Ordered partitions
You are encouraged to solve this task according to the task description, using any language you may know.
In this task we want to find the ordered partitions into fixed-size blocks. This task is related to Combinations in that it has to do with discrete mathematics and moreover a helper function to compute combinations is (probably) needed to solve this task.
should generate all distributions of the elements in into blocks of respective size .
Example 1: would create:
{({1, 2}, {}, {3, 4}), ({1, 3}, {}, {2, 4}), ({1, 4}, {}, {2, 3}),
({2, 3}, {}, {1, 4}), ({2, 4}, {}, {1, 3}), ({3, 4}, {}, {1, 2})}
Example 2: would create:
{({1}, {2}, {3}), ({1}, {3}, {2}), ({2}, {1}, {3}), ({2}, {3}, {1}),
({3}, {1}, {2}), ({3}, {2}, {1})}
Note that the number of elements in the list is
and the number of elements remains the same regardless of how the argument is permuted (i.e. the multinomial coefficient). Also, creates the permutations of and thus there would be elements in the list.
Note: Do not use functions that are not in the standard library of the programming language you use. Your file should be written so that it can be executed on the command line and by default outputs the result of . If the programming language does not support polyvariadic functions pass a list as an argument.
Notation
Remarks on the used notation for the task in order to understand it easierly.
denotes the set of consecutive numbers from to , e.g. if . is the function that takes as its sole argument a set of natural numbers and computes the sum of the numbers, e.g. . are the arguments — natural numbers — that the sought function receives.
D
Using code from this lexicographical Combination. <lang d>import std.stdio, std.algorithm, std.range, std.array, std.conv ; import comblex ;
alias int[] iRNG ;
auto setDiff(iRNG s, iRNG c) { return array(setDifference(s,c)) ; }
iRNG[][] orderPart(iRNG blockSize ...) {
iRNG sum = array(iota(1, 1 + reduce!"a+b"(blockSize))) ;
iRNG[][] p(iRNG s, iRNG b) {
if(b.length == 0) return [[]] ;
iRNG[][] res ;
foreach(c ; Comb.On(s, b[0]))
foreach(r ; p(setDiff(s, c), b[1..$]))
res ~= (c.dup ~ r) ;
return res ;
}
return p(sum, blockSize) ;
}
void main(string[] args) {
auto b = args.length > 1 ? array(map!(to!int)(args[1..$])) : [2,0,2] ;
foreach(p;orderPart(b))
writeln(p) ;
}</lang> Output:
[[1, 2], , [3, 4]] [[1, 3], , [2, 4]] [[1, 4], , [2, 3]] [[2, 3], , [1, 4]] [[2, 4], , [1, 3]] [[3, 4], , [1, 2]]
Haskell
<lang haskell>comb :: Int -> [a] -> a comb 0 _ = [[]] comb _ [] = [] comb k (x:xs) = [ x:cs | cs <- comb (k-1) xs ] ++ comb k xs
partitions :: [Int] -> [[[Int]]] partitions xs = p [1..sum xs] xs
where p _ [] = [[]]
p xs (k:ks) = [ cs:rs | cs <- comb k xs, rs <- p (xs `minus` cs) ks ]
minus xs ys = [ x | x <- xs, not $ x `elem` ys ]
main = do putStrLn $ show $ partitions [2,0,2]</lang>
An alternative where minus is not needed anymore because comb now not only
keeps the chosen elements but also the not chosen elements together in a tuple.
<lang haskell>comb :: Int -> [a] -> [([a],[a])] comb 0 xs = [([],xs)] comb _ [] = [] comb k (x:xs) = [ (x:cs,zs) | (cs,zs) <- comb (k-1) xs ] ++
[ (cs,x:zs) | (cs,zs) <- comb k xs ]
partitions :: [Int] -> [[[Int]]] partitions xs = p [1..sum xs] xs
where p _ [] = [[]]
p xs (k:ks) = [ cs:rs | (cs,zs) <- comb k xs, rs <- p zs ks ]
main = do putStrLn $ show $ partitions [2,0,2]</lang>
J
Brute force approach:
<lang j>require'stats' partinds=: (0 +/\@,}:) +&.> i.&.> partitions=: [: ~. partinds /:~@:{L:0"1 [: perm +/</lang>
This generates all permutations and then discards the unwanted possibilities.
Note that the most complicated bit of coding here was complicated because of the need to support empty partitions.
A more efficient approach is possible, but since the typical case is simpler (two partitions, neither of which are empty), I am not sure if efficiency can be important here.
Lua
A pretty verbose solution. Maybe somebody can replace with something terser/better. <lang lua>--- Create a list {1,...,n}. local function range(n)
local res = {}
for i=1,n do
res[i] = i
end
return res
end
--- Return true if the element x is in t. local function isin(t, x)
for _,x_t in ipairs(t) do if x_t == x then return true end end return false
end
--- Return the sublist from index u to o (inclusive) from t. local function slice(t, u, o)
local res = {}
for i=u,o do
res[#res+1] = t[i]
end
return res
end
--- Compute the sum of the elements in t. -- Assume that t is a list of numbers. local function sum(t)
local s = 0 for _,x in ipairs(t) do s = s + x end return s
end
--- Generate all combinations of t of length k (optional, default is #t). local function combinations(m, r)
local function combgen(m, n)
if n == 0 then coroutine.yield({}) end
for i=1,#m do
if n == 1 then coroutine.yield({m[i]})
else
for m0 in coroutine.wrap(function() combgen(slice(m, i+1, #m), n-1) end) do
coroutine.yield({m[i], unpack(m0)})
end
end
end
end
return coroutine.wrap(function() combgen(m, r) end)
end
--- Generate a list of partitions into fized-size blocks. local function partitions(...)
local function helper(s, ...)
local args = {...}
if #args == 0 then return {% templatetag openvariable %}{% templatetag closevariable %} end
local res = {}
for c in combinations(s, args[1]) do
local s0 = {}
for _,x in ipairs(s) do if not isin(c, x) then s0[#s0+1] = x end end
for _,r in ipairs(helper(s0, unpack(slice(args, 2, #args)))) do
res[#res+1] = {{unpack(c)}, unpack(r)}
end
end
return res
end
return helper(range(sum({...})), ...)
end
-- Print the solution io.write "[" local parts = partitions(2,0,2) for i,tuple in ipairs(parts) do
io.write "("
for j,set in ipairs(tuple) do
io.write "{"
for k,element in ipairs(set) do
io.write(element)
if k ~= #set then io.write(", ") end
end
io.write "}"
if j ~= #tuple then io.write(", ") end
end
io.write ")"
if i ~= #parts then io.write(", ") end
end io.write "]" io.write "\n"</lang>
Python
<lang python>from itertools import combinations
def partitions(*args):
def p(s, *args):
if not args: return [[]]
res = []
for c in combinations(s, args[0]):
s0 = [x for x in s if x not in c]
for r in p(s0, *args[1:]):
res.append([c] + r)
return res
s = range(sum(args))
return p(s, *args)
print partitions(2, 0, 2)</lang>
An equivalent but terser solution. <lang python>from itertools import combinations as comb
def partitions(*args):
def minus(s1, s2): return [x for x in s1 if x not in s2]
def p(s, *args):
if not args: return [[]]
return [[c] + r for c in comb(s, args[0]) for r in p(minus(s, c), *args[1:])]
return p(range(1, sum(args) + 1), *args)
print partitions(2, 0, 2)</lang>
Tcl
<lang tcl>package require Tcl 8.5 package require struct::set
- Selects all k-sized combinations from a list.
- "Borrowed" from elsewhere on RC
proc selectCombinationsFrom {k l} {
if {$k == 0} {return {}} elseif {$k == [llength $l]} {return [list $l]}
set all {}
set n [expr {[llength $l] - [incr k -1]}]
for {set i 0} {$i < $n} {} {
set first [lindex $l $i]
incr i
if {$k == 0} {
lappend all $first
} else { foreach s [selectCombinationsFrom $k [lrange $l $i end]] { lappend all [list $first {*}$s] }
} } return $all
}
- Construct the partitioning of a given list
proc buildPartitions {lst n args} {
# Base case when we have no further partitions to process
if {[llength $args] == 0} {
return [list [list $lst]]
}
set result {}
set c [selectCombinationsFrom $n $lst]
if {[llength $c] == 0} {set c [list $c]}
foreach comb $c {
# Sort necessary for "nice" order set rest [lsort -integer [struct::set difference $lst $comb]] foreach p [buildPartitions $rest {*}$args] { lappend result [list $comb {*}$p] }
} return $result
}
- Wrapper that assembles the initial list and calls the partitioner
proc partitions args {
set sum [tcl::mathop::+ {*}$args]
set startingSet {}
for {set i 1} {$i <= $sum} {incr i} {
lappend startingSet $i
}
return [buildPartitions $startingSet {*}$args]
}</lang> Demonstration code: <lang tcl>puts [partitions 1 1 1] puts [partitions 2 2] puts [partitions 2 0 2] puts [partitions 2 2 0]</lang> Output:
{1 2 3} {1 3 2} {2 1 3} {2 3 1} {3 1 2} {3 2 1}
{{1 2} {3 4}} {{1 3} {2 4}} {{1 4} {2 3}} {{2 3} {1 4}} {{2 4} {1 3}} {{3 4} {1 2}}
{{1 2} {} {3 4}} {{1 3} {} {2 4}} {{1 4} {} {2 3}} {{2 3} {} {1 4}} {{2 4} {} {1 3}} {{3 4} {} {1 2}}
{{1 2} {3 4} {}} {{1 3} {2 4} {}} {{1 4} {2 3} {}} {{2 3} {1 4} {}} {{2 4} {1 3} {}} {{3 4} {1 2} {}}