Consider a 3-SAT problem where the formula is in conjunctive normal form and we restrict the Boolean formulas such that the number of clauses in the formula is equal to the number of variables. Is this problem still NP-hard?

For example, this formula has $3$ variables and has $3$ clauses $(\lnot x_1 \cap \lnot x_2 \cap \lnot x_3 ) \cup (\lnot x_1 \cap \lnot x_3) \cup (\lnot x_2 \cap \lnot x_3)$,

and the following formula has three variables but has only two clauses $(\lnot x_1 \cap \lnot x_2 \cap \lnot x_3 ) \cup (\lnot x_2 \cap \lnot x_3)$.

Consider the 3-SAT problem where the formula is in conjunctive normal form and we restrict the Boolean formulas such that the number of clauses in the formula is equal to the number of variables. Is this problem still NP-hard?

For example, this formula has $3$ variables and has $3$ clauses $(\lnot x_1 \vee \lnot x_2 \vee \lnot x_3 ) \wedge (\lnot x_1 \vee \lnot x_3) \wedge (\lnot x_2 \vee \lnot x_3)$,

and the following formula has three variables but has only two clauses $(\lnot x_1 \vee \lnot x_2 \vee \lnot x_3 ) \wedge (\lnot x_2 \vee \lnot x_3)$.

Consider a SAT problem where the formula is in conjunctive normal form and we restrict the Boolean formulas such that the number of clauses in the formula is equal to the number of variables. Is this problem still NP-hard?

For example, this formula has $3$ variables and has $3$ clauses $(\lnot x_1 \cup \lnot x_2 ) \cap (\lnot x_1 \cup \lnot x_3) \cup (\lnot x_2 \cup \lnot x_3)$,

and the following formula has three variables but has only two clauses $(\lnot x_1 \cup \lnot x_2 ) \cup (\lnot x_2 \cup \lnot x_3)$.

Consider a 3-SAT problem where the formula is in conjunctive normal form and we restrict the Boolean formulas such that the number of clauses in the formula is equal to the number of variables. Is this problem still NP-hard?

For example, this formula has $3$ variables and has $3$ clauses $(\lnot x_1 \cap \lnot x_2 \cap \lnot x_3 ) \cup (\lnot x_1 \cap \lnot x_3) \cup (\lnot x_2 \cap \lnot x_3)$,

and the following formula has three variables but has only two clauses $(\lnot x_1 \cap \lnot x_2 \cap \lnot x_3 ) \cup (\lnot x_2 \cap \lnot x_3)$.

Consider a SAT problems where the formula is in conjunctive normal form and we restrict the Boolean formulas such that the number of clauses in the formula is equal to the number of variables, is this problem still NP-hard?

E.g. This formula has $3$ variables and has three clauses $(\lnot x_1 \cup \lnot x_2 \cup ) \cap (\lnot x_1 \cup \lnot x_3) \cup (\lnot x_2 \cup \lnot x_3)$.

and the following formula has three variables but has only two clauses $(\lnot x_1 \cup \lnot x_2 \cup ) \cup (\lnot x_2 \cup \lnot x_3)$.

Consider a SAT problem where the formula is in conjunctive normal form and we restrict the Boolean formulas such that the number of clauses in the formula is equal to the number of variables. Is this problem still NP-hard?

For example, this formula has $3$ variables and has $3$ clauses $(\lnot x_1 \cup \lnot x_2 ) \cap (\lnot x_1 \cup \lnot x_3) \cup (\lnot x_2 \cup \lnot x_3)$,

and the following formula has three variables but has only two clauses $(\lnot x_1 \cup \lnot x_2 ) \cup (\lnot x_2 \cup \lnot x_3)$.