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.

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One comment

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

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