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I asked a question about initializing a 2 dimensional array yesterday, this is the link: How to implement this C++ source in python?

There is a problem in the answer, a friend mentioned a way:

G = [[0]*11]*11

But in this way, when I change the G[0][0] to 2, all the G[i][0](0<=i<11) will all change to 2, but I don't know why?

Supplement:

This is what I thought: The 0 or other number is immutable, so we change one of them, the others will not be changed. But the list [0, 0 ,0 ,.....] is mutable, so when we [0, 0, ...] * 11, all the [0, 0, ...] list will be the same, as is function is True. am I right?

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Because you have 11 references to the same list.

G = [[0] * 11 for x in range(11)]
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The *11 notation makes 11 references to the same object. If the object is immutable you don't notice, because any attempt to change it changes the reference to a different object. When the object is mutable you can modify it, like assigning to a member of a list; since all the references are to the same object, all of them get modified at the same time.

Mutable/immutable might seem to change things, but it doesn't - Python is being consistent in both cases. Consider this example:

G[0] = [3]*11

You'll see that G[1] has not changed.

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sorry, I don't know"Mutable/immutable might seem to change things, but it doesn't - Python is being consistent in both cases" this means? why"since all the references are to the same object, all of them get modified at the same time"

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