NFA to DFA conversion -


when converting nfa dfa there may result image below... question is, necessary write state {4} it's going 0 state? mean without showing input symbol 1 of {4} same picture below right? or no?

enter image description here

it’s matter of convention. personally, prefer not clutter dfa unnecessary states, since dfas obtained via transformation nfas tend become quite complex anyway, , since it’s deterministic know non-displayed transition must invalid.

however, i’ve experienced many people in academia teach / use other convention, , require transitions explicitly shown. when working ta (tutor) i’ve had discussion professor – wanted tutors deduct points on final tests missing transitions in dfas convinced him deducting points unfair.


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