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given is a reduction from 4-SAT to 5-SAT. How is it possible to describe such a function? I found some informations about reduction 3-SAT to 4-SAT here, but it can't help me so much.

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    $\begingroup$ Can you define your version of $k$-SAT? $\endgroup$ – Yuval Filmus May 21 at 10:29
  • $\begingroup$ k-SAT = { F | F is a satisfiable Boolean formula in k-CNF } $\endgroup$ – samTT May 21 at 10:31
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    $\begingroup$ What’s $k$-CNF for you? $\endgroup$ – Yuval Filmus May 21 at 10:57
  • $\begingroup$ Can you also explain what you don't understand in the linked answer? The reduction from 4-SAT to 5-SAT is essentially the same as the reduction from 3-SAT to 4-SAT. $\endgroup$ – Yuval Filmus May 21 at 14:54
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    $\begingroup$ To make @YuvalFilmus's question more specific, does "$k$-CNF" mean that each clause has at most $k$ literals, or exactly $k$ literals? $\endgroup$ – David Richerby May 21 at 21:45
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Here is a translation of the answer how to prove 4-SAT CNF is NP-complete to the current situation.


Suppose an instance of 4-SAT over variables $x_1,x_2,\cdots, x_m$ is given as a boolean formula $$f=c_1\land c_2\land \cdots\land c_m,$$ where $c_i$ is a disjunction that has exactly 4 literals for all $i$.

Introduce a new variable $s$. Suppose $c_i=w\lor x \lor y \lor z$ for some literal $w,x,y,z$. Let $$\begin{aligned} c_{i+}&=w\lor x \lor y \lor z\lor s\\ c_{i-}&=w\lor x \lor y \lor z\lor \neg s. \end{aligned}$$

  • If $c_i$ can be satisfied by an assignment, then $c_{i+}\land c_{i-}$ is satisfied by the same assignment plus $s=0$ (or $s=1$).
  • If $c_{i+}\land c_{i-}$ can be satisfied by an assignment, the same assignment without considering $s$ must satisfy $c_i$ since one of $s$ and $\neg s$ must be false.

Construct an instance of 5-SAT over variables $x_1,x_2,\cdots, x_m, s$, $$g=c_{1+}\land c_{1-}\land c_{2+}\land c_{2-}\land\cdots\land c_{m+}\land c_{m-}.$$

Because of the relation between $c_i$ and $c_{i+}\land c_{i-}$,

  • If $f$ is satisfied by an assignment, then $g$ is satisfied by the same assignment plus $s=0$.
  • Conversely, if $g$ is satisfied by an assignment, then $f$ is satisfied by the same assignment without considering $s$.

The above means the transformation from $f$ to $g$ is a reduction from 4-SAT to 5-SAT. It runs in polynomial time.


For simplicity of explanation, the above restricts $k$-SAT to CNF formulas with exactly $k$ literals without duplicates. Thanks to a similar simple padding technique, it is does not matter to the construction of reduction above whether we allow at most $k$ literals with or without duplicates when we define $k$-SAT.

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  • $\begingroup$ very nice answer, Thank you very much! $\endgroup$ – samTT May 22 at 11:38
  • $\begingroup$ @samTT, welcome. $\endgroup$ – Apass.Jack May 22 at 13:12
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Since you did not specify exactly the two languages, let us suppose that 4-SAT is a CNF of this type: (A ∨ B ∨ C ∨ D) ∧ (A¯ ∨ C¯ ∨ F ∨ S) ∧ (E ∨ G ∨ H¯ ∨ Q) ∧ .... A way to reduce this instance of k-SAT into an instance of (k+1)-SAT is this: you can add inside every clause a further literal (for example Z) so the first clause will become (A ∨ B ∨ C ∨ D ∨ Z), the second clause will become (A¯ ∨ C¯ ∨ F ∨ S ∨ Z), the third clause will become (E ∨ G ∨ H¯ ∨ Q ∨ Z) and so on. It is easy to observe that by assigning the value 0 to the variable Z, you get an instance of 5-SAT which is verified if and only if the starting 4-SAT instance is verified.

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  • $\begingroup$ You cannot assign a value to a variable — that's not how you specify a $k$-SAT instance. $\endgroup$ – Yuval Filmus May 21 at 14:53
  • $\begingroup$ @Yuval Filmus you are right, i meant that the adding of Z to every clause is trivial to the satisfabiality of the problem. $\endgroup$ – Yamar69 May 21 at 16:58

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