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
   “”
   

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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
   

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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
   

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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
   

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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
   

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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
   

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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
   

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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
   

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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
   

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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
    

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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
    

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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
    

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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
    

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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
    

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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
    

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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
    

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