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
360 Assembly
<lang 360asm>* 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</lang>
- 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
<lang Lisp>(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)))</lang>
ActionScript
<lang ActionScript>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;
}</lang>
Ada
Using character sets
<lang Ada>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; </lang>
Using quantified expressions
<lang Ada>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; </lang>
- Output:
FALSE TRUE TRUE FALSE
ALGOL 68
<lang algol68># 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
)</lang>
- Output:
"Big fjiords vex quick waltz nymph" is a pangram! "The quick brown fox jumps over a lazy dog" is a pangram!
APL
<lang apl>
a←'abcdefghijklmnopqrstuvwxyz' A←'ABCDEFGHIJKLMNOPQRSTUVWXYZ' Panagram←{∧/ ∨⌿ 2 26⍴(a,A) ∊ ⍵} Panagram 'This should fail'
0
Panagram 'The quick brown fox jumps over the lazy dog'
1 </lang>
AppleScript
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:
<lang AppleScript>use framework "Foundation" -- ( for case conversion function )
-- PANGRAM CHECK -------------------------------------------------------------
-- 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 length of filter(charUnUsed, "abcdefghijklmnopqrstuvwxyz") = 0
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</lang>
- Output:
<lang AppleScript>{false, true}</lang>
ATS
<lang 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] *)
(* ****** ****** *) </lang>
An alternate implementation that makes a single pass through the string:
<lang ATS>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
} </lang>
AutoHotkey
<lang 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</lang>
AutoIt
<lang 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 </lang>
AWK
Solution using string-operations
<lang AWK>#!/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;
}</lang>
- Output:
1 0
Solution using associative arrays and split
<lang AWK># 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($0) }
function testPangram(str, c,i,S,H,hit,miss) {
print str ## split( toupper(str), S, "") for (c in S) { H[ S[c] ]++ #print c, S[c], H[ S[c] ] ## } for (i=1; i<=ac; i++) { c = A[i] #printf("%2d %c : %4d\n", i, c, H[c] ) ## if (H[c]) { hit=hit c } else { miss=miss c } } print "# hit:",hit, "# miss:",miss, "." ## if (miss) return 0 return 1
}</lang>
- Output:
# Checking for all 26 chars in 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' : The quick brown fox jumps over the lazy dog. # hit: ABCDEFGHIJKLMNOPQRSTUVWXYZ # miss: . 1 The five boxing wizards dump quickly. # hit: ABCDEFGHIKLMNOPQRSTUVWXYZ # miss: J . 0 Heavy boxes perform waltzes and jigs # hit: ABDEFGHIJLMNOPRSTVWXYZ # miss: CKQU . 0 The quick onyx goblin jumps over the lazy dwarf. # hit: ABCDEFGHIJKLMNOPQRSTUVWXYZ # miss: . 1 Pack my box with five dozen liquor jugs # hit: ABCDEFGHIJKLMNOPQRSTUVWXYZ # miss: . 1
BASIC
<lang qbasic>DECLARE FUNCTION IsPangram! (sentence AS STRING)
DIM x AS STRING
x = "My dog has fleas." GOSUB doIt x = "The lazy dog jumps over the quick brown fox." GOSUB doIt x = "Jackdaws love my big sphinx of quartz." GOSUB doIt x = "What's a jackdaw?" GOSUB doIt
END
doIt:
PRINT IsPangram!(x), x RETURN
FUNCTION IsPangram! (sentence AS STRING)
'returns -1 (true) if sentence is a pangram, 0 (false) otherwise DIM l AS INTEGER, s AS STRING, t AS INTEGER DIM letters(25) AS INTEGER
FOR l = 1 TO LEN(sentence) s = UCASE$(MID$(sentence, l, 1)) SELECT CASE s CASE "A" TO "Z" t = ASC(s) - 65 letters(t) = 1 END SELECT NEXT
FOR l = 0 TO 25 IF letters(l) < 1 THEN IsPangram! = 0 EXIT FUNCTION END IF NEXT
IsPangram! = -1
END FUNCTION</lang>
- Output:
0 My dog has fleas. -1 The quick brown fox jumps over the lazy dog. -1 Jackdaws love my big sphinx of quartz. 0 What's a jackdaw?
Sinclair ZX81 BASIC
Works (just) with the 1k RAM model. The "37" that crops up a couple of times stops being a mystery if we remember that the ZX81 character code for A
is 38 and that strings (like arrays) are indexed from 1, not from 0.
<lang basic> 10 LET A$="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</lang>
- 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
Batch File
<lang dos>@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</lang>
- 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 . . .
BBC BASIC
<lang bbcbasic> 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$</lang>
- Output:
"The quick brown fox jumped over the lazy dog" is not a pangram "The five boxing wizards jump quickly" is a pangram
Befunge
Reads the sentence to test from stdin.
<lang befunge>>~>:65*`!#v_:"`"`48*v>g+04p1\4p ^#*`\*93\`0<::-"@"-*<^40!%2g4:_ "pangram."<v*84<_v#-":"g40\" a" >>:#,_55+,@>"ton">48*>"si tahT"</lang>
- Input:
The quick brown fox jumps over the lazy dog.
- Output:
That is a pangram.
Bracmat
<lang bracmat>(isPangram=
k
. low$!arg:?arg
& a:?k & whl ' ( @(!arg:? !k ?) & chr$(1+asc$!k):?k:~>z ) & !k:>z &
);</lang> 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
<lang 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</lang>
Alternative version:
<lang brat>pangram? = { sentence |
sentence.downcase.dice.unique.select(:alpha?).length == 26
}</lang>
C
<lang 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; }</lang>
Using bitmask
Assumes an execution environment using the ASCII character set (will invoke undefined behavior on other systems).
<lang c>#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; }</lang>
- 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)
<lang csharp>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()); }
}</lang>
Any version of C# language and .NET Framework
<lang csharp>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(); } }
}</lang>
C++
<lang cpp>#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; }
} </lang>
Ceylon
<lang 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)``"); }
}</lang>
Clojure
<lang lisp>(defn pangram? [s]
(let [letters (into #{} "abcdefghijklmnopqrstuvwxyz")] (= (->> s .toLowerCase (filter letters) (into #{})) letters)))</lang>
COBOL
<lang 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.</lang>
CoffeeScript
<lang 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}"
</lang>
Common Lisp
<lang 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))))</lang>
Component Pascal
BlackBox Component Builder <lang oberon2> 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.
</lang>
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?:> $FALSE
Crystal
Copied and modified from the Ruby version.
<lang ruby>def pangram?(sentence)
('a'..'z').all? {|c| sentence.downcase.includes?(c) }
end
p pangram?("not a pangram") p pangram?("The quick brown fox jumps over the lazy dog.")</lang>
false true
D
ASCII Bitmask version
<lang d>bool isPangram(in string text) pure nothrow @safe @nogc {
uint bitset;
foreach (immutable c; text) { if (c >= 'a' && c <= 'z') bitset |= (1u << (c - 'a')); else if (c >= 'A' && c <= 'Z') bitset |= (1u << (c - 'A')); }
return bitset == 0b11_11111111_11111111_11111111;
}
void main() {
assert("the quick brown fox jumps over the lazy dog".isPangram); assert(!"ABCDEFGHIJKLMNOPQSTUVWXYZ".isPangram); assert(!"ABCDEFGHIJKL.NOPQRSTUVWXYZ".isPangram); assert("ABC.D.E.FGHI*J/KL-M+NO*PQ R\nSTUVWXYZ".isPangram);
}</lang>
Unicode version
<lang d>import std.string, std.traits, std.uni;
// Do not compile with -g (debug info). enum Alphabet : dstring {
DE = "abcdefghijklmnopqrstuvwxyzßäöü", EN = "abcdefghijklmnopqrstuvwxyz", SV = "abcdefghijklmnopqrstuvwxyzåäö"
}
bool isPangram(S)(in S s, dstring alpha = Alphabet.EN) pure /*nothrow*/ if (isSomeString!S) {
foreach (dchar c; alpha) if (indexOf(s, c) == -1 && indexOf(s, std.uni.toUpper(c)) == -1) return false; return true;
}
void main() {
assert(isPangram("the quick brown fox jumps over the lazy dog".dup, Alphabet.EN)); assert(isPangram("Falsches Üben von Xylophonmusik quält jeden größeren Zwerg"d, Alphabet.DE)); assert(isPangram("Yxskaftbud, ge vår wczonmö iqhjälp"w, Alphabet.SV));
}</lang>
Delphi
<lang Delphi>program PangramChecker;
{$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.</lang>
E
<lang e>def isPangram(sentence :String) {
return ("abcdefghijklmnopqrstuvwxyz".asSet() &! sentence.toLowerCase().asSet()).size() == 0
}</lang>
&!
is the “but-not” or set difference operator.
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). <lang edsac>
[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. </lang>
- Output:
Y
Elixir
<lang 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}"</lang>
- Output:
true The quick brown fox jumps over the lazy dog. false ABCDEFGHIJKLMNOPQRSUVWXYZ
Erlang
<lang Erlang>-module(pangram). -export([is_pangram/1]).
is_pangram(String) ->
ordsets:is_subset(lists:seq($a, $z), ordsets:from_list(string:to_lower(String))).</lang>
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: <lang fsharp>let isPangram (str: string) = (set['a'..'z'] - set(str.ToLower())).IsEmpty</lang>
Factor
<lang factor>: pangram? ( str -- ? )
[ "abcdefghijklmnopqrstuvwxyz" ] dip >lower diff length 0 = ;
"How razorback-jumping frogs can level six piqued gymnasts!" pangram? .</lang>
Forth
<lang 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</lang>
Fortran
<lang fortran>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</lang> Example: <lang fortran>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</lang>
- Output:
This is a sentence. F The five boxing wizards jumped quickly. T
FreeBASIC
<lang 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</lang>
- 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
Fōrmulæ
In this page you can see the solution of this task.
Fōrmulæ programs are not textual, visualization/edition of programs is done showing/manipulating structures but not text (more info). 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 transportation effects more than visualization and edition.
The option to show Fōrmulæ programs and their results is showing images. Unfortunately images cannot be uploaded in Rosetta Code.
Go
<lang 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 }</lang>
- 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
<lang 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!"</lang>
HicEst
<lang 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</lang>
Icon and Unicon
A panagram procedure: <lang Icon>procedure panagram(s) #: return s if s is a panagram and fail otherwise if (map(s) ** &lcase) === &lcase then return s end</lang>
And a main to drive it: <lang Icon>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</lang>
Io
<lang 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</lang>
Ioke
<lang ioke>Text isPangram? = method(
letters = "abcdefghijklmnopqrstuvwxyz" chars text = self lower chars letters map(x, text include?(x)) reduce(&&)
)</lang>
Here is an example of it's use in the Ioke REPL:
<lang ioke> 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</lang>
J
Solution: <lang j>require 'strings' isPangram=: (a. {~ 97+i.26) */@e. tolower</lang>
Example use: <lang j> isPangram 'The quick brown fox jumps over the lazy dog.' 1
isPangram 'The quick brown fox falls over the lazy dog.'
0</lang>
Java
<lang java5>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 }
}</lang>
- Output:
true false true false false true false
JavaScript
ES5
Iterative
<lang javascript>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</lang>
ES6
Functional
<lang JavaScript>(() => {
'use strict';
// isPangram :: String -> Bool let isPangram = s => { let lc = s.toLowerCase();
return 'abcdefghijklmnopqrstuvwxyz' .split() .filter(c => lc.indexOf(c) === -1) .length === 0; };
// TEST return [ 'is this a pangram', 'The quick brown fox jumps over the lazy dog' ].map(isPangram);
})();</lang>
- Output:
[false, true]
jq
<lang 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</lang>
- 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. <lang Julia>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</lang>
- 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
<lang k>lcase : _ci 97+!26 ucase : _ci 65+!26 tolower : {@[x;p;:;lcase@n@p:&26>n:ucase?/:x]} panagram: {&/lcase _lin tolower x}</lang>
Example: <lang k> panagram "The quick brown fox jumps over the lazy dog" 1
panagram "Panagram test"
0</lang>
Kotlin
<lang scala>// 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 ${if (isPangram(candidate)) "a" else "not a"} pangram")
}</lang>
- Output:
'The quick brown fox jumps over the lazy dog' is a pangram 'New job: fix Mr. Gluck's hazy TV, PDQ!' is a pangram 'A very bad quack might jinx zippy fowls' is a pangram 'A very mad quack might jinx zippy fowls' is not a pangram
Liberty BASIC
<lang lb>'Returns 0 if the string is NOT a pangram or >0 if it IS a pangram string$ = "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</lang>
Logo
<lang 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</lang>
Lua
<lang 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")</lang>
Maple
<lang 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; </lang>
- Usage:
<lang>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.");</lang>
- Output:
true true false
Mathematica
<lang Mathematica>pangramQ[msg_]:=Complement[CharacterRange["a", "z"], Characters[ToLowerCase[msg]]]=== {}</lang> 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: <lang Mathematica>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]]]]</lang>
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
<lang 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</lang>
- Output:
<lang MATLAB>isPangram('The quick brown fox jumps over the lazy dog.')
ans =
1</lang>
MATLAB / Octave
<lang matlab>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</lang>
- Output:
>>isPangram('The quick brown fox jumps over the lazy dog.') ans = 1
min
<lang min>"abcdefghijklmnopqrstuvwxyz" "" split =alphabet ('alphabet dip lowercase (swap match) prepend all?) :pangram?
"The quick brown fox jumps over the lazy dog." pangram? puts</lang>
MiniScript
<lang 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</lang>
- 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
<lang ocaml>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"); </lang>
- 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
NetRexx
NetRexx's verify
built–in method is all you need!
<lang NetRexx>/* 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
</lang>
- 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
<lang 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) </lang>
Nim
<lang 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: ")</lang> Example usage:
Sentence: The quick brown fox jumps over the lazy dog true
Objeck
<lang objeck> 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 } }
} </lang>
OCaml
<lang 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</lang>
<lang ocaml>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.";
- </lang>
- Output:
false -- this is a sentence true -- The quick brown fox jumps over the lazy dog.
Oz
<lang 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."}}</lang>
PARI/GP
<lang parigp>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.")</lang>
Pascal
See Delphi
Perl
Get an answer with a module, or without. <lang perl>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';
}</lang>
- Output:
Yes Yes No No
Phix
<lang 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", ".!$\"AbCdEfghijklmnoprqstuvwxyz", "THE FIVE BOXING WIZARDS DUMP QUICKLY.", "THE FIVE BOXING WIZARDS JUMP QUICKLY.", "HEAVY BOXES PERFORM WALTZES AND JIGS.", "PACK MY BOX WITH FIVE DOZEN LIQUOR JUGS.", "Big fjiords vex quick waltz nymph", "The quick onyx goblin jumps over the lazy dwarf.", "no"}
for i=1 to length(checks) do
string ci = checks[i] integer {r,ch} = pangram(ci) printf(1,"%-50s - %s\n",{ci,iff(r?"yes":"no "&ch)})
end for</lang>
- Output:
The quick brown fox jumped over the lazy dog - no s The quick brown fox jumps over the lazy dog - yes .!$"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
PHP
<lang php>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', '
';</lang>
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 <lang php>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( $txt ) ? "Yes" : "No", PHP_EOL, PHP_EOL;
} </lang>
- Output:
"The quick brown fox jumps over the lazy dog." Yes "The brown fox jumps over the lazy dog." No "ABCDEFGHIJKL.NOPQRSTUVWXYZ" No "ABC.D.E.FGHI*J/KL-M+NO*PQ R STUVWXYZ" Yes "How vexingly quick daft zebras jump" Yes "Is hotdog?" No "How razorback-jumping frogs can level six piqued gymnasts!" Yes
PicoLisp
<lang PicoLisp>(de isPangram (Str)
(not (diff '`(chop "abcdefghijklmnopqrstuvwxyz") (chop (lowc Str)) ) ) )</lang>
PL/I
<lang PL/I> test_pangram: procedure options (main);
is_pangram: procedure() returns (bit(1) aligned);
declare text character (200) varying; declare c character (1);
get edit (text) (L); put skip list (text);
text = lowercase(text);
do c = 'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z'; if index(text, c) = 0 then return ('0'b); end; return ('1'b);
end is_pangram;
put skip list ('Please type a sentence');
if is_pangram() then put skip list ('The sentence is a pangram.'); else put skip list ('The sentence is not a pangram.');
end test_pangram; </lang>
- Output:
Please type a sentence the quick brown fox jumps over the lazy dog The sentence is a pangram.
PowerShell
Cyrillic test sample borrowed from Raku.
<lang PowerShell> function Test-Pangram ( [string]$Text, [string]$Alphabet = 'abcdefghijklmnopqrstuvwxyz' )
{ $Text = $Text.ToLower() $Alphabet = $Alphabet.ToLower() $IsPangram = @( $Alphabet.ToCharArray() | Where-Object { $Text.Contains( $_ ) } ).Count -eq $Alphabet.Length return $IsPangram }
Test-Pangram 'The quick brown fox jumped over the lazy dog.' Test-Pangram 'The quick brown fox jumps over the lazy dog.' Test-Pangram 'Съешь же ещё этих мягких французских булок, да выпей чаю' 'абвгдежзийклмнопрстуфхцчшщъыьэюяё' </lang>
- Output:
False True True
A faster version can be created using .Net HashSet to do what the F# version does: <lang PowerShell> Function Test-Pangram ( [string]$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 $alSet.Count -eq 0 # any alphabet letters still remaining?
} </lang>
Prolog
Works with SWI-Prolog
<lang Prolog>pangram(L) :- numlist(0'a, 0'z, Alphabet), forall(member(C, Alphabet), member(C, L)).
pangram_example :- L1 = "the quick brown fox jumps over the lazy dog", ( pangram(L1) -> R1= ok; R1 = ko), format('~s --> ~w ~n', [L1,R1]),
L2 = "the quick brown fox jumped over the lazy dog", ( pangram(L2) -> R2 = ok; R2 = ko), format('~s --> ~w ~n', [L2, R2]). </lang>
- Output:
?- pangram_example. the quick brown fox jumps over the lazy dog --> ok the quick brown fox jumped over the lazy dog --> ko true.
PureBasic
<lang PureBasic>Procedure IsPangram_fast(String$)
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")</lang>
Python
Using set arithmetic: <lang python>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: ')) )</lang>
- Output:
Sentence: The quick brown fox jumps over the lazy dog True
R
Using the built-in R vector "letters": <lang R>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="") }
}
</lang>
- 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>
- 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") </lang>
Raku
(formerly Perl 6) <lang perl6>constant Eng = set 'a' .. 'z'; constant Cyr = set <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>; 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;</lang>
- Output:
True False False True True
Retro
<lang Retro>'abcdefghijklmnopqrstuvwxyz 'FULL s:const '__________________________ 'TEST s:const
- s:pangram? (s-f)
'__________________________ &TEST #26 copy s:to-lower [ c:letter? ] s:filter [ dup $a - &TEST + store ] s:for-each &TEST &FULL s:eq? ;
</lang>
REXX
<lang REXX>/*REXX program verifies if an entered/supplied string (sentence) is a pangram. */ @abc= 'ABCDEFGHIJKLMNOPQRSTUVWXYZ' /*a list of all (Latin) capital letters*/
do forever; say /*keep promoting 'til null (or blanks).*/ say '──────── Please enter a pangramic sentence (or a blank to quit):'; say pull y /*this also uppercases the Y variable.*/ if y= then leave /*if nothing entered, then we're done.*/ absent= space( translate( @abc, , y), 0) /*obtain a list of any absent letters. */ if absent== then say "──────── Sentence is a pangram." else say "──────── Sentence isn't a pangram, missing: " absent say end /*forever*/
say '──────── PANGRAM program ended. ────────' /*stick a fork in it, we're all done. */</lang>
- output:
──────── Please enter a pangramic sentence (or a blank to quit): The quick brown fox jumped over the lazy dog. ◄■■■■■■■■■■ user input. ──────── Sentence isn't a pangram, missing: S ──────── Please enter a pangramic sentence (or a blank to quit): The quick brown fox JUMPS over the lazy dog!!! ◄■■■■■■■■■■ user input. ──────── Sentence is a pangram. ──────── Please enter a pangramic sentence (or a blank to quit): ◄■■■■■■■■■■ user input (null or some blanks). ──────── PANGRAM program ended. ────────
Ring
<lang ring> pangram = 0 s = "The quick brown fox jumps over the lazy dog." see "" + pangram(s) + " " + s + nl
s = "My dog has fleas." see "" + pangram(s) + " " + s + nl
func pangram str
str = lower(str) for i = ascii("a") to ascii("z") bool = substr(str, char(i)) > 0 pangram = pangram + bool next pan = (pangram = 26) return pan
</lang>
Ruby
<lang ruby>def pangram?(sentence)
('a'..'z').all? {|chars| sentence.downcase.include? (chars) }
end
p pangram?('this is a sentence') # ==> false p pangram?('The quick brown fox jumps over the lazy dog.') # ==> true</lang>
Run BASIC
<lang runbasic>s$ = "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</lang>
1 The quick brown fox jumps over the lazy dog. 0 My dog has fleas.
Rust
<lang 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); }
}</lang>
Scala
<lang scala>def is_pangram(sentence: String) = sentence.toLowerCase.filter(c => c >= 'a' && c <= 'z').toSet.size == 26 </lang>
<lang scala> 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 </lang>
Seed7
<lang 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;</lang>
- Output:
FALSE TRUE TRUE FALSE
Sidef
<lang ruby>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);</lang>
- Output:
true false false true true
Smalltalk
<lang smalltalk>!String methodsFor: 'testing'! isPangram ^((self collect: [:c | c asUppercase]) select: [:c | c >= $A and: [c <= $Z]]) asSet size = 26 </lang>
<lang smalltalk> 'The quick brown fox jumps over the lazy dog.' isPangram </lang>
SNOBOL4
<lang SNOBOL4> 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</lang>
- 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
<lang 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</lang> Swift 2.0:
<lang swift>func isPangram(str: String) -> Bool {
let (char, alph) = (Set(str.characters), "abcdefghijklmnopqrstuvwxyz".characters) return !alph.contains {!char.contains($0)}
}</lang>
Tcl
<lang 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</lang>
TI-83 BASIC
<lang ti83b>:Prompt Str1
- For(L,1,26
- If not(inString(Str1,sub("ABCDEFGHIJKLMNOPQRSTUVWXYZ",L,1))
- L=28
- End
- If L<28
- Disp "IS A PANGRAM"</lang>
(not tested yet)
TUSCRIPT
<lang 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 </lang>
- Output:
pangram: The quick brown fox jumps over the lazy dog no pangram: the quick brown fox falls over the lazy dog
TXR
<lang 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].*/</lang>
- 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 $? # successful termination 0
UNIX Shell
<lang bash>function pangram? {
local alphabet=abcdefghijklmnopqrstuvwxyz local string="$*" string="${string,,}" while -n "$string" && -n "$alphabet" ; do local ch="${string%%${string#?}}" string="${string#?}" alphabet="${alphabet/$ch}" done -z "$alphabet"
}</lang>
Ursala
<lang Ursala>
- import std
is_pangram = ^jZ^(!@l,*+ @rlp -:~&) ~=`A-~ letters </lang> example usage: <lang Ursala>
- cast %bL
test =
is_pangram* <
'The quick brown fox jumps over the lazy dog', 'this is not a pangram'>
</lang>
- Output:
<true,false>
VBA
The function pangram() in the VBScript section below will do just fine.
Here is an alternative version:
<lang vb> 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 </lang>
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
<lang vb>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</lang>
Invocation
<lang vb>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")"</lang>
Wren
<lang ecmascript>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))")
}</lang>
- 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
<lang 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 = $3FFFFFF; ]; \Pangram
int Sentence, I; [Sentence:=
["The quick brown fox jumps over the lazy dog.", "Pack my box with five dozen liquor jugs.", "Now is the time for all good men to come to the aid of their country."];
for I:= 0 to 3-1 do
[Text(0, if Pangram(Sentence(I)) then "yes" else "no"); CrLf(0); ];
]</lang>
- Output:
yes yes no
Yabasic
<lang Yabasic>sub isPangram$(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</lang>
zkl
<lang zkl>var letters=["a".."z"].pump(String); //-->"abcdefghijklmnopqrstuvwxyz" fcn isPangram(text){(not (letters-text.toLower()))}</lang>
- 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
- String manipulation
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