10
J, 1 byte
Courtesy of ngn
{
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'tis called Catalogue…
6
Python 3, 56 bytes
def f(M,*l):M and[f(M[1:],*l,x)for x in M[0]]or print(l)
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No itertools. This is one of those weird functions that prints. Thanks to Unrelated String for -2 bytes with def.
6
Haskell, 7 bytes
mapM id
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Built-in, 8 bytes
sequence
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Less boring, 33 bytes
Out-golfed by xnor's answer. Go upvote that instead!
f[]=[[]]
f(h:t)=[i:j|i<-h,j<-f t]
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5
Haskell, 23 bytes
foldr((<*>).map(:))[[]]
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Without using mapM or sequence or the like.
4
Python 2, 60 59 bytes
f=lambda A,*x:[[v]+u for v in A for u in x and f(*x)or[[]]]
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1 byte thx to ovs.
Look Ma! No itertools!
3
Ruby, 20 bytes
->i,*j{i.product *j}
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3
Perl 6, 8 4 bytes
&[X]
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Simple reduce by cross product. It would be nice if I could return the meta-operator by itself, but I've never figured out how to do that. Turns out it works for cross product?
3
K (oK), 31 15 bytes
{x@'/:+!(#:)'x}
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3
Brachylog, 1 byte
ẋ
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A slightly less boring generator solution:
Brachylog, 2 bytes
∋ᵐ
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3
Japt, 2 bytes
A simple reduction by Cartesian product.
rï
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3
Jelly, 4 2 bytes
Œp
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Output is "pretty printed" in the TIO link
3
Python 2, 58 bytes
lambda l:l.index(max(l,key=lambda s:(-s.count(0),sum(s))))
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2
Getting the solutions from @Delfad0r's program
I extended @Delfad0r's program to output solutions.
It also gives intermediate results, so you get output like this:
Solving n = 8:
a(8) >= 9
a(8) >= 10
a(8) >= 11
a(8) >= 12
a(8) >= 13
o
. o
. o o
. o o .
o o . o o
o o o o . .
o . . o o o .
o . . o . o o o
a(8) = 13
...
2
R, 52 bytes
function(l)order(lengths(l)-(m=sapply(l,sum)),-m)[1]
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Minimizes the number of toothbrushes, then maximizes the pieces of candy, since this is apparently what OP is asking.
If we want to first maximize the amount of candy, then minimize the number of toothbrushes (seems more plausible…) it becomes
R, 51 bytes
function(l)order(m&...
2
RUBY, 24 bytes 18 bytes
n=0;10.times{puts n+=1}
thanks to @A_
10.times{|n|p n+1}
2
Zsh, 30 bytes
for l;a=($^a\ ${^${=l}})
<<<$a
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Split = and cartesian product ^ for each element. Our base case adds an extra space, which cleanly separates our output lists.
2
Retina, 54 39 32 bytes
This took really long to write (as I tried to golf it too) (unsurprisingly, all the golfing ideas only appeared after I finally posted this). Takes input as a list of strings of alphanumeric characters and underscores (whatever \w matches in your universe), each preceded by a semicolon (I assume characters are as acceptable in Retina ...
1
05AB1E, 3 (or 6?) bytes
.ȉ
Outputs [[a,b],[c,d],[d,e]] in the format [[[a,c],d], [[a,c],e], ...].
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If this format is not allowed (I see some other answers use it, though) and it should be [[a,c,d], [a,c,e], ...], it would be this instead:
.»â}€˜
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Two other formats are also possible: [[a,[c,d]], [a,[c,e]], ...] (3 bytes) or ...
1
R, 54 47 bytes
f=function(x)split(j<-expand.grid(x),1:nrow(j))
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If you consider data.frame rows iterable:
R, 27 bytes
f=function(x)expand.grid(x)
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1
APL(NARS), chars 11, bytes 22
{,↑(∘.,)/⍵}
test for product of sets:
f←{,↑(∘.,)/⍵}
⎕fmt (1 2)(3 4)
┌2────────────┐
│┌2───┐ ┌2───┐│
││ 1 2│ │ 3 4││
│└~───┘ └~───┘2
└∊────────────┘
⎕fmt f (1 2)(3 4)
┌4──────────────────────────┐
│┌2───┐ ┌2───┐ ┌2───┐ ┌2───┐│
││ 1 3│ │ 1 4│ │ 2 3│ │ 2 4││
│└~───┘ └~───┘ └~───┘ └~───┘2
└∊──────────────────────────┘
⎕...
1
Charcoal, 22 bytes
ＩＥΠＥθＬιＥθ§λ÷ιΠ∨Ｅ…θμＬν¹
Try it online! Link is to verbose version of code. Explanation:
θ Input list
Ｅ Map over elements
ι Current element
Ｌ Length
Π Product
Ｅ Map over implicit range
θ ...
1
Japt, 2 bytes
rï
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Reduces input by combination with initial value of 1st element
Duplicate of @Shaggy answer, I was solving this while he just posted the same solution. I hope I can leave my answer too because it's awesome
1
JavaScript (Node.js), 52 bytes
Returns a list of lists.
f=([a,...b],o=[])=>a?a.flatMap(x=>f(b,[...o,x])):[o]
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JavaScript (V8), 52 bytes
Prints the results.
f=([a,...b],o)=>a?a.map(x=>f(b,o?o+[,x]:x)):print(o)
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1
Julia 1.0, 34 bytes
No imports used, iterators in Base has this. It actually makes a lazy form of this, but to print them all it will collect each one.
println.(Iterators.product(l...))
where l is the list of lists.
1
Erlang, 57 bytes
c([]) -> [[]];
c([H|T]) -> [[X|Y] || X <- H, Y <- c(T)].
1
Mathematica, 180 172 bytes
ToExpression[StringReplace[a,{"["->"{","]"->"}"," "->"=","True"->"-10^-9","False"->"1"}]];(If[Length[#]==1,#[[1]],-1]&@@Flatten[Position[#,Min[#]]]&/@{Total/@Streets})[[1]]
We assume the input is already stored as text in the variable a. If the reformatting code is neglected (input is already stored in the ...
1
APL(NARS), 149 chars, 298 bytes
r←f w;n;s;i;k
(n s)←w⋄r←⍬⋄→0×⍳s<3⋄i←1
→0×⍳n<k←2÷⍨(i×i×s-2)-i×s-4⋄r←r,k⋄i+←1⋄→2
h←{0=≢b←((v←↑⍵)=+/¨a)/a←{0=≢⍵:⊂⍬⋄m,(⊂1⌷⍵),¨m←∇1↓⍵}f⍵:v⍴1⋄k←↑⍋≢¨b⋄k⊃b}
if not find solutions "0=≢b" than return for (n s) input, n times 1;
else it would return the sum of s numbers that has less addend...
test:
h 1 3
1
h 2 8
1 1
...
1
Japt, 12 10 9 bytes
É&1-U)>Uq
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Port of Dennis' Jelly answer again. - 1 thanks to @Shaggy.
Only top voted, non community-wiki answers of a minimum length are eligible
