Drawing a tree from an array
$begingroup$
Given a possibly nested, non-empty array of single-digit positive integers (not guaranteed unique), output the ASCII-art representation as a tree, using the box-drawing characters ┌ ┴ ┐ ─ │ ┬ ┼
. (These were copied from Code Page 437, but you can use any equivalent representation).
Every integer of the array should be a leaf of the tree. Elements the same level deep in the array should be present at the same level of the tree. All elements should be separated by enough whitespace to be distinct (up to you to determine how wide, minimum of one space between).
For example, given array [[1, [2]], [3, [4, 5]]]
, output the following tree
┌─┴─┐
┌┴┐ ┌┴─┐
1 │ 3 ┌┴┐
2 4 5
For array [1, 2, 3]
the tree could look like
┌─┼─┐
1 2 3
But the array [[1, 2, 3]]
would look like
│
┌─┼─┐
1 2 3
While the array [1, [1, [1, [1]]]]
could look like
┌─┴┐
1 ┌┴─┐
1 ┌┴┐
1 │
1
As a more complicated example, [1, [[[2, 3], 4], 5]]
could be
┌┴───┐
1 ┌─┴┐
┌─┴┐ 5
┌┴┐ 4
2 3
or several other variations.
- Input and output can be given by any convenient method.
- You can print it to STDOUT or return it as a function result.
- Either a full program or a function are acceptable.
- Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately.
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
code-golf ascii-art data-structures
$endgroup$
|
show 9 more comments
$begingroup$
Given a possibly nested, non-empty array of single-digit positive integers (not guaranteed unique), output the ASCII-art representation as a tree, using the box-drawing characters ┌ ┴ ┐ ─ │ ┬ ┼
. (These were copied from Code Page 437, but you can use any equivalent representation).
Every integer of the array should be a leaf of the tree. Elements the same level deep in the array should be present at the same level of the tree. All elements should be separated by enough whitespace to be distinct (up to you to determine how wide, minimum of one space between).
For example, given array [[1, [2]], [3, [4, 5]]]
, output the following tree
┌─┴─┐
┌┴┐ ┌┴─┐
1 │ 3 ┌┴┐
2 4 5
For array [1, 2, 3]
the tree could look like
┌─┼─┐
1 2 3
But the array [[1, 2, 3]]
would look like
│
┌─┼─┐
1 2 3
While the array [1, [1, [1, [1]]]]
could look like
┌─┴┐
1 ┌┴─┐
1 ┌┴┐
1 │
1
As a more complicated example, [1, [[[2, 3], 4], 5]]
could be
┌┴───┐
1 ┌─┴┐
┌─┴┐ 5
┌┴┐ 4
2 3
or several other variations.
- Input and output can be given by any convenient method.
- You can print it to STDOUT or return it as a function result.
- Either a full program or a function are acceptable.
- Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately.
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
code-golf ascii-art data-structures
$endgroup$
$begingroup$
[1,[[[2,3],4],5]]
could be an interesting test case since it needs to have the root artificially extend so the right subtree doesn't collide with the left subtree.
$endgroup$
– Poke
Feb 13 at 20:39
$begingroup$
@Poke Added as an example. There are several possible variations for that test case.
$endgroup$
– AdmBorkBork
Feb 13 at 21:09
2
$begingroup$
The first example for that test case can't be right. That suggests that the s second element next to the1
is an array of 3 items:[2,3]
,4
, and5
. But 4 and 5 are not adjacent.
$endgroup$
– Draco18s
Feb 13 at 21:34
4
$begingroup$
That looks like[1, [[[2, 3]], [4], 5]]
to me.
$endgroup$
– Neil
Feb 13 at 21:45
$begingroup$
Which (if any) of these alternative input formats would be acceptable?
$endgroup$
– Οurous
Feb 13 at 22:02
|
show 9 more comments
$begingroup$
Given a possibly nested, non-empty array of single-digit positive integers (not guaranteed unique), output the ASCII-art representation as a tree, using the box-drawing characters ┌ ┴ ┐ ─ │ ┬ ┼
. (These were copied from Code Page 437, but you can use any equivalent representation).
Every integer of the array should be a leaf of the tree. Elements the same level deep in the array should be present at the same level of the tree. All elements should be separated by enough whitespace to be distinct (up to you to determine how wide, minimum of one space between).
For example, given array [[1, [2]], [3, [4, 5]]]
, output the following tree
┌─┴─┐
┌┴┐ ┌┴─┐
1 │ 3 ┌┴┐
2 4 5
For array [1, 2, 3]
the tree could look like
┌─┼─┐
1 2 3
But the array [[1, 2, 3]]
would look like
│
┌─┼─┐
1 2 3
While the array [1, [1, [1, [1]]]]
could look like
┌─┴┐
1 ┌┴─┐
1 ┌┴┐
1 │
1
As a more complicated example, [1, [[[2, 3], 4], 5]]
could be
┌┴───┐
1 ┌─┴┐
┌─┴┐ 5
┌┴┐ 4
2 3
or several other variations.
- Input and output can be given by any convenient method.
- You can print it to STDOUT or return it as a function result.
- Either a full program or a function are acceptable.
- Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately.
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
code-golf ascii-art data-structures
$endgroup$
Given a possibly nested, non-empty array of single-digit positive integers (not guaranteed unique), output the ASCII-art representation as a tree, using the box-drawing characters ┌ ┴ ┐ ─ │ ┬ ┼
. (These were copied from Code Page 437, but you can use any equivalent representation).
Every integer of the array should be a leaf of the tree. Elements the same level deep in the array should be present at the same level of the tree. All elements should be separated by enough whitespace to be distinct (up to you to determine how wide, minimum of one space between).
For example, given array [[1, [2]], [3, [4, 5]]]
, output the following tree
┌─┴─┐
┌┴┐ ┌┴─┐
1 │ 3 ┌┴┐
2 4 5
For array [1, 2, 3]
the tree could look like
┌─┼─┐
1 2 3
But the array [[1, 2, 3]]
would look like
│
┌─┼─┐
1 2 3
While the array [1, [1, [1, [1]]]]
could look like
┌─┴┐
1 ┌┴─┐
1 ┌┴┐
1 │
1
As a more complicated example, [1, [[[2, 3], 4], 5]]
could be
┌┴───┐
1 ┌─┴┐
┌─┴┐ 5
┌┴┐ 4
2 3
or several other variations.
- Input and output can be given by any convenient method.
- You can print it to STDOUT or return it as a function result.
- Either a full program or a function are acceptable.
- Any amount of extraneous whitespace is acceptable, so long as the characters line up appropriately.
Standard loopholes are forbidden.- This is code-golf so all usual golfing rules apply, and the shortest code (in bytes) wins.
code-golf ascii-art data-structures
code-golf ascii-art data-structures
edited Feb 14 at 13:36
AdmBorkBork
asked Feb 13 at 18:39
AdmBorkBorkAdmBorkBork
28k468241
28k468241
$begingroup$
[1,[[[2,3],4],5]]
could be an interesting test case since it needs to have the root artificially extend so the right subtree doesn't collide with the left subtree.
$endgroup$
– Poke
Feb 13 at 20:39
$begingroup$
@Poke Added as an example. There are several possible variations for that test case.
$endgroup$
– AdmBorkBork
Feb 13 at 21:09
2
$begingroup$
The first example for that test case can't be right. That suggests that the s second element next to the1
is an array of 3 items:[2,3]
,4
, and5
. But 4 and 5 are not adjacent.
$endgroup$
– Draco18s
Feb 13 at 21:34
4
$begingroup$
That looks like[1, [[[2, 3]], [4], 5]]
to me.
$endgroup$
– Neil
Feb 13 at 21:45
$begingroup$
Which (if any) of these alternative input formats would be acceptable?
$endgroup$
– Οurous
Feb 13 at 22:02
|
show 9 more comments
$begingroup$
[1,[[[2,3],4],5]]
could be an interesting test case since it needs to have the root artificially extend so the right subtree doesn't collide with the left subtree.
$endgroup$
– Poke
Feb 13 at 20:39
$begingroup$
@Poke Added as an example. There are several possible variations for that test case.
$endgroup$
– AdmBorkBork
Feb 13 at 21:09
2
$begingroup$
The first example for that test case can't be right. That suggests that the s second element next to the1
is an array of 3 items:[2,3]
,4
, and5
. But 4 and 5 are not adjacent.
$endgroup$
– Draco18s
Feb 13 at 21:34
4
$begingroup$
That looks like[1, [[[2, 3]], [4], 5]]
to me.
$endgroup$
– Neil
Feb 13 at 21:45
$begingroup$
Which (if any) of these alternative input formats would be acceptable?
$endgroup$
– Οurous
Feb 13 at 22:02
$begingroup$
[1,[[[2,3],4],5]]
could be an interesting test case since it needs to have the root artificially extend so the right subtree doesn't collide with the left subtree.$endgroup$
– Poke
Feb 13 at 20:39
$begingroup$
[1,[[[2,3],4],5]]
could be an interesting test case since it needs to have the root artificially extend so the right subtree doesn't collide with the left subtree.$endgroup$
– Poke
Feb 13 at 20:39
$begingroup$
@Poke Added as an example. There are several possible variations for that test case.
$endgroup$
– AdmBorkBork
Feb 13 at 21:09
$begingroup$
@Poke Added as an example. There are several possible variations for that test case.
$endgroup$
– AdmBorkBork
Feb 13 at 21:09
2
2
$begingroup$
The first example for that test case can't be right. That suggests that the s second element next to the
1
is an array of 3 items: [2,3]
,4
, and 5
. But 4 and 5 are not adjacent.$endgroup$
– Draco18s
Feb 13 at 21:34
$begingroup$
The first example for that test case can't be right. That suggests that the s second element next to the
1
is an array of 3 items: [2,3]
,4
, and 5
. But 4 and 5 are not adjacent.$endgroup$
– Draco18s
Feb 13 at 21:34
4
4
$begingroup$
That looks like
[1, [[[2, 3]], [4], 5]]
to me.$endgroup$
– Neil
Feb 13 at 21:45
$begingroup$
That looks like
[1, [[[2, 3]], [4], 5]]
to me.$endgroup$
– Neil
Feb 13 at 21:45
$begingroup$
Which (if any) of these alternative input formats would be acceptable?
$endgroup$
– Οurous
Feb 13 at 22:02
$begingroup$
Which (if any) of these alternative input formats would be acceptable?
$endgroup$
– Οurous
Feb 13 at 22:02
|
show 9 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Python 3, 400 393 390 bytes
L=len
S,*K=' ┴┼│123456789'
def T(x):
try:return[str(x+0)]
except:
z=[*map(T,x)];q=max(map(L,z))
for p in z:p+=[S*L(p[0])]*(q-L(p))
b=[S.join(a)for a in zip(*z)];t=b[0];l=L(t);s=0;e=L(z);r=[S]*l
if e<2:return['│'.center(l),*b]
for i in range(l):
if t[i]in K:s+=1;r[i]='┬┌┐'[(s<e)-(s>1)]
elif 0<s<e:r[i]='─'
c=l//2;r[c]=K[r[c]=='┬'];return[''.join(r),*b]
Returns a list of strings from top to bottom.
EDIT 1: Trimmed 7 bytes by avoiding duplication of ┴┼
(net save of 2 bytes), cutting out 0 from the one string, changing how drawing characters are selected from ┬┌┐
(use <
instead of ==
), and replacing a L(z)
I missed with e
EDIT 2: -2 bytes thanks to ovs and -1 byte thanks to Kevin Cruijssen
Try it online!
Ungolfed
def layer(item):
if isinstance(item, int):
return [str(item)]
else:
subs = [layer(sub) for sub in item]
longest = max(map(len, subs))
for sub in subs:
sub += [' ' * len(sub[0])] * (longest - len(sub))
below = [' '.join(l) for l in zip(*subs)]
top = below[0]
l = len(top)
if len(subs) == 1:
return ['│'.center(l), *below]
seen = 0
expected = len(subs)
builder = [' '] * l
for i in range(l):
c = top[i]
if c in '┴┼│123456789':
seen += 1
if seen == 1:
builder[i] = '┌'
elif seen == expected:
builder[i] = '┐'
else:
builder[i] = '┬'
elif 0 < seen < expected:
builder[i] = '─'
center = l // 2
if builder[center] == '┬':
builder[center] = '┼'
else:
builder[center] = '┴'
return [''.join(builder), *below]
Builds a tree from the leaves up, one layer at a time.
$endgroup$
2
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:S,*K=' ┴┼│123456789'
.
$endgroup$
– ovs
Feb 14 at 10:12
1
$begingroup$
e==1
can bee<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at themax(map(L,z))
in that case anyway.)
$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
add a comment |
$begingroup$
Clean, 544 506 bytes
Escapes are used to avoid invalid UTF-8 on SE/TIO but counted as one byte as they're valid literals
import StdEnv,Data.List;::T=I Int|L[T];$l#m= @l#k=map maxList(transpose m)=flatlines[[last[' ':[(_|v<0|w<[j]|q>hd w|q<last w|any((==)q)w|q==j='305'='302'|q==j='301'='304'='277'='332'='263'=toChar v+'0')0\[v,r,j:w]<-m|r/2==p&&q>=hd w&&q<=last w]]\q<-[0..k!!3]]\p<-[0..k!!1]];@(L l)#p=twice(p=[[v,r+1:[e+sum([2\[v:_]<-i|0<=v]++zipWith(c j=j!!2-c!!3)t(takeWhile([z:_]=v+z< -1)(tl t)))-x!!1\e<-x]]\i<-inits p&t<-tails p&[v,r:x]<-p])(concatMap@l)#g=[g\[_,2,g:_]<-p]=[[-1,0,(hd g+last g)/2:g]:p];@(I i)=[[i,0,0,0]];
Try it online!
Takes input in the format L[I 3, L[I 4, I 5], I 2]..
Connects the trees from the bottom up, left to right, then adjusts the distances from right to left.
Prettified, sort-of:
import StdEnv, Data.List;
:: T = I Int | L [T];
$ l
#m = @l
#k = map maxList (transpose m)
= flatlines [
[
last[
' ':
[
if(v < 0)
if(w < [j])
if(q > hd w)
if(q < last w)
if(any ((==) q) w)
(
if(q == j)
'305'
'302'
)(
if(q == j)
'301'
'304'
)
'277'
'332'
'263'
(toChar v + '0')
\ [v, r, j: w] <- m
| r/2 == p && q >= hd w && q <= last w
]
]
\ q <- [0..k!!3]
]
\p<-[0..k!!1]
];
@ (L l)
#p = twice
( p
= [
[
v, r + 1:
map
(
(+)
(
sum [2 \ [v: _] <- i| 0 <= v]
+ sum (
zipWith
(
[_, _, _, c: _] [_, _, j: _] = j - c
)
t
(
takeWhile ([v: _] = v < 0) (tl t)
)
) * (1 - sign (v + 1))
- x!!1
)
)
x
]
\ i <- inits p
& t <- tails p
& [v, r: x] <- p
]
)
(concatMap @ l)
#g = [g \ [_, 2, g: _] <- p]
=[[-1, 0, (hd g + last g)/2: g]: p];
@ (I i) = [[i, 0, 0, 0]];
$endgroup$
add a comment |
$begingroup$
Charcoal, 127 123 bytes
↶≔⟦⟦θ⟧⟧ηFη«≔⊟ιζ¿⁼Iζ⪫⟦ζ⟧ω⊞υ⊞OιζFLζ⊞η⁺ι⟦⊖Lζκ§ζκ⟧»Wυ«≔⌊υι≔Φυ¬⁼κιυJ±⊗Lυ⊘⊖LιI⊟ιWι«≔⊟ιζ¿ζ«←§┐┬‹ζ⊟ιW⁼KKψ←─≔⁰ι»¿⊟ι┌¶┴¦│
Try it online! Link is to verbose version of code. Explanation:
↶
Change the default drawing direction to up since we don't draw anything to the right.
≔⟦⟦θ⟧⟧η
The first step is to convert the nested array representation into an index representation which is a list of all the entries together with the indices of the subarrays, e.g. for the input q=[1, [[[2, 3]], [4], 5]]
the 5
is q[1][2]
and therefore the list we want is 1, 2
. We start with a single entry to process which is a list containing a list of the current indices (i.e. none so far) and the original input.
Fη«
Loop over the arrays as we process them. (Conveniently Charcoal will continue to iterate over a list if you push to it during iteration.)
≔⊟ιζ
Get the next array to process.
¿⁼Iζ⪫⟦ζ⟧ω
Is this actually a scalar rather than an array?
⊞υ⊞Oιζ
If so, then the list we had actually belongs on the final list of lists of indices.
FLζ
Otherwise, loop over each element in this array...
⊞η⁺ι⟦⊖Lζκ§ζκ⟧»
... and save it with its new index list so far for further processing. The maximum index of the array is also saved which is used to special-case the last element of the array.
Wυ«
We are now ready to loop over the list of index lists. However, the list is not in lexicographical order, so we can't iterate it directly.
≔⌊υι
Find the next element in lexicographical order.
≔Φυ¬⁼κιυ
Remove it from the list.
J±⊗Lυ⊘⊖Lι
Jump to the position of the scalar in the output. We can calculate this given that we can keep count of the number of scalars we output and we also know the number of entries in its index list.
I⊟ι
Actually print the scalar.
Wι«
Loop over the entries in the index list. Again, this isn't simple iteration, because the entries come in pairs, and we also need to be able to break out of the loop.
≔⊟ιζ
Extract the next index from the list.
¿ζ«
If this is not the first element in the list...
←§┐┬‹ζ⊟ι
... then print ┐
or ┬
depending on whether this is the last element in the list...
W⁼KKψ←─
... and print enough ─
s to fill up to the previous entry at this level...
≔⁰ι»
... and clear the variable to break out of the loop since we're done here.
¿⊟ι┌¶┴
Otherwise if this is (the first element of) a multiple-element list then print the ┌┴
, leaving the cursor above the ┴
to deal with this level's parent.
¦│
Otherwise if this is a 1-element list then just print a │
and move up a line to deal with this level's parent.
$endgroup$
add a comment |
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3 Answers
3
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3 Answers
3
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active
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active
oldest
votes
$begingroup$
Python 3, 400 393 390 bytes
L=len
S,*K=' ┴┼│123456789'
def T(x):
try:return[str(x+0)]
except:
z=[*map(T,x)];q=max(map(L,z))
for p in z:p+=[S*L(p[0])]*(q-L(p))
b=[S.join(a)for a in zip(*z)];t=b[0];l=L(t);s=0;e=L(z);r=[S]*l
if e<2:return['│'.center(l),*b]
for i in range(l):
if t[i]in K:s+=1;r[i]='┬┌┐'[(s<e)-(s>1)]
elif 0<s<e:r[i]='─'
c=l//2;r[c]=K[r[c]=='┬'];return[''.join(r),*b]
Returns a list of strings from top to bottom.
EDIT 1: Trimmed 7 bytes by avoiding duplication of ┴┼
(net save of 2 bytes), cutting out 0 from the one string, changing how drawing characters are selected from ┬┌┐
(use <
instead of ==
), and replacing a L(z)
I missed with e
EDIT 2: -2 bytes thanks to ovs and -1 byte thanks to Kevin Cruijssen
Try it online!
Ungolfed
def layer(item):
if isinstance(item, int):
return [str(item)]
else:
subs = [layer(sub) for sub in item]
longest = max(map(len, subs))
for sub in subs:
sub += [' ' * len(sub[0])] * (longest - len(sub))
below = [' '.join(l) for l in zip(*subs)]
top = below[0]
l = len(top)
if len(subs) == 1:
return ['│'.center(l), *below]
seen = 0
expected = len(subs)
builder = [' '] * l
for i in range(l):
c = top[i]
if c in '┴┼│123456789':
seen += 1
if seen == 1:
builder[i] = '┌'
elif seen == expected:
builder[i] = '┐'
else:
builder[i] = '┬'
elif 0 < seen < expected:
builder[i] = '─'
center = l // 2
if builder[center] == '┬':
builder[center] = '┼'
else:
builder[center] = '┴'
return [''.join(builder), *below]
Builds a tree from the leaves up, one layer at a time.
$endgroup$
2
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:S,*K=' ┴┼│123456789'
.
$endgroup$
– ovs
Feb 14 at 10:12
1
$begingroup$
e==1
can bee<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at themax(map(L,z))
in that case anyway.)
$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
add a comment |
$begingroup$
Python 3, 400 393 390 bytes
L=len
S,*K=' ┴┼│123456789'
def T(x):
try:return[str(x+0)]
except:
z=[*map(T,x)];q=max(map(L,z))
for p in z:p+=[S*L(p[0])]*(q-L(p))
b=[S.join(a)for a in zip(*z)];t=b[0];l=L(t);s=0;e=L(z);r=[S]*l
if e<2:return['│'.center(l),*b]
for i in range(l):
if t[i]in K:s+=1;r[i]='┬┌┐'[(s<e)-(s>1)]
elif 0<s<e:r[i]='─'
c=l//2;r[c]=K[r[c]=='┬'];return[''.join(r),*b]
Returns a list of strings from top to bottom.
EDIT 1: Trimmed 7 bytes by avoiding duplication of ┴┼
(net save of 2 bytes), cutting out 0 from the one string, changing how drawing characters are selected from ┬┌┐
(use <
instead of ==
), and replacing a L(z)
I missed with e
EDIT 2: -2 bytes thanks to ovs and -1 byte thanks to Kevin Cruijssen
Try it online!
Ungolfed
def layer(item):
if isinstance(item, int):
return [str(item)]
else:
subs = [layer(sub) for sub in item]
longest = max(map(len, subs))
for sub in subs:
sub += [' ' * len(sub[0])] * (longest - len(sub))
below = [' '.join(l) for l in zip(*subs)]
top = below[0]
l = len(top)
if len(subs) == 1:
return ['│'.center(l), *below]
seen = 0
expected = len(subs)
builder = [' '] * l
for i in range(l):
c = top[i]
if c in '┴┼│123456789':
seen += 1
if seen == 1:
builder[i] = '┌'
elif seen == expected:
builder[i] = '┐'
else:
builder[i] = '┬'
elif 0 < seen < expected:
builder[i] = '─'
center = l // 2
if builder[center] == '┬':
builder[center] = '┼'
else:
builder[center] = '┴'
return [''.join(builder), *below]
Builds a tree from the leaves up, one layer at a time.
$endgroup$
2
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:S,*K=' ┴┼│123456789'
.
$endgroup$
– ovs
Feb 14 at 10:12
1
$begingroup$
e==1
can bee<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at themax(map(L,z))
in that case anyway.)
$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
add a comment |
$begingroup$
Python 3, 400 393 390 bytes
L=len
S,*K=' ┴┼│123456789'
def T(x):
try:return[str(x+0)]
except:
z=[*map(T,x)];q=max(map(L,z))
for p in z:p+=[S*L(p[0])]*(q-L(p))
b=[S.join(a)for a in zip(*z)];t=b[0];l=L(t);s=0;e=L(z);r=[S]*l
if e<2:return['│'.center(l),*b]
for i in range(l):
if t[i]in K:s+=1;r[i]='┬┌┐'[(s<e)-(s>1)]
elif 0<s<e:r[i]='─'
c=l//2;r[c]=K[r[c]=='┬'];return[''.join(r),*b]
Returns a list of strings from top to bottom.
EDIT 1: Trimmed 7 bytes by avoiding duplication of ┴┼
(net save of 2 bytes), cutting out 0 from the one string, changing how drawing characters are selected from ┬┌┐
(use <
instead of ==
), and replacing a L(z)
I missed with e
EDIT 2: -2 bytes thanks to ovs and -1 byte thanks to Kevin Cruijssen
Try it online!
Ungolfed
def layer(item):
if isinstance(item, int):
return [str(item)]
else:
subs = [layer(sub) for sub in item]
longest = max(map(len, subs))
for sub in subs:
sub += [' ' * len(sub[0])] * (longest - len(sub))
below = [' '.join(l) for l in zip(*subs)]
top = below[0]
l = len(top)
if len(subs) == 1:
return ['│'.center(l), *below]
seen = 0
expected = len(subs)
builder = [' '] * l
for i in range(l):
c = top[i]
if c in '┴┼│123456789':
seen += 1
if seen == 1:
builder[i] = '┌'
elif seen == expected:
builder[i] = '┐'
else:
builder[i] = '┬'
elif 0 < seen < expected:
builder[i] = '─'
center = l // 2
if builder[center] == '┬':
builder[center] = '┼'
else:
builder[center] = '┴'
return [''.join(builder), *below]
Builds a tree from the leaves up, one layer at a time.
$endgroup$
Python 3, 400 393 390 bytes
L=len
S,*K=' ┴┼│123456789'
def T(x):
try:return[str(x+0)]
except:
z=[*map(T,x)];q=max(map(L,z))
for p in z:p+=[S*L(p[0])]*(q-L(p))
b=[S.join(a)for a in zip(*z)];t=b[0];l=L(t);s=0;e=L(z);r=[S]*l
if e<2:return['│'.center(l),*b]
for i in range(l):
if t[i]in K:s+=1;r[i]='┬┌┐'[(s<e)-(s>1)]
elif 0<s<e:r[i]='─'
c=l//2;r[c]=K[r[c]=='┬'];return[''.join(r),*b]
Returns a list of strings from top to bottom.
EDIT 1: Trimmed 7 bytes by avoiding duplication of ┴┼
(net save of 2 bytes), cutting out 0 from the one string, changing how drawing characters are selected from ┬┌┐
(use <
instead of ==
), and replacing a L(z)
I missed with e
EDIT 2: -2 bytes thanks to ovs and -1 byte thanks to Kevin Cruijssen
Try it online!
Ungolfed
def layer(item):
if isinstance(item, int):
return [str(item)]
else:
subs = [layer(sub) for sub in item]
longest = max(map(len, subs))
for sub in subs:
sub += [' ' * len(sub[0])] * (longest - len(sub))
below = [' '.join(l) for l in zip(*subs)]
top = below[0]
l = len(top)
if len(subs) == 1:
return ['│'.center(l), *below]
seen = 0
expected = len(subs)
builder = [' '] * l
for i in range(l):
c = top[i]
if c in '┴┼│123456789':
seen += 1
if seen == 1:
builder[i] = '┌'
elif seen == expected:
builder[i] = '┐'
else:
builder[i] = '┬'
elif 0 < seen < expected:
builder[i] = '─'
center = l // 2
if builder[center] == '┬':
builder[center] = '┼'
else:
builder[center] = '┴'
return [''.join(builder), *below]
Builds a tree from the leaves up, one layer at a time.
edited Feb 14 at 17:30
answered Feb 13 at 21:34
BeefsterBeefster
2,8001346
2,8001346
2
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:S,*K=' ┴┼│123456789'
.
$endgroup$
– ovs
Feb 14 at 10:12
1
$begingroup$
e==1
can bee<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at themax(map(L,z))
in that case anyway.)
$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
add a comment |
2
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:S,*K=' ┴┼│123456789'
.
$endgroup$
– ovs
Feb 14 at 10:12
1
$begingroup$
e==1
can bee<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at themax(map(L,z))
in that case anyway.)
$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
2
2
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
I added the test-cases to your TIO link Try it online!
$endgroup$
– pizzapants184
Feb 14 at 2:52
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:
S,*K=' ┴┼│123456789'
.$endgroup$
– ovs
Feb 14 at 10:12
$begingroup$
Nice answer! You can shorten this by two bytes by assigning the space to a variable like this:
S,*K=' ┴┼│123456789'
.$endgroup$
– ovs
Feb 14 at 10:12
1
1
$begingroup$
e==1
can be e<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at the max(map(L,z))
in that case anyway.)$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
$begingroup$
e==1
can be e<2
to save a byte (I don't think it ever can be 0, since the challenge states the input is non-empty - and empty inputs would have already failed at the max(map(L,z))
in that case anyway.)$endgroup$
– Kevin Cruijssen
Feb 14 at 14:44
add a comment |
$begingroup$
Clean, 544 506 bytes
Escapes are used to avoid invalid UTF-8 on SE/TIO but counted as one byte as they're valid literals
import StdEnv,Data.List;::T=I Int|L[T];$l#m= @l#k=map maxList(transpose m)=flatlines[[last[' ':[(_|v<0|w<[j]|q>hd w|q<last w|any((==)q)w|q==j='305'='302'|q==j='301'='304'='277'='332'='263'=toChar v+'0')0\[v,r,j:w]<-m|r/2==p&&q>=hd w&&q<=last w]]\q<-[0..k!!3]]\p<-[0..k!!1]];@(L l)#p=twice(p=[[v,r+1:[e+sum([2\[v:_]<-i|0<=v]++zipWith(c j=j!!2-c!!3)t(takeWhile([z:_]=v+z< -1)(tl t)))-x!!1\e<-x]]\i<-inits p&t<-tails p&[v,r:x]<-p])(concatMap@l)#g=[g\[_,2,g:_]<-p]=[[-1,0,(hd g+last g)/2:g]:p];@(I i)=[[i,0,0,0]];
Try it online!
Takes input in the format L[I 3, L[I 4, I 5], I 2]..
Connects the trees from the bottom up, left to right, then adjusts the distances from right to left.
Prettified, sort-of:
import StdEnv, Data.List;
:: T = I Int | L [T];
$ l
#m = @l
#k = map maxList (transpose m)
= flatlines [
[
last[
' ':
[
if(v < 0)
if(w < [j])
if(q > hd w)
if(q < last w)
if(any ((==) q) w)
(
if(q == j)
'305'
'302'
)(
if(q == j)
'301'
'304'
)
'277'
'332'
'263'
(toChar v + '0')
\ [v, r, j: w] <- m
| r/2 == p && q >= hd w && q <= last w
]
]
\ q <- [0..k!!3]
]
\p<-[0..k!!1]
];
@ (L l)
#p = twice
( p
= [
[
v, r + 1:
map
(
(+)
(
sum [2 \ [v: _] <- i| 0 <= v]
+ sum (
zipWith
(
[_, _, _, c: _] [_, _, j: _] = j - c
)
t
(
takeWhile ([v: _] = v < 0) (tl t)
)
) * (1 - sign (v + 1))
- x!!1
)
)
x
]
\ i <- inits p
& t <- tails p
& [v, r: x] <- p
]
)
(concatMap @ l)
#g = [g \ [_, 2, g: _] <- p]
=[[-1, 0, (hd g + last g)/2: g]: p];
@ (I i) = [[i, 0, 0, 0]];
$endgroup$
add a comment |
$begingroup$
Clean, 544 506 bytes
Escapes are used to avoid invalid UTF-8 on SE/TIO but counted as one byte as they're valid literals
import StdEnv,Data.List;::T=I Int|L[T];$l#m= @l#k=map maxList(transpose m)=flatlines[[last[' ':[(_|v<0|w<[j]|q>hd w|q<last w|any((==)q)w|q==j='305'='302'|q==j='301'='304'='277'='332'='263'=toChar v+'0')0\[v,r,j:w]<-m|r/2==p&&q>=hd w&&q<=last w]]\q<-[0..k!!3]]\p<-[0..k!!1]];@(L l)#p=twice(p=[[v,r+1:[e+sum([2\[v:_]<-i|0<=v]++zipWith(c j=j!!2-c!!3)t(takeWhile([z:_]=v+z< -1)(tl t)))-x!!1\e<-x]]\i<-inits p&t<-tails p&[v,r:x]<-p])(concatMap@l)#g=[g\[_,2,g:_]<-p]=[[-1,0,(hd g+last g)/2:g]:p];@(I i)=[[i,0,0,0]];
Try it online!
Takes input in the format L[I 3, L[I 4, I 5], I 2]..
Connects the trees from the bottom up, left to right, then adjusts the distances from right to left.
Prettified, sort-of:
import StdEnv, Data.List;
:: T = I Int | L [T];
$ l
#m = @l
#k = map maxList (transpose m)
= flatlines [
[
last[
' ':
[
if(v < 0)
if(w < [j])
if(q > hd w)
if(q < last w)
if(any ((==) q) w)
(
if(q == j)
'305'
'302'
)(
if(q == j)
'301'
'304'
)
'277'
'332'
'263'
(toChar v + '0')
\ [v, r, j: w] <- m
| r/2 == p && q >= hd w && q <= last w
]
]
\ q <- [0..k!!3]
]
\p<-[0..k!!1]
];
@ (L l)
#p = twice
( p
= [
[
v, r + 1:
map
(
(+)
(
sum [2 \ [v: _] <- i| 0 <= v]
+ sum (
zipWith
(
[_, _, _, c: _] [_, _, j: _] = j - c
)
t
(
takeWhile ([v: _] = v < 0) (tl t)
)
) * (1 - sign (v + 1))
- x!!1
)
)
x
]
\ i <- inits p
& t <- tails p
& [v, r: x] <- p
]
)
(concatMap @ l)
#g = [g \ [_, 2, g: _] <- p]
=[[-1, 0, (hd g + last g)/2: g]: p];
@ (I i) = [[i, 0, 0, 0]];
$endgroup$
add a comment |
$begingroup$
Clean, 544 506 bytes
Escapes are used to avoid invalid UTF-8 on SE/TIO but counted as one byte as they're valid literals
import StdEnv,Data.List;::T=I Int|L[T];$l#m= @l#k=map maxList(transpose m)=flatlines[[last[' ':[(_|v<0|w<[j]|q>hd w|q<last w|any((==)q)w|q==j='305'='302'|q==j='301'='304'='277'='332'='263'=toChar v+'0')0\[v,r,j:w]<-m|r/2==p&&q>=hd w&&q<=last w]]\q<-[0..k!!3]]\p<-[0..k!!1]];@(L l)#p=twice(p=[[v,r+1:[e+sum([2\[v:_]<-i|0<=v]++zipWith(c j=j!!2-c!!3)t(takeWhile([z:_]=v+z< -1)(tl t)))-x!!1\e<-x]]\i<-inits p&t<-tails p&[v,r:x]<-p])(concatMap@l)#g=[g\[_,2,g:_]<-p]=[[-1,0,(hd g+last g)/2:g]:p];@(I i)=[[i,0,0,0]];
Try it online!
Takes input in the format L[I 3, L[I 4, I 5], I 2]..
Connects the trees from the bottom up, left to right, then adjusts the distances from right to left.
Prettified, sort-of:
import StdEnv, Data.List;
:: T = I Int | L [T];
$ l
#m = @l
#k = map maxList (transpose m)
= flatlines [
[
last[
' ':
[
if(v < 0)
if(w < [j])
if(q > hd w)
if(q < last w)
if(any ((==) q) w)
(
if(q == j)
'305'
'302'
)(
if(q == j)
'301'
'304'
)
'277'
'332'
'263'
(toChar v + '0')
\ [v, r, j: w] <- m
| r/2 == p && q >= hd w && q <= last w
]
]
\ q <- [0..k!!3]
]
\p<-[0..k!!1]
];
@ (L l)
#p = twice
( p
= [
[
v, r + 1:
map
(
(+)
(
sum [2 \ [v: _] <- i| 0 <= v]
+ sum (
zipWith
(
[_, _, _, c: _] [_, _, j: _] = j - c
)
t
(
takeWhile ([v: _] = v < 0) (tl t)
)
) * (1 - sign (v + 1))
- x!!1
)
)
x
]
\ i <- inits p
& t <- tails p
& [v, r: x] <- p
]
)
(concatMap @ l)
#g = [g \ [_, 2, g: _] <- p]
=[[-1, 0, (hd g + last g)/2: g]: p];
@ (I i) = [[i, 0, 0, 0]];
$endgroup$
Clean, 544 506 bytes
Escapes are used to avoid invalid UTF-8 on SE/TIO but counted as one byte as they're valid literals
import StdEnv,Data.List;::T=I Int|L[T];$l#m= @l#k=map maxList(transpose m)=flatlines[[last[' ':[(_|v<0|w<[j]|q>hd w|q<last w|any((==)q)w|q==j='305'='302'|q==j='301'='304'='277'='332'='263'=toChar v+'0')0\[v,r,j:w]<-m|r/2==p&&q>=hd w&&q<=last w]]\q<-[0..k!!3]]\p<-[0..k!!1]];@(L l)#p=twice(p=[[v,r+1:[e+sum([2\[v:_]<-i|0<=v]++zipWith(c j=j!!2-c!!3)t(takeWhile([z:_]=v+z< -1)(tl t)))-x!!1\e<-x]]\i<-inits p&t<-tails p&[v,r:x]<-p])(concatMap@l)#g=[g\[_,2,g:_]<-p]=[[-1,0,(hd g+last g)/2:g]:p];@(I i)=[[i,0,0,0]];
Try it online!
Takes input in the format L[I 3, L[I 4, I 5], I 2]..
Connects the trees from the bottom up, left to right, then adjusts the distances from right to left.
Prettified, sort-of:
import StdEnv, Data.List;
:: T = I Int | L [T];
$ l
#m = @l
#k = map maxList (transpose m)
= flatlines [
[
last[
' ':
[
if(v < 0)
if(w < [j])
if(q > hd w)
if(q < last w)
if(any ((==) q) w)
(
if(q == j)
'305'
'302'
)(
if(q == j)
'301'
'304'
)
'277'
'332'
'263'
(toChar v + '0')
\ [v, r, j: w] <- m
| r/2 == p && q >= hd w && q <= last w
]
]
\ q <- [0..k!!3]
]
\p<-[0..k!!1]
];
@ (L l)
#p = twice
( p
= [
[
v, r + 1:
map
(
(+)
(
sum [2 \ [v: _] <- i| 0 <= v]
+ sum (
zipWith
(
[_, _, _, c: _] [_, _, j: _] = j - c
)
t
(
takeWhile ([v: _] = v < 0) (tl t)
)
) * (1 - sign (v + 1))
- x!!1
)
)
x
]
\ i <- inits p
& t <- tails p
& [v, r: x] <- p
]
)
(concatMap @ l)
#g = [g \ [_, 2, g: _] <- p]
=[[-1, 0, (hd g + last g)/2: g]: p];
@ (I i) = [[i, 0, 0, 0]];
edited Feb 15 at 5:52
answered Feb 15 at 5:24
ΟurousΟurous
7,41611136
7,41611136
add a comment |
add a comment |
$begingroup$
Charcoal, 127 123 bytes
↶≔⟦⟦θ⟧⟧ηFη«≔⊟ιζ¿⁼Iζ⪫⟦ζ⟧ω⊞υ⊞OιζFLζ⊞η⁺ι⟦⊖Lζκ§ζκ⟧»Wυ«≔⌊υι≔Φυ¬⁼κιυJ±⊗Lυ⊘⊖LιI⊟ιWι«≔⊟ιζ¿ζ«←§┐┬‹ζ⊟ιW⁼KKψ←─≔⁰ι»¿⊟ι┌¶┴¦│
Try it online! Link is to verbose version of code. Explanation:
↶
Change the default drawing direction to up since we don't draw anything to the right.
≔⟦⟦θ⟧⟧η
The first step is to convert the nested array representation into an index representation which is a list of all the entries together with the indices of the subarrays, e.g. for the input q=[1, [[[2, 3]], [4], 5]]
the 5
is q[1][2]
and therefore the list we want is 1, 2
. We start with a single entry to process which is a list containing a list of the current indices (i.e. none so far) and the original input.
Fη«
Loop over the arrays as we process them. (Conveniently Charcoal will continue to iterate over a list if you push to it during iteration.)
≔⊟ιζ
Get the next array to process.
¿⁼Iζ⪫⟦ζ⟧ω
Is this actually a scalar rather than an array?
⊞υ⊞Oιζ
If so, then the list we had actually belongs on the final list of lists of indices.
FLζ
Otherwise, loop over each element in this array...
⊞η⁺ι⟦⊖Lζκ§ζκ⟧»
... and save it with its new index list so far for further processing. The maximum index of the array is also saved which is used to special-case the last element of the array.
Wυ«
We are now ready to loop over the list of index lists. However, the list is not in lexicographical order, so we can't iterate it directly.
≔⌊υι
Find the next element in lexicographical order.
≔Φυ¬⁼κιυ
Remove it from the list.
J±⊗Lυ⊘⊖Lι
Jump to the position of the scalar in the output. We can calculate this given that we can keep count of the number of scalars we output and we also know the number of entries in its index list.
I⊟ι
Actually print the scalar.
Wι«
Loop over the entries in the index list. Again, this isn't simple iteration, because the entries come in pairs, and we also need to be able to break out of the loop.
≔⊟ιζ
Extract the next index from the list.
¿ζ«
If this is not the first element in the list...
←§┐┬‹ζ⊟ι
... then print ┐
or ┬
depending on whether this is the last element in the list...
W⁼KKψ←─
... and print enough ─
s to fill up to the previous entry at this level...
≔⁰ι»
... and clear the variable to break out of the loop since we're done here.
¿⊟ι┌¶┴
Otherwise if this is (the first element of) a multiple-element list then print the ┌┴
, leaving the cursor above the ┴
to deal with this level's parent.
¦│
Otherwise if this is a 1-element list then just print a │
and move up a line to deal with this level's parent.
$endgroup$
add a comment |
$begingroup$
Charcoal, 127 123 bytes
↶≔⟦⟦θ⟧⟧ηFη«≔⊟ιζ¿⁼Iζ⪫⟦ζ⟧ω⊞υ⊞OιζFLζ⊞η⁺ι⟦⊖Lζκ§ζκ⟧»Wυ«≔⌊υι≔Φυ¬⁼κιυJ±⊗Lυ⊘⊖LιI⊟ιWι«≔⊟ιζ¿ζ«←§┐┬‹ζ⊟ιW⁼KKψ←─≔⁰ι»¿⊟ι┌¶┴¦│
Try it online! Link is to verbose version of code. Explanation:
↶
Change the default drawing direction to up since we don't draw anything to the right.
≔⟦⟦θ⟧⟧η
The first step is to convert the nested array representation into an index representation which is a list of all the entries together with the indices of the subarrays, e.g. for the input q=[1, [[[2, 3]], [4], 5]]
the 5
is q[1][2]
and therefore the list we want is 1, 2
. We start with a single entry to process which is a list containing a list of the current indices (i.e. none so far) and the original input.
Fη«
Loop over the arrays as we process them. (Conveniently Charcoal will continue to iterate over a list if you push to it during iteration.)
≔⊟ιζ
Get the next array to process.
¿⁼Iζ⪫⟦ζ⟧ω
Is this actually a scalar rather than an array?
⊞υ⊞Oιζ
If so, then the list we had actually belongs on the final list of lists of indices.
FLζ
Otherwise, loop over each element in this array...
⊞η⁺ι⟦⊖Lζκ§ζκ⟧»
... and save it with its new index list so far for further processing. The maximum index of the array is also saved which is used to special-case the last element of the array.
Wυ«
We are now ready to loop over the list of index lists. However, the list is not in lexicographical order, so we can't iterate it directly.
≔⌊υι
Find the next element in lexicographical order.
≔Φυ¬⁼κιυ
Remove it from the list.
J±⊗Lυ⊘⊖Lι
Jump to the position of the scalar in the output. We can calculate this given that we can keep count of the number of scalars we output and we also know the number of entries in its index list.
I⊟ι
Actually print the scalar.
Wι«
Loop over the entries in the index list. Again, this isn't simple iteration, because the entries come in pairs, and we also need to be able to break out of the loop.
≔⊟ιζ
Extract the next index from the list.
¿ζ«
If this is not the first element in the list...
←§┐┬‹ζ⊟ι
... then print ┐
or ┬
depending on whether this is the last element in the list...
W⁼KKψ←─
... and print enough ─
s to fill up to the previous entry at this level...
≔⁰ι»
... and clear the variable to break out of the loop since we're done here.
¿⊟ι┌¶┴
Otherwise if this is (the first element of) a multiple-element list then print the ┌┴
, leaving the cursor above the ┴
to deal with this level's parent.
¦│
Otherwise if this is a 1-element list then just print a │
and move up a line to deal with this level's parent.
$endgroup$
add a comment |
$begingroup$
Charcoal, 127 123 bytes
↶≔⟦⟦θ⟧⟧ηFη«≔⊟ιζ¿⁼Iζ⪫⟦ζ⟧ω⊞υ⊞OιζFLζ⊞η⁺ι⟦⊖Lζκ§ζκ⟧»Wυ«≔⌊υι≔Φυ¬⁼κιυJ±⊗Lυ⊘⊖LιI⊟ιWι«≔⊟ιζ¿ζ«←§┐┬‹ζ⊟ιW⁼KKψ←─≔⁰ι»¿⊟ι┌¶┴¦│
Try it online! Link is to verbose version of code. Explanation:
↶
Change the default drawing direction to up since we don't draw anything to the right.
≔⟦⟦θ⟧⟧η
The first step is to convert the nested array representation into an index representation which is a list of all the entries together with the indices of the subarrays, e.g. for the input q=[1, [[[2, 3]], [4], 5]]
the 5
is q[1][2]
and therefore the list we want is 1, 2
. We start with a single entry to process which is a list containing a list of the current indices (i.e. none so far) and the original input.
Fη«
Loop over the arrays as we process them. (Conveniently Charcoal will continue to iterate over a list if you push to it during iteration.)
≔⊟ιζ
Get the next array to process.
¿⁼Iζ⪫⟦ζ⟧ω
Is this actually a scalar rather than an array?
⊞υ⊞Oιζ
If so, then the list we had actually belongs on the final list of lists of indices.
FLζ
Otherwise, loop over each element in this array...
⊞η⁺ι⟦⊖Lζκ§ζκ⟧»
... and save it with its new index list so far for further processing. The maximum index of the array is also saved which is used to special-case the last element of the array.
Wυ«
We are now ready to loop over the list of index lists. However, the list is not in lexicographical order, so we can't iterate it directly.
≔⌊υι
Find the next element in lexicographical order.
≔Φυ¬⁼κιυ
Remove it from the list.
J±⊗Lυ⊘⊖Lι
Jump to the position of the scalar in the output. We can calculate this given that we can keep count of the number of scalars we output and we also know the number of entries in its index list.
I⊟ι
Actually print the scalar.
Wι«
Loop over the entries in the index list. Again, this isn't simple iteration, because the entries come in pairs, and we also need to be able to break out of the loop.
≔⊟ιζ
Extract the next index from the list.
¿ζ«
If this is not the first element in the list...
←§┐┬‹ζ⊟ι
... then print ┐
or ┬
depending on whether this is the last element in the list...
W⁼KKψ←─
... and print enough ─
s to fill up to the previous entry at this level...
≔⁰ι»
... and clear the variable to break out of the loop since we're done here.
¿⊟ι┌¶┴
Otherwise if this is (the first element of) a multiple-element list then print the ┌┴
, leaving the cursor above the ┴
to deal with this level's parent.
¦│
Otherwise if this is a 1-element list then just print a │
and move up a line to deal with this level's parent.
$endgroup$
Charcoal, 127 123 bytes
↶≔⟦⟦θ⟧⟧ηFη«≔⊟ιζ¿⁼Iζ⪫⟦ζ⟧ω⊞υ⊞OιζFLζ⊞η⁺ι⟦⊖Lζκ§ζκ⟧»Wυ«≔⌊υι≔Φυ¬⁼κιυJ±⊗Lυ⊘⊖LιI⊟ιWι«≔⊟ιζ¿ζ«←§┐┬‹ζ⊟ιW⁼KKψ←─≔⁰ι»¿⊟ι┌¶┴¦│
Try it online! Link is to verbose version of code. Explanation:
↶
Change the default drawing direction to up since we don't draw anything to the right.
≔⟦⟦θ⟧⟧η
The first step is to convert the nested array representation into an index representation which is a list of all the entries together with the indices of the subarrays, e.g. for the input q=[1, [[[2, 3]], [4], 5]]
the 5
is q[1][2]
and therefore the list we want is 1, 2
. We start with a single entry to process which is a list containing a list of the current indices (i.e. none so far) and the original input.
Fη«
Loop over the arrays as we process them. (Conveniently Charcoal will continue to iterate over a list if you push to it during iteration.)
≔⊟ιζ
Get the next array to process.
¿⁼Iζ⪫⟦ζ⟧ω
Is this actually a scalar rather than an array?
⊞υ⊞Oιζ
If so, then the list we had actually belongs on the final list of lists of indices.
FLζ
Otherwise, loop over each element in this array...
⊞η⁺ι⟦⊖Lζκ§ζκ⟧»
... and save it with its new index list so far for further processing. The maximum index of the array is also saved which is used to special-case the last element of the array.
Wυ«
We are now ready to loop over the list of index lists. However, the list is not in lexicographical order, so we can't iterate it directly.
≔⌊υι
Find the next element in lexicographical order.
≔Φυ¬⁼κιυ
Remove it from the list.
J±⊗Lυ⊘⊖Lι
Jump to the position of the scalar in the output. We can calculate this given that we can keep count of the number of scalars we output and we also know the number of entries in its index list.
I⊟ι
Actually print the scalar.
Wι«
Loop over the entries in the index list. Again, this isn't simple iteration, because the entries come in pairs, and we also need to be able to break out of the loop.
≔⊟ιζ
Extract the next index from the list.
¿ζ«
If this is not the first element in the list...
←§┐┬‹ζ⊟ι
... then print ┐
or ┬
depending on whether this is the last element in the list...
W⁼KKψ←─
... and print enough ─
s to fill up to the previous entry at this level...
≔⁰ι»
... and clear the variable to break out of the loop since we're done here.
¿⊟ι┌¶┴
Otherwise if this is (the first element of) a multiple-element list then print the ┌┴
, leaving the cursor above the ┴
to deal with this level's parent.
¦│
Otherwise if this is a 1-element list then just print a │
and move up a line to deal with this level's parent.
edited Feb 28 at 9:19
answered Feb 24 at 22:19
NeilNeil
83k745179
83k745179
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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$begingroup$
[1,[[[2,3],4],5]]
could be an interesting test case since it needs to have the root artificially extend so the right subtree doesn't collide with the left subtree.$endgroup$
– Poke
Feb 13 at 20:39
$begingroup$
@Poke Added as an example. There are several possible variations for that test case.
$endgroup$
– AdmBorkBork
Feb 13 at 21:09
2
$begingroup$
The first example for that test case can't be right. That suggests that the s second element next to the
1
is an array of 3 items:[2,3]
,4
, and5
. But 4 and 5 are not adjacent.$endgroup$
– Draco18s
Feb 13 at 21:34
4
$begingroup$
That looks like
[1, [[[2, 3]], [4], 5]]
to me.$endgroup$
– Neil
Feb 13 at 21:45
$begingroup$
Which (if any) of these alternative input formats would be acceptable?
$endgroup$
– Οurous
Feb 13 at 22:02