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Intuitively $A \le_p B$ means that a polynomial-time algorithm for $B$ can be used to solve $A$ in polynomial-time, not vice-versa. That said $f$ is not required to be surjective, think for example of $A=\Sigma^*$ and $B=\{0,1\}$. Suppose that $f(x) = 0$ (i.e., $f$ is the constant function). Simulate the algorithm with $y=1$ and notice that it will reject ...

1

We start with the assumption that there is a non-trivial language $A$ which is in $\mathrm{coNP}$, but not $\mathrm{coNP}$-complete. We note that it is rather straight-forward to see that if $\mathrm{P} = \mathrm{coNP}$, then every non-trivial language in $\mathrm{coNP}$ is $\mathrm{coNP}$-complete. Conversely, Ladner's theorem states that if $\mathrm{P} \... 2 Alright, we are given a language$C \in \mathrm{P}$and a language$A \in \mathrm{NP}$, and we are promised that$C \leq_p A$. What can we conclude? First a side remark: The reduction$C \leq_p A$tells us very little since we know$C \in \mathrm{P}$. The only extra information we get is that if$A$is empty, then so is$C$, and if$A = \Sigma^*$, then$C = \...

2

Option 1 is odd. We don't know whether a polynomial algorithm can be constructed. The "passing all the neighbors at a distance of less than 6 from each vertex in $G$" is unclear. Any algorithm solving the TSP problem must visit all vertices. Therefore it must also visit all neighbors of each vertex and, in particular, all neighbors at distance less ...

2

1 is wrong. The problem is NP-Complete by an easy reduction $f$ from the clique problem. Let $\langle G,k \rangle$ with $G=(V,E)$ be an instance of (the decision version of) clique. If $k$ is odd then the reduction is the identity function, i.e., $f(\langle G,k \rangle)=\langle G,k \rangle$. Otherwise create a new graph $G=(V', E')$ with $V' = V \cup \{v\}$...

1

Saying that a "language reaches the state of reject" is meaningless. Languages don't have states. Turing machines and automatons do. That said, your language is undecidable. Given $\langle M,w \rangle$ you can compute the pair $\langle M',w \rangle$ where $M'$ behaves exactly as $M$ except for the following: every transition that halts the machine ...

2

From what I understand in the question, there is a language that accepts Turing machines that go into an endless loop. We do not say a language "accepts" something. Rather the language contains all pairs of turing machines and words, where the TM runs forever (which I take "infinite loop" to mean) on the word. Also, note that you can ...

0

Let $X = \{ x_1, x_2, \dots, x_n \}$ be the (multi-)set of elements in your subset-sum instance and let $t$ be the target value. Create the undirected graph $G=(V,E)$ where $V=\{a,b\} \cup X$ and $E = ( \{a,b\} \times X) \cup \{(a,b)\}$. The weight of edge $(a, x_i)$, for $x_i \in X$, is $x_i$. The weight of edge $(b, z)$ for $z \in \{a\} \cup X$ is $0$. ...

1

The main "key" here is to understand how a formula is structured. Assuming the formulas of TAUTOLOGY are in CNF, then there exists a polynomial algorithm: Let $\varphi=\bigwedge_{i=1}^{k}\limits C_i$ where each $C_i$ is described as an $\lor$ of multiple literals. Notice, that in order for an assignment $\rho$ to satisfy $\varphi$, we need $\rho$ ...

2

If $\mathrm{FPT}=W[1]$, then we will not directly get an answer to $\mathrm{P}$ vs. $\mathrm{NP}$ (or at least, not that we know). The assumption that $\mathrm{FPT} \neq W[1]$ is a weaker claim than $\mathrm{P}\neq \mathrm{NP}$: If $\mathrm{P} = \mathrm{NP}$, then there is polynomial time algorithm for the $W[1]$-complete problem $k$-Clique, so $\mathrm{FPT}=... 2 Because$k$-CLIQUE is W[1]-hard, your result would imply that FPT = W[1], which by the result of Downey and Fellows [1] implies that the Exponential Time Hypothesis (ETH) is false, i.e., that 3-SAT can be solved in$2^{o(n)}$time. But ETH is stronger than P not equal to NP, so its falsification doesn't settle P = NP (but of course, if true, ETH would imply ... 0$P$is closed under complement. The rest is up to you. 0 It is 2 that is correct. Here is an alternative explanation: Notice that any language in$P$has to be$P$-complete. In addition to that, if$P=NP$then$NPC=P$-complete. Combining both arguments implies that$NP=P=P\text{-complete}=NPC$. Therefore, if$NP\neq NPC$then$P\neq NP$. 1 I assume that by "verses" you mean boolean formulas. The language is definitely in CO-NP since a "no" certificate is a satisfying assignment. Currently we don't know whether FALSE is in P nor whether it is in NP. 1 A clique$C$on$k$vertices has$\binom{k}{2}$edges, i.e., all unordered pairs$\{a,b\}$such that$a$and$b$are distinct vertices in$C$. The number of such pairs is$\frac{k (k-1)}{2} = \binom{k}{2}$. You can count these by focusing on the number of ordered pairs first. To make an ordered pair$(a,b)$you have$k$choices for$a$and$k-1$choice for$...

1

$L$ is given as part of the input, and can be e.g., $n/2$, where $n$ is the number of items. Then, iterating over ${n\choose L}={n\choose \frac{n}2}$ is exponential. Note that it doesn't matter whether $L$ is given in unary or binary, since $n$ is given in unary (as a list of the different items).

1

Theorem. Any finite language L is in P. Proof. Let M be a Turing machine which has all strings of L on its tape. When given an input it checks whether the input is on its tape. This is O(1) time clearly. Theorem. If and only if P=NP, then every L in P (including thus all finite languages) are NP-Complete. Proof. Recall that a language is NP-Complete iff ...

1

You are right that that doesn't prove the algorithm is polynomial, as it just gives the certificate. But it is very easy to see that (for example with a program in a C-like language, given a reasonable representation of the graphs) the check is very easy to do (is polynomial). The part of coming up with a polynomial verifier given a reasonable certificate (...

2

The second approach is technically the correct interperation. When we say "a verifier gets a certificate $c$ that looks like [...] and computes something", then we actually mean that this verifier checks whether $c$ is of the required format (e.g, check that $c$ is representing a subset of nodes from the graph), and if it isn't it will immediately ...

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