4 ms·

I think you are confusing commutativity in general with commutativity of binary operations.

by jonsen 8y ago

BeetleB 8y ago
We are now lost in a semantic discussion, as it is clear that all parties agree on the content, just not the terminology. In that spirit, I will disagree. In mathematics, commutativity is always about two operands. My textbook on floating point arithmetic[1] (probably the most famous one) states that addition and multiplication in floating point is commutative. They use examples similar to yours as an example of associativity being violated, and point out that commutativity is preserved. Putting FP aside: In mathematics, no one disputes that addition is commutative. Yet, they all agree that for infinite sums, if you rearrange the order of the terms, you can get different results. They don't say that addition is commutative only for finite sums. They say it is commutative for all addition, and that associativity is the property that fails for infinite sums. See the discussion here[2] for example: >The commutative property of addition is that a+b=b+a. Technically, it applies only to sums of two numbers! >Actually, commutivity does hold for infinite sums. Glossing over a few technicalities, commutivity says that you get the same answer whenever you interchange any two terms in a sum. Although I'll grant that when I look at some other answers in StackExchage, they do generalize to "repeated use of commutativity", and that if commutativity is applied a finite number of times, the sum is preserved, but not if applied an infinite number of times. The common refrain in both cases, though, is that commutativity applies only to two operands. [1] https://www.springer.com/us/book/9780817647056 https://www.springer.com/us/book/9780817647056 [2] https://math.stackexchange.com/questions/646665/why-does-commutativity-of-addition-fail-for-infinite-sums https://math.stackexchange.com/questions/646665/why-does-com...
- jonsen 8y ago
  Take subtraction. The binary operator - is not commutative in the sense that a - b is not equal to b - a. But a - b commutes to - b + a. Any number of +or-with-operand can be arranged in any order. This is commutativity in general. See the start of Chapter 1 in Part 1: https://onlinebooks.library.upenn.edu/webbin/book/lookupid?key=olbp36404 https://onlinebooks.library.upenn.edu/webbin/book/lookupid?k... This is a very practical semantics of commutativity when programming. It’s unfortunately lost in most modern math expositions.
  - 0815test 8y ago
    But this relies on associativity - without associativity, you just don't have a consistent definition of e.g. (a + b + c + d) but only, e.g. + + + / \ / \ / \ + + a + a + / \ / \ / \ / \ a b c d + d b + / \ / \ b c c d and others as well. There would be practically no point in trying to find a definition of "commutative" for these structures that don't involve associativity.
    - jonsen 8y ago
      -6-7-8 == -8-7-6 == -7-8-6 == ... How does this rely on associativity? It does not rely on parenthetical associativity. The operator is of course associated with an operand, but that’s not what is meant by associativity.
      - dragonwriter 8y ago
        > How does this rely on associativity? It does not rely on parenthetical associativity. It's exactly the combination of commutivity of addition, associativity of addition, and equivalence of subtraction with addition of the additive inverse, that allows the general rearrangement of symbols you are discussing.
      - deleted 8y ago
        [deleted]
    - dllthomas 8y ago
      I think there may be a mismatch in terminology here. Above, "associativity" was used in the sense of "the property that any re-association produces the same answer". It's used that way when talking about properties of an operation, often in an algebraic context. I think you're using it in the sense of "an understanding of how operands should be associated". It's often used that way when describing a language in practice, like "(+) is right-associative". If we're going to be working with expressions like (a + b + c), then whenever + is not (sense 1) associative we clearly need some understanding about what that expression means. But we can restrict ourselves to dealing with fully-parenthesized expressions and not need any sort of "this associates to the left", and still properties like associativity and commutativity can be interesting. In the case of your trees, associativity means any trees with the same ordering in the leaves (with an in-order traversal) must be equivalent. Commutativity means any trees that differ by swapping the left/right children of a parent node must be equivalent. You can have neither property, either property, or both properties.

I think you are confusing commutativity in general with commutativity of binary operations.

by jonsen 8y ago

I think you are confusing commutativity in general with commutativity of binary operations.

We are now lost in a semantic discussion, as it is clear that all parties agree on the content, just not the terminology. In that spirit, I will disagree. In mathematics, commutativity is always about two operands. My textbook on floating point arithmetic[1] (probably the most famous one) states that addition and multiplication in floating point is commutative. They use examples similar to yours as an example of associativity being violated, and point out that commutativity is preserved. Putting FP aside: In mathematics, no one disputes that addition is commutative. Yet, they all agree that for infinite sums, if you rearrange the order of the terms, you can get different results. They don't say that addition is commutative only for finite sums. They say it is commutative for all addition, and that associativity is the property that fails for infinite sums. See the discussion here[2] for example: >The commutative property of addition is that a+b=b+a. Technically, it applies only to sums of two numbers! >Actually, commutivity does hold for infinite sums. Glossing over a few technicalities, commutivity says that you get the same answer whenever you interchange any two terms in a sum. Although I'll grant that when I look at some other answers in StackExchage, they do generalize to "repeated use of commutativity", and that if commutativity is applied a finite number of times, the sum is preserved, but not if applied an infinite number of times. The common refrain in both cases, though, is that commutativity applies only to two operands. [1] https://www.springer.com/us/book/9780817647056 https://www.springer.com/us/book/9780817647056 [2] https://math.stackexchange.com/questions/646665/why-does-commutativity-of-addition-fail-for-infinite-sums https://math.stackexchange.com/questions/646665/why-does-com...

jonsen 8y ago
Take subtraction. The binary operator - is not commutative in the sense that a - b is not equal to b - a. But a - b commutes to - b + a. Any number of +or-with-operand can be arranged in any order. This is commutativity in general. See the start of Chapter 1 in Part 1: https://onlinebooks.library.upenn.edu/webbin/book/lookupid?key=olbp36404 https://onlinebooks.library.upenn.edu/webbin/book/lookupid?k... This is a very practical semantics of commutativity when programming. It’s unfortunately lost in most modern math expositions.
- 0815test 8y ago
  But this relies on associativity - without associativity, you just don't have a consistent definition of e.g. (a + b + c + d) but only, e.g. + + + / \ / \ / \ + + a + a + / \ / \ / \ / \ a b c d + d b + / \ / \ b c c d and others as well. There would be practically no point in trying to find a definition of "commutative" for these structures that don't involve associativity.
  - jonsen 8y ago
    -6-7-8 == -8-7-6 == -7-8-6 == ... How does this rely on associativity? It does not rely on parenthetical associativity. The operator is of course associated with an operand, but that’s not what is meant by associativity.
    - dragonwriter 8y ago
      > How does this rely on associativity? It does not rely on parenthetical associativity. It's exactly the combination of commutivity of addition, associativity of addition, and equivalence of subtraction with addition of the additive inverse, that allows the general rearrangement of symbols you are discussing.
    - deleted 8y ago
      [deleted]
  - dllthomas 8y ago
    I think there may be a mismatch in terminology here. Above, "associativity" was used in the sense of "the property that any re-association produces the same answer". It's used that way when talking about properties of an operation, often in an algebraic context. I think you're using it in the sense of "an understanding of how operands should be associated". It's often used that way when describing a language in practice, like "(+) is right-associative". If we're going to be working with expressions like (a + b + c), then whenever + is not (sense 1) associative we clearly need some understanding about what that expression means. But we can restrict ourselves to dealing with fully-parenthesized expressions and not need any sort of "this associates to the left", and still properties like associativity and commutativity can be interesting. In the case of your trees, associativity means any trees with the same ordering in the leaves (with an in-order traversal) must be equivalent. Commutativity means any trees that differ by swapping the left/right children of a parent node must be equivalent. You can have neither property, either property, or both properties.