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Please may explain in below exam how do we compute the red highlighted part in "Problem 3.a"?


in * TF "Intelligent Systems" by (770 points)
Also why from state {p} there is not a path to {q} since its transition relation result is p' XOR q' = (!p'&q') | (p'&!q')?
There is a transition missing from {p} to {q}, you are right, the figure has been fixed meanwhile.
Noted with thanks

1 Answer

+1 vote

You just consider the transition relation ((q⊕p′)∨(p∨q′))∧(p′⊕q′) and use the constant values, i.e.,

p=0 and q=0 : ((0⊕p′)∨(0∨q′))∧(p′⊕q′) = (p′∨q′)∧(p′⊕q′) = p′⊕q′
p=0 and q=1 : ((1⊕p′)∨(0∨q′))∧(p′⊕q′) = (!p′∨q′)∧(p′⊕q′) = (p'?0:q') = !p′∧q′
p=1 and q=0 : ((0⊕p′)∨(1∨q′))∧(p′⊕q′) = p′⊕q′
p=1 and q=1 : ((1⊕p′)∨(1∨q′))∧(p′⊕q′) = p′⊕q′
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