text:
Output the full name of titin
Titin is a protein found in muscle tissue. It is a pretty big protein, and its full chemical name is giant as well. Comprising 189819 characters, the full name of titin would take about 3.5 hours to pronounce. Your task is to output this entire name!
Details
The full name can be found here: Gist (this file does not include the trailing newline)
The MD5 hash of the output is ec73f8229f7f8f2e542c316a41cbfff4 (without a trailing newline) or 628c6eb157eb1de675cfc5b91a8149ce (with a trailing newline).

You may not access an external resource, be it a file or an internet location.
You must output only the exact text from the pastebin to STDOUT or STDERR. Nothing else should be output to your chosen output stream; leading and trailing whitespace are acceptable.
You may not assume there will be any input, but you may assume that there will be no input if needed.
You may output anything to the other output stream (STDOUT if you chose to output to STDERR and vice versa)
Standard loopholes apply

This is a code-golf challenge, so the shortest program in bytes wins!
EDIT
This is NOT a duplicate of the Rick Astley "Never Gonna Give You Up" challenge. Besides the tags, they're not even similar. In that one, output is a multi-line song with a few sections that repeats the first two words of every line. In this one, output is a fairly compressible string with 189819 letters. Please don't close-vote this as a dupe of that.

It also takes 3.5 hours to copy-paste as well :))

At first glance, I thought we had to output the full name of Tintin, which didn't seem that hard.

But as usual, we can create a file as long as we count it in the bytecount?

@Arnauld I would presume that may be a bit too easy :P Dennis was saying too that he kept typing that instead of Titin :P

@GregMartin Yes, you may. I don't know if you need to count extra bytes for each created file though; I'll need to check Meta for that.

"the full name of titin would take about 3.5 hours to pronounce" Not only would, it's been done: https://www.youtube.com/watch?v=NFR-ADakI-c

Can you import in-built Python modules, or does this count as an external resource?

@techydesigner It depends. What module do you plan on importing? Basic utilities are allowed but anything specifically meant to print protein names, if one exists, would be disallowed.

Decompression modules (such as base65536 (which is not in-built) or zlib (which is in-built))

@techydesigner I will say that those are allowed.

I am surprised that nobody has made a Jelly answer with a base-250 number.

Possible duplicate of We're no strangers to code golf, you know the rules, and so do I

@DownChristopher How is the difference not apparent? In that challenge, the output is a multi-line song with a part that repeats the first two words every line for a bit. In this one, the output is a fairly compressible word that has 189819 characters... I don't see how they're even similar except for that they're both KC challenges. Not every KC challenge is automatically a dupe of the Rickroll challenge.

ProteinData["TTNIsoformN2A"]. Ugh why does Titin have 2 Isoforms

@SIGSEGV ProteinData is just trying to be annoying >_>

@HyperNeutrino Now caaaaarefully make the lists... Oh crap it crashed

@DownChristopher Most [kolmogorov-complexity] problems are similar, as they're all "output this specific string." Since there's a tag for the type, it's most definitely not a dupe since it is a different string (and not by a few chars either).

@Challenger5 The youtube account no longer exists.

@mbomb007 That's strange, it was working just a few days ago.

I thought Mr. Osas have a long name.

You could probably do this in O(1) time, bruteforcing all the possible 189819 length strings, computing the hash, and comparing. I don't know if the hash is unique, though.

@Thunda I don't think that's O(1); I think that would be O(n!) and that would take a long time. The hash would likely not be unique.

I'm disappointed that the bash command look titin didn't work.  :P

Bash + sed + paq8kx v7, 11682 bytes
The submission consists of two files: an arbitrarily named Bash script (23 bytes) and a paq8kx archive named a (11657 bytes, plus 1 byte for the extra file, plus 1 byte for the specific filename).
Bash script
paq8kx -d a>t
sed -fd b

a (base64 encoded)
[encoded data, 15544 characters]

Verification
$ bash titin.sh | md5sum
ec73f8229f7f8f2e542c316a41cbfff4  -

How it works
The first line of the Bash script – paq8kx -d a>t – decompresses the archive a, redirecting its output to t. The redirection is required as paq8kx prints information to STDOUT, which should not be part of the final output. Said information will look as follows.
Extracting ./b 26926 -> done
Extracting ./d 307 -> done
Time 15.36 sec, used 1929185730 bytes of memory

This creates a sed script d with the following content.
s/T/methionyl/g
s/S/cysteinyl/g
s/R/histidyl/g
s/Q/tryptophyl/g
s/P/phenylalanyl/g
s/O/glutaminyl/g
s/N/tyrosyl/g
s/M/asparaginyl/g
s/L/arginyl/g
s/K/aspartyl/g
s/J/alanyl/g
s/I/isoleucyl/g
s/H/leucyl/g
s/G/glycyl/g
s/F/prolyl/g
s/E/seryl/g
s/D/threonyl/g
s/C/lysyl/g
s/B/glutamyl/g
s/A/valyl/g
s/..$/ine/g

It also creates a large (printable) binary stream that encodes the different parts of titin's name.
TDDOJFDPDOFHOEAAAHBG<26886 more uppercase letters>EHGMBPGEKEJDAMIRILEI

Finally, the second line – sed -fd b – executes the sed script d on the file b, printing the desired output to STDOUT.

Watch out, Mathematica is coming!

JavaScript (ES6), 16840 16825 16814 bytes
Saved a couple of bytes by:

inserting the last acid prefix in the packed string as suggested by ETHproductions
choosing the next character group in a slightly more efficient way when there's a tie

Assumes Windows-1252 character encoding. Outputs without a trailing newline.
Below is a simplified version without any actual data.
[...Array(256)].map((_,i)=>i<52|i>126&i<161|i==92|i==96?0:s=s.split(String.fromCharCode(i)).join(`...dictionary...`.slice(j,j+=2)),j=0,s=`...packed_data...`)&&[...s].map(c=>'methion|threon|glutamin|alan|prol|phen|leuc|ser|val|glutam|glyc|histid|isoleuc|tryptoph|argin|aspart|lys|asparagin|tyros|cystein'.split`|`[c.charCodeAt()-32]).join`yl`+'ine'

And here is the full version:
Try it online!
The TIO link includes some additional code to print the MD5 of the output rather than printing the output itself:
ec73f8229f7f8f2e542c316a41cbfff4
How?
This was compressed by:

Mapping the 20 distinct amino acids to ASCII characters 32 to 51.
Repeatedly replacing the most frequent 2-character group by a new ASCII character in the following ranges:

52-91
93-95
97-126
161-255

Which leaves us with a dictionary of 336 bytes (168 entries of 2 characters) and a final compressed string of 16161 bytes.
The above code is doing the exact opposite.

Very, very nice. I was going to do something similar but had to leave for a couple of hours. I think you can save another couple bytes by including the last ylisoleuc in the compressed data and just doing +'ine'.

@ETHproductions Ah yes, you're right. I'll do that tomorrow. (I've just switched to another computer and don't have the compression code at my fingertips anymore.)

I actually thought about doing just the same, only up to 126; but actually I had more important things to do these holidays. Very nice work! * bows *

What's wrong with 0-31, 127-160?

Mathematica, 319 + 14971 = 15290 bytes
For[p={};a={t={{{{{tyros,asparagin},prol},{ser,threon}},{{lys,alan},{glutam,val}}},{{{{phen,{methion,cystein}},argin},{aspart,{glutamin,{histid,tryptoph}}}},{leuc,{,glyc}}}},r=Join@@IntegerDigits[BinaryReadList@"b",2,8]},r!={},If[AtomQ@a,p=Join[p,{a,yl}/.{,yl}->{iso}];a=t,a=a[[1+#&@@r]];r=Rest@r]];##~Print~leucine&@@p

Incredibly, this is exactly one byte more than this bzip2 answer. Can somebody find a couple of bytes to save?!
This is a program that prints the full name of titin to STDOUT. We use the same structure described in Jörg Hülsermann's PHP answer, and additionally note that all the words except for "iso" end in "yl", which we add separately each time unless inappropriate (this is what p=Join[p,{a,yl}/.{,yl}->{iso}] does). We store the data in a 14971 -byte file called "b" in the working directory (hex dump is here), which is converted to the list of its corresponding bits by Join@@IntegerDigits[BinaryReadList@"b",2,8].
That list of bits has been determined by Huffman coding, which is a lovely data compression scheme that takes relative frequencies into account; it requires the decoding template, stored here as the binary tree t, as well as the raw list of bits. Inexplicably, Mathematica doesn't have a Huffman decoding (or encoding) built-in, so that's what the For-loop implements.

Never would've expected to see a Mathematica answer here, nice! How did you get away with not using strings for the words? (Also, if I let this run, how long would it be expected to take?)

(It takes a bit less than 2 minutes on my machine.) All those non-strings tyros, asparagin, etc. are interpreted as undefined variables, which can still be manipulated by a hopeful Mathematica; fortunately, Print is perfectly happy to print the names of undefined variables, and also (super conveniently) prints whatever arguments you give it without intervening spaces, thus doing away with needing to 'StringJoin' anything.

What about the Mathematica function IUPACName["Titin"]? ;)

Also, I feel that the amino acids are a built in in mathematica (citation needed)

@BetaDecay Trust Mathematica to have a builtin...

What about ProteinData["TTNIsoformN2A"]?

Bash + sed + xz, 13322 13290 12785 12706 12694 12687 12681 bytes
sed 1d $0|unxz|sed /`sed 's:[A-T]:yl/g;s/&/:g'<<<AvalBglutamClysDthreonEserFprolGglycHleucIisoleucJalanKaspartLarginMasparaginNtyrosOglutaminPphenylalanQtryptophRhistidScysteinTmethion`yl/g\;s/yl$/ine/
<12479 bytes of binary data>

Try it online!

I think this is using LZMA, right? As a side note, I did a similar test with paq8l which is compressing the uppercase letter string to 11877 bytes.

Yes, it's LZMA. Brotli does better, but not as good as paq8l.

Pip, 27747 27081 19701 19229 bytes
Take that, Bubblegum!
O("alan argin asparagin aspart cystein glutam glutamin glyc histid isoleuc leuc lys methion phenylalan prol ser threon tryptoph tyros val"^sA*Y"..."R^(PA3,68)."xyz{|}~
 !"Y"uttscfkmovgsmmoshiwnqtwjgioettdwnmvmfisgnjdgkrvik nrdrerjikomezastkvktssdewmdogtvtwweidsrineogsiwedkngwfotnsdtnomowrmsqemikiksoidiorntgkmtniwswowirrwt"<>2)J"yl""ine"

where "..." is a very long string. Here's the full code as a Gist.
Try it online! (Results are truncated.)
Strategy: split the name on yl, and encode each piece as one of the lowercase letters d through w. Use their code points (between 100 and 119) to index into the list of pieces (which is of length 20, so the indices are taken mod 20 automatically). Join on yl, output, and tack ine on the end.
Further compression: find the most common pair of letters in the encoded string, and replace it with an unused ASCII character. Repeat for all ASCII characters # through c, xyz{|}~!, space, and newline. The code takes this twice-compressed string and replaces the characters that aren't d through w with their corresponding two-character strings. Goto paragraph 1.

PHP 18731 Bytes
echo strtr(bzdecompress(base64_decode("$base64encodedstring")),
[A=>alanyl,B=>arginyl,C=>aspartyl,D=>asparaginyl,E=>cysteinyl,F=>glutaminyl
,G=>glutamyl,H=>glycyl,I=>histidyl,J=>isoleucyl,K=>leucyl,L=>lysyl
,M=>methionyl,N=>phenyl,O=>prolyl,P=>seryl,Q=>tryptophyl,R=>threonyl
,S=>tyrosyl,T=>valyl,U=>isoleucine]);');

Try it online!
Create the Base64 string
Replacement of all words with uppercase chars in the range A-U
after that bzip2 and base64 encoding
$a=preg_replace(["#alanyl#","#arginyl#","#aspartyl#","#asparaginyl#","#cysteinyl#","#glutaminyl#","#glutamyl#","#glycyl#","#histidyl#",
"#isoleucyl#","#leucyl#","#lysyl#","#methionyl#","#phenyl#","#prolyl#","#seryl#","#tryptophyl#","#threonyl#","#tyrosyl#","#valyl#","#isoleucine#"],range("A","U"),$titin);
$z=base64_encode(bzcompress($a));

PHP, 19920 Bytes
After replace the words with numbers and create a very long decimal number I can only read the word "leucine"
I have 20 words found that I have replace with numbers
After that i use bzcompress and the last action was base64 encoding
$g="base64String"; # see online version
echo $o=strtr(bzdecompress(base64_decode($g)),array_flip(
["methionyl"=>9999,"cysteinyl"=>9990,"histidyl"=>9909,"tryptophyl"=>9090
,"phenyl"=>9009,"glutaminyl"=>9000,"tyrosyl"=>899,"asparaginyl"=>889
,"arginyl"=>888,"aspartyl"=>800,"iso"=>70,"glycyl"=>79
,"prolyl"=>78,"seryl"=>77,"threonyl"=>6,"lysyl"=>5
,"alanyl"=>4,"glutamyl"=>3,"valyl"=>2,"leucyl"=>1]));

Online Version gzinflate
The words and count
[methionyl] => 337
[threonyl] => 2083
[glutaminyl] => 724
[alanyl] => 2304
[prolyl] => 1834
[phenyl] => 672
[leucyl] => 3351
[seryl] => 1910
[valyl] => 2414
[glutamyl] => 2324
[glycyl] => 1731
[histidyl] => 391
[iso] => 1658
[tryptophyl] => 401
[arginyl] => 1405
[aspartyl] => 1430
[lysyl] => 2199
[asparaginyl] => 902
[tyrosyl] => 829
[cysteinyl] => 356

("19 words" should be "20 words")

@GregMartin Thank You

I bet you can save a few bytes by flipping the array manually and omit the quotes for the words. Why the weird numbers?

@Titus Yes but I found a better way to make it. The first way was my first thinking and i have try to convert these numbers additional to base 36 but this was not a good idea

Bubblegum, 19655 bytes
0000000: e2 e5 7a 4c bf 5d 00 36 99 4a ef 2b 08 5d b1 93  ..zL.].6.J.+.]..
0000010: 69 cb 7a 61 51 8c 3c 7c 04 41 de 7e ea ed 68 3c  i.zaQ.<|.A.~..h<
0000020: ca 5c e1 c8 6c 5b b5 31 2d 0b 4e e2 8e 52 06 ec  .\..l[.1-.N..R..
0000030: 5e e6 da d5 f2 f5 d3 8f 83 a3 51 0b 45 5d 19 c6  ^.........Q.E]..
.
.
.

Here's a full hexdump.
You have to paste the code manually, but you can Try it online! (only shows the first 128 KiB)

... Well, at least it's shorter than just straightforward printing it...

When the output is cut because of a single word.

Yea, bubblegum got beaten!

Can´t You save about 35% by using 5 bit to 8 bit encoding?

@Titus I'm afraid I don't understand.

would be easier to help if you´d say what you´re doing. I assumed you did the same translations as most of the others.

@Titus Bubblegum only understands plain DEFLATE and LZMA encoding.

bzip2, 15287+2 bytes
BZh91AY&SY...

Full source (hex-encoded). Run with bzip2 -c. Where do I pick up my "best programmer" trophy?

I'm afraid bzip2 doesn't qualify as a programming language.

@Steadybox Peter Taylor's idea of what "qualifies as a programming language" is a flawed one (see Bubblegum). In any case, it's not endorsed in the question.

Well, yeah, Bubblegum stretches the rules quite a bit to fit that definition.

I would say that this is not a valid answer because it's just a compressed file; bzip2 is not an actual programming language, it's a program itself.

@HyperNeutrino CPython is just a program as well. On PPCG, programming languages are defined by their implementations, which are (usually) programs themselves.

@Challenger5 You have a fair point, I see what you mean. But then where do we draw the line between programs and languages?

@Steadybox https://codegolf.meta.stackexchange.com/a/10426/8478

@HyperNeutrino I don't think it's necessary to. The size of the stupidly-compressed output makes a useful goalpost, if nothing else. If you've done some clever stuff in a KC challenge, but you still haven't beat bzip2, that's a good sign you haven't found the right approach yet.

@Sneftel That's a good point. I agree with you.

On the basis of the upvotes on Geobits' comment on my answer about deliberately making languages to create/exploit loopholes, Bubblegum answers should get at least 12 downvotes, and I don't see any reason why this answer shouldn't also.

R, 17684 bytes
E="/Td6WFoAAA..." # base64 string reduced for better readability
U=unlist(lapply(match(utf8ToInt(E),c(22:47,54:79,5:14,0,4)+43)-1,function(x)x%/%2^(5:0)%%2))
D=memDecompress(as.raw(sapply(split(U,0:(length(U)-1)%/%8),function(x)sum(x*2^(7:0)))),'x',T)
for(i in 1:21)D=gsub(LETTERS[i],paste0(el(strsplit("phenylalan valylthreon isoleuc leuc val glutam alan lys threon ser prol glyc aspart argin asparagin tyros glutamin tryptoph histid cystein methion"," ")),"yl")[i],D)
cat(D,"isoleucine",sep='')

Try it online!
The approach is similar to other answers :

Remove the suffix "isoleucine" from the original string
Replace each occurrencies of the following tokens with an uppercase letter in the range A-U :
        TOKEN   REPLACEMENT
 phenylalanyl             A
valylthreonyl             B
    isoleucyl             C
       leucyl             D
        valyl             E
     glutamyl             F
       alanyl             G
        lysyl             H
     threonyl             I
        seryl             J
       prolyl             K
       glycyl             L
     aspartyl             M
      arginyl             N
  asparaginyl             O
      tyrosyl             P
   glutaminyl             Q
   tryptophyl             R
     histidyl             S
    cysteinyl             T
    methionyl             U

obtaining a string of 26610 characters
compress the string using the R builtin memCompress(string,"xz") obtaining a raw vector of length 12936 (bytes)
encode the vector to base64 (because I don't know a better way to store these bytes into the script) obtaining a string of 17248 characters

The code does exactly the opposite (from points 4 to 1) and prints the string to stdout. The TIO captures this outputs, save it to a file and compute the md5 hash to verify that is correct.
Note: we can't compute md5 directly on string because we need a package that is not installed on TIO.

saved 37 bytes thanks to J.Doe

There are only 20 amino acids. I noticed you've got some that don't exist: valylthreon (really valyl + threonyl), phen (really just phenylalanyl), lyc (doesn't exist) and one extra isoleuc are in your code.

@J.Doe: Yes, some of them are "pairs", replacing them gave me some advantages in compression (e.g. phenylalanyl is repeated 672 times, so replacing the pair instead of replacing the 2 separatedly saves some bytes). But isoleuc repeated and lycyl are definitely leftovers of some tests, thanks for pointing it out ! :)

Sorry, I didn't explain that comment well. I mean that "phenylalanyl" is a single amino acid. They will always occur together. Confusingly, so is "alanyl", so replacement order matters!

@J.Doe : ah, so I can remove phenyl from the tokens, thanks !

PHP, 19851 18639 bytes
204 bytes code + 19647 bytes base64 data
207 bytes code + 18432 bytes base64 data
for(;$c=bzuncompress(base64_decode("[encoded data, 18432 characters]"))[$i++];)echo[val,glutam,alan,lys,threon,ser,prol,glyc,leuc,isoleuc,aspart,argin,asparagin,tyros,glutamin,phen,tryptoph,histid,cystein,methion][ord($c)],$i<27598?yl:ine;

or, without the data:
for(;$c=bzuncompress(base64_decode(""))[$i++];)echo[val,glutam,alan,lys,threon,ser,prol,glyc,leuc,isoleuc,aspart,argin,asparagin,tyros,glutamin,phen,tryptoph,histid,cystein,methion][ord($c)],$i<27598?yl:ine;

I took the words ordered by number of appearances (descending), replaced them with bytes 0..19 (isoleucyl and isoleucine share code 9) followed by gzdeflate gzcompress and base64_encode. In PHP:
echo base64_encode(bzcompress(strtr($o,[asparaginyl=>chr(12),glutaminyl=>chr(14),tryptophyl=>chr(16),cysteinyl=>chr(18),isoleucyl=>chr(9),methionyl=>chr(19),aspartyl=>chr(10),glutamyl=>chr(1),histidyl=>chr(17),threonyl=>chr(4),arginyl=>chr(11),tyrosyl=>chr(13),alanyl=>chr(2),glycyl=>chr(7),leucyl=>chr(8),phenyl=>chr(15),prolyl=>chr(6),lysyl=>chr(3),seryl=>chr(5),valyl=>chr(0),isoleucine=>chr(9)])));

Moving from A..T to 0..19 did not only save 4 bytes of code (65=>), but also 3 bytes of encoded data (with gzdeflate; I didn´t test again).

replacements, words and count:
 0 val          # 2414
 1 glutam       # 2324
 2 alan         # 2304
 3 lys          # 2199
 4 threon       # 2083
 5 ser          # 1910
 6 prol         # 1834
 7 glyc         # 1731
 8 leuc         # 1694
 9 isoleuc      # 1657+1
10 aspart       # 1430
11 argin        # 1405
12 asparagin    #  902
13 tyros        #  829
14 glutamin     #  724
15 phen         #  672
16 tryptoph     #  401
17 histid       #  391
18 cystein      #  356
19 methion      #  337

18432 Bytes bzcompress against  19647  Bytes gzdeflate only for the base64 string if this help you

Python 2, 18409 18403 bytes
(with no libraries)
n=sum(95**i*(ord(v)-32)for i,v in enumerate(r'''~'W/P58:`Ote}|QRokHosM/W%~zh%Ab*ZI~i4?=Z/$_-Y*6D{ EGAP\ENvC,4=v$YQ#Z=<)e^5!p5*|K/Ap|.?CyO*51Ladt8c`it>w~1w2@{8.kb%;2n}Xvp8d{/GE4_c'8dO\)Qu(5G8[::X-(cMkgKeUli@0HJNa@Y84p^~:8'Jn #BoNyJn#5KAy[Lb} K{i=TlKzA^#EBWd>,|Ut!V.y5GCiXrO?~'dGYY;Pg6A|ub<ASiLi_@gL?Fb14=!x4-HE@4>3<_thTv&~8X%T2@;gaQy8Ao=DAnI50|7N6_lh=~7gF-@;CmFz:]?aBr~]K}eoFh{nG!4&:T>MSKxB6{N0TJ<ig@Md(ix?*\$;E0H5<f/M7EubRYLA1`A_Cuu0g"]!#w,Hf/uH2(a~~41\D.w"pg|{vO=.a)(xEqiW4/h7GIc3=vVKH[{7gTMpubkk5*YJ=b7Cn*MAmqsrTU%RE53E9{u^6@\FH&00={e2?EaeKu6>X3*B4g=b7'OW;p@2-+IoWPJVsg"`D-6yiwE)2Rd|"tM!|LVJ&gGLQGdv8WXAnT[me5~+$(~cJYxBx%UgO^^Eo|?fE/6~@wb`;7pw pHln{[PyjV}bfbbB`JBo$mt'@jiPZ.Rq"fpBe'IDiqCtV?JP80jZuxl#-,V&5oQ<.{/kn>bR!G;2sK#w8qXxQ-1VloKc+D5*EF3 qt*t9S\fq?^T54>\THI0#:l?guj}p&,yWNV1TZP^O.(BW8YK}?sD6IV"haFpfNZV>[CA<$tWVru5JiN+[j)KZyfg-axjd?E70,ae5(<)GR:9B?k0~ju#YoZb9CglDy<d9qSx\$5d[9scc8Bpc}X_NU^&K@- 0k* D'Pyy kI8&^:5=cTclE};`2r,^)&(t'f]^o-ZPHG"q^q;D^H[rFh1}O9qUEYWeg6F@y2!OwUnG2<)7qinEFJxa+%!b`yTu>oQOa+dhh{W4Ud"<FHl(r%Wjn>la[zgz"S:AubNNuYIjj*:FB1Aco.zjI~QHqy{93] Po^k3 yj/yL7'}lf24=R/`osaJHa9OF5%l*$MiBV=`=m_H=sJd/G)Ew2k-X|g$~E)s;s.R55x+6pmQIct0,ToJkoSu?D3@`O=cO+ZWg8FfsVhq?D^AB2|OzY~e;/9M~KDP)GMyC%v].1.u_'6A"y=|HE2h~GQjwW7j;1%KZ4fe8&x.{clSU'}rUsX+(q*mzzDtMwNZZa$c_a?SsIN*tA^{<Q\c'\(-M$&7douJ^@uhr##?|z\zoq$#POF|sbvI#*+IPeq<'EG'eTOhyd8M^i*|-h$aVa}.9r@\O6P-T<B!B<5d@r/|O](J8TW=Wjt=rS^H)qZ-la/"|" D49PxtBPmv;ir<:w7~x@jf[q_:z3e@p.+5/>)(NYqe]0#cP+7U<cglKWTR-JS4-;C_Ha#*N1IGV;,(GA5;9F<4ZJ0^u,]B{DNDIc{uu8<M\2~X3?`Ym7.ij%:86}8~!2w7oy8!#_k%/hQyn)DLBOqE7w]w8@d[lSE4:)w\'&*>#?b|ty2'z(!HNJ(oHkZ fTbR{WC^?]4wkPqT+&ka;XL(N?7*[{a$(@**GFl\4Do84*&DI\TL"HR:NKiXXtC(5?vYHm'g0J*w+(8?M4d:? 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#lCm1nto$MzWg'/h0h!S\Bx/Pz~=n1Ejn[lycN.cz'kGA5]$d{+Tjeja72,^)ODF@sR"AxH_.w[lmt:Moa38Y_EU]QGO,:tBCtPy$\<nNSl='M}B)i<o~q/Wj1fpk$vU!F3`q^kG5f%u,J*$C)Ru>^R$\nJ:S_WC=(x2BAlQXJ:n1c?K+]*(],K$3l(F/w3*Nm+xZa?r@56@hBAL['\kh8nIG:.cmLziq,."(SAAddoMBCo4!cqp5k54T Ix|QE0\',2j#]X2x9l-L#`X(/o8'*CVM_FkOD0&glK!\g5XC(N7I`P3o|LlSvz572xjsA_2eDAFDxO|Rk8,m]TL4Lt]/|nCU.xPG$i$Xz\Y<:tv+E%Gl`A&},3(BzDRLJC?&<`fB`d,jsch+ib4Fl?eMGcU RpQ1`l4u$:pHe)ihm{#L"NbN @S@BUI-{V)'@6Ug?ZMK]**fL'''))
r='isoleucine'
while n:d=n%20;n=n/20;r='cystein methion threon glutamin glutam alan prol phen leuc ser val glyc histid isoleuc tryptoph argin aspart lys asparagin tyros'.split()[d]+'yl'+r
print r

Creates a number, n, by reading a raw string using characters between ' ' and '~' as digits in base-95. Decodes that as a base-20 number using parts of the full name of tintin as the digits adding 'yl' each time, and initialising with the trailing 'isoleucine' and prints the result.

Python 2, 14627 bytes

Encode the amino acids as bytes, from \x00 to \x13
Special treatment for phenyl which only arises as phenylalanyl.
No special treatment for iso in isoleucyl, just a different group than leucyl.
Special treatment for the -ine.
encode the string with bz2
custom base 216 encoding of the bz2 binary string (218 printable characters in cp1252 excluding ' and \ for ease of use in a string literal.) encoding 31 bytes in 32 printable characters.

Source with cut down string s:
#coding:cp1252
a='glutamin tryptoph leuc val aspart isoleuc glutam methion alan glyc phenylalan prol argin threon asparagin cystein lys histid tyros ser'.split()
p=[chr(x)for x in range(32,256) if x not in(39,92,127,129,141,143,144,157)]
s='ª9W¾½ÍWÕiì7.{ç5æÄ‚¯g×;NíbdXÔùP¯ ...'
d=''
for i in range(0,14241,32):
 c=s[i:i+32];n=i=0
 for c in c:n+=p.index(c)*216**i;i+=1
 for _ in s[:31]:n,c=divmod(n,256);d+=chr(c)
print'yl'.join(a[ord(c)]for c in d.decode('bz2'))+'ine'

Pascal (FPC), 27917 bytes
{$H+}const s='<Compressed_string>';var c:char;a:array['A'..'T']of string=('methionyl','threonyl','glutaminyl','alanyl','prolyl','phenyl','leucyl','seryl','valyl','glutamyl','glycyl','histidyl','isoleucyl','tryptophyl','arginyl','aspartyl','lysyl','asparaginyl','tyrosyl','cysteinyl');begin for c in s do write(a[c]);write('isoleucine')end.

Try it online!
Compressed string
Hash print of the output
Compressed string, along with count of each string ending with yl, was made using this:
type part=record
      name:AnsiString;
      count:QWord;
     end;
var s:AnsiString;
    c:QWord=0;
    a:array[0..50] of part;
    i,j:word;
    p:word;
begin
  read(s);
  p:=pos('yl',s);
  while p>0 do begin
    c:=c+p;
    //writeln(c);
    j:=0;
    while j<i do begin
      if a[j].name=s[1..p+1] then begin
        Inc(a[j].count);
        break;
      end;
      j:=j+1;
    end;
    if j=i then begin
      a[i].name:=s[1..p+1];
      i:=i+1;
    end;
    write(chr(j+ord('A')));
    Delete(s,1,p+1);
    p:=pos('yl',s);
  end;
  writeln(' + isoleucine');
  for j:=0 to i-1 do writeln(j:3,chr(j+ord('A')):2,' ',a[j].name:20,' ',a[j].count);
end.

Deadfish~, 1103473 bytes
View source here
By far the longest answer I've ever done, and probably one of the longest (non-Unary) on Code Golf. Definitely longer than all my other answers and questions put together.

"probably one of the longest on Code Golf" no, I don't think it is

@hakr14 Dammit, forgot about Unary....

JavaScript (Node.js), 16071 bytes
_=>[...Array(27598)].map(_=>'isoleuc|methion|threon|glutamin|alan|prol|phen|leuc|ser|val|glutam|glyc|histid|tryptoph|argin|aspart|lys|asparagin|tyros|cystein'.split`|`[a/=20n,a%20n],a=0n,[...`-}Þ÷...Ý¦û´`].map(i=>[i=i.charCodeAt(),i-=i<127?34:68,a=a*188n+BigInt(i<0?66+4*i:i)])).join`yl`+'ine'

Try it online!
Generator used

Python, 32148 bytes
import zlib, base64
exec(zlib.decompress(base64.b64decode('[encoded data, 32084 characters]')))

Explanation: All that nonsense inside the exec() statement is actually the compressed version of print('methionyl...'
As per the OP's instructions, the hash of the output is ec73f8229f7f8f2e542c316a41cbfff4
This program uses the inbuilt Python modules zlib and base64

You coukd at least post the code without the data here.

@Titus I edited it in.

Is it really shorter to put the print(' and ') inside the compressed string and call exec rather than just to call print and only compress the string of interest?

pastebin link is now dead

... and this isn't a valid answer anymore given it has no working code link and should be deleted.

@pppery I can verify that this once worked since I am pretty sure I tested the code back in 2017. In any case, I generated what should be the original code (same byte count, same output hash, goes 'eJzN...zYN'). Thanks for bringing this to my attention.


text:
Make a string from a string!
Instructions
Your goal is to write code that makes a "string" from a string in the least amount of bytes possible, as per the rules of code-golf. Standard loophole rules apply.
To make a string, you have to assume the string contains only alphabetical characters and basic punctuation such as the period, comma, colons and semicolons, exclamation marks, and question marks. If not, your program can return your language's false value (0, false, False, etc.) instead of making the string.
Spaces and the aforementioned basic punctuation are ignored in the output, so you can strip them before you make the string.
The string will move up or down based on if it has a higher value than the previous letter's position; if it's the same, it stays.
Test Cases
In this example, the string goes down because the characters are of a higher value than the previous one.
ABC

A
 B
  C

This example is like the previous but also shows that case doesn't matter when deciding to go down or up.
AbC

A
 b
  C

The next example goes up as well.
ABCDEFEDCBA

A         A
 B       B
  C     C
   D   D
    E E
     F

This test case includes same-letter conditions.
Roof fluff

   ff
 oo  l ff
R     u

The final test case is very complex.
The quick, brown fox jumped over the lazy dog.

             f       d           d g
  e   c b o n o j m e o e   e a y o
 h q i k r w   x u p   v r h l z
T   u                     t

@ETHproductions Specifics are quite different

Only difference is the lack of connecting lines and the input restriction/freedoms, which I don't even understand. It's a trivial modification in many cases.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Greedily partition the list of combinations with repetition
First, a few definitions:

Given n and k, consider the sorted list of multisets, where for each multiset we choose k numbers from {0, 1, ..., n-1} with repetitions.

For example, for n=5 and k=3, we have:

[(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 0, 4), (0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 1, 4), (0, 2, 2), (0, 2, 3), (0, 2, 4), (0, 3, 3), (0, 3, 4), (0, 4, 4), (1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 1, 4), (1, 2, 2), (1, 2, 3), (1, 2, 4), (1, 3, 3), (1, 3, 4), (1, 4, 4), (2, 2, 2), (2, 2, 3), (2, 2, 4), (2, 3, 3), (2, 3, 4), (2, 4, 4), (3, 3, 3), (3, 3, 4), (3, 4, 4), (4, 4, 4)]

A part is a list of multisets with the property that the size of the intersection of all multisets in the part is at least k-1.  That is we take all the multisets and intersect them (using multiset intersection) all at once.  As examples, [(1, 2, 2), (1, 2, 3), (1, 2, 4)] is a
part as its intersection is of size 2, but [(1, 1, 3),(1, 2, 3),(1, 2, 4)] is not, because its intersection is of size 1.

Task
Your code should take two arguments n and k. It should then greedily go through these multisets in sorted order and output the parts of the list. For the case n=5, k=3, the correct partitioning is:
(0, 0, 0), (0, 0, 1), (0, 0, 2), (0, 0, 3), (0, 0, 4)
(0, 1, 1), (0, 1, 2), (0, 1, 3), (0, 1, 4)
(0, 2, 2), (0, 2, 3), (0, 2, 4)
(0, 3, 3), (0, 3, 4)
(0, 4, 4)
(1, 1, 1), (1, 1, 2), (1, 1, 3), (1, 1, 4)
(1, 2, 2), (1, 2, 3), (1, 2, 4)
(1, 3, 3), (1, 3, 4)
(1, 4, 4)
(2, 2, 2), (2, 2, 3), (2, 2, 4)
(2, 3, 3), (2, 3, 4)
(2, 4, 4)
(3, 3, 3), (3, 3, 4)
(3, 4, 4), (4, 4, 4)

Here is another example for n = 4, k = 4.
(0, 0, 0, 0), (0, 0, 0, 1), (0, 0, 0, 2), (0, 0, 0, 3)
(0, 0, 1, 1), (0, 0, 1, 2), (0, 0, 1, 3)
(0, 0, 2, 2), (0, 0, 2, 3)
(0, 0, 3, 3)
(0, 1, 1, 1), (0, 1, 1, 2), (0, 1, 1, 3)
(0, 1, 2, 2), (0, 1, 2, 3)
(0, 1, 3, 3)
(0, 2, 2, 2), (0, 2, 2, 3)
(0, 2, 3, 3), (0, 3, 3, 3)
(1, 1, 1, 1), (1, 1, 1, 2), (1, 1, 1, 3)
(1, 1, 2, 2), (1, 1, 2, 3)
(1, 1, 3, 3)
(1, 2, 2, 2), (1, 2, 2, 3)
(1, 2, 3, 3), (1, 3, 3, 3)
(2, 2, 2, 2), (2, 2, 2, 3)
(2, 2, 3, 3), (2, 3, 3, 3)
(3, 3, 3, 3)

Clarification of what greedy means:  For each multiset in turn we look to see if it can be added to the existing part. If it can we add it. If it can't we start a new part.  We look at the multisets in sorted order as in the example given above.
Output
You can output the partitioning in any sensible format you like.  However, multisets should be written horizontally on one line. That is an individual multiset should not be written vertically or spread over several lines.  You can choose how you separate the representation of parts in the output.
Assumptions
We can assume that n >= k > 0.

@LuisMendo I just made a mistake.  I meant multisets should be written horizontally on one line.

In the first test case, why is (0, 4, 4) by itself? Given your description, I would think its "part" would be (0, 4, 4), (1, 4, 4), (2, 4, 4), (3, 4, 4), (4, 4, 4). Similarly for (0, 0, 3, 3) in the second test case.

@GregMartin Because of the greediness of the method. You are right that it will in general be suboptimal.  The minimal number of parts you can get by a non greedy method is an interesting if hard question,

Oh, you literally mean that once the very next term doesn't match the "active" part, then that part is closed forever. Ok.

Jelly, 26 25 bytes
œ&µL‘<⁴ȧ⁹ȯ
œċµç\L€=⁴œṗµḊ’

Full program which prints a representation of a list of lists, each list being a part, e.g. for n=5, k=3:
[[[0, 0, 0], [0, 0, 1], [0, 0, 2], [0, 0, 3], [0, 0, 4]], [[0, 1, 1], [0, 1, 2], [0, 1, 3], [0, 1, 4]], [[0, 2, 2], [0, 2, 3], [0, 2, 4]], [[0, 3, 3], [0, 3, 4]], [0, 4, 4], [[1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 1, 4]], [[1, 2, 2], [1, 2, 3], [1, 2, 4]], [[1, 3, 3], [1, 3, 4]], [1, 4, 4], [[2, 2, 2], [2, 2, 3], [2, 2, 4]], [[2, 3, 3], [2, 3, 4]], [2, 4, 4], [[3, 3, 3], [3, 3, 4]], [[3, 4, 4], [4, 4, 4]]]

Note: the representation used removes the redundant [ and ] around lists of length 1.
Try it online! or see a pretty print version (cost 3 bytes)
How?
œ&µL‘<⁴ȧ⁹ȯ - Link 1, conditional multi-set intersection: list x, list y
œ&         - multi-set intersection(x, y)
  µ        - monadic chain separation (call that i)
   L       - length(i)
    ‘      - increment
     <     - less than?:
      ⁴    -     2nd program input, k
       ȧ   - logical and with:
        ⁹  -     link's right argument, y (y if i is too short, else 0)
         ȯ - logical or (y if i is too short, else i)

œċµç\L€=⁴œṗµḊ’ - Main link: n, k
œċ             - combinations with replacement(n, k) (sorted since n implies [1,n])
  µ            - monadic chain separation (call that w)
         œṗ    - partition w at truthy indexes of:
   ç\          -     reduce w with last link (1) as a dyad
     L€        -     length of €ach
        ⁴      -     2nd program input, k
       =       -     equal (vectorises)
           µ   - monadic chain separation
            Ḋ  - dequeue (since the result will always start with an empty list)
             ’ - decrement (vectorises) (since the Natural numbers were used by œċ)

This is great. Thank you.

MATL, 34 bytes
vi:qiZ^!S!Xu!"@!&vt1&dXasq?0&Y)0cb

Parts are separated by a line containing whitespace.
Try it online!
Explanation
Disclaimer: this method seems to work (and it does in the test cases), but I don't have a proof that it always does
Multisets are sorted, both internally (i.e. each multiset has non-decreasing entries) and externally (i.e. multiset M comes before multiset N if M precedes N lexicographically).
To compute the multiset intersection, the sorted multisets are arranged as rows of a matrix and consecutive differences are computed along each column. If all columns except at most one have all differences equal to zero, the multisets belong to the same part.
This test would give a false negative result for multisets like (1,2,3) and (2,3,4): even if 2, 3 are common entries, they would not be detected as such because they are in non-matching columns.
However, this doesn't seem to be a problem, at least in the test cases. It appears that a test between multisets like 1,2,3 and 2,3,4 never actually has to be done, because some intermediate multiset gave a negative result and so they already are in different parts. If this is indeed true, the reason no doubt has to do with the fact that the multisets are sorted.
I don't have a proof of this, though. It just seems to work.
v           % Concatenate stack vertically: gives an empty array. This will
            % grow into the first part
i:q         % Input n. Push [0 1 ... n-1]
i           % Input k
Z^          % Cartesian power. Each Cartesian tuple is on a row
!S!         % Sort each row
Xu          % Unique rows. This gives all multisets, sorted, each on a row
!           % Transpose
"           % For each column
  @!        %   Push current multiset as a row
  &v        %   Vertically concatenate with the part so far
  t         %   Duplicate
  1&d       %   Consecutive differences along each column
  Xas       %   Number of columns that contain at least one non-zero entry
  q?        %   If that number is not 1 (this means that the current
            %   multiset should begin a new part)
    0&Y)    %     Push last row, then the array with the remaining rows.
            %     Said array is a part, which we now know is complete
    0c      %     Push character 0. This will be shown as a line containing
            %     a space. This is used as a separator between parts.
    b       %     Bubble up. This moves the loose row to the top. This row
            %     is the beginning of a new part
            %   Implicitly end if
            % Implicitly end for
            % Implicitly display

It's very impressive.

I'm trying to understand if the method you describe will always work. I see in that in the n=k=4 case we have a new part start at (0, 0, 3, 3), the vectorised consecutive difference of that and the previous multi-set, (0, 0, 2, 3) only has one difference, so how does the "part so far" make this work? (or equivalently what was the prior step result that got used instead of (0, 0, 2, 3)?)

Ah, I see you perform a consecutive difference. Yes this should always work! You are literally looking for the points at which more than one item changes, but rather than multi-set intersection, simply vectorised intersection - which will work since the mutli-sets are sorted to start with.

@JonathanAllan Yes, it's consecutive difference rather than intersection. I still don't see it clear that it will always work, but if you say so... :-)

MATLAB, 272 bytes
function g(n,k);l=unique(sort(nchoosek(repmat(0:n-1,1,k),k),2),'rows');p=zeros(0,k);for i=1:size(l,1)p=[p;l(i,:)];a=0;for j=1:size(p,1)for m=1:size(p,1)b=0;for h=1:k if(p(j,h)==p(m,h))b=b+1;end;end;if(b<k-1)a=1;end;end;end;if(a)fprintf('\n');p=l(i,:);end;disp(l(i,:));end;

Output:
>> g(5,3)
 0     0     0

 0     0     1

 0     0     2

 0     0     3

 0     0     4

 0     1     1

 0     1     2

 0     1     3

 0     1     4

 0     2     2

 0     2     3

 0     2     4

 0     3     3

 0     3     4

 0     4     4

 1     1     1

 1     1     2

 1     1     3

 1     1     4

 1     2     2

 1     2     3

 1     2     4

 1     3     3

 1     3     4

 1     4     4

 2     2     2

 2     2     3

 2     2     4

 2     3     3

 2     3     4

 2     4     4

 3     3     3

 3     3     4

 3     4     4

 4     4     4

>> g(4,4)
 0     0     0     0

 0     0     0     1

 0     0     0     2

 0     0     0     3

 0     0     1     1

 0     0     1     2

 0     0     1     3

 0     0     2     2

 0     0     2     3

 0     0     3     3

 0     1     1     1

 0     1     1     2

 0     1     1     3

 0     1     2     2

 0     1     2     3

 0     1     3     3

 0     2     2     2

 0     2     2     3

 0     2     3     3

 0     3     3     3

 1     1     1     1

 1     1     1     2

 1     1     1     3

 1     1     2     2

 1     1     2     3

 1     1     3     3

 1     2     2     2

 1     2     2     3

 1     2     3     3

 1     3     3     3

 2     2     2     2

 2     2     2     3

 2     2     3     3

 2     3     3     3

 3     3     3     3

Two empty lines between different parts.
Ungolfed:
function g(n,k);
l=unique(sort(nchoosek(repmat(0:n-1,1,k),k),2),'rows');
p=zeros(0,k);
for i=1:size(l,1)
    p=[p;l(i,:)];
    a=0;
    for j=1:size(p,1)
        for m=1:size(p,1)
            b=0;
            for h=1:k
                if(p(j,h)==p(m,h))
                    b=b+1;
                end;
            end;
                if(b<k-1)
                    a=1;
                end;
        end;
    end;
    if(a)
        fprintf('\n');
        p=l(i,:);
    end;
    disp(l(i,:));
end;

Explanation:
First we find all the multisets with brute force:
l=unique(sort(nchoosek(repmat(0:n-1,1,k),k),2),'rows');

repmat(0:n-1, 1, k) repeats the vector of values from 0 to n-1 k times.
nchoosek(x, k) returns a matrix containing all k-combinations of the repeated vector.
sort(x, 2) sorts all the k-combinations, and then unique(x, 'rows') removes all duplicates.
p=zeros(0,k); creates an empty matrix with k columns. We'll use it as a stack. On each iteration of the outernmost for loop, we first add the current multiset to said stack: p=[p;l(i,:)];.
Then we check if the intersection of all multisets in the stack is at least k-1 long with the following code (we can't use MATLAB's intersect command to check for the intersections, because it returns a set, but we need a multiset):
a=0;
for j=1:size(p,1)
    for m=1:size(p,1)
        b=0;
        for h=1:k
            if(p(j,h)==p(m,h))
                b=b+1;
            end;
        end;
        if(b<k-1)
            a=1;
        end;
    end;
end;

Now, if the intersection is long enough, a == 0, otherwise a == 1.
If the intersection isn't long enough, we print a newline and empty the stack:
if(a)
    fprintf('\n');
    p=l(i,:); % Only the current multiset will be left in the stack.
end;

Then we just print the current multiset:
disp(l(i,:));

Looks like you cracked it!  Could you explain your method?

@Lembik I added an explanation.

PHP, 245 Bytes
for(;$i<($n=$argv[1])**$m=$argv[2];$i++){for($a=[],$v=$i;$v|count($a)<$m;$v=$v/$n^0)array_unshift($a,$v%$n);sort($a);in_array($a,$r)?:$r[]=$a;}foreach($r as$k=>$v)$k&&count(array_diff_assoc($x[$c][0],$v))<2?$x[$c][]=$v:$x[++$c][]=$v;print_r($x);

Try it online!
Expanded
for(;$i<($n=$argv[1])**$m=$argv[2];$i++){ # loop till $argv[1]**$argv[2]
    for($a=[],$v=$i;$v|count($a)<$m;$v=$v/$n^0)
    array_unshift($a,$v%$n); # create base n array
    sort($a); #sort array
    in_array($a,$r)?:$r[]=$a; # if sorted array is not in result add it
}
foreach($r as$k=>$v)
    $k&& # > first item and
    count(array_diff_assoc($x[$c][0],$v))<2 # if difference is only 1 item between actual item and first item in last storage item
    ?$x[$c][]=$v # add item in last storage array
    :$x[++$c][]=$v; # make a new last storage array
print_r($x); # Output as array

Output as String
foreach($x as$y){$p=[];
foreach($y as$z){$p[]=$o="(".join(",",$z).")";}
    echo join(", ",$p)."\n";
}

n>15 for more precision
for($i=0;$i<bcpow($argv[1],$argv[2]);$i=bcadd($i,1)){
    for($a=[],$v=$i;$v|count($a)<$argv[2];$v=bcdiv($v,$argv[1]))
    array_unshift($a,bcmod($v,$argv[1]));
    sort($a);
    in_array($a,$r)?:$r[]=$a;
}

This seems to work!  But what do you mean by more precision?

@Lembik the short version gives back 0 for (16**16-1)%16 and the long version work only with the precision that is needed for n>15 bcmod(bcsub(bcpow(16,16),1),16) is 15 http://php.net/manual/en/ref.bc.php

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Simulate my keyboard
I have an issue with my keyboard (more likely an issue with my browser). Some times when I click on a new text field and begin typing all my text comes out backwards.  After a very short amount of time (for the purpose of the challenge we will say it is 10 milliseconds) it resumes typing forwards again.  It seems that at first my cursor does not move when I begin to type so all the letters get inserted at the beginning of the textfield.
For example if I were typing python and I typed pyt in the first 10 milliseconds the result would be
hontyp

Task
Your job is to simulate this behavior in as few bytes as possible.
Input
You may take input in any reasonable method. You may not however take the input as two separate fields.  The 10 millisecond delay should start when the program begins to allow input.  That is you may have preprocessing to do before input is allowed, which should not be counted towards the 10 milliseconds.
Some leniency is permitted in the timing as long as it averages 10 milliseconds and is off by more than 1 milliseconds no more than 1% of the time on a reasonably speced computer.
Output
You may output the string in whatever method you see fit.  You may either update live as the user is typing or output the result after you are finished taking input (either by a newline or end of file ^D).
Test cases
Some of these were generated by this CMC in The Nineteenth Byte
First 10  | after   | result
bra         in-flak   in-flakarb
ap          ple       plepa
code-       golf      golf-edoc
no          ne        neon
we          st        stew
ai          med       media
draw        back      backward

So answers should figure out themselves what characters were input in the first 10ms? How will this work in languages where the input is frozen as soon as the program starts?

@Luke 1) Yes. 2) If that is the only way they can take input then they cannot compete.

^ That probably rules out a lot of languages

It is unfortunate.  I don't want this challenge to rule our languages, however I really don't want it to become a trivial string manipulation challenge.

So many people have keyboard problems on this site xD

10 milliseconds is a very short time. Did you really mean that?

@Neil You are probably right, but it is what it is, I'm not in a position to change it at this point.

@WheatWizard Can you clarify what is meant by two separate fields? Am I allowed to use two instances of the command to gather input, one of those being timed?

VimScript, 48 47 bytes
Vim has some nice builtins for this. Requires Vim 8.
se ri|cal timer_start(10,{->execute("se ri&")})

Expects text entered in insert mode, a very reasonable input for VimScript. Explanation:
se ri " short for 'set revins', or 'reverse insert'
cal timer_start(10, " Start a timer that fires after 10ms
                 {->execute("se ri&")}) " Give it a lambda that unsets ri

My initial approach, se ri|sl10m|se ri&, doesn't work because the text is buffered until after the sleep command, and I couldn't find a way around this.

HTML (JS), 96 bytes

<input id=i oninput=i.d=i.d||Date.now()+1e3;Date.now()<i.d&&(i.selectionEnd=i.selectionStart=0)>

Starts timing for 1s from the first keypress to give you a chance to see it in action.

JavaScript (ES6), 88 bytes
p=s='',t=+new Date,document.onkeyup=k=>console.log(+new Date-t>9?(s+=k.key)+p:p=k.key+p)

Demo
The way the snippet is working, you'll need to click inside its frame so that it gets focus after clicking on the Run button and before you can start typing.
There's no way to do that in less that 10ms. So, I've set the delay to 1 second in this demo -- and even so, you'd better hurry.

p=s='',t=+new Date,document.onkeyup=k=>console.log(+new Date-t>999?(s+=k.key)+p:p=k.key+p)

Processing, 167 165 146 bytes
String a="",b="";long m=0;void setup(){size(9,9);}void draw(){m=m<1?millis():m;println(b+a);}void keyTyped(){if(millis()-m>9)b+=key;else a=key+a;}

Takes input as keystrokes and outputs to the console. It updates every frame.
Explanation
String a="",b="";                // a contains the reversed (in the 10 milliseconds) string, b contains the text that is typed after the 10 milliseconds.
long m=0;                        // Store the start milliseconds in variable m
void setup() {
  size(9,9);                     // set the dimensions of the window
}

void draw() {                    // loop forever
  m=m<1?millis():m;              // update m
  println(b+a);                  // print the string b+a
}

void keyTyped() {                // this function gets called whenever a key is typed
  if(millis() - m) >9)           // if over 9 milliseconds have elapsed then
    b+=key;                      //   add the key to the end of b
  else                           // else (if less than 10 milliseconds have elapsed)
    a=key+a;                     //   prepend the key to a
}

@WheatWizard Say the input was asdf + ghjk. If everything was in one variable, then after the 10 milliseconds, we would have fdsa in that one string. After that, when g is typed, the string will prepend the g and become gfdsa. Then when the h is typed, I cannot prepend h to the variable (because then the input is still reversed), I will need to use substrings and have to calculate the length of that String variable. This  will cost a lot of bytes.

use ternary to reduce bytes

@Mr.Alien In order to use a ternary, I would need to deal with the same variable, but here, I am use two different variables (namely a and b). Even if I assign the result to a temp variable, the resulting bytecount is still larger.

My bad, I thought it's a JavaScript answer x) .. realized when I saw type casting in your answer...

Python 2, 112 106
import msvcrt as m,time
a=b=''
T=time.time
t=T()
while 1:c=m.getch();exec'ba=+c=+cb'[T()-t>1::2];print a+b

If you want the output to be flush, without flooding with lines, use print'\r'+a+b, instead
You'll need to kill the process to stop it

C - 169 bytes
Well C is not the best for this as it has no platform-independent way of doing this. Works under MSVC.
#include<time.h>
#include<conio.h>
l,k,b[9];main(a){for(a=clock();clock()-a<100;(k=_getch())?b[l++]=k:0);for(;(k=_getch())-13;k?putchar(k):0);for(;l;putchar(b[--l]));}

Bit of a hacky approach, if you can type more than 9 characters in 10 ms, this will crash. Updates after a newline character.

I'm not sure if I can mash 9 random characters in 10ms...


text:
Fermat's factorization helper
We'd like to factorize a semiprime \$N\$. The goal of this challenge is to find two small integers \$u\$ and \$v\$ such that \$uvN\$ can be trivially factorized with Fermat's method, thus allowing to easily deduct the factors of \$N\$.
The task
Given a semiprime \$N\$ and a positive integer \$k\$, we define \$x\$ and \$y\$ as:
$$x=\lceil\sqrt{kN}\rceil$$
$$y=x^2-kN$$
Step #1 - Find \$k\$
You first need to find the smallest possible value of \$k\$ such that \$y\$ is a square number (aka perfect square).
This allows to factorize \$kN\$ with a single iteration of Fermat's factorization method. More concretely, this immediately leads to:
$$kN=(x+\sqrt{y})\times(x-\sqrt{y})$$
(Update: this sequence is now published as A316780)
Step #2 - Factorize \$k\$
You then have to find the two positive integers \$u\$ and \$v\$ such that:
$$uv=k$$
$$cu=x+\sqrt{y}$$
$$dv=x-\sqrt{y}$$
where \$c\$ and \$d\$ are the prime factors of \$N\$.
Summary
Your task is to write a program or function that takes \$N\$ as input and prints or outputs \$u\$ and \$v\$ in any order and any reasonable format.
Example
Let's consider \$N = 199163\$
Step #1
The smallest possible value of \$k\$ is \$40\$, which gives:
$$x = \lceil(\sqrt{40 \times 199163})\rceil = 2823$$
$$y = 2823^2 - 40 \times 199163 = 7969329 - 7966520 = 2809 = 53^2$$
$$kN = (2823 + 53) \times (2823 - 53)$$
$$kN = 2876 \times 2770$$
Step #2
The correct factorization of \$k\$ is \$k = 4 \times 10\$, because:
$$kN = 2876 \times 2770$$
$$kN = (719 \times 4) \times (277 \times 10)$$
$$N = 719 \times 277$$
So, the correct answer would be either \$[ 4, 10 ]\$ or \$[ 10, 4 ]\$.
Rules

It is not required to strictly apply the two steps described above. You're free to use any other method, as long as it finds the correct values of \$u\$ and \$v\$.
You must support all values of \$uvN\$ up to the native maximum size of an unsigned integer in your language.
The input is guaranteed to be a semiprime.
This is code-golf, so the shortest answer in bytes wins.
Standard loopholes are forbidden.

Test cases
N          | k    | Output
-----------+------+------------
143        | 1    | [   1,  1 ]
2519       | 19   | [   1, 19 ]
199163     | 40   | [   4, 10 ]
660713     | 1    | [   1,  1 ]
4690243    | 45   | [   9,  5 ]
11755703   | 80   | [  40,  2 ]
35021027   | 287  | [   7, 41 ]
75450611   | 429  | [ 143,  3 ]
806373439  | 176  | [   8, 22 ]
1355814601 | 561  | [  17, 33 ]
3626291857 | 77   | [   7, 11 ]
6149223463 | 255  | [  17, 15 ]
6330897721 | 3256 | [  74, 44 ]

Example implementation
In the snippet below, the \$f\$ function is an ungolfed implementation which takes \$N\$ as input and returns \$u\$ and \$v\$.
For illustrative purposes only, the snippet also includes the \$g\$ function which takes \$N\$, \$u\$ and \$v\$ as input and computes the factors of \$N\$ in \$O(1)\$.

f = N => {
  for(k = 1;; k++) {
    x = Math.ceil(Math.sqrt(k * N));
    y = x * x - k * N;
    ySqrt = Math.round(Math.sqrt(y));

    if(ySqrt * ySqrt == y) {
      p = x + ySqrt;

      for(u = 1;; u++) {
        if(!(p % u) && !(N % (p / u))) {
          v = k / u;
          return [ u, v ];
        }
      }
    }
  }
}

g = (N, u, v) => {
  x = Math.ceil(Math.sqrt(u * v * N));
  y = x * x - u * v * N;
  ySqrt = Math.round(Math.sqrt(y));
  p = x + ySqrt;
  q = x - ySqrt;
  return [ p / u, q / v ];
}

[
  143, 2519, 199163, 660713, 4690243, 11755703,
  35021027, 75450611, 806373439, 1355814601,
  3626291857, 6149223463, 6330897721
]
.map(N => {
  [u, v] = f(N);
  [c, d] = g(N, u, v);
  console.log(
    'N = ' + N + ', ' +
    'u = ' + u + ', ' +
    'v = ' + v + ', ' +
    'N = ' + c + ' * ' + d
  );
});

Are we guaranteed that the input N will in fact be a semiprime?

@GregMartin Yes you are.

Mathematica, 81 79 bytes
Thanks to Martin Ender for saving 2 bytes!
(c=Ceiling;For[j=0;z=E,c@z>z,p=(x=c@Sqrt[j+=#])+{z=Sqrt[x^2-j],-z}];p/#~GCD~p)&

Pure function taking a semiprime as input and returning an ordered pair of positive integers. The For loop implements the exact procedure described in the question (using # for the input in place of n), with x as defined there, although we store j = k*n instead of k itself and z=Sqrt[y] instead of y itself. We also compute p={x+z,x-z} inside the For loop, which ends up saving one byte (on like the seventh try). Then the two desired factors are (x+z)/GCD[#,x+z] and (x-z)/GCD[#,x-z], which the concise expression p/#~GCD~p computes directly as an ordered pair.
Curiosities: we want to loop until z is an integer; but since we're going to use Ceiling already in the code, it saves two bytes over !IntegerQ@z to define c=Ceiling (which costs four bytes, as Mathematica golfers know) and then test whether c@z>z. We have to initialize z to something, and that something had better not be an integer so that the loop can start; fortunately, E is a concise choice.

NB: I do not accept the shortest answer anymore as I used to do in my first code-golf challenges. I usually just let accepted answers as-is unless they are outgolfed, for the sake of consistency. This is what happened here.

no worries, thanks for the note :)

JavaScript (ES7), 86 81 bytes
n=>(g=k=>(y=(n*k)**.5+1|0,y+=(y*y-n*k)**.5)%1?g(k+1):n*u++%y?g(k):[--u,k/u])(u=1)

Edit: Saved 4 bytes thanks to @Arnauld.

APL (Dyalog Unicode), 60 49 45 bytes
{×1|w←⍵∨(⌈+.5*⍨×⍨-⍨⌈×⌈).5*⍨⍵×⍺:⍺∇1+⍵⋄w,⍵÷w}∘1

Try it online!
-11b thanks to Adám, -4b courtesy of ngn along with some excellent tips from both on The APL Orchard
Explanation:
First ever APL answer so might be able to be golfed further! Explanation below is for the 49b solution. Approach pretty much just follows the algorithm in the original post, but uses the GCD to compute the factors once k, x and y are known.
{×1|y←.5*⍨(×⍨x←⌈.5*⍨⍵×⍺)-⍵×⍺:⍺∇1+⍵⋄w,⍵÷w←⍵∨x+y}∘1 ⍝
{                                             }∘1 ⍝ dfn; ⍺ is left argument (N), ⍵ is right argument acting as a counter for k incremented each recursion from 1
          (  x←⌈.5*⍨⍵×⍺)-⍵×⍺                      ⍝ Computes x as per question for current k
    y←.5*⍨(×⍨x         )                          ⍝ Computes sqrt(y) from x as per question (×⍨x = x^2). Note that y here actually represents sqrt(y) from the question as we do not need to store actual y
 ×1|y                                             ⍝ Use 1 modulo y and 'direction' to return 0 if y is a whole number, else 1
                            :     ⋄               ⍝ 'guard' - if the evaluation of the left side of : is true, performs the statement immediately following it, otherwise the next statement (separated by ⋄)
                             ⍺∇1+⍵                ⍝ if guard is true, recurse with the same left argument of N, and a right argument of the new k (k+1)
                                   w,⍵÷w←⍵∨x+y    ⍝ if guard is false, computes gcd (∨) of k and (x+y) as one factor (w), and k÷w as the second factor

Python 2, 127 121 117 111 107 104 101 99 bytes
 -1 byte thanks to Neil & -3 bytes thanks to ovs
N=input()
k=u=1;p=m=.5
while p%1:p=1+(k*N)**m//1;p+=(p*p-k*N)**m;k+=1
while N*u%p:u+=1
print~-k/u,u

Try it Online!
Curiosities:
p is  initialized to .5 so that the loop condition will be true on the first iteration. Note that it is shorter to store p (as x + sqrt(y)) than it is to store each of x and y separately.

x*x instead of x**2?

@Neil Yes, of course. Thanks

Axiom, 131 115 bytes
v(x)==floor(x^.5)::INT;r(n)==(k:=0;repeat(k:=k+1;x:=1+v(k*n);y:=v(x*x-k*n);x^2-y^2=k*n=>break);[w:=gcd(k,x+y),k/w])

The function that would resolve question is r(n) above. ungolf and test
vv(x)==floor(x^.5)::INT

--(x-y)*(x+y)=k*n
rr(n)==
  k:=0
  repeat
     k:=k+1
     x:=1+vv(k*n)
     y:=vv(x*x-k*n)
     x^2-y^2=k*n=>break
  [w:=gcd(k,x+y),k/w]

(4) -> [[i,r(i)] for i in [143,2519,199163,660713,4690243,11755703]]
   (4)
   [[143,[1,1]], [2519,[1,19]], [199163,[4,10]], [660713,[1,1]],
    [4690243,[9,5]], [11755703,[40,2]]]
                                                      Type: List List Any

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/