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BFS

Tiffany
November 09, 2022

 BFS

ECE 590

Tiffany

November 09, 2022
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Transcript

  1. BFS to find any string accepted 42 Queue 0 Start

    by putting start state+ info into queue Visited is empty set Visited Curr We add extra info to our queue. For this problem, a string “”
  2. BFS to find any string accepted 43 Queue 0 Visited

    Curr How we get newinfo from v + info is problem specific Here, we concatenate info with the symbol on the edge “” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  3. BFS to find any string accepted 44 Queue 0 Visited

    Curr “” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  4. BFS to find any string accepted 45 Queue 0 Visited

    Curr 1 2 3 0 “” “a” “b” “c” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  5. BFS to find any string accepted 46 Queue Visited Curr

    1 2 3 0 “b” “c” “a” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  6. BFS to find any string accepted 47 Queue Visited Curr

    1 2 3 0 1 2 1 1 “a” “b” “c” “aa” “ab” “ac” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  7. BFS to find any string accepted 48 Queue Visited Curr

    2 3 0 1 2 1 1 “b” “c” “aa” “ab” “ac” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  8. BFS to find any string accepted 49 Queue Visited Curr

    2 3 0 1 2 1 1 2 1 4 3 “b” “c” “aa” “ab” “ac” “bb” “ba” “bc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  9. BFS to find any string accepted 50 Queue Visited Curr

    3 0 1 2 1 1 2 1 4 3 “c” “aa” “ab” “ac” “bb” “ba” “bc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  10. BFS to find any string accepted 51 Queue Visited Curr

    3 0 1 2 1 1 2 1 4 3 0 2 3 3 “c” “aa” “ab” “ac” “bb” “ba” “bc” “ca” “cb” “cc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  11. BFS to find any string accepted 52 Queue Visited Curr

    0 1 2 1 1 2 1 4 3 0 2 3 3 “aa” “ab” “ac” “bb” “ba” “bc” “ca” “cb” “cc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  12. BFS to find any string accepted 53 Queue Visited Curr

    0 1 1 1 2 1 4 3 0 2 3 3 “ab” “ac” “bb” “ba” “bc” “ca” “cb” “cc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  13. BFS to find any string accepted 54 Queue Visited Curr

    0 1 1 2 1 4 3 0 2 3 3 “ac” “bb” “ba” “bc” “ca” “cb” “cc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  14. BFS to find any string accepted 55 Queue Visited Curr

    0 1 1 2 4 3 0 2 3 3 “bb” “ba” “bc” “ca” “cb” “cc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  15. BFS to find any string accepted 56 Queue Visited Curr

    0 1 2 4 3 0 2 3 3 For this problem, goal = accepting state, The info is the string to get us there. “bb” “bc” “ca” “cb” “cc” Repeat until goal is met curr, info = dequeue from queue if curr is not in visited add curr to visited for each v where curr->v enqueue v, newinfo
  16. 57 Queue Visited Curr 0 1 2 4 3 0

    2 3 3 BFS will get us the shortest string to get To the accepting state. More generally, it gives a result with the fewest Edges along the path “bb” “bc” “ca” “cb” “cc” BFS to find any string accepted