Discrete - Relation Properties and Test 2

Created by shockerluke 

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16 terms

Reflexive

iff Ax((x,x) eR)
e.g. S={1,2,3,4}, R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)} (i.e., the <= relation, is reflexive)

Irreflexive

iff Ax((x,x) !eR)
e.g. S={1,2,3,4}, R = {(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)} (i.e., the < relation, is irreflexive)

Symmetric

iff AxAy((x,y)eR -> (y,x)eR)
e.g. S={1,2,3,4},: R={(2,1), (1,2), (2,3), (3,4), (3,2), (4,3)}
This relation is "elements whose absolute difference is 1."

Antisymmetric

iff AxAy((x,y)eR /\ (y,x)eR) -> (x=y)
e.g. S={1,2,3,4}, R = {(1,1), (1,2), (1,3), (1,4), (2,2), (2,3), (2,4), (3,3), (3,4), (4,4)} (i.e., the <= relation)

Asymmetric

iff AxAy((x,y)eR -> (y,x)!eR)
E.g., the relation < ("less than") = {(1,2), (1,3), (1,4), (2,3), (2,4), (3,4)}

Transitive

iff AxAyAz(((x,y)eR /\ (y,z)eR) -> (x,z)eR)
E.g., <=, <, >=, > (think Hypothetical Syllogism)

Implications of Properties

Reflexive -> !Asymmetric
Asymmetric -> Antisymmetric
!Antisymmetric -> !Asymmetric

Solving Linear Homogeneous Recurrence Relations

r^2 - c(sub1) r - c(sub2) = 0
a(sub1)r^n + a(sub2)r(sub2)^n

Solving Linear Homogeneous Recurrence Relations with a single root

r^2 - c(sub1)r - c(sub2) = 0 produces two rs of same val.
a(sub1)r^n + a(sub2)nr^n

r permutations, no repetition

n! / (n - r)!

r combinations, no repetition

n! / [ r! (n-r)! ]

r permutations, repetition ok

n^r

r combinations, repetition ok

(n + r -1)! / [ r! (n-1)! ]

recurrence 1 + 2 + 3 + 4 + 5...

try n(n + 1) / 2

to prove a closed form is a solution

plug closed form into recursive form

Partition Trick for multisets

Given a set of n elements, the number of multisets of size r that can be formed is C(n-1+r, r)=C(n-1+r, n-1).
The n-1 is the number of partitions

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