# Symmetric and Cyclic Assumptions

It is quite noticeable than this word – “WLOG” (=Without Loss Of Generality) is strewn across the net and books …. This post shall convey its real meaning.

(1) If a function – $f(x,y,z)$ is symmetric

$\implies f(x,y,z) = f(x,z,y) = f(y,x,z) = f(y,z,x) = f(z,x,y) = f(z,y,x)$.

If for some permutation – $f(a,b,c)$ with $a \ge b \ge c$

Since the function is symmetric,

$f(a,b,c) = f(x,y,z)$

Therefore , Proving,

$f(x,y,z) \ge P \Leftrightarrow f(a,b,c) \ge P$ with $a \ge b \ge c$

Thus a pairwise order can be assumed in symmetric functions..

(2) If a function – $f(x,y,z)$ is cyclic

Since the function is only cyclic , we only have –

$f(x,y,z) = f(y,z,x) = f(z,x,y)$

We can either have –

$x \ge y,z$

\$latex y \ge z,x\$

$z\ge x,y$

Let $(x,y,z) = (a,b,c)$ with $a \ge b,c$

$(y,z,x) = (a',b',c')$ with $a' \ge b',c'$

$(z,x,y) = (a'',b'',c'')$ with $a'' \ge b'',c''$

Thus , proving ,

$f(x,y,z) \ge P \Leftrightarrow f(a,b,c) \ge P$

where $(a,b,c)$ is any cyclic permutation of $(x,y,z)$ with $a \ge b,c$

P.S. Thanks to Tejs Mathias for making me understand this.

### One comment

1. can someone please tell why atleast one latex code doesnt get converted every post!!!!????