# Challenge description

In this challenge, we only consider love and hate as feelings. If we want to utter a feeling expression of order N, we alternate between these two (starting with 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.


The pattern follows for every positive integer N. Given N, output the correspoinding feeling expression of order N.

# Notes

• Full stop (.) at the end of the expression is mandatory,
• Trailing and leading whitespaces (including newlines) are permitted,
• Output for a non-positive or non-integer N is undefined,
• This is a challenge, so make your code as short as possible!
• Aug 31, 2016 at 10:16
• Quite confused. So is order the input, and expression the output? Aug 31, 2016 at 13:43
• @Whothehellisthat Yes, exactly. (Welcome to PPCG! :)) Aug 31, 2016 at 13:46
• @Whothehellisthat: Yes. You may take input through stdin, though it's often shorter to define a method (function), as you can see in the submissions below. Aug 31, 2016 at 14:23
• I hate that I love this question and its answers! Sep 2, 2016 at 11:35

# Python, 54 bytes

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

• Ported to Haskell: f n=take(12*n-5)(cycle"I hate that I love that ")++"it." (56 bytes)
– xnor
Sep 1, 2016 at 2:06

## CJam, 36 bytes

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


Try it online!

### Explanation

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."


# C, 837675 74 bytes

Thanks to @Leaky Nun for saving 11 bytes and adding 4 bytes!
Thanks to @YSC for saving a byte!

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


Try it on Ideone

• i=0;while(n--) -> for(i=0;n--;) saves 1 char.
– YSC
Aug 31, 2016 at 11:39

## Javascript (ES6), 7573 70 bytes

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


Saved 2 bytes thanks to Neil
Saved 3 bytes thanks to Whothehellisthat

### Test

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))

• Save 3 bytes: ['I hate','I love'][i&1] -> i&1?'I love':'I hate' Aug 31, 2016 at 14:19
• @Whothehellisthat - Thanks! I missed that one. Aug 31, 2016 at 14:26

# 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);
}

• Why not remove the whitespace? c=i->for(...) Aug 31, 2016 at 11:24
• I simply forgot. Aug 31, 2016 at 12:47
• 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). Aug 31, 2016 at 12:55
• @KevinCruijssen You tell me this everytime ;) fine Aug 31, 2016 at 13:13
• @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.) Aug 31, 2016 at 13:37

## Mathematica, 63 bytes

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

• I'm loving the head extraction there, really clever. Aug 31, 2016 at 13:46
• @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. ;) Aug 31, 2016 at 13:51

# 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.'

• I need an explanation for this. Aug 31, 2016 at 22:12

# 05AB1E, 3432 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!

# 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.)

• Neat use of the for loop and %%. +1 Aug 31, 2016 at 14:11

# 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 .

• You can save four more by dropping l: retina.tryitonline.net/… Aug 31, 2016 at 13:21
• @MartinEnder: I think I ninja'd you to that edit :P Aug 31, 2016 at 13:38
• Timestamps say you were 9 seconds late. :P Aug 31, 2016 at 13:38

# 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;
}

• function(c){for(d="",b=0;b<c;++b)d+=(b%2?"I love ":"I hate ")+(b==c-1?"it.":"that ");return d} Aug 31, 2016 at 10:28

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


## 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.


# 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 ## Perl, 6254 50 bytes $_="I xe tx "x$_;s/tx$/it./;s/x/++$.%4?hat:lov/ge  (credit to @Ton Hospel) ### 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+$/it./; say  • Hi and welcome to PPCG. Nice answer. Here is a shorter solution though (54 bytes) : perl -pE '$_="I xe that "x$_;s/x/$@++&1?lov:hat/ge;s/\w+.$/it./'. – Dada Sep 1, 2016 at 0:07 • 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 Sep 1, 2016 at 10:16 • $@ 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.
Sep 1, 2016 at 10:47
• 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. Sep 1, 2016 at 10:54
• Nice solution. You can gain a few more bytes by using $|-- as a toggle instead of $@++%2 Sep 1, 2016 at 15:19

# 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)

• 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 love ((i<$1))&&echo that||echo it. ";}. By the way, we usually label such solutions as Bash + coreutils, because the use of seq. Sep 2, 2016 at 13:46 # ///, 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! • You could add /T/that / to the beginning and replace all instances of that  with T. Sep 21, 2016 at 22:24 ## 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] • It works out better to loop instead; see my alternative R solution. – JDL Aug 31, 2016 at 14:06 ## 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. # 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  # 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); # 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, 8786 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. # Thue, 100 bytes @::=::: >##::=>$.
\$::=~I hate that I love
>.#::=>&#
&::=~that
>#<::=~I hate it.
>.<::=~it.
::=
>@<


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

## 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)

# ><> (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!

# Lua , 75 Bytes

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

• 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. Sep 2, 2016 at 14:15
• 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. Sep 21, 2016 at 0:54

# ///, 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!

## 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.


## 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."))