Objective-C random alphabet no repeat

Question

i made a random for A-Z. The random letter is shown in a label. everything works fine. But the letter should not repeat till every letter from A-Z is called. I´am new in xcode an need a litte help. heres my code in the .m file.

NSString *letters = @"ABCDEFGHIJKLMNOPQRSTUVWXYZ";
-(NSString *) randomStringWithLength:(int) len {

NSMutableString *randomString = [NSMutableString stringWithCapacity: len];

for (int i=26; i<len; i++) {
    [randomString appendFormat: @"%C", [letters characterAtIndex: arc4random() % [letters length]]]; buchstabeAusgabe.text = randomString;
}

return randomString;}



-(void)neuerGenerator {

int text = rand() %26;

switch (text) {
    case 0:
        buchstabeAusgabe.text =@"A";
        break;
    case 1:
        buchstabeAusgabe.text =@"B";
        break;
    case 2:
        buchstabeAusgabe.text =@"C";
        break;
    case 3:
        buchstabeAusgabe.text =@"D";
        break;
    case 4:
        buchstabeAusgabe.text =@"E";
        break;
    case 5:
        buchstabeAusgabe.text =@"F";
        break;
    case 6:
        buchstabeAusgabe.text =@"G";
        break;
    case 7:
        buchstabeAusgabe.text =@"H";
        break;
    case 8:
        buchstabeAusgabe.text =@"I";
        break;
    case 9:
        buchstabeAusgabe.text =@"J";
        break;
    case 10:
        buchstabeAusgabe.text =@"K";
        break;
    case 11:
        buchstabeAusgabe.text =@"L";
        break;
    case 12:
        buchstabeAusgabe.text =@"M";
        break;
    case 13:
        buchstabeAusgabe.text =@"N";
        break;
    case 14:
        buchstabeAusgabe.text =@"O";
        break;
    case 15:
        buchstabeAusgabe.text =@"P";
        break;
    case 16:
        buchstabeAusgabe.text =@"Q";
        break;
    case 17:
        buchstabeAusgabe.text =@"R";
        break;
    case 18:
        buchstabeAusgabe.text =@"S";
        break;
    case 19:
        buchstabeAusgabe.text =@"T";
        break;
    case 20:
        buchstabeAusgabe.text =@"U";
        break;
    case 21:
        buchstabeAusgabe.text =@"V";
        break;
    case 22:
        buchstabeAusgabe.text =@"W";
        break;
    case 23:
        buchstabeAusgabe.text =@"X";
        break;
    case 24:
        buchstabeAusgabe.text =@"Y";
        break;
    case 25:
        buchstabeAusgabe.text =@"Z";
        break;

    default:
        break;
}}

Answer 1

instead of the switch, perhaps store the alphabet in an NSMutableArray. When a letter is taken, remove it from the array. Instead of %26 do %[array count]. To look up the item in array, use [array objectAtIndex:index] where index is the random number.

I am not on XCode at the moment, but I'll try to write out the full code:

- (NSString *) randomStringWithLength:(int) len andAlphabet: (NSString *) alphabet {
    NSMutableArray *alphabetArrayMut = [[self arrayFromString: alphabet] mutableCopy];
    NSMutableString *resultString = [NSMutableString stringWithString:@""];
    while([alphabetArrayMut count]&&[resultString length]<len){
        int index = rand() % [alphabetArrayMut count];
        NSString *charToAdd = [alphabetArrayMut objectAtIndex:index];
        [resultString appendString:charToAdd];
        [alphabetArrayMut removeObjectAtIndex:index];
    }
    return [resultString copy];
}

- (NSArray *) arrayFromString: (NSString *) string{
    NSMutableArray *characters = [[NSMutableArray alloc] initWithCapacity:[string length]];
    for (int i=0; i < [string length]; i++) {
        NSString *ichar  = [NSString stringWithFormat:@"%c", [string characterAtIndex:i]];
        [characters addObject:ichar];
    }
    return [characters copy];
}

Note that it is probably a lot easier to use recursion. Unfortunately, I am not on my mac at the moment, so I can't test it:

- (NSString *) randomStringWithLength:(int) len andAlphabet: (NSString *) alphabet {
    if(len <= 0 || ![alphabet count]){ // base case
         return @"";
    }
    int index = rand() % [alphabet count];
    NSString *chosenLetter = [alphabet substringWithRange:NSMakeRange(index, 1)];
    NSString *newAlphabet = [alphabet stringByReplacingCharactersInRange:NSMakeRange(index, 1) withString:@""];
    NSString *resultString = [NSString stringWithFormat:@"%@%@",chosenLetter,[self randomStringWithLength:len-1,newAlphabet];
    return resultString;
}

Answer 2

because c string is terminal by '\0', we need 27 bytes.

    NSString *alp = @"ABCDEFGHIJKLMNOPQRSTUVWXYZ";

    char cstr[27];

    [alp getCString:cstr maxLength:27 encoding:NSUTF8StringEncoding];

    // do i from 25 to 0. to 1 is ok, also
    for (int i = alp.length - 1; i >= 0; --i) {
        int mark = arc4random_uniform(i);
        char temp = cstr[i];
        cstr[i] = cstr[mark];
        cstr[mark] = temp;
    }

    NSString *str = [NSString stringWithUTF8String:cstr];

    NSLog(@"%@", str);

Answer 3

Lots of different ways to do this. My suggestion would be to use a mutable array:

Add this statement to your .h file:

NSMutableArray *randomLetters;

And then add this method to your .m file:

(Code edited to clean up a ton of typos and minor mistakes)

- (NSString *) randomLetter;
{
  if (randomLetters == nil)
    randomLetters = [NSMutableArray arrayWithCapacity: 26];
  if (randomLetters.count == 0)
  {
    for (unichar aChar = 'A';  aChar <= 'Z'; aChar++)
    {
      [randomLetters addObject: [NSString stringWithCharacters: &aChar length: 1]];
    }
  }
  NSUInteger randomIndex = arc4random_uniform((u_int32_t)randomLetters.count);
  NSString *result =  randomLetters[randomIndex];
  [randomLetters removeObjectAtIndex: randomIndex];
  return result;
}

(Disclaimer: I typed that code out in the SO editor. I haven't tried to compile it, so it may contain minor typos.)

The method randomLetter will give you a single, non-repeating random letter every time you call it, until the array of remaining random letters is empty. At that point it will repopulate the array with the full alphabet and start over.

The random number generator arc4random_uniform() gives much better results that rand(), and doesn't suffer from "modulo bias" (link) like the expression rand()%range does.

Note that it is possible for the above method to give you the last random letter (an "a", for example") then on the next call, repopulate the array, and give you another "a" from the newly populated array. However, the odds of that happening are only 1 in 26.

You could tweak the above code so it remembers the last character it gives you and doesn't give you that same character twice in a row if that's important.

You could pretty easily change it slightly so that it would give you letters one at a time until it's empty and then return nil, and then write a separate method to fill it. That way you could get exactly 26 non-repeating characters and know when you about to repeat with another set of 26 characters.

Answer 4

Given these methods:

- (NSString *)stringOfRandomLettersWithLength:(NSUInteger)length {

    if (length > 26) {
        return nil;
    }

    NSMutableString *stringOfRandomLetters = [NSMutableString stringWithCapacity:length];
    NSArray *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" ];
    NSMutableArray *unusedLetters = [letters mutableCopy];
    NSString *randomLetter;

    for (int i=0; i<length; i++) {
        randomLetter = [self randomItemFromArray:unusedLetters];
        [unusedLetters removeObject:randomLetter];
        [stringOfRandomLetters appendString:randomLetter];
    }

    return stringOfRandomLetters;
}

- (NSString *)randomItemFromArray:(NSArray *)items {

    if (items.count < 1) {
        return nil;
    }

    return items[ arc4random_uniform((u_int32_t)items.count) ];
}

You could create a string of random, distinct letters like this:

NSString *label = [self stringOfRandomLettersWithLength:26];
NSLog(@"label= %@", label);

In the console you'd see something like this:

label= YGRHCXTFDZLKNPAIEOJSUQWVMB