BFS

ECE 590

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