Am I an engineer?

Question

Recently, an electrical engineer studying traffic light timings was fined $500 by the state of Oregon for referring to himself as an engineer.

Given a 2 letter string as input, representing a US state, output:

I am not an engineer if the state is Oregon (OR)
I am an engineer if the state is any other US state
What is an engineer? given any other input

Output may not contain any leading whitespace, but may contain as much trailing whitespace as you wish.

You can assume the input will always be 2 uppercase letters.

Here is a list of all 50 US state abbreviations:

AL, AK, AZ, AR, CA, CO, CT, DE, FL, GA, HI, ID, IL, IN, IA, KS, KY, LA, ME,
MD, MA, MI, MN, MS, MO, MT, NE, NV, NH, NJ, NM, NY, NC, ND, OH, OK, OR, PA,
RI, SC, SD, TN, TX, UT, VT, VA, WA, WV, WI, WY

Scoring

This is code-golf, so fewest bytes in each language wins!

Side note: If you concatenate the whole string together, you get "...OKORPARISCSDTN..." ;-) — Mateen Ulhaq, Commented Jun 4, 2017 at 6:42
Mats Järlström wouldn't be allowed to present himself as an engineer in Texas, either. — Ben Voigt, Commented Jun 5, 2017 at 17:22

Mayube · Accepted Answer · 2017-06-02 12:11:14Z

17

C#, 311 309 240 237 222 195 184 183 bytes

s=>s=="OR"?"I am not an engineer":"MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY".Contains(s)?"I am an engineer":"What is an engineer?";

Try it online!

Saved 2 bytes by adding the space before an to b

~~-69 (huehue)~~ -72 bytes thanks to TheLethalCoder

-15 bytes thanks to TotallyHuman's genius states string

-38 bytes cos more string compression

Ungolfed:

public static string a(string s)
{
    var b = " an engineer";
    if (s == "OR")
    {
        return "I am not" + b;
    }
    else
    {
        if ("MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY".Contains(s))
        {
            return "I am" + b;
        }
        else
        {
            return $"What is{b}?";
        }
    }
}

edited Jun 2, 2017 at 12:11

answered Jun 2, 2017 at 10:00

Mayube

11.7k2 gold badges48 silver badges92 bronze badges

\$\begingroup\$ Use ..."+b instead of interpolated strings, change if..else if...else to return s=="OR"?....:System.Arr...` i.e. use a ternary. Use u.Contains instead of Array.Exists. I think if you use spaces instead of commas then .Split() with no params will work. \$\endgroup\$
– TheLethalCoder
Commented Jun 2, 2017 at 10:28
\$\begingroup\$ declare b and u in the same line and even though you have to use explicit type string should still save you a byte \$\endgroup\$
– LiefdeWen
Commented Jun 2, 2017 at 10:31
\$\begingroup\$ @StefanDelport u is an array not string. However, you can save bytes by setting u as a string and using .Contains on it. \$\endgroup\$
– TheLethalCoder
Commented Jun 2, 2017 at 10:35
\$\begingroup\$ 240 bytes: s=>{var b=" an engineer";return s=="OR"?"I am not"+b:"AL AK AZ AR CA CO CT DE FL GA HI ID IL IN IA KS KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC ND OH OK PA RI SC SD TN TX UT VT VA WA WV WI WY".Contains(s)?"I am"+b:$"What is{b}?";};. Think I've removed all the relevant whitespace. (Untested) \$\endgroup\$
– TheLethalCoder
Commented Jun 2, 2017 at 10:38
1

\$\begingroup\$ You can golf this by defining " an engineer" as a separate variable to save some bytes. \$\endgroup\$
– AAM111
Commented Jun 2, 2017 at 23:29

| Show 5 more comments

Arnauld · Accepted Answer · 2017-06-02 13:13:54Z

11

JavaScript (ES6), 182 bytes

s=>['I am'+(x=' an engineer'),`What is${x}?`,'I am not'+x][s=='OR'?2:'MNNMLATNAKALARAZCACOCTDEFLGAHIIAIDILINKSKYMAMDMEMIMOMSMTNCNDNENHNJNVNYOHOKPARISCSDTXUTVAVTWAWIWVWY'.search(s)&1]

Demo

let f =

s=>['I am'+(x=' an engineer'),`What is${x}?`,'I am not'+x][s=='OR'?2:'MNNMLATNAKALARAZCACOCTDEFLGAHIIAIDILINKSKYMAMDMEMIMOMSMTNCNDNENHNJNVNYOHOKPARISCSDTXUTVAVTWAWIWVWY'.search(s)&1]

console.log(f("OR"))
console.log(f("NY"))
console.log(f("TX"))
console.log(f("YO"))

edited Jun 2, 2017 at 13:13

answered Jun 2, 2017 at 11:34

Arnauld

195k20 gold badges179 silver badges645 bronze badges

3

\$\begingroup\$ Fails for e.g. LA :/ \$\endgroup\$
– Christoph
Commented Jun 2, 2017 at 13:07
\$\begingroup\$ @Christoph Should be OK now. Thanks for reporting this! \$\endgroup\$
– Arnauld
Commented Jun 2, 2017 at 13:14
\$\begingroup\$ Thought it was missing OR for moment - silly! \$\endgroup\$
– Chas Brown
Commented Jun 3, 2017 at 2:33
\$\begingroup\$ Use codegolf.stackexchange.com/a/124164/76323 to optimize \$\endgroup\$
– l4m2
Commented Mar 9, 2018 at 6:19

Add a comment |

MD XF · Accepted Answer · 2018-03-11 18:23:05Z

8

C, 215 208 190 bytes

^{-7 thanks to Cool Guy}

#define z" an engineer"
#define f(s)!strcmp(s,"OR")?"I am not"z:strstr("MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY",s)?"I am"z:"What is"z"?"

Used @totallyhuman's "genus string".

How it works:

"string"z automatically concatenates "string" with z (" an engineer"). Yes, C does that.
!strcmp(s,"OR") compares the string against "OR".
?"I am not"z returns "I am not an engineer" if true. Otherwise...
:strstr(...,s) checks if @totallyhuman's genius string contains the provided string.
?"I am"z returns "I am an engineer" if so, and...
:"What is"z"?") returns "What is an engineer?" otherwise.

Try it online!

edited Mar 11, 2018 at 18:23

answered Jun 2, 2017 at 18:25

MD XF

13.9k4 gold badges65 silver badges104 bronze badges

1

\$\begingroup\$ 206 bytes: #define z " an engineer" f(char*s){!strcmp(s,"OR")?puts("I am not"z):strstr("MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY",s)?puts("I am"z):puts("What is"z);} \$\endgroup\$
– Spikatrix
Commented Jun 3, 2017 at 9:31
\$\begingroup\$ @CoolGuy That won't work for "What is an engineer?" But I got it to work in 209 bytes. Thanks! \$\endgroup\$
– MD XF
Commented Jun 3, 2017 at 16:49
\$\begingroup\$ You can return the string instead of printing it \$\endgroup\$
– l4m2
Commented Mar 9, 2018 at 16:03
\$\begingroup\$ @l4m2 Did something similar. \$\endgroup\$
– MD XF
Commented Mar 11, 2018 at 18:23

Add a comment |

Felipe Nardi Batista · Accepted Answer · 2017-06-02 12:59:06Z

6

Python 2, 228 168 bytes

lambda s:('What is%s?','I am'+' not'*(s=='OR')+'%s')[s in'MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH ORIA UT WVT WIL WY']%' an engineer'

Try it online!

edited Jun 2, 2017 at 12:59

answered Jun 2, 2017 at 11:15

Felipe Nardi Batista

2,4211 gold badge14 silver badges32 bronze badges

2

\$\begingroup\$ The state encoding is hilarious. \$\endgroup\$
– Ramon Snir
Commented Jun 2, 2017 at 16:50

Add a comment |

totallyhuman · Accepted Answer · 2017-06-02 11:44:24Z

5

Python 2, 192 186 182 178 176 bytes

Could probably compress the state string more.

lambda s,e=' an engineer':'I am'+' not'*(s=='OR')+e if s in'MINCALAZ SCT FL GA WIAKSD ME MD MA MNMS MOKY MTNE NVTX NH NJ NY ND COHIDE OR PARIL UT VA WA WV WY'else'What is%s?'%e

Try it online!

edited Jun 2, 2017 at 11:44

answered Jun 2, 2017 at 11:18

totallyhuman

17.3k3 gold badges32 silver badges88 bronze badges

2

\$\begingroup\$ You only use u in if s in u can you note use it directly instead of declaring it? \$\endgroup\$
– TheLethalCoder
Commented Jun 2, 2017 at 11:20
1

\$\begingroup\$ @totallyhuman that string compression is pretty amazing, nice one! \$\endgroup\$
– Mayube
Commented Jun 2, 2017 at 11:23
2

\$\begingroup\$ @totallyhuman VALA WAZ NCA COH CTX SDE FL GA HID WIL MIN IAKSC KY ME MD MA MNMS MOK MTNE NH NJ NY ND PARI UT NVT WV WY is the smallest I can make it \$\endgroup\$
– Mayube
Commented Jun 2, 2017 at 11:27
2

\$\begingroup\$ 'MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH ORIA UT WVT WIL WY' \$\endgroup\$
– Felipe Nardi Batista
Commented Jun 2, 2017 at 11:29
2

\$\begingroup\$ Whoops, sorry it took me so long. Seems I have lost as copying that string doesn't make my answer better than the one using it. Please stop upvoting this and upvote the other better answer. \$\endgroup\$
– totallyhuman
Commented Jun 2, 2017 at 12:49

| Show 3 more comments

Olivier Grégoire · Accepted Answer · 2018-09-17 14:05:27Z

5

Java (JDK 10), 184 bytes

s->s.format(s.equals("OR")?"I am not%s":"MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY".contains(s)?"I am%s":"What is%s?"," an engineer")

Try it online!

I apologize for reusing the compressed string: I couldn't find anything better by myself... :(

edited Sep 17, 2018 at 14:05

answered Jun 2, 2017 at 13:18

Olivier Grégoire

14.4k3 gold badges32 silver badges56 bronze badges

Add a comment |

Emigna · Accepted Answer · 2017-06-02 12:23:50Z

4

05AB1E, 104 103 101 bytes

„€À€ˆ„I€Ü‚.•~Zµ∞/'—¶[@øl•Œ2ù.•9₆\|&×ƒΩ#àöF},cΓ páe;ð²∍₆jÌn#dÅ?DvĆ8A•2ôìuIå©è¹„ORQi'€–}„€¤©É)ðýª'?®_×J

Try it online!

Old 104 byte version in case that is more easily improved.

„€À€ˆ„I€Ü‚.•ÂkXñ…ΓVt€Ïè∍‡Λi„2¶’að=–½6™oÑþÁāõgO·ð~
λ†₃›;â&ÄFv¾‡1~ǝQa«;cS•u2ôIå©è¹„ORQi'€–}„€¤©É)ðýª'?®_×J

Not happy with the compression nor the special case for ?.

edited Jun 2, 2017 at 12:23

answered Jun 2, 2017 at 11:59

Emigna

53k5 gold badges42 silver badges162 bronze badges

\$\begingroup\$ OK what is this? Did you just re-order the states or something? \$\endgroup\$
– Erik the Outgolfer
Commented Jun 2, 2017 at 12:05
\$\begingroup\$ @EriktheOutgolfer: The 104 byte version is just a compression of the numbers corresponding to the characters positions in the alphabet (A=1,C=3 ...). The 103 byte version does the same for some states and merge some states that end in the same letter that a new state starts with. I'm sure this can still be improved and I'll add a better explanation when I've golfed it more. \$\endgroup\$
– Emigna
Commented Jun 2, 2017 at 12:12
1

\$\begingroup\$ Nice answer! I like the .•~Zµ∞/'—¶[@øl•Œ2ù you've used for some of the states, instead of just compressing all the states and doing the 2ô. And nice approach overall. (Glad I saw your answer before attempting something myself, because it definitely wouldn't have been as short..) \$\endgroup\$
– Kevin Cruijssen
Commented Sep 17, 2018 at 14:35

Add a comment |

JayCe · Accepted Answer · 2018-09-18 15:34:41Z

4

R, 109 96 bytes

function(x)sub("#"," an engineer",c("I am not#","I am#","What is#?"))[2-x%in%state.abb+!x=="OR"]

Try it online!

13 bytes thanks to J.Doe - through use of regex and indexation.

edited Sep 18, 2018 at 15:34

answered Sep 17, 2018 at 14:22

JayCe

3,01312 silver badges17 bronze badges

1

\$\begingroup\$ 97 bytes \$\endgroup\$
– J.Doe
Commented Sep 17, 2018 at 15:01
1

\$\begingroup\$ @J.Doe sub is much better, thank you! \$\endgroup\$
– JayCe
Commented Sep 17, 2018 at 16:54
\$\begingroup\$ 96 bytes 😉 \$\endgroup\$
– J.Doe
Commented Sep 18, 2018 at 13:47
\$\begingroup\$ @J.Doe Of course! \$\endgroup\$
– JayCe
Commented Sep 18, 2018 at 15:34

Add a comment |

Vern DeHaven · Accepted Answer · 2020-01-24 23:58:26Z

F#, 222 211 bytes

let f v=let(a,b)=if"OR"=v then("I am not",".")elif"WALAKSCARINMNVTNCTX NHIDE MOHIL COKY MSD PAZ WIA WVA FL GA MA MD ME MI MT NE ND NJ NY UT WY".Contains v then("I am",".")else("What is","?")in a+" an engineer"+b

Try it online!

Expanded:

let f v =
    let (a,b) =
        if "OR" = v then ("I am not",".")
        elif "WALAKSCARINMNVTNCTX NHIDE MOHIL COKY MSD PAZ WIA WVA FL GA MA MD ME MI MT NE ND NJ NY UT WY".Contains v then ("I am",".")
        else ("What is","?")
    a + " an engineer" + b

Given a two-letter state v passed to the function f, build a tuple (a, b) representing the head and tail of the "engineer" sentence.

Feel free to use the "compressed state string" freely; it's one whole byte shorter than the MINCALA one...

Community · Accepted Answer · 2020-06-17 09:04:33Z

CJam, 143 bytes

"ORIA MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY"q#g"I am not 
I am 
What is?"N/=)" an engineer"\

Try it online! or as a test suite

Explanation

"ORIA...."       e# Push a string in which every state is a substring, but no non-state is
                 e# a substring.
q                e# Read the input.
#                e# Find the index of the input in the string. (-1 if not found)
g                e# Signum of the index: -1 for negative, 0 for 0, 1 for positive.
"I am.... "      e# Push this string. Note the trailing space on the first two lines of it.
N/               e# Split it on newlines.
=                e# Get the string at index given by the signum.
)                e# Pull out the last character.
" an engineer"\  e# Push " an engineer" and bring the other character to the TOS.
                 e# Implicit output.

Since Oregon (OR) is at the start of the string, finding the signum of the index of the input in that will be -1 if not found, 0 if OR, 1 if any other state. Which string to print can be decided by that.

Community · Accepted Answer · 2020-06-17 09:04:33Z

Japt, 136 135 131 129 128 bytes

More savings may be available by experimenting with the order of the state abbreviations - I'll come back to that in a while.

`mnnmlãGLÏz¯¬ct¸flgaá[9¨kyµmçpCijmsmtnhnvnyn¬kpÂÉcsdk¡x©vavt°±wvwy`ò øUv)?"I am {¥"OR"?"not ":P}"+` à¨\ `:`Wt   à¨\?

Try it online

Explanation

We take a compressed string of the lowercased abbreviations, decompress it, and split it into an array of 2 character strings using the ò method.
We then use the ø method to see if the array contains Uv, which is the input string converted to lowercase.
If we so, we build our output string, beginning with "I am "
Checking if the input string ¥ (is equal to) "OR" allows us to append either "not " or the empty string variable P.
And then we decompress and append the compressed string "an engineer".
If the input was not found in the array, then we output the decompression of compressed string "What is an engineer?".

decorator-factory · Accepted Answer · 2018-09-17 18:56:59Z

3

Python 3, 180 179 178 bytes

def f(s):e=" not"*(s=="OR")+" an engineer";return"I am"+e if s in"PALAKSCAZ CTNMINCOR FL GA MDE ME MND MA MSD MOKY NE NH NJ NY WA OHID UTX MTNVARIA WIL WVT WY"else"What is"+e+"?"

Try it online!

edited Sep 17, 2018 at 18:56

answered Sep 17, 2018 at 15:14

decorator-factory

1316 bronze badges

\$\begingroup\$ Yep, it's correct: Try it online! \$\endgroup\$
– decorator-factory
Commented Sep 17, 2018 at 15:32
\$\begingroup\$ 166 bytes. I'm sure the string can also be tightened up a liitle more by reusing states \$\endgroup\$
– Jo King ♦
Commented Sep 17, 2018 at 22:53
\$\begingroup\$ Thanks @JoKing. It resembles the Python 2 answer too much at this point IMO :). I've been trying to compress the string, but the shortest I could get was 92 symbols. I've written a short script to do that, but it can only cut it down to 94 characters, and it doesn't compress optimally. I also think that in this particular challenge the magic string itself should not be included in the byte count. \$\endgroup\$
– decorator-factory
Commented Sep 18, 2018 at 18:50

Add a comment |

Jörg Hülsermann · Accepted Answer · 2017-06-02 14:39:21Z

2

PHP, 188 Bytes

$e=" an engineer";echo strpos(_TNNMLAALAKAZARCACOCTDEFLGAHIIDILINIAKSKYMEMDMAMIMNMSMOMTNENVNHNJNYNCNDOHOKORPARISCSDTXUTVTVAWAWVWIWY,$argn)&1?"I am".($argn!="OR"?"":" not").$e:"What is$e?";

Try it online!

edited Jun 2, 2017 at 14:39

answered Jun 2, 2017 at 12:09

Jörg Hülsermann

13.3k3 gold badges15 silver badges36 bronze badges

\$\begingroup\$ in_array(...) to strpos(_AKALARAZCACOCTDEFLGAHIIAIDILINKSKYLAMAMDMEMIMNMOMSMTNCNDNENHNJNMNVNYOHOKPARISCSDTNTXUTVAVTWAWIWVWY,$argn)&1. \$\endgroup\$
– Christoph
Commented Jun 2, 2017 at 12:54
\$\begingroup\$ @Christoph works not for an Input for example of LA \$\endgroup\$
– Jörg Hülsermann
Commented Jun 2, 2017 at 13:03
\$\begingroup\$ Hm right :/ maybe we can reorder the string a bit .. I'll try. Also invalidates the js answer btw. \$\endgroup\$
– Christoph
Commented Jun 2, 2017 at 13:08
\$\begingroup\$ _NMMNINTNRIHIMIWISCNCCTMTUTVTWVNVFLILCAALGAIAMAPAVAWACOLAMOAKARKSMSNHOHOKORAZDEIDKYMEMDNENJNYNDSDTXWY works \$\endgroup\$
– Christoph
Commented Jun 2, 2017 at 13:24
\$\begingroup\$ @Christoph Or my order work too. \$\endgroup\$
– Jörg Hülsermann
Commented Jun 2, 2017 at 14:40

Add a comment |

Arthur Rump · Accepted Answer · 2017-06-02 20:27:23Z

2

C#, 178 bytes

s=>(s=="OR"?"I am notx":"MINCALA MSCTNMNVAKY WAZ PARIA FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY".Contains(s)?"I amx":"What isx?").Replace("x"," an engineer");

Run in C# Pad

Golf based on Mayube's solution; new here, so I don't have enough rep to comment.

edited Jun 2, 2017 at 20:27

answered Jun 2, 2017 at 20:23

Arthur Rump

1815 bronze badges

\$\begingroup\$ Might want to edit in a link in to his solution :) If you click share on his solution, it gives you a link straight to it. \$\endgroup\$
– Stephen
Commented Jun 2, 2017 at 20:26
\$\begingroup\$ @StephenS Thanks for the tip! \$\endgroup\$
– Arthur Rump
Commented Jun 2, 2017 at 20:27

Add a comment |

bartavelle · Accepted Answer · 2017-07-03 12:24:56Z

2

Haskell, 220 214 210 209 bytes

s(a:b:c)=[a,b]:s(b:c)
s _=[]
a="I am "
e="an engineer "
i"OR"=a++"not "++e
i x|x`elem`s"MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY"=a++e
i _="What is "++e++"?"

Try it online!

edited Jul 3, 2017 at 12:24

answered Jun 2, 2017 at 11:40

bartavelle

1,3118 silver badges9 bronze badges

\$\begingroup\$ otherwise = True, 1<2 is even shorter \$\endgroup\$
– BlackCap
Commented Jun 2, 2017 at 14:45
\$\begingroup\$ Facepalm, thanks! \$\endgroup\$
– bartavelle
Commented Jun 2, 2017 at 14:51
\$\begingroup\$ You have included OR in the long string in the second definition of i \$\endgroup\$
– BlackCap
Commented Jun 2, 2017 at 15:02
\$\begingroup\$ I stole the long string from the other answers. OR isn't part of it though? \$\endgroup\$
– bartavelle
Commented Jun 2, 2017 at 16:26
1

\$\begingroup\$ No, not anymore :) \$\endgroup\$
– BlackCap
Commented Jun 2, 2017 at 16:28

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Javascript 204

s=>{h=/^(A[LKZR]|C[AOT]|DE|FL|[GPLV]A|[HR]I|I[DLNA]|K[SY]|M[EDAINSOT]|N[EVHJMYCD]|O[HK]|S[CD]|T[NX]|[VU]T|W[AVIY]|(OR))$/.exec(s);return(!h?"What is ":"I am "+(h[2]?"not ":""))+"an engineer"+(!h?'?':'')}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 2, 2017 at 16:51

martin

1911 gold badge1 silver badge6 bronze badges

1

\$\begingroup\$ Pure optimize s=>(!(h=/(A[LKZR]|C[AOT]|DE|FL|[GPLV]A|[HR]I|I[DLNA]|K[SY]|M[EDAINSOT]|N[EVHJMYCD]|O[HKR]|S[CD]|T[NX]|[VU]T|W[AVIY])/.test(s))?"What is":"I am"+(s=='OR'?" not":""))+" an engineer"+['?'[+h]] \$\endgroup\$
– l4m2
Commented Mar 9, 2018 at 6:15
\$\begingroup\$ @l4m2 nice, that's got my answer beat. \$\endgroup\$
– martin
Commented Mar 9, 2018 at 6:27

Add a comment |

Community · Accepted Answer · 2020-06-17 09:04:33Z

1

AWK, 189 bytes

/A[LKZR]|C[AOT]|DE|FL|[GPV]A|HI|I[DLNA]|KS|KY|LA|M[EDAINSOT]|N[EVHJMYCD]|O[HKR]|RI|SC|SD|TN|TX|UT|VT|W[AVIY]/{print"I am "($0~OR?"not ":"")"an engineer";exit}
{print"What is an engineer?"}

If input matches a regex containing all the state abbreviations, print "I am an engineer" with a 'not' inserted in the middle if the state is Oregon, then exit.

If the input does not match the regex, it must not be a US state abbreviation.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered Jun 3, 2017 at 8:53

Arc676

3011 silver badge7 bronze badges

Add a comment |

Mateen Ulhaq · Accepted Answer · 2017-06-04 09:41:58Z

1

Python 3, 238 bytes

def f(x):s=x in('ALAKAZARCACOCTDEFLGAHIIDILINIAKSKYLAMEMDMAMIMNMSMOMTNENVNHNJNMNYNCNDOHOKORPARISCSDTNTXUTVTVAWAWVWIWY'[i:i+2]for i in range(0,100,2));o=x=='OR';q=(1-o)*(1-s);return q*'What is'+(1-q)*('I am'+o*' not')+' an engineer'+q*'?'

Explanation

No compression techniques used.

def f(x):
    # Check if State
    s = x in ('ALAK...WIWY'[i:i+2]
              for i in range(0, 100, 2))

    # Check if Oregon
    o = x == 'OR'

    # Check if neither Oregon nor State
    q = (1-o) * (1-s)

    # Construct output string
    return q * 'What is' + \
        (1-q) * ('I am' + o * ' not') + \
        ' an engineer' + \
        q * '?'

edited Jun 4, 2017 at 9:41

answered Jun 4, 2017 at 6:32

Mateen Ulhaq

1,97319 silver badges33 bronze badges

\$\begingroup\$ It's Oregon, not Ohio. \$\endgroup\$
– L3viathan
Commented Jun 4, 2017 at 9:24
1

\$\begingroup\$ @L3viathan Not sure how I mess that up after seeing Oregon everywhere due to Life is Strange... ¯\_(ツ)_/¯ \$\endgroup\$
– Mateen Ulhaq
Commented Jun 4, 2017 at 9:43

Add a comment |

JWH · Accepted Answer · 2017-06-05 16:57:30Z

1

Java, 173 bytes

s->(!"MINCALARIA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY OR".contains(s)?"What is":"I am"+(s.equals("OR")?" not":""))+" an engineer"

edited Jun 5, 2017 at 16:57

answered Jun 2, 2017 at 22:23

JWH

112 bronze badges

3

\$\begingroup\$ Welcome to PPCG! All entries must be a full program or a function; I believe you can turn this into a valid lambda expression (and therefore function) by adding s-> to the beginning. \$\endgroup\$
– ETHproductions
Commented Jun 2, 2017 at 23:01

Add a comment |

recursive · Accepted Answer · 2018-03-08 23:36:44Z

1

Stax, 100 bytes

This language postdates the challenge. But the author (me) didn't see it until now.

éë&W≈#W¬π█▐╜╣╟◙√a☻∞ZrπU♫ÿô♠▌⌠Që≡AûpI⌡ÄNAç¶µ↑╝╣òøΩ.¬É]╩Æ↓d∩é¡2Å²eB┼¼▬5∟┤sW♠♂↑q▐WMï╝|Ñ↑╫+3¼↔îûvlLΩ∟┬oë

Run and debug it

answered Mar 8, 2018 at 23:36

recursive

10.4k20 silver badges36 bronze badges

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l4m2 · Accepted Answer · 2018-03-09 16:13:28Z

1

JavaScript ES6, 175171 Bytes

x=>[`What is${e=" an engineer"}?`,`I am${x=="OR"?" not"+e:e}`][+!!'MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH ORIA UT WVT WIL WY'.match(x)]

Joined lots of good

Or 152 bytes on ISO encoding

Generator:

'x=>[`What is${e=" an engineer"}?`,`I am${x=="OR"?" not"+e:e}`][+!!btoa`*`.match(x)]'.replace('*',atob('MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH ORIA UT WVT WIL WY '.replace(/ /g,'/')))

edited Mar 9, 2018 at 16:13

answered Mar 9, 2018 at 15:47

l4m2

26.2k2 gold badges23 silver badges98 bronze badges

\$\begingroup\$ It might be possible to rearrange the string so using btoa to expand \$\endgroup\$
– l4m2
Commented Mar 9, 2018 at 15:50

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l4m2 · Accepted Answer · 2018-03-09 22:03:09Z

1

C (gcc), 0+176 bytes

-Dz"=an engineer" -Df(s)*s-79|1[s]-82?strstr("MINCALA=MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY",s)?"I am"z:"What is"z"?":"I am not"z

Try it online!

pure translate

edited Mar 9, 2018 at 22:03

answered Mar 9, 2018 at 18:05

l4m2

26.2k2 gold badges23 silver badges98 bronze badges

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mazzy · Accepted Answer · 2018-09-17 17:26:24Z

Powershell, 175 bytes

(('I am'+' not'*!($i='ORIA MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY'.IndexOf($args))+($e=' an engineer')),"What is$e`?")[!++$i]

Test script:

$f = {

$e=' an engineer'
$i='ORIA MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY'.IndexOf($args)
(('I am'+' not'*!$i+$e),"What is$e`?")[!++$i]

# Important! OR is a first state in the modified @totallyhuman's genuis string

}

@(
    ,('OR', 'I am not an engineer')
    ,('AL', 'I am an engineer')
    ,('IL', 'I am an engineer')
    ,('ZZ', 'What is an engineer?')
) | % {
    $s,$e = $_
    $r = &$f $s
    "$($r-eq$e): $r"
}

Output:

True: I am not an engineer
True: I am an engineer
True: I am an engineer
True: What is an engineer?

Notts90 · Accepted Answer · 2017-06-02 12:17:25Z

0

Python 3, 236 182 181 bytes

lambda s:'I am not'+e if s=='OR'else'I am'+e if s in'MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH RIA UT WVT WIL WY'else'What is%s?'%e
e=' an engineer'

Try it online!

TIO contains test cases for all the states.
-54 bytes thanks to the string compression

edited Jun 2, 2017 at 12:17

answered Jun 2, 2017 at 11:31

Notts90

1,2512 gold badges9 silver badges20 bronze badges

\$\begingroup\$ You can use the string compression from the other answers to save bytes \$\endgroup\$
– TheLethalCoder
Commented Jun 2, 2017 at 11:31
1

\$\begingroup\$ MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH ORIA UT WVT WIL WY is better compression \$\endgroup\$
– Mayube
Commented Jun 2, 2017 at 11:38

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mkst · Accepted Answer · 2017-06-02 18:23:06Z

q/kdb+, 174 bytes

Solution:

{a:" an engineer?";$[(#:)l:ss["ORIA MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY";x];-1_"I am",$[l~(),0;" not";""],a;"What is",a]}

Explanation:

  {
  // save string into variable a
  a:" an engineer?";
  // try to find the input x in the condensed string, save location in variable l, $ is a if/else
  $[(#:)l:ss["ORIA MINCALA MSCTNMNVAKY WAZ PAR FL GA NHID COKSD ME MDE MA MTX NE NJ NY ND MOH UT WVT WIL WY";x];
  // found a match, check if it was at index 0 ('OR') and inject " not" if so, drop the '?' off the end
  -1_"I am",$[l~(),0;" not";""],a;
  // otherwise return 'What is an engineer?'
  "What is",a]
  }

Notes:

Used the 'compressed' string from other answers, struggling to find a way to bring it into a single line in order to avoid assignment of the a variable (yet still add the ? when input is not a state.

Neil · Accepted Answer · 2017-06-02 19:33:14Z

Retina, 175 bytes

..
I am $&~MINCALAZ SCT FL GA WIAKSD ME MD MA MNMS MOKY MTNE NVTX NH NJ NY ND COHIDE PARIL UT VA WA WV WY
(..)~.*\1.*
~
OR~.*
not ~
I am ..~.*
What is ~?
~
an engineer

Try it online! Hopefully I've appropriated the best state list. Explanation:

..
I am $&~MINCALAZ SCT FL GA WIAKSD ME MD MA MNMS MOKY MTNE NVTX NH NJ NY ND COHIDE PARIL UT VA WA WV WY

Start building up the result. Also, insert the state list for use by the next stage.

(..)~.*\1.*
~

If it's one of the 49 states, delete the state and list.

OR~.*
not ~

If it's Oregon, replace the state with not, and delete the list.

I am ..~.*
What is ~?

If it's anything else, replace everything with the other output.

~
an engineer

Add these word in last to avoid repetition.

Domii · Accepted Answer · 2017-06-03 09:53:08Z

0

Crystal, 232 207 205 Bytes

i="ORALAKAZARCACOCTDEFLGAHIIDILINIAKSKYLAMEMDMAMIMNMSMOMTNENVNHNJNMNYNCNDOHOKPARISCSDTNTXUTVTVAWAWVWIWY".split(/(..)/).index ARGV[0];p (i ?"I am ": "What is ")+(i==1 ?"not ": "")+"an engineer"+(i ?"": "?")

Try it online. (slightly modified due to an issue of ARGV)

edited Jun 3, 2017 at 9:53

answered Jun 3, 2017 at 0:32

Domii

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cat · Accepted Answer · 2017-06-03 12:24:20Z

Factor, 135 bytes

USE: usa-cities [ " an engineer"swap [ "I am"swap string>state OR = 
[ " not"append ] when ""] [ 2drop "What is" "?"] recover surround ]

Python 2, 213 211 194 Bytes

c="uTXnMSCORIDEwVAKYmTNHILfLAZpALmNEmOKSDwINCARmEwAnJnDmAmIAgAwYcTnVToHnYmD"
r=(lambda x:x in c or x[0].lower()+x[1]in c)(i)^1
u="What is "*r+("I am "+"not "*(i=="OR"))*(r^1)+"an engineer"+"?"*r

Try It Online

Stuff i'm working on shortening:

(i=="OR")
or x[0].lower()+x[1]in c

Update:

Saved 2 Bytes by replacing s=not r with s=r^1
Separated header and footer of code

edited Jun 4, 2017 at 0:51

answered Jun 3, 2017 at 22:20

Raffi

1476 bronze badges

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Community · Accepted Answer · 2020-06-17 09:04:33Z

Ruby, 164 bytes

->s{r="What is an engineer?"
j=379
"##(*Q0'7q;N>%*$o(.F%#&'#&#5##%$%+%5%)5r@#P,B353*/%".bytes{|i|s==(j+=i-34).to_s(36).upcase&&r="I am not"[0,i==41?8:4]+r[7,12]}
r}

Uses a run length encoding, therefore magic string is 50 bytes long, one per state. In order to build this, it was first necessary to put the state codes in alphabetical order of state code, rather than state name. Unfortunately 7 bytes are needed to convert the base36 representation of j from a lowercase statecode into an uppercase statecode.

Ungolfed in test program

f=->s{                                                  #s is the input string.
  r="What is an engineer?"                              #Set r to "What is an engineer?"
  j=379                                                 #Set j to one less than 380, which in base36 becomes AK, the first statecode alphabetically
  "##(*Q0'7q;N>%*$o(.F%#&'#&#5##%$%+%5%)5r@#P,B353*/%". #Iterate through the run length encoded string 
  bytes{|i|                                             #(each character represents how far in base 36 each state code is from the previous one)
    s==(j+=i-34).to_s(36).upcase&&                      #take the ascii value of the character and subtract 34 (example #=35-34=1) and add to j. Convert j to base36 to get a state code.
      r="I am not"[0,i==41?8:4]+r[7,12]                 #if the state code matches s, modify r. Take the first 4 characters of "I am not" (first 8 in the case of OR where i==41) 
  }                                                     #and add r[7,12]==" an engineer" (12 characters of the existing value of r, starting at character 7 
r}                                                      #return r

%w{AL AK AZ AR CA CO CT DE FL GA HI ID IL IN IA KS KY LA ME MD MA MI MN MS MO MT NE NV NH NJ NM NY NC ND OH OK OR PA RI SC SD TN TX UT VT VA WA WV WI WY XX}.map{|i|p [i,f[i]]}

Am I an engineer?

Scoring

38 Answers 38

C#, 311 309 240 237 222 195 184 183 bytes

JavaScript (ES6), 182 bytes

Demo

C, 215 208 190 bytes

Python 2, 228 168 bytes

Python 2, 192 186 182 178 176 bytes

Java (JDK 10), 184 bytes

05AB1E, 104 103 101 bytes

R, 109 96 bytes

F#, 222 211 bytes

CJam, 143 bytes

Explanation

Japt, 136 135 131 129 128 bytes

Explanation

Python 3, 180 179 178 bytes

PHP, 188 Bytes

C#, 178 bytes

Haskell, 220 214 210 209 bytes

Javascript 204

AWK, 189 bytes

Python 3, 238 bytes

Explanation

Java, 173 bytes

Stax, 100 bytes

JavaScript ES6, 175171 Bytes

Or 152 bytes on ISO encoding

C (gcc), 0+176 bytes

Powershell, 175 bytes

Python 3, 236 182 181 bytes

q/kdb+, 174 bytes

Retina, 175 bytes

Crystal, 232 207 205 Bytes

Factor, 135 bytes

Python 2, 213 211 194 Bytes

Ruby, 164 bytes

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