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The thing I find the most difficult is the transitive part. The only way that I know of is to see if the relation is transitive, but I have to do check if it is transitive for every relation possible on a given matrices. It gets quite cumbersome when you have a matrix of 5 elements or more. So is there some sort of shortcut?

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closed as not a real question by Chris Eagle, dtldarek, Henry T. Horton, tomasz, Cameron Buie Dec 9 '12 at 3:40

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What are you actually trying to do? –  Chris Eagle Dec 9 '12 at 0:08
    
Finding the smallest partial order for S={(1,5) (2,3) (2,4) (3,3) (4,3) (5,1)} –  xeonphi Dec 9 '12 at 0:09
    
What on earth does that mean? –  Chris Eagle Dec 9 '12 at 0:10
    
If you are trying to find the smallest partial order containing some relation, look at closure operations. –  dtldarek Dec 9 '12 at 0:11
    
it defines a relation matrix –  xeonphi Dec 9 '12 at 0:11

1 Answer 1

Draw a picture. Make a dot for each of the elements. Whenever you know that $a\prec b$, draw an arrow from $b$ to $a$.

Except that if you know that $a\prec b$ and $b\prec c$, don't bother drawing an arrow for $a\prec c$, because you'll be able to see at a glance that $a\prec c$. Also don't bother to draw the arrow from $a$ to itself, because you'll know they are there anyway.

If you have $a\prec b$ and $b\prec a$ for some $a$ and $b$, this will be apparent, and to make the order into a partial order, you merge $a$ and $b$ into a single dot. Similarly any cycle of arrows $a\prec b\prec c\prec\cdots\prec a$ collapses down to a single point in a partial order.

Then look at the picture. Maximal and minimal elements will be obvious; elements that can't be compared will be obvious; relations implied by transitivity will be obvious.

After you absorb the picture, try drawing it again, and this time when $a\prec b$, instead of drawing an arrow from $b$ to $a$, just try drawing $b$ above $a$ on the paper. For example, here's when we have $a\prec 1, 0\prec c, b\prec 1, c\prec a, 0\prec b$:

some partial order

Clearly 1 is maximal, 0 is minimal, and $b$ is not comparable with $a$ or $c$. Also it's obvious that $0\prec a$ by transitivity.

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+1. Nice answer, and nice approach! –  amWhy Dec 9 '12 at 1:54
    
@amWhy: I'm not sure it really helps the OP in this case, however. –  MJD Dec 9 '12 at 1:57

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