Given no input write a program or a function that outputs or returns the following string:
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)
"(<>.".∞ηJÀ24£.∞
Explanation
"(<>." # push this string
# STACK: "(<>."
.∞ # intersected mirror
# STACK: "(<>.<>)"
η # compute prefixes
J # join to string
# STACK: "((<(<>(<>.(<>.<(<>.<>(<>.<>)"
À # rotate left
24£ # take the first 24 chars
# STACK: "(<(<>(<>.(<>.<(<>.<>(<>."
.∞ # intersected mirror
# STACK: "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
Didn't have access to the website, had to figure this all out manually -.-
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Most of the characters are relatively easy: For the brackets im executing
([]["fill"]+[])[13] // '('
([]["fill"]+[])[14] // ')'
The full stop is gotten by making js create a scientific notation number like 1.1e+21 And taking the dot character after converting to a string
The square brackets are harder, we have to execute the function to create an italic html object string and steal the angle brackets. The main trick in this is getting the 'c' for the word italics, which requires building up another function to take the c from 'function'
There's a lot of scope for improvement here, mostly on which functions are used to get the square brackets. I also recall an easier way to get the character 'c' if italics is better, but i'd have to look though my old files to find it
console.log( (!![]+[][(![]+[])[+[]]+([![]]+[][[]])[+!+[]+[+[]]]+(![]+[])[+!+[]+!+[]]+(![]+[])[+!+[]+!+[]]])[+!+[]+(+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+!+[]+[])]
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pΙ2○3V?hG]‘Γ
Explanation:
pΙ2○3V?hG]‘ Push compressed string of the first half of the string (this is the kind of thing this compressor was made for)
Γ mirror, swap chars & palindromize with one character overlap
s=>"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
-2 bytes thanks to Kevin Cruijssen
RIP C#
; isn't counted for lambda-answers. And () can be replaced with a single character (unused null object). Relevant meta post for the second.
\$\endgroup\$
Jun 20, 2017 at 11:08
s=> is perfectly fine.
\$\endgroup\$
Jun 20, 2017 at 11:13
3? 4? bytes thanks to @KritixiLithos
i(<>.<>)òÙxlHÄ$xGòxÍî
i(<>.<>) 'insert (<>.<>)
ò ò 'recursively
Ù 'duplicate this (bottom line) down
x 'delete the first character
l 'and break when there's only one character left
H 'go to the top of the buffer
Ä 'duplicate that line up
$x 'delete the last character
G 'and go back to the bottom of the buffer
x 'delete the extra ) left by the loop
Íî 'and remove all newlines
dkHxògJ is 1 byte shorter than dkVHgJ0x.
\$\endgroup\$
Jun 20, 2017 at 11:35
lX for xl too :D
\$\endgroup\$
Jun 20, 2017 at 11:41
cat(substring('(<>.<>)',-4:6,2:12),sep='')
Fairly simple program that takes advantage of the way substring works in R. So substring('(<>.<>)',-4:6,2:12) produces the following vector
> substring('(<>.<>)',-4:6,2:12)
[1] "(<" "(<>" "(<>." "(<>.<" "(<>.<>" "(<>.<>)" "<>.<>)" ">.<>)"
[9] ".<>)" "<>)" ">)"
cat with an empty separator outputs it to STDOUT in the required format.
Ｆ⁶…(<>.<>)⁺²ι‖ＢＯ⁷
Try it online! Link is to verbose version of code. Uses the new Slice operator.
I know, it's a boring solution.
say "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
America is great
"(<>.<>)"aᶠckb
"(<>.<>)" # string
aᶠ # get all Affixes (pre- and suffixes)
c # Concatenate
kb # remove last (Knife) and first (Behead)
ᶠ into a ᵘ.
\$\endgroup\$
Aug 16, 2019 at 5:18
2 4 bytes less than a simple puts() solution. There is sure to be some elegant recursive solution, but so far the overhead becomes too large at every attempt.
f(i){for(i=1;i++<12;)printf("%.*s",i,"(<>.<>)"+i/7*i%7);}
I have never done a codegolf, so this is my first one. Feedback is highly appreciated.
++++[>+++++<-]>[>++>+++>+++>++>++[<]>-]>.>.<.>.>++.<<.>.>.>>++++++.<<<<.>.>.>>.<<<.<.>.>.>>.<<<.>.<<.>.>.>>.<<<.>.>+.<<.>.>>.<<<.>.>.<.>>.<<<.>.>.>.<<<.>.>.<<.>.>.<.>.
I went for the most obvious solution in my opinion. First, I set cells 1-5 to one of the letters "().<>". Then I just go to the right cells and output the character.
UPDATE: I changed the order in which the characters appear on the "tape" making the program way shorter and more efficient.
UPDATE 2: Just revisited my post after a while and realized, that using a shorter sequence for setting up the cells I could save some bytes.
g=(i=2)=>i>12?'':'(<>.<>)'.slice(i-14,i)+g(i+1)
f=(i=11)=>i?f(i-1)+'(<>.<>)'.slice(i-13,i+1):''
// Both of the above work and are the same size
document.write('<pre>Actual g(): ' + g() + '\nActual f(): ' + f() + '\nExpected: (<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)</pre>');Uses the same method as Rod's Python answer.
""<>"(<>.<>)"~StringDrop~i~Table~{i,-5,5}
Ａ(<>ι(<ιι.ι.<ι.ιι.‖Ｂ
Thanks to Destructible Lemon for noticing a pattern (-4).
AST:
Program
├Ａ: Assign
│├'(<>': String '(<>'
│└ι: Identifier ι
├Print
│└'(<': String '(<'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└'.<': String '.<'
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
├Print
│└ι: Identifier ι
├Print
│└ι: Identifier ι
├Print
│└'.': String '.'
└‖Ｂ: Reflect butterfly
└Multidirectional
What ‖Ｂ does is basically visually palindromize canvas.
-a.
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Jun 20, 2017 at 18:59
Prints in there, not sure what you mean.
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Jun 22, 2017 at 8:56
I know this is lame but it's much shorter than my other solution...
f(){puts("(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)");}
Here's a version where I'm trying to be somewhat tricky...
char*e="(<>.<>)<>.<>)>.<>).<>)<>)>)";f(i,n){for(n=2;n<8;n++)for(i=0;i<n;i++)putchar(e[i]);puts(&e[i]);}
A recursive version that's fully tricky...
char*e="(<>.<>)";main(i,j){if(i<6){for(++i,j=0;j<i;putchar(e[j++]));main(i);}else if(i<12)printf(e+i++-6),main(i);}
A crowd of blank stares watching Batman in honor of Adam West...
s=>{string b="NanananaNanaBATMAN!NanaBATMAN!NANANanaBATMAN!NANAnanaNanaBATMAN!NANABATMAN!NanaBATMAN!NANABATMAN!naNABATMAN!NANABATMAN!naNANAnaNANABATMAN!naNANANABATMAN!naNABATMAN!naNANAnanaNA",r="",t;for(int i=0;i<176;)if(b[i]=='B'){r+="<>";i+=7;}else{t=b.Substring(i,4);if(t=="Nana")r+="(";else if(t=="nana")r+="<";else if(t=="NAna")r+=">";else if(t=="NANA")r+=".";else r+=")";i+=4;}return r;}
raze"(<>.<>)"{(y-5)_x}/:(!)11
-2 bytes thanks to streetster
EDIT for explanation:
Language is right to left interpreted.
How it works
raze"(<>.<>)"{(y-5)_x}/:(!)11
(!)11 /yields indices [0-10], right parameter of function (y)
/: /each right: loop right parameter of a dyadic function
{ } /function
"(<>.<>)" /left parameter of function (x)
(y-5) /subtract 5 from y (indices)
_ /remove first y chars from x, or last y chars if negative
raze /flatten string output to produce the final string
_x for 1 byte, and change til 11 to (!)11 for another 1 byte saving = 2 bytes saved :)
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map take[2..7]++map drop[1..5]>>=($"(<>.<>)")
First Haskell answer to beat the hardcoded solution!
The way this answer works is by building a list of functions to be applied to the string (<>.<>). First we build the left and center with
map take[2..7]
which gives us all the prefixes from size two to seven. Then we build the right with
map drop[1..5]
which gives us all the suffixes from size six to two.
Once we have the list of functions we use a monadic bind (>>=) which is just concatMap but shorter. The function we concatmap with is ($"(<>.<>)") which applies the input to the string (<>.<>).
This makes the string.
x->"(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
Boring, but there isn't any way to make this shorter in Java.. Just initializing a temp String is already 11 bytes.. (String t="something";), and using substring a couple of times certainly costs too many bytes..
Shortest alternative to a literal return is probably this (58 bytes):
x->"(<(x(x.(x.<(x.x(x.x)x.x)>.x).x)x)>)".replace("x","<>")
=>) if you want to add it.
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print''.join("(<>.<>)"[max(0,n-5):n+2]for n in range(11))
I say it looks like a gang of owls peeking out behind their leader. Just saying.
Golfing in progress.
>++++++++[<+++++>-]>++++++[<++++++++++>-]>++++++[<++++++++++>-]<++>>++++++++[<+++++>-]<++++++<<<.>.<.>.>.<<.>.>.>.<<<.>.>.>.<<.<.>.>.>.<<.>.<<.>.>.>.<<.>.<<+.>.>.>.<<.>.<<.>>.>.<<.>.<<.>>>.<<.>.<<.>.>.<<.>>.<<.
import Data.List
import Control.Arrow
concat$tail$init$uncurry(++)$inits&&&tail.tails$"(<>.<>)"
This is far longer than the 49 bytes necessary for a string literal with the output, but the best I could do to utilise the structure. As usual, I love arrows, and inits &&& tails does produce a tuple of the list of leading substrings and the list of trailing substrings of the input. Then those two tuple elements are put together in one list by passing the tuple to ++, and that list is concatenated to one large string. The tail and init calls avoid duplicating the (<>.<>) in the middle (one generated by inits, the other by tails) and drop the unwanted parenthesis from the start and end, taking only substrings of length 2 or more into account.
“(<>.<>)”ḣJœ|ṫJ$ḊṖ
“(<>.<>)”ḣJœ|ṫJ$ḊṖ Main link. No arguments.
“(<>.<>)” Set the argument and return value to s := "(<>.<>)".
J Yield all indices of s, i.e., [1, 2, 3, 4, 5, 6, 7].
ḣ Dyadic head; yield s's prefixes of lengths 1 to 7.
$ Combine the two links to the left into a chain.
J Indices; yield [1, 2, 3, 4, 5, 6, 7].
ṫ Dyadic tail; yield s's postfixes of lengths 1 to 7.
œ| Multiset union; concatenate the results to both sides,
discarding the copy of s. (s is both a prefix and a postfix.)
Ḋ Dequeue; remove the first prefix, i.e., "(".
Ṗ Pop; remove the last postfix, i.e., ")".
main(n){while(write(++n<13,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
This terminates only after timing out on TIO because the write to standard input fails, but it will terminate in a terminal. The program relies on a specific order of evaluation (undefined behavior) and won't work with, e.g., gcc.
Thanks to @Steadybox for an idea that saved a byte!
$ cat crowd.c
main(n){while(write(++n<12,n/7*(n-7)+"(<>.<>)",n/7?14-n:n));}
$ clang -o crowd crowd.c 2> /dev/null
$ ./crowd 0> /dev/null
(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)
main(n){while(++n<13)write(1,n/7*(n-7)+"(<>.<>)",n/7?14-n:n);}
At the cost of one more byte, the solution becomes a lot more portable.
(< is missing there.
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Jun 22, 2017 at 9:00
couldn't resist beating the current python 2 answer
print "(<(<>(<>.(<>.<(<>.<>(<>.<>)<>.<>)>.<>).<>)<>)>)"
(<>.<>):"(<" + "(<>" + "(<>." + ... + "(<>.<>)" + "<>.<>)" + ">.<>)" + ... + ">)"\$\endgroup\$