Fisher-Yates 算法解释？

Question

I'm wondering if some of you understand how the Fisher-Yates shuffle works and can explain it to me. so I found this Fisher-Yates Shuffle code online:

public function Main() {
var tempArray:Array = [ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 ]
ShuffleArray(tempArray);
trace(tempArray);
}
public function ShuffleArray(input:Array)
{
for (var i:int = input.length-1; i >=0; i--)
{
var randomIndex:int = Math.floor(Math.random()*(i+1));
var itemAtIndex:Object = input[randomIndex];
input[randomIndex] = input[i];
input[i] = itemAtIndex;
}
}

That code works perfectly but I'm still confused

1. I changed the loop to "input.length" and it doesn't work well, I still got "0" values sometimes. I have no idea why should I use "input.length-1" instead of "input.length"
2. At the randomize section, why should I randomize the index from 0 to the value (i+1), why don't we just randomize it from 0 to (i) instead?

If some of you understand it, can you please explain it to me? Thank you so much

Answer 1

1. Js's array index starts at 0, so array a with length n 's last element is a[n -1] .

2. Math.random return a value from 0 to 0.9999...., but not include 1(range at [0, 1) ), so Math.random()* (i + 1) , would have a value from 0 to i + 0.999999...... , but not i + 1 (range [0, i+1) ), and use Math.floor to cut the dot parts to get a Integer , so we get a number in range [0, i] .

Answer 2

let me explain with an example with negation lets says the array size is 10.

1)if we use index.length line 3 in the for loop will read

input[randomIndex] = input[i] i.e.
input[randomIndex] = input[10];

but since javascript has 0 based arrays ,it has values from index 0 to 9 .index 10 will be out of bounds .Hence we should shuffle from last element(index 9 only)

2)for your second question if we use i instead of i+1. lets say you are in the 1st iteration of the loop for index 9(will hold true for other iterations also). Here i is 9 as seen above .we want 9th index to be shuffled from any one of the indices from 0 to 9 Math.random will return from 0 to .999 and Math.floor will lower bound it so in our case,maximum value will be .999 * (9+1) = 9.99 .Math.floor will lower bound it to 9.So range is [0,9]

incase we used i maximum possible value would be 8 ie, Range[0,8]

Hence we use i+1 since we want values from [0,9]