# Tag Info

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Translating Code to Mathematics Given a (more or less) formal operational semantics you can translate an algorithm's (pseudo-)code quite literally into a mathematical expression that gives you the result, provided you can manipulate the expression into a useful form. This works well for additive cost measures such as number of comparisons, swaps, statements,...

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Before I answer your general question, let me first take a step back, give some history background, and answer a preliminary question: Do non-computable functions even exist? [notational note: we can relate any function $f$ with a language $L_f=\{ (x,y) \mid y=f(x) \}$ and then discuss the decidability of $L_f$ rather than the computability of $f$] ...

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It all comes from undecidability of the halting problem. Suppose we have a "perfect" dead code function, some Turing Machine M, and some input string x, and a procedure that looks something like this: Run M on input x; print "Finished running input"; If M runs forever, then we delete the print statement, since we will never reach it. If M doesn't run ...

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I feel like i am memorizing the proofs rather than learn how to prove You can't learn "how to prove". "Proving" is not a mechanical process, but rather a creative one where you have to invent a new technique to solve a given problem. A professional mathematician could spend their entire life attempting to prove a given statement and ...

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Ultimately, you'll need a mathematical proof of correctness. I'll get to some proof techniques for that below, but first, before diving into that, let me save you some time: before you look for a proof, try random testing. Random testing As a first step, I recommend you use random testing to test your algorithm. It's amazing how effective this is: in my ...

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Execution Counts of Statements There is another method, championed by Donald E. Knuth in his The Art of Computer Programming series. In contrast to translating the whole algorithm into one formula, it works independently from the code's semantics on the "putting things together" side and allows to go to a lower level only when necessary, starting from an "...

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A problem $Q$ is NP-complete if it is both NP-hard and in NP. This means that you need to disprove one of these two. Under the assumption that P $\neq$ NP, you can give a polynomial time algorithm solving $Q$. Rarer, under the assumption that graph isomorphism is not NP-hard, you can show that $Q$ is polytime reducible to graph isomorphism. You show that $... 15 Leveraging a known nearby problem When faced with a problem that feels hard, it is often a good idea to try to search for a similar problem that is already proven hard. Or, perhaps you can immediately see that a problem is very similar to a known problem. Example problem Consider a problem $$\text{DOUBLE-SAT} = \{ \varphi \mid \varphi \text{ is a ... 14 A practical approach that in many examples works [but not always, I know] is trying to find the nesting structure of the strings in the language. "Nested dependencies" have to be generated at the same time in different parts of the string. Also we have the basic toolbox: concatenation: S\to S_1S_2 if you can split the language in two consecutive ... 14 The most common technique is to prove that the problem is hard for some class. For example, consider the problem HALT in which we are given a Turing machine T and a number n (encoded in binary, i.e., in the usual way), and the task is to decide whether T halts within n steps. HALT is EXPTIME-hard. If HALT were in P then P=EXPTIME (using the EXPTIME-... 14 The simplest case I know of an algorithm that exists, though it is not known which algorithm, concerns finite state automata. The quotient L_1/L_2 of a language L_1 by a language L_2 is defined as L_1/L_2=\{x \mid \exists y\in L_2 \text{ such that } xy\in L_1\}. It is easily proved that regular set are closed under quotient by an arbitrary set. In ... 14 Yes, it's possible. The classic example is the fact that any comparison-based sorting algorithm requires \Omega(n\log n) comparisons to sort a list of length n. However, lower bounds seem to be much harder to prove than upper bounds. To prove that there's a sorting algorithm that requires O(n\log n) comparisons, you just need to exhibit such an ... 14 This is a twist on jmite's answer that circumvents the potential confusion about non-termination. I'll give a program that always halts itself, may have dead code but we can not (always) algorithmically decide if it has. Consider the following class of inputs for the dead-code identifier: simulateMx(n) { simulate TM M on input x for n steps if M did ... 14 I will use the following simple sorting algorithm as an example: repeat: if there are adjacent items in the wrong order: pick one such pair and swap else break To prove the correctness I use two steps. First I show that the algorithm always terminates. Then I show that the solution where it terminates is the one I want. For the first point,... 13 One useful tool is Rice's theorem. Here is what it says: Let \emptyset \subsetneq P \subsetneq \mathcal{P} a non-trivial set of partially computable unary functions and \varphi a Gödel numbering of \mathcal{P}. Then the index set of P \qquad I_P = \{ i \in \mathbb{N} \mid \varphi_i \in P \} is not recursive. You find it also ... 13 A stronger version of the Ogden's condition (OC) is the Bader-Moura’s condition (BMC) A language L\subseteq \Sigma^* satisfies BMC if there exists a constant n such that if z \in L and we label in it "distinguished" positions d(z) and e(z) "excluded" positions, with d(z) > n^{e(z)+1}, then we may write z = uvwxy such ... 12 Summations Often one encounters a recurrence of the form$$ T(n) = T(n-1) + f(n), $$where f(n) is monotone. In this case, we can expand$$ T(n) = T(c) + \sum_{m=c+1}^n f(m), $$and so given a starting value T(c), in order to estimate T(n) we need to estimate the sum f(c+1) + \cdots + f(m). Non-decreasing f(n) When f(n) is monotone non-... 12 Elementary methods Finite automata (possibly nondeterministic, with empty transitions). Regular expressions. Right (or Left, but not both) linear equations, like X = KX + L where K and L are regular. Regular (Type 3) grammar. Operations preserving regular languages (Boolean operations, product, star, shuffle, morphisms, inverses of morphisms, reversal,... 12 You are asking for a constructive proof of the Lesser limited principle of omniscience (LLPO), which states (in one of its forms) that for a decidable proposition P on natural numbers$$(\forall n \in \mathbb{N} \,.\, P(n)) \lor \lnot \forall n \in \mathbb{N} \,.\, P(n).$$That's exactly your problem. It is well known that LLPO is not provable ... 12 To expand on Hendrick's original comment, consider this problem Given an integer$n\ge 0$is there a run of$n$or more consecutive 7s in the decimal expansion of$\pi$? This problem is decidable, since one of two cases may obtain: There is an integer$N$for which the decimal expansion of$\pi$contains a run of$N$consecutive 7s, but no longer run. ... 12 Oracles are a very general formalization of the idea, "If I could solve$X$efficiently, I could use that to solve$Y$efficiently." I accept that it sounds a bit silly to go as far as "If I could solve problem$X$in constant time, I could use that to solve$Y$efficiently" but, actually, that doesn't make any real difference at the ... 12 Self application is not a necessary ingredient of the proof In a nutshell If there is a Turing machine$H$that solves the halting problem, then from that machine we can build another Turing machine$L$with a halting behavior (halting characteristic function) that cannot be the halting behavior of any Turing machine. The paradox built on the self applied ... 12 The error in your argument is the claim Nothing is known a priori about the function$f$, (...) so it is necessary to plug in all$2^n$values. , which is simply false. I will demonstrate why it isn't necessary to plug in all possible values and sketch how to arrive at an$O^*((2^k-1)^{n/k})$time algorithm, where$k$is the number of clauses.$O^*\$ '...

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