चुनौती का वर्णन

इस चुनौती में, हम केवल विचार करना loveऔर hateभावनाओं के रूप में। यदि हम आदेश की भावना को व्यक्त करना चाहते हैं N, तो हम इन दोनों के बीच वैकल्पिक शुरुआत करते हैं hate:

order | expression

1       I hate it.
2       I hate that I love it.
3       I hate that I love that I hate it.
4       I hate that I love that I hate that I love it.

पैटर्न हर सकारात्मक पूर्णांक के लिए अनुसरण करता है N। यह देखते हुए N, ऑर्डर की अभिव्यक्ति को महसूस करने वाली आउटपुट बनाता है N।

.अभिव्यक्ति के अंत में पूर्ण विराम ( ) अनिवार्य है,
अनुगामी और प्रमुख व्हाट्सएप (नई सूचियों सहित) की अनुमति है,
एक गैर-सकारात्मक या गैर-पूर्णांक के लिए आउटपुट N अपरिभाषित है,
यह एक कोड-गोल्फ चुनौती है, इसलिए अपने कोड को यथासंभव छोटा बनाएं!

code-golf string

— shooqie
स्रोत

पायथन, 54 बाइट्स

lambda n:("I hate that I love that "*n)[:12*n-5]+"it."

अजगर 2: Ideone!
पायथन 3: Ideone it!

— लीक नन
स्रोत

हास्केल में f n=take(12*n-5)(cycle"I hate that I love that ")++"it."

— पोर्ट किया गया

15

CJam , 36 बाइट्स

ri{"hatlov"3/='IS@"e that "}/2<"it."

इसे ऑनलाइन आज़माएं!

व्याख्या

ri            e# Read input and convert to integer N.
{             e# For each i from 0 to N-1...
  "hatlov"3/  e#   Push ["hat" "lov"].
  =           e#   Use i as a cyclic index into this array to alternate between
              e#   "hat" and "lov".
  'IS         e#   Push character 'I' and a space.
  @           e#   Pull "hat" or "lov" on top of it.
  "e that "   e#   Push "e that "
}/
2<            e#   Truncate the last "e that " to just "e ".
"it."         e#   Push "it."

— मार्टिन एंडर
स्रोत

7

सी, 83 76 75 74 बाइट्स

11 बाइट बचाने और 4 बाइट जोड़ने के लिए @Leaky नन को धन्यवाद!
एक बाइट को बचाने के लिए @YSC को धन्यवाद!

i;f(n){for(i=0;n--;)printf("I %se %s",i++%2?"lov":"hat",n?"that ":"it.");}

Ideone पर इसे आज़माएं

— betseg
स्रोत

1

i=0;while(n--)-> for(i=0;n--;)1 चार बचाता है।

— YSC

6

जावास्क्रिप्ट (ईएस 6), 75 73 70 बाइट्स

n=>[...Array(n)].map((_,i)=>i&1?'I love':'I hate').join` that `+' it.'

नील को धन्यवाद देकर 2 बाइट्स बचाए
Saved 3 bytes thanks to Whothehellisthat

परीक्षा

let f =
n=>[...Array(n)].map((_,i)=>i&1?'I love':'I hate').join` that `+' it.'

console.log(f(1))
console.log(f(2))
console.log(f(3))
console.log(f(4))

स्निपेट का विस्तार करें

— Arnauld
स्रोत

Save 3 bytes: ['I hate','I love'][i&1] -> i&1?'I love':'I hate'

— Whothehellisthat

@Whothehellisthat - Thanks! I missed that one.

— Arnauld

5

Java 8, 91 bytes

i->{for(int j=0;j++<i;)System.out.printf("I %se %s",j%2>0?"hat":"lov",j<i?"that ":"it.");};

Ungolfed Test Program

public static void main(String[] args) {
    Consumer<Integer> c = i -> {
        for (int j = 0; j++ < i;) {
            System.out.printf("I %se %s", j % 2 > 0 ? "hat" : "lov", j < i ? "that " : "it.");
        }
    };

    c.accept(1);
    c.accept(2);
    c.accept(3);
}

— Shaun Wild
स्रोत

Why not remove the whitespace? c=i->for(...)

— shooqie

I simply forgot.

— Shaun Wild

Nice, you've beat me to it. +1 And probably shorter than my answer would be. PS: I would indicate this as 'Java 8' instead of just 'Java'. Not mandatory however, just personal preference from me since I usually write my answers in Java 7 (and since Java 9 is incoming).

— Kevin Cruijssen

@KevinCruijssen You tell me this everytime ;) fine

— Shaun Wild

@SeanBean Well, usually I already have my Java 7 answer, and you post a shorter answer, which can then most of the time be golfed even more. ;P (This time I can't find anything to make it shorter, though. But perhaps someone else is able to do so.)

— Kevin Cruijssen

5

गणितज्ञ, 63 बाइट्स

"I love"["I hate"][[#~Mod~2]]&~Array~#~Riffle~" that "<>" it."&

— मार्टिन एंडर
स्रोत

1

I'm loving the head extraction there, really clever.

— A Simmons

@ASimmons It's actually quite an old trick (which I didn't even come up with myself), but I definitely don't get to use it often enough. ;)

— Martin Ender

5

Jelly, 25 bytes

“ṅɠT5“£ẏkg⁷»ṁj“¥ıQ»ṙ1“it.

Try it online!

Explanation

                             Input: n, a number.
“ṅɠT5“£ẏkg⁷»                 Compressed string array [' I hate', ' I love']
            ṁ                Cycle for n repetitions.
             j“¥ıQ»          Join by compressed string ' that'.
                   ṙ1        Rotate left once: move the initial space to the end.
                     “it.    Implicitly print the result, then print 'it.'

— Lynn
स्रोत

I need an explanation for this.

— Steven H.

4

05AB1E, 34 32 27 bytes

Saved 5 bytes thanks to Adnan.

“I«¢€Š I„Î€Š “×¹12*5-£…it.«

Explanation

“I«¢€Š I„Î€Š “               # "I hate that I love that "
              ×              # repeat input nr of times
               ¹12*5-£       # take the first input*12-5 chars of string above
                      …it.«  # append "it."
                             # implicitly display

Try it online!

— Emigna
स्रोत

4

R, 79 bytes

n=scan();for(i in n:1)cat(if((i+n)%%2)"I love"else"I hate",if(i>1)"that "else"it.")

Luckily in R, the default separator for cat is a space.

(Edited from original 73 byte version which didn't quite solve the problem.)

— JDL
स्रोत

Neat use of the for loop and %%. +1

— Billywob

2

Retina, 42 38 bytes

Thanks to Leaky Nun for helping me golf it!

11
1I love n
1
I hate n
n$
it.
n
that

Input is taken in unary.

Try it online!

Explanation

11
1I love n

Replace every pair of 1s with 1I love n.

1
I hate n

Replace the remaining 1s with I hate n.

n$
it.
n
that

Replace the n at the end of the line with it. and every other n with that .

— Business Cat
स्रोत

You can save four more by dropping l: retina.tryitonline.net/…

— Martin Ender

@MartinEnder: I think I ninja'd you to that edit :P

— Business Cat

1

Timestamps say you were 9 seconds late. :P

— Martin Ender

1

Javascript (ES5), 99 94 bytes

function(c){for(d="",b=0;b<c;++b)d+=(b%2?"I love ":"I hate ")+(b==c-1?"it.":"that ");return d}

Saved 5 bytes thanks to Leaky Nun.

OLD 99 byte solution:

function(c){for(d="",b=0;b<c;++b)d+=(0==b%2?"I hate":"I love")+" "+(b==c-1?"it.":"that ");return d}

Another 98 byte solution:

function(d){for(c=[],b=0;b<d;++b)c.push(["love","hate"][b%2]);return"I "+c.join(" that I ")+" it"}

My code before minification:

function a(n){
  var hate="I hate",love="I love",it="it ",that="that ",out="";
  for(var i=0;i<n;++i){out+=(i%2==0?hate:love)+" "+(i==n-1?it+".":that)}return out;
}

— Paul Schmitz
स्रोत

1

function(c){for(d="",b=0;b<c;++b)d+=(b%2?"I love ":"I hate ")+(b==c-1?"it.":"that ");return d}

— Leaky Nun

1

Haskell, 70 bytes

g 1="I hate";g n=g(n-1)++" that "++cycle["I love",g 1]!!n
(++" it.").g

— Damien
स्रोत

1

PowerShell v2+, 64 bytes

((1..$args[0]|%{('I love','I hate')[$_%2]})-join' that ')+' it.'

Rather straightforward. Loops from 1 up to the input $args[0], each iteration placing either 'I love' or 'I hate' on the pipeline, based on a pseudo-ternary for modulo-2 (i.e., it alternates back and forth, starting with 'I hate'). Those strings are encapsulated in parens and -joined with ' that ' to smush them together, then string concatenation ' it.' at the end.

Test Cases

PS C:\Tools\Scripts\golfing> 1..5|%{.\hate-love-conundrum.ps1 $_}
I hate it.
I hate that I love it.
I hate that I love that I hate it.
I hate that I love that I hate that I love it.
I hate that I love that I hate that I love that I hate it.

— AdmBorkBork
स्रोत

1

php, 64 62 bytes

<?=str_pad("",$argv[1]*12-5,"I hate that I love that ")."it.";

Unfortunately I couldn't work out a way to avoid repeating the " that I ", or at least no way to do it in less than 7 bytes.

edit: saved 2 bytes thanks to @Jörg Hülsermann

— user59178
स्रोत

1

Perl, 62 54 50 bytes

$_="I xe tx "x$_;s/tx $/it./;s/x/++$.%4?hat:lov/ge

(credit to @Ton Hospel)

Demo: http://ideone.com/zrM27p

Previous solutions:

$_="I xe that "x$_;s/x/$@++&1?lov:hat/ge;s/\w+.$/it./

(credit to @Dada)

Run with perl -pE '$_="I xe that "x$_;s/x/$@++&1?lov:hat/ge;s/\w+.$/it./'

First solution (only this was mine)

for$x(1..<>){$_.=' I '.($x%2?hat:lov).'e that'}s/\w+$/it./;say

In parts:

for $x (1..<>) {
   $_ .= ' I '.($x % 2 ? hat : lov).'e that'
}
s/\w+$/it./;
say

Demo: http://ideone.com/mosnVz

— Al.G.
स्रोत

Hi and welcome to PPCG. Nice answer. Here is a shorter solution though (54 bytes) : perl -pE '$_="I xe that "x$_;s/x/$@++&1?lov:hat/ge;s/\w+.$/it./'.

— Dada

What does this part mean $@++&1? For @+ perldoc says "holds the offsets of the ends of the last successful submatches in the currently active dynamic scope" which doesn't make much sense for me. As I understand, you use this array in scalar context ($@+ - are you dereferencing it?) to get the number of elements and then add (+) the matched string (&1). No no no I knew I shouldn't have posted to PPCG it's too obfuscated :D

— Al.G.

$@ is just a scalar (I could have used $x or any other scalar), ++ is the increment operator, and &1 is roughly the same as %2. So it's basically the same as $x++%2.

— Dada

So you're using @ for scalar variable name; &1 for "and"ing the last bit to check if it's even (and not a backreference as I thought). Ok understood now, thanks.

— Al.G.

Nice solution. You can gain a few more bytes by using $|-- as a toggle instead of $@++%2

— Ton Hospel

1

Bash + coreutils, 106 bytes:

for i in `seq $1`;{ printf "I %s %s " `((i%2>0))&&echo hate||echo love` `((i==$1))&&echo it.||echo that`;}

Simply creates a sequence starting at 1 up to and including the input integer using the seq built-in, and then iterates through it one by one, first outputting hate if the value of the iteration variable, i, is not divisible by 2 and love otherwise. In the same iteration, it then chooses to output that if i is not equal to the input value, and it. otherwise.

Try It Online! (Ideone)

— R. Kap
स्रोत

Better put the command substitutions directly in printf's string and use no format specifiers. Is pointless to compare whether i%2 is greater than 0. I you reverse the commands in the list, you can use less than comparison instead of i==$1: for i in `seq $1`;{ printf "I `((i%2))&&echo hat||echo lov`e `((i<$1))&&echo that||echo it.` ";}. By the way, we usually label such solutions as Bash + coreutils, because the use of seq.

— manatwork

1

///, 60 57 bytes

/!/I hate //T/that //
/it.//00/!TI love T//Tit./it.//0/!/

-3 bytes thanks to m-chrzan

Input in unary with trailing new line.

Try it online!

— acrolith
स्रोत

You could add /T/that / to the beginning and replace all instances of that with T.

— m-chrzan

0

R, 92 90 bytes

An R adaptation of @Leaky Nun's python answer. Working with strings in R is tedious as always.

n=scan();cat(rep(strsplit("I hate that I love that ","")[[1]],n)[6:(n*12)-5],"it.",sep="")

This could probably be golfed further though.

Edit: saved 2 bytes by changing:

[1:((n*12)-5)]to [6:(n*12)-5]

— Billywob
स्रोत

It works out better to loop instead; see my alternative R solution.

— JDL

0

C, 96 Bytes

c;f(i){printf("I hate");for(;c<i+1/2-1;c++)printf(" that I %s",c&1?"hate":"love");puts(" it.");}

I didn't see the above solution from Releasing Helium Nuclei which is better.

— cleblanc
स्रोत

0

MATL, 37 bytes

:"2@o3]Qv'hate I that it. love'Ybw)Zc

Try it online!

Explanation

The code is based on the following mapping from numbers to strings:

0: 'love'
1: 'hate'
2: 'I'
3: 'that'
4: 'it.'

The program pushes numbers to the string in groups of three: 2, 0, 3; then 2, 1, 3; then 2, 0, 3; ... as many times as the input n. After that, the final 3 is transformed into a 4, the mapping is applied to transform numbers to strings, and the strings are joined using space as separator.

:                         % Input n implicitly. Push range [1 2 ... n]
"                         % For each
  2                       %   Push 2
  @o                      %   Iteration index modulo 2: pushes 0 or 1
  3                       %   Push 3
]                         % End
Q                         % Add 1 to the last 3
v                         % Concatenate stack contents into a numeric column vector
'hate I that it. love'    % Push this string
Yb                        % Split at spaces. Gives a cell array of five strings
w                         % Swap to move numeric vector to top
)                         % Index cell array of strings with that vector. Indexing
                          % is 1-based and modular, so 0 refers to the last string.
                          % This gives a cell array of (repeated) strings
Zc                        % Join those strings by spaces. Display implicitly

— Luis Mendo
स्रोत

0

JavaScript (ES6), 68 bytes

f=(n,i)=>n?(i?' that I ':'I ')+(i&1?'love':'hate')+f(n-1,-~i):' it.'

document.write('<pre>'+[ 1, 2, 3, 4, 10, 100 ].map(c=>c+': '+f(c)).join`\n`);

Expand snippet

— user81655
स्रोत

0

C#, 85 83 bytes

string f(int n,int m=2)=>"I "+(1>m%2?"hat":"lov")+(m>n?"e it.":"e that "+f(n,m+1));

Recursively constructs the string, using an optional parameter to keep track of which hate/love and how many to append.

-2 bytes from this tip for checking the evenness/oddness of a number.

No optional parameter solution, 87 86 84 bytes

string h(int n)=>"I "+(0<n?"hat":"lov")+(2>n&-2<n?"e it.":"e that "+h(0<n?~n+2:~n));

This one does the same thing except determines which hate/love to append based on whether the parameter is positive or negative. Each iteration the parameter approaches zero, alternating sign.

— milk
स्रोत

0

Thue, 100 bytes

@::=:::
>##::=>$.
$::=~I hate that I love 
>.#::=>&#
&::=~that 
>#<::=~I hate it.
>.<::=~it.
::=
>@<

Takes input as unary. (A string of n #s)

— MegaTom
स्रोत

0

Pyke, 36 bytes

2/"I hate ""I love "]*"that "J"it."+

Try it here!

2/                                   -    input/2
                     *               -   ^ * v
  "I hate ""I love "]                -    ["I hate ", "I love "]
                      "that "J       -  "that ".join(^)
                              "it."+ - ^+"it."

Also 36 bytes

12*5+.daෆ   ű   l5d+12"I ":Q*<"it."+

Try it here! (Link uses X instead of I, this should work for the same amount of bytes offline where you can literally use those bytes. Online \r gets automatically replaced with \n)

— Blue
स्रोत

0

><> (Fish), 82 Bytes

' I'oov.2coo<;oooo' it.'<
'ahv'v>'et'
oop26<^ooooo^o' that I '^!?:-1oo
'ol^'^>'ev'

Doubt it's very efficient, but it seems to more or less work. Input is via the starting stack, which makes the score 85 Bytes if you include the size of the -v argument required to do so.

Try it online!

— Callum Kerr
स्रोत

0

Lua , 75 Bytes

n=io.read();print(('I hate that I love that '):rep(n):sub(1,n*12-5)..'it.')

— Jörg Hülsermann
स्रोत

1

Instead of static methods better use instance methods: ('I hate that I love that '):rep(n):sub(1,n*12-5). And would look nicer if you concatenate "it." to the end, because print() outputs its parameters separated by tab.

— manatwork

1

The ';' between io.read() and print is unneeded, and arg[2] is a valid input method for lua scripts, which is the first command line argument.

— ATaco

0

///, 68 bytes

/@/1\/#love //%/hate//#/that I //%1i/% i//I %1/I % //1@#% //@/I %1it.

Input in unary - add more 1s in the last section.

Try it online!

— m-chrzan
स्रोत

0

dc, 75 bytes

?[1-lbP[I lov]P]sa[e that ]sb[[e it.]Pq]sc[1-[I hat]Pd2%1=ad1>clbPlxx]dsxx

We're really just printing one chunk of string at a time here, and not leaving any garbage on the stack. This is great, we don't need to waste any bytes dealing with a register for our counter.

?                              #Input, works as our decrement counter
[1-lbP[I lov]P]sa              #Macro 'a' decrements our counter, prints string 'b', 
                               #then prints 'I lov'
[e that ]sb                    #String 'b'
[[e it.]Pq]sc                  #Macro 'c' prints the tail end of our message and quits
[1-                            #I'll unfold our main macro here; first step is to 
                               #decrement our counter
   [I hat]P                    #We always start by hating 'it,' no conditional needed
           d2%1=a              #Check the oddness of our counter; execute 'a' if odd
                 d1>c          #Check our counter; If we're done, execute 'c'
                     lbPlxx]   #Otherwise, print 'b' again and start this macro ('x') over
dsxx                           #Stores the above macro as 'x' and runs it.

— brhfl
स्रोत

0

Julia, 91 Bytes

Thought I add a julia solution:

f(N)=string("I hate ",join(["that I love that I hate " for _ in 1:N-1])[1:12*(N-1)],"it."))

— nyro_0
स्रोत

नफरत / प्यार करना

चुनौती का वर्णन

टिप्पणियाँ

पायथन, 54 बाइट्स

CJam , 36 बाइट्स

व्याख्या

सी, 83 76 75 74 बाइट्स

जावास्क्रिप्ट (ईएस 6), 75 73 70 बाइट्स

परीक्षा

Java 8, 91 bytes

Ungolfed Test Program

गणितज्ञ, 63 बाइट्स

Jelly, 25 bytes

Explanation

05AB1E, 34 32 27 bytes

R, 79 bytes

Retina, 42 38 bytes

Javascript (ES5), 99 94 bytes

Haskell, 70 bytes

PowerShell v2+, 64 bytes

Test Cases

php, 64 62 bytes

Perl, 62 54 50 bytes

Previous solutions:

First solution (only this was mine)

Bash + coreutils, 106 bytes:

///, 60 57 bytes

R, 92 90 bytes

C, 96 Bytes

MATL, 37 bytes

Explanation

JavaScript (ES6), 68 bytes

C#, 85 83 bytes

No optional parameter solution, 87 86 84 bytes

Thue, 100 bytes

Pyke, 36 bytes

><> (Fish), 82 Bytes

Lua , 75 Bytes

///, 68 bytes

dc, 75 bytes

Julia, 91 Bytes