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Introduction

On March 24th, 2015 @isaacg golfed his Pyth answer from 44 to 42 bytes. Since a crossed out 44 (44) looks a lot like a regular 44, @Optimizer made the following comment:

striked out 44 is still normal 44 :(

After that, on October 21st, 2015, @Doorknob♦ golfed his Ruby answer from 44 to 40 (and later 38) bytes and added the following part to his answer, with a link to that original comment of @Optimizer:

crossed out 44 is still regular 44 ;(

This was the start of an answer-chaining meme, where every crossed out 44 (and in some occasions 4 or 444) linked back to the previous one.

Then on April 8th, 2017 (I'm not sure if this was the first answer to do so, but it's the earliest one I could find), @JonathanAllan golfed his Python answer from 44 to 39. He however used <s>&nbsp;44&nbsp;</s> so the 44 would look like this:  44 , and added the following to his answer:

Crossed out &nbsp;44&nbsp; is no longer 44 :)

And that was basically the (beginning of the) end of the meme.

Challenge

As for this challenge: Given a list of positive integers and a date, output the list comma-and-space separated where every number except for the last one is placed between <s>...</s> tags.

In addition, if any of the crossed out numbers is in the sequence [4, 44, 444, 4444, ...] (A00278 on oeis.org):

  • If the date is before April 8th, 2017: Also output the exact (all lowercase and with semicolon emoticon) text crossed out N is still regular N ;( (N being the crossed out number from the sequence) on a second line.
  • If the date is April 8th, 2017 or later: The crossed out number N from the sequence should have the leading and trailing &nbsp; added. No need for any additional lines of output.

Examples:

Input: list = [50, 48, 44, 41], date = January 1st, 2017
Output:

<s>50</s>, <s>48</s>, <s>44</s>, 41 crossed out 44 is still regular 44 ;(

Input: list = [500, 475, 444, 301, 248], date = June 2nd, 2018
Output:

<s>500</s>, <s>475</s>, <s>&nbsp;444&nbsp;</s>, <s>301</s>, 248

Challenge rules:

  • You can assume the input-list is a sorted list from largest to smallest, only containing positive integers. In reality a byte-count can also go up due to bug-fixes, but for the sake of this challenge we pretend it only goes down.
  • You can assume only a single number from the sequence [4, 44, 444, 4444, ...] is present in the input-list (if any).
  • The output format is either printed to STDOUT, or returned as a string (or character array/list/2D-array if that's preferable). A trailing newline is of course optional.
  • The output format is strict. <s>...</s> is mandatory; &nbsp;...&nbsp; is mandatory; ", " (comma and space) is mandatory; and \ncrossed out ... is still regular ... ;( exactly is mandatory (on a separated line).
  • You may take the input-date as date-objects; timestamps; loose integers for year, month, and day; a single number in the format yyyyMMdd; integer days since December 31st, 1899 (which would be 42832 for April 8th, 2017); or any other reasonable input-format. The date if-statement isn't the major part of this challenge.
  • The input integer-list can also be a list of strings if you want.
  • You don't have to add the <sup>...</sup> tags to the crossed out ... is still regular ... ;( line as is usually done with the actual meme answers.
  • You can assume the input-list will never contain byte-counts outside the [1, 50000] range (so you'll only have these five {4, 44, 444, 4444, 44444} to worry about).

General rules:

  • This is , so shortest answer in bytes wins.
    Don't let code-golf languages discourage you from posting answers with non-codegolfing languages. Try to come up with an as short as possible answer for 'any' programming language.
  • Standard rules apply for your answer, so you are allowed to use STDIN/STDOUT, functions/method with the proper parameters and return-type, full programs. Your call.
  • Default Loopholes are forbidden.
  • If possible, please add a link with a test for your code.
  • Also, adding an explanation for your answer is highly recommended.

Test cases:

Input: [50, 48, 44, 41] and January 1st, 2017 Output: <s>50</s>, <s>48</s>, <s>44</s>, 41 crossed out 44 is still regular 44 ;( Input: [500, 475, 444, 301, 248] and June 2nd, 2018 Output: <s>500</s>, <s>475</s>, <s>&nbsp;444&nbsp;</s>, <s>301</s>, 248 Input: [8, 6, 4] and December 5th, 2017 Output: <s>8</s>, <s>6</s>, 4 Input: [8, 6, 4, 3, 2] and September 15th, 2015 Output: <s>8</s>, <s>6</s>, <s>4</s>, <s>3</s>, 2 crossed out 4 is still regular 4 ;( Input: [119, 99, 84, 82, 74, 60, 51, 44, 36, 34] and February 29th, 2016 Output: <s>119</s>, <s>99</s>, <s>84</s>, <s>82</s>, <s>74</s>, <s>60</s>, <s>51</s>, <s>44</s>, <s>36</s>, 34 crossed out 44 is still regular 44 ;( Input: [404, 123, 44] and March 4th, 2016 Output: <s>404</s>, <s>123</s>, 44 Input: [4, 3] and April 8th, 2017 Output: <s>&nbsp;4&nbsp;</s>, 3 Input: [44] and October 22nd, 2017 Output: 44 Input: [50000, 44444, 1500] and August 1st, 2018 Output: <s>50000</s>, <s>&nbsp;44444&nbsp;</s>, 1500 Input: 50, 38, 23 and September 8th, 2001 Output: <s>50</s>, <s>38</s>, 23
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  • \$\begingroup\$ Your last test case is missing its date? \$\endgroup\$ Commented Aug 8, 2018 at 8:47
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    \$\begingroup\$ @Neil Ah oops.. Fixed. Date doesn't really matter for that one, but added one regardless. Thanks for noticing. \$\endgroup\$ Commented Aug 8, 2018 at 8:49
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    \$\begingroup\$ I can confirm that I had not noticed anyone use the non-breaking space for this specific purpose and that was why I wrote the "no longer" text. \$\endgroup\$ Commented Aug 8, 2018 at 9:16
  • \$\begingroup\$ (I can also confirm that I had used it prior to using it with 44 - I believe that is when I first used them) \$\endgroup\$ Commented Aug 8, 2018 at 9:35
  • \$\begingroup\$ As we can take input as days since 1889-12-31, can we also take days since 2017-04-08? \$\endgroup\$ Commented Aug 8, 2018 at 11:49

8 Answers 8

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Jelly,  444 , 94, 93 bytes

Ñȧ$“&nbsp;”,¤j$€io.ɗ¦@Ṗj@€“<s>“</s>”oj⁾, ⁴>⁽A€ “¢⁻$gẆẠ⁷Ṭ]ḳṁṛż?=çỊI×V»Ỵjṭ⁷ẋǬȧẠƲ ṖḟÐḟ”4Ḣµñ³,Ç

A full program. The inputs are a list of strings and a date taken as integer days since January the first 1970 (making 17264 April the eighth 2017)

Try it online!

How?

Ñȧ$“&nbsp;”,¤j$€io.ɗ¦@Ṗj@€“<s>“</s>”oj⁾, - Link 1: L = list of characters ("4...4") OR integer (0), - R = list of lists of characters (the strings provided to the program) $ - last 2 links as a monad: Ñ - call next Link (2) as a monad - ...gets: is date input to program greater than 2017-04-07? ȧ - AND (if so gets the value of L, else 0), say X ¦@ - sparse application (with swa@pped @rguments)... Ṗ - ...with right argument = popped R (without it's rightmost entry) ɗ - ...to: last 3 links as a dyad i - first index of X in popped R (0 if no found, so 0->0) . - literal 0.5 o - OR (change any 0 to 0.5) - ...i.e. index of "4...4" if L was one or 0.5, an invalid index $€ - ...do: for €ach... last 2 links as a monad: ¤ - nilad followed by link(s) as a nilad: “&nbsp;” - literal list of characters = "&nbsp;" , - pair (with itself) = ["&nbsp;", "&nbsp;"] j - join (with the item) e.g.: "&nbsp;444&nbsp;" or ["&nbsp;", 0, "&nbsp;"] “<s>“</s>” - literal list of lists of characters = ["<s>", "</s>"] j@€ - for €ach... join (with swa@pped @rguments) o - OR with R (vectorises, so adds the popped entry back onto the right-side) ⁾, - literal list of characters = ", " j - join ⁴>⁽A€ - Link 2: greater than 2017-04-07? ⁴ - program's 4th argument (2nd input) ⁽A€ - literal 17263 (days(2017-04-07 - 1970-01-01)) > - greater than? “¢⁻$gẆẠ⁷Ṭ]ḳṁṛż?=çỊI×V»Ỵjṭ⁷ẋǬȧẠƲ - Link 3: L = list of characters ("4...4") OR integer (0) “¢⁻$gẆẠ⁷Ṭ]ḳṁṛż?=çỊI×V» - compressed list of characters = "crossed out \n is still regular \n ;(" Ỵ - split at newlines = ["crossed out ", " is still regular ", " ;("] j - join with L ⁷ - literal newline character ṭ - tack (add to the front) Ʋ - last 4 links as a monad: Ç - call last Link (2) as a monad ¬ - NOT Ạ - All (1 if L is "4...4", 0 if L is 0) ȧ - AND ẋ - repeat (i.e. get the list of characters to print or an empty list) ṖḟÐḟ”4Ḣµñ³,Ç - Main Link: list of strings, integer (days since 1970-01-01) Ṗ - pop (list of strings without it's rightmost entry) Ðḟ - filter discard if: ḟ - filter discard any which are in... ”4 - ...literal character '4' Ḣ - head (yields 0 if list is now empty) µ - new monadic chain, call that X ³ - program's 3rd argument (1st input) - call that Y) ñ - call next Link (1) as a dyad (i.e. f1(X, Y)) Ç - call last Link (3) as a monad (ie. f3(X)) , - pair - implicit (smashing) print
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  • 1
    \$\begingroup\$ You forgot to remove the input from another challenge in your TIO. ;) Nice answer regardless. Would you mind adding an explanation? And rofl at that 444-byte count. I don't believe it one bit that you started with that, unless you added comments. ;p \$\endgroup\$ Commented Aug 12, 2018 at 9:18
  • \$\begingroup\$ Opps, thanks. I'll add an explanation shortly. \$\endgroup\$ Commented Aug 12, 2018 at 10:16
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    \$\begingroup\$ 444 version no comments, numbers rather than compressed strings \$\endgroup\$ Commented Aug 12, 2018 at 10:17
  • \$\begingroup\$ ...ah the 444 version has a bug - its using days since 1969-12-31 not since 1970-01-01 as the golfed version does >_< (naive fix is still 444) \$\endgroup\$ Commented Aug 12, 2018 at 12:31
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Python 2, 208 204  203 201 197 bytes

Takes input as a list of strings, and an int of yyyymmDD

def f(l,d): A=a=d>20170407;r=[] for n in l[:-1]:x=set(n)=={'4'};S='&nbsp;'*x*a;r+=['<s>'+S+n+S+'</s>'];A=x*n or A print', '.join(r+l[-1:])+'\ncrossed out %s is still regular %s ;('%(A,A)*(a*A<A)

Try it online!

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3
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Excel VBA, 217 bytes

VBE immediate window function that takes input array from range [A:A], and date from range [B1] and outputs to the console.

c=[Count(A:A)]:d=[B1]>42832:For i=1To c-1:n=Cells(i,1):l=InStr(44444,n):s=IIf(d*l,"&nbsp;",""):v=IIf((d=0)*l,n,v):?"<s>"s;""&n;s"</s>, ";:Next:?""&Cells(i,1):?IIf(v,"crossed out "&v &" is still regular "&v &" ;(","");

Ungolfed and Commented

c=[Count(A:A)] '' Get numer of elements d=[B1]>42832 '' Check if date is after 7 Apr 2017, For i=1To c-1 '' Iterate over index n=Cells(i,1) '' Get array val at index l=InStr(44444,n) '' Check if val is all 4s s=IIf(d*l,"&nbsp;","") '' If after 7 Aug 2017, and All 4s, let `s` be "&nbsp;" v=IIf((d=0)*l,n,v) '' If all 4s, and not after date, let v hold n, else hold v ?"<s>"s;""&n;s"</s>, "; '' Print striked vales, with "&nbsp;", if applicable Next '' Loop ?""&Cells(i,1) '' Print last value in array '' (below) Print meme, if needed ?IIf(v,"crossed out "&v &" is still regular "&v &" ;(","");

-2 bytes for changing date format to YYYYMMDD

-1 byte for comparing to 42832 (int value for 07 Apr 2017), Thanks @Neil

-2 bytes for removing 1, from the InStr statement, Thanks @SeaDoggie01

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  • \$\begingroup\$ I guess For i=1To~-c instead of For i=1To c-1 isn't possible in Excel VBA? If I recall correctly Bitwise not is Not instead of ~, or are both possible? (Note that I know next to nothing of Excel VBA, so I'm probably saying something stupid here. ;p) \$\endgroup\$ Commented Aug 8, 2018 at 13:05
  • \$\begingroup\$ @KevinCruijssen Yeah, bitwise not is Not , so no ~ :( \$\endgroup\$ Commented Aug 8, 2018 at 13:07
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    \$\begingroup\$ I think you can save more bytes by changing the date format back to date and comparing against 42832 (the integer value of 2017-04-07). \$\endgroup\$ Commented Aug 8, 2018 at 15:24
  • \$\begingroup\$ I think you can save 2 bytes by removing the "1," from the instr (it's optional) \$\endgroup\$ Commented Sep 11, 2018 at 19:25
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Retina 0.8.2, 130 bytes

$ ;42833 O`;.{5} \b(4+),(?=.*;42833;) &$1&, & &nbsp; .{12}$ \b(4+),.* $&¶crossed out $1 is still regular $1 ;( (.+?), <s>$1</s>,

Try it online! Link includes test cases. Uses Excel date stamps (days since 1899-12-31 but including 1900-02-49). 141 bytes for a version that takes ISO dates:

$ ;2017-04-08 O`;.{10} \b(4+),(?=.*;2017-04-08;) &$1&, & &nbsp; .{22}$ \b(4+),.* $&¶crossed out $1 is still regular $1 ;( (.+?), <s>$1</s>,

Try it online! Link includes test cases. Explanation:

$ ;2017-04-08

Append the cut-off date to the input.

O`;.{10}

Sort the dates. If the given date is on or after the cut-off date then the first date will be the cut-off date.

\b(4+),(?=.*;2017-04-08;) &$1&, & &nbsp;

In that case, wrap the 4+ in &nbsp; (using two stages as it saves a byte).

.{22}$

Delete the dates as they have done their job.

\b(4+),.* $&¶crossed out $1 is still regular $1 ;(

If there's an unspaced 4+, then append the meme.

(.+?), <s>$1</s>,

Strike out all of the obsolete byte counts.

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8
  • \$\begingroup\$ You can save 5 bytes by taking the date-format without -. \$\endgroup\$ Commented Aug 8, 2018 at 9:11
  • \$\begingroup\$ @KevinCruijssen I could save even more bytes by asking for it as an Excel date (days since 1899-12-31). How far can I go with this? \$\endgroup\$ Commented Aug 8, 2018 at 10:13
  • \$\begingroup\$ Hmm.. You know what, that's also fine by me. How the if-statement for the date April 8th, 2017 is done I don't really care. It's not the major part of this challenge. \$\endgroup\$ Commented Aug 8, 2018 at 11:12
  • \$\begingroup\$ That would be 42832 for April 8th, 2017 I assume? \$\endgroup\$ Commented Aug 8, 2018 at 11:18
  • \$\begingroup\$ @KevinCruijssen My Excel says that's the 7th. \$\endgroup\$ Commented Aug 8, 2018 at 11:33
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Ruby, 208 184 180 bytes

TIO-test

Thanks for @KevinCruijssen for saving 2 bytes!

->n,d{*h,t=n;i=20170408>d;f=?\n;h.map{|u|o=u.digits|[]==[4];f=o&&i ?f+"crossed out #{u} is still regular #{u} ;(":f;o&&!i ?"<s>&nbsp;#{u}&nbsp;</s>":"<s>#{u}</s>"}*", "+", #{t}"+f}

It's a lambda function that takes a list of numbers and an integer as a date in the format of YYYYmmdd.

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  • \$\begingroup\$ Would it be possible to add a TIO-link with test code? And switching formats so it's an integer input would indeed save some bytes. :) \$\endgroup\$ Commented Aug 9, 2018 at 12:21
  • \$\begingroup\$ @KevinCruijssen Added link ;) \$\endgroup\$ Commented Aug 9, 2018 at 12:27
  • \$\begingroup\$ Thanks! +1 from me. You can save 2 bytes by removing the spaces after the ?. Also, for your TIO, you can split your actual submission code and test code with the header/footer, like this. :) \$\endgroup\$ Commented Aug 9, 2018 at 12:41
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Haskell, 227 bytes

i[]=1>0;i('4':s)=i s;i(_:_)=0>1 f l d=m l++c where a=head$filter i l;t d|d<42832=("","\ncrossed out "++a++" is still regular "++a++" ;(")|1>0=("&nbsp;","");(b,c)=t d;w n|i n=b++n++b|1>0=n;m[n]=n;m(x:s)="<s>"++w x++"</s>, "++m s

Try it online! Run f with list l and date d. 42832 is the changing date.

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  • \$\begingroup\$ Would it be possible to add a TIO-link with test code? \$\endgroup\$ Commented Aug 9, 2018 at 12:21
  • \$\begingroup\$ @KevinCruijssen in Russia it's banned \$\endgroup\$ Commented Aug 9, 2018 at 15:38
  • \$\begingroup\$ Oh, didn't knew that. Well, I don't know Haskell too well, otherwise I would have tested it myself a bit. But the code itself seems logical, so +1 from me nonetheless. \$\endgroup\$ Commented Aug 9, 2018 at 16:39
  • \$\begingroup\$ @KevinCruijssen you can copy code to TIO, make link and add it to post \$\endgroup\$ Commented Aug 9, 2018 at 20:12
  • \$\begingroup\$ @KevinCruijssen TIO is banned by mistake, but I'm too lazy to use proxy/contact ISP about it \$\endgroup\$ Commented Aug 9, 2018 at 20:18
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JavaScript (Node.js), 173 bytes

a=>d=>a.map(x=>--i<1?x:`<s>${(b=/^4+$/.exec(x)?(c=x,d)<14915808e5?n=[]:"&nbsp;":"")+x+b}</s>`,n="",c=0,i=a.length).join`, `+(n&&` crossed out ${c} is still regular ${c} ;(`)

Try it online!

Using curry syntax f(array)(js_timestamp)

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JavaScript, 194 bytes

(a,d,l=a.pop(),n=a.find(x=>/^4+$/.exec(x)),s=a.map(n=>`<s>${n}</s>, `).join``+l)=>d<1491609600?n?s+` crossed out ${n} is still regular ${n} ;(`:s:s.replace(/>(4+)</g,(_,m)=>`>&nbsp;${m}&nbsp;<`) 

f= (a,d,l=a.pop(),n=a.find(x=>/^4+$/.exec(x)),s=a.map(n=>`<s>${n}</s>, `).join``+l)=>d<1491609600?n?s+` crossed out ${n} is still regular ${n} ;(`:s:s.replace(/>(4+)</g,(_,m)=>`>&nbsp;${m}&nbsp;<`) const date = s => Date.parse(s) / 1000 for(const output of [	f([50, 48, 44, 41], date('January 1, 2017')),	f([500, 475, 444, 301, 248], date('June 2, 2018')),	f([8, 6, 4], date('December 5, 2017')),	f([8, 6, 4, 3, 2], date('September 15, 2015')),	f([119, 99, 84, 82, 74, 60, 51, 44, 36, 34], date('February 29, 2016')),	f([404, 123, 44], date('March 4, 2016')),	f([4, 3], date('April 8, 2017')),	f([44], date('October 22, 2017')),	f([50000, 44444, 1500], date('August 1, 2018')),	f([50, 38, 23], date('September 8, 2001')), ]) console.log(output)

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