Letter substitutions termination

Question

Given:

• A char string S length l containing only characters from 'a' to 'z'
• A set of ordered substitution rules R (in the form X->Y) where x, y are single letters from 'a' to 'z' (eg, 'a' -> ' e' could be a valid rule but 'ce'->'abc' would never be a valid rule)

When a rule r in R is applied on S, all letters of S which are equal to the left side of the rule r would be replaced by the letter in the right side of r, if the rule r cause any replacement in S, r is called triggered rule.

Flowchart (Algorithm) :
(1) Alternately apply all rules in R (following the order of rules in R) on S.
(2) While (there exists any 'triggered rule' DURING (1) ) : repeat (1)
(3) Terminate

The question is: Is there any way to determine if with a given string S and set R, the algorithm would terminate or not (running forever)

Example1 : (manually executed)

S = 'abcdef' R = { 'a'->'b' , 'b' -> 'c' }
(the order is implied the order of appearance from left to right of each rule)

Ater running algorithm on S and R:
(1.1): 'abcdef' --> 'bbcdef' --> 'cccdef'
(2.1): repeat (1) because there are 2 replacements during the (1.1)
(1.2): 'cccdef'
(2.2): continue to (3) because there is no replacement during the (1.2)
(3) : terminate the algorithm
=> The algorithm terminate with the given S and R
Example2:

S = 'abcdef' R = { 'a'->'b' , 'b' -> 'a' }
(the order is implied the appearance order from left to right of each rule)

Ater running algorithm on S and R:
(1.1): 'abcdef' --> 'bbcdef' --> 'abcdef'
(2.1): repeat (1) because there are 2 replacements during the (1.1)
(1.2): 'abcdef --> 'bbcdef' --> 'abcdef'
(2.2): repeat (1) because there are 2 replacements during the (1.2)
(1.3): ...... that would be alike (1.1) forever....
The step (3) (terminate) is never reached.
=> The algorithm won't terminate with the given S and R.

• I worked on this and found no efficient algorithm for the question "if the algorithm halts".

• First idea came to my mind was to "find cycle" of letters which are in triggered rules but the number of rules may be too large for this idea to be ideal.

• The second one is to propose a "threshold" for the time of the repeat, if the threshold is exceeded, we conclude the algorithm would not terninate.

• The "threshold" could be choosen randomly, (as long as it big enough) - this approach is not really compelling.

• I am thinking that if there is any upper bound for the "threshold" which ensures that we always get the right answer. And I came up with threshold = 26 where 26 is the number of letter from 'a' to 'z' - but I can't prove that it true (or not). (I hope that It would be something like Bellman-Ford algorithm which determines negative cycle in a fixed number of step,..)

• How about you? Please help me find the answer (this is not a homework)

  Thankyou for reading.

Answer 1

One simple way to think about solving this is to consider a string of length 1 and see if the problem can loop for any given starting letter. Since the string's length is never changing, and applying a rule applies to each character in S independently, it suffices to consider just a string of length 1.

Now, start with a state diagram with 26 states - 1 for each letter of the alphabet. Now, for your state transitions, consider this process:

Apply the transitions from R 1 at a time in order, until you reach the end of R. If from a particular state (letter), you do not ever reach a new letter, you know that if you reach the starting letter, you terminate. Otherwise, after applying the entire sequence of R, you will end up with a new letter. This will be your new state.

Note that all state transitions are deterministic because we apply the entire sequence of R, not just the individual transitions. If we applied the individual transitions, we might get confused, because we might have a -> b, b->a, a->c. When looking at the individual operations, we might think there are two possible transitions from a (either to b or to c), but really, considering the entire sequence, we see definitively that a transitions to c.

You will be done creating your state diagram after considering the next states of each starting letter. Creating the entire state diagram in this manner requires 26 * |R| operations. If the state diagram contains a loop, then if the string S contains any of the letters in the loop, then it fails to halt, otherwise it will halt.

Alternatively, if you just consider halting after 26 iterations through the entire sequence from R, you can use that as well.