What problem is NP-hard but not NP-complete? #
Regex Equivalence.
A regular expression is a way of specifying a pattern that will match some set of strings (and not match all other strings).
Rules:
- A literal character matches that character. So the regex “a” matches the string “a” but no other string.
- A sequence of regular expressions connected directly matches those patterns in that order. So “ab” is two literal characters and matches “a” then “b”.
- A sequence of regular expressions connected by “|” matches either option, so “a|b” matches strings in the set {“a”, “b”}.
- We can use parens for clarity. So “(a|b)(cd)” applies all the previous rules to match strings in the set {“acd”, “bcd”}.
- An expression followed by a star “*”, matches 0 or more of the previous item. That’s one character or a set of parens. This lets us match infinite sets with finite regex.
Why is Regex Equivalence NOT in NP? #
To be NP-complete, a problem must be NP-hard AND it must be in NP.
To be in NP, a “Yes” answer must have a short (polynomial-length) certificate—a proof that we can verify in polynomial time. For 3SAT, the certificate is just the satisfying variable assignment. You plug it in, check the clauses, and you’re done.
But what is the certificate for Regex Equivalence? If I propose the correct
answer is “Yes, these two massive regular expressions match the exact same set
of strings,” how do I prove it to you? Because of the * operator, the set of
matched strings might be infinite. There is no known short certificate you can
hand over to prove two regular expressions are perfectly equivalent in general.
Because we can’t verify it in polynomial time, it’s not in NP. It ends up being in a larger, harder complexity class called PSPACE.
Why is Regex Equivalence NP-Hard? (The Reduction) #
Even though it isn’t in NP, we can prove it is NP-hard by reducing 3SAT to it in polynomial time. Specifically, we will reduce 3SAT to Regex Non-Equivalence.
The Goal:
Given a 3SAT formula $\phi$ with $n$ variables and $m$ clauses, we want to construct two regular expressions, $R_1$ and $R_2$, such that $\phi$ is satisfiable if and only if $R_1 \not\equiv R_2$.
The Setup:
Let any assignment of the $n$ variables be represented as a binary string of
length $n$. For example, if $n=4$, the string 1011 means $x_1=True, x_2=False,
x_3=True, x_4=True$.
Constructing $R_1$:
$R_1$ will simply be the regex that matches all possible variable assignments of length $n$.
R_1 = (0|1)(0|1)...(0|1) // Repeated n times
Constructing $R_2$:
$R_2$ will be a regex designed to match all falsifying assignments—any string that makes the 3SAT formula evaluate to False.
How does a 3SAT formula become false? It happens if at least one clause is false. Let’s look at a single clause: $C_1 = (x_1 \vee \neg x_2 \vee x_4)$. This clause is false only if $x_1=0$, $x_2=1$, and $x_4=0$. The variable $x_3$ can be anything.
So, the regex matching all assignments that falsify $C_1$ is:
R_{C1} = 01(0|1)0
We do this for every clause. Generating the regex for a single clause takes
$O(n)$ time. Now, we combine all the clause regexes with the OR (|) operator
to create $R_2$:
R_2 = R_{C1} | R_{C2} | R_{C3} | ... | R_{Cm}
Because generating each clause takes $O(n)$ time, generating $R_2$ takes $O(m \cdot n)$ time. This is a polynomial time reduction!
The Conclusion: $R_2$ matches all strings that break at least one clause.
- If the 3SAT formula is satisfiable, there is at least one valid assignment (string) that $R_2$ does not match. Therefore, $R_2$ does not match all possible strings, meaning $R_1 \not\equiv R_2$.
- If the 3SAT formula is unsatisfiable, every possible string breaks at least one clause, so $R_2$ matches everything. Therefore, $R_1 \equiv R_2$.
Because we can successfully map 3SAT into a Regex Equivalence question in polynomial time, Regex Equivalence is at least as hard as 3SAT, making it NP-hard!
What about the hardest problems in P? (P-Completeness) #
If NP-Complete problems are the “hardest problems in NP” (likely impossible to solve fast), P-Complete problems are the “hardest problems in P” (likely impossible to parallelize). They represent inherently sequential tasks where you must wait for step $i$ to finish before starting step $i+1$.
Technical Note: To prove P-Completeness, we can’t use polynomial-time reductions. If we did, every problem in P would reduce to every other problem in P (because the reduction itself could just solve the problem!). Instead, we use Log-Space Reductions. The translation algorithm is only allowed a logarithmic amount of memory (just enough for a few loop counters/pointers), ensuring the reduction can’t accidentally solve the problem.
1. The Canonical P-Complete Problem: Circuit Value Problem (CVP) #
The Problem: You are given a boolean circuit made of AND and OR gates (Monotone CVP). The bottom layer has fixed boolean inputs (True/False). Does the final top-level gate output True?
Why it’s in P: It’s trivial to solve. You just do a topological sort and evaluate the gates from bottom to top. It takes $O(n)$ time. Why it’s hard to parallelize: Look at a deep, narrow circuit. Gate 100 relies on Gate 99, which relies on Gate 98. Even with 1,000 CPU cores, you can’t speed this up. It is the ultimate bottleneck.
2. A Thematic Problem: Horn-SAT #
Since we are talking about SAT, let’s look at a restricted version. The Problem: Exactly like Boolean Satisfiability, but every clause can have at most one positive literal.
- Valid: $(\neg x_1 \vee \neg x_2 \vee x_3)$
- Valid: $(\neg x_4 \vee \neg x_5)$
- Valid: $(x_6)$
- INVALID: $(x_1 \vee \neg x_2 \vee x_3)$ (Two positive literals)
Why it’s in P: A Horn clause is just an IF-THEN implication in disguise. By De Morgan’s Laws, $(\neg x_1 \vee \neg x_2 \vee x_3)$ is exactly the same as: $(x_1 \wedge x_2) \rightarrow x_3$
We can solve this in linear time using Unit Propagation:
- Find any clause with just a single positive literal (e.g., $x_6 = True$).
- Plug that into your other clauses (knocking over the dominos).
- If you ever prove a contradiction, it’s Unsatisfiable. Otherwise, Satisfiable.
3. Hardware $\leftrightarrow$ Software (Proving Horn-SAT is P-Complete) #
We can prove Horn-SAT is P-Complete by reducing CVP to it, and vice versa!
CVP $\rightarrow$ Horn-SAT (Logic can simulate Hardware) #
We create a variable for every gate. We want the Horn formula to be Unsatisfiable if the circuit outputs True.
- For an AND gate $g$ with inputs $y, z$: We need $y \text{ AND } z \rightarrow g$. We write the Horn clause: $(\neg x_y \vee \neg x_z \vee x_g)$.
- For an OR gate $g$ with inputs $y, z$: We need $y \rightarrow g$ and $z \rightarrow g$. We write two Horn clauses: $(\neg x_y \vee x_g)$ and $(\neg x_z \vee x_g)$.
- The Output Trick: Add one final clause demanding the output wire is False: $(\neg x_{out})$.
If the circuit naturally evaluates to True, our forward implications force $x_{out}$ to True, violating the final clause and creating a contradiction! (Log-space because we just loop over gates and print clauses).
Horn-SAT $\rightarrow$ CVP (Hardware can unroll Algorithms) #
We can’t write a “loop” in a circuit, but we can unroll the Unit Propagation algorithm into physical layers.
- Build $n$ layers of the circuit.
- Wire up AND/OR gates to represent the implications (e.g., if $(x_1 \wedge x_2) \rightarrow x_3$, wire $x_1$ and $x_2$ from layer $t$ into an AND gate, and feed it into $x_3$ at layer $t+1$).
- At the final layer $n$, check for any contradictions (e.g., a purely negative clause like $\neg x_4 \vee \neg x_7$ failing). Wire all contradiction checks into a massive OR gate.
4. Bridge to NP: Reducing Horn-SAT $\rightarrow$ 3SAT #
Can we reduce a problem in P to an NP-Complete problem? Yes. Because $P \subseteq NP$, we can always use a hard tool to solve an easy problem.
We just use the standard SAT $\rightarrow$ 3SAT reduction:
- Pad short clauses: $(x_1) \rightarrow (x_1 \vee x_1 \vee x_1)$
- Split long clauses with dummy variables: $(\neg x_1 \vee \neg x_2 \vee \neg x_3 \vee x_4) \rightarrow (\neg x_1 \vee \neg x_2 \vee y_1) \wedge (\neg y_1 \vee \neg x_3 \vee x_4)$
Notice that the split clauses are still valid Horn clauses! We just proved that Horn-3SAT is also P-Complete. This requires only log-space to track the dummy variable counter.
5. Class Exercise: Reduce 3SAT $\rightarrow$ Horn-SAT #
For the rest of the lecture, we are going to do the reduction in reverse. We are going to reduce an NP-Complete problem to a P problem. If we do this, we prove P = NP, revolutionize computer science, break all modern encryption, and win a million dollars.
The Goal: Convert this standard 3SAT clause into a set of valid Horn clauses: $$ (x_1 \vee x_2 \vee x_3) $$
Instructor notes / Trap warnings for student ideas:
- Try De Morgan’s Law? $\neg(\neg x_1 \wedge \neg x_2) \rightarrow x_3$. A Horn clause requires a conjunction of positive variables on the left. The left side here is an OR statement in disguise.
- Try Dummy Variables? $(x_1 \vee y) \wedge (x_2 \vee \neg y \vee x_3)$. You got rid of a positive literal in the second clause, but look at the first clause: $(x_1 \vee y)$. It still has two positive literals. It’s not a Horn clause.
The Conceptual Takeaway: Horn clauses represent Determinism. “If A and B, then C.” There are no choices. It’s a straight line of falling dominos. 3SAT represents Nondeterminism (Branching). “Either A is true, or B is true, or C is true.” You have to guess which path to pursue.
Trying to reduce 3SAT to Horn-SAT is the act of trying to force a universe of branching, exponential choices into a single, deterministic straight line.
