Pangram checker
From Rosetta Code
You are encouraged to solve this task according to the task description, using any language you may know.
Write a function or method to check a sentence to see if it is a pangram or not and show its use.
A pangram is a sentence that contains all the letters of the English alphabet at least once, for example: The quick brown fox jumps over the lazy dog.
[edit] Ada
Using Sets
with Ada.Text_IO; use Ada.Text_IO;
with Ada.Strings.Maps; use Ada.Strings.Maps;
with Ada.Characters.Handling; use Ada.Characters.Handling;
procedure pangram is
function ispangram(txt: String) return Boolean is
lowtxt : String := To_Lower(txt);
letset,txtset : Character_Set;
begin
letset := To_Set("abcdefghijklmnopqrstuvwxyz");
txtset := To_Set(lowtxt);
return (letset-txtset)=Null_Set;
end ispangram;
begin
put_line(Boolean'Image(ispangram("This is a test")));
put_line(Boolean'Image(ispangram("The quick brown fox jumps over the lazy dog")));
put_line(Boolean'Image(ispangram("NOPQRSTUVWXYZ abcdefghijklm")));
put_line(Boolean'Image(ispangram("abcdefghijklopqrstuvwxyz"))); --Missing m, n
end pangram;
Output:
FALSE TRUE TRUE FALSE
[edit] ALGOL 68
Works with: ALGOL 68 version Standard - no extensions to language used
Works with: ALGOL 68G version Any - tested with release 1.18.0-9h.tiny
Works with: ELLA ALGOL 68 version Any (with appropriate job cards)
# init pangram: #
INT la = ABS "a", lz = ABS "z";
INT ua = ABS "A", uz = ABS "Z";
IF lz-la+1 > bits width THEN
put(stand error, "Exception: insufficient bits in word for task");
stop
FI;
PROC is a pangram = (STRING test)BOOL: (
BITS a2z := BIN(ABS(2r1 SHL (lz-la))-1); # assume: ASCII & Binary #
FOR i TO UPB test WHILE
INT c = ABS test[i];
IF la <= c AND c <= lz THEN
a2z := a2z AND NOT(2r1 SHL (c-la))
ELIF ua <= c AND c <= uz THEN
a2z := a2z AND NOT(2r1 SHL (c-ua))
FI;
# WHILE # a2z /= 2r0 DO
SKIP
OD;
a2z = 2r0
);
main:(
[]STRING test list = (
"Big fjiords vex quick waltz nymph",
"The quick brown fox jumps over a lazy dog",
"A quick brown fox jumps over a lazy dog"
);
FOR key TO UPB test list DO
STRING test = test list[key];
IF is a pangram(test) THEN
print(("""",test,""" is a pangram!", new line))
FI
OD
)
Output:
"Big fjiords vex quick waltz nymph" is a pangram! "The quick brown fox jumps over a lazy dog" is a pangram!
[edit] AutoHotkey
Gui, -MinimizeBox
Gui, Add, Edit, w300 r5 vText
Gui, Add, Button, x105 w100 Default, Check Pangram
Gui, Show,, Pangram Checker
Return
GuiClose:
ExitApp
Return
ButtonCheckPangram:
Gui, Submit, NoHide
Loop, 26
If Not InStr(Text, Char := Chr(64 + A_Index)) {
MsgBox,, Pangram, Character %Char% is missing!
Return
}
MsgBox,, Pangram, OK`, this is a Pangram!
Return
[edit] BASIC
Works with: QBasic
DECLARE FUNCTION IsPangram! (sentence AS STRING)
DIM x AS STRING
x = "My dog has fleas."
GOSUB doIt
x = "The lazy dog jumps over the quick brown fox."
GOSUB doIt
x = "Jackdaws love my big sphinx of quartz."
GOSUB doIt
x = "What's a jackdaw?"
GOSUB doIt
END
doIt:
PRINT IsPangram!(x), x
RETURN
FUNCTION IsPangram! (sentence AS STRING)
'returns -1 (true) if sentence is a pangram, 0 (false) otherwise
DIM l AS INTEGER, s AS STRING, t AS INTEGER
DIM letters(25) AS INTEGER
FOR l = 1 TO LEN(sentence)
s = UCASE$(MID$(sentence, l, 1))
SELECT CASE s
CASE "A" TO "Z"
t = ASC(s) - 65
letters(t) = 1
END SELECT
NEXT
FOR l = 0 TO 25
IF letters(l) < 1 THEN
IsPangram! = 0
EXIT FUNCTION
END IF
NEXT
IsPangram! = -1
END FUNCTION
Output:
0 My dog has fleas. -1 The quick brown fox jumps over the lazy dog. -1 Jackdaws love my big sphinx of quartz. 0 What's a jackdaw?
[edit] C
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
int isPangram( char *string )
{
static char *alphabet="abcdefghijklmnopqrstuvwxyz";
char wasused[26], *sp, *ap;
int j;
for (j=0; j<26; j++) wasused[j]=0;
for (sp=string;*sp; sp++) {
ap = strchr(alphabet, tolower(*sp));
if (ap != NULL) {
wasused[ap-alphabet] = 1;
}
}
for (j=0; (j<26) && wasused[j]; j++);
// if (j<26) printf("Missing character %c\n", alphabet[j]);
return (j==26);
}
main( int argc, char *argv[])
{
char *pangramTxt;
if (argc < 2) {
printf("usage %s (text to check for pangram)\n", argv[0]);
exit(1);
}
pangramTxt = argv[1];
printf("%s\n",pangramTxt);
printf("Is pangram? %s\n", (isPangram(pangramTxt)?"Yes":"No"));
return 0;
}
Usage:
>pangram "The quick brown fox jumps over lazy dogs." Is Pangram? Yes
[edit] C++
#include <algorithm>
#include <cctype>
#include <string>
using namespace std;
const string kAlphabet("abcdefghijklmnopqrstuvwxyz");
bool is_pangram(string s) {
// Convert to lower case.
transform(s.begin(), s.end(), s.begin(), (int(*)(int))tolower);
// Convert to a sorted sequence of unique characters.
sort(s.begin(), s.end());
s.erase(
unique(s.begin(), s.end()),
s.end());
// Remove non-alphabetic characters.
s.erase(
set_intersection(
s.begin(), s.end(),
kAlphabet.begin(), kAlphabet.end(),
s.begin()),
s.end());
// For a pangram, we're left with a..z.
return s == kAlphabet;
}
[edit] C#
using System;
using System.Linq;
class Program
{
public static string alphabet = "abcdefghijklmnopqrstuvwxyz";
static bool CheckPangram(string str)
{
var chars = str.ToLower().ToList();
return alphabet.ToList().All(x => chars.Contains(x));
}
static void Main(string[] args)
{
Console.WriteLine(CheckPangram("the quick brown fox jumps over the lazy dog"));
Console.WriteLine(CheckPangram("this is not a pangram"));
}
}
[edit] Common Lisp
(defun pangramp (s)
(null (set-difference
(loop for c from (char-code #\A) upto (char-code #\Z) collect (code-char c))
(coerce (string-upcase s) 'list))))
[edit] D
There are several ways to write this function, this is a low-level one.
/*pure*/ nothrow bool isPangram(string txt) {
pure nothrow static uint fullBitSet() { // compile-time function
uint result;
foreach (c; "abcdefghijklmnopqrstuvwxyz")
result |= (1u << (c - 'a'));
return result;
}
enum uint FULL_BIT_SET = fullBitSet(); // 67_108_863;
uint charset;
foreach (c; txt) {
if (c >= 'a' && c <= 'z')
charset |= (1u << (c - 'a'));
else if (c >= 'A' && c <= 'Z')
charset |= (1u << (c - 'A'));
}
return charset == FULL_BIT_SET;
}
void main() {
assert(isPangram("the quick brown fox jumps over the lazy dog")); // true
assert(!isPangram("the quick brown fox jumped over the lazy dog")); // false, no s
assert(isPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ")); // true
assert(!isPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ")); // false, no r
assert(!isPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ")); // false, no m
assert(isPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ")); // true
assert(!isPangram("")); // false
}
[edit] Clojure
(defn pangram? [s]
(let [letters (into #{} "abcdefghijklmnopqrstuvwxyz")]
(= (->> s .toLowerCase (filter letters) (into #{})) letters)))
[edit] E
def isPangram(sentence :String) {
return ("abcdefghijklmnopqrstuvwxyz".asSet() &! sentence.toLowerCase().asSet()).size() == 0
}
&! is the “but-not” or set difference operator.
[edit] F#
If the difference between the set of letters in the alphabet and the set of letters in the given string (after conversion to lower case) is the empty set then every letter appears somewhere in the given string:
let isPangram (str: string) = (set['a'..'z'] - set(str.ToLower())).IsEmpty
[edit] Factor
Translation of: E
: pangram? ( str -- ? )
[ "abcdefghijklmnopqrstuvwxyz" ] dip >lower diff length 0 = ;
"How razorback-jumping frogs can level six piqued gymnasts!" pangram? .
[edit] Forth
: pangram? ( addr len -- ? )
0 -rot bounds do
i c@ 32 or [char] a -
dup 0 26 within if
1 swap lshift or
else drop then
loop
1 26 lshift 1- = ;
s" The five boxing wizards jump quickly." pangram? . \ -1
[edit] Fortran
Works with: Fortran version 90 and later
module pangram
implicit none
private
public :: is_pangram
character (*), parameter :: lower_case = 'abcdefghijklmnopqrstuvwxyz'
character (*), parameter :: upper_case = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
contains
function to_lower_case (input) result (output)
implicit none
character (*), intent (in) :: input
character (len (input)) :: output
integer :: i
integer :: j
output = input
do i = 1, len (output)
j = index (upper_case, output (i : i))
if (j /= 0) then
output (i : i) = lower_case (j : j)
end if
end do
end function to_lower_case
function is_pangram (input) result (output)
implicit none
character (*), intent (in) :: input
character (len (input)) :: lower_case_input
logical :: output
integer :: i
lower_case_input = to_lower_case (input)
output = .true.
do i = 1, len (lower_case)
if (index (lower_case_input, lower_case (i : i)) == 0) then
output = .false.
exit
end if
end do
end function is_pangram
end module pangram
Example:
program test
use pangram, only: is_pangram
implicit none
character (256) :: string
string = 'This is a sentence.'
write (*, '(a)') trim (string)
write (*, '(l1)') is_pangram (string)
string = 'The five boxing wizards jumped quickly.'
write (*, '(a)') trim (string)
write (*, '(l1)') is_pangram (string)
end program test
Output:
This is a sentence.
F
The five boxing wizards jumped quickly.
T
[edit] Haskell
import Data.Char (toLower)
import Data.List ((\\))
pangram :: String -> Bool
pangram = null . (['a' .. 'z'] \\) . map toLower
main = print $ pangram "How razorback-jumping frogs can level six piqued gymnasts!"
[edit] Icon and Unicon
[edit] Icon
A panagram procedure:
procedure panagram(s) #: return s if s is a panagram and fail otherwise
if (map(s) ** &lcase) === &lcase then return s
end
And a main to drive it:
procedure main(arglist)
if *arglist > 0 then
every ( s := "" ) ||:= !arglist || " "
else
s := "The quick brown fox jumps over the lazy dog."
writes(image(s), " -- is")
writes(if not panagram(s) then "n't")
write(" a panagram.")
end
[edit] Unicon
This Icon solution works in Unicon.
[edit] Ioke
Text isPangram? = method(
letters = "abcdefghijklmnopqrstuvwxyz" chars
text = self lower chars
letters map(x, text include?(x)) reduce(&&)
)
Here is an example of it's use in the Ioke REPL:
iik> "The quick brown fox jumps over the lazy dog" isPangram?
"The quick brown fox jumps over the lazy dog" isPangram?
+> true
iik> "The quick brown fox jumps over the" isPangram?
"The quick brown fox jumps over the" isPangram?
+> false
[edit] J
Solution:
require 'strings'
isPangram=: (a. {~ 97+i.26) */@e. tolower
Example use:
isPangram 'The quick brown fox jumps over the lazy dog.'
1
isPangram 'The quick brown fox falls over the lazy dog.'
0
[edit] Java
Works with: Java version 1.5+
public class Pangram {
public static boolean isPangram(String test){
boolean pangram = true;
String lcTest = test.toLowerCase();
for(char a = 'a'; a <= 'z'; a++){
pangram &= lcTest.contains("" + a);
}
return pangram;
}
public static void main(String[] args){
System.out.println(isPangram("the quick brown fox jumps over the lazy dog"));//true
System.out.println(isPangram("the quick brown fox jumped over the lazy dog"));//false, no s
System.out.println(isPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ"));//true
System.out.println(isPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ"));//false, no r
System.out.println(isPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ"));//false, no m
System.out.println(isPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ"));//true
System.out.println(isPangram(""));//false
}
}
Output:
true false true false false true false
[edit] JavaScript
Translation of: Java
function is_pangram(str) {
var s = str.toLowerCase();
// sorted by frequency ascending (http://en.wikipedia.org/wiki/Letter_frequency)
var letters = "zqxjkvbpygfwmucldrhsnioate";
for (var i = 0; i < 26; i++)
if (s.indexOf(letters.charAt(i)) == -1)
return false;
return true;
}
print(is_pangram("is this a pangram")); // false
print(is_pangram("The quick brown fox jumps over the lazy dog")); // true
[edit] Logo
to remove.all :s :set
if empty? :s [output :set]
if word? :s [output remove.all butfirst :s remove first :s :set]
output remove.all butfirst :s remove.all first :s :set
end
to pangram? :s
output empty? remove.all :s "abcdefghijklmnopqrstuvwxyz
end
show pangram? [The five boxing wizards jump quickly.] ; true
[edit] Lua
require"lpeg"
S, C = lpeg.S, lpeg.C
function ispangram(s)
return #(C(S(s)^0):match"abcdefghijklmnopqrstuvwxyz") == 26
end
print(ispangram"waltz, bad nymph, for quick jigs vex")
print(ispangram"bobby")
print(ispangram"long sentence")
[edit] MATLAB
function trueFalse = isPangram(string)
%This works by histogramming the ascii character codes for lower case
%letters contained in the string (which is first converted to all
%lower case letters). Then it finds the index of the first letter that
%is not contained in the string (this is faster than using the find
%without the second parameter). If the find returns an empty array then
%the original string is a pangram, if not then it isn't.
trueFalse = isempty(find( histc(lower(string),(97:122))==0,1 ));
end
Sample Output:
isPangram('The quick brown fox jumps over the lazy dog.')
ans =
1
[edit] OCaml
let pangram str =
let ar = Array.make 26 false in
String.iter (function
| 'a'..'z' as c -> ar.(Char.code c - Char.code 'a') <- true
| _ -> ()
) (String.lowercase str);
Array.fold_left ( && ) true ar
let check str =
Printf.printf " %b -- %s\n" (pangram str) str
let () =
check "this is a sentence";
check "The quick brown fox jumps over the lazy dog.";
;;
outputs:
false -- this is a sentence true -- The quick brown fox jumps over the lazy dog.
[edit] Oz
declare
fun {IsPangram Xs}
{List.sub
{List.number &a &z 1}
{Sort {Map Xs Char.toLower} Value.'<'}}
end
in
{Show {IsPangram "The quick brown fox jumps over the lazy dog."}}
[edit] Perl
use List::MoreUtils 'all';
sub pangram {all {$_[0] =~ /$_/i} 'a' .. 'z';}
print "Yes.\n" if pangram 'Cozy lummox gives smart squid who asks for job pen.';
[edit] Perl 6
sub pangram($s) {
Set.new("a".."z").subsetorequal($s.comb);
}
[edit] PicoLisp
(de isPangram (Str)
(not
(diff
'`(chop "abcdefghijklmnopqrstuvwxyz")
(chop (lowc Str)) ) ) )
[edit] PHP
function pangram($str){
$alphabet = Array("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z");
foreach($alphabet as $val){
if(stripos($str,$val) === false){
return false;
}
}
return true;
}
echo pangram("The quick brown fox jumps over the lazy dog"); // 1
echo pangram("Am I a pangram?"); // 0
[edit] PL/I
test_pangram: procedure options (main);
is_pangram: procedure() returns (bit(1) aligned);
declare text character (200) varying;
declare c character (1);
get edit (text) (L);
put skip list (text);
text = lowercase(text);
do c = 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k',
'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u',
'v', 'w', 'x', 'y', 'z';
if index(text, c) = 0 then return ('0'b);
end;
return ('1'b);
end is_pangram;
put skip list ('Please type a sentence');
if is_pangram() then
put skip list ('The sentence is a pangram.');
else
put skip list ('The sentence is not a pangram.');
end test_pangram;
Output:
Please type a sentence
the quick brown fox jumps over the lazy dog
The sentence is a pangram.
[edit] Prolog
Works with SWI-Prolog
pangram(L) :-
numlist(0'a, 0'z, Alphabet),
forall(member(C, Alphabet), member(C, L)).
pangram_example :-
L1 = "the quick brown fox jumps over the lazy dog",
( pangram(L1) -> R1= ok; R1 = ko),
format('~s --> ~w ~n', [L1,R1]),
L2 = "the quick brown fox jumped over the lazy dog",
( pangram(L2) -> R2 = ok; R2 = ko),
format('~s --> ~w ~n', [L2, R2]).
output
?- pangram_example.
the quick brown fox jumps over the lazy dog --> ok
the quick brown fox jumped over the lazy dog --> ko
true.
[edit] PureBasic
Procedure IsPangram_fast(String$)
String$ = LCase(string$)
char_a=Asc("a")
; sets bits in a variable if a letter is found, reads string only once
For a = 1 To Len(string$)
char$ = Mid(String$, a, 1)
pos = Asc(char$) - char_a
check.l | 1 << pos
Next
If check & $3FFFFFF = $3FFFFFF
ProcedureReturn 1
EndIf
ProcedureReturn 0
EndProcedure
Procedure IsPangram_simple(String$)
String$ = LCase(string$)
found = 1
For a = Asc("a") To Asc("z")
; searches for every letter in whole string
If FindString(String$, Chr(a), 0) = 0
found = 0
EndIf
Next
ProcedureReturn found
EndProcedure
Debug IsPangram_fast("The quick brown fox jumps over lazy dogs.")
Debug IsPangram_simple("The quick brown fox jumps over lazy dogs.")
Debug IsPangram_fast("No pangram")
Debug IsPangram_simple("No pangram")
[edit] Python
Using set arithmetic:
import string, sys
if sys.version_info[0] < 3:
input = raw_input
def ispangram(sentence, alphabet=string.ascii_lowercase):
alphaset = set(alphabet)
return not alphaset - set(sentence.lower())
print ( ispangram(input('Sentence: ')) )
Sample output:
Sentence: The quick brown fox jumps over the lazy dog True
[edit] R
Using the built-in R vector "letters":
checkPangram <- function(sentence){
my.letters <- tolower(unlist(strsplit(sentence, "")))
is.pangram <- all(letters %in% my.letters)
if (is.pangram){
cat("\"", sentence, "\" is a pangram! \n", sep="")
} else {
cat("\"", sentence, "\" is not a pangram! \n", sep="")
}
}
Sample output:
s1 <- "The quick brown fox jumps over the lazy dog"
s2 <- "The quick brown fox jumps over the sluggish dog"
checkPangram(s1)
"The quick brown fox jumps over the lazy dog" is a pangram!
checkPangram(s2)
"The quick brown fox jumps over the sluggish dog" is not a pangram!
[edit] Ruby
def pangram?(sentence)
unused_letters = ('a'..'z').to_a - sentence.downcase.chars.to_a
unused_letters.empty?
end
p pangram?('this is a sentence') # ==> false
p pangram?('The quick brown fox jumps over the lazy dog.') # ==> true
[edit] Scala
def is_pangram(sentence: String) = sentence.toLowerCase.filter(c => c >= 'a' && c <= 'z').toSet.size == 26
scala> is_pangram("This is a sentence")
res0: Boolean = false
scala> is_pangram("The quick brown fox jumps over the lazy dog")
res1: Boolean = true
[edit] Smalltalk
!String methodsFor: 'testing'!
isPangram
^((self collect: [:c | c asUppercase]) select: [:c | c >= $A and: [c <= $Z]]) asSet size = 26
'The quick brown fox jumps over the lazy dog.' isPangram
[edit] SNOBOL4
Works with: Macro Spitbol Works with: Snobol4+ Works with: CSnobol
define('pangram(str)alfa,c') :(pangram_end)
pangram str = replace(str,&ucase,&lcase)
alfa = &lcase
pgr_1 alfa len(1) . c = :f(return)
str c :s(pgr_1)f(freturn)
pangram_end
define('panchk(str)tf') :(panchk_end)
panchk output = str
tf = 'False'; tf = pangram(str) 'True'
output = 'Pangram: ' tf :(return)
panchk_end
* # Test and display
panchk("The quick brown fox jumped over the lazy dogs.")
panchk("My girl wove six dozen plaid jackets before she quit.")
panchk("This 41-character string: it's a pangram!")
end
Output:
The quick brown fox jumped over the lazy dogs. Pangram: True My girl wove six dozen plaid jackets before she quit. Pangram: True This 41-character string: it's a pangram! Pangram: False
[edit] Tcl
proc pangram? {sentence} {
set letters [regexp -all -inline {[a-z]} [string tolower $sentence]]
expr {
[llength [lsort -unique $letters]] == 26
}
}
puts [pangram? "This is a sentence"]; # ==> false
puts [pangram? "The quick brown fox jumps over the lazy dog."]; # ==> true
[edit] Ursala
#import std
is_pangram = ^jZ^(!@l,*+ @rlp -:~&) ~=`A-~ letters
example usage:
#cast %bL
test =
is_pangram* <
'The quick brown fox jumps over the lazy dog',
'this is not a pangram'>
output:
<true,false>
[edit] VBScript
[edit] Implementation
function pangram( s )
dim i
dim sKey
dim sChar
dim nOffset
sKey = "abcdefghijklmnopqrstuvwxyz"
for i = 1 to len( s )
sChar = lcase(mid(s,i,1))
if sChar <> " " then
if instr(sKey, sChar) then
nOffset = asc( sChar ) - asc("a") + 1
if nOffset > 1 then
sKey = left(sKey, nOffset - 1) & " " & mid( sKey, nOffset + 1)
else
sKey = " " & mid( sKey, nOffset + 1)
end if
end if
end if
next
pangram = ( ltrim(sKey) = vbnullstring )
end function
function eef( bCond, exp1, exp2 )
if bCond then
eef = exp1
else
eef = exp2
end if
end function
[edit] Invocation
wscript.echo eef(pangram("a quick brown fox jumps over the lazy dog"), "is a pangram", "is not a pangram")
wscript.echo eef(pangram(""), "is a pangram", "is not a pangram")"
Categories: Programming Tasks | Solutions by Programming Task | Ada | ALGOL 68 | AutoHotkey | BASIC | C | C++ | C sharp | Common Lisp | D | Clojure | E | F Sharp | Factor | Forth | Fortran | Haskell | Icon | Unicon | Ioke | J | Java | JavaScript | Logo | Lua | MATLAB | OCaml | Oz | Perl | Perl 6 | PicoLisp | PHP | PL/I | Prolog | PureBasic | Python | R | Ruby | Scala | Smalltalk | SNOBOL4 | Tcl | Ursala | VBScript
