Utilizing the random function in a loop can result in unpredictable and unexpected values

Question

Utilizing the random function in a loop can result in unpredictable and unexpected values

Why is it that I get numbers ranging from 17 to 70,000 in the three console.log statements?

However, in the loop, y always seems to fall between 200 and 800. Why is that?

console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)

let y=0;
for (let x = 0; x < Math.floor(Math.random()*5000)*17; x++) {
  y++
}
console.log("y " + y)

Does random() work differently in for loops? Is there an alternative approach to achieve the desired output?

javascript

Answer 1

Answer №1

One reason for this is that the Math.... in your loop is evaluated each time it runs.

When you use it as the condition to stop looping, it means that typically, it takes around 200 to 800 attempts for your entire expression to evaluate to a value less than the current x. The higher the iteration, the more likely it is to exit the loop.

You might have intended to:

console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)

let y=0;
for (let x = 0, iterations = Math.floor(Math.random()*5000)*17; x < iterations; x++) {
    y++
}
console.log("y " + y)

This will create the iterations variable only once and use it throughout the full loop cycle.

If you want to understand what’s happening in the original code, run the following:

let y = 0;
let limit;
for (let x = 0; x < (limit = Math.floor(Math.random() * 5000) * 17); x++) {
  console.log(`${x} < ${limit} : ${x<limit}`);
  y++
}
console.log("y " + y, 'last limit was', limit);

Answer 2

One reason for this is that the Math.... in your loop is evaluated each time it runs.

When you use it as the condition to stop looping, it means that typically, it takes around 200 to 800 attempts for your entire expression to evaluate to a value less than the current x. The higher the iteration, the more likely it is to exit the loop.

You might have intended to:

console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)

let y=0;
for (let x = 0, iterations = Math.floor(Math.random()*5000)*17; x < iterations; x++) {
    y++
}
console.log("y " + y)

This will create the iterations variable only once and use it throughout the full loop cycle.

If you want to understand what’s happening in the original code, run the following:

let y = 0;
let limit;
for (let x = 0; x < (limit = Math.floor(Math.random() * 5000) * 17); x++) {
  console.log(`${x} < ${limit} : ${x<limit}`);
  y++
}
console.log("y " + y, 'last limit was', limit);

Answer 3

Answer №2

While your for loop may not directly result in a normal distribution, it does lead to an interesting pattern of outcomes resembling one. The ending condition being based on a random number introduces variability into the loop's termination point, making it unpredictable.

This setup creates a scenario akin to a bell curve, where certain values are more likely to be reached than others. The chances of the loop ending within a specific range, such as 200 to 800, are notably higher compared to extreme values outside of this range.

The underlying logic involves the incrementing value x and its relationship with the randomly generated comparison number. Initially, hitting the target can be challenging due to low x values requiring a very small random number. However, as x grows, the likelihood of meeting the condition increases since each roll has a chance to match the current x value.

Eventually, the probability of success rises as it becomes statistically more probable for a sufficiently low random number to occur and surpass the increasing x value. This trend continues until the loop inevitably meets its end based on the randomness of the generator.

Answer 4

While your for loop may not directly result in a normal distribution, it does lead to an interesting pattern of outcomes resembling one. The ending condition being based on a random number introduces variability into the loop's termination point, making it unpredictable.

This setup creates a scenario akin to a bell curve, where certain values are more likely to be reached than others. The chances of the loop ending within a specific range, such as 200 to 800, are notably higher compared to extreme values outside of this range.

The underlying logic involves the incrementing value x and its relationship with the randomly generated comparison number. Initially, hitting the target can be challenging due to low x values requiring a very small random number. However, as x grows, the likelihood of meeting the condition increases since each roll has a chance to match the current x value.

Eventually, the probability of success rises as it becomes statistically more probable for a sufficiently low random number to occur and surpass the increasing x value. This trend continues until the loop inevitably meets its end based on the randomness of the generator.

Answer 5

Answer №3

Imagine being asked this question during a job interview! 😄 What's even more intriguing is the ability to retrieve the generated value as demonstrated in the code snippet below:

console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)

let y=0;
for (let x = 0; b = Math.floor(Math.random()*5000)*17; x++) {
    console.log(x, y, b)
    y++
}
console.log("y " + y, b)

It turns out that the strange behavior is due to the fact that 'b' evaluates to zero during one iteration because it is considered a falsy value. With the precision of the numbers and how they are generated, any number smaller than 0.0002 can lead to unpredictable outcomes with such calculations. This serves as a reminder that relying solely on Math.floor(Math.random()*X) for generating random numbers may not be the most secure approach.

Answer 6

Imagine being asked this question during a job interview! 😄 What's even more intriguing is the ability to retrieve the generated value as demonstrated in the code snippet below:

console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)
console.log("RND " + Math.floor(Math.random()*5000)*17)

let y=0;
for (let x = 0; b = Math.floor(Math.random()*5000)*17; x++) {
    console.log(x, y, b)
    y++
}
console.log("y " + y, b)

It turns out that the strange behavior is due to the fact that 'b' evaluates to zero during one iteration because it is considered a falsy value. With the precision of the numbers and how they are generated, any number smaller than 0.0002 can lead to unpredictable outcomes with such calculations. This serves as a reminder that relying solely on Math.floor(Math.random()*X) for generating random numbers may not be the most secure approach.

Utilizing the random function in a loop can result in unpredictable and unexpected values

Answer №1

Answer №2

Answer №3

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