Generate a Random Number in the Range 1 - 10

Generating a Random Number between 1 and 10 Java

As the documentation says, this method call returns "a pseudorandom, uniformly distributed int value between 0 (inclusive) and the specified value (exclusive)". This means that you will get numbers from 0 to 9 in your case. So you've done everything correctly by adding one to that number.

Generally speaking, if you need to generate numbers from min to max (including both), you write

random.nextInt(max - min + 1) + min

Generate a random number in the range 1 - 10

If by numbers between 1 and 10 you mean any float that is >= 1 and < 10, then it's easy:

select random() * 9 + 1

This can be easily tested with:

# select min(i), max(i) from (
    select random() * 9 + 1 as i from generate_series(1,1000000)
) q;
       min       |       max
-----------------+------------------
 1.0000083274208 | 9.99999571684748
(1 row)

If you want integers, that are >= 1 and < 10, then it's simple:

select trunc(random() * 9 + 1)

And again, simple test:

# select min(i), max(i) from (
    select trunc(random() * 9 + 1) as i from generate_series(1,1000000)
) q;
 min | max
-----+-----
   1 |   9
(1 row)

generate a random number between 1 and 10 in c

You need a different seed at every execution.

You can start to call at the beginning of your program:

srand(time(NULL));

Note that % 10 yields a result from 0 to 9 and not from 1 to 10: just add 1 to your % expression to get 1 to 10.

Generating random whole numbers in JavaScript in a specific range

There are some examples on the Mozilla Developer Network page:

/**
 * Returns a random number between min (inclusive) and max (exclusive)
 */
function getRandomArbitrary(min, max) {
    return Math.random() * (max - min) + min;
}

/**
 * Returns a random integer between min (inclusive) and max (inclusive).
 * The value is no lower than min (or the next integer greater than min
 * if min isn't an integer) and no greater than max (or the next integer
 * lower than max if max isn't an integer).
 * Using Math.round() will give you a non-uniform distribution!
 */
function getRandomInt(min, max) {
    min = Math.ceil(min);
    max = Math.floor(max);
    return Math.floor(Math.random() * (max - min + 1)) + min;
}

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Here's the logic behind it. It's a simple rule of three:

Math.random() returns a Number between 0 (inclusive) and 1 (exclusive). So we have an interval like this:

[0 .................................... 1)

Now, we'd like a number between min (inclusive) and max (exclusive):

[0 .................................... 1)
[min .................................. max)

We can use the Math.random to get the correspondent in the [min, max) interval. But, first we should factor a little bit the problem by subtracting min from the second interval:

[0 .................................... 1)
[min - min ............................ max - min)

This gives:

[0 .................................... 1)
[0 .................................... max - min)

We may now apply Math.random and then calculate the correspondent. Let's choose a random number:

                Math.random()
                    |
[0 .................................... 1)
[0 .................................... max - min)
                    |
                    x (what we need)

So, in order to find x, we would do:

x = Math.random() * (max - min);

Don't forget to add min back, so that we get a number in the [min, max) interval:

x = Math.random() * (max - min) + min;

That was the first function from MDN. The second one, returns an integer between min and max, both inclusive.

Now for getting integers, you could use round, ceil or floor.

You could use Math.round(Math.random() * (max - min)) + min, this however gives a non-even distribution. Both, min and max only have approximately half the chance to roll:

min...min+0.5...min+1...min+1.5   ...    max-0.5....max
└───┬───┘└────────┬───────┘└───── ... ─────┘└───┬──┘   ← Math.round()
   min          min+1                          max

With max excluded from the interval, it has an even less chance to roll than min.

With Math.floor(Math.random() * (max - min +1)) + min you have a perfectly even distribution.

min.... min+1... min+2 ... max-1... max.... max+1 (is excluded from interval)
|        |        |         |        |        |
└───┬───┘└───┬───┘└─── ... ┘└───┬───┘└───┬───┘   ← Math.floor()
   min     min+1               max-1    max

You can't use ceil() and -1 in that equation because max now had a slightly less chance to roll, but you can roll the (unwanted) min-1 result too.

how to randomly print 1 to 10 in C

The simplest way is to use the modulo operator to cut down the result to the range you want. Doing rand() % 10 will give you a number from 0 to 9, if you add 1 to it, i.e. 1 + (rand() % 10), you'll get a number from 1 to 10 (inclusive).

And before others complain, this may dilute the random distribution, nevertheless, it should work fine for simple purposes.

Generate random integers between 0 and 9

Try random.randrange:

from random import randrange
print(randrange(10))

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