Interval Notations

Input:

Two integer parameters a and b (where a<b and the difference is at least 2)

Output:

Output or return this text, where a and b are filled in:

(a,b) = ]a,b[ = {a<x<b}   = {a<x&&x<b}   = a+1..b-1
[a,b) = [a,b[ = {a<=x<b}  = {a<=x&&x<b}  = a..b-1
(a,b] = ]a,b] = {a<x<=b}  = {a<x&&x<=b}  = a+1..b
[a,b] = [a,b] = {a<=x<=b} = {a<=x&&x<=b} = a..b


Challenge rules:

• I/O is flexible. Can be printed to STDOUT, returned as a string/character-array, etc. Can be inputted as two integers, decimals, strings (not sure why since you need to calculate the a+1 and b-1, but be my guest..), etc.
• Any amount of leading and/or trailing new-lines are allowed, and any amount of trailing and/or leading spaces for each line is allowed.
• The spaces at the equal signs (including those to align them in the same column) are mandatory, spaces between the other characters are not allowed.
• a+1 and b-1 are replaced with the correct values after these calculations.
• You are not allowed to use ≤ instead of <=.
• You are not allowed to use & instead of &&.
• You are allowed to output the numbers with .0 (as long as it's consistent, and not more than one decimal zero).
• You can assume a is at least 2 lower than b (in order for (a,b) to be correct).
• The lines should be output in the order shown.

Example:

Input: a=-5, b=10
Output:

(-5,10) = ]-5,10[ = {-5<x<10}   = {-5<x&&x<10}   = -4..9
[-5,10) = [-5,10[ = {-5<=x<10}  = {-5<=x&&x<10}  = -5..9
(-5,10] = ]-5,10] = {-5<x<=10}  = {-5<x&&x<=10}  = -4..10
[-5,10] = [-5,10] = {-5<=x<=10} = {-5<=x&&x<=10} = -5..10


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.

PS: For those who had seen this challenge in the Sandbox when it was still a challenge with a and b hard-coded, I've changed it to an input challenge to prevent boring hard-coded and encoded answers like we usually see with KC challenges.

• So for a=5, b=6 still output (5,6) = ... = 6...5? – l4m2 Apr 4 '18 at 11:03
• @l4m2 Ah.. hadn't thought about that. You can assume the a and b will always be at least 2 apart. I'll edit the challenge. – Kevin Cruijssen Apr 4 '18 at 11:18
• It turned out to be much more fun than I was expecting. Nice challenge! – Arnauld Apr 4 '18 at 11:54
• @DigitalTrauma Sorry, but this exact order. – Kevin Cruijssen Apr 5 '18 at 7:02

Stax, 74 bytes

ÉyU≤₧pΔz▀σ┬♪•a≤☻Σ╕←k►¬╗Ö)ßâL╫§▐ƒ┼°╚íS3:Y¶7]7♂e╖à╙ô≥;M0h8♦Oún┼ë←B╠╫║┌♂α▲╚


Run and debug it

This uses stax's string templates heavily. Unpacked, ungolfed, and commented, it looks like this.

Y                   save second input in Y register (first is already in X)
.)].([|*            cross product of ")]" and "(["; this produces [")(", ")[", "](", "]["]
{                   begin block to map over interval delimiters
E"x,y"a++       push delimiters separately, then wrap them around the inputs    e.g. "(-5,10)"
c"(])["|t         copy last value, then replace parentheses with braces           e.g. "]-5,10["
ih'=              push half the iteration index and "="                           e.g. 0 "="
|;'=              push iteration parity (alternating 0 and 1) and "="             e.g. 0 "=" 0 "="
"{x<*x<*y}"   multiply each equal sign by its occurrence, and template        e.g. "{-5<x<10}"
c'x.x&:mR         copy last value, then replace "x" with "x&&x"                   e.g. "{-5<x&&x<10}"
yvih xi|e         calculate final bounds offsets                                  e.g. -5 1 10 -1
"+..+"          add inputs to offsets, and embed in template                    e.g. "-4..9"
5l                combine last 5 values into array
m                   map [")(", ")[", "](", "]["] using block
:<                  left-align grid colums to add extra spaces
m" = "*             for each row, join with " = " and output


Run this one

JavaScript (ES6), 184182181 180 bytes

Takes input in currying syntax (a)(b). Returns an array of 4 strings.

a=>b=>[1,2,3,4].map(k=>'31,23 = 31,23 = {10x72}4{10x&&x72}45..6'.replace(/\d/g,(n,i)=>[(+n?k<3:k&1)?'<':'<=',a,b,'][)([[]('[(i*17^k*718)%9],'   = '.slice(k/2),a+k%2,b-(k<3)][n%7]))


Try it online!

How?

For each row k with 1 ≤ k ≤ 4, we start with the following template:

"31,23 = 31,23 = {10x72}4{10x&&x72}45..6"


and replace each decimal digit n at position i according to the following table:

  n  | Replaced with           | Code
-----+-------------------------+------------------------------------------
0,7 | comparison operator     | (+n ? k < 3 : k & 1) ? '<' : '<='
1  | a                       | a
2  | b                       | b
3  | interval bound          | '][)([[]('[(i * 17 ^ k * 718) % 9]
4  | a substring of '   = '  | '   = '.slice(k / 2)
5  | either 'a' or 'a + 1'   | a + k % 2
6  | either 'b' or 'b - 1'   | b - (k < 3)


Python 2, 225203 195 bytes

a,b=input()
for d in 0,1:
for m in 0,1:k=a+','+b;o='{'+a+'<'+m*'=';c='x<'+d*'='+b+'}'+'  '[m+d:];print' = '.join(['(['[m]+k+')]'[d],']['[m]+k+'[]'[d],o+c,o+'x&&'+c,a+1-m+'..'+b-1+d])


Try it online!

m4, 194

Seems like a job for a macro processor. Not sure if m4 meets our standards for a programming language. It does have looping ability and arithmetic eval, so I assume its close to the mark.

define(l,$1a,b$2 = $3a,b$4 = {a<$5x<$6b} $8= {a<$5x&&x<$6b}$8= $7')dnl l((',)',],[,,,incr(a)..decr(b), ') l([,)',[,[,=,,a..decr(b), ') l((',],],],,=,incr(a)..b, ') l([,],[,],=,=,a..b,)  This is my first non-trivial look at m4, so I suspect there are more golfing opportunities I've missed. Inputs are passed using -D macro definitions at the command line. Not sure if anything needs to be added to the score for these, as as far as I can tell this is the only way to meaningfully pass parameters: $ m4 -Da=-5 -Db=10 intnot.m4
(-5,10) = ]-5,10[ = {-5<x<10}   = {-5<x&&x<10}   = -4..9
[-5,10) = [-5,10[ = {-5<=x<10}  = {-5<=x&&x<10}  = -5..9
(-5,10] = ]-5,10] = {-5<x<=10}  = {-5<x&&x<=10}  = -4..10
[-5,10] = [-5,10] = {-5<=x<=10} = {-5<=x&&x<=10} = -5..10 $ Thanks @Dennis for adding m4 to TIO so quickly! • I don't see why m4 wouldn't be a valid programming language to answer in, but what do I know. ;) As for the -D arguments it's completely fine. In my challenge description I stated I/O is completely flexible, so whether you take input as STDIN, function parameters, program command-line arguments, compiler-flags, reading a file, or whatever other input-method you can think of is completely up to you. – Kevin Cruijssen Apr 5 '18 at 15:39 Python 2, 187 bytes t=a,b=input() for j in 1,0: for i in 1,0:print"%%s%d,%d%%s = "%t*2%('[('[i],'])'[j],'[]'[i],']['[j])+"{%d<%sx%%s<%s%d}%s = "%(a,'='[i:],'='[j:],b,' '*(i+j))*2%('','&&x')+a+i+'..'+b-j  Try it online! Java (JDK 10), 251 bytes a->b->("(a,b)q]a,b[q{a<x<b} q{a<x&&x<b} q"+-~a+".."+~-b+"\n[a,b)q[a,b[q{a<=x<b} q{a<=x&&x<b} qa.."+~-b+"\n(a,b]q]a,b]q{a<x<=b} q{a<x&&x<=b} q"+-~a+"..b\n[a,b]q[a,b]q{a<=x<=b}q{a<=x&&x<=b}qa..b").replace("a",a+"").replace("b",b+"").replace("q"," = ")  Try it online! Credits • I'm still golfing... – Olivier Grégoire Apr 4 '18 at 12:14 • Almost done with golfing? I got a suggestion to remove 85 bytes. ;) – Kevin Cruijssen Apr 4 '18 at 12:38 • @KevinCruijssen I'm done but I removed far less than 85 bytes... – Olivier Grégoire Apr 4 '18 at 12:45 • 251 bytes with three boring .replace. – Kevin Cruijssen Apr 4 '18 at 12:46 • Yeah, basically my first solution using replace indeed, instead of formatting. Nice. Boring indeed, but nice! :-) – Olivier Grégoire Apr 4 '18 at 12:48 Perl 5, 181 bytes I thought this would have worked out a lot shorter... $_="sd,ds = sd,ds = {dsxsd}s= {dsx&&xsd}s= d..d
"x4;s!s!("()][<<   [)[[<=<  (]]]<<=  [][]<=<= "=~s/[<= ]+/$&$&/gr=~/ +|<=|./g)[$-++]!ge;s/d/$F[$x%2]+{8,1,9,-1,19,-1,28,1}->{$x++}/ge


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Explanation

Originally this was using a printf format string, but just having s and d was shorter when combined with s///.

First the format string is built into $_ and quadruplicated, then all ss are replaced with the corresponding bracket, <, <= or spaces, depending on the replacement index. I hoped to save some more bytes with the duplication of the last 5 chars of each block, but this only ended up saving 2 bytes. The resultant string is split up into elements of spaces, <= or single characters. Finally all ds are replaced with the desired number which is adjusted based on the index of the current replacement via a hash key. JavaScript, 190 189 bytes x=>y=>(0)7]0[7{1<x<2} 7{1<52} 73..4 [0)7[0[7{18x<26{185261..4 (0]7]0]7{1<x826{1<5=263..2 [0]7[0]7{18x82}7{185=2}71..2.replace(/\d/g,d=>[[x,y],x,y,x+1,y-1,x&&x<,} = , = ,<=][d])  Try it online Python 2, 277199193 189 bytes a,b=input() for i in 4,3,2,1:x,y=i%2,i>2;e='=';p=a+','+b;print'(['[x]+p+'])'[y],e,']['[x]+p+']['[y],e,2*('{%d<%s<%s%d} %s= '%(a,e*x+'%sx',e[y:],b,i/2*' '))%('','x&&')+a+1-x+'..'+b-y  Try it online! • 2*('{%d<%s<%s%d} %s= '%(a,e*x+'%sx',e[y:],b,i/2*' '))%('','x&&') -> 2*('{%d<%s%%sx<%s%d} %s= '%(a,e*x,e[y:],b,i/2*' '))%('','x&&') saves two :) – Jonathan Allan Apr 4 '18 at 19:47 Excel, 399 bytes ="("&A1&","&B1&") = ]"&A1&","&B1&"[ = {"&A1&"<x<"&B1&"} = {"&A1&"<x&&x<"&B1&"} = "&A1+1&".."&B1-1&" ["&A1&","&B1&") = ["&A1&","&B1&"[ = {"&A1&"<=x<"&B1&"} = {"&A1&"<=x&&x<"&B1&"} = "&A1&".."&B1-1&" ("&A1&","&B1&"] = ]"&A1&","&B1&"] = {"&A1&"<x<="&B1&"} = {"&A1&"<x&&x<="&B1&"} = "&A1+1&".."&B1&" ["&A1&","&B1&"] = ["&A1&","&B1&"] = {"&A1&"<=x<="&B1&"} = {"&A1&"<=x&&x<="&B1&"} = "&A1&".."&B1  Nothing particularly interesting here. C (gcc), 224 237 bytes f(a,b,c,m,n,o){for(c=0;++c<5;printf("%c%d,%d%c = %c%d,%d%c = {%d<%sx<%s%d}%*s= {%d<%sx&&x<%s%d}%*s= %d..%d\n","[("[m],a,b,"])"[n],"[]"[m],a,b,"]["[n],a,"="+m,"="+n,b,o,"",a,"="+m,"="+n,b,o,"",a+m,b-n)){m=c%2;n=c<3;o=3-c/2;}}  Try it online! Moving the "<[=]" into the format string allowed me to remove the array altogether. Also, moving printf() into the for loop saved a semicolon. Original answer f(a,b,c,m,n,o){char*e[]={"<=","<"};for(c=0;++c<5;){m=c%2;n=c<3;o=3-c/2;printf("%c%d,%d%c = %c%d,%d%c = {%d%sx%s%d}%*s= {%d%sx&&x%s%d}%*s= %d..%d\n","[("[m],a,b,"])"[n],"[]"[m],a,b,"]["[n],a,e[m],e[n],b,o,"",a,e[m],e[n],b,o,"",a+m,b-n);}}  Try it online! There's nothing particularly noteworthy here: I used the usual tricks for shaving the size of the function (hoisting int autos into the function header, using K&R style, indexing into string constants.) Storing the {"<=","<"} into an array proved more size-efficient than inserting a format specifier for this challenge as it was used more than once. Javascript, 273258 232 bytes Thanks Kevin Cruijssen for saving me 15 bytes ! e=>f=>'(j)h]j[h{a<x<b} h{a<x&&x<b} hc..d\n[j)h[j[h{a<=x<b} h{a<=x&&x<b} ha..d\n(j]h]j]h{a<x<=b} h{a<x&&x<=b} hc..b\n[j]h[j]h{a<=x<=b}h{a<=x&&x<=b}ha..b'.replace(/a|b|c|d|h|j/g,m=>{return{a:e,b:f,c:e+1,d:f-1,h:" = ",j:e+","+f}[m]})  Try it online Thanks for TFeld for giving me that idea, saving around 60 bytes from my original answer. Ungolfed : e => f => '(j)h]j[h{a<x<b} h{a<x&&x<b} hc..d' + '\n[j)h[j[h{a<=x<b} h{a<=x&&x<b} ha..d' + '\n(j]h]j]h{a<x<=b} h{a<x&&x<=b} hc..b' + '\n[j]h[j]h{a<=x<=b}h{a<=x&&x<=b}ha..b' .replace(/a|b|c|d|h|j/g, m=>{ return {a:e,b:f,c:e+1,d:f-1,h:" = ",j:e+","+f}[m] } )  Javascript (original answer), 340 bytes (a,b)=>alert((${a},${b}) = ]${a},${b}[ = {${a}<x<${b}} = {${a}<x&&x<${b}} =${a+1}..${b-1}\n[${a},${b}) = [${a},${b}[ = {${a}<=x<${b}} = {${a}<=x&&x<${b}} =${a}..${b-1}\n(${a},${b}] = ]${a},${b}] = {${a}<x<=${b}} = {${a}<x&&x<=${b}} =${a+1}..${b}\n[${a},${b}] = [${a},${b}] = {${a}<=x<=${b}} = {${a}<=x&&x<=${b}} =${a}..${b}\n)  • The new-line at {a<=x&&\nx<=b} shouldn't be there, and you're missing the first (a,b) = . As for some golfing: instead of alert you can just return the result. (m)=> can be m=>. (e,f)=> can be e=>f=>. And you can use g directly instead of creating the let g=: m=>{return{a:e,b:f,c:e+1,d:f-1}[m]}. Try it online 258 bytes. And probably more can be golfed, I'm not too skilled in JS.. – Kevin Cruijssen Apr 4 '18 at 9:43 • Well you look more skilled than I am. Thanks for the tips, i'll add them to the answer – The random guy Apr 4 '18 at 9:46 • Well, I'm mostly someone who golfs in Java (or Whitespace), and these golfing tips also apply to Java answers, which is mostly why I know them: (m)-> to m->; (e,f)-> to e->f->, and var g=new int[]{...}return g[m] to return new int[]{...}[m]. ;) Also, I've been active on this SO for little over two years now, so I've seen JS answers use these kind of things before as well. – Kevin Cruijssen Apr 4 '18 at 9:49 Retina, 216 bytes .+$&()][<<  ¶$&[)[[<=< ¶$&(]]]<<= ¶$&[][]<=<= ((.*),(.*))(.)(.)(.)(.)(<=?)(<=?)( *)$4$1$5 = $6$1$7 = {$2$8x$9$3}$10 = {$2$8x&&x$9$3}$10 =$2_$8..$3_$9 _<= \d+_$&*___
T<_ _+<|\.-_+<
___<
_
__<
-1
_(_*)
$.1 -0 0  Try it online! Explanation: .+$&()][<<  ¶$&[)[[<=< ¶$&(]]]<<= ¶$&[][]<=<= ((.*),(.*))(.)(.)(.)(.)(<=?)(<=?)( *)$4$1$5 = $6$1$7 = {$2$8x$9$3}$10 = {$2$8x&&x$9$3}$10 =$2_$8..$3_$9  Build up the main bulk of the result. _<=  If the variable is involved in a loose inequality, then the value is inclusive, so we can delete the placeholder. \d+_$&*___


Convert the value to unary and add 2.

T<_ _+<|\.-_+<


Remove the placeholder for a strict lower inequality or a negative strict upper inequality. It's still had 2 added, but 1 will be subtracted later, giving the desired result.

___<
_


Subtract 2 from the other non-zero strict inequalities, restoring the original value, from which 1 will be subtracted later.

__<
-1


Change a strict upper inequality of 0 to -1.

_(_*)
\$.1


Subtract 1 from the remaining strict inequalities and convert to decimal.

-0
0


Fix up another edge case.

Python 3, 180 bytes:

lambda a,b:[eval('f"'+"{%r[j]}{a},{b}{%r[i]} = "*2%('([',')]','][','[]')+"{{{a}<{'='[:j]}x%s<{'='[:i]}{b}}}{'  '[i+j:]} = "*2%('','&&x')+'{a+j}..{b-i}"')for i in(1,0)for j in(1,0)]


explanation

Basically builds an f-string that is evaluated in a list comprehension. Old-style % string interpolation is used to delay evaluating the expressions until the f-string is evaluated.

lambda a,b:[
eval(
'f"' +                                                            # f-string prefix
"{%r[j]}{a},{b}{%r[i]} = "*2%('([',')]','][','[]') +              # first two terms
"{{{a}<{'='[:j]}x%s<{'='[:i]}{b}}}{'  '[i+j:]} = "*2%('','&&x') + # second two terms
'{a+j}..{b-i}"'                                                   # last term
)
for i in(1,0)for j in(1,0)
]


The first part of the string, 'f"', will become the prefix for the f-string.

The second part of the string builds the format string for the first two interval expressions. %r is used to save from needing to put quotes in the format, i.e., "{%r[j]}" is the same as "{'%s'[j]}". When the f-string is evaluated, the correct bracket is selected.

The third part of the string builds the next two interval expressions.

The last part formats the "a..b" part of the f-string.

The assembled f-string looks like: f"{'(['[j]}{a},{b}{')]'[i]} = .... = {a+j}..{b-i}"

When the f-string is evaluated, all the expressions in braces {} are are replaced by their value. So, {a} gets replaced by the value of a, and {'(['[j]} gets replaced by ( if j is 0 or [ if j is 1.

SOGL V0.12, 110 bytes

<ŗŗ}”⁴
"{ŗ<ŗx³
W}↔b ,e++Κ+²
4∫2\f»¹Aa{Ƨ[(²a{Ƨ[]²ba{ =*}eο+++:³⁴;³&&x⁴a_beh+H⁶;!+ƨ.+Κο++++}⁰№Iā;0E{┼ē4=‽"δY↑æ‘┼


Try it Here!

Python 3, 248 bytes

def f(a,b):
l=[['(',')','[','<',1],['[',']',']','<=',0]]
r=[0,1]
for i in r:
for j in r:
print(('%s%d,%d%s='*2+'{%d%sx%s%d}={%d%sx&&x%s%d}=%d..%d')%((l[j][0],a,b,l[i][1],l[1-j][2],a,b,l[i][2])+(a,l[j][3],l[i][3],b)*2+(a+l[j][4],b-l[i][4])))


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