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The paper I'm looking at: Fast approximate string matching with finite automata (2009)

Explanation of the algorithm (from my understanding anyway):

A word is inputted into the automaton and from each state, a number of possible actions can be taken depending on the current symbol/character being looked at at position $pos$. These "actions" include:

  • Staying at the current state and deleting the character currently being pointed at by $pos$.
  • Taking a transition to a next state and substituting the current character at $pos$ with the transition label used.
  • Taking a transition to a next state and inserting the transition label at $pos$.

The action is decides to take is based on a cost function $f=g+h$ (where the lowest $f$ is chosen) for each action.

$g = 0$ if no edit operation is performed on the input word (i.e. the transition label taken to a next state, is the same as the current character at pos) and $g=1$ otherwise.

$h$ is the maximum between two heuristic functions $h(2)$ and $h(\infty)$.

  • $h(2)$ looks at the transition labels that can be reached within two transitions from that state (note, not the current state) and compares those reachable transitions to the next two characters in the input word after $pos$. $h(2)$ gets $+1$ added for each character in the input word that does not match any of the characters that the state $s$ can reach within two transitions.

  • $h(\infty)$ looks at the transition labels that can be reached until the end of the path from that state and compares that to the rest of the input word after pos. As with $h(2)$, $+1$ for each character in the input word that can't be matched to any of the reachable characters from state $s$.

Eventually this algorithm will lead to some accepting state where the path up to that accepting state is the closest accepted word to the input.

I think I understand the algorithm correctly since I tried to implement it from my understanding in Java (I'm doing it for my project, although my project only handles regular expressions of a small alphabet of around 3 different symbols as opposed to a natural language dictionary) and it seems to work to some extent and works fine for the example given in the paper, however my problem is is that I don't understand what priority is used in the case where $f$ values tie for $f=g+h$ (tied $f$ values are actually common).

Taken from page 3 of the paper

The example used in the paper, inputs the word "dat" into the automaton, what I don't understand is why the algorithm goes from state 0 to state 3 (I circled this part in red) when $f=g+h$ is also 1 for the state 0 choice (I circled this part in blue).

The paper mentions

"ties were resolved in the priority queue by giving priority to the highest word position."

but I don't understand what is meant by this.

It also mentioned:

"In our experiments, the strategy of breaking ties so that the node that has advanced farthest in the word to be matched is given priority has proven to be far superior to any of the tiebreaking strategies we explored."

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Those two statements are correlated: "giving priority to the highest word position" is the same as "the node that has advanced farthest in the word is given priority".

This appears to be why the algorithm selects state three. State zero does not result in an advancement in word position, whereas the transition to state three is further in the word position and so satisfies their heuristic of highest word position and thus gets priority for first expansion.

To see why this heuristic results in better results, examine the node zero. Deleting 'd' you see all the letters remaining are in the reachable nodes from zero, which means $h = 0$, and we only performed 1 operation, so $f = 1$. However in order to get to the acceptance state along the shortest path we have to take another operation to go to 3 while adding 'c' back, then we transition through the states 5, and 7 making no changes. Obviously the fastest (and shortest) path is to simply swap 'd' to 'c' and transition to state three first, which the heuristic suggests in this case.

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