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Meat math

A Mark Stivers cartoon of 5/21/11 (found via Funny Times):

This is an ambiguity in context, discussed in this blog here:

If you order a hot dog in an eating establishment, you’ll automatically get the sausage in a bun. But if you go to a grocery store to buy hot dogs, they don’t come with buns; these you have to buy separately.

So strictly speaking, hotdog-1 + bun = hotdog-2 (and we can’t conclude that bun = 0, even conceding the ambiguity of + and =).

(Hamburger works the same way.)

A related ambiguity, from the posting linked to above:

If you go to an Olneyville N.Y. System place and order a wiener roll, you expect to get a roll (or bun) with a wiener in it, a wiener in a roll. On the other hand, if you go to a grocery story and WIENER ROLLS are on the shopping list, you’ll be looking for just the buns, and if you want some sort of hot-dog-oid items to put in those buns, that’s a separate trip to a different aisle.

So wiener roll is ambiguous, with the appropriate meaning determined by context.

This entry was posted on June 17, 2012 at 7:31 am and is filed under Ambiguity, Language and food, Linguistics in the comics, Pragmatics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

3 Responses to “Meat math”

  1. arnold zwicky Says:
    June 17, 2012 at 9:36 am | Reply

    An exchange on Facebook:

    Frank McQuarry: One of the requirements in Abstract Algebra for a set to be a group under a binary operator is that the group is closed under that operator. That is to say when the operator is applied to any two members of a set, the result is also a member of the set.

    Thus {hotdog, hotdog, bun} is a group under +.

    Julian Lander: Well, no. While I realize that the joke is hotdog + bun = hotdog, you only get a group if hotdog + hotdog is a member of the set. It does work as a group if buns are considered to be zero, so that what you’ve got is the plain old group of integers with hotdogs as the item being counted. Once you start including buns explicitly, you lose uniqueness of the identity element: both 1 hotdog + 0 hotdog = 1 hotdog and 1 hotdog + 1 bun = 1 hotdog. That implies (if we have a group), that a bun is the zero. (Yes, I like proofs by contradiction, although I feel that constructive proofs have more value.)

    It’s hard to disentangle this discussion, since it’s not clear what the expression hotdog stands for. If the elements of the group are two, HOTDOG and BUN, then HOTDOG + HOTDOG must be an element, either HOTDOG (in which case + is idempotent, and that’s fine) or BUN, but in either case the results require a non-standard interpretation of +.

    The alternative is that the elements of the group are hotdogs and buns, in which case hotdog-x + bun-y is a hotdog, either hotdog-x (in which case bun-y is a zero) or hotdog-z (different from hotdog-x); hotdog-z is at least consistent with the intended interpretation of +. But then what are hotdog-x + hotdog-x and hotdog-x + hotdog-z? This is where the mind starts boggling.

    Group theory is not a good place to go.

  2. Greg Lee Says:
    June 17, 2012 at 12:21 pm | Reply

    Evidently definite anaphora is not sensitive to the “ambiguity” of “hotdog”. “I bought hotdogs at the market and served them to my kids” has the most probably interpretation that the “them” refers to hotdogs-with-buns and the antecedent “hotdog” refers to hogdogs-without-buns. I imagine one could find other similar instances where the name of a dish is taken from its principle ingredient.

    • Ellen K. Says:
      June 18, 2012 at 6:55 am | Reply

      I’m inclined to say “them” refers to the hotdogs without buns, and it is inferred (but not stated) that they are served in buns, simply because that’s the most common serving method and no serving method is indicated. Still, it’s an inference, and what’s directly stated is only that the meat was served.

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