Title
Simple reduction example: \(\probIndSet\) to \(\probClique\)
Recall the the \(k\)-clique problem (clique = set of nodes with edges between all pairs): \[ \probClique = \setbuild{\encoding{G, k}}{ G \text{ is an undirected graph with a } k\text{-clique}} \]
We now introduce a closely related problem, the \(k\)-independent set problem (independent set = set of nodes with edges between no pairs): \[ \probIndSet = \setbuild{\encoding{G, k}}{ G \text{ is an undirected graph with a } k\text{-independent set}} \]
Given input \(\encoding{G_i, k}\) for \(\probIndSet\),
how can we convert it to input \(\encoding{G_c, k}\) for \(\probClique\) such that:\[ \encoding{G_i, k} \in \probIndSet \iff \encoding{G_c, k} \in \probClique \]
Recall: Example reduction to bound complexity of \(\probClique\)
from itertools import combinations as subsets
def reduction_from_independent_set_to_clique(G: Graph[Node], k: int) -> tuple[Graph[Node], int]:
V, E = G
Ec = [ {u,v} for (u,v) in subsets(V,2)
if {u,v} not in E and u!=v ]
Gc = (V, Ec)
return (Gc, k)
# Hypothetical polynomial-time algorithm for Clique
def clique_algorithm(G: Graph[Node], k: int) -> bool:
raise NotImplementedError
# Polynomial-time algorithm for IndependentSet
# that calls clique_algorithm as a subroutine
def independent_set(G: Graph[Node], k: int) -> bool:
Gp, kp = reduction_from_independent_set_to_clique(G, k)
return clique_algorithm(Gp, kp)
- We conclude that \(\probIndSet \le^\P \probClique\).
- \(\probClique\) is at least as hard as \(\probIndSet\).
- Since
reduction_from_independent_set_to_cliquealso happens to work in the other direction
(just complement again!), we can conclude also: \(\probClique \le^\P \probIndSet\).
Implications of \(A \le^\P B\)
Let’s now formalize what we mean by “\(B\) is at least as hard as \(A\)”
Theorem: If \(A \le^\P B\) and \(B \in \P\), then \(A \in \P\). (“if \(B\) is easy, then \(A\) is easy”)
Proof:
Since \(B \in \P\), there is an algorithm \(M_B\) for \(B\) running in time \(O(n^k)\) for some constant \(k\).
Since \(A \le^\P B\), there is a reduction \(f\) from \(A\) to \(B\) running in time \(O(n^c)\) for some constant \(c\).
We can construct an algorithm \(M_A\) deciding \(A\) simply by composing \(f\) and \(M_B\): \[ M_A(x) = M_B(f(x)) \]
What is the time complexity of \(M_A\)? Let \(n = |x|\).
- \(O(n^c)\) for \(f(x)\), plus \(O(m^k)\) for \(M_B(f(x))\), where \(m = |f(x)|\).
- Since \(m = O(|x|^c)\), the full time complexity is: \[ O(n^c) + O(O(|x|^c)^k) \fragment{= O(n^c + n^{ck})} \fragment{= O(n^{ck})} \fragment{\quad \implies \quad A \in \P} \]
Example gadget construction
\[ \phi = (x \lor x \lor y) \land (\overline{x} \lor \overline{y} \lor \overline{y}) \land (\overline{x} \lor y \lor y) \hspace{6em} \]
Is this formula satisfiable?
Yes: \(x = 0, y = 1\).
\(\encoding{\phi} \in \probThreeSAT\) iff we can pick
one literal from each clause
to set to “1” without conflicts!
Reductions between problems of different types: \(\probThreeSAT\)
Theorem: \(\probThreeSAT \le^\P \probIndSet\).
Proof:
Given a 3-CNF formula \(\phi\): \[ \phi = (a_1 \lor b_1 \lor c_1) \land (a_2 \lor b_2 \lor c_2) \land \ldots \land (a_k \lor b_k \lor c_k) \] we produce a pair \((G, k)\) such that: \[ \encoding{\phi} \in \probThreeSAT \; \iff \; \encoding{G, k} \in \probIndSet \]
Set \(k\) equal to the number of clauses in \(\phi\).
- Create one node in \(G\) for each literal in \(\phi\).
- Create edge {u, v} in \(G\) if and only if:
- \(u\) and \(v\) correspond to literals in the same clause (triple), or
- \(u\) is labeled \(x\) and \(v\) is labeled \(\overline{x}\) for some variable \(x\) (or vice versa).
- \(u\) and \(v\) correspond to literals in the same clause (triple), or
We can generate these edges in polynomial time by looping over all pairs of nodes (\(O(|V|^2)\))
and checking the these two conditions by scanning through \(\phi\). (Python code in lecture notes)
Reductions between problems of different types: \(\probThreeSAT\)
Proof that \(\encoding{\phi} \in \probThreeSAT \; \iff \; \encoding{G, k} \in \probIndSet\):
- \(\phi\) satisfiable \(\implies\) \(G\) has a \(k\)-independent set.
- Suppose \(\phi(w) = 1\) (\(w\) is a satisfying assignment)
- Then \(w\) makes at least one literal in each clause true.
- Construct \(k\)-independent set \(S\) by selecting one node from each clause labeled by a true literal. (break ties arbitrarily)
- For every \(u, v \in S\):
- in different triples (so not connected by edge type 1)
- not labeled by variables \(a\) and \(\overline{a}\) respectively (so not connected by edge type 2)
- So \(u\) and \(v\) are not connected by an edge, so \(S\) is a \(k\)-independent set in \(G\)!
- \(G\) has a \(k\)-independent set \(\implies\) \(\phi\) satisfiable
- Given an \(k\)-independent set \(S\), choose assignment \(w\) by setting variable \(a\) to true if some node in \(S\) is labeled \(a\) and false if labeled \(\overline{a}\).
- If some neither literal \(a\) nor \(\overline{a}\) appears in \(S\), set \(a\) arbitrarily.
- Every node in \(S\) must correspond to a different clause due to edge type \(1\).
- Since there are \(k\) clauses and \(|S| = k\), each clause must have at least one true literal marked.
- The edges in \(G\) between nodes for “conflicting” literals mean \(w\) does not set \(a\) and \(\overline{a}\) to true at the same time for any \(a\), so \(w\) is a satisfying assignment for \(\phi\)!
- Given an \(k\)-independent set \(S\), choose assignment \(w\) by setting variable \(a\) to true if some node in \(S\) is labeled \(a\) and false if labeled \(\overline{a}\).
Reductions don’t themselves solve problems!
- A reduction \(f\) is just a (polynomial-time) algorithm
- It enables solving problem \(A\) in terms of an algorithm for problem \(B\).
- However, \(f\) does not itself solve either problem!
- It transforms an instance of problem \(A\) (e.g., a formula for \(\probThreeSAT\))
into an instance of problem \(B\) (e.g., a graph for \(\probIndSet\)). - It preserves the “yes” or “no” answer to the original problem
(but does not provide the answer itself).
- It transforms an instance of problem \(A\) (e.g., a formula for \(\probThreeSAT\))