Hfrxdjt

Can a virus destroy the BIOS of a modern computer?

Notepad++ delete until colon for every line with replace all

OP Amp not amplifying audio signal

Processor speed limited at 0.4 Ghz

How to remove border from elements in the last row?

Did 'Cinema Songs' exist during Hiranyakshipu's time?

Salesman text me from his personal phone

Does Dispel Magic work on Tiny Hut?

Why was the shrink from 8″ made only to 5.25″ and not smaller (4″ or less)

My singleton can be called multiple times

Is it possible to create a QR code using text?

What reasons are there for a Capitalist to oppose a 100% inheritance tax?

Implication of namely

How obscure is the use of 令 in 令和?

Is there a hemisphere-neutral way of specifying a season?

Different meanings of こわい

Are British MPs missing the point, with these 'Indicative Votes'?

Use of noexpand in the implementation of TextOrMath for eTeX

Pact of Blade Warlock with Dancing Blade

My ex-girlfriend uses my Apple ID to login to her iPad, do I have to give her my Apple ID password to reset it?

How could indestructible materials be used in power generation?

What is the fastest integer factorization to break RSA?

How to prevent "they're falling in love" trope

What is the opposite of "eschatology"?

Decidable problems for which no concrete decision procedure is known

complexity of decision problems vs computing functionsCan all decision problems reduce to undecidable?Efficiently decidable logicsWhat are some results for non-trivial lower bounds for the time complexity of decision problems?How many decidable decision problems are there?How many decision problems do exist?Are poly-reductions to PSPACE problems for following problems are known?Is there a generic procedure to produce (hard enough) decision problems?A TSP to HamCycle ReductionDecision problems involving finite automata

1

$begingroup$

I am looking for an example of decidable problems the decision procedures of which are unknown. I believe someone mentioned one to me once, and I also have read somewhere, but my memory is corrupted. I suppose the decidability of these problems are proved non-constructively, so that they have this kind of counter-intuitive property.

I think that in graph theory, there are a number of problems in this trait. I am wondering if you are aware of any of these.

time-complexity decision-problem

share|cite|improve this question

asked 7 hours ago

Jason HuJason Hu

303110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I doubt there is any such kind of decidable problem, because of the definition of decidable. There are lots of decidable problems for which people do not care enough to figure out or specify a decision algorithm in its full detail. However, I would not be willing to call those situations as without known concrete decision procedure. It is more like no explicit implementation in a programming language has been done. Or is that your real intention?
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  

add a comment | 

1

$begingroup$

I am looking for an example of decidable problems the decision procedures of which are unknown. I believe someone mentioned one to me once, and I also have read somewhere, but my memory is corrupted. I suppose the decidability of these problems are proved non-constructively, so that they have this kind of counter-intuitive property.

I think that in graph theory, there are a number of problems in this trait. I am wondering if you are aware of any of these.

time-complexity decision-problem

share|cite|improve this question

asked 7 hours ago

Jason HuJason Hu

303110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I doubt there is any such kind of decidable problem, because of the definition of decidable. There are lots of decidable problems for which people do not care enough to figure out or specify a decision algorithm in its full detail. However, I would not be willing to call those situations as without known concrete decision procedure. It is more like no explicit implementation in a programming language has been done. Or is that your real intention?
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  

add a comment | 

$begingroup$

I am looking for an example of decidable problems the decision procedures of which are unknown. I believe someone mentioned one to me once, and I also have read somewhere, but my memory is corrupted. I suppose the decidability of these problems are proved non-constructively, so that they have this kind of counter-intuitive property.

I think that in graph theory, there are a number of problems in this trait. I am wondering if you are aware of any of these.

time-complexity decision-problem

share|cite|improve this question

asked 7 hours ago

Jason HuJason Hu

303110

$endgroup$

share|cite|improve this question

asked 7 hours ago

Jason HuJason Hu

303110

share|cite|improve this question

asked 7 hours ago

Jason HuJason Hu

303110

asked 7 hours ago

Jason HuJason Hu

303110

asked 7 hours ago

Jason HuJason Hu

303110

2 Answers
2

active

oldest

votes

2

$begingroup$

Here is a trivial example of such a problem: consider any yes-no question with an unknown answer, e.g. $P=NP$. For any $n in mathbbN$, let $A(n)$ be true iff $P=NP$. Then $A$ is decidable by one of two decision procedures:

1. Always return "true".




2. Always return "false".

But you need to solve $P=NP$ to know which one.

share|cite|improve this answer

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah I really like this sort of construction.
  $endgroup$
  – Jason Hu
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, some people believe $P=NP$ is undecidable. Can you prove that there is a Turing machine that will either prove or disprove $P=NP$? So the meaning of "let $A(n)$ be true iff $P=NP$" is somewhat ambiguous. What I am saying is that I would like to require that a valid or, you say, interesting example should not contain a part that is not known to be decidable from the ground, for example, ZFC system.
  $endgroup$
  – Apass.Jack
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Has $P=NP$ been proved to be falsifiable in ZFC system? It looks like I am determined to confuse myself.
  $endgroup$
  – Apass.Jack
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack This construction doesn't need $P=NP$ to be decidable. Taking an undecidable (in some system, e.g. in ZFC) statement would actually work a bit better: then no explicit decision procedure can ever be proven to work (in that system), as opposed to just not having one now. I am not sure exactly what you mean by "falsifiable" in the second comment.
  $endgroup$
  – Alexey Romanov
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack This might be a bit confusing because there are two different senses of "(un)decidable" involved: for logical formulas and for decision problems.
  $endgroup$
  – Alexey Romanov
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

For the record, I found the one in graph theory, which is called graph minor theorem, or Robertson–Seymour theorem.

https://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem

Indeed, this theorem is proved non-constructively.

share|cite|improve this answer

answered 7 hours ago

Jason HuJason Hu

303110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The truth for formidden minors is much more complicated and refined than this answer. Check is there an algorithm that finds the forbidden minors?. Note that Robertson–Seymour theorem is a theorem, not a decidable problem. It might mention a decidable problem, but it is not a decidable problem by itself. I recommend that you state the problem, the more formally the better. Explain why the problem is decidable, which might not be obvious at all and why there is no algorithm found for it, which might not be obvious at all again.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In short, I am saying this answer cannot be counted as correct.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack I don't understand the theorem in its full details. I am search for an example of such kind as a concrete example, and it stops there. from wiki, it says that "As a result, the theorem proves that the problem can be solved in polynomial time, but does not provide a concrete polynomial-time algorithm for solving it. Such proofs of polynomiality are non-constructive: they prove polynomiality of problems without providing an explicit polynomial-time algorithm." I suppose it says that the poly algorithm is just unknown. what's your concern?
  $endgroup$
  – Jason Hu
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are talking about polynomial-time instead of decidable. These two concepts are very different. They are plenty of problems that are decidable but without polynomial-time algorithm. Did you mean to ask "decision problems that are solvable in polynomial-time for which no concrete polynomial-time algorithm is known"? That sounds like an interesting question (which might have been asked before) that makes much more sense.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JasonHu Compare statements: 1. There's no known explicit algorithm for solving problem $X$. 2. The proof of R-S theorem doesn't provide an explicit polynomial-time algorithm for solving problem $X$. The Wikipedia only says something like 2 ("like" because as Apass.Jack says, you haven't specified the problem yet). Does this imply 1?
  $endgroup$
  – Alexey Romanov
  3 hours ago
  

  
  
  
  

 | 
show 1 more comment

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "419"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcs.stackexchange.com%2fquestions%2f106370%2fdecidable-problems-for-which-no-concrete-decision-procedure-is-known%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

Here is a trivial example of such a problem: consider any yes-no question with an unknown answer, e.g. $P=NP$. For any $n in mathbbN$, let $A(n)$ be true iff $P=NP$. Then $A$ is decidable by one of two decision procedures:

1. Always return "true".




2. Always return "false".

But you need to solve $P=NP$ to know which one.

share|cite|improve this answer

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah I really like this sort of construction.
  $endgroup$
  – Jason Hu
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, some people believe $P=NP$ is undecidable. Can you prove that there is a Turing machine that will either prove or disprove $P=NP$? So the meaning of "let $A(n)$ be true iff $P=NP$" is somewhat ambiguous. What I am saying is that I would like to require that a valid or, you say, interesting example should not contain a part that is not known to be decidable from the ground, for example, ZFC system.
  $endgroup$
  – Apass.Jack
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Has $P=NP$ been proved to be falsifiable in ZFC system? It looks like I am determined to confuse myself.
  $endgroup$
  – Apass.Jack
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack This construction doesn't need $P=NP$ to be decidable. Taking an undecidable (in some system, e.g. in ZFC) statement would actually work a bit better: then no explicit decision procedure can ever be proven to work (in that system), as opposed to just not having one now. I am not sure exactly what you mean by "falsifiable" in the second comment.
  $endgroup$
  – Alexey Romanov
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack This might be a bit confusing because there are two different senses of "(un)decidable" involved: for logical formulas and for decision problems.
  $endgroup$
  – Alexey Romanov
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

2

$begingroup$

Here is a trivial example of such a problem: consider any yes-no question with an unknown answer, e.g. $P=NP$. For any $n in mathbbN$, let $A(n)$ be true iff $P=NP$. Then $A$ is decidable by one of two decision procedures:

1. Always return "true".




2. Always return "false".

But you need to solve $P=NP$ to know which one.

share|cite|improve this answer

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah I really like this sort of construction.
  $endgroup$
  – Jason Hu
  3 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In fact, some people believe $P=NP$ is undecidable. Can you prove that there is a Turing machine that will either prove or disprove $P=NP$? So the meaning of "let $A(n)$ be true iff $P=NP$" is somewhat ambiguous. What I am saying is that I would like to require that a valid or, you say, interesting example should not contain a part that is not known to be decidable from the ground, for example, ZFC system.
  $endgroup$
  – Apass.Jack
  2 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Has $P=NP$ been proved to be falsifiable in ZFC system? It looks like I am determined to confuse myself.
  $endgroup$
  – Apass.Jack
  2 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack This construction doesn't need $P=NP$ to be decidable. Taking an undecidable (in some system, e.g. in ZFC) statement would actually work a bit better: then no explicit decision procedure can ever be proven to work (in that system), as opposed to just not having one now. I am not sure exactly what you mean by "falsifiable" in the second comment.
  $endgroup$
  – Alexey Romanov
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack This might be a bit confusing because there are two different senses of "(un)decidable" involved: for logical formulas and for decision problems.
  $endgroup$
  – Alexey Romanov
  1 hour ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Here is a trivial example of such a problem: consider any yes-no question with an unknown answer, e.g. $P=NP$. For any $n in mathbbN$, let $A(n)$ be true iff $P=NP$. Then $A$ is decidable by one of two decision procedures:

1. Always return "true".




2. Always return "false".

But you need to solve $P=NP$ to know which one.

share|cite|improve this answer

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

$endgroup$

share|cite|improve this answer

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

answered 3 hours ago

Alexey RomanovAlexey Romanov

2,2521114

1

$begingroup$

For the record, I found the one in graph theory, which is called graph minor theorem, or Robertson–Seymour theorem.

https://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem

Indeed, this theorem is proved non-constructively.

share|cite|improve this answer

answered 7 hours ago

Jason HuJason Hu

303110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The truth for formidden minors is much more complicated and refined than this answer. Check is there an algorithm that finds the forbidden minors?. Note that Robertson–Seymour theorem is a theorem, not a decidable problem. It might mention a decidable problem, but it is not a decidable problem by itself. I recommend that you state the problem, the more formally the better. Explain why the problem is decidable, which might not be obvious at all and why there is no algorithm found for it, which might not be obvious at all again.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In short, I am saying this answer cannot be counted as correct.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack I don't understand the theorem in its full details. I am search for an example of such kind as a concrete example, and it stops there. from wiki, it says that "As a result, the theorem proves that the problem can be solved in polynomial time, but does not provide a concrete polynomial-time algorithm for solving it. Such proofs of polynomiality are non-constructive: they prove polynomiality of problems without providing an explicit polynomial-time algorithm." I suppose it says that the poly algorithm is just unknown. what's your concern?
  $endgroup$
  – Jason Hu
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are talking about polynomial-time instead of decidable. These two concepts are very different. They are plenty of problems that are decidable but without polynomial-time algorithm. Did you mean to ask "decision problems that are solvable in polynomial-time for which no concrete polynomial-time algorithm is known"? That sounds like an interesting question (which might have been asked before) that makes much more sense.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JasonHu Compare statements: 1. There's no known explicit algorithm for solving problem $X$. 2. The proof of R-S theorem doesn't provide an explicit polynomial-time algorithm for solving problem $X$. The Wikipedia only says something like 2 ("like" because as Apass.Jack says, you haven't specified the problem yet). Does this imply 1?
  $endgroup$
  – Alexey Romanov
  3 hours ago
  

  
  
  
  

 | 
show 1 more comment

1

$begingroup$

For the record, I found the one in graph theory, which is called graph minor theorem, or Robertson–Seymour theorem.

https://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem

Indeed, this theorem is proved non-constructively.

share|cite|improve this answer

answered 7 hours ago

Jason HuJason Hu

303110

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  The truth for formidden minors is much more complicated and refined than this answer. Check is there an algorithm that finds the forbidden minors?. Note that Robertson–Seymour theorem is a theorem, not a decidable problem. It might mention a decidable problem, but it is not a decidable problem by itself. I recommend that you state the problem, the more formally the better. Explain why the problem is decidable, which might not be obvious at all and why there is no algorithm found for it, which might not be obvious at all again.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  In short, I am saying this answer cannot be counted as correct.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Apass.Jack I don't understand the theorem in its full details. I am search for an example of such kind as a concrete example, and it stops there. from wiki, it says that "As a result, the theorem proves that the problem can be solved in polynomial time, but does not provide a concrete polynomial-time algorithm for solving it. Such proofs of polynomiality are non-constructive: they prove polynomiality of problems without providing an explicit polynomial-time algorithm." I suppose it says that the poly algorithm is just unknown. what's your concern?
  $endgroup$
  – Jason Hu
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You are talking about polynomial-time instead of decidable. These two concepts are very different. They are plenty of problems that are decidable but without polynomial-time algorithm. Did you mean to ask "decision problems that are solvable in polynomial-time for which no concrete polynomial-time algorithm is known"? That sounds like an interesting question (which might have been asked before) that makes much more sense.
  $endgroup$
  – Apass.Jack
  4 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  @JasonHu Compare statements: 1. There's no known explicit algorithm for solving problem $X$. 2. The proof of R-S theorem doesn't provide an explicit polynomial-time algorithm for solving problem $X$. The Wikipedia only says something like 2 ("like" because as Apass.Jack says, you haven't specified the problem yet). Does this imply 1?
  $endgroup$
  – Alexey Romanov
  3 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

For the record, I found the one in graph theory, which is called graph minor theorem, or Robertson–Seymour theorem.

https://en.wikipedia.org/wiki/Robertson%E2%80%93Seymour_theorem

Indeed, this theorem is proved non-constructively.

share|cite|improve this answer

answered 7 hours ago

Jason HuJason Hu

303110

$endgroup$

share|cite|improve this answer

answered 7 hours ago

Jason HuJason Hu

303110

answered 7 hours ago

Jason HuJason Hu

303110

answered 7 hours ago

Jason HuJason Hu

303110