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## New answers tagged decision-problem

5

Let $M$ be a $n \times n$ matrix. And I am searching a word $l_ll_2 \dots l_k$ of length $k$. Now first we search $l_1$ in $M$. It takes $O(n^2)$ time. For each successful search, we look at all eight direction (up to down, down to up, left to right, right to left, diagonal-left-up, diagonal-left-down, diagonal-right-up and diagonal-right-down) and check ...

15

Your language is in P. Suppose that the matrix is $n\times n$ and that the words have total length $\ell$. Each word can start at at most $n^2$ positions and be written in $O(1)$ many orientations, for a total of $O(n^2)$ possible placements. Checking each one costs at most $O(m)$, where $m$ is the length of the word. In total, we obtain an algorithm whose ...

1

The language consists of ternary relations $R_1, \dots, R_4$, which represent the clauses as follows: $(x,y,z)\in R_1$ iff there is a clause $(x\lor y\lor z)$; $(x,y,z)\in R_2$ iff there is a clause $(x\lor y\lor \neg z)$; $(x,y,z)\in R_3$ iff there is a clause $(x\lor \neg y\lor \neg z)$; $(x,y,z)\in R_4$ iff there is a clause $(\neg x\lor \neg y\lor \neg ... 2 Nice question. Notations and terms$M$or$N$means a Turing machine (TM), whose specification may or may not given.$\langle M\rangle$is the description of$M$according to a predefined effective encoding scheme for TMs.$L(M)$is the language recognized by$M$, i.e., the set of words accepted by$M$. At least that is what I have seen everywhere. ... 3 The statement in CRLS is not wrong in any case; an algorithm that runs in$O(n)$time also runs in$O(n^2)$time. Of course, it would be more precise to state the running time as$O(n)$if this were true, so why doesn't CLRS do this? First off, this depends on the encoding chosen for$G$. If an adjacency matrix is used, a graph with$V$vertices always has ... 9 If I understand you correctly, your question is — why a solution can be encoded by a natural number and a problem with a real number. (I assume that you understand the next phase of the proof which is based on the difference between sets of cardinality$\mathfrak c$and$\aleph_0$.) The reason lies in set theory, more specifically in the cardinality ... 11 Reformulating in a more mathematically precise way, what the lecturer is trying to say is this: Any algorithm can be (uniquely) encoded as a finite string of bits, and any finite string of bits (uniquely) encodes a program; hence, there is a bijection between$\mathbb{N}$and the set of algorithms, so both are countable sets. Conversely, having fixed an ... 6 Every algorithm can be described by a finite string, and so there are only countably many algorithms. In contrast, we can describe every decision problem as an infinite decimal in base 2, and moreover this is a surjective mapping: every number in$[0,1]\$ can be "decoded" into a decision problem. Therefore there are uncountably many decision problems. The ...

0

As Konstantin Vladimirov said in the comment: Consider you have this program. Take any while-program and put any instruction 1000 times just before halt. Next run your program and ask if this instruction will be reached 1000 times. Bingo, halting is solved.

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