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Comparison of two nodes fails about nir HOT 8 CLOSED

neuromorphs avatar neuromorphs commented on July 24, 2024
Comparison of two nodes fails

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Comments (8)

sheiksadique avatar sheiksadique commented on July 24, 2024

In the NIRTorch implementation of Graph and Nodes each node has a name attribute attached to it.

But the equality of two nodes is determined by the element/torch module it holds which makes it easier to deal with.

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matjobst avatar matjobst commented on July 24, 2024

In the end it boils down to what equality should mean for nodes. Are two nodes equal when all their attributes are the same? When they have the same connectivity? Or if they are the identical object?

In my opinion, we should use the identity, meaning they are the same object. That prevents this:

a = nir.Input(shape=np.array([1]))
b = nir.Input(shape=np.array([1]))
a == b   # True

The implementation of using identity for comparisons would be very simple, just overwriting the default eq function:

@dataclass(eq=False)
class NIRNode:
    """Base superclass of Neural Intermediate Representation Unit (NIR).

    All NIR primitives inherit from this class, but NIRNodes should never be
    instantiated.
    """
    def __eq__(self, other):
        return self is other

With this, the following works as I would expect:

a = nir.Input(shape=np.array([1]))
b = nir.Input(shape=np.array([1]))
c = nir.Output(shape=np.array([1]))
a == a   # True
a == b   # False
a == c   # False

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Jegp avatar Jegp commented on July 24, 2024

NIR is explicitly declarative, meaning that a NIR node declares a type of computation. In that sense, I would actually argue that two NIR nodes, doing the same thing should be equal.

That said, I also agree that we two nodes in a graph should not be equal. But, the simplest way to achieve that would be to consider a "node in a graph" differently from simply a "node" in that nodes in a specific graph have names. So the comparison would really be (name1, node1) == (name2, node2).

Would you agree with that distinction @matjobst?

If so, the only remaining question is how to deal with the implementation. I can imagine two ways to address that

  1. Allowing a == b to be true, and relying on the position/label in the graph to test for object differences
  2. Adding the name attribute in to each NIRNode, in which case equality solves the problem since a == b will automatically check for the name

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sheiksadique avatar sheiksadique commented on July 24, 2024

Where do we need to compare two nodes for equality in practice? We will need to answer this to be able to clearly define equality operation. In general I agree with @Jegp

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DylanMuir avatar DylanMuir commented on July 24, 2024

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matjobst avatar matjobst commented on July 24, 2024

@Jegp I agree with this distinction. I currently can not think of a use case where you would need the first type of equality.
For the second one, I agree with @DylanMuir and would add that a simple statement such as "node in graph" requires equality checks according to the second equality type.
The implementation of the second type does not actually require us to use IDs, see my previous suggestion in PR #30 :

The implementation of using identity for comparisons would be very simple, just overwriting the default eq function:

@dataclass(eq=False)
class NIRNode:
    """Base superclass of Neural Intermediate Representation Unit (NIR).

    All NIR primitives inherit from this class, but NIRNodes should never be
    instantiated.
    """
    def __eq__(self, other):
        return self is other

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Jegp avatar Jegp commented on July 24, 2024

I see the point, and I agree that we should expect collisions to happen often. @matjobst would you be up for giving this a go? Perhaps in #30 ?

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stevenabreu7 avatar stevenabreu7 commented on July 24, 2024

closed with #30

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