CODEWITHSUNDEEP
c#
Aurelius Schnitzler
Aurelius Schnitzler

Reputation: 531

How to create a random increment that is a multiple of 5?

I create a random Time randomTime between two TimeSpans using

Random random = new Random();
TimeSpan start = TimeSpan.FromHours(StundenS);
TimeSpan end = TimeSpan.FromHours(StundenE);
int maxMinutes = (int)((end - start).TotalMinutes);
int minutes = random.Next(maxMinutes);
TimeSpan randomTime = start.Add(TimeSpan.FromMinutes(minutes));

If StundenS = 10 and StundenE = 11, the result could be 10:51, which is correct. Yet how do I make it change in 5 minute increments instead of 1 minute increment, so 10:51 is forbidden, 10:50 is not and 10:55 is not either?

EDIT: I solved it with

minutes = (int)(Math.Round((double)minutes / 5) * 5);
TimeSpan randomTime = start.Add(TimeSpan.FromMinutes(minutes));

Upvotes: 3

Views: 2809

Answers (2)

coding4life
coding4life

Reputation: 39

You could give the random.Next an array which consists of increments of 5s.

int[]  array = new int {5,10,15,...,55};
int minutes = random.Next(0, array.Length);
int randomnumber = array[minutes];

Upvotes: 0

BradleyDotNET
BradleyDotNET

Reputation: 61359

Reduce your range and multiply.

It is very easy to get a random integer between two values (say, 0 and 10). In such a case 10 discrete values are possible outcomes.

Now that I can only get one of 10 numbers, I can multiply to get the correct scale. So to get any multiple of 10 from 0 to 100 I could write:

int rand = random.Next(0, 10);
int scaledRand = rand * 10;

If I then wanted values between 50 and 150 (still that 100 range notice) you simply add:

int finalRand = scaledRand + 50;

Similarly, you have a range of 60 with multiples of 5. Dividing gives a requirement of 12 discrete values so the code:

int rand = random.Next(0, 12);
int scaledRand = rand * 5;

Gives you the multiples you want.

Upvotes: 5

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