Acronyms can really obviously narrow your message sensors

Question

Because of this, you need a better way of working out if a phrase is an acronym of a word. You also think it would be worthwhile to see if the phrase and word in question are recursive acronyms.

Your task:

Given a word and then a phrase separated by a line, output if the phrase is an acronym and then if it is a recursive acronym. (The phrase contains what it stands for)

• The input will compose of alphabetical characters as well as spaces.
• Your program shouldn't be case sensitive.

Example Input/Output:

Case 1:

Input:

Acronyms
Acronyms can really obviously narrow your message sensors

Output:

True 
True

Case 2:

Input:

FAQ
frequently asked questions

Output:

True 
False

Case 3:

Input:

foo
bar baz

Output:

False
False

Case 4:

Input:

GNU
GNU is not Unix

Output:

False
False

Case 5:

Input:

Aha
A huge Aha

Output:

True
True

Answer 1

Pyth, 19 18

&pqJrz0hCKcrw0)}JK

This prints the result in a rather odd format, like: TrueFalse.

You can try it online or run the Test Suite.

Explanation:

&pqJrz0hCKcrw0)}JK      :
    rz0    rw0          : read two lines of input, and convert each to lower case
          c   )         : chop the second input on whitespace
   J     K              : store the first line in J and the chopped second line in K
  q    hC               : zip K and take the first element, check if it is the same as J
 p                      : print and return this value
&              }JK      : and the value with whether J is in K, implicit print

Answer 2

Python 3, 89

Saved a bunch of bytes thanks to SOPython.

a=input().lower()
d=input().lower().split()
h=tuple(a)==next(zip(*d))
print(h,h&(a in d))

The most complicated part of this solution is h=tuple(a)==next(zip(*d)).
This unpacks the list d into zip and then calls next to return a tuple of the first element of each iterable passed into zip which is then compared against a tuple of each letter in a (tuple(a)).

Answer 3

CJam, 21 20 bytes

qeuN/)S/_:c2$s=_p*&,

Try this fiddle in the CJam interpreter or verify all test cases at once.

How it works

qeu                  e# Read from STDIN and convert to uppercase.
   N/                e# Split at linenfeeds.
     )S/             e# Pop the second line form the array.
      S/             e# Split it at spaces.
        _:c          e# Push a copy and keep on the initial of each word.
           2$s       e# Push a copy of the line array and flatten it.
                     e# This pushes the first line.
              =      e# Check for equality.
               _p    e# Print a copy of the resulting Boolean.
                 *   e# Repeat the word array 1 or 0 times.
                  &  e# Intersect the result with the line array.
                   , e# Push the length of the result (1 or 0).

Answer 4

Scala, 135 110 108 bytes

val Array(x,y)=args.map(_.toLowerCase)
val z=y.split(" ").map(_(0)).mkString
print(z==x,z==x&&y.contains(z))

Saved a few bytes by using command line arguments (thanks to J Atkin for the hint), putting the booleans out as a tupel, using mkString instead of new String and print instead of println.

EDIT: Misinterpreted the question, and had to reimplement the solution

Answer 5

Haskell, 81 80 bytes

import Data.Char
f[a,b]|c<-words b=(a==map(!!0)c,elem a c)
p=f.lines.map toLower

The output format is not strictly defined, so I return a pair of booleans, e.g. p "Aha\na huge arm" -> (True,False).

Answer 6

Python 3, 106 bytes

Well, at least it beat Scala ;)

x=input().lower()
y=input().lower().split()
g=all(x[i]==y[i][0]for i in range(len(y)))
print(g,g&(x in y))

Answer 7

AppleScript, 302 301 297 293 Bytes

Aw, hell yeah. Not even bothered that I lose, this is competitive for AppleScript.

set x to(display dialog""default answer"")'s text returned
set y to(display dialog""default answer"")'s text returned's words
set n to y's items's number
repeat n
try
if not y's item n's character 1=(x as text)'s character n then return{false,false}
end
set n to n-1
end
return{true,x is in y}

Outputs as:

{true, false}

Or whatever the answer happens to be.

Answer 8

Ruby, 77 74 bytes

b=gets.chop.upcase
a=gets.upcase
p c=a.scan(/\b\w/)*''==b,c&&a.include?(b)

Answer 9

PHP, 120 bytes

Not being case sensitive is a lot of weight (26 bytes). Passed all test cases:

foreach($c=explode(' ',strtolower($argv[2]))as$l)$m.=$l[0];var_dump($x=$m==$a=strtolower($argv[1]),$x&&in_array($a,$c));

Outputs two bool values in this form:

bool(true)
bool(false)

Reads two arguments from the command line, like:

a.php Acronyms "Acronym can really obviously narrow your message sensors"

Ungolfed

$acronym = strtolower($argv[1]);
$words = strtolower($argv[2]);
$words = explode(' ', $words);

foreach($words as $word) {
    $letters .= $word[0];
}

$isAcronym = $letters == $acronym;

var_dump(
    $isAcronym,
    $isAcronym && in_array($acronym, $words)
);

Answer 10

Ruby, 52 bytes

p !!gets[/^#{x=gets.scan(/\b\w/)*""}$/i],!! ~/#{x}/i

Example:

$ echo "Aha
A huge AHA" | ruby acronym.rb
true
true

Answer 11

Matlab, 90 bytes

function z=f(r,s)
z=[sum(regexpi(s(regexpi(s,'(?<=(\s|^))\S')),r))>0 nnz(regexpi(s,r))>0];

Example (note that Matlab displays true/ false as 1/ 0):

>> f('Aha', 'A huge Aha')
ans =
     1     1

Answer 12

JavaScript ES6, 95 92 bytes

(a,b)=>[(r=eval(`/^${a}$/i`)).test((s=b.split` `).map(c=>c[0]).join``),s.some(c=>r.test(c))]

Input both strings as parameters. Outputs an array with two values: one for each boolean.

Answer 13

GNU sed, 118 bytes

Requires -r flag, included in score as +1. Note that I'm using \b for a word-boundary match, even though I can't find this documented in GNU sed. It Works For Me...

N
h
s/^(.*)\n.*\b\1\b.*/True/i
/\n/s/.*/False/
x
:
s/(.)(.*\n)\1[^ ]* ?/\2/i
t
/../s/.*/False/
/F/h
/F/!s/.*/True/
G

Expanded:

#!/bin/sed -rf

N
h

# Is it recursive?
s/^(.*)\n.*\b\1\b.*/True/i
# If no replacement, there's still a newline => false
/\n/s/.*/False/

x

# Is it an acronym?
# Repeatedly consume one letter and corresponding word
:
s/(.)(.*\n)\1[^ ]* ?/\2/i
t
# If more than just \n remain, then false
/../s/.*/False/
# And falsify recursive, too
/F/h
# !False => true
/F/!s/.*/True/

G

Answer 14

Groovy, 91 bytes

a=args*.toLowerCase()
println([a[1].split()*.charAt(0).join("")==a[0],a[1].contains(a[0])])

Output format is [bool, bool].This takes it's input from the command line args.

Answer 15

Lua 5.3, 182 bytes

a=""b=io.read c=a.lower d=a.reverse e=d(c(b()))f=~e:len()h=a.sub g=d(c(b()))for m in g:gmatch"[^ ]+"do f=-~f if h(e,f,f)~=h(m,~0)then break end k=k or m==e end f=f>~1print(f,f and k)

Answer 16

R, 93 bytes

a=tolower(readLines(,2));cat(a[1]==gsub("([^ ])\\w* ?","\\1",a[2]),a[1]%in%scan(t=a[2],w=""))

Usage:

> a=tolower(readLines(,2));cat(a[1]==gsub("([^ ])\\w* ?","\\1",a[2]),a[1]%in%scan(t=a[2],w=""))
Aha
A huge Aha
Read 3 items
TRUE TRUE
> a=tolower(readLines(,2));cat(a[1]==gsub("([^ ])\\w* ?","\\1",a[2]),a[1]%in%scan(t=a[2],w=""))
Acronyms
Acronyms can really obviously narrow your message sensors
Read 8 items
TRUE TRUE
> a=tolower(readLines(,2));cat(a[1]==gsub("([^ ])\\w* ?","\\1",a[2]),a[1]%in%scan(t=a[2],w=""))
FAQ
frequently asked questions
Read 3 items
TRUE FALSE
> a=tolower(readLines(,2));cat(a[1]==gsub("([^ ])\\w* ?","\\1",a[2]),a[1]%in%scan(t=a[2],w=""))
foo
bar baz
Read 2 items
FALSE FALSE

Answer 17

awk 137 bytes

awk 'BEGIN{T="True";F="False"}NR*NF<2{a=tolower($1)}END{for(i=1;i<=NF;i++)b=b substr(tolower($i),1,1);print(a==b?T:F)"\n"(a==tolower($1)?T:F)}'

• Initialize T="True";F="False" to simplify output.
• NR*NF<2{a=tolower($1)}: set a only if the first line only has one field.
• END{...}: assuming only two lines...
  • for(i=1;i<=NF;i++)b=b substr(tolower($i),1,1): construct recursive acronym.
  • print(a==b?T:F)"\n"(a==tolower($1)?T:F): print the output of both comparisons, a==b and a==tolower($1).

If anyone knows how to optimize the recursive acronym construction, feel free to suggest.

Answer 18

SpecBAS - 144 bytes

1 INPUT a$,b$: LET a$=UP$(a$),b$=UP$(b$),d$="": DIM c$(SPLIT b$,NOT " ")
2 FOR EACH w$ IN c$(): LET d$=d$+w$(1): NEXT w$
3 TEXT d$=a$'POS(a$,b$)>0

Converting the 2 x inputs to uppercase saves characters vs. lowercase conversion. Can now have multiple assignments done in one LET statement, which also helps. And TEXT saves one character over PRINT.

Uses 1/0 to show true/false (the apostrophe just moves output to the next line).

Answer 19

Perl5, 90 bytes

($a,$b)=map{chomp;lc}<>;print((join"",map{substr($_,0,1)}split/ /,$b)ne $a?0:($b=~$a?2:1))

cheating a bit: 0 = all false, 1 = one true, 2 = both true. I'm not a golfer, but am upset perl is missing while browsing!

($a,$b)=map{chomp;lc}<>;              # get the two lines as lowercase
print((                               #
join"",map{substr($_,0,1)}split/ /,$b # collapse first letters of secondline
     ) ne $a  ? 0 : ( $b=~$a ? 2 : 1))# 0 nothing, 1 not recursive, or 2 

Answer 20

JavaScript (ES6), 89 96 95 bytes

(a,b)=>[p=(a=a[l='toLowerCase']())==(c=b[l]().split` `).map(x=>x[0]).join``,p&&c.some(x=>x==a)]

Shucks...I thought I had it all sorted out, but apparently I was wrong.

This defines an anonymous function which takes input as two strings, and returns and array of two boolean items. The first item is calculated by comparing the the first string in all lowercase with the first char of each word in the second string. The second item is calculated simply by checking if the second string contains the first.

Here's another solution for the second item; 2 bytes shorter, but very few browsers support it:

p&&c.includes(a)

Answer 21

JavaScript (ES6) 93

(w,s)=>s[L='toLowerCase'](w=w[L](z=y='')).replace(/\w+/g,v=>y+=v[z=z||v==w,0])&&[y=y==w,y&&z]

Test running the snippet below in any EcmaScript 6 compliant browser

f=(w,s)=>s[L='toLowerCase'](w=w[L](z=y='')).replace(/\w+/g,v=>y+=v[z=z||v==w,0])&&[y=y==w,y&&z]

// TEST

out=x=>O.innerHTML+=x+'\n';

;[
 ['Acronyms', 'Acronyms can really obviously narrow your message sensors', true, true]
,['FAQ', 'frequently asked questions', true, false]
,['foo', 'bar baz', false, false]
,['GNU', 'GNU is not Unix', false, false]
,['Aha', 'A huge Aha', true, true]
,['Lolcat', 'Laughing over lolcat captions and tearing.', true, true]
,['ABCDE', 'Another basic clearly defined example.', true, false]
,['GNU', 'Gnus nettle unicorns', true, false]
,['PHP', 'PHP Hypertext Preprocessor', true, true]
].forEach(([a,b,c,d]) => (
  [r,s]=f(a,b), 
  out(((r==c && s==d)?'OK ':'KO ') + a + ',' + b + ' -> ' + f(a,b))
))

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Answer 22

Pyke (was untitled when posted), (noncompetitive), 20 bytes

l1c"jFh)J"iQl1qDji{&

You can find the source code here, language is completely unstable (first test challenge for it) so don't expect it to work in the future (commit 8)

Or 18 bytes (stableish)

l1idcmhsRl1jqDji{&

Try it here!