In Latex, in math mode, if I want to express summation over a range I can use the following expression \sum_{from}^{to}. I can do the same for the product.

What is the name of the symbol that does this for XOR or concatenation?

If I do:


I don't get the i=0 and 7 parts below and above the symbol, respectively, but to the right instead, like this:


With summation (\sum_{i=0}^7), that is not the case - they appear below and below in the output:


How can I make XOR, or concatenation (II) larger and with indices below and above the symbol?

3 Answers 3


You want to use \bigoplus instead of \oplus.

I've never seen concatenation done that way. Addition and XOR are commutative operations so it makes sense to sum over a set (or take the exclusive OR of a set). Concatenation is not like that. I think I would explicitly write out the concatenation. That said, you can use \bigparallel from the stmaryrd package.

\[x_1\concat x_2\concat\dotsb\concat x_n\]
\[\bigparallel_{i=1}^n x_i\]

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  • I saw these guys using concatenation summation: schneier.com/skein.pdf, pages 17,18.
    – axel22
    Jun 14, 2011 at 9:25
  • @axel22: Interesting. I'd never seen that before. Let me update my answer.
    – TH.
    Jun 14, 2011 at 9:34
  • 2
    I don’t see why the non-commutativity of the concatenation should disqualify it for this notation. In functional programming, a reduction of character strings via concatenation is a well-defined operation, and defined in the same way as a sum. We’re not talking about sets here, we’re talking about (well-ordered) sequences. Jun 14, 2011 at 12:19
  • 2
    @Konrad: Fair enough. One might write $\sum_{x\in S}x$. A similar expression would be meaningless for concatenation. That's all I was saying. In this case, you're right that there's a canonical well order imposed by the indexing.
    – TH.
    Jun 14, 2011 at 12:39

The usual way to get a larger \oplus symbol that takes limits above and below in display math mode is with \bigoplus. However this symbol might appear too big; a not-so-large symbol can be obtained by


For a concatenation big symbol one can do a similar thing:


Now \bigconc will behave like \sum:

\[ \bigconc_{i=0}^{3} X_{i} \]

You can use the \DeclareMathOperator* command that defines operator with super/subscripts above/below itself:

\[ \OPLUS^a_b \]

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