# rearranging arguments without using FA

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So you want to re-arrange the order of arguments of your selbri but don't want to use FA, and can't figure out which cmavo from SE to use and in which order to achieve the desired result? This page will tell you how to.

Imagine that you have five selbri with different number of places: x1 broda, x1 x2 brode, x1 x2 x3 brodi, x1 x2 x3 x4 brodo, and x1 x2 x3 x4 x5 brodu. Then here's how to rearrange them in the order you want.

Note that in some cases, more than one possibility exists, even if you rule out non-shortest-length things such as sticking in se se somewhere. In such a case, the alphabetically method picked first.

#### How to use this table

Imagine that you want to use cusku but want to put the quotation after the audience. Then instead of saying mi cusku fi le patfu be mi fe lu na go'i li'u, you'd consider that you want to convert x1 x2 x3 x4 cusku into x1 x3 x2 x4 XXX cusku, where XXX is the desired sequence of SE cmavo.

The table is sorted so that the last sumti place starts at x1 and rises. Since cusku takes four arguments, look up the table under x1 x2 x3 x4 brodo and look for sequences ending with x4 (which will be toward the end since x4 is the largest-numbered argument for four-argument selbri) and find x1 x3 x2 x4 se te se brodo, and you'll know that you could also say mi setese cusku le patfu be mi lu na go'i li'u. Simple!

You could also ignore final places and look for x1 x3 x2 and find se te se brodi; as you see, the sequence of SE is the same. Goodbye FA :)

You could also use this table in the other direction. For example, What does it mean? asks what xelseltervelterklama means. By using the full table (not reproduced here), one can see that x5 x1 x4 x3 x2 xe se te ve te brodu is the same as x1 x2 x3 x4 x5 brodu, or in other words, x1 of xe se te ve te brodu is x5 of brodu, and so on. So xe se te ve te klama has order x5, x1, x4, x3, x2 of klama, or in other words:

1. vehicle
2. go-er
3. route
4. origin
5. destination

(The version chosen in this table for this particular reordering is te ve te xe se brodu. Showing that the two are equivalent is left as an exercise to the reader.)

• x1 broda

#### x1 x2 brode

• x1 x2 brode
• x2 x1 se brode

#### x1 x2 x3 brodi

• x3 x2 x1 te brodi
• x2 x3 x1 se te brodi
• x3 x1 x2 te se brodi
• x1 x3 x2 se te se brodi
• x2 x1 x3 se brodi
• x1 x2 x3 brodi

#### x1 x2 x3 x4 brodo

• x4 x3 x2 x1 se te se ve brodo
• x3 x4 x2 x1 te se ve brodo
• x4 x2 x3 x1 ve brodo
• x2 x4 x3 x1 se ve brodo
• x3 x2 x4 x1 te ve brodo
• x2 x3 x4 x1 se te ve brodo
• x4 x3 x1 x2 ve se te brodo
• x3 x4 x1 x2 se ve se te brodo
• x4 x1 x3 x2 ve se brodo
• x1 x4 x3 x2 se ve se brodo
• x3 x1 x4 x2 te ve se brodo
• x1 x3 x4 x2 se te ve se brodo
• x4 x2 x1 x3 ve te brodo
• x2 x4 x1 x3 se ve te brodo
• x4 x1 x2 x3 ve te se brodo
• x1 x4 x2 x3 se ve te se brodo
• x1 x4 x2 x3 ve te se ve brodo
• x2 x1 x4 x3 se te ve te brodo
• x1 x2 x4 x3 te ve te brodo
• x3 x2 x1 x4 te brodo
• x2 x3 x1 x4 se te brodo
• x3 x1 x2 x4 te se brodo
• x1 x3 x2 x4 se te se brodo
• x2 x1 x3 x4 se brodo
• x1 x2 x3 x4 brodo

#### x1 x2 x3 x4 x5 brodu

• x5 x4 x3 x2 x1 se ve se xe brodu
• x4 x5 x3 x2 x1 ve se xe brodu
• x5 x3 x4 x2 x1 se te ve se xe brodu
• x3 x5 x4 x2 x1 te ve se xe brodu
• x4 x3 x5 x2 x1 ve se te xe brodu
• x3 x4 x5 x2 x1 se ve se te xe brodu
• x5 x4 x2 x3 x1 se ve te se xe brodu
• x4 x5 x2 x3 x1 ve te se xe brodu
• x5 x2 x4 x3 x1 te ve te xe brodu
• x2 x5 x4 x3 x1 se te ve te xe brodu
• x4 x2 x5 x3 x1 ve te xe brodu
• x2 x4 x5 x3 x1 se ve te xe brodu
• x5 x3 x2 x4 x1 se te se xe brodu
• x3 x5 x2 x4 x1 te se xe brodu
• x5 x2 x3 x4 x1 xe brodu
• x2 x5 x3 x4 x1 se xe brodu
• x3 x2 x5 x4 x1 te xe brodu
• x2 x3 x5 x4 x1 se te xe brodu
• x4 x3 x2 x5 x1 se te se ve xe brodu
• x3 x4 x2 x5 x1 te se ve xe brodu
• x4 x2 x3 x5 x1 ve xe brodu
• x2 x4 x3 x5 x1 se ve xe brodu
• x3 x2 x4 x5 x1 te ve xe brodu
• x2 x3 x4 x5 x1 se te ve xe brodu
• x5 x4 x3 x1 x2 xe se ve brodu
• x4 x5 x3 x1 x2 se xe se ve brodu
• x5 x3 x4 x1 x2 xe se te ve brodu
• x3 x5 x4 x1 x2 se xe se te ve brodu
• x4 x3 x5 x1 x2 se te xe se ve brodu
• x3 x4 x5 x1 x2 te xe se ve brodu
• x5 x4 x1 x3 x2 xe se ve te brodu
• x4 x5 x1 x3 x2 se xe se ve te brodu
• x5 x1 x4 x3 x2 te ve te xe se brodu
• x1 x5 x4 x3 x2 se te ve te xe se brodu
• x4 x1 x5 x3 x2 ve te xe se brodu
• x1 x4 x5 x3 x2 se ve te xe se brodu
• x5 x3 x1 x4 x2 xe se te brodu
• x3 x5 x1 x4 x2 se xe se te brodu
• x5 x1 x3 x4 x2 xe se brodu
• x1 x5 x3 x4 x2 se xe se brodu
• x3 x1 x5 x4 x2 te xe se brodu
• x1 x3 x5 x4 x2 se te xe se brodu
• x4 x3 x1 x5 x2 ve xe se te brodu
• x3 x4 x1 x5 x2 se ve xe se te brodu
• x4 x1 x3 x5 x2 ve xe se brodu
• x1 x4 x3 x5 x2 se ve xe se brodu
• x3 x1 x4 x5 x2 te ve xe se brodu
• x1 x3 x4 x5 x2 se te ve xe se brodu
• x5 x4 x2 x1 x3 xe te se ve brodu
• x4 x5 x2 x1 x3 se xe te se ve brodu
• x5 x2 x4 x1 x3 xe te ve brodu
• x2 x5 x4 x1 x3 se xe te ve brodu
• x4 x2 x5 x1 x3 te xe te ve brodu
• x2 x4 x5 x1 x3 se te xe te ve brodu
• x5 x4 x1 x2 x3 se ve se xe te brodu
• x4 x5 x1 x2 x3 ve se xe te brodu
• x5 x1 x4 x2 x3 xe te ve se brodu
• x1 x5 x4 x2 x3 se xe te ve se brodu
• x4 x1 x5 x2 x3 te xe te ve se brodu
• x1 x4 x5 x2 x3 se te xe te ve se brodu
• x5 x2 x1 x4 x3 xe te brodu
• x2 x5 x1 x4 x3 se xe te brodu
• x5 x1 x2 x4 x3 xe te se brodu
• x1 x5 x2 x4 x3 se xe te se brodu
• x2 x1 x5 x4 x3 se te xe te brodu
• x1 x2 x5 x4 x3 te xe te brodu
• x4 x2 x1 x5 x3 ve xe te brodu
• x2 x4 x1 x5 x3 se ve xe te brodu
• x4 x1 x2 x5 x3 ve xe te se brodu
• x1 x4 x2 x5 x3 se ve xe te se brodu
• x2 x1 x4 x5 x3 se te ve xe te brodu
• x1 x2 x4 x5 x3 te ve xe te brodu
• x5 x3 x2 x1 x4 se te se xe ve brodu
• x3 x5 x2 x1 x4 te se xe ve brodu
• x5 x2 x3 x1 x4 xe ve brodu
• x2 x5 x3 x1 x4 se xe ve brodu
• x3 x2 x5 x1 x4 te xe ve brodu
• x2 x3 x5 x1 x4 se te xe ve brodu
• x5 x3 x1 x2 x4 xe ve se te brodu
• x3 x5 x1 x2 x4 se xe ve se te brodu
• x5 x1 x3 x2 x4 xe ve se brodu
• x1 x5 x3 x2 x4 se xe ve se brodu
• x3 x1 x5 x2 x4 te xe ve se brodu
• x1 x3 x5 x2 x4 se te xe ve se brodu
• x5 x2 x1 x3 x4 xe ve te brodu
• x2 x5 x1 x3 x4 se xe ve te brodu
• x5 x1 x2 x3 x4 xe ve te se brodu
• x1 x5 x2 x3 x4 se xe ve te se brodu
• x2 x1 x5 x3 x4 se te xe ve te brodu
• x1 x2 x5 x3 x4 te xe ve te brodu
• x3 x2 x1 x5 x4 te ve xe ve brodu
• x2 x3 x1 x5 x4 se te ve xe ve brodu
• x3 x1 x2 x5 x4 te se ve xe ve brodu
• x1 x3 x2 x5 x4 se te se ve xe ve brodu
• x2 x1 x3 x5 x4 se ve xe ve brodu
• x1 x2 x3 x5 x4 ve xe ve brodu
• x4 x3 x2 x1 x5 se te se ve brodu
• x3 x4 x2 x1 x5 te se ve brodu
• x4 x2 x3 x1 x5 ve brodu
• x2 x4 x3 x1 x5 se ve brodu
• x3 x2 x4 x1 x5 te ve brodu
• x2 x3 x4 x1 x5 se te ve brodu
• x4 x3 x1 x2 x5 ve se te brodu
• x3 x4 x1 x2 x5 se ve se te brodu
• x4 x1 x3 x2 x5 ve se brodu
• x1 x4 x3 x2 x5 se ve se brodu
• x3 x1 x4 x2 x5 te ve se brodu
• x1 x3 x4 x2 x5 se te ve se brodu
• x4 x2 x1 x3 x5 ve te brodu
• x2 x4 x1 x3 x5 se ve te brodu
• x4 x1 x2 x3 x5 ve te se brodu
• x1 x4 x2 x3 x5 se ve te se brodu
• x2 x1 x4 x3 x5 se te ve te brodu
• x1 x2 x4 x3 x5 te ve te brodu
• x3 x2 x1 x4 x5 te brodu
• x2 x3 x1 x4 x5 se te brodu
• x3 x1 x2 x4 x5 te se brodu
• x1 x3 x2 x4 x5 se te se brodu
• x2 x1 x3 x4 x5 se brodu
• x1 x2 x3 x4 x5 brodu

#### Final note

Probably, all possible conversions are covered in here, but if you see one that's missing, or believe you've found a shorter conversion for a given place structure, then feel free to leave comments on the Talk page.

#### Random musings

The largest number of SE cmavo required for varying number of places is as follows:

• 1 place - no cmavo (only one order possible)
• 2 places - 1 cmavo (only one conversion possible with two places: se brode)
• 3 places - 3 cmavo (only one conversion requires three: se te se brodi)
• 4 places - 4 cmavo (e.g. se te se ve brodo for order x4 x3 x2 x1)
• 5 places - 6 cmavo (e.g. se te ve te xe se brodu for order x1 x5 x4 x3 x2)

#### Questions and Answers

• Is there a formula for computing the "maximum minimum" number of SE cmavo for conversions among a given number of place structures? For example, what is the smallest number of SE cmavo required for the re-arrangement of 7 places for the case requiring the largest number of SE cmavo? (Counting sexida as one SE cmavo for such purposes since you'll need subscripted SE cmavo once you get beyond 5.)
• Avital:
Yes. This is a standard question in algebraic combinatorics, or equivalently a group theory question. Stated mathematically, the question is, given the group S_n - the symmetric group on \$n\$ letters, and the generating set A = {(1 2), (1 3), ... (1 n)}, what is the maximal length of the elements of S_n according to the generating set A? Hmm... Now that I think of it, this isn't such a simple question. Let me think about it some more.
• Is there a formula for computing the number of variants of equal shortest lenth possible for a given conversion? For example, se te se brodi can also be te se te brodi, which is two variants. Some conversions of brodu have 24 variants! (All of those require all six cmavo. Does the number of variants depend on the number of SE cmavo required for that particular conversion?)
• Avital:
This question is also related to algebraic combinatorics and group theory, and yet again I do not have an immediate answer.