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Do I need to consider instance restrictions when showing a language is in P?
Decision problem and algorithmShowing a language is in co-NPCDo Turing machines assume something infinite at some point?Is deciding whether the language of a TM contains all strings of length 4 computable?Proving that a language is not RecursiveUndecidable vs Unsolvable?In which direction should I carry out a reduction when proving decidability/recognizability?The defining property of problems in NPIs there a polynomial time algorithm to determine whether an 'up down' language is 'emptible'?Define the following problem as a language and prove that it is undecidable with a reduction from the halting problem.
$begingroup$
I have already shown that 3-colorable for an unrestricted graph is in NP, but I was thinking about the similar language defined as the set of all acyclic $G$, where $G$ such that $G$ is 3-colorable. In my proposition of an algorithm in P, I wasn't sure if my algorithm needs to verify that $G$ indeed contains no cycle or if I assume all inputs are instances of this language (assume all inputs are acyclic graphs). In general, I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time. I'm still learning so I am confused about this part and this has been bugging me for a while.
If this question is too confusing, consider 3SAT. Must I show that verifying the input is indeed a 3CNF is in P to conclude, or can I assume we are only considering inputs within an instance?
turing-machines time-complexity terminology
New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
$begingroup$
I have already shown that 3-colorable for an unrestricted graph is in NP, but I was thinking about the similar language defined as the set of all acyclic $G$, where $G$ such that $G$ is 3-colorable. In my proposition of an algorithm in P, I wasn't sure if my algorithm needs to verify that $G$ indeed contains no cycle or if I assume all inputs are instances of this language (assume all inputs are acyclic graphs). In general, I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time. I'm still learning so I am confused about this part and this has been bugging me for a while.
If this question is too confusing, consider 3SAT. Must I show that verifying the input is indeed a 3CNF is in P to conclude, or can I assume we are only considering inputs within an instance?
turing-machines time-complexity terminology
New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I have already shown that 3-colorable for an unrestricted graph is in NP, but I was thinking about the similar language defined as the set of all acyclic $G$, where $G$ such that $G$ is 3-colorable. In my proposition of an algorithm in P, I wasn't sure if my algorithm needs to verify that $G$ indeed contains no cycle or if I assume all inputs are instances of this language (assume all inputs are acyclic graphs). In general, I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time. I'm still learning so I am confused about this part and this has been bugging me for a while.
If this question is too confusing, consider 3SAT. Must I show that verifying the input is indeed a 3CNF is in P to conclude, or can I assume we are only considering inputs within an instance?
turing-machines time-complexity terminology
New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have already shown that 3-colorable for an unrestricted graph is in NP, but I was thinking about the similar language defined as the set of all acyclic $G$, where $G$ such that $G$ is 3-colorable. In my proposition of an algorithm in P, I wasn't sure if my algorithm needs to verify that $G$ indeed contains no cycle or if I assume all inputs are instances of this language (assume all inputs are acyclic graphs). In general, I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time. I'm still learning so I am confused about this part and this has been bugging me for a while.
If this question is too confusing, consider 3SAT. Must I show that verifying the input is indeed a 3CNF is in P to conclude, or can I assume we are only considering inputs within an instance?
turing-machines time-complexity terminology
turing-machines time-complexity terminology
New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 3 hours ago
Jahar Mehru
New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Jahar MehruJahar Mehru
133
133
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New contributor
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jahar Mehru is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time.
Very nice question!
What you are talking about is best characterized as promise problem, "a generalization of a decision problem where the input is promised to belong to a particular subset of all possible inputs". The convention way to handle promise problem is there are no requirements on the output if the input does not belong to the promise. In particular, if you want to show a promise problem is in P of promise problems, your algorithm do not need to check whether the input is valid or not and your algorithm can behave arbitrarily if the input is invalid.
For an in-depth discussion on promise problems, you are encouraged to read Oded Goldreich's exposition [On Promise Problems], July 11, 2005.
However, if you want to show a promise problem is in P, your algorithm must check whether the input is valid or not and, if the input is invalid, a.k.a as a noninstance, output 0. Here P stands for the complexity class that have been talked about the most, a.k.a. P of decision problems, i.e. PTIME or DTIME$(n^O(1))$ as defined at Wikipedia, or the complexity class P as defined in section 34.1 Polynomial time of the popular textbook introduction to algorithm by CLRS, version 3.
Take 3SAT for an example. An algorithm that show 3SAT is in P (of decision problems) should check whether each clause is a conjunctive normal form with at most three literals and many other restrictions. The algorithm should output 0 if it finds the input is not a valid instance of 3SAT.
It is easy to check whether a problem instance is a valid instance or not in almost all decision problems that have been discussed. It is so easy and so common that people have become so sloppy (or efficient) that this verification step is usually skipped or even forgotten in the specification of an algorithm. That might be the source of your confusion.
I would recommend beginners to write clearly this verification steps for the first few occasions before joining the common practice.
$endgroup$
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
add a comment |
$begingroup$
The answer depends on exactly what problem you're solving.
If your goal is to produce an algorithm that correctly solves the problem on the restricted instances, then it's kind of up to you whether or not you check. It feels more robust to check the input but it's perfectly reasonable not to, and that puts you in the realm of promise problems. Here, the "user" promises that the input is valid, and you just have to determine whether the answer is yes or no.
If your goal is to produce an algorithm that decides whether the input is, e.g., a satisfiable 3CNF then, yes, you do need to check that the input has the properties it's supposed to have.
Your example of 3-colourability for acyclic graphs shows that there can be a big difference between the two approaches. Every acyclic graph is 3-colourable (even 2-colourable) so the algorithm for the promise problem is just output "yes". For the non-promise version, you need to check that your input is a valid graph representation, then check that the graph is acyclic, and only say "yes" if the input passes both tests.
In practical terms, most descriptions of graph algorithms tend to assume that the input is a valid encoding of a graph – it gets a bit tedious writing "Check the input is a valid encoding of a graph" as the first line of every algorithm!
$endgroup$
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
add a comment |
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2 Answers
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2 Answers
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$begingroup$
I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time.
Very nice question!
What you are talking about is best characterized as promise problem, "a generalization of a decision problem where the input is promised to belong to a particular subset of all possible inputs". The convention way to handle promise problem is there are no requirements on the output if the input does not belong to the promise. In particular, if you want to show a promise problem is in P of promise problems, your algorithm do not need to check whether the input is valid or not and your algorithm can behave arbitrarily if the input is invalid.
For an in-depth discussion on promise problems, you are encouraged to read Oded Goldreich's exposition [On Promise Problems], July 11, 2005.
However, if you want to show a promise problem is in P, your algorithm must check whether the input is valid or not and, if the input is invalid, a.k.a as a noninstance, output 0. Here P stands for the complexity class that have been talked about the most, a.k.a. P of decision problems, i.e. PTIME or DTIME$(n^O(1))$ as defined at Wikipedia, or the complexity class P as defined in section 34.1 Polynomial time of the popular textbook introduction to algorithm by CLRS, version 3.
Take 3SAT for an example. An algorithm that show 3SAT is in P (of decision problems) should check whether each clause is a conjunctive normal form with at most three literals and many other restrictions. The algorithm should output 0 if it finds the input is not a valid instance of 3SAT.
It is easy to check whether a problem instance is a valid instance or not in almost all decision problems that have been discussed. It is so easy and so common that people have become so sloppy (or efficient) that this verification step is usually skipped or even forgotten in the specification of an algorithm. That might be the source of your confusion.
I would recommend beginners to write clearly this verification steps for the first few occasions before joining the common practice.
$endgroup$
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
add a comment |
$begingroup$
I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time.
Very nice question!
What you are talking about is best characterized as promise problem, "a generalization of a decision problem where the input is promised to belong to a particular subset of all possible inputs". The convention way to handle promise problem is there are no requirements on the output if the input does not belong to the promise. In particular, if you want to show a promise problem is in P of promise problems, your algorithm do not need to check whether the input is valid or not and your algorithm can behave arbitrarily if the input is invalid.
For an in-depth discussion on promise problems, you are encouraged to read Oded Goldreich's exposition [On Promise Problems], July 11, 2005.
However, if you want to show a promise problem is in P, your algorithm must check whether the input is valid or not and, if the input is invalid, a.k.a as a noninstance, output 0. Here P stands for the complexity class that have been talked about the most, a.k.a. P of decision problems, i.e. PTIME or DTIME$(n^O(1))$ as defined at Wikipedia, or the complexity class P as defined in section 34.1 Polynomial time of the popular textbook introduction to algorithm by CLRS, version 3.
Take 3SAT for an example. An algorithm that show 3SAT is in P (of decision problems) should check whether each clause is a conjunctive normal form with at most three literals and many other restrictions. The algorithm should output 0 if it finds the input is not a valid instance of 3SAT.
It is easy to check whether a problem instance is a valid instance or not in almost all decision problems that have been discussed. It is so easy and so common that people have become so sloppy (or efficient) that this verification step is usually skipped or even forgotten in the specification of an algorithm. That might be the source of your confusion.
I would recommend beginners to write clearly this verification steps for the first few occasions before joining the common practice.
$endgroup$
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
add a comment |
$begingroup$
I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time.
Very nice question!
What you are talking about is best characterized as promise problem, "a generalization of a decision problem where the input is promised to belong to a particular subset of all possible inputs". The convention way to handle promise problem is there are no requirements on the output if the input does not belong to the promise. In particular, if you want to show a promise problem is in P of promise problems, your algorithm do not need to check whether the input is valid or not and your algorithm can behave arbitrarily if the input is invalid.
For an in-depth discussion on promise problems, you are encouraged to read Oded Goldreich's exposition [On Promise Problems], July 11, 2005.
However, if you want to show a promise problem is in P, your algorithm must check whether the input is valid or not and, if the input is invalid, a.k.a as a noninstance, output 0. Here P stands for the complexity class that have been talked about the most, a.k.a. P of decision problems, i.e. PTIME or DTIME$(n^O(1))$ as defined at Wikipedia, or the complexity class P as defined in section 34.1 Polynomial time of the popular textbook introduction to algorithm by CLRS, version 3.
Take 3SAT for an example. An algorithm that show 3SAT is in P (of decision problems) should check whether each clause is a conjunctive normal form with at most three literals and many other restrictions. The algorithm should output 0 if it finds the input is not a valid instance of 3SAT.
It is easy to check whether a problem instance is a valid instance or not in almost all decision problems that have been discussed. It is so easy and so common that people have become so sloppy (or efficient) that this verification step is usually skipped or even forgotten in the specification of an algorithm. That might be the source of your confusion.
I would recommend beginners to write clearly this verification steps for the first few occasions before joining the common practice.
$endgroup$
I was wondering if my algorithm has to decide whether the input is of the desired instance ON TOP OF actually showing the properties of the language can be done in polynomial time.
Very nice question!
What you are talking about is best characterized as promise problem, "a generalization of a decision problem where the input is promised to belong to a particular subset of all possible inputs". The convention way to handle promise problem is there are no requirements on the output if the input does not belong to the promise. In particular, if you want to show a promise problem is in P of promise problems, your algorithm do not need to check whether the input is valid or not and your algorithm can behave arbitrarily if the input is invalid.
For an in-depth discussion on promise problems, you are encouraged to read Oded Goldreich's exposition [On Promise Problems], July 11, 2005.
However, if you want to show a promise problem is in P, your algorithm must check whether the input is valid or not and, if the input is invalid, a.k.a as a noninstance, output 0. Here P stands for the complexity class that have been talked about the most, a.k.a. P of decision problems, i.e. PTIME or DTIME$(n^O(1))$ as defined at Wikipedia, or the complexity class P as defined in section 34.1 Polynomial time of the popular textbook introduction to algorithm by CLRS, version 3.
Take 3SAT for an example. An algorithm that show 3SAT is in P (of decision problems) should check whether each clause is a conjunctive normal form with at most three literals and many other restrictions. The algorithm should output 0 if it finds the input is not a valid instance of 3SAT.
It is easy to check whether a problem instance is a valid instance or not in almost all decision problems that have been discussed. It is so easy and so common that people have become so sloppy (or efficient) that this verification step is usually skipped or even forgotten in the specification of an algorithm. That might be the source of your confusion.
I would recommend beginners to write clearly this verification steps for the first few occasions before joining the common practice.
edited 56 mins ago
answered 3 hours ago
Apass.JackApass.Jack
12.8k1939
12.8k1939
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
add a comment |
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
$begingroup$
This is EXACTLY what I was looking for. Thank you so much! My mind is now at ease.
$endgroup$
– Jahar Mehru
2 hours ago
add a comment |
$begingroup$
The answer depends on exactly what problem you're solving.
If your goal is to produce an algorithm that correctly solves the problem on the restricted instances, then it's kind of up to you whether or not you check. It feels more robust to check the input but it's perfectly reasonable not to, and that puts you in the realm of promise problems. Here, the "user" promises that the input is valid, and you just have to determine whether the answer is yes or no.
If your goal is to produce an algorithm that decides whether the input is, e.g., a satisfiable 3CNF then, yes, you do need to check that the input has the properties it's supposed to have.
Your example of 3-colourability for acyclic graphs shows that there can be a big difference between the two approaches. Every acyclic graph is 3-colourable (even 2-colourable) so the algorithm for the promise problem is just output "yes". For the non-promise version, you need to check that your input is a valid graph representation, then check that the graph is acyclic, and only say "yes" if the input passes both tests.
In practical terms, most descriptions of graph algorithms tend to assume that the input is a valid encoding of a graph – it gets a bit tedious writing "Check the input is a valid encoding of a graph" as the first line of every algorithm!
$endgroup$
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
add a comment |
$begingroup$
The answer depends on exactly what problem you're solving.
If your goal is to produce an algorithm that correctly solves the problem on the restricted instances, then it's kind of up to you whether or not you check. It feels more robust to check the input but it's perfectly reasonable not to, and that puts you in the realm of promise problems. Here, the "user" promises that the input is valid, and you just have to determine whether the answer is yes or no.
If your goal is to produce an algorithm that decides whether the input is, e.g., a satisfiable 3CNF then, yes, you do need to check that the input has the properties it's supposed to have.
Your example of 3-colourability for acyclic graphs shows that there can be a big difference between the two approaches. Every acyclic graph is 3-colourable (even 2-colourable) so the algorithm for the promise problem is just output "yes". For the non-promise version, you need to check that your input is a valid graph representation, then check that the graph is acyclic, and only say "yes" if the input passes both tests.
In practical terms, most descriptions of graph algorithms tend to assume that the input is a valid encoding of a graph – it gets a bit tedious writing "Check the input is a valid encoding of a graph" as the first line of every algorithm!
$endgroup$
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
add a comment |
$begingroup$
The answer depends on exactly what problem you're solving.
If your goal is to produce an algorithm that correctly solves the problem on the restricted instances, then it's kind of up to you whether or not you check. It feels more robust to check the input but it's perfectly reasonable not to, and that puts you in the realm of promise problems. Here, the "user" promises that the input is valid, and you just have to determine whether the answer is yes or no.
If your goal is to produce an algorithm that decides whether the input is, e.g., a satisfiable 3CNF then, yes, you do need to check that the input has the properties it's supposed to have.
Your example of 3-colourability for acyclic graphs shows that there can be a big difference between the two approaches. Every acyclic graph is 3-colourable (even 2-colourable) so the algorithm for the promise problem is just output "yes". For the non-promise version, you need to check that your input is a valid graph representation, then check that the graph is acyclic, and only say "yes" if the input passes both tests.
In practical terms, most descriptions of graph algorithms tend to assume that the input is a valid encoding of a graph – it gets a bit tedious writing "Check the input is a valid encoding of a graph" as the first line of every algorithm!
$endgroup$
The answer depends on exactly what problem you're solving.
If your goal is to produce an algorithm that correctly solves the problem on the restricted instances, then it's kind of up to you whether or not you check. It feels more robust to check the input but it's perfectly reasonable not to, and that puts you in the realm of promise problems. Here, the "user" promises that the input is valid, and you just have to determine whether the answer is yes or no.
If your goal is to produce an algorithm that decides whether the input is, e.g., a satisfiable 3CNF then, yes, you do need to check that the input has the properties it's supposed to have.
Your example of 3-colourability for acyclic graphs shows that there can be a big difference between the two approaches. Every acyclic graph is 3-colourable (even 2-colourable) so the algorithm for the promise problem is just output "yes". For the non-promise version, you need to check that your input is a valid graph representation, then check that the graph is acyclic, and only say "yes" if the input passes both tests.
In practical terms, most descriptions of graph algorithms tend to assume that the input is a valid encoding of a graph – it gets a bit tedious writing "Check the input is a valid encoding of a graph" as the first line of every algorithm!
answered 1 hour ago
David RicherbyDavid Richerby
68.5k15103194
68.5k15103194
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
add a comment |
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
I had almost finished adding more explanation when I clicked the link "there is an update to this answer ...". Oops, all my edit were gone. I had to write again. Of course, thanks for your update to my answer! (I will delete this comment shortly.)
$endgroup$
– Apass.Jack
46 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
$begingroup$
@Apass.Jack Oh no! I hate it when that happens. I try to remember to copy everything to the clipboard before clicking that notification but I often forget. It really ougth to warn you that you'll lose all your edits if you click that appealing-looking link. I'm sorry I stepped on your toes!
$endgroup$
– David Richerby
37 mins ago
add a comment |
Jahar Mehru is a new contributor. Be nice, and check out our Code of Conduct.
Jahar Mehru is a new contributor. Be nice, and check out our Code of Conduct.
Jahar Mehru is a new contributor. Be nice, and check out our Code of Conduct.
Jahar Mehru is a new contributor. Be nice, and check out our Code of Conduct.
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