I just came across this little problem on UVA's Online Judge and thought, that it may be a good candidate for a little code-golf.
The problem:
You are to design a program to assist an architect in drawing the skyline of a city given the locations of the buildings in the city. To make the problem tractable, all buildings are rectangular in shape and they share a common bottom (the city they are built in is very flat). The city is also viewed as two-dimensional. A building is specified by an ordered triple (Li, Hi, Ri) where Li and Ri are left and right coordinates, respectively, of building i and Hi is the height of the building.
In the diagram below buildings are shown on the left with triples
The output should consist of the vector that describes the skyline as shown in the example above. In the skyline vector (v1, v2, v3, ... vn) , the vi such that i is an even number represent a horizontal line (height). The vi such that i is an odd number represent a vertical line (x-coordinate). The skyline vector should represent the "path" taken, for example, by a bug starting at the minimum x-coordinate and traveling horizontally and vertically over all the lines that define the skyline. Thus the last entry in the skyline vector will be a 0. The coordinates must be separated by a blank space.
If I will not count declaration of provided (test) buildings and including all spaces and tab characters, my solution, in Python, is 223 characters long.
Here is the condensed version:
B=[[1,11,5],[2,6,7],[3,13,9],[12,7,16],[14,3,25],[19,18,22],[23,13,29],[24,4,28]]
# Solution.
R=range
v=[0 for e in R(max([y[2] for y in B])+1)]
for b in B:
for x in R(b[0], b[2]):
if b[1]>v[x]:
v[x]=b[1]
p=1
k=0
for x in R(len(v)):
V=v[x]
if p and V==0:
continue
elif V!=k:
p=0
print "%s %s" % (str(x), str(V)),
k=V
I think that I didn't made any mistake but if so - feel free to criticize me.
I don't have much reputation, so I will pay only 100 for a bounty - I am curious, if anyone could try to solve this in less than .. lets say, 80 characters. Solution posted by cobbal is 101 characters long and currently it is the best one.
I thought, that 80 characters is a sick limit for this kind of problem. cobbal, with his 46 character solution totaly amazed me - though I must admit, that I spent some time reading his explanation before I partially understood what he had written.
cobbal
I'm just starting to learn J, so here goes my first attempt at golf:
although I'm sure someone who actually knows the language well could shorten this by quite a bit
An explanation of the code:
NB. list numbers up to right bound of the building
([: i. {:) 14 3 25
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
NB. compare to left bound of building and then multiply by height
(1&{ * {. <: [: i. {:) 14 3 25
0 0 0 0 0 0 0 0 0 0 0 0 0 0 3 3 3 3 3 3 3 3 3 3 3
NB. apply to each row of b, note how shorter entries are padded with 0s
(1&{ * {. <: [: i. {:)"1 b
0 11 11 11 11 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 6 6 6 6 6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
...
NB. collapse to find max, add a 0 to the end for rotate later, assign to s
] s =: 0 ,~ >./ (1&{ * {. <: [: i. {:)"1 b
0 11 11 13 13 13 13 13 13 0 0 0 7 7 7 7 3 3 3 18 18 18 3 13 13 13 13 13 13 0
NB. rotate s right 1 and then compare to s to find where height changes
s ~: _1 |. s
0 1 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1
NB. find indices of all differences
I. s ~: _1 |. s
1 3 9 12 16 19 22 23 29
NB. pair each index with the height of the building there
(,. {&s) I. s ~: _1 |. s
1 11
3 13
9 0
...
NB. and finally flatten the list
, (,. {&s) I. s ~: _1 |. s
1 11 3 13 9 0 12 7 16 3 19 18 22 3 23 13 29 0
Python, 89 characters, also using Triptych's 5001 cheat:
B=[[1,11,5],[2,6,7],[3,13,9],[12,7,16],[14,3,25],[19,18,22],[23,13,29],[24,4,28]]
x=o=0
while x<5001:
n=max([H for L,H,R in B if L<=x<R]+[0])
if n-o:print x,n,
o=n;x+=1
Replacing 5001 by max(map(max,B))+1 to allow (almost) arbitrarily large cities leaves 102 characters.
Revision History:
saved two chars as described in John Pirie's comment
saved one char as MahlerFive suggested
Python: 115 characters
Like the OP, I'm not including the declaration of the data, but I am counting whitespace.
D = [(1,11,5), (2,6,7), (3,13,9), (12,7,16),
(14,3,25), (19,18,22), (23,13,29), (24,4,28)]
P=[max([0]+[h for s,h,e in D if s<=x<e])for x in range(5001)]
for i,x in enumerate(P[1:]):
if x!=P[i]:print i+1,x,
Note that I am using the link provided by the OP as the exact definition of the problem. For instance, I cheat a bit by assuming there is not building coordinate over 5000, and that all coordinates are positive integers. The original post is not tightly constrained enough for this to be fun, in my opinion.
edit: thanks to John Pirie for the tip about collapsing the list construction into the printing for loop. How'd I miss that?!
edit: I changed range(1+max(zip(*D)[2])) to range(5001) after deciding the use the exact definition given in the original problem. The first version would handle buildings of arbitrary positive integers (assuming it all fit into memory).
edit: Realized I was overcomplicating things. Check my revisions.
BTW - I have a hunch there's a much more elegant, and possibly shorter, way to do this. Somebody beat me!
Base64 encoded, I used this site to encode it. Decode to a .com file. The program reads stdin until an EOF, which is a Ctrl-Z when reading from the console, and then outputs the result to stdout.
EDIT: The source code:
mov bp,10
add dh,10h
mov es,dx
mov ds,dx
xor ax,ax
xor di,di
mov cx,8000h
rep stosw
mov ah,0bh
int 21h
cmp al,255
mov si,offset l9
je l1
mov si,offset l11
l1:
call l7
mov di,cx
call l7
mov bx,cx
call l7
sub cx,di
add di,di
l2:
cmp bx,[di]
jbe l3
mov [di],bx
l3:
add di,2
loop l2
jmp l1
l4:
xor cx,cx
l5:
cwd
div bp
push dx
inc cx
or ax,ax
jnz l5
mov dl,bh
mov ah,2
int 21h
mov bh,44
l6:
pop dx
add dl,48
mov ah,2
int 21h
loop l6
ret
l7:
call si
cmp al,10
jae l7
db 0fh, 0b6h, 0c8h
l8:
call si
cmp al,10
jae ret
mov ah,0
xchg cx,ax
mul bp
add cx,ax
jmp l8
l9:
mov ah,0bh
int 21h
cmp al,255
jne l12
mov ah,8
int 21h
l10:
sub al,48
ret
l11:
mov ah,1
int 21h
cmp al,26
jne l10
mov si,offset l12
ret
l12:
xor si,si
xor di,di
mov bh,32
l13:
lodsw
cmp ax,di
je l14
mov di,ax
lea ax,[si-2]
shr ax,1
call l4
mov ax,di
call l4
l14:
or si,si
jne l13
int 20h
Compiled, as usual for me, using A86.
Python with 133 chars, memory and time efficient, no restrictions on data input
D = [(1,11,5), (2,6,7), (3,13,9), (12,7,16), (14,3,25), (19,18,22), (23,13,29), (24,4,28)]
l,T=0,zip(*D)
for x,h in map(lambda x:(x,max([y for a,y,b in D if a<=x<b]or[0])),sorted(T[0]+T[2])):
if h!=l: print x,h,
l=h
explanation:
lambda x:(x,max([y for a,y,b in D if a<=x<b]or[0])
returns the position and the height at position x.
Now loop over the sorted coordinate list compiled by sorted(zip(*D)[0]+zip(*D)[2]) and output if the height changes.
the second version isn't as efficient as the one above and has a coordinate limit but only uses 115 chars:
for x in range(100):
E=[max([y for a,y,b in D if a<=(x-i)<b]+[0])for i in(0,1)]
if E[0]-E[1]:print x,E[0],
98 characters of J, tacitly defined (no variable names!):
The tacit definition just can't compete with using s=:… to eliminate redundancy, though.
Whenever I answer a question with J, I try to take the time to explain what's going on, because I think others might enjoy a look into the workings of an alien language.
NB. The first, second, and last element of a vector
({. 0{b), (1 { 0{b), ({: 0{b)
1 11 5
NB. Count from 0 to (last element of vector)-1
i. {: 0{b
0 1 2 3 4
NB. Booleans: first element of vector less than or equal to (above)?
({. <: [:i.{:) 0{b
0 1 1 1 1
NB. Multiply by second element of vector
(1&{ * {.<:[:i.{:) 0{b
0 11 11 11 11
NB. Stack up results for each vector, then find maximum by column
>./ (1&{*{.<:[:i.{:) " 1 b
0 11 11 13 13 13 13 13 13 0 0 0 7 7 7 7 3 3 3 18 18 18 3 13 13 13 13 13 13
NB. Identify leaders and make table
|: (,. (~: _1 & |.)) >./(1&{*{.<:[:i.{:)"1 b
0 11 11 13 13 13 13 13 13 0 0 0 7 7 7 7 3 3 3 18 18 18 3 13 13 13 13 13 13
1 1 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0
NB. Rotate left
|: 1 |. (,.(~:_1&|.))>./(1&{*{.<:[:i.{:)"1 b
11 11 13 13 13 13 13 13 0 0 0 7 7 7 7 3 3 3 18 18 18 3 13 13 13 13 13 13 0
1 0 1 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 0 0 1 1 0 0 0 0 0 1
NB. 1-based index and first element, when last element is true
|: ({:"1 # >: @ i. @ # ,. {."1) 1&|.(,.(~:_1&|.))>./(1&{*{.<:[:i.{:)"1 b
1 3 9 12 16 19 22 23 29
11 13 0 7 3 18 3 13 0
NB. Flatten
, ({:"1#>:@i.@#,.{."1)1&|.(,.(~:_1&|.))>./(1&{*{.<:[:i.{:)"1 b
1 11 3 13 9 0 12 7 16 3 19 18 22 3 23 13 29 0
NB. Rearrange for tacit verb
([:,@({:"1#>:@i.@#,.{."1)[:1&|.[:(,.(~:_1&|.))[:>./(1&{*{.<:[:i.{:)"1) b
1 11 3 13 9 0 12 7 16 3 19 18 22 3 23 13 29 0
2 C# answers - way too long, but I'd love to see better?
LINQ approach (135 chars excluding array line):
var a=new[]{new[]{1,11,5},new[]{2,6,7},new[]{3,13,9},new[]{12,7,16},new[]{14,3,25},new[]{19,18,22},new[]{23,13,29},new[]{24,4,28}};
int l=0,y,x=-1;while(++x<5001){var b=a.Where(c=>c[0]<=x&&c[2]>x);if((y=b.Any()?b.Max(c=>c[1]):0)!=l)Console.Write(x+", "+(l=y)+", ");}
Or a non-LINQ answer (179 185 chars excluding array line):
var a={1,11,5,2,6,7,3,13,9,12,7,16,13,3,25,19,18,22,23,13,29,24,4,28};
var b=new int[5000];int i=-1,j,z;while(++i<a.Length)for(j=a[i*3];j<a[i*3+2];j++)if((z=a[i*3+1])>b[j])b[j]=z;i=-1;z=0;while(++i<5000)if(b[i]!=z)Console.Write(i+", "+(z=b[i])+", ");
if((y=a.Where(c=>c[0]<=x&&c[2]>x).Max(c=>(int?)c[1])??0)!=l) saves you a few characters
– JimmyJul 6 '09 at 22:20
"Console.Write" c#'s achilles heal of golf :) You have 3 references to [x,y] in the latter solution. replacing with [x*3+y] costs 6 characters but a.GetLength(0) becomes a.Length. the first [i*3+0] can lose th plus zero though so you net save ywo characters. you can lift the definition of j to where you define z.
– ShuggyCoUkJul 18 '09 at 2:46
not sure if you allow yourself var in the non Linq version, shaves one character of the declaration of b.
– ShuggyCoUkJul 18 '09 at 2:51
Nice; feel free to hack either/both in ;-p
– Marc Gravell♦Jul 18 '09 at 7:05
Code is condensed (few lines to code) which is good for the tournament (time is the scarcest resource), and seems correct (I don't know python, but I think I understand the code).
Your solution basically paints the city skyline in a buffer and then outputs the contents of the buffer in the required format.
The extra info you ommited from the problem is that there will be at most 5000 buildings and the horizontal positions will smaller than 10.000. That implies that memory does not seem a problem in your case (40kb for the skyline assuming 32bit architecture, plus 45kb for the building description - optional, you could paint the skyline in the read loop). The algorithm is linear in the number of buildings so it is fast.
With toughter memory constraints you could go for a one-pass algorithm, but I believe that in this case it would perform slower and be much more complex to implement (more of your time, more CPU time)
Now you should consider really reading the input in the given format and using that data for your computations instead of a prestored data array.
BTW, is python a valid language now in ACM contests?
Valid Languages = Java, C, C++, C#. As of 2008, problems may not have a judge-written C# solution.
– glasntJul 1 '09 at 3:02
2
TomatoSandwich is right - Python is not a supported language for ACM contests - I just used it, because with it, I could write condensed code. I also omitted information about test cases, because SO doesn't have means of providing them automatically, so I assumed, that algorithm written for test data will work for all data
– zeroDivisibleJul 1 '09 at 4:05
2
The info about test data is important. If your skyline coordinates instead of ranging to 10.000 could range to 2^32-1 (max 32bit integer) then your algorithm would be flawed as it would require 2^34 bytes or 16Gb of memory to store the v variable (array from 0 to 2^32-1 of 32bit integers). The same goes for buildings, if the number was big enough you would have to refactor into a one-pass algorithm. Those are really important items to evaluate an algorithm, not only correctness but viability of the implementation.
– David Rodríguez - dribeasJul 1 '09 at 5:29
what happened with Pascal? have they dropped it?
– fortranJul 9 '09 at 9:42
h x=maximum$0:[y|(l,y,r)<-b,l<=x,x<r]
main=putStr$unwords[show x++" "++show(h x)|x<-[1..9999],h x/=h(x-1)]
String formatting seems to be where Haskell falls behind the Python solutions. Having to use an extra 5 characters to write 'main=' doesn't help either, but perhaps it shouldn't be included, the C#/Java solutions would be massive if their code had to demonstrate the complete program :)
Haskell: 76 characters(no string formatting & no main)
h x=maximum$0:[y|(l,y,r)<-b,l<=x,x<r]
print[(x,h x)|x<-[1..9999],h x/=h(x-1)]
Looking back at the original problem it requires that you to read the input from a file, so I thought it would be interesting to see how many characters that adds.
Haskell: 149 characters (full solution)
main=interact f
f i=unwords[show x++" "++show(h x)|x<-[1..9999],h x/=h(x-1)] where
h x=maximum$0:[y|[l,y,r]<-b,l<=x,x<r]
b=map(map read.words)$lines i
Below is what the full solution looks like with more descriptive variable names and type signatures where possible.
main :: IO ()
main = interact skyline
skyline :: String -> String
skyline input =
unwords [show x ++ " " ++ show (heightAt x) |
x <- [1..9999], heightAt x /= heightAt (x-1)]
where heightAt :: Int -> Int
heightAt x = maximum $ 0 : [h | [l,h,r] <- buildings, l <= x, x < r]
buildings :: [[Int]]
buildings = map (map read . words) $ lines input
Rereading the UVA rules, we're not limited to a max X of 5000, but rather 5000 buildings. X and Y values up to (and including) 9999 are allowed.
Also, apparently only C, C++, C#, and Java are officially recognized languages, so I did mine up in Java. The numbers are only space separated, but commas could be put back in (at a cost of two more total chars). Totalling 153 chars (excluding the array line):
The logic is pretty straightforward. The only things that make the flow a little wonky are variable reuse and nonstandard placement of post-increment. Generates:
