i used arc4random to create a random number, is there a way to tell arc4random to begin at for example -5 instead of 0? because i want to create a random number in the range of
-3,4 to 4,3, im not that good in iOS developing yet, so what other possiblities do i have if that wont work with arc4random (Links are appreciated if theres a guide or something like that)
First of all, use arc4random_uniform to get a random number in the desired absolute range (for -3 to 4 it would be 7): arc4random_uniform(7).
You might also see the form arc4random() % max, but that will introduce a modulo bias making the distribution less random, arc4random_uniform is prefered.
Afterwards, adjust your lower bound:
arc4random_uniform(7) - 3
apple docs
Since arc4random() % n returns an integer from 0..n-1, arc4random() % (n - k) + k returns one from k..n-1. Plug in k=-5. Is that what you need?
If you want it to start at -5 instead of 0, just subtract 5.
If you want to create a random number in the range of -3 to 4, just create a number from 0 to 7 and subtract 3 from it.
The arc4random function gives you a number from zero to 4,294,967,295. To change that to a number from 0 to 7, just divide by 613,566,756. (Or use arc4random_uniform to avoid any bias.)
Related
I don't have much coding experience so I don't really know of an efficient alternative to modulo, the issue I have is that I want to have the same funcionality but witouth it ever returning zero if that makes sense.
So I have an arbritary value % 8 and I want my results to go (1,2,3,4,5,6,7,8,1,2,3,etc)
any help or push in the right direction would be appreciated.
I assume you're trying to make indices from 1 to 8 loop. For zero-based offsets from 0 to 7 this would be trivial by using i % 8; consider simply making your table zero-based.
For one-based indices, the simplest way to go is to first subtract 1 to make it zero-based, then apply the modulo to wrap around, then add 1 to make it one-based again: ((i - 1) % 8) + 1.
So I have an arbritary value % 8 and I want my results to go
(1,2,3,4,5,6,7,8,1,2,3,etc)
local result = value % 8 + 1
This is a simple maths problem. If one arrithmetic operator doesn't give you the desired result, use or add others to your formula.
HI I know there have been may question about this here but I wasn't able to find a detailed enough answer, Wikipedia has two examples of ISIN and how is their checksum calculated.
The part of calculation that I'm struggling with is
Multiply the group containing the rightmost character
The way I understand this statement is:
Iterate through each character from right to left
once you stumble upon a character rather than digit record its position
if the position is an even number double all numeric values in even position
if the position is an odd number double all numeric values in odd position
My understanding has to be wrong because there are at least two problems:
Every ISIN starts with two character country code so position of rightmost character is always the first character
If you omit the first two characters then there is no explanation as to what to do with ISINs that are made up of all numbers (except for first two characters)
Note
isin.org contains even less information on verifying ISINs, they even use the same example as Wikipedia.
I agree with you; the definition on Wikipedia is not the clearest I have seen.
There's a piece of text just before the two examples that explains when one or the other algorithm should be used:
Since the NSIN element can be any alpha numeric sequence (9 characters), an odd number of letters will result in an even number of digits and an even number of letters will result in an odd number of digits. For an odd number of digits, the approach in the first example is used. For an even number of digits, the approach in the second example is used
The NSIN is identical to the ISIN, excluding the first two letters and the last digit; so if the ISIN is US0378331005 the NSIN is 037833100.
So, if you want to verify the checksum digit of US0378331005, you'll have to use the "first algorithm" because there are 9 digits in the NSIN. Conversely, if you want to check AU0000XVGZA3 you're going to use the "second algorithm" because the NSIN contains 4 digits.
As to the "first" and "second" algorithms, they're identical, with the only exception that in the former you'll multiply by 2 the group of odd digits, whereas in the latter you'll multiply by 2 the group of even digits.
Now, the good news is, you can get away without this overcomplicated algorithm.
You can, instead:
Take the ISIN except the last digit (which you'll want to verify)
Convert all letters to numbers, so to obtain a list of digits
Reverse the list of digits
All the digits in an odd position are doubled and their digits summed again if the result is >= 10
All the digits in an even position are taken as they are
Sum all the digits, take the modulo, subtract the result from 0 and take the absolute value
The only tricky step is #4. Let's clarify it with a mini-example.
Suppose the digits in an odd position are 4, 0, 7.
You'll double them and get: 8, 0, 14.
8 is not >= 10, so we take it as it is. Ditto for 0. 14 is >= 10, so we sum its digits again: 1+4=5.
The result of step #4 in this mini-example is, therefore: 8, 0, 5.
A minimal, working implementation in Python could look like this:
import string
isin = 'US4581401001'
def digit_sum(n):
return (n // 10) + (n % 10)
alphabet = {letter: value for (value, letter) in
enumerate(''.join(str(n) for n in range(10)) + string.ascii_uppercase)}
isin_to_digits = ''.join(str(d) for d in (alphabet[v] for v in isin[:-1]))
isin_sum = 0
for (i, c) in enumerate(reversed(isin_to_digits), 1):
if i % 2 == 1:
isin_sum += digit_sum(2*int(c))
else:
isin_sum += int(c)
checksum_digit = abs(- isin_sum % 10)
assert int(isin[-1]) == checksum_digit
Or, more crammed, just for functional fun:
checksum_digit = abs( - sum(digit_sum(2*int(c)) if i % 2 == 1 else int(c)
for (i, c) in enumerate(
reversed(''.join(str(d) for d in (alphabet[v] for v in isin[:-1]))), 1)) % 10)
I'm trying to find the last digit of the result of any number raised to any power, using binomial theorem, not modulus or something. Please explain me why last digit of a number's unit number raised to a power is same as the original number raised to the same power using binomial theorem.
Ex. XV^Y = V^Y
Also, I found out that each integer each its cyclicity and I understand that. But I'm confused since:
17^8 = 7^8 = 7^4 since 8 is a multiple of 4.
But why not 7^2 = 7^8 as well? 8 is also a multiple of 2.
It's because of the last digit that you are raising to a power several times and not about the power.
7^1=...7 <=
7^2=...9
7^3=...3
7^4=...1
7^5=...7 <=
7^6=...9
7^7=...3
7^8=...1
7^9=...7 <=
Say you have a number x=t*10+u, where t is the "tens" and u is the units, so e.g. 1234=123*10+4. The binomial theorem states: x^n = sum{k=0,...,n} (t*10)^(n-k)*u^k. As long as (n-k)>0, the summand will be a multiple of 10. You should be able to figure it out from there.
I have this values that i've logged:
label.frame.size.height :18.000000, label.font.lineHeight: 17.895000
if i use roundf() like:
roundf(label.frame.size.height / label.font.lineHeight) // answer: 1
while with ceil()
ceil(label.frame.size.height / label.font.lineHeight) // answer: 2
but when computed manually: answer is 1.00586756
I wonder whats the best and more reliable(generally) between this two. Why is everybody using ceil() to determine the number of lines of UILabel?
In the case of number of lines each letter after the limit a line could display should be taken to next line so .005 is also significant this .005 part of the text should carry to next line. So it is better to use ceil() rather than roundf( ). In roundf( ) a value will be significant only when it is greater or equal to its half value)
ceil()
The C library function ceil(x) returns the smallest integer value greater than or equal to x.
I still dont understand why must of the people use ceil() when computing the number of line since roundf() is more accurate..
But when talking about computing for the number of line.. i look to me that 'roundf()' is indeed more accurate, but since its number of lines.. decimal values are not significant..
Computing what is the image:
54 / 17.895000 = 3.01760268
And numberOflines = 3
if we use roundf() answer would be 3 as well
while if ceil() is already 4
therefore using floor() or simply converting the result to int will do the work:
int result = (int)floor(answer);
//or
int result = (int)answer;
About my question, i think roundf() to the work for me for computing number of lines generally..
I'm making a class that will compute the number of line base from this values, and will be used by the whole app.
What is the best way to generate random numbers using Objective-C on iOS?
If I use (int)((double) rand() / ((double)(RAND_MAX) + (double) 1) * 5.0) to generate a number from 0 to 4, every time I start the program on the iPhone it generates the same numbers to start off with.
There is a very similar question here on StackOverFlow. Here is one of the better solutions (no need for seeding):
int r = arc4random() % 5;
i use
#define RANDOM_SEED() srandom(time(NULL))
#define RANDOM_INT(__MIN__, __MAX__) ((__MIN__) + random() % ((__MAX__+1) - (__MIN__)))
so you can give a min and max
You should seed the random number generator with the current time.
srand(time(0));
How random do you need? If you want random enough for crypto, then use SecRandomCopyBytes().
Call srand() at the start of your program, it'll reseed random number generator
Simple function for random number generation:
int r = arc4random() % 42; // generate number up to 42 (limit)