Removed)" wrote:
> Can someone tell me of a good algorithm to partition a set of n
> elements into m non-empty, disjoint subsets, where each subset has
> size k?

and later wrote in a separate post
> Also, if I do not care about the number of subsets, what is a good
> algorithm to partition a set of n elements into non-empty, disjoint
> subsets of size k?

and then execrably wrote [ie, top-posted]:
> Not quite what I'm looking for. I would like a list of all partitions
> with each partition having k or less elements, not just one instance.

when Aaron Watters wrote:
> Something like this, or am I missing something?
>
> def partition(List, n, m, k):

[snip Watters' program to partition set of n elements
into m non-empty disjoint subsets, each of size k]

So if n=3 and k=2 and set S={a,b,c}, you would want a list
like the following?
1. {a},{b},{c}
2. {a,b},{c}
3. {a,c},{b}
4. {b,c},{a}

So it appears you need to do 2 things -

(1) Produce a list of additive partitions of n with numbers
not exceeding k. In the S={a,b,c} example, the list contains
(1,1,1) and (2,1). I think you can make the list with work
O(T(n,k)). For values of T(n,k) (ie, "number of partitions of
n in which the greatest part is k, 1<=k<=n") see
http://www.research.att.com/~njas/sequences/A008284 .

(2) For each list from (1), fill in elements from S into each
of the subsets, in canonical order. (Eg, after filling in (1,1,1)
to make {a},{b},{c}, don't produce {a},{c},{b} etc.) See a
combinatorial algorithms book, eg Reingold, for this part.
http://www.amazon.com/gp/product/013...96722?n=283155

-jiw

Edward Elliott
Guest
Posts: n/a

 05-02-2006
(E-Mail Removed) wrote:

> For the list [1,2,3,4], I'm looking for the following for k = 2:
>
> [[1,2], [3,4]]
> [[1,3], [2,4]]
> [[1,4], [2,3]]

That's not what you asked for the first time. You said you wanted "m
non-empty, disjoint subsets", but the subsets above are clearly not all
disjoint; only those on the same line are. It sounds like what you really
want is the "powerset of non-empty, disjoint partitions of size k".

Boris Borcic
Guest
Posts: n/a

 05-02-2006
(E-Mail Removed) wrote:
> For the list [1,2,3,4], I'm looking for the following for k = 2:
>
> [[1,2], [3,4]]
> [[1,3], [2,4]]
> [[1,4], [2,3]]
>
> for k = 3:
>
> [[1,2,3], [4]]
> [[1,2,4], [3]]
> [[1,3,4], [2]]
> [[2,3,4], [1]]
>

def pickk(k,N,m=0) :
if k==1 :
return ((n,) for n in range(m,N))
else :
return ((n,)+more for more in pickk(k-1,N,m+1)
for n in range(m,more[0]))

def partitionN(k,N) :
subsets = [frozenset(S) for S in pickk(k,N)]
def exhaust(rest,bound=0) :
if len(rest) < k :
if rest :
yield [sorted(rest)]
else :
yield []
for newbound in range(bound,len(subsets)) :
S = subsets[newbound]
if rest >= S :
sS = [sorted(S)]
for restpart in exhaust(rest-S,newbound+1) :
yield sS+restpart
return exhaust(set(range(N)))

partition = lambda k,S : [[[S[j] for j in S0] for S0 in P0]
for P0 in partitionN(k,len(S))]

>>> partition(2,[1,2,3,4])

[[[1, 2], [3, 4]], [[1, 3], [2, 4]], [[2, 3], [1, 4]]]
>>> partition(3,[1,2,3,4])

[[[1, 2, 3], [4]], [[1, 2, 4], [3]], [[1, 3, 4], [2]], [[2, 3, 4], [1]]]

CAVEAT : insufficiently tested, not proved correct, uncommented, provided as is

Boris Borcic
Guest
Posts: n/a

 05-02-2006
I wrote:
>
> def pickk(k,N,m=0) :
> if k==1 :
> return ((n,) for n in range(m,N))
> else :
> return ((n,)+more for more in pickk(k-1,N,m+1)
> for n in range(m,more[0]))
>
> def partitionN(k,N) :
> subsets = [frozenset(S) for S in pickk(k,N)]

while it doesn't otherwise change results, changing this line to

subsets = [frozenset(S) for S in sorted(pickk(k,N))]

provides everything nicely ordered (e.g. lexicographically)

> def exhaust(rest,bound=0) :
> if len(rest) < k :
> if rest :
> yield [sorted(rest)]
> else :
> yield []
> for newbound in range(bound,len(subsets)) :
> S = subsets[newbound]
> if rest >= S :
> sS = [sorted(S)]
> for restpart in exhaust(rest-S,newbound+1) :
> yield sS+restpart
> return exhaust(set(range(N)))
>
> partition = lambda k,S : [[[S[j] for j in S0] for S0 in P0]
> for P0 in partitionN(k,len(S))]
>
> >>> partition(2,[1,2,3,4])

> [[[1, 2], [3, 4]], [[1, 3], [2, 4]], [[2, 3], [1, 4]]]
> >>> partition(3,[1,2,3,4])

> [[[1, 2, 3], [4]], [[1, 2, 4], [3]], [[1, 3, 4], [2]], [[2, 3, 4], [1]]]
>
>
> CAVEAT : insufficiently tested, not proved correct, uncommented,
> provided as is