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Pangram checker
From Rosetta Code
Pangram checker is a programming task. Visitors like you are encouraged to solve it according to the task description, using any language they may happen to 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.
Contents |
[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] 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] E
def isPangram(sentence :String) {
return ("abcdefghijklmnopqrstuvwxyz".asSet() &! sentence.toLowerCase().asSet()).size() == 0
}
&! is the “but-not” or set difference operator.
[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] 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
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] 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] 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] 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] Ruby
def pangram?(sentence)
unused_letters = ('a'..'z').to_a - str.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] 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")"
