I have two text boxes and user can input 2 positive integers (Using Objective-C). The goal is to return a random value between the two numbers.
I've used "man arc4random" and still can't quite wrap my head around it. I've came up with some code but it's buggy.
float lowerBound = lowerBoundNumber.text.floatValue;
float upperBound = upperBoundNumber.text.floatValue;
float rndValue;
//if lower bound is lowerbound < higherbound else switch the two around before randomizing.
if(lowerBound < upperBound)
{
rndValue = (((float)arc4random()/0x100000000)*((upperBound-lowerBound)+lowerBound));
}
else
{
rndValue = (((float)arc4random()/0x100000000)*((lowerBound-upperBound)+upperBound));
}
Right now if I put in the values 0 and 3 it seems to work just fine. However if I use the numbers 10 and 15 I can still get values as low as 1.0000000 or 2.000000 for "rndValue".
Do I need to elaborate my algorithm or do I need to change the way I use arc4random?
You could simply use integer values like this:
int lowerBound = ...
int upperBound = ...
int rndValue = lowerBound + arc4random() % (upperBound - lowerBound);
Or if you mean you want to include float number between lowerBound and upperBound? If so please refer to this question: https://stackoverflow.com/a/4579457/1265516
The following code include the minimum AND MAXIMUM value as the random output number:
- (NSInteger)randomNumberBetween:(NSInteger)min maxNumber:(NSInteger)max
{
return min + arc4random_uniform((uint32_t)(max - min + 1));
}
Update:
I edited the answer by replacing arc4random() % upper_bound with arc4random_uniform(upper_bound) as #rmaddy suggested.
And here is the reference of arc4random_uniform for the details.
Update2:
I updated the answer by inserting a cast to uint32_t in arc4random_uniform() as #bicycle indicated.
-(int) generateRandomNumberWithlowerBound:(int)lowerBound
upperBound:(int)upperBound
{
int rndValue = lowerBound + arc4random() % (upperBound - lowerBound);
return rndValue;
}
You should avoid clamping values with mod (%) if you can, because even if the pseudo-random number generator you're using (like arc4random) is good at providing uniformly distributed numbers in its full range, it may not provide uniformly distributed numbers within the restricted modulo range.
You also don't need to use a literal like 0x100000000 because there is a convenient constant available in stdint.h:
(float)arc4random() / UINT32_MAX
That will give you a random float in the interval [0,1]. Note that arc4random returns an integer in the interval [0, 2**32 - 1].
To move this into the interval you want, you just add your minimum value and multiply the random float by the size of your range:
lowerBound + ((float)arc4random() / UINT32_MAX) * (upperBound - lowerBound);
In the code you posted you're multiplying the random float by the whole mess (lowerBound + (upperBound - lowerBound)), which is actually just equal to upperBound. And that's why you're still getting results less than your intended lower bound.
Objective-C Function:
-(int)getRandomNumberBetween:(int)from to:(int)to
{
return (int)from + arc4random() % (to-from+1);
}
Swift:
func getRandomNumberBetween(_ from: Int, to: Int) -> Int
{
return Int(from) + arc4random() % (to - from + 1)
}
Call it anywhere by:
int OTP = [self getRandomNumberBetween:10 to:99];
NSLog(#"OTP IS %ld",(long)OTP);
NSLog(#"OTP IS %#",[NSString stringWithFormat #"%ld",(long)OTP]);
For Swift:
var OTP: Int = getRandomNumberBetween(10, to: 99)
In Swift:
let otp = Int(arc4random_uniform(6))
Try this.
Related
I'm using a for-loop to determine whether the long double is an int. I have it set up that the for loop loops another long double that is between 2 and final^1/2. Final is a loop I have set up that is basically 2 to the power of 2-10 minus 1. I am then checking if final is an integer. My question is how can I get only the final values that are integers?
My explanation may have been a bit confusing so here is my entire loop code. BTW I am using long doubles because I plan on increasing these numbers very largely.
for (long double ld = 1; ld<10; ld++) {
long double final = powl(2, ld) - 1;
//Would return e.g. 1, 3, 7, 15, 31, 63...etc.
for (long double pD = 2; pD <= powl(final, 0.5); pD++) {
//Create new long double
long double newFinal = final / pD;
//Check if new long double is int
long int intPart = (long int)newFinal;
long double newLong = newFinal - intPart;
if (newLong == 0) {
NSLog(#"Integer");
//Return only the final ints?
}
}
}
Just cast it to an int and subtract it from itself?
long double d;
//assign a value to d
int i = (int)d;
if((double)(d - i) == 0) {
//d has no fractional part
}
As a note... because of the way floating point math works in programming, this == check isn't necessarily the best thing to do. Better would be to decide on a certain level of tolerance, and check whether d was within that tolerance.
For example:
if(fabs((double)(d - i)) < 0.000001) {
//d's fractional part is close enough to 0 for your purposes
}
You can also use long long int and long double to accomplish the same thing. Just be sure you're using the right absolute value function for whatever type you're using:
fabsf(float)
fabs(double)
fabsl(long double)
EDIT... Based on clarification of the actual problem... it seems you're just trying to figure out how to return a collection from a method.
-(NSMutableArray*)yourMethodName {
NSMutableArray *retnArr = [NSMutableArray array];
for(/*some loop logic*/) {
// logic to determine if the number is an int
if(/*number is an int*/) {
[retnArr addObject:[NSNumber numberWithInt:/*current number*/]];
}
}
return retnArr;
}
Stick your logic into this method. Once you've found a number you want to return, stick it into the array using the [retnArr addObject:[NSNumber numberWithInt:]]; method I put up there.
Once you've returned the array, access the numbers like this:
[[arrReturnedFromMethod objectAtIndex:someIndex] intValue];
Optionally, you might want to throw them into the NSNumber object as different types.
You can also use:
[NSNumber numberWithDouble:]
[NSNumber numberWithLongLong:]
And there are matching getters (doubleValue,longLongValue) to extract the number. There are lots of other methods for NSNumber, but these seem the most likely you'd want to be using.
Not to get confused with the NSString sizeWithFont method that returns a CGSize, what I'm looking for is a method that returns an NSString constrained to a certain CGSize. The reason I want to do this is so that when drawing text with Core Text, I can get append an ellipses (...) to the end of the string. I know NSString's drawInRect method does this for me, but I'm using Core Text, and kCTLineBreakByTruncatingTail truncates the end of each line rather than the end of the string.
There's this method that I found that truncates a string to a certain width, and it's not that hard to change it to make it work for a CGSize, but the algorithm is unbelievably slow for long strings, and is practically unusable. (It took over 10 seconds to truncate a long string). There has to be a more "computer science"/mathematical algorithm way to do this faster. Anyone daring enough to try to come up with a faster implementation?
Edit: I've managed to make this in to a binary algorithm:
-(NSString*)getStringByTruncatingToSize:(CGSize)size string:(NSString*)string withFont:(UIFont*)font
{
int min = 0, max = string.length, mid;
while (min < max) {
mid = (min+max)/2;
NSString *currentString = [string substringWithRange:NSMakeRange(min, mid - min)];
CGSize currentSize = [currentString sizeWithFont:font constrainedToSize:CGSizeMake(size.width, MAXFLOAT)];
if (currentSize.height < size.height){
min = mid + 1;
} else if (currentSize.height > size.height) {
max = mid - 1;
} else {
break;
}
}
NSMutableString *finalString = [[string substringWithRange:NSMakeRange(0, min)] mutableCopy];
if(finalString.length < self.length)
[finalString replaceCharactersInRange:NSMakeRange(finalString.length - 3, 3) withString:#"..."];
return finalString;
}
The problem is that this sometimes cuts the string too short when it has room to spare. I think this is where that last condition comes in to play. How do I make sure it doesn't cut off too much?
Good news! There is a "computer science/mathematical way" to do this faster.
The example you link to does a linear search: it just chops one character at a time from the end of the string until it's short enough. So, the amount of time it takes will scale linearly with the length of the string, and with long strings it will be really slow, as you've discovered.
However, you can easily apply a binary search technique to the string. Instead of starting at the end and dropping off one character at a time, you start in the middle:
THIS IS THE STRING THAT YOU WANT TO TRUNCATE
^
You compute the width of "THIS IS THE STRING THAT". If it is too wide, you move your test point to the midpoint of the space on the left. Like this:
THIS IS THE STRING THAT YOU WANT TO TRUNCATE
^ |
On the other hand, if it isn't wide enough, you move the test point to the midpoint of the other half:
THIS IS THE STRING THAT YOU WANT TO TRUNCATE
| ^
You repeat this until you find the point that is just under your width limit. Because you're dividing your search area in half each time, you'll never need to compute the width more than log2 N times (where N is the length of the string) which doesn't grow very fast, even for very long strings.
To put it another way, if you double the length of your input string, that's only one additional width computation.
Starting with Wikipedia's binary search sample, here's an example. Note that since we're not looking for an exact match (you want largest that will fit) the logic is slightly different.
int binary_search(NSString *A, float max_width, int imin, int imax)
{
// continue searching while [imin,imax] is not empty
while (imax >= imin)
{
/* calculate the midpoint for roughly equal partition */
int imid = (imin + imax) / 2;
// determine which subarray to search
float width = ComputeWidthOfString([A substringToIndex:imid]);
if (width < max_width)
// change min index to search upper subarray
imin = imid + 1;
else if (width > max_width )
// change max index to search lower subarray
imax = imid - 1;
else
// exact match found at index imid
return imid;
}
// Normally, this is the "not found" case, but we're just looking for
// the best fit, so we return something here.
return imin;
}
You need to do some math or testing to figure out what's the right index at the bottom, but it's definitely imin or imax, plus or minus one.
I currently have code in objective C that can pull out an integer's most significant digit value. My only question is if there is a better way to do it than with how I have provided below. It gets the job done, but it just feels like a cheap hack.
What the code does is that it takes a number passed in and loops through until that number has been successfully divided to a certain value. The reason I am doing this is for an educational app that splits a number up by it's value and shows the values added all together to produce the final output (1234 = 1000 + 200 + 30 + 4).
int test = 1;
int result = 0;
int value = 0;
do {
value = input / test;
result = test;
test = [[NSString stringWithFormat:#"%d0",test] intValue];
} while (value >= 10);
Any advice is always greatly appreciated.
Will this do the trick?
int sigDigit(int input)
{
int digits = (int) log10(input);
return input / pow(10, digits);
}
Basically it does the following:
Finds out the number of digits in input (log10(input)) and storing it in 'digits'.
divides input by 10 ^ digits.
You should now have the most significant number in digits.
EDIT: in case you need a function that get the integer value at a specific index, check this function out:
int digitAtIndex(int input, int index)
{
int trimmedLower = input / (pow(10, index)); // trim the lower half of the input
int trimmedUpper = trimmedLower % 10; // trim the upper half of the input
return trimmedUpper;
}
i am not sure about my english, but i need to get the unit digit of an integer.
WITHOUT complex algorithm but with some API or another trick.
for example :
int a= 53;
int b=76;
this i add because i almost always dont "meet the quality standards" to post! its drive me crazy! please , fix it ! it took me 10 shoots to post this,and other issue also.
i need to get a=3 and b=6 in a simple smart way.
same about the other digit.
thanks a lot .
here is how to split the number into parts
int unitDigit = a % 10; //is 3
int tens= (a - unitDigit)/10; //is 53-3=50 /10 =5
You're looking for % operator.
a=a%10;//divides 'a' by 10, assigns remainder to 'a'
WARNING
here is how to divine the number into parts
int unitDigit = a % 10; //is 3
int tens= (a - unitDigit)/10; //is 53-3=50 /10 =5
this answer is totally incorrect. It may work only in a number of cases. For example try to get the first digit of 503 via this way
It seems the simplest answer (but not very good in performance):
int a = ...;
int digit = [[[NSString stringWithFormat:#"%d", a] substringToIndex:1] intValue]; //or use substringWithRange to get ANY digit
Modulo operator will help you (as units digit is a reminder when number is divided by 10):
int unitDigit = a % 10;
The following code "gets" the digits of a given number and counts how many of them divide the number exactly.
int findDigits(long long N){
int count = 0;
long long newN = N;
while(newN) // kinda like a right shift
{
int div = newN % 10;
if (div != 0)
if (N % div == 0) count++;
newN = newN / 10;
}
return count;
}
In Xcode /Objective-C for the iPhone.
I have a float with the value 0.00004876544. How would I get it to display to two decimal places after the first significant number?
For example, 0.00004876544 would read 0.000049.
I didn't run this through a compiler to double-check it, but here's the basic jist of the algorithm (converted from the answer to this question):
-(float) round:(float)num toSignificantFigures:(int)n {
if(num == 0) {
return 0;
}
double d = ceil(log10(num < 0 ? -num: num));
int power = n - (int) d;
double magnitude = pow(10, power);
long shifted = round(num*magnitude);
return shifted/magnitude;
}
The important thing to remember is that Objective-C is a superset of C, so anything that is valid in C is also valid in Objective-C. This method uses C functions defined in math.h.