Legacy Documentation: Version 4.3
Unity Manual>User Guide>Creating Gameplay>Adding Random Gameplay Elements
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Adding Random Gameplay Elements

Randomly chosen items or values are important in many games. This sections shows how you can use Unity's built-in random functions to implement some common game mechanics.

Choosing a Random Item from an Array

Picking an array element at random boils down to choosing a random integer between zero and the array's maximum index value (which is equal to the length of the array minus one). This is easily done using the built-in Random.Range function:-

var element = myArray[Random.Range(0, myArray.Length)];

Note that Random.Range returns a value from a range that includes the first parameter but excludes the second, so using myArray.Length here gives the correct result.

Choosing Items with Different Probabilities

Sometimes, you need to choose items at random but with some items more likely to be chosen than others. For example, an NPC may react in several different ways when it encounters a player:-

You can visualise these different outcomes as a paper strip divided into sections each of which occupies a fraction of the strip's total length. The fraction occupied is equal to the probability of that outcome being chosen. Making the choice is equivalent to picking a random point along the strip's length (say by throwing a dart) and then seeing which section it is in.

In the script, the paper strip is actually an array of floats that contain the different probabilities for the items in order. The random point is obtained by multiplying Random.value by the total of all the floats in the array (they need not add up to 1; the significant thing is the relative size of the different values). To find which array element the point is "in", firstly check to see if it is less than the value in the first element. If so, then the first element is the one selected. Otherwise, subtract the first element's value from the point value and compare that to the second element and so on until the correct element is found. In code, this would look something like the following:- 

function Choose(probs: float[]) {
	var total = 0;

	for (elem in probs) {
		total += elem;
	}

	var randomPoint = Random.value * total;

	for (i = 0; i < probs.Length; i++) {
		if (randomPoint < probs[i])
			return i;
		else
			randomPoint -= probs[i];
	}

	return probs.Length - 1;
}

Note that the final return statement is necessary because Random.value can return a result of 1. In this case, the search will not find the random point anywhere. Changing the line

if (randomPoint &lt; probs[i])

...to a less-than-or-equal test would avoid the extra return statement but would also allow an item to be chosen occasionally even when its probability is zero.

Shuffling a List

A common game mechanic is to choose from a known set of items but have them arrive in random order. For example, a deck of cards is typically shuffled so they are not drawn in a predictable sequence. You can shuffle the items in an array by visiting each element and swapping it with another element at a random index in the array:- 

function Shuffle(deck: int[]) {
	for (i = 0; i < deck.Length; i++) {
		var temp = deck[i];
		var randomIndex = Random.Range(0, deck.Length);
		deck[i] = deck[randomIndex];
		deck[randomIndex] = temp;
	}
}

Choosing from a Set of Items Without Repetition

A common task is to pick a number of items randomly from a set without picking the same one more than once. For example, you may want to generate a number of NPCs at random spawn points but be sure that only one NPC gets generated at each point. This can be done by iterating through the items in sequence, making a random decision for each as to whether or not it gets added to the chosen set. As each item is visited, the probability of its being chosen is equal to the number of items still needed divided by the number still left to choose from.

As an example, suppose that ten spawn points are available but only five must be chosen. The probability of the first item being chosen will be 5 / 10 or 0.5. If it is chosen then the probability for the second item will be 4 / 9 or 0.44 (ie, four items still needed, nine left to choose from). However, if the first was not chosen then the probability for the second will be 5 / 9 or 0.56 (ie, five still needed, nine left to choose from). This continues until the set contains the five items required. You could accomplish this in code as follows:- 

var spawnPoints: Transform[];

function ChooseSet(numRequired: int) {
	var result = new Transform[numRequired];

	var numToChoose = numRequired;

	for (numLeft = spawnPoints.Length; numLeft > 0; numLeft--) {
		// Adding 0.0 is simply to cast the integers to float for the division.
		var prob = numToChoose + 0.0 / numLeft + 0.0;

		if (Random.value <= prob) {
			numToChoose--;
			result[numToChoose] = spawnPoints[numLeft - 1];

			if (numToChoose == 0)
				break;
		}
	}

	return result;
}

Note that although the selection is random, items in the chosen set will be in the same order they had in the original array. If the items are to be used one at a time in sequence then the ordering can make them partly predictable, so it may be necessary to shuffle the array before use.

Random Points in Space

A random point in a cubic volume can be chosen by setting each component of a Vector3 to a value returned by Random.value:-

var randVec = Vector3(Random.value, Random.value, Random.value);

This gives a point inside a cube with sides one unit long. The cube can be scaled simply by multiplying the X, Y and Z components of the vector by the desired side lengths. If one of the axes is set to zero, the point will always lie within a single plane. For example, picking a random point on the "ground" is usually a matter of setting the X and Z components randomly and setting the Y component to zero.

When the volume is a sphere (ie, when you want a random point within a given radius from a point of origin), you can use Random.insideUnitSphere multiplied by the desired radius:-

var randWithinRadius = Random.insideUnitSphere * radius;

Note that if you set one of the resulting vector's components to zero, you will *not* get a correct random point within a circle. Although the point is indeed random and lies within the right radius, the probability is heavily biased toward the edge of the circle and so points will be spread very unevenly. You should use Random.insideUnitCircle for this task instead:-

var randWithinCircle = Random.insideUnitCircle * radius;

Page last updated: 2013-07-17

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