Pangram checker

Pangram checker
You are encouraged to solve this task according to the task description, using any language you may know.

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.

Task

Write a function or method to check a sentence to see if it is a   pangram   (or not)   and show its use.

Related tasks

•   determine if a string has all the same characters
•   determine if a string has all unique characters

11l

F is_pangram(sentence)
   R Set(sentence.lowercase().filter(ch -> ch C ‘a’..‘z’)).len == 26

L(sentence) [‘The quick brown fox jumps over the lazy dog.’,
             ‘The quick brown fox jumped over the lazy dog.’]
   print(‘'#.' is #.a pangram’.format(sentence, ‘not ’ * !is_pangram(sentence)))

Output:

'The quick brown fox jumps over the lazy dog.' is a pangram
'The quick brown fox jumped over the lazy dog.' is not a pangram

360 Assembly

*        Pangram RC                11/08/2015
PANGRAM  CSECT
         USING  PANGRAM,R12
         LR     R12,R15
BEGIN    LA     R9,SENTENCE
         LA     R6,4
LOOPI    LA     R10,ALPHABET       loop on sentences
         LA     R7,26
LOOPJ    LA     R5,0               loop on letters
         LR     R11,R9
         LA     R8,60
LOOPK    MVC    BUFFER+1(1),0(R10) loop in sentence
         CLC    0(1,R10),0(R11)    if alphabet[j=sentence[i]
         BNE    NEXTK
         LA     R5,1               found
NEXTK    LA     R11,1(R11)         next character
         BCT    R8,LOOPK
         LTR    R5,R5              if found
         BNZ    NEXTJ
         MVI    BUFFER,C'?'        not found
         B      PRINT
NEXTJ    LA     R10,1(R10)         next letter
         BCT    R7,LOOPJ
         MVC    BUFFER(2),=CL2'OK'
PRINT    MVC    BUFFER+3(60),0(R9)
         XPRNT  BUFFER,80
NEXTI    LA     R9,60(R9)          next sentence
         BCT    R6,LOOPI
RETURN   XR     R15,R15
         BR     R14
ALPHABET DC     CL26'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
SENTENCE DC     CL60'THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.'
         DC     CL60'THE FIVE BOXING WIZARDS DUMP QUICKLY.'
         DC     CL60'HEAVY BOXES PERFORM WALTZES AND JIGS.'
         DC     CL60'PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS.'
BUFFER   DC     CL80' '
         YREGS
         END    PANGRAM

Output:

OK THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.
?J THE FIVE BOXING WIZARDS DUMP QUICKLY.
?C HEAVY BOXES PERFORM WALTZES AND JIGS.
OK PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS.

ACL2

(defun contains-each (needles haystack)
   (if (endp needles)
       t
       (and (member (first needles) haystack)
            (contains-each (rest needles) haystack))))

(defun pangramp (str)
   (contains-each (coerce "abcdefghijklmnopqrstuvwxyz" 'list)
                  (coerce (string-downcase str) 'list)))

Action!

Library: Action! Tool Kit

INCLUDE "D2:CHARTEST.ACT" ;from the Action! Tool Kit

DEFINE CHAR_COUNT="26"

BYTE FUNC IsPangram(CHAR ARRAY t)
  BYTE ARRAY tab(CHAR_COUNT)
  BYTE i,c

  FOR i=0 TO CHAR_COUNT-1
  DO tab(i)=0 OD

  FOR i=1 TO t(0)
  DO
    c=ToLower(t(i))
    IF c>='a AND c<='z THEN
      tab(c-'a)=1
    FI
  OD

  FOR i=0 TO CHAR_COUNT-1
  DO
    IF tab(i)=0 THEN
      RETURN (0)
    FI
  OD
RETURN (1)

PROC Test(CHAR ARRAY t)
  BYTE res

  res=IsPangram(t)
  PrintF("""%S"" is ",t)
  IF res=0 THEN
    Print("not ")
  FI
  PrintE("a pangram.")
  PutE()
RETURN

PROC Main()
  Put(125) PutE() ;clear screen
  Test("The quick brown fox jumps over the lazy dog.")
  Test("QwErTyUiOpAsDfGhJkLzXcVbNm")
  Test("Not a pangram")
  Test("")
RETURN

Output:

Screenshot from Atari 8-bit computer

"The quick brown fox jumps over the lazy dog." is a pangram.

"QwErTyUiOpAsDfGhJkLzXcVbNm" is a pangram.

"Not a pangram" is not a pangram.

"" is not a pangram.

ActionScript

Works with: ActionScript version 2.0

function pangram(k:string):Boolean {
  var lowerK:String = k.toLowerCase();
  var has:Object = {}

  for (var i:Number=0; i<=k.length-1; i++) {
    has[lowerK.charAt(i)] = true;
  }

  var result:Boolean = true;

  for (var ch:String='a'; ch <= 'z'; ch=String.fromCharCode(ch.charCodeAt(0)+1)) {
      result = result && has[ch]
  }

  return result || false;
}

Ada

Using character 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
     (Is_Subset(To_Set(Span => ('a','z')), To_Set(To_Lower(txt))));

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;

Using quantified expressions

with Ada.Text_IO; use Ada.Text_IO;
with Ada.Characters.Handling; use Ada.Characters.Handling;
procedure pangram is

  function ispangram(txt : in String) return Boolean is
     (for all Letter in Character range 'a'..'z' =>
         (for some Char of txt => To_Lower(Char) = Letter));

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

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!

APL

    a←'abcdefghijklmnopqrstuvwxyz' ⍝ or ⎕ucs 96 + ⍳26 in GNU/Dyalog
    A←'ABCDEFGHIJKLMNOPQRSTUVWXYZ' ⍝ or ⎕ucs 64 + ⍳26, or just ⎕a in Dyalog

    Pangram ← {∧/ ∨⌿ 2 26⍴(a,A) ∊ ⍵}
    Pangram 'This should fail'
0
    Pangram 'The quick brown fox jumps over the lazy dog'
1

AppleScript

AppleScriptObjC

Out of the box, AppleScript lacks many library basics – no regex, no higher order functions, not even string functions for mapping to upper or lower case.

From OSX 10.10 onwards, we can, however, use ObjC functions from AppleScript by importing the Foundation framework. We do this below to get a toLowerCase() function. If we also add generic filter and map functions, we can write and test a simple isPangram() function as follows:

use framework "Foundation" -- ( for case conversion function )

--------------------- PANGRAM CHECKER --------------------

-- isPangram :: String -> Bool
on isPangram(s)
    script charUnUsed
        property lowerCaseString : my toLower(s)
        on |λ|(c)
            lowerCaseString does not contain c
        end |λ|
    end script

    0 = length of filter(charUnUsed, ¬
        "abcdefghijklmnopqrstuvwxyz")
end isPangram


--------------------------- TEST -------------------------
on run
    map(isPangram, {¬
        "is this a pangram", ¬
        "The quick brown fox jumps over the lazy dog"})

    --> {false, true}
end run


-------------------- GENERIC FUNCTIONS -------------------

-- filter :: (a -> Bool) -> [a] -> [a]
on filter(f, xs)
    tell mReturn(f)
        set lst to {}
        set lng to length of xs
        repeat with i from 1 to lng
            set v to item i of xs
            if |λ|(v, i, xs) then set end of lst to v
        end repeat
        return lst
    end tell
end filter


-- map :: (a -> b) -> [a] -> [b]
on map(f, xs)
    tell mReturn(f)
        set lng to length of xs
        set lst to {}
        repeat with i from 1 to lng
            set end of lst to |λ|(item i of xs, i, xs)
        end repeat
        return lst
    end tell
end map


-- Lift 2nd class handler function into
-- 1st class script wrapper
-- mReturn :: Handler -> Script
on mReturn(f)
    if class of f is script then
        f
    else
        script
            property |λ| : f
        end script
    end if
end mReturn


-- toLower :: String -> String
on toLower(str)
    set ca to current application
    ((ca's NSString's stringWithString:(str))'s ¬
        lowercaseStringWithLocale:(ca's NSLocale's currentLocale())) as text
end toLower

Output:

{false, true}

---

Core language

Contrary to the impression given above, AppleScript is perfectly capable of handling this task very simply and without the need for imported libraries.

on isPangram(txt)
    set alphabet to "abcedfghijklmnopqrstuvwxyz"
    ignoring case -- The default, but ensure it here.
        repeat with letter in alphabet
            if (txt does not contain letter) then return false
        end repeat
    end ignoring

    return true
end isPangram

local result1, result2
set result1 to isPangram("The Quick Brown Fox Jumps Over The Lazy Dog")
set result2 to isPangram("This is not a pangram")
return {result1, result2}

Output:

{true, false}

Arturo

chars: map 97..122 => [to :string to :char &]
pangram?: function [sentence][
    every? chars 'ch ->
        in? ch sentence
]

print pangram? "this is a sentence"
print pangram? "The quick brown fox jumps over the lazy dog."

Output:

false
true

ATS

(* ****** ****** *)
//
#include
"share/atspre_staload.hats"
#include
"share/HATS/atspre_staload_libats_ML.hats"
//
(* ****** ****** *)
//
fun
letter_check
(
cs: string, c0: char
) : bool = cs.exists()(lam(c) => c0 = c)
//
(* ****** ****** *)

fun
Pangram_check
  (text: string): bool = let
//
val
alphabet = "abcdefghijklmnopqrstuvwxyz"
val
((*void*)) = assertloc(length(alphabet) = 26)
//
in
  alphabet.forall()(lam(c) => letter_check(text, c) || letter_check(text, toupper(c)))
end // end of [Pangram_check]

(* ****** ****** *)

implement
main0 () =
{
//
val
text0 = "The quick brown fox jumps over the lazy dog."
//
val-true = Pangram_check(text0)
val-false = Pangram_check("This is not a pangram sentence.")
//
} (* end of [main0] *)

(* ****** ****** *)

An alternate implementation that makes a single pass through the string:

fn is_pangram{n:nat}(s: string(n)): bool = loop(s, i2sz(0)) where {
  val letters: arrayref(bool, 26) = arrayref_make_elt<bool>(i2sz(26), false)
  fn check(): bool = loop(0) where {
    fun loop{i:int | i >= 0 && i <= 26}(i: int(i)) =
      if i < 26 then
        if letters[i] then loop(i+1) else
        false
      else true
  }
  fun add{c:int}(c: char(c)): void =
    if (c >= 'A') * (c <= 'Z') then letters[char2int1(c) - char2int1('A')] := true else
    if (c >= 'a') * (c <= 'z') then letters[char2int1(c) - char2int1('a')] := true
  fun loop{i:nat | i <= n}.<n-i>.(s: string(n), i: size_t(i)): bool =
    if string_is_atend(s, i) then check() else
    begin
      add(s[i]);
      loop(s, succ(i))
    end
}

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

AutoIt

Pangram("The quick brown fox jumps over the lazy dog")
Func Pangram($s_String)
For $i = 1 To 26
	IF Not StringInStr($s_String, Chr(64 + $i)) Then
		Return MsgBox(0,"No Pangram", "Character " & Chr(64 + $i) &" is missing")
	EndIf
Next
Return MsgBox(0,"Pangram", "Sentence is a Pangram")
EndFunc

AWK

Solution using string-operations

#!/usr/bin/awk -f
BEGIN {
   allChars="ABCDEFGHIJKLMNOPQRSTUVWXYZ";
   print isPangram("The quick brown fox jumps over the lazy dog.");
   print isPangram("The quick brown fo.");
}

function isPangram(string) {
    delete X;
    for (k=1; k<length(string); k++) {
        X[toupper(substr(string,k,1))]++;  # histogram
    }
    for (k=1; k<=length(allChars); k++) {
        if (!X[substr(allChars,k,1)]) return 0;
    }
    return 1;
}

Output:

1
0

Solution using associative arrays and split

Works with: gawk version 4.1.0

Works with: mawk version 1.3.3

# usage: awk -f pangram.awk -v p="The five boxing wizards dump quickly." input.txt
#
# Pangram-checker, using associative arrays and split
BEGIN {
  alfa="ABCDEFGHIJKLMNOPQRSTUVWXYZ"; ac=split(alfa,A,"")
  print "# Checking for all",ac,"chars in '" alfa "' :"

  print testPangram("The quick brown fox jumps over the lazy dog.");
  print testPangram(p);
}

{ print testPangram($0) }

function testPangram(str,   c,i,S,H,hit,miss) {
    print str  						##
    split( toupper(str), S, "")
    for (c in S) {
      H[ S[c] ]++
     #print c, S[c], H[ S[c] ]				##
    }
    for (i=1; i<=ac; i++) {
      c = A[i]
     #printf("%2d %c : %4d\n", i, c, H[c] )  		##
      if (H[c]) { hit=hit c } else { miss=miss c }
    }
    print "# hit:",hit, "# miss:",miss, "."		##
    if (miss) return 0
    return 1
}

Output:

# Checking for all 26 chars in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' :
The quick brown fox jumps over the lazy dog.
# hit: ABCDEFGHIJKLMNOPQRSTUVWXYZ # miss:  .
1
The five boxing wizards dump quickly.
# hit: ABCDEFGHIKLMNOPQRSTUVWXYZ # miss: J .
0
Heavy boxes perform waltzes and jigs
# hit: ABDEFGHIJLMNOPRSTVWXYZ # miss: CKQU .
0
The quick onyx goblin jumps over the lazy dwarf.
# hit: ABCDEFGHIJKLMNOPQRSTUVWXYZ # miss:  .
1
Pack my box with five dozen liquor jugs
# hit: ABCDEFGHIJKLMNOPQRSTUVWXYZ # miss:  .
1

BASIC

Applesoft BASIC

 100  P$ = "11111111111111111111111111"
 110  FOR Q = 1 TO 3
 120      READ S$
 130      GOSUB 200"IS PANGRAM?
 140      PRINT  MID$ ("NO  YES ",P * 4 + 1,4)S$
 150  NEXT Q
 160  END
 170  DATA"THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG."
 180  DATA"THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG."
 190  DATA"THE FIVE BOXING WIZARDS JUMP QUICKLY."
 200  P = 0:L =  LEN (S$): IF  NOT L THEN  RETURN 
 210  F$ = "00000000000000000000000000"
 220  FOR I = 1 TO L
 230      C =  ASC ( MID$ (S$,I,1)):C = C - 32 * (C > 95): IF C > 64 AND C < 91 THEN J = C - 64:F$ =  MID$ (F$,1,J - 1) + "1" +  MID$ (F$,J + 1):P = F$ = P$: IF P THEN  RETURN 
 240  NEXT I
 250  RETURN

Output:

YES THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.
NO  THE QUICK BROWN FOX JUMPED OVER THE LAZY DOG.
YES THE FIVE BOXING WIZARDS JUMP QUICKLY.

BaCon

This can be done in a one-liner.

DEF FN Pangram(x) = IIF(AMOUNT(UNIQ$(EXPLODE$(EXTRACT$(LCASE$(x), "[^[:alpha:]]", TRUE), 1))) = 26, TRUE, FALSE)

PRINT Pangram("The quick brown fox jumps over the lazy dog.")
PRINT Pangram("Jackdaws love my big sphinx of quartz.")
PRINT Pangram("My dog has fleas.")
PRINT Pangram("What's a jackdaw?")
PRINT Pangram("The five boxing wizards jump quickly")

Output:

1
1
0
0
1

BASIC256

function isPangram$(texto$)
    longitud = Length(texto$)
    if longitud < 26 then return "is not a pangram"
    t$ = lower(texto$)
    print "'"; texto$; "' ";
    for i = 97 to 122
        if instr(t$, chr(i)) = 0 then return "is not a pangram"
    next i
    return "is a pangram"
end function

print isPangram$("The quick brown fox jumps over the lazy dog.")   # --> true
print isPangram$("The quick brown fox jumped over the lazy dog.")  # --> false
print isPangram$("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ")          # --> true

BBC BASIC

      FOR test% = 1 TO 2
        READ test$
        PRINT """" test$ """ " ;
        IF FNpangram(test$) THEN
          PRINT "is a pangram"
        ELSE
          PRINT "is not a pangram"
        ENDIF
      NEXT test%
      END

      DATA "The quick brown fox jumped over the lazy dog"
      DATA "The five boxing wizards jump quickly"

      DEF FNpangram(A$)
      LOCAL C%
      A$ = FNlower(A$)
      FOR C% = ASC("a") TO ASC("z")
        IF INSTR(A$, CHR$(C%)) = 0 THEN = FALSE
      NEXT
      = TRUE

      DEF FNlower(A$)
      LOCAL A%, C%
      FOR A% = 1 TO LEN(A$)
        C% = ASCMID$(A$,A%)
        IF C% >= 65 IF C% <= 90 MID$(A$,A%,1) = CHR$(C%+32)
      NEXT
      = A$

Output:

"The quick brown fox jumped over the lazy dog" is not a pangram
"The five boxing wizards jump quickly" is a pangram

String manipulation is expensive, especially in loops, so it may be better to buffer the string and use character values:

DEFFNisPangram(text$)
  LOCAL size%,text%,char%,bits%
  size%=LENtext$
  IF size%<27 THEN =FALSE:REM too few characters
  DIM text% LOCAL size%:REM BB4W and RISC OS 5 only
  $text%=text$:REM buffer the string
  FOR text%=text% TO text%+size%-1:REM each character
    char%=?text% OR 32:REM to lower case
    IF 96<char% AND char%<123 THEN bits%=bits% OR 1<<(char%-97):REM set ordinal bit
    IF bits%=&3FFFFFF THEN =TRUE:REM all ordinal bits set
  NEXT text%
=FALSE

Commodore BASIC

10 rem detect model for title display
20 mx=peek(213): if mx=21 or mx=39 or mx=79 then 50:rem pet, vic, c64
30 mx=peek(238): if mx=39 or mx=79 then 50: rem c128
40 mx=39:color 4,1:rem assume plus/4 or c-16
50 if mx=21 then poke 36879,30:rem fix color on vic-20
60 print chr$(147);chr$(14);chr$(18);"**";:for i=2 to (mx-15)/2:print " ";:next
70 print "Pangram Checker";
80 for i=(mx-15)/2+16 to mx-2: print " ";: next: print "**"
100 read s$
110 if len(s$)=0 then end
120 gosub 1000:print
130 print "'"s$"' is";
140 if p=0 then print " not";
150 print " a pangram."
160 goto 100
500 data "The quick brown fox jumps over the lazy dog."
510 data "The quick brown fox jumped over the lazy dog."
520 data "The five boxing wizards jump quickly."
530 data
900 rem pangram checker
1000 if f=0 then f=1:dim seen(25),a(2):a(0)=65:a(1)=97:a(2)=193:goto 1020
1010 for i=0 to 25:seen(i)=0:next
1020 for i=1 to len(s$)
1030 : c=asc(mid$(s$,i))
1040 : for a = 0 to 2
1050 :   if c>=a(a) and c<=a(a)+25 then seen(c-a(a))=seen(c-a(a))+1
1060 : next a
1070 next i
1080 p=-1
1090 for i=0 to 25
1100 : if seen(i)=0 then p=0:i=25
1110 next i
1120 return

Output:

**                               Pangram Checker                              **


'The quick brown fox jumps over the lazy dog.' is a pangram.

'The quick brown fox jumped over the lazy dog.' is not a pangram.

'The five boxing wizards jump quickly.' is a pangram.

ready.

Chipmunk Basic

The Applesoft BASIC solution works without any changes.

FreeBASIC

' FB 1.05.0 Win64

Function isPangram(s As Const String) As Boolean
  Dim As Integer length = Len(s)
  If length < 26 Then Return False
  Dim p As String = LCase(s)
  For i As Integer = 97 To 122
    If Instr(p, Chr(i)) = 0 Then Return False
  Next
  Return True
End Function

Dim s(1 To 3) As String = _
{ _
 "The quick brown fox jumps over the lazy dog", _
 "abbdefghijklmnopqrstuVwxYz", _ '' no c!
 "How vexingly quick daft zebras jump!" _
}

For i As Integer = 1 To 3:
  Print "'"; s(i); "' is "; IIf(isPangram(s(i)), "a", "not a"); " pangram"
  Print
Next

Print
Print "Press nay key to quit"
Sleep

Output:

'The quick brown fox jumps over the lazy dog' is a pangram

'abbdefghijklmnopqrstuVwxYz' is not a pangram

'How vexingly quick daft zebras jump!' is a pangram

Liberty BASIC

'Returns 0 if the string is NOT a pangram or >0 if it IS a pangram
string$ = "The quick brown fox jumps over the lazy dog."

Print isPangram(string$)

Function isPangram(string$)
    string$ = Lower$(string$)
    For i = Asc("a") To Asc("z")
        isPangram = Instr(string$, chr$(i))
        If isPangram = 0 Then Exit Function
    Next i
End Function

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")

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?

Run BASIC

s$ = "The quick brown fox jumps over the lazy dog."
Print pangram(s$);" ";s$

s$ = "My dog has fleas."
Print pangram(s$);" ";s$

function pangram(str$)
  str$  = lower$(str$)
  for i = asc("a") to asc("z")
      pangram = pangram + (instr(str$, chr$(i)) <> 0)
  next i
pangram = (pangram = 26)
end function

1 The quick brown fox jumps over the lazy dog.
0 My dog has fleas.

Sinclair ZX81 BASIC

Works (just) with the 1k RAM model. The "37" that crops up a couple of times stops being a mystery if we remember that the ZX81 character code for A is 38 and that strings (like arrays) are indexed from 1, not from 0.

 10 LET A$="ABCDEFGHIJKLMNOPQRSTUVWXYZ"
 20 LET L=26
 30 INPUT P$
 40 IF LEN P$<26 THEN GOTO 170
 50 FAST
 60 LET C=1
 70 IF P$(C)<"A" OR P$(C)>"Z" THEN GOTO 120
 80 IF A$(CODE P$(C)-37)=" " THEN GOTO 120
 90 LET A$(CODE P$(C)-37)=" "
100 LET L=L-1
110 IF L=0 THEN GOTO 150
120 IF C=LEN P$ THEN GOTO 170
130 LET C=C+1
140 GOTO 70
150 PRINT "PANGRAM"
160 GOTO 180
170 PRINT "NOT A PANGRAM"
180 SLOW

Input:

THE QUICK BROWN FOX JUMPS OVER THE LAZY DOG.

Output:

PANGRAM

Input:

AND DARK THE SUN AND MOON, AND THE ALMANACH DE GOTHA

Output:

NOT A PANGRAM

uBasic/4tH

Proc _ShowPangram ("The quick brown fox jumps over the lazy dog.")
Proc _ShowPangram ("QwErTyUiOpAsDfGhJkLzXcVbNm")
Proc _ShowPangram ("Not a pangram")

End

_ShowPangram                           ' demonstrate the Pangram() function
  Param (1)
  Print Show (a@);Tab (50);Show (Iif (FUNC(_Pangram (a@)), "A pangram", "Not a pangram"))
Return

_Pangram
  Param (1)                            ' pangram candidate
  Local (3)

  b@ = 0                               ' reset the bitmap

  For d@ = 0 To Len(a@) -1             ' parse the string
    c@ = Peek (a@, d@)                 ' get current character
    If (c@ > Ord ("A") - 1) * (c@ < Ord ("Z") + 1) Then c@ = c@ + 32
    If (c@ > Ord ("a") - 1) * (c@ < Ord ("z") + 1) Then b@ = OR(b@, 2^(c@ - Ord ("a")))
  Next                                 ' update the bitmap
Return (b@ = 67108863)                 ' all bits set?

Output:

The quick brown fox jumps over the lazy dog.      A pangram
QwErTyUiOpAsDfGhJkLzXcVbNm                        A pangram
Not a pangram                                     Not a pangram

0 OK, 0:156

Yabasic

sub isPangram$(t$, l1$)
	local lt, ll, r$, i, cc, ic

	if numparams = 1 then
		l1$ = "abcdefghijklmnopqrstuvwxyz"
	end if

	t$ = lower$(t$)
	ll = len(l1$)
	for i = 1 to ll
		r$ = r$ + " "
	next
	lt = len(t$)
	cc = asc("a")

	for i = 1 to lt
		ic = asc(mid$(t$, i, 1)) - cc + 1
		if ic > 0 and ic <= ll then
			mid$(r$, ic, 1) = chr$(ic + cc - 1)
		end if
	next i

	if l1$ = r$ then return "true" else return "false" end if

end sub

print isPangram$("The quick brown fox jumps over the lazy dog.")   // --> true
print isPangram$("The quick brown fox jumped over the lazy dog.")  // --> false
print isPangram$("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ")          // --> true

Batch File

@echo off
setlocal enabledelayedexpansion

	%===The Main Thing===%
call :pangram "The quick brown fox jumps over the lazy dog."
call :pangram "The quick brown fox jumped over the lazy dog."
echo.
pause
exit /b 0

	%===The Function===%
:pangram
set letters=abcdefgihjklmnopqrstuvwxyz
set cnt=0
set inp=%~1
set str=!inp: =!

:loop
set chr=!str:~%cnt%,1!
if "!letters!"=="" (
	echo %1 is a pangram^^!
	goto :EOF
)
if "!chr!"=="" (
	echo %1 is not a pangram.
	goto :EOF
)
set letters=!letters:%chr%=!
set /a cnt+=1
goto loop

Output:

"The quick brown fox jumps over the lazy dog." is a pangram!
"The quick brown fox jumped over the lazy dog." is not a pangram.

Press any key to continue . . .

BCPL

get "libhdr"

// Test if s is a pangram. The ASCII character set is assumed.
let pangram(s) = valof
$(  let letters = vec 25
    for i=0 to 25 do letters!i := false
    for i=1 to s%0 do
    $(  let c = (s%i | 32) - 'a'
        if c >= 0 & c < 26 then
            letters!c := true
    $)
    for i=0 to 25 unless letters!i resultis false
    resultis true
$)

// Display s and whether or not it is a pangram.
let check(s) be
$(  writes(s)
    writes(" -> ")
    test pangram(s)
        then writes("yes*N")
        else writes("no*N")
$)

let start() be
$(  check("The quick brown fox jumps over the lazy dog.")
    check("The five boxing wizards dump quickly.")
$)

Output:

The quick brown fox jumps over the lazy dog. -> yes
The five boxing wizards dump quickly. -> no

Befunge

Reads the sentence to test from stdin.

>~>:65*`!#v_:"`"`48*v>g+04p1\4p
^#*`\*93\`0<::-"@"-*<^40!%2g4:_
"pangram."<v*84<_v#-":"g40\" a"
>>:#,_55+,@>"ton">48*>"si tahT"

Input:

The quick brown fox jumps over the lazy dog.

Output:

That is a pangram.

Bracmat

(isPangram=
  k
.   low$!arg:?arg
  & a:?k
  &   whl
    ' ( @(!arg:? !k ?)
      & chr$(1+asc$!k):?k:~>z
      )
  & !k:>z
  &
);

Some examples:

isPangram$("the Quick brown FOX jumps over the lazy do")
no
isPangram$("the Quick brown FOX jumps over the lazy dog")
yes
isPangram$"My dog has fleas."
no
isPangram$"The quick brown fox jumps over the lazy dog."
yes
isPangram$"Jackdaws love my big sphinx of quartz."
yes
isPangram$"What's a jackdaw?"
no
isPangram$"Lynx c.q. vos prikt bh: dag zwemjuf!"
yes

Brat

pangram? = { sentence |
  letters = [: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]

    sentence.downcase!

    letters.reject! { l |
      sentence.include? l
    }

  letters.empty?
}

p pangram? 'The quick brown fox jumps over the lazy dog.' #Prints true
p pangram? 'Probably not a pangram.'  #Prints false

Alternative version:

pangram? = { sentence |
  sentence.downcase.dice.unique.select(:alpha?).length == 26
}

C

#include <stdio.h>

int is_pangram(const char *s)
{
	const char *alpha = ""
		"abcdefghjiklmnopqrstuvwxyz"
		"ABCDEFGHIJKLMNOPQRSTUVWXYZ";

	char ch, wasused[26] = {0};
	int total = 0;

	while ((ch = *s++) != '\0') {
		const char *p;
		int idx;

		if ((p = strchr(alpha, ch)) == NULL)
			continue;

		idx = (p - alpha) % 26;

		total += !wasused[idx];
		wasused[idx] = 1;
		if (total == 26)
			return 1;
	}
	return 0;
}

int main(void)
{
	int i;
	const char *tests[] = {
		"The quick brown fox jumps over the lazy dog.",
		"The qu1ck brown fox jumps over the lazy d0g."
	};

	for (i = 0; i < 2; i++)
		printf("\"%s\" is %sa pangram\n",
			tests[i], is_pangram(tests[i])?"":"not ");
	return 0;
}

Using bitmask

Assumes an execution environment using the ASCII character set (will invoke undefined behavior on other systems).

#include <stdio.h>

int pangram(const char *s)
{
	int c, mask = (1 << 26) - 1;
	while ((c = (*s++)) != '\0') /* 0x20 converts lowercase to upper */
		if ((c &= ~0x20) <= 'Z' && c >= 'A')
			mask &= ~(1 << (c - 'A'));

	return !mask;
}

int main()
{
	int i;
	const char *s[] = {	"The quick brown fox jumps over lazy dogs.",
				"The five boxing wizards dump quickly.",  };

	for (i = 0; i < 2; i++)
		printf("%s: %s\n", pangram(s[i]) ? "yes" : "no ", s[i]);

	return 0;
}

Output:

yes: The quick brown fox jumps over lazy dogs.
no : The five boxing wizards dump quickly.

C#

C# 3.0 or higher (.NET Framework 3.5 or higher)

using System;
using System.Linq;

static class Program
{
    static bool IsPangram(this string text, string alphabet = "abcdefghijklmnopqrstuvwxyz")
    {
        return alphabet.All(text.ToLower().Contains);
    }

    static void Main(string[] arguments)
    {
        Console.WriteLine(arguments.Any() && arguments.First().IsPangram());
    }
}

Any version of C# language and .NET Framework

using System;

namespace PangrammChecker
{
    public class PangrammChecker
    {
        public static bool IsPangram(string str)
        {
            bool[] isUsed = new bool[26];
            int ai = (int)'a';
            int total = 0;
            for (CharEnumerator en = str.ToLower().GetEnumerator(); en.MoveNext(); )
            {
                int d = (int)en.Current - ai;
                if (d >= 0 && d < 26)
                    if (!isUsed[d])
                    {
                        isUsed[d] = true;
                        total++;
                    }
            }
            return (total == 26);
        }
    }

    class Program
    {
        static void Main(string[] args)
        {
            string str1 = "The quick brown fox jumps over the lazy dog.";
            string str2 = "The qu1ck brown fox jumps over the lazy d0g.";
            Console.WriteLine("{0} is {1}a pangram", str1,
                PangrammChecker.IsPangram(str1)?"":"not ");
            Console.WriteLine("{0} is {1}a pangram", str2,
                PangrammChecker.IsPangram(str2)?"":"not ");
            Console.WriteLine("Press Return to exit");
            Console.ReadLine();
        }
    }
}

C++

#include <algorithm>
#include <cctype>
#include <string>
#include <iostream>

const std::string alphabet("abcdefghijklmnopqrstuvwxyz");

bool is_pangram(std::string s)
{
    std::transform(s.begin(), s.end(), s.begin(), ::tolower);
    std::sort(s.begin(), s.end());
    return std::includes(s.begin(), s.end(), alphabet.begin(), alphabet.end());
}

int main()
{
    const auto examples = {"The quick brown fox jumps over the lazy dog",
                           "The quick white cat jumps over the lazy dog"};

    std::cout.setf(std::ios::boolalpha);
    for (auto& text : examples) {
        std::cout << "Is \"" << text << "\" a pangram? - " << is_pangram(text) << std::endl;
    }
}

Ceylon

shared void run() {

	function pangram(String sentence) =>
 		let(alphabet = set('a'..'z'),
			letters = set(sentence.lowercased.filter(alphabet.contains)))
 		letters == alphabet;

 	value sentences = [
 		"The quick brown fox jumps over the lazy dog",
 		"""Watch "Jeopardy!", Alex Trebek's fun TV quiz game.""",
 		"Pack my box with five dozen liquor jugs.",
 		"blah blah blah"
 	];
 	for(sentence in sentences) {
 		print("\"``sentence``\" is a pangram? ``pangram(sentence)``");
 	}
}

Clojure

(defn pangram? [s]
  (let [letters (into #{} "abcdefghijklmnopqrstuvwxyz")]
    (= (->> s .toLowerCase (filter letters) (into #{})) letters)))

CLU

pangram = proc (s: string) returns (bool)
    letters: array[bool] := array[bool]$fill(0,26,false)
    for c: char in string$chars(s) do
        if c>='a' & c<='z' then
            c := char$i2c(char$c2i(c) - 32)
        end
        if c>='A' & c<='Z' then
            letters[char$c2i(c) - 65] := true
        end
    end
    for seen: bool in array[bool]$elements(letters) do
        if ~seen then return(false) end
    end
    return(true)
end pangram

start_up = proc ()
    po: stream := stream$primary_output()
    examples: array[string] := array[string]$[
        "The quick brown fox jumps over the lazy dog.",
        "The five boxing wizards dump quickly.",
        "abcdefghijklmnopqrstuvwxyz"
    ]

    for example: string in array[string]$elements(examples) do
        stream$puts(po, "\"" || example || "\" is")
        if ~pangram(example) then
            stream$puts(po, " not")
        end
        stream$putl(po, " a pangram.")
    end
end start_up

Output:

"The quick brown fox jumps over the lazy dog." is a pangram.
"The five boxing wizards dump quickly." is not a pangram.
"abcdefghijklmnopqrstuvwxyz" is a pangram.

COBOL

       identification division.
       program-id. pan-test.
       data division.
       working-storage section.
       1 text-string pic x(80).
       1 len binary pic 9(4).
       1 trailing-spaces binary pic 9(4).
       1 pangram-flag pic x value "n".
        88 is-not-pangram value "n".
        88 is-pangram value "y".
       procedure division.
       begin.
           display "Enter text string:"
           accept text-string
           set is-not-pangram to true
           initialize trailing-spaces len
           inspect function reverse (text-string)
           tallying trailing-spaces for leading space
               len for characters after space
           call "pangram" using pangram-flag len text-string
           cancel "pangram"
           if is-pangram
               display "is a pangram"
           else
               display "is not a pangram"
           end-if
           stop run
           .
       end program pan-test.

       identification division.
       program-id. pangram.
       data division.
       1 lc-alphabet pic x(26) value "abcdefghijklmnopqrstuvwxyz".
       linkage section.
       1 pangram-flag pic x.
        88 is-not-pangram value "n".
        88 is-pangram value "y".
       1 len binary pic 9(4).
       1 text-string pic x(80).
       procedure division using pangram-flag len text-string.
       begin.
           inspect lc-alphabet converting
               function lower-case (text-string (1:len))
               to space
           if lc-alphabet = space
               set is-pangram to true
           end-if
           exit program
           .
       end program pangram.

CoffeeScript

is_pangram = (s) ->
  # This is optimized for longish strings--as soon as all 26 letters
  # are encountered, we will be done.  Our worst case scenario is a really
  # long non-pangram, or a really long pangram with at least one letter
  # only appearing toward the end of the string.
  a_code = 'a'.charCodeAt(0)
  required_letters = {}
  for i in [a_code...a_code+26]
    required_letters[String.fromCharCode(i)] = true

  cnt = 0
  for c in s
    c = c.toLowerCase()
    if required_letters[c]
      cnt += 1
      return true if cnt == 26
      delete required_letters[c]
  false

do ->
  tests = [
    ["is this a pangram", false]
    ["The quick brown fox jumps over the lazy dog", true]
  ]

  for test in tests
    [s, exp_value] = test
    throw Error("fail") if is_pangram(s) != exp_value
    # try long strings
    long_str = ''
    for i in [1..500000]
      long_str += s
    throw Error("fail") if is_pangram(long_str) != exp_value
    console.log "Passed tests: #{s}"

Comal

0010 FUNC pangram#(s$) CLOSED
0020   FOR i#:=ORD("A") TO ORD("Z") DO
0030     IF NOT (CHR$(i#) IN s$ OR CHR$(i#+32) IN s$) THEN RETURN FALSE
0040   ENDFOR i#
0050   RETURN TRUE
0060 ENDFUNC
0070 //
0080 WHILE NOT EOD DO
0090   READ s$
0100   PRINT "'",s$,"' is ",
0110   IF NOT pangram#(s$) THEN PRINT "not ",
0120   PRINT "a pangram"
0130 ENDWHILE
0140 END
0150 DATA "The quick brown fox jumps over the lazy dog."
0160 DATA "The five boxing wizards dump quickly."

Output:

'The quick brown fox jumps over the lazy dog.' is a pangram
'The five boxing wizards dump quickly.' is not a pangram

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))))

Component Pascal

BlackBox Component Builder

MODULE BbtPangramChecker;
IMPORT StdLog,DevCommanders,TextMappers;

PROCEDURE Check(str: ARRAY OF CHAR): BOOLEAN;
CONST
	letters = 26;
VAR
	i,j: INTEGER;
	status: ARRAY letters OF BOOLEAN;
	resp : BOOLEAN;
BEGIN
	FOR i := 0 TO LEN(status) -1 DO status[i] := FALSE END;

	FOR i := 0 TO LEN(str) -  1 DO
		j := ORD(CAP(str[i])) - ORD('A');
		IF (0 <= j) & (25 >= j) & ~status[j] THEN status[j] := TRUE END
	END;

	resp := TRUE;
	FOR i := 0 TO LEN(status) - 1 DO;
		resp := resp & status[i]
	END;
	RETURN resp;
END Check;

PROCEDURE Do*;
VAR
	params: DevCommanders.Par;
	s: TextMappers.Scanner;
BEGIN
	params := DevCommanders.par;
	s.ConnectTo(params.text);
	s.SetPos(params.beg);
	s.Scan;
	WHILE (~s.rider.eot) DO
		IF (s.type = TextMappers.char) & (s.char = '~') THEN
			RETURN
		ELSIF (s.type # TextMappers.string) THEN
			StdLog.String("Invalid parameter");StdLog.Ln
		ELSE
			StdLog.Char("'");StdLog.String(s.string + "' is pangram?:> ");
			StdLog.Bool(Check(s.string));StdLog.Ln
		END;
		s.Scan
	END
END Do;

END BbtPangramChecker.

Execute: ^Q BbtPangramChecker.Do "The quick brown fox jumps over the lazy dog"~
^Q BbtPangramChecker.Do "abcdefghijklmnopqrstuvwxyz"~
^Q BbtPangramChecker.Do "A simple text"~

Output:

'The quick brown fox jumps over the lazy dog' is pangram?:>  $TRUE
'abcdefghijklmnopqrstuvwxyz' is pangram?:>  $TRUE
'A simple text' is pangram?:>  $FALSE

Cowgol

include "cowgol.coh";

sub pangram(str: [uint8]): (r: uint8) is
    var letters: uint8[26];
    MemZero(&letters[0], 26);

    loop
        var chr := [str];
        if chr == 0 then break; end if;
        str := @next str;
        chr := (chr | 32) - 'a';
        if chr >= 26 then continue; end if;
        letters[chr] := letters[chr] | 1;
    end loop;

    r := 1;
    chr := 0;
    while chr < 26 loop
        r := r & letters[chr];
        if r == 0 then break; end if;
        chr := chr + 1;
    end loop;
end sub;

var yesno: [uint8][] := {": no\n", ": yes\n"};
var test: [uint8][] := {
    "The quick brown fox jumps over the lazy dog.",
    "The five boxing wizards dump quickly."
};

var i: @indexof test := 0;
while i < @sizeof test loop
    print(test[i]);
    print(yesno[pangram(test[i])]);
    i := i + 1;
end loop;

Output:

The quick brown fox jumps over the lazy dog.: yes
The five boxing wizards dump quickly.: no

Crystal

Copied and modified from the Ruby version.

def pangram?(sentence)
  ('a'..'z').all? {|c| sentence.downcase.includes?(c) }
end

p pangram?("not a pangram")
p pangram?("The quick brown fox jumps over the lazy dog.")

false
true

D

ASCII Bitmask version

bool isPangram(in string text) pure nothrow @safe @nogc {
    uint bitset;

    foreach (immutable c; text) {
        if (c >= 'a' && c <= 'z')
            bitset |= (1u << (c - 'a'));
        else if (c >= 'A' && c <= 'Z')
            bitset |= (1u << (c - 'A'));
    }

    return bitset == 0b11_11111111_11111111_11111111;
}

void main() {
    assert("the quick brown fox jumps over the lazy dog".isPangram);
    assert(!"ABCDEFGHIJKLMNOPQSTUVWXYZ".isPangram);
    assert(!"ABCDEFGHIJKL.NOPQRSTUVWXYZ".isPangram);
    assert("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ".isPangram);
}

Unicode version

import std.string, std.traits, std.uni;

// Do not compile with -g (debug info).
enum Alphabet : dstring {
    DE = "abcdefghijklmnopqrstuvwxyzßäöü",
    EN = "abcdefghijklmnopqrstuvwxyz",
    SV = "abcdefghijklmnopqrstuvwxyzåäö"
}

bool isPangram(S)(in S s, dstring alpha = Alphabet.EN)
pure /*nothrow*/ if (isSomeString!S) {
    foreach (dchar c; alpha)
       if (indexOf(s, c) == -1 && indexOf(s, std.uni.toUpper(c)) == -1)
            return false;
    return true;
}

void main() {
    assert(isPangram("the quick brown fox jumps over the lazy dog".dup, Alphabet.EN));
    assert(isPangram("Falsches Üben von Xylophonmusik quält jeden größeren Zwerg"d, Alphabet.DE));
    assert(isPangram("Yxskaftbud, ge vår wczonmö iqhjälp"w, Alphabet.SV));
}

Delphi

program PangramChecker;

{$APPTYPE CONSOLE}

uses StrUtils;

function IsPangram(const aString: string): Boolean;
var
  c: char;
begin
  for c := 'a' to 'z' do
    if not ContainsText(aString, c) then
      Exit(False);

  Result := True;
end;

begin
  Writeln(IsPangram('The quick brown fox jumps over the lazy dog')); // true
  Writeln(IsPangram('Not a panagram')); // false
end.

Draco

proc nonrec pangram(*char s) bool:
    ulong letters;
    char c;
    byte b;
    byte A = pretend('a', byte);
    byte Z = pretend('z', byte);

    letters := 0L0;
    while
        c := s*;
        s := s + 1;
        c /= '\e'
    do
        b := pretend(c, byte) | 32;
        if b >= A and b <= Z then
            letters := letters | 0L1 << (b-A)
        fi
    od;
    letters = 0x3FFFFFF
corp

proc nonrec test(*char s) void:
    writeln("\"", s, "\": ",
            if pangram(s) then "yes" else "no" fi)
corp

proc nonrec main() void:
    test("The quick brown fox jumps over the lazy dog.");
    test("The five boxing wizards jump quickly.");
    test("Not a pangram")
corp

Output:

"The quick brown fox jumps over the lazy dog.": yes
"The five boxing wizards jump quickly.": yes
"Not a pangram": no

E

def isPangram(sentence :String) {
    return ("abcdefghijklmnopqrstuvwxyz".asSet() &! sentence.toLowerCase().asSet()).size() == 0
}

&! is the “but-not” or set difference operator.

EasyLang

func pangr s$ .
   len d[] 26
   for c$ in strchars s$
      c = strcode c$
      if c >= 97 and c <= 122
         c -= 32
      .
      if c >= 65 and c <= 91
         d[c - 64] = 1
      .
   .
   for h in d[]
      s += h
   .
   return s
.
repeat
   s$ = input
   until s$ = ""
   print s$
   if pangr s$ = 26
      print "  --> pangram"
   .
   print ""
.
input_data
This is a test.
The quick brown fox jumps over the lazy dog.
The quick brown fox jumped over the lazy dog.
QwErTyUiOpAsDfGhJkLzXcVbNm

EDSAC order code

The program includes a test string (at the end). If the program is running in the EdsacPC simulator, the user can enter another string by storing it in a text file, making that file the active file, and clicking Reset. The string must be terminated by a blank row of tape (represented by '.' in EdsacPC).

 [Pangram checker for Rosetta Code.
  EDSAC program, Initial Orders 2.]

 [Outline: Make a table, one entry per 5-bit character code.
  Initialize entry for each letter to 1.
  When a letter is read, convert its entry to 0.]

 [Subroutine to read string from the input and
    store it with character codes in low 5 bits.
  String is terminated by blank row of tape, which is stored.
  Input: 0F holds address of string in address field (bits 1..11).
  21 locations; workspace: 4F]
            T   56 K
  GKA3FT17@AFA18@T7@I4FA4FUFS19@G12@S20@G16@T4FA7@A2FE4@T4FEFUFP8FPD

 [*************** Rosetta Code task ***************
  Subroutine to test whether string is a pangram.
  Input:  0F = address of string, characters in low 5 bits,
               terminated by blank row of tape.
  Output: 1F = (number of missing letters) - 1.
  87 memory locations; workspace 4F.]
            T   88 K
            G      K
            A    3 F  [make and plant link for return]
            T   48 @
       [Fill letter table with 1's.
        The code is a bit neater if we work backwards.]
            A   54 @  [index of last entry]
      [3]   A   51 @  [make T order for table entry]
            T    6 @  [plant in code]
            A   53 @  [acc := 1]
      [6]   T      F  [table entry := 1]
            A    6 @  [dec address in table]
            S    2 F
            S   51 @  [finished table?]
            E    3 @  [loop back if not]
       [Set non-letters to 0, except blank row := -1]
            T    4 F  [clear acc]
            T   66 @  [figures shift]
            T   70 @  [letters shift]
            T   73 @  [carriage return]
            T   75 @  [space]
            T   79 @  [line feed]
            S   53 @  [acc := -1]
            T   71 @  [blank row of tape]
       [Loop to read characters from string.
        Terminated by blank row of tape.
        Assume acc = 0 here.]
            A      F  [load address of string]
            A   49 @  [make order to read first char]
     [21]   T   22 @  [plant in code]
     [22]   A      F  [char to acc]
            L      D  [shift to address field]
            A   50 @  [make A order for this char in table]
            U   28 @  [plant in code]
            A   52 @  [convert to T order]
            T   31 @  [plant in code]
     [28]   A      F  [load table entry]
            G   35 @  [jump out if it's -1, i.e. blank row]
            T    4 F  [clear acc]
     [31]   T      F  [table entry := 0 to flag that letter is present]
            A   22 @  [inc address in input string]
            A    2 F
            G   21 @  [back to read next char]
       [Get total of table entries, again working backwards.
        The number of missing letters is (total + 1).]
     [35]   T    4 F  [clear acc]
            T    1 F  [initialize total := 0]
            A   54 @  [index of last entry]
     [38]   A   50 @  [make A order for table entry]
            T   41 @  [plant in code]
            A    1 F  [load total so far]
     [41]   A      F  [add table entry]
            T    1 F  [update total]
            A   41 @  [load A order]
            S    2 F  [dec address]
            S   50 @  [finished table?]
            E   38 @  [loop back if not]
            T    4 F  [clear acc before exit]
     [48]   E      F
  [Constants]
     [49]   A      F  [to make A order referring to input]
     [50]   A   55 @  [to make A order referring to table]
     [51]   T   55 @  [to make T order referring to table]
     [52]   O      F  [add to A order to convert to T order]
     [53]   P      D  [constant 1]
     [54]   P   31 F  [to change address by 31]
  [Table]
     [55]   PFPFPFPFPFPFPFPFPFPFPF
     [66]   PFPFPFPF               [11 = figures shift]
     [70]   PF                     [15 = letters shift]
     [71]   PFPF                   [16 = blank row of tape]
     [73]   PFPF                   [18 = carriage return]
     [75]   PFPFPFPF               [20 = space]
     [79]   PFPFPFPFPFPFPFPF       [24 = line feed]

 [Main routine to demonstrate pangram-checking subroutine]
            T  200 K
            G      K
  [Constants]
      [0]   P   25 @  [address for input string]
      [1]   N      F  [letter N]
      [2]   Y      F  [letter Y]
      [3]   K 2048 F  [letter shift]
      [4]   @      F  [carriage return]
      [5]   &      F  [line feed]
      [6]   K 4096 F  [null char]
 [Enter with acc = 0]
      [7]   O    3 @  [set letters shift]
      [8]   A      @  [load address of input]
            T      F  [pass to input subroutine in 0F]
     [10]   A   10 @  [call input subroutine, doesn't change 0F]
            G   56 F
     [12]   A   12 @  [call pangram subroutine]
            G   88 F
       [We could print the number of missing letters,
        but we'll just print 'Y' or 'N'.]
            A    1 F  [load (number missing) - 1]
            E   18 @  [jump if not pangram]
            O    2 @  [print 'Y']
            G   19 @  [exit]
     [18]   O    1 @  [print 'N']
     [19]   O    4 @  [print CR, LF]
            O    5 @
            O    6 @  [print null to flush printer buffer]
            Z      F  [stop]
            T      F  [on Reset, clear acc]
            E    8 @  [and test another string]
     [25]             [input string goes here]
            E    7 Z  [define entry point]
            P      F  [acc = 0 on entry]
THE!QUICK!BROWN!FOX!JUMPS!OVER!THE!LAZY!DOG.

Output:

Y

Elixir

defmodule Pangram do
  def checker(str) do
    unused = Enum.to_list(?a..?z) -- to_char_list(String.downcase(str))
    Enum.empty?(unused)
  end
end

text = "The quick brown fox jumps over the lazy dog."
IO.puts "#{Pangram.checker(text)}\t#{text}"
text = (Enum.to_list(?A..?Z) -- 'Test') |> to_string
IO.puts "#{Pangram.checker(text)}\t#{text}"

Output:

true    The quick brown fox jumps over the lazy dog.
false   ABCDEFGHIJKLMNOPQRSUVWXYZ

Erlang

-module(pangram).
-export([is_pangram/1]).

is_pangram(String) ->
  ordsets:is_subset(lists:seq($a, $z), ordsets:from_list(string:to_lower(String))).

Excel

LAMBDA

With the following lambda bound to the name ISPANGRAM in the Excel Workbook Name Manager:

(See LAMBDA: The ultimate Excel worksheet function)

Works with: Office 265 Betas 2010

ISPANGRAM
=LAMBDA(s,
    LET(
        abc, CHARS(LOWER("abcdefghijklmnopqrstuvwxyz")),
        AND(
            LAMBDA(c,
                ISNUMBER(SEARCH(c, s, 1))
            )(
                abc
            )
        )
    )
)

And assuming that the name CHARS is also bound in the Name Manager

to the generic (String -> Array Char) lambda:

CHARS
=LAMBDA(s,
    MID(s, ROW(INDIRECT("1:" & LEN(s))), 1)
)

Output:

	fx	=ISPANGRAM(A2)
	A	B
1	Test strings	Verdicts
2	The quick brown fox jumps over the lazy dog	TRUE
3	Is this a pangram	FALSE
4	How vexingly quick daft zebras jump!	TRUE
5	The five boxing wizards jumped quickly.	TRUE

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

Factor

Translation of: E

: pangram? ( str -- ? )
    [ "abcdefghijklmnopqrstuvwxyz" ] dip >lower diff length 0 = ;

"How razorback-jumping frogs can level six piqued gymnasts!" pangram? .

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

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

Frink

s = "The quick brown fox jumps over the lazy dog."
println["\"$s\" is" + (isPangram[s] ? "" : " not") + " a pangram."]

isPangram[s] :=
{
   charSet = toSet[charList[lc[s]]]
   for c = "a" to "z"
      if ! charSet.contains[c]
         return false

   return true
}

Output:

"The quick brown fox jumps over the lazy dog." is a pangram.

FutureBasic

include "NSLog.incl"

local fn IsPangram( pangramString as CFStringRef ) as BOOL
  NSUInteger  i, count
  BOOL        result
  
  CFStringRef   lcPanStr = fn StringLowerCaseString( pangramString )
  CFMutableSetRef mutSet = fn MutableSetWithCapacity( 0 )
  
  count = len(lcPanStr)
  for i = 0 to count - 1
    if ( fn CharacterSetCharacterIsMember( fn CharacterSetLowercaseLetterSet, fn StringCharacterAtIndex( lcPanStr, i ) ) )
      MutableSetAddObject( mutSet, fn StringWithFormat( @"%c", fn StringCharacterAtIndex( lcPanStr, i ) ) )
    end if
  next
  if fn SetCount( mutSet ) >= 26 then result = YES else result = NO
end fn = result


CFStringRef testStr, trueStr, falseStr
CFArrayRef  array

trueStr  = @"Is a pangram"
falseStr = @"Not a pangram"

array = @[¬
@"My dog has fleas.",¬
@"The quick brown fox jumps over the lazy do.",¬
@"The quick brown fox jumped over the lazy dog.",¬
@"The quick brown fox jumps over the lazy dog.",¬
@"Jackdaws love my big sphinx of quartz.",¬
@"What's a jackdaw?",¬
@"Watch \"Jeopardy!\", Alex Trebek's fun TV quiz game.",¬
@"Pack my box with five dozen liquor jugs.",¬
@"This definitely is not a pangram.",¬
@"This is a random long sentence just for testing purposes."]

for testStr in array
  if ( fn IsPangram( testStr ) )
    NSLog( @"%13s : %@", fn StringUTF8String( trueStr ), testStr ) else NSLog( @"%s : %@", fn StringUTF8String( falseStr ), testStr )
  end if
next

HandleEvents

Output:

Not a pangram : My dog has fleas.
Not a pangram : The quick brown fox jumps over the lazy do.
Not a pangram : The quick brown fox jumped over the lazy dog.
 Is a pangram : The quick brown fox jumps over the lazy dog.
 Is a pangram : Jackdaws love my big sphinx of quartz.
Not a pangram : What's a jackdaw?
 Is a pangram : Watch "Jeopardy!", Alex Trebek's fun TV quiz game.
 Is a pangram : Pack my box with five dozen liquor jugs.
Not a pangram : This definitely is not a pangram.
Not a pangram : This is a random long sentence just for testing purposes.

Fōrmulæ

Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text. Moreover, there can be multiple visual representations of the same program. Even though it is possible to have textual representation —i.e. XML, JSON— they are intended for storage and transfer purposes more than visualization and edition.

Programs in Fōrmulæ are created/edited online in its website.

In this page you can see and run the program(s) related to this task and their results. You can also change either the programs or the parameters they are called with, for experimentation, but remember that these programs were created with the main purpose of showing a clear solution of the task, and they generally lack any kind of validation.

Solution

Test cases

Go

package main

import "fmt"

func main() {
    for _, s := range []string{
        "The quick brown fox jumps over the lazy dog.",
        `Watch "Jeopardy!", Alex Trebek's fun TV quiz game.`,
        "Not a pangram.",
    } {
        if pangram(s) {
            fmt.Println("Yes:", s)
        } else {
            fmt.Println("No: ", s)
        }
    }
}

func pangram(s string) bool {
	var missing uint32 = (1 << 26) - 1
	for _, c := range s {
		var index uint32
		if 'a' <= c && c <= 'z' {
			index = uint32(c - 'a')
		} else if 'A' <= c && c <= 'Z' {
			index = uint32(c - 'A')
		} else {
			continue
		}

		missing &^= 1 << index
		if missing == 0 {
			return true
		}
	}
	return false
}

Output:

Yes: The quick brown fox jumps over the lazy dog.
Yes: Watch "Jeopardy!", Alex Trebek's fun TV quiz game.
No:  Not a pangram.

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!"

HicEst

PangramBrokenAt("This is a Pangram.") ! => 2 (b is missing)
PangramBrokenAt("The quick Brown Fox jumps over the Lazy Dog") ! => 0 (OK)

FUNCTION PangramBrokenAt(string)
   CHARACTER string, Alfabet="abcdefghijklmnopqrstuvwxyz"
   PangramBrokenAt = INDEX(Alfabet, string, 64)
   ! option 64: verify = 1st letter of string not in Alfabet
END

Icon and Unicon

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

Insitux

(function pangram? sentence
  (let prepped (-> sentence lower-case to-vec))
  (all? prepped (map char-code (range 97 123))))

(pangram? "The five boxing wizards jump quickly.")

Io

Sequence isPangram := method(
    letters := " " repeated(26)
    ia := "a" at(0)
    foreach(ichar,
        if(ichar isLetter,
            letters atPut((ichar asLowercase) - ia, ichar)
        )
    )
    letters contains(" " at(0)) not     // true only if no " " in letters
)

"The quick brown fox jumps over the lazy dog." isPangram println    // --> true
"The quick brown fox jumped over the lazy dog." isPangram println   // --> false
"ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ" isPangram println           // --> true

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

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

Java

Works with: Java version 1.5+

public class Pangram {
    public static boolean isPangram(String test){
        for (char a = 'A'; a <= 'Z'; a++)
            if ((test.indexOf(a) < 0) && (test.indexOf((char)(a + 32)) < 0))
                return false;
        return true;
    }

    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

JavaScript

ES5

Iterative

function isPangram(s) {
    var letters = "zqxjkvbpygfwmucldrhsnioate"
    // sorted by frequency ascending (http://en.wikipedia.org/wiki/Letter_frequency)
    s = s.toLowerCase().replace(/[^a-z]/g,'')
    for (var i = 0; i < 26; i++)
        if (s.indexOf(letters[i]) < 0) return false
    return true
}

console.log(isPangram("is this a pangram"))  // false
console.log(isPangram("The quick brown fox jumps over the lazy dog"))  // true

ES6

Functional

(() => {
    "use strict";

    // ----------------- PANGRAM CHECKER -----------------

    // isPangram :: String -> Bool
    const isPangram = s =>
        0 === "abcdefghijklmnopqrstuvwxyz"
        .split("")
        .filter(c => -1 === s.toLowerCase().indexOf(c))
        .length;

    // ---------------------- TEST -----------------------
    return [
        "is this a pangram",
        "The quick brown fox jumps over the lazy dog"
    ].map(isPangram);
})();

Output:

[false, true]

jq

def is_pangram:
  explode
  | map( if 65 <= . and . <= 90 then . + 32 # uppercase
         elif 97 <= . and . <= 122 then .   # lowercase
         else empty
         end )
  | unique
  | length == 26;

# Example:
"The quick brown fox jumps over the lazy dog" | is_pangram

Output:

$ jq -M -n -f pangram.jq
true

Julia

makepangramchecker creates a function to test for pangramity based upon the contents of its input string, allowing one to create arbitrary pangram checkers.

function makepangramchecker(alphabet)
    alphabet = Set(uppercase.(alphabet))
    function ispangram(s)
        lengthcheck = length(s) ≥ length(alphabet)
        return lengthcheck && all(c in uppercase(s) for c in alphabet)
    end
    return ispangram
end

const tests = ["Pack my box with five dozen liquor jugs.",
                "The quick brown fox jumps over a lazy dog.",
                "The quick brown fox jumps\u2323over the lazy dog.",
                "The five boxing wizards jump quickly.",
                "This sentence contains A-Z but not the whole alphabet."]

is_english_pangram = makepangramchecker('a':'z')

for s in tests
    println("The sentence \"", s, "\" is ", is_english_pangram(s) ? "" : "not ", "a pangram.")
end

Output:

The sentence "Pack my box with five dozen liquor jugs." is a pangram.
The sentence "The quick brown fox jumps over a lazy dog." is a pangram.
The sentence "The quick brown fox jumps⌣over the lazy dog." is a pangram.
The sentence "The five boxing wizards jump quickly." is a pangram.
The sentence "This sentence contains A-Z but not the whole alphabet." is not a pangram.

K

Works with: Kona

lcase   : _ci 97+!26
ucase   : _ci 65+!26
tolower : {@[x;p;:;lcase@n@p:&26>n:ucase?/:x]}
panagram: {&/lcase _lin tolower x}

Example:

  panagram "The quick brown fox jumps over the lazy dog"
1
  panagram "Panagram test"
0

Works with: ngn/k

isPangram:0=#(`c$"a"+!26)^_:

isPangram"This is a test"
0
isPangram"The quick brown fox jumps over the lazy dog."
1

Kotlin

// version 1.0.6

fun isPangram(s: String): Boolean {
    if (s.length < 26) return false
    val t = s.toLowerCase()
    for (c in 'a' .. 'z')
        if (c !in t) return false
    return true
}

fun main(args: Array<String>) {
   val candidates = arrayOf(
       "The quick brown fox jumps over the lazy dog",
       "New job: fix Mr. Gluck's hazy TV, PDQ!",
       "A very bad quack might jinx zippy fowls",
       "A very mad quack might jinx zippy fowls"   // no 'b' now!
   )
   for (candidate in candidates)
       println("'$candidate' is ${if (isPangram(candidate)) "a" else "not a"} pangram")
}

Output:

'The quick brown fox jumps over the lazy dog' is a pangram
'New job: fix Mr. Gluck's hazy TV, PDQ!' is a pangram
'A very bad quack might jinx zippy fowls' is a pangram
'A very mad quack might jinx zippy fowls' is not a pangram

Ksh

#!/bin/ksh

# Pangram checker

#	# Variables:
#
alphabet='abcdefghijklmnopqrstuvwxyz'

typeset -a strs
strs+=( 'Mr. Jock, TV quiz PhD., bags few lynx.' )
strs+=( 'A very mad quack might jinx zippy fowls.' )

#	# Functions:
#

#	# Function _ispangram(str) - return 0 if str is a pangram
#
function _ispangram {
	typeset _str ; typeset -l _str="$1"
	typeset _buff ; _buff="${alphabet}"
	typeset _i ; typeset -si _i

	for ((_i=0; _i<${#_str} && ${#_buff}>0; _i++)); do
		_buff=${_buff/${_str:${_i}:1}/}
	done
	return ${#_buff}
}

 ######
# main #
 ######

typeset -si i
for ((i=0; i<${#strs[*]}; i++)); do
	_ispangram "${strs[i]}"
	if (( ! $? )); then
		print "${strs[i]}   <<< IS A PANGRAM."
	else
		print "${strs[i]} <<< Is not a pangram."
	fi
done

Output:

Mr. Jock, TV quiz PhD., bags few lynx.   <<< IS A PANGRAM.
A very mad quack might jinx zippy fowls. <<< Is not a pangram.

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

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")

Maple

#Used built-in StringTools package
is_pangram := proc(str)
	local present := StringTools:-LowerCase~(select(StringTools:-HasAlpha, StringTools:-Explode(str)));
	local alphabets := {"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"};
	present := convert(present, set);
	return evalb(present = alphabets);
end proc;

Usage:

is_pangram("The quick brown fox jumps over the lazy dog.");
is_pangram("The 2 QUIck brown foxes jumped over the lazy DOG!!");
is_pangram(""The quick brown fox jumps over the lay dog.");

Output:

true
true
false

Mathematica/Wolfram Language

pangramQ[msg_]:=Complement[CharacterRange["a", "z"], Characters[ToLowerCase[msg]]]=== {}

Usage:

pangramQ["The quick brown fox jumps over the lazy dog."]
True

Or a slightly more verbose version that outputs the missing characters if the string is not a pangram:

pangramQ[msg_] :=
 Function[If[# === {}, Print["The string is a pangram!"],
    Print["The string is not a pangram. It's missing the letters " <>
      ToString[#]]]][
  Complement[CharacterRange["a", "z"], Characters[ToLowerCase[msg]]]]

Usage:

pangramQ["The quick brown fox jumps over the lazy dog."]
The string is a pangram!

pangramQ["Not a pangram"]
The string is not a pangram. It's missing the letters {b, c, d, e, f, h, i, j, k, l, q, s, u, v, w, x, y, z}

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

Output:

isPangram('The quick brown fox jumps over the lazy dog.')

ans =

     1

MATLAB / Octave

function trueFalse = isPangram(string)
    % X is a histogram of letters
    X = sparse(abs(lower(string)),1,1,128,1);
    trueFalse = full(all(X('a':'z') > 0));
end

Output:

>>isPangram('The quick brown fox jumps over the lazy dog.')
ans = 1

min

Works with: min version 0.19.3

"abcdefghijklmnopqrstuvwxyz" "" split =alphabet
('alphabet dip lowercase (swap match) prepend all?) :pangram?

"The quick brown fox jumps over the lazy dog." pangram? puts

MiniScript

sentences = ["The quick brown fox jumps over the lazy dog.",
    "Peter Piper picked a peck of pickled peppers.",
    "Waltz job vexed quick frog nymphs."]

alphabet = "abcdefghijklmnopqrstuvwxyz"

pangram = function (toCheck)
    sentence = toCheck.lower
    fail = false
    for c in alphabet
        if sentence.indexOf(c) == null then return false
    end for
    return true
end function

for sentence in sentences
    if pangram(sentence) then
        print """" + sentence + """ is a Pangram"
    else
        print """" + sentence + """ is not a Pangram"
    end if
end for

Output:

"The quick brown fox jumps over the lazy dog." is a Pangram
"Peter Piper picked a peck of pickled peppers." is not a Pangram
"Waltz job vexed quick frog nymphs." is a Pangram

ML

mLite

fun to_locase s = implode ` map (c_downcase) ` explode s

fun is_pangram
	(h :: t, T) =
		let
			val flen = len (filter (fn c = c eql h) T)
		in
			if (flen = 0) then
				false
			else
				is_pangram (t, T)
		end
|	([], T) = true
| 	S = is_pangram (explode "abcdefghijklmnopqrstuvwxyz", explode ` to_locase S)

fun is_pangram_i
	(h :: t, T) =
		let
			val flen = len (filter (fn c = c eql h) T)
		in
			if (flen = 0) then
				false
			else
				is_pangram (t, T)
		end
|	([], T) = true
| 	(A,S) = is_pangram (explode A, explode ` to_locase S)

fun test (f, arg, res, ok, notok) = if (f arg eql res) then ("'" @ arg @ "' " @ ok) else ("'" @ arg @ "' " @ notok)
fun test2 (f, arg, res, ok, notok) = if (f arg eql res) then ("'" @ ref (arg,1) @ "' " @ ok) else ("'" @ ref (arg,1) @ "' " @ notok)

;
println ` test (is_pangram, "The quick brown fox jumps over the lazy dog", true, "is a pangram", "is not a pangram");
println ` test (is_pangram, "abcdefghijklopqrstuvwxyz", true, "is a pangram", "is not a pangram");
val SValphabet = "abcdefghijklmnopqrstuvwxyzåäö";
val SVsentence = "Yxskaftbud, ge vår wczonmö iq hjälp";
println ` test2 (is_pangram_i, (SValphabet, SVsentence), true, "is a Swedish pangram", "is not a Swedish pangram");

Output:

'The quick brown fox jumps over the lazy dog' is a pangram
'abcdefghijklopqrstuvwxyz' is not a pangram
'Yxskaftbud, ge vår wczonmö iq hjälp' is a Swedish pangram

Modula-2

MODULE Pangrams;
FROM InOut IMPORT WriteString, WriteLn;
FROM Strings IMPORT Length;

(* Check if a string is a pangram *)
PROCEDURE pangram(s: ARRAY OF CHAR): BOOLEAN;
    VAR letters: ARRAY [0..25] OF BOOLEAN;
        i: CARDINAL;
BEGIN
    FOR i := 0 TO 25 DO letters[i] := FALSE; END;
    FOR i := 0 TO Length(s)-1 DO
        IF (s[i] >= 'A') AND (s[i] <= 'Z') THEN
            letters[ORD(s[i]) - ORD('A')] := TRUE;
        ELSIF (s[i] >= 'a') AND (s[i] <= 'z') THEN
            letters[ORD(s[i]) - ORD('a')] := TRUE;
        END;
    END;
    FOR i := 0 TO 25 DO
        IF NOT letters[i] THEN
            RETURN FALSE;
        END;
    END;
    RETURN TRUE;
END pangram;

PROCEDURE example(s: ARRAY OF CHAR);
BEGIN
    WriteString("'");
    WriteString(s);
    WriteString("' is ");
    IF NOT pangram(s) THEN
        WriteString("not ");
    END;
    WriteString("a pangram.");
    WriteLn();
END example;

BEGIN
    example("The quick brown fox jumps over the lazy dog");
    example("The five boxing wizards dump quickly");
    example("abcdefghijklmnopqrstuvwxyz");
END Pangrams.

Output:

'The quick brown fox jumps over the lazy dog' is a pangram.
'The five boxing wizards dump quickly' is not a pangram.
'abcdefghijklmnopqrstuvwxyz' is a pangram.

NetRexx

NetRexx's verify built–in method is all you need!

/* NetRexx */
options replace format comments java crossref savelog symbols nobinary

A2Z = 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'

pangrams = create_samples

loop p_ = 1 to pangrams[0]
  pangram = pangrams[p_]
  q_ = A2Z.verify(pangram.upper) -- <= it basically all happens in this function call!
  say pangram.left(64)'\-'
  if q_ == 0 then -
    say ' [OK, a pangram]'
  else -
    say ' [Not a pangram.  Missing:' A2Z.substr(q_, 1)']'
  end p_

method create_samples public static returns Rexx

  pangrams = ''

  x_ = 0
  x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick brown fox jumps over a lazy dog.'    -- best/shortest pangram
  x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick brown fox jumps over the lazy dog.'  -- not as short but at least it's still a pangram
  x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick brown fox jumped over the lazy dog.' -- common misquote; not a pangram
  x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'The quick onyx goblin jumps over the lazy dwarf.'
  x_ = x_ + 1; pangrams[0] = x_; pangrams[x_] = 'Bored? Craving a pub quiz fix? Why, just come to the Royal Oak!' -- (Used to advertise a pub quiz in Bowness-on-Windermere)

  return pangrams

Output:

The quick brown fox jumps over a lazy dog.                       [OK, a pangram]
The quick brown fox jumps over the lazy dog.                     [OK, a pangram]
The quick brown fox jumped over the lazy dog.                    [Not a pangram.  Missing: S]
The quick onyx goblin jumps over the lazy dwarf.                 [OK, a pangram]
Bored? Craving a pub quiz fix? Why, just come to the Royal Oak!  [OK, a pangram]

NewLISP

(context 'PGR)                              ;; Switch to context (say namespace) PGR
(define (is-pangram? str)
    (setf chars (explode (upper-case str))) ;; Uppercase + convert string into a list of chars
    (setf is-pangram-status true)           ;; Default return value of function
    (for (c (char "A") (char "Z") 1 (nil? is-pangram-status)) ;; For loop with break condition
        (if (not (find (char c) chars))     ;; If char not found in list, "is-pangram-status" becomes "nil"
            (setf is-pangram-status nil)
        )
    )
    is-pangram-status                       ;; Return current value of symbol "is-pangram-status"
)
(context 'MAIN)                             ;; Back to MAIN context

;; - - - - - - - - - -

(println (PGR:is-pangram? "abcdefghijklmnopqrstuvwxyz"))  ;; Print true
(println (PGR:is-pangram? "abcdef"))  ;; Print nil
(exit)

Nim

import rdstdin

proc isPangram(sentence: string, alphabet = {'a'..'z'}): bool =
  var sentset: set[char] = {}
  for c in sentence: sentset.incl c
  alphabet <= sentset

echo isPangram(readLineFromStdin "Sentence: ")

Example usage:

Sentence: The quick brown fox jumps over the lazy dog
true

Objeck

Translation of: Java

bundle Default {
  class Pangram {
    function : native : IsPangram(test : String) ~ Bool {
      for(a := 'A'; a <= 'Z'; a += 1;) {
        if(test->Find(a) < 0 & test->Find(a->ToLower()) < 0) {
          return false;
        };
      };

      return true;
    }

    function : Main(args : String[]) ~ Nil {
      IsPangram("the quick brown fox jumps over the lazy dog")->PrintLine(); # true
      IsPangram("the quick brown fox jumped over the lazy dog")->PrintLine(); # false, no s
      IsPangram("ABCDEFGHIJKLMNOPQRSTUVWXYZ")->PrintLine(); # true
      IsPangram("ABCDEFGHIJKLMNOPQSTUVWXYZ")->PrintLine(); # false, no r
      IsPangram("ABCDEFGHIJKL.NOPQRSTUVWXYZ")->PrintLine(); # false, no m
      IsPangram("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ")->PrintLine(); # true
      IsPangram("")->PrintLine(); # false
    }
  }
}

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.";
;;

Output:

false -- this is a sentence
true -- The quick brown fox jumps over the lazy dog.

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."}}

PARI/GP

pangram(s)={
  s=vecsort(Vec(s),,8);
  for(i=97,122,
    if(!setsearch(s,Strchr(i)) && !setsearch(s,Strchr(i-32)),
      return(0)
    )
  );
  1
};

pangram("The quick brown fox jumps over the lazy dog.")
pangram("The quick brown fox jumps over the lazy doe.")

Pascal

See Delphi

Perl

Get an answer with a module, or without.

use strict;
use warnings;
use feature 'say';

sub pangram1 {
    my($str,@set) = @_;
    use List::MoreUtils 'all';
    all { $str =~ /$_/i } @set;
}

sub pangram2 {
    my($str,@set) = @_;
    '' eq (join '',@set) =~ s/[$str]//gir;
}

my @alpha = 'a' .. 'z';

for (
    'Cozy Lummox Gives Smart Squid Who Asks For Job Pen.',
    'Crabby Lummox Gives Smart Squid Who Asks For Job Pen.'
) {
    say pangram1($_,@alpha) ? 'Yes' : 'No';
    say pangram2($_,@alpha) ? 'Yes' : 'No';
}

Output:

Yes
Yes
No
No

Phix

function pangram(string s)
sequence az = repeat(false,26)
integer count = 0
    for i=1 to length(s) do
        integer ch = lower(s[i])
        if ch>='a'
        and ch<='z'
        and not az[ch-96] then
            count += 1
            if count=26 then return {true,0} end if
            az[ch-96] = true
        end if
    end for
    return {false,find(false,az)+96}
end function

sequence checks = {"The quick brown fox jumped over the lazy dog",
                   "The quick brown fox jumps over the lazy dog",
                   ".!$\"AbCdEfghijklmnoprqstuvwxyz",
                   "THE FIVE BOXING WIZARDS DUMP QUICKLY.",
                   "THE FIVE BOXING WIZARDS JUMP QUICKLY.",
                   "HEAVY BOXES PERFORM WALTZES AND JIGS.",
                   "PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS.",
                   "Big fjiords vex quick waltz nymph",
                   "The quick onyx goblin jumps over the lazy dwarf.",
                   "no"}
for i=1 to length(checks) do
    string ci = checks[i]
    integer {r,ch} = pangram(ci)
    printf(1,"%-50s - %s\n",{ci,iff(r?"yes":"no "&ch)})
end for

Output:

The quick brown fox jumped over the lazy dog       - no s
The quick brown fox jumps over the lazy dog        - yes
.!$"AbCdEfghijklmnoprqstuvwxyz                     - yes
THE FIVE BOXING WIZARDS DUMP QUICKLY.              - no j
THE FIVE BOXING WIZARDS JUMP QUICKLY.              - yes
HEAVY BOXES PERFORM WALTZES AND JIGS.              - no c
PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS.           - yes
Big fjiords vex quick waltz nymph                  - yes
The quick onyx goblin jumps over the lazy dwarf.   - yes
no                                                 - no a

Phixmonti

include ..\Utilitys.pmt

def pangram?
	lower "abcdefghijklmnopqrstuvwxyz" swap remove len not nip
enddef

"The quick brown fox jumps over the lazy dog." pangram?
"This is a test" pangram?
"NOPQRSTUVWXYZ  abcdefghijklm" pangram?
"abcdefghijklopqrstuvwxyz" pangram?

pstack

Output:

[1, 0, 1, 0]

=== Press any key to exit ===

PHP

Translation of: D

function isPangram($text) {
    foreach (str_split($text) as $c) {
        if ($c >= 'a' && $c <= 'z')
            $bitset |= (1 << (ord($c) - ord('a')));
        else if ($c >= 'A' && $c <= 'Z')
            $bitset |= (1 << (ord($c) - ord('A')));
    }
    return $bitset == 0x3ffffff;
}

$test = array(
    "the quick brown fox jumps over the lazy dog",
    "the quick brown fox jumped over the lazy dog",
    "ABCDEFGHIJKLMNOPQSTUVWXYZ",
    "ABCDEFGHIJKL.NOPQRSTUVWXYZ",
    "ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ"
);

foreach ($test as $str)
    echo "$str : ", isPangram($str) ? 'T' : 'F', '</br>';

the quick brown fox jumps over the lazy dog : T
the quick brown fox jumped over the lazy dog : F
ABCDEFGHIJKLMNOPQSTUVWXYZ : F
ABCDEFGHIJKL.NOPQRSTUVWXYZ : F
ABC.D.E.FGHI*J/KL-M+NO*PQ R STUVWXYZ : T

Using array

function is_pangram( $sentence ) {

    // define "alphabet"
    $alpha = range( 'a', 'z' );

    // split lowercased string into array
    $a_sentence = str_split( strtolower( $sentence ) );

    // check that there are no letters present in alpha not in sentence
    return empty( array_diff( $alpha, $a_sentence ) );

}

$tests = array(
    "The quick brown fox jumps over the lazy dog.",
    "The brown fox jumps over the lazy dog.",
    "ABCDEFGHIJKL.NOPQRSTUVWXYZ",
    "ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ",
    "How vexingly quick daft zebras jump",
    "Is hotdog?",
    "How razorback-jumping frogs can level six piqued gymnasts!"
);

foreach ( $tests as $txt ) {
    echo '"', $txt, '"', PHP_EOL;
    echo is_pangram( $txt ) ? "Yes" : "No", PHP_EOL, PHP_EOL;
}

Output:

"The quick brown fox jumps over the lazy dog."
Yes

"The brown fox jumps over the lazy dog."
No

"ABCDEFGHIJKL.NOPQRSTUVWXYZ"
No

"ABC.D.E.FGHI*J/KL-M+NO*PQ R
STUVWXYZ"
Yes

"How vexingly quick daft zebras jump"
Yes

"Is hotdog?"
No

"How razorback-jumping frogs can level six piqued gymnasts!"
Yes

Picat

go =>
   S1 = "The quick brown fox jumps over the lazy dog",
   S2 = "The slow brown fox jumps over the lazy dog",
   println([S1, is_pangram(S1)]),
   println([S2, is_pangram(S2)]),
   nl,
   println("With missing chars:"),
   println([S1, is_pangram2(S1)]),
   println([S2, is_pangram2(S2)]),
   nl.

% Check if S is a pangram and get the missing chars
is_pangram(S) = P =>
   Lower = S.to_lowercase,
   Alpha = [chr(I+96) : I in 1..26],
   foreach(A in Alpha) membchk(A,Lower) end -> P = true ; P = false.

% Check if S is a pangram and get the missing chars (if any)
is_pangram2(S) = [pangram=cond(Missing==[],true,false),missing=Missing] =>
   Lower = S.to_lowercase,
   Missing = [A : A in [chr(I+96) : I in 1..26], not membchk(A,Lower)].

Output:

[The quick brown fox jumps over the lazy dog,true]
[The slow brown fox jumps over the lazy dog,false]

With missing chars:
[The quick brown fox jumps over the lazy dog,[pangram = true,missing = []]]
[The slow brown fox jumps over the lazy dog,[pangram = false,missing = cikq]]

PicoLisp

(de isPangram (Str)
   (not
      (diff
         '`(chop "abcdefghijklmnopqrstuvwxyz")
         (chop (lowc Str)) ) ) )

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.

PowerShell

Cyrillic test sample borrowed from Raku.

Works with: PowerShell version 2

function Test-Pangram ( [string]$Text, [string]$Alphabet = 'abcdefghijklmnopqrstuvwxyz' )
    {
    $Text = $Text.ToLower()
    $Alphabet = $Alphabet.ToLower()

    $IsPangram = @( $Alphabet.ToCharArray() | Where-Object { $Text.Contains( $_ ) } ).Count -eq $Alphabet.Length

    return $IsPangram
    }

Test-Pangram 'The quick brown fox jumped over the lazy dog.'
Test-Pangram 'The quick brown fox jumps over the lazy dog.'
Test-Pangram 'Съешь же ещё этих мягких французских булок, да выпей чаю' 'абвгдежзийклмнопрстуфхцчшщъыьэюяё'

Output:

False
True
True

A faster version can be created using .Net HashSet to do what the F# version does:

Function Test-Pangram ( [string]$Text, [string]$Alphabet = 'abcdefghijklmnopqrstuvwxyz' )
{
    $alSet   = [Collections.Generic.HashSet[char]]::new($Alphabet.ToLower())
    $textSet = [Collections.Generic.HashSet[char]]::new($Text.ToLower())

    $alSet.ExceptWith($textSet)    # remove text chars from the alphabet

    return $alSet.Count -eq 0    # any alphabet letters still remaining?
}

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.

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 alphaset <= set(sentence.lower())

print ( ispangram(input('Sentence: ')) )

Output:

Sentence: The quick brown fox jumps over the lazy dog
True

Quackery

  [ dup char A char [ within
    swap char a char { within
    or ]                        is letter  ( c --> b )

  [ 0 26 of swap witheach
      [ dup letter iff
          [ 1 unrot lower
            char a - poke ]
        else drop ]
    0 swap find 26 = ]          is pangram ( $ --> b )

  $ "This is a sentence." pangram echo cr                     ( 0 )
  $ "The five boxing wizards jumped quickly." pangram echo cr ( 1 )

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="")
  }
}

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!

Racket

#lang racket
(define (pangram? str)
  (define chars (regexp-replace* #rx"[^a-z]+" (string-downcase str) ""))
  (= 26 (length (remove-duplicates (string->list chars)))))
(pangram? "The quick Brown Fox jumps over the Lazy Dog")

Raku

(formerly Perl 6)

constant Eng = set 'a' .. 'z';
constant Cyr = (set 'а' .. 'ё') (-) (set 'ъ', 'ѐ');
constant Hex = set 'a' .. 'f';

sub pangram($str, Set $alpha = Eng) {
  $alpha ⊆ $str.lc.comb
}

say pangram("The quick brown fox jumps over the lazy dog.");
say pangram("My dog has fleas.");
say pangram("My dog has fleas.", Hex);
say pangram("My dog backs fleas.", Hex);
say pangram "Съешь же ещё этих мягких французских булок, да выпей чаю", Cyr;

Output:

True
False
False
True
True

Retro

'abcdefghijklmnopqrstuvwxyz 'FULL s:const
'__________________________ 'TEST s:const
:s:pangram? (s-f)
  '__________________________ &TEST #26 copy
  s:to-lower [ c:letter? ] s:filter
  [ dup $a - &TEST + store ] s:for-each
  &TEST &FULL s:eq? ;

REXX

/*REXX program  verifies  if an  entered/supplied  string  (sentence)  is a pangram.    */
@abc= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ'               /*a list of all (Latin) capital letters*/

    do forever;    say                           /*keep promoting 'til null (or blanks).*/
    say '──────── Please enter a pangramic sentence   (or a blank to quit):';      say
    pull y                                       /*this also uppercases the  Y variable.*/
    if y=''  then leave                          /*if nothing entered,  then we're done.*/
    absent= space( translate( @abc, , y), 0)     /*obtain a list of any absent letters. */
    if absent==''  then say  "──────── Sentence is a pangram."
                   else say  "──────── Sentence isn't a pangram, missing: "    absent
    say
    end   /*forever*/

say '──────── PANGRAM program ended. ────────'   /*stick a fork in it,  we're all done. */

output:

──────── Please enter a pangramic sentence   (or a blank to quit):
The quick brown fox jumped over the lazy dog.      ◄■■■■■■■■■■ user input.
──────── Sentence isn't a pangram, missing:  S


──────── Please enter a pangramic sentence   (or a blank to quit):
The quick brown fox JUMPS over the lazy dog!!!     ◄■■■■■■■■■■ user input.
──────── Sentence is a pangram.


──────── Please enter a pangramic sentence   (or a blank to quit):
                                                   ◄■■■■■■■■■■ user input   (null  or  some blanks).

──────── PANGRAM program ended. ────────

Ring

pangram = 0
s = "The quick brown fox jumps over the lazy dog."
see "" + pangram(s) + " " + s + nl

s = "My dog has fleas."
see "" + pangram(s) + " " + s + nl

func pangram str
     str  = lower(str)
     for i = ascii("a") to ascii("z")
             bool = substr(str, char(i)) > 0
             pangram = pangram + bool
     next
     pan = (pangram = 26)
     return pan

RPL

Structurally programmed

RPL code	Comment
≪ # 0h SWAP 1 OVER SIZE FOR j DUP j DUP SUB NUM IF DUP 97 ≥ OVER 122 ≤ AND THEN 32 - END IF DUP 65 ≥ OVER 90 ≤ AND THEN 2 SWAP 65 - ^ R→B ROT OR SWAP ELSE DROP END NEXT DROP # 3FFFFFFh == ≫ ‘PANG?’ STO	PANG? ( "sentence" → boolean ) alpha = 0; loop for j = 1 to length(sentence) c = ascii(sentence[j]) if c lowercase then make it uppercase if c in ("A".."Z") then alpha &= 2^(c-65) end loop return alpha & 3FFFFFFh

Idiomatically optimized for HP-28S

RPL code	Comment
≪ # 7FFFFFEh RCLF OVER NOT AND STOF SWAP 1 OVER SIZE FOR j DUP j DUP SUB NUM DUP 97 ≥ 95 63 IFTE - 1 MAX 28 MIN SF NEXT DROP RCLF OVER AND == ≫ ‘PANG?’ STO	PANG? ( "sentence" → boolean ) clear flags 1 to 28 loop for j = 1 to length(sentence) c = ascii(sentence[j]) reduce c to a value between 1 and 28 and set related flag return 1 if all flags between 2 and 27 are set

To run on more recent models, the following sequence in line 2

RCLF OVER NOT AND STOF

must be replaced by

RCLF DUP 2 GET 3 PICK NOT AND 2 SWAP PUT STOF

and RCLF in the last line by RCLF 2 GET.

"The quick brown fox jumps over the lazy dog" PANG?
"The quick brown fox jumped over the lazy dog" PANG?

Output:

2: 1
1: 0

Ruby

def pangram?(sentence)
  s = sentence.downcase
  ('a'..'z').all? {|char| s.include? (char) }
end

p pangram?('this is a sentence')  # ==> false
p pangram?('The quick brown fox jumps over the lazy dog.')  # ==> true

Rust

#![feature(test)]

extern crate test;

use std::collections::HashSet;

pub fn is_pangram_via_bitmask(s: &str) -> bool {

    // Create a mask of set bits and convert to false as we find characters.
    let mut mask = (1 << 26) - 1;

    for chr in s.chars() {
        let val = chr as u32 & !0x20; /* 0x20 converts lowercase to upper */
        if val <= 'Z' as u32 && val >= 'A' as u32 {
            mask = mask & !(1 << (val - 'A' as u32));
        }
    }

    mask == 0
}

pub fn is_pangram_via_hashset(s: &str) -> bool {

    // Insert lowercase letters into a HashSet, then check if we have at least 26.
    let letters = s.chars()
        .flat_map(|chr| chr.to_lowercase())
        .filter(|&chr| chr >= 'a' && chr <= 'z')
        .fold(HashSet::new(), |mut letters, chr| {
            letters.insert(chr);
            letters
        });

    letters.len() == 26
}

pub fn is_pangram_via_sort(s: &str) -> bool {

    // Copy chars into a vector, convert to lowercase, sort, and remove duplicates.
    let mut chars: Vec<char> = s.chars()
        .flat_map(|chr| chr.to_lowercase())
        .filter(|&chr| chr >= 'a' && chr <= 'z')
        .collect();

    chars.sort();
    chars.dedup();

    chars.len() == 26
}

fn main() {

    let examples = ["The quick brown fox jumps over the lazy dog",
                    "The quick white cat jumps over the lazy dog"];

    for &text in examples.iter() {
        let is_pangram_sort = is_pangram_via_sort(text);
        println!("Is \"{}\" a pangram via sort? - {}", text, is_pangram_sort);

        let is_pangram_bitmask = is_pangram_via_bitmask(text);
        println!("Is \"{}\" a pangram via bitmask? - {}",
                 text,
                 is_pangram_bitmask);

        let is_pangram_hashset = is_pangram_via_hashset(text);
        println!("Is \"{}\" a pangram via bitmask? - {}",
                 text,
                 is_pangram_hashset);
    }
}

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

Seed7

$ include "seed7_05.s7i";

const func boolean: isPangram (in string: stri) is func
  result
    var boolean: isPangram is FALSE;
  local
    var char: ch is ' ';
    var set of char: usedChars is (set of char).value;
  begin
    for ch range lower(stri) do
      if ch in {'a' .. 'z'} then
        incl(usedChars, ch);
      end if;
    end for;
    isPangram := usedChars = {'a' .. 'z'};
  end func;

const proc: main is func
  begin
    writeln(isPangram("This is a test"));
    writeln(isPangram("The quick brown fox jumps over the lazy dog"));
    writeln(isPangram("NOPQRSTUVWXYZ  abcdefghijklm"));
    writeln(isPangram("abcdefghijklopqrstuvwxyz"));  # Missing m, n
  end func;

Output:

FALSE
TRUE
TRUE
FALSE

Sidef

Translation of: Raku

define Eng = 'a'..'z';
define Hex = 'a'..'f';
define Cyr = %w(а б в г д е ж з и й к л м н о п р с т у ф х ц ч ш щ ъ ы ь э ю я ё);

func pangram(str, alpha=Eng) {
    var lstr = str.lc;
    alpha.all {|c| lstr.contains(c) };
}

say pangram("The quick brown fox jumps over the lazy dog.");
say pangram("My dog has fleas.");
say pangram("My dog has fleas.", Hex);
say pangram("My dog backs fleas.", Hex);
say pangram("Съешь же ещё этих мягких французских булок, да выпей чаю", Cyr);

Output:

true
false
false
true
true

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

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

Swift

import Foundation

let str = "the quick brown fox jumps over the lazy dog"

func isPangram(str:String) -> Bool {
    let stringArray = Array(str.lowercaseString)
    for char in "abcdefghijklmnopqrstuvwxyz" {
        if (find(stringArray, char) == nil) {
            return false
        }
    }
    return true
}

isPangram(str) // True
isPangram("Test string") // False

Swift 2.0:

func isPangram(str: String) -> Bool {
  let (char, alph) = (Set(str.characters), "abcdefghijklmnopqrstuvwxyz".characters)
  return !alph.contains {!char.contains($0)}
}

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

TI-83 BASIC

:Prompt Str1
:For(L,1,26
:If not(inString(Str1,sub("ABCDEFGHIJKLMNOPQRSTUVWXYZ",L,1))
:L=28
:End
:If L<28
:Disp "IS A PANGRAM"

(not tested yet)

TUSCRIPT

$$ MODE TUSCRIPT,{}
alfabet="abcdefghijklmnopqrstuvwxyz"
sentences = *
DATA The quick brown fox jumps over the lazy dog
DATA the quick brown fox falls over the lazy dog
LOOP s=sentences
 getchars      =STRINGS    (s," {&a} ")
 sortchars     =ALPHA_SORT (getchars)
 reducechars   =REDUCE     (sortchars)
 chars_in_s    =EXCHANGE   (reducechars," '  ")
 IF (chars_in_s==alfabet) PRINT "   pangram: ",s
 IF (chars_in_s!=alfabet) PRINT "no pangram: ",s
ENDLOOP

Output:

   pangram: The quick brown fox jumps over the lazy dog
no pangram: the quick brown fox falls over the lazy dog

TXR

@/.*[Aa].*&.*[Bb].*&.*[Cc].*&.*[Dd].*& \
  .*[Ee].*&.*[Ff].*&.*[Gg].*&.*[Hh].*& \
  .*[Ii].*&.*[Jj].*&.*[Kk].*&.*[Ll].*& \
  .*[Mm].*&.*[Nn].*&.*[Oo].*&.*[Pp].*& \
  .*[Qq].*&.*[Rr].*&.*[Ss].*&.*[Tt].*& \
  .*[Uu].*&.*[Vv].*&.*[Ww].*&.*[Xx].*& \
  .*[Yy].*&.*[Zz].*/

Run:

$ echo "The quick brown fox jumped over the lazy dog." | txr is-pangram.txr -
$echo $? # failed termination
1
$ echo "The quick brown fox jumped over the lazy dogs." | txr is-pangram.txr -
$ echo $?   # successful termination
0

UNIX Shell

Works with: Bourne Again SHell

Works with: Korn Shell

Works with: Z Shell

function is_pangram {
  typeset alphabet=abcdefghijklmnopqrstuvwxyz
  typeset -l string=$*
  while [[ -n $string && -n $alphabet ]]; do
    typeset ch=${string%%${string#?}}
    string=${string#?}
    alphabet=${alphabet/$ch}
  done
  [[ -z $alphabet ]]
}

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>

VBA

The function pangram() in the VBScript section below will do just fine.

Here is an alternative version:

Function pangram2(s As String) As Boolean
    Const sKey As String = "abcdefghijklmnopqrstuvwxyz"
    Dim sLow As String
    Dim i As Integer

    sLow = LCase(s)
    For i = 1 To 26
      If InStr(sLow, Mid(sKey, i, 1)) = 0 Then
        pangram2 = False
        Exit Function
      End If
    Next
    pangram2 = True
End Function

Invocation e.g. (typed in the Immediate window):

print pangram2("the quick brown dog jumps over a lazy fox")
print pangram2("it is time to say goodbye!")

VBScript

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

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")"

VTL-2

10 I=1
20 :I)=0
30 I=I+1
40 #=26>I*20
50 ?="Enter sentence: ";
60 C=$
70 #=C=13*120
80 C=C<97*32+C-96
90 #=C>27*60
100 :C)=1
110 #=60
120 ?=""
130 I=1
140 N=0
150 N=N+:I)
160 I=I+1
170 #=26>I*150
180 #=N=26*200
190 ?="not ";
200 ?="a pangram"

Output:

#=1
Enter sentence: The quick brown fox jumps over the lazy dog.
a pangram

OK
#=1
Enter sentence: This is not a pangram.
not a pangram

OK

Wren

Library: Wren-str

import "./str" for Str

var isPangram = Fn.new { |s|
    s = Str.lower(s)
    var used = List.filled(26, false)
    for (cp in s.codePoints) {
        if (cp >= 97 && cp <= 122) used[cp-97] = true
    }
    for (u in used) if (!u) return false
    return true
}

var candidates = [
    "The quick brown fox jumps over the lazy dog.",
    "New job: fix Mr. Gluck's hazy TV, PDQ!",
    "Peter Piper picked a peck of pickled peppers.",
    "Sphinx of black quartz, judge my vow.",
    "Foxy diva Jennifer Lopez wasn’t baking my quiche.",
    "Grumpy wizards make a toxic stew for the jovial queen."
]

System.print("Are the following pangrams?")
for (candidate in candidates) {
    System.print("  %(candidate) -> %(isPangram.call(candidate))")
}

Output:

Are the following pangrams?
  The quick brown fox jumps over the lazy dog. -> true
  New job: fix Mr. Gluck's hazy TV, PDQ! -> true
  Peter Piper picked a peck of pickled peppers. -> false
  Sphinx of black quartz, judge my vow. -> true
  Foxy diva Jennifer Lopez wasn’t baking my quiche. -> true
  Grumpy wizards make a toxic stew for the jovial queen. -> false

XPL0

include c:\cxpl\codes;          \intrinsic 'code' declarations
string 0;                       \use zero-terminated strings

func StrLen(Str);               \Return number of characters in an ASCIIZ string
char Str;
int  I;
for I:= 0 to -1>>1-1 do
        if Str(I) = 0 then return I;

func Pangram(S);
char S;
int  A, I, C;
[A:= 0;
for I:= 0 to StrLen(S)-1 do
        [C:= S(I);
        if C>=^A & C<=^Z then C:= C or $20;
        if C>=^a & C<=^z then [C:= C - ^a;  A:= A or 1<<C];
        ];
return A = $3FFFFFF;
]; \Pangram

int Sentence, I;
[Sentence:=
    ["The quick brown fox jumps over the lazy dog.",
     "Pack my box with five dozen liquor jugs.",
     "Now is the time for all good men to come to the aid of their country."];
for I:= 0 to 3-1 do
    [Text(0, if Pangram(Sentence(I)) then "yes" else "no");
    CrLf(0);
    ];
]

Output:

yes
yes
no

zkl

var letters=["a".."z"].pump(String); //-->"abcdefghijklmnopqrstuvwxyz"
fcn isPangram(text){(not (letters-text.toLower()))}

Output:

isPangram("The quick brown fox jumps over the lazy dog.")
True
isPangram("Pack my box with five dozen liquor jugs.")
True
isPangram("Now is the time for all good men to come to the aid of their country.")
False