Python set: isdisjoint()

[masiarek]

https://docs.google.com/document/d/1jabPCNDI-OtCBxaEP1HnbpsJsjy4PWhvdDvN73mI0gA/edit?usp=sharing
context Python set: isdisjoint()

I am not expecting 2 to be in end result of this program
# test symmetric difference
s1 = {0, 1, 2, 3}
s2 = {2, 3, 4}
s3 = {2, 5}

print(s1 ^ s2 ^ s3) # result {0, 1, 2, 4, 5} - why is 2 in the end result? what am i missing?

 

[masiarek]

maybe the keyword here '*two* sets'

 

[blamocur]

Here is a general statement (provable by induction) : [imath]x \in \wedge_k s_k[/imath] if and only if [imath]x[/imath] belongs to an odd number of [imath]s_k[/imath]. I.e., every time you add another set containing [imath]x[/imath] to a symmetric difference the following happens:

• [imath]x[/imath] is added to the result if it was not already there
• [imath]x[/imath] is removed from the result if it was already there.

 

[Dr.Peterson]

https://docs.google.com/document/d/1jabPCNDI-OtCBxaEP1HnbpsJsjy4PWhvdDvN73mI0gA/edit?usp=sharing
context Python set: isdisjoint()

I am not expecting 2 to be in end result of this program
# test symmetric difference
s1 = {0, 1, 2, 3}
s2 = {2, 3, 4}
s3 = {2, 5}

print(s1 ^ s2 ^ s3) # result {0, 1, 2, 4, 5} - why is 2 in the end result? what am i missing?

maybe the keyword here '*two* sets'View attachment 30156

That;'s right.

Just evaluate one step at a time: s1^s2^s3 = (s1^s2)^s3, so, first, what is s1^s2?

 

[masiarek]

Thank you - now I get it.