{"paper_meta":{"paper_id":"arxiv:0706.2035","title":"0706.2035","authors":[],"year":null,"published":null,"updated":null,"venue_or_source":"arxiv-papers-shard","primary_category":"cs","secondary_categories":[],"doi":null,"license":"unknown","source_type":"pdf+shard","source_url":null,"pdf_url":null},"abstract":{"raw":"","cleaned":"","offsets":[]},"full_paper_text":{"raw_ordered_text":"arXiv:0706.2035v1 [cs.CC] 14 Jun 2007\nCritique of Feinstein’s Proof that P ̸= NP\nKyle Sabo\nRyan Schmitt\nMichael Silverman\nJune 14, 2007\nAbstract\nWe examine a proof by Craig Alan Feinstein that P ̸= NP. We\npresent counterexamples to claims made in his paper and expose a flaw\nin the methodology he uses to make his assertions. The fault in his\nargument is the incorrect use of reduction. Feinstein makes incorrect\nassumptions about the complexity of a problem based on the fact that\nthere is a more complex problem that can be used to solve it. His\npaper introduces the terminology “imaginary processor” to describe\nhow it is possible to beat the brute force reduction he offers to solve\nthe Subset-Sum problem. The claims made in the paper would not\nbe validly established even were imaginary processors to exist.\n1\nIntroduction\nIn this paper we analyze the argument set forth by Craig Alan Feinstein in\nhis paper, ‘A New and Elegant Argument that P ̸= NP’ [Fei07].We present\nhis argument and a counterargument by using his reasoning to “prove” a\nclearly trivial problem to be non-polynomially difficult.\n2\nFeinstein’s Argument\nFeinstein argues that P ̸= NP through the use of “imaginary processors.”\nThe author begins by introducing the problem of searching for a record out\nof n unsorted records. We state this as the following formal language.\nDefinition 2.1 Find-Record = {⟨R, r, i⟩| r = ri ∈{r1, r2, ..., rn} = R}\nFeinstein then claims that Find-Record is Θ(n). However, he continues\nby arguing that using multiple processors to search subsets of R it is possible\nto achieve a better runtime. The author’s definition of “multiple processors”\n1\n\nis one such that if there were a Find-Record algorithm using n processors,\nit would be possible to find the record in Θ(1) time.\nIn this way, Feinstein’s use of multiple processors is intuitively similar to\na non-deterministic Turing Machine. Each branch of the computation would\nuse another processor. However, the author never makes this comparison\nand it is possible, therefore, that his model of computation has some extra\nproperties that a traditional nondeterministic Turing Machine does not. We\nuse Feinstein’s notation of multiple processors in order to not bias the reader\nto disagree with his argument during its presentation.\nWe will see that\nunder every possible reasonable definition of multiple processor computation\nFeinstein’s arguments fail to prove his claim.\nFeinstein notes that using n processors to solve Find-Record is expen-\nsive, and that the best solution, he argues, is to instead optimize the sum\nof processors used and computations per processor. This, under his nota-\ntion, implies that the best solution is to use Θ(√n) processors, each doing\nΘ(√n) computations. Furthermore, he implies that the only way to beat\nthe efficiency of brute force is to use these multiple processors.\nFeinstein then describes the Subset-Sum problem. In his paper, the\nSubset-Sum problem is the task of deciding if for a given set of integers,\nS (||S|| = n), and another integer, x, it is possible to find set S′ (||S′|| =\nm ≤n) ⊆S such that the sum of all the elements in S′ is equal to x. We\ncan state this alternatively as:\nDefinition 2.2\nSubset-Sum =\n \n \n ⟨S, x, y⟩| y =\n \n \n \n1\nif there is a set S′ ⊆S | x =\nX\nw∈S′\nw\n0\notherwise\n \n \n \nTo solve the Subset-Sum problem, one can simply check all possible sub-\nsets, which would take Θ(2n) time, however, Feinstein notes there are faster\nsolutions, such as the Meet-In-The-Middle algorithm which is O(\n√\n2n).\nFeinstein argues that it is possible to beat the brute force approach of Find-\nRecord because there is an inherent mathematical structure to Subset-\nSum. That is, there is some property the Subset-Sum problem has that\nthe Find-Record problem does not have.\nEssentially, Feinstein argues that Subset-Sum is a search for a record\nwhere instead of a special number or label, the record is a subset with\nthe special property of having a sum equal to the desired value x. This\n2\n\nspecial record is a specific subset or set of subsets of S out of all possible\nsubsets. There are clearly 2n subsets, where again n is the size of S, and\ntherefore the Find-Record search through each subset will take Θ(2n).\nMore formally, the paper offers a non-polynomial time reduction to Find-\nRecord from Subset-Sum, though the paper never formally defines the\nreduction. Given an instance of Subset-Sum Feinstein describes how to\nenumerate all subsets and their sums, Θ(2n), and then searches that list for\na record that has a sum of x. However, as Feinstein notes, it is clear the\nalgorithm Meet-In-The-Middle has a runtime of O(\n√\n2n).\nTo reconcile this, Feinstein states there must be some inherent math-\nematical structure to Subset-Sum which allows for a faster solution than\nusing the reduction to Find-Record. Using the reduction above, in essence\nFeinstein implies that the since the Find-Record search is Θ(e), where e is\nthe number of elements to search through, and that since S has 2n subsets,\nwhere n is the size of S, it is impossible to beat the runtime O(2n) without\nsomehow searching the 2n subsets of S in faster than linear time.\nSince there is in fact a way to solve Subset-Sum faster than such a\nlinear search through the 2n subsets of S, and since, according to Feinstein,\nthe only way to go faster is to search through the 2n subsets in less than\n2n operations, there must be some aspect of the Meet-In-The-Middle\nalgorithm that recreates multiple processors which Feinstein calls imaginary\nprocessors. That is, since checking every possible subset would take longer\nthan Meet-In-The-Middle there must be some way that Meet-In-The-\nMiddle achieves this speedup.\nSince there is an inherent mathematical\nstructure to Subset-Sum, according to Feinstein, Meet-In-The-Middle\ncan create these “imaginary processors” that give the same boost in com-\nputation that using extra real processors does for Find-Record.\nFeinstein then states that there is a computational penalty to creating\nthese Θ(√r) imaginary processors, where r is the number of records to search\nthrough and we have each of the √r imaginary processors searching √r\nelements. He argues that because each creation requires a sorting operation,\nthe amount of time it takes to create each “imaginary processor,” is Θ(p)\nwhere p is the number of processors. Therefore, he argues, we are bound\nagain by minimizing the product of the number of ‘imaginary processors,’\ntimes the number of elements processed by each, giving us a lower bound\nof Θ(\n√\n2n).\nFeinstein states that since Subset-Sum is now bounded to\nΘ(\n√\n2n), the best solution for an NP complete problem requires more than\npolynomially many computations and therefore P ̸= NP. This bound is\nbased entirely on Feinstein’s idea that it is too expensive to create more\nimaginary processors and that these imaginary processors are the only way\n3\n\nto get a speedup.\n3\nBrief Summary\nIn the earlier section we present Feinstein’s argument in as rigorous and\nformal a way as possible without changing the spirit or meaning of his ar-\ngument. We now reword his argument to the equivalent and more simple\nlogically valid statements.\n1. A brute force search for Subset-Sum takes Θ(2n) unless we use mul-\ntiple processors. This is because we have a non-polynomial time re-\nduction from Subset-Sum to Find-Record and records cannot be\nsolved faster than linear time without multiple processors.\n2. Meet-In-The-Middle is Θ(\n√\n2n)\n3. Meet-In-The-Middle must simulate multiple processors by creating\nimaginary processors to achieve better than Θ(2n) because of 1 and 2.\n4. The only way to get a faster runtime for Subset-Sum without multiple\nprocessors is to use imaginary processors.\n5. The best runtime possible for Subset-Sum with imaginary processors\nis Θ(\n√\n2n) due to penalties of creating new imaginary processors.\n6. Since 4 and 5, P ̸= NP.\n4\nCounterargument\nThe fault in Feinstein’s paper lies in neglecting proof of some nontrivial\nstatements that are implied by his argument. Most simply, the paper fails\nto prove that the reduction given in argument 1 above is the most efficient\nreduction. The author argues that since the reduction shown to exist in\nargument 1 above is Θ(2n), any such algorithm that has faster than Θ(2n)\nruntime for Subset-Sum must be imitating some more powerful form of\ncomputation. This is simply not true. There could in concept be a more\nefficient reduction that could even be polynomial time. Without proof that\nthe reduction in 3 is the most efficient such reduction, it is impossible to\nmake the author’s claim.\nEven so, let’s assume the reduction in 3 is optimal. The author’s argu-\nment then relies on the fact that the non-polynomial solution of brute force\n4\n\nhas a non-polynomial runtime to imply that the problem the non-polynomial\ntime algorithm being used to solve must somehow be using imaginary pro-\ncessors.\nIf Feinstein were correct, then many things use imaginary processors.\nFor instance, consider checking if the first number in a list of n numbers is\n3, where n varies on input, called First-Three. We offer a non-polynomial\nreduction from First-Three to records. We can take the first number of\nour input list, and put it in a list of n random but non-3 elements, and\nthen we can repeat putting the element into new random lists n! times.\nThis algorithm is a non-polynomial reduction from First-Three to Find-\nRecord. Therefore, according to Feinstein’s reasoning, First-Three not\nonly cannot be done in polynomial time, but to do better than the speed of\nthe reduction it must use imaginary processors. However, it seems clear that\nthe code that outputs true if the first element is 3, does not use imaginary\nprocessors, and is constant time. If Feinstein were correct, First-Three\ncould not be solved this way.\nReferences\n[Fei07] C.A. Feinstein. A New and Elegant Argument that P is not equal\nto NP. 2007. arXiv:cs/0607093v2\n5","paragraphs":[{"paragraph_id":"p1","order":1,"text":"arXiv:0706.2035v1 [cs.CC] 14 Jun 2007\nCritique of Feinstein’s Proof that P ̸= NP\nKyle Sabo\nRyan Schmitt\nMichael Silverman\nJune 14, 2007\nAbstract\nWe examine a proof by Craig Alan Feinstein that P ̸= NP. We\npresent counterexamples to claims made in his paper and expose a flaw\nin the methodology he uses to make his assertions. The fault in his\nargument is the incorrect use of reduction. Feinstein makes incorrect\nassumptions about the complexity of a problem based on the fact that\nthere is a more complex problem that can be used to solve it. His\npaper introduces the terminology “imaginary processor” to describe\nhow it is possible to beat the brute force reduction he offers to solve\nthe Subset-Sum problem. The claims made in the paper would not\nbe validly established even were imaginary processors to exist.\n1\nIntroduction\nIn this paper we analyze the argument set forth by Craig Alan Feinstein in\nhis paper, ‘A New and Elegant Argument that P ̸= NP’ [Fei07].We present\nhis argument and a counterargument by using his reasoning to “prove” a\nclearly trivial problem to be non-polynomially difficult.\n2\nFeinstein’s Argument\nFeinstein argues that P ̸= NP through the use of “imaginary processors.”\nThe author begins by introducing the problem of searching for a record out\nof n unsorted records. We state this as the following formal language.\nDefinition 2.1 Find-Record = {⟨R, r, i⟩| r = ri ∈{r1, r2, ..., rn} = R}\nFeinstein then claims that Find-Record is Θ(n). However, he continues\nby arguing that using multiple processors to search subsets of R it is possible\nto achieve a better runtime. The author’s definition of “multiple processors”\n1"},{"paragraph_id":"p2","order":2,"text":"is one such that if there were a Find-Record algorithm using n processors,\nit would be possible to find the record in Θ(1) time.\nIn this way, Feinstein’s use of multiple processors is intuitively similar to\na non-deterministic Turing Machine. Each branch of the computation would\nuse another processor. However, the author never makes this comparison\nand it is possible, therefore, that his model of computation has some extra\nproperties that a traditional nondeterministic Turing Machine does not. We\nuse Feinstein’s notation of multiple processors in order to not bias the reader\nto disagree with his argument during its presentation.\nWe will see that\nunder every possible reasonable definition of multiple processor computation\nFeinstein’s arguments fail to prove his claim.\nFeinstein notes that using n processors to solve Find-Record is expen-\nsive, and that the best solution, he argues, is to instead optimize the sum\nof processors used and computations per processor. This, under his nota-\ntion, implies that the best solution is to use Θ(√n) processors, each doing\nΘ(√n) computations. Furthermore, he implies that the only way to beat\nthe efficiency of brute force is to use these multiple processors.\nFeinstein then describes the Subset-Sum problem. In his paper, the\nSubset-Sum problem is the task of deciding if for a given set of integers,\nS (||S|| = n), and another integer, x, it is possible to find set S′ (||S′|| =\nm ≤n) ⊆S such that the sum of all the elements in S′ is equal to x. We\ncan state this alternatively as:\nDefinition 2.2\nSubset-Sum ="},{"paragraph_id":"p3","order":3,"text":"⟨S, x, y⟩| y ="},{"paragraph_id":"p4","order":4,"text":"1\nif there is a set S′ ⊆S | x =\nX\nw∈S′\nw\n0\notherwise"},{"paragraph_id":"p5","order":5,"text":"To solve the Subset-Sum problem, one can simply check all possible sub-\nsets, which would take Θ(2n) time, however, Feinstein notes there are faster\nsolutions, such as the Meet-In-The-Middle algorithm which is O(\n√\n2n).\nFeinstein argues that it is possible to beat the brute force approach of Find-\nRecord because there is an inherent mathematical structure to Subset-\nSum. That is, there is some property the Subset-Sum problem has that\nthe Find-Record problem does not have.\nEssentially, Feinstein argues that Subset-Sum is a search for a record\nwhere instead of a special number or label, the record is a subset with\nthe special property of having a sum equal to the desired value x. This\n2"},{"paragraph_id":"p6","order":6,"text":"special record is a specific subset or set of subsets of S out of all possible\nsubsets. There are clearly 2n subsets, where again n is the size of S, and\ntherefore the Find-Record search through each subset will take Θ(2n).\nMore formally, the paper offers a non-polynomial time reduction to Find-\nRecord from Subset-Sum, though the paper never formally defines the\nreduction. Given an instance of Subset-Sum Feinstein describes how to\nenumerate all subsets and their sums, Θ(2n), and then searches that list for\na record that has a sum of x. However, as Feinstein notes, it is clear the\nalgorithm Meet-In-The-Middle has a runtime of O(\n√\n2n).\nTo reconcile this, Feinstein states there must be some inherent math-\nematical structure to Subset-Sum which allows for a faster solution than\nusing the reduction to Find-Record. Using the reduction above, in essence\nFeinstein implies that the since the Find-Record search is Θ(e), where e is\nthe number of elements to search through, and that since S has 2n subsets,\nwhere n is the size of S, it is impossible to beat the runtime O(2n) without\nsomehow searching the 2n subsets of S in faster than linear time.\nSince there is in fact a way to solve Subset-Sum faster than such a\nlinear search through the 2n subsets of S, and since, according to Feinstein,\nthe only way to go faster is to search through the 2n subsets in less than\n2n operations, there must be some aspect of the Meet-In-The-Middle\nalgorithm that recreates multiple processors which Feinstein calls imaginary\nprocessors. That is, since checking every possible subset would take longer\nthan Meet-In-The-Middle there must be some way that Meet-In-The-\nMiddle achieves this speedup.\nSince there is an inherent mathematical\nstructure to Subset-Sum, according to Feinstein, Meet-In-The-Middle\ncan create these “imaginary processors” that give the same boost in com-\nputation that using extra real processors does for Find-Record.\nFeinstein then states that there is a computational penalty to creating\nthese Θ(√r) imaginary processors, where r is the number of records to search\nthrough and we have each of the √r imaginary processors searching √r\nelements. He argues that because each creation requires a sorting operation,\nthe amount of time it takes to create each “imaginary processor,” is Θ(p)\nwhere p is the number of processors. Therefore, he argues, we are bound\nagain by minimizing the product of the number of ‘imaginary processors,’\ntimes the number of elements processed by each, giving us a lower bound\nof Θ(\n√\n2n).\nFeinstein states that since Subset-Sum is now bounded to\nΘ(\n√\n2n), the best solution for an NP complete problem requires more than\npolynomially many computations and therefore P ̸= NP. This bound is\nbased entirely on Feinstein’s idea that it is too expensive to create more\nimaginary processors and that these imaginary processors are the only way\n3"},{"paragraph_id":"p7","order":7,"text":"to get a speedup.\n3\nBrief Summary\nIn the earlier section we present Feinstein’s argument in as rigorous and\nformal a way as possible without changing the spirit or meaning of his ar-\ngument. We now reword his argument to the equivalent and more simple\nlogically valid statements.\n1. A brute force search for Subset-Sum takes Θ(2n) unless we use mul-\ntiple processors. This is because we have a non-polynomial time re-\nduction from Subset-Sum to Find-Record and records cannot be\nsolved faster than linear time without multiple processors.\n2. Meet-In-The-Middle is Θ(\n√\n2n)\n3. Meet-In-The-Middle must simulate multiple processors by creating\nimaginary processors to achieve better than Θ(2n) because of 1 and 2.\n4. The only way to get a faster runtime for Subset-Sum without multiple\nprocessors is to use imaginary processors.\n5. The best runtime possible for Subset-Sum with imaginary processors\nis Θ(\n√\n2n) due to penalties of creating new imaginary processors.\n6. Since 4 and 5, P ̸= NP.\n4\nCounterargument\nThe fault in Feinstein’s paper lies in neglecting proof of some nontrivial\nstatements that are implied by his argument. Most simply, the paper fails\nto prove that the reduction given in argument 1 above is the most efficient\nreduction. The author argues that since the reduction shown to exist in\nargument 1 above is Θ(2n), any such algorithm that has faster than Θ(2n)\nruntime for Subset-Sum must be imitating some more powerful form of\ncomputation. This is simply not true. There could in concept be a more\nefficient reduction that could even be polynomial time. Without proof that\nthe reduction in 3 is the most efficient such reduction, it is impossible to\nmake the author’s claim.\nEven so, let’s assume the reduction in 3 is optimal. The author’s argu-\nment then relies on the fact that the non-polynomial solution of brute force\n4"},{"paragraph_id":"p8","order":8,"text":"has a non-polynomial runtime to imply that the problem the non-polynomial\ntime algorithm being used to solve must somehow be using imaginary pro-\ncessors.\nIf Feinstein were correct, then many things use imaginary processors.\nFor instance, consider checking if the first number in a list of n numbers is\n3, where n varies on input, called First-Three. We offer a non-polynomial\nreduction from First-Three to records. We can take the first number of\nour input list, and put it in a list of n random but non-3 elements, and\nthen we can repeat putting the element into new random lists n! times.\nThis algorithm is a non-polynomial reduction from First-Three to Find-\nRecord. Therefore, according to Feinstein’s reasoning, First-Three not\nonly cannot be done in polynomial time, but to do better than the speed of\nthe reduction it must use imaginary processors. However, it seems clear that\nthe code that outputs true if the first element is 3, does not use imaginary\nprocessors, and is constant time. If Feinstein were correct, First-Three\ncould not be solved this way.\nReferences\n[Fei07] C.A. Feinstein. A New and Elegant Argument that P is not equal\nto NP. 2007. arXiv:cs/0607093v2\n5"}],"pages":[{"page":1,"text":"arXiv:0706.2035v1 [cs.CC] 14 Jun 2007\nCritique of Feinstein’s Proof that P ̸= NP\nKyle Sabo\nRyan Schmitt\nMichael Silverman\nJune 14, 2007\nAbstract\nWe examine a proof by Craig Alan Feinstein that P ̸= NP. We\npresent counterexamples to claims made in his paper and expose a flaw\nin the methodology he uses to make his assertions. The fault in his\nargument is the incorrect use of reduction. Feinstein makes incorrect\nassumptions about the complexity of a problem based on the fact that\nthere is a more complex problem that can be used to solve it. His\npaper introduces the terminology “imaginary processor” to describe\nhow it is possible to beat the brute force reduction he offers to solve\nthe Subset-Sum problem. The claims made in the paper would not\nbe validly established even were imaginary processors to exist.\n1\nIntroduction\nIn this paper we analyze the argument set forth by Craig Alan Feinstein in\nhis paper, ‘A New and Elegant Argument that P ̸= NP’ [Fei07].We present\nhis argument and a counterargument by using his reasoning to “prove” a\nclearly trivial problem to be non-polynomially difficult.\n2\nFeinstein’s Argument\nFeinstein argues that P ̸= NP through the use of “imaginary processors.”\nThe author begins by introducing the problem of searching for a record out\nof n unsorted records. We state this as the following formal language.\nDefinition 2.1 Find-Record = {⟨R, r, i⟩| r = ri ∈{r1, r2, ..., rn} = R}\nFeinstein then claims that Find-Record is Θ(n). However, he continues\nby arguing that using multiple processors to search subsets of R it is possible\nto achieve a better runtime. The author’s definition of “multiple processors”\n1"},{"page":2,"text":"is one such that if there were a Find-Record algorithm using n processors,\nit would be possible to find the record in Θ(1) time.\nIn this way, Feinstein’s use of multiple processors is intuitively similar to\na non-deterministic Turing Machine. Each branch of the computation would\nuse another processor. However, the author never makes this comparison\nand it is possible, therefore, that his model of computation has some extra\nproperties that a traditional nondeterministic Turing Machine does not. We\nuse Feinstein’s notation of multiple processors in order to not bias the reader\nto disagree with his argument during its presentation.\nWe will see that\nunder every possible reasonable definition of multiple processor computation\nFeinstein’s arguments fail to prove his claim.\nFeinstein notes that using n processors to solve Find-Record is expen-\nsive, and that the best solution, he argues, is to instead optimize the sum\nof processors used and computations per processor. This, under his nota-\ntion, implies that the best solution is to use Θ(√n) processors, each doing\nΘ(√n) computations. Furthermore, he implies that the only way to beat\nthe efficiency of brute force is to use these multiple processors.\nFeinstein then describes the Subset-Sum problem. In his paper, the\nSubset-Sum problem is the task of deciding if for a given set of integers,\nS (||S|| = n), and another integer, x, it is possible to find set S′ (||S′|| =\nm ≤n) ⊆S such that the sum of all the elements in S′ is equal to x. We\ncan state this alternatively as:\nDefinition 2.2\nSubset-Sum =\n \n \n ⟨S, x, y⟩| y =\n \n \n \n1\nif there is a set S′ ⊆S | x =\nX\nw∈S′\nw\n0\notherwise\n \n \n \nTo solve the Subset-Sum problem, one can simply check all possible sub-\nsets, which would take Θ(2n) time, however, Feinstein notes there are faster\nsolutions, such as the Meet-In-The-Middle algorithm which is O(\n√\n2n).\nFeinstein argues that it is possible to beat the brute force approach of Find-\nRecord because there is an inherent mathematical structure to Subset-\nSum. That is, there is some property the Subset-Sum problem has that\nthe Find-Record problem does not have.\nEssentially, Feinstein argues that Subset-Sum is a search for a record\nwhere instead of a special number or label, the record is a subset with\nthe special property of having a sum equal to the desired value x. This\n2"},{"page":3,"text":"special record is a specific subset or set of subsets of S out of all possible\nsubsets. There are clearly 2n subsets, where again n is the size of S, and\ntherefore the Find-Record search through each subset will take Θ(2n).\nMore formally, the paper offers a non-polynomial time reduction to Find-\nRecord from Subset-Sum, though the paper never formally defines the\nreduction. Given an instance of Subset-Sum Feinstein describes how to\nenumerate all subsets and their sums, Θ(2n), and then searches that list for\na record that has a sum of x. However, as Feinstein notes, it is clear the\nalgorithm Meet-In-The-Middle has a runtime of O(\n√\n2n).\nTo reconcile this, Feinstein states there must be some inherent math-\nematical structure to Subset-Sum which allows for a faster solution than\nusing the reduction to Find-Record. Using the reduction above, in essence\nFeinstein implies that the since the Find-Record search is Θ(e), where e is\nthe number of elements to search through, and that since S has 2n subsets,\nwhere n is the size of S, it is impossible to beat the runtime O(2n) without\nsomehow searching the 2n subsets of S in faster than linear time.\nSince there is in fact a way to solve Subset-Sum faster than such a\nlinear search through the 2n subsets of S, and since, according to Feinstein,\nthe only way to go faster is to search through the 2n subsets in less than\n2n operations, there must be some aspect of the Meet-In-The-Middle\nalgorithm that recreates multiple processors which Feinstein calls imaginary\nprocessors. That is, since checking every possible subset would take longer\nthan Meet-In-The-Middle there must be some way that Meet-In-The-\nMiddle achieves this speedup.\nSince there is an inherent mathematical\nstructure to Subset-Sum, according to Feinstein, Meet-In-The-Middle\ncan create these “imaginary processors” that give the same boost in com-\nputation that using extra real processors does for Find-Record.\nFeinstein then states that there is a computational penalty to creating\nthese Θ(√r) imaginary processors, where r is the number of records to search\nthrough and we have each of the √r imaginary processors searching √r\nelements. He argues that because each creation requires a sorting operation,\nthe amount of time it takes to create each “imaginary processor,” is Θ(p)\nwhere p is the number of processors. Therefore, he argues, we are bound\nagain by minimizing the product of the number of ‘imaginary processors,’\ntimes the number of elements processed by each, giving us a lower bound\nof Θ(\n√\n2n).\nFeinstein states that since Subset-Sum is now bounded to\nΘ(\n√\n2n), the best solution for an NP complete problem requires more than\npolynomially many computations and therefore P ̸= NP. This bound is\nbased entirely on Feinstein’s idea that it is too expensive to create more\nimaginary processors and that these imaginary processors are the only way\n3"},{"page":4,"text":"to get a speedup.\n3\nBrief Summary\nIn the earlier section we present Feinstein’s argument in as rigorous and\nformal a way as possible without changing the spirit or meaning of his ar-\ngument. We now reword his argument to the equivalent and more simple\nlogically valid statements.\n1. A brute force search for Subset-Sum takes Θ(2n) unless we use mul-\ntiple processors. This is because we have a non-polynomial time re-\nduction from Subset-Sum to Find-Record and records cannot be\nsolved faster than linear time without multiple processors.\n2. Meet-In-The-Middle is Θ(\n√\n2n)\n3. Meet-In-The-Middle must simulate multiple processors by creating\nimaginary processors to achieve better than Θ(2n) because of 1 and 2.\n4. The only way to get a faster runtime for Subset-Sum without multiple\nprocessors is to use imaginary processors.\n5. The best runtime possible for Subset-Sum with imaginary processors\nis Θ(\n√\n2n) due to penalties of creating new imaginary processors.\n6. Since 4 and 5, P ̸= NP.\n4\nCounterargument\nThe fault in Feinstein’s paper lies in neglecting proof of some nontrivial\nstatements that are implied by his argument. Most simply, the paper fails\nto prove that the reduction given in argument 1 above is the most efficient\nreduction. The author argues that since the reduction shown to exist in\nargument 1 above is Θ(2n), any such algorithm that has faster than Θ(2n)\nruntime for Subset-Sum must be imitating some more powerful form of\ncomputation. This is simply not true. There could in concept be a more\nefficient reduction that could even be polynomial time. Without proof that\nthe reduction in 3 is the most efficient such reduction, it is impossible to\nmake the author’s claim.\nEven so, let’s assume the reduction in 3 is optimal. The author’s argu-\nment then relies on the fact that the non-polynomial solution of brute force\n4"},{"page":5,"text":"has a non-polynomial runtime to imply that the problem the non-polynomial\ntime algorithm being used to solve must somehow be using imaginary pro-\ncessors.\nIf Feinstein were correct, then many things use imaginary processors.\nFor instance, consider checking if the first number in a list of n numbers is\n3, where n varies on input, called First-Three. We offer a non-polynomial\nreduction from First-Three to records. We can take the first number of\nour input list, and put it in a list of n random but non-3 elements, and\nthen we can repeat putting the element into new random lists n! times.\nThis algorithm is a non-polynomial reduction from First-Three to Find-\nRecord. Therefore, according to Feinstein’s reasoning, First-Three not\nonly cannot be done in polynomial time, but to do better than the speed of\nthe reduction it must use imaginary processors. However, it seems clear that\nthe code that outputs true if the first element is 3, does not use imaginary\nprocessors, and is constant time. If Feinstein were correct, First-Three\ncould not be solved this way.\nReferences\n[Fei07] C.A. Feinstein. A New and Elegant Argument that P is not equal\nto NP. 2007. arXiv:cs/0607093v2\n5"}]},"section_tree":[],"assets":{"figures":[],"tables":[],"images":[]},"math_expressions":{"inline":[],"block":[{"equation_id":"eq1","equation_number":null,"raw_text":"Critique of Feinstein’s Proof that P ̸= NP","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq2","equation_number":null,"raw_text":"We examine a proof by Craig Alan Feinstein that P ̸= NP. We","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq3","equation_number":null,"raw_text":"his paper, ‘A New and Elegant Argument that P ̸= NP’ [Fei07].We present","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq4","equation_number":null,"raw_text":"Feinstein argues that P ̸= NP through the use of “imaginary processors.”","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq5","equation_number":null,"raw_text":"Definition 2.1 Find-Record = {⟨R, r, i⟩| r = ri ∈{r1, r2, ..., rn} = R}","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq6","equation_number":null,"raw_text":"S (||S|| = n), and another integer, x, it is possible to find set S′ (||S′|| =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq7","equation_number":null,"raw_text":"Subset-Sum =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq8","equation_number":null,"raw_text":"⟨S, x, y⟩| y =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq9","equation_number":null,"raw_text":"if there is a set S′ ⊆S | x =","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq10","equation_number":null,"raw_text":"polynomially many computations and therefore P ̸= NP. This bound is","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"},{"equation_id":"eq11","equation_number":null,"raw_text":"6. Since 4 and 5, P ̸= NP.","raw_latex":null,"normalized_latex":null,"display_type":"block_candidate"}],"equation_refs":[]},"symbol_table":[],"references":{"in_text_markers":[],"bibliography":[]},"claims":[],"flow_graph":{"nodes":[],"edges":[]},"quality":{"text_char_count":9906,"parse_confidence":0.5,"equation_parse_rate_proxy":0.55,"citation_resolution_rate_proxy":0.0,"structure_coverage_proxy":0.0,"asset_coverage_proxy":0.0,"accepted_for_training":true}}