Regular Expressions with repeated characters - regex

Question

I need to write a regular expression that can detect a string that contains only the characters x,y, and z, but where the characters are different from their neighbors.
Here is an example
xyzxzyz = Pass
xyxyxyx = Pass
xxyzxz = Fail (repeated x)
zzzxxzz = Fail (adjacent characters are repeated)
I thought that this would work ((x|y|z)?)*, but it does not seem to work. Any suggestions?
EDIT
Please note, I am looking for an answer that does not allow for look ahead or look behind operations. The only operations allowed are alternation, concatenation, grouping, and closure

Answer 1

Usually for this type of question, if the regex is not simple enough to be derived directly, you can start from drawing a DFA and derive a regex from there.
You should be able to derive the following DFA. q1, q2, q3, q4 are end states, with q1 also being the start state. q5 is the failed/trap state.
There are several methods to find Regular Expression for a DFA. I am going to use Brzozowski Algebraic Method as explained in section 5 of this paper:
For each state qi, the equation Ri is a union of terms: for a transition a from qi to qj, the term is aRj. Basically, you will look at all the outgoing edges from a state. If Ri is a final state, λ is also one of the terms.
Let me quote the identities from the definition section of the paper, since they will come in handy later (λ is the empty string and ∅ is the empty set):
(ab)c = a(bc) = abc
λx = xλ = x
∅x = x∅ = ∅
∅ + x = x
λ + x* = x*
(λ + x)* = x*
Since q5 is a trap state, the formula will end up an infinite recursion, so you can drop it in the equations. It will end up as empty set and disappear if you include it in the equation anyway (explained in the appendix).
You will come up with:
R1 = xR2 + yR3 + zR4 + λ
R2 = + yR3 + zR4 + λ
R3 = xR2 + + zR4 + λ
R4 = xR2 + yR3 + λ
Solve the equation above with substitution and Arden's theorem, which states:
Given an equation of the form X = AX + B where λ ∉ A, the equation has the solution X = A*B.
You will get to the answer.
I don't have time and confidence to derive the whole thing, but I will show the first few steps of derivation.
Remove R4 by substitution, note that zλ becomes z due to the identity:
R1 = xR2 + yR3 + (zxR2 + zyR3 + z) + λ
R2 = + yR3 + (zxR2 + zyR3 + z) + λ
R3 = xR2 + + (zxR2 + zyR3 + z) + λ
Regroup them:
R1 = (x + zx)R2 + (y + zy)R3 + z + λ
R2 = zxR2 + (y + zy)R3 + z + λ
R3 = (x + zx)R2 + zyR3 + z + λ
Apply Arden's theorem to R3:
R3 = (zy)*((x + zx)R2 + z + λ)
= (zy)*(x + zx)R2 + (zy)*z + (zy)*
You can substitute R3 back to R2 and R1 and remove R3. I leave the rest as exercise. Continue ahead and you should reach the answer.
Appendix
We will explain why trap states can be discarded from the equations, since they will just disappear anyway. Let us use the state q5 in the DFA as an example here.
R5 = (x + y + z)R5
Use identity ∅ + x = x:
R5 = (x + y + z)R5 + ∅
Apply Arden's theorem to R5:
R5 = (x + y + z)*∅
Use identity ∅x = x∅ = ∅:
R5 = ∅
The identity ∅x = x∅ = ∅ will also take effect when R5 is substituted into other equations, causing the term with R5 to disappear.

Answer 2

This should do what you want:
^(?!.*(.)\1)[xyz]*$
(Obviously, only on engines with lookahead)
The content itself is handled by the second part: [xyz]* (any number of x, y, or z characters). The anchors ^...$ are here to say that it has to be the entirety of the string. And the special condition (no adjacent pairs) is handled by a negative lookahead (?!.*(.)\1), which says that there must not be a character followed by the same character anywhere in the string.

Answer 3

I've had an idea while I was walking today and put it on regex and I have yet to find a pattern that it doesn't match correctly. So here is the regex :
^((y|z)|((yz)*y?|(zy)*z?))?(xy|xz|(xyz(yz|yx|yxz)*y?)|(xzy(zy|zx|zxy)*z?))*x?$
Here is a fiddle to go with it!
If you find a pattern mismatch tell me I'll try to modify it! I know it's a bit late but I was really bothered by the fact that I couldn't solve it.

Answer 4

I understand this is quite an old question and has an approved solution as well. But then I am posting 1 more possible and quick solution for the same case, where you want to check your regular expression that contains consecutive characters.
Use below regular expression:
String regex = "\\b\\w*(\\w)\\1\\1\\w*";
Listing possible cases that above expression returning the result.
Case 1: abcdddd or 123444
Result: Matched
Case 2: abcd or 1234
Result: Unmatched
Case 3: &*%$$$ (Special characters)
Result: Unmatched
Hope this will be helpful...
Thanks:)