how to generate random numbers from 1 to 4 and 8 in swift
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how to generate random numbers from 1 to 4 and 8 in swift
I'm making an iOS SpriteKit game right now for first time I'm developing any app and it's tough challenge for me because it's really a big project for first time. My game is a Dice game and I defined move as 4 number. So when ever I touch the screen Player moves 4 blocks, But now I want to add Dice to the Game and I need Numbers from 1 to 4 and 8. So it will be 1,2,3,4 and 8 Numbers in the Dice. I know in Swift we can get Random Numbers with "arc4random" but how do I get numbers 1 to 4 and also 8 can I do it with arc4random and if possible I need 4 and 8 numbers to come 20% more often.
Any Help will be really Helpful. Thanks.
See also Generate random numbers with a given distribution
– Martin R
Jul 1 at 10:58
thanks for suggestion
– ramesh sanghar
Jul 1 at 11:00
2 Answers
2
Put your sides into an array and use arc4random_uniform to perform the roll:
sides
arc4random_uniform
let sides = [1, 2, 3, 4, 8]
let roll = sides[Int(arc4random_uniform(UInt32(sides.count)))]
if possible I need 4 and 8 numbers to come 20% more often
That means, 4 and 8 should come up 6 times for every time the others come up 5. So put 5 of each of 1-3 in your array, and 6 each of 4 and 8:
4
8
1
3
4
8
let sides = [1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 8, 8, 8, 8, 8, 8]
let roll = sides[Int(arc4random_uniform(UInt32(sides.count)))]
Also, check out MartinR's randomNumber(probabilities:) function.
randomNumber(probabilities:)
You'd use his function like this:
let sides = [1, 2, 3, 4, 8]
let probabilities = [1.0, 1.0, 1.0, 1.2, 1.2]
let rand = randomNumber(probabilities: probabilities)
let roll = sides[rand]
thank you very much I going to try it
– ramesh sanghar
Jul 1 at 10:56
StackOverFlow and people here are really helpful this is the first time I got help from here first time I didn't know how to ask for it :)
– ramesh sanghar
Jul 1 at 10:59
@rameshsanghar don't forget to mark the right answer as accepted by pressing the ✓ sign on the left side
– Somebody
Jul 1 at 11:04
You can generate a random number from 1-8, and if 5,6,7 comes, you can run the random number function again until an acceptable value.
Its kinda a workaround, but it works because the dataset is small.
thank you so much help I got from here :)
– ramesh sanghar
Jul 1 at 10:57
you can in theory cause your game to freeze with this. What if you have the very unfortunate luck of always randomly getting 5? You will never break out of your loop then.
– Knight0fDragon
Jul 1 at 14:48
@Knight0fDragon The number of iterations required to get a result follows a geometric distribution with parameter 5/8. The expected number of attempts is 1.6, i.e., on average it will take less than 2 attempts. The probability of requiring 30 or more attempts is 1.6631753199735166e-13, less than one in a trillion. You quite literally stand a much better chance of being hit by lightning than having this happen, by several orders of magnitude. If you get 30 or more 5's in a row, you need a new PRNG.
– pjs
Jul 3 at 4:13
@pjs that is still hire than zero, what is your point? Need to look up what “in theory” means I guess.
– Knight0fDragon
Jul 3 at 11:04
@Knight0fDragon Acceptance/rejection techniques are common in random variate generation because their performance can sometimes be better than O(1) alternatives. Marsaglia's polar method for generating Gaussian random variables is a concrete example. It's conceptually the same as the Box-Muller method, but is preferred to the latter because generating x/y pairs over a square and throwing away those that don't fall inside a circumscribed circle is faster than evaluating sines and cosines. The ziggurat method, also a rejection method, is one of the fastest algorithms for Gaussians.
– pjs
Jul 3 at 12:11
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This might be helpful for you - stackoverflow.com/questions/32552336/…
– Md Rashed Pervez
Jul 1 at 10:45