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A Weird Soviet Method to Partially Solve the Perebor Problem
The document discusses a Soviet method by Alexander Razborov for partially solving the Perebor problem, which relates to NP vs. P complexity theory. The author argues that NP cannot be converted to P and proposes using cellular automata and Boolean algebra to demonstrate that 'not' operations are unnecessary in polynomial time algorithms. Additionally, the document provides various links to external resources for deeper understanding of the mentioned concepts.
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Similar to A Weird Soviet Method to Partially Solve the Perebor Problem
A Weird Soviet Method to Partially Solve the Perebor Problem
- 1. A Weird SovietMethod to Partially Solve the Perebor Problem How is this method related to my NP vs P solution Sing Kuang Tan [email protected] 17 August 2021
- 2. Perebor Problems • Pereborproblems are difficult problems which current methods can only solve it by brute force algorithm • https://drdoane.com/the-perebor-problem/ • Perebor problems are the NP (Non-Deterministic Polynomial time complexity) problem • Can we convert NP into P (Polynomial time complexity) problem? • My answer is NO. • My solution to NP vs P problem is NP≠P Link to my paper https://www.slideshare.net/SingKuangTan/brief-np-vspexplain-249524831 Prove Np not equal P using Markov Random Field and Boolean Algebra Simplification https://vixra.org/abs/2105.0181
- 3. Alexander Razborov (fromSoviet Union) • He developed the method to partially solve Perebor Problem. • Bio: https://en.wikipedia.org/wiki/Alexander_Razborov • In 1985, Razborov has proved that Clique problem of a graph requires a Non- Deterministic Polynomial time complexity (NP) algorithm if the algorithm is implemented using Boolean algebra consist of only “And” and “Or” operations • As there is no “Not” operation, so it only partially proves that NP≠P • Later I will show how I circumvent the problem of no “Not” operation • https://blog.computationalcomplexity.org/2005/09/circuit-complexity-and-p-versus-np.html • https://www.cs.utexas.edu/~danama/courses/adv-comp/scribe5.pdf • A. A. Razborov, Lower bounds on the monotone complexity of some Boolean functions, Dokl. Akad. Nauk. SSSR, 281(4):798-801, 1985 • At that time, Soviet is one of the country leading in the problem on NP vs P • Soviet Song “Our Army” LOL • https://www.youtube.com/watch?v=5yYQIa-4rLE • Enable subtitles
- 4. Clique Problem • https://en.wikipedia.org/wiki/Clique_problem 4-cliquein this 7 vertices graph can be found using brute force “Or” operation “And” operation This digital circuit is used to detect whether a 3-clique exists in this 4 vertices graph This circuit has only “And” and “Or” operations (In literature, this is called a monotone circuit)
- 5. Cellular Automata • Whatis Cellular Automata? • https://mathworld.wolfram.com/CellularAutomaton.ht ml • https://natureofcode.com/book/chapter-7-cellular- automata/ • https://towardsdatascience.com/simple-but-stunning- animated-cellular-automata-in-python-c912e0c156a9 • https://towardsdatascience.com/algorithmic-beauty-an- introduction-to-cellular-automata-f53179b3cf8f • Animation • https://www.youtube.com/watch?v=3MJ8deSCOCE
- 6. “Not” operations areunnecessary • I will show using “Not” operations are unnecessary for implementation of any algorithms using Boolean algebra • I will use Cellular Automata as an Universal Computing model • A Polynomial time (P) Cellular Automata can be implemented without any “Not” operation First layer is the input of an algorithm Last layer is the output of an algorithm Cellular Automata can implement any algorithm Different algorithm can be implemented by changing the rule
- 7. Cellular Automata canEmulate any Computer Algorithm An algorithm implemented on a modern computer Can also be implemented using Cellular Automata A Polynomial time complexity (P) algorithm remains Polynomial time (P) after implemented on Cellular Automata A Non-Deterministic Polynomial time complexity (NP) algorithm remains Non-Deterministic Polynomial time (NP) after implemented on Cellular Automata
- 8. Simple Cellular AutomataExample 5 bit binary input 1 bit binary output
- 9. Simple Cellular AutomataExample Each intermediate output is dependent on input Each output is dependent on intermediate outputs • This looks like a convolutional neural network • Neural Network can be represented by multi-layer Boolean algebra • Multi-layer Boolean algebra simplification gives us insight on how to train a neural network
- 10. Simply Cellular AutomataExample a1,1 a1,2 a1,3 a1,4 a1,5 a2,2 a2,3 a2,4 a3,3 I give a variable name to each cell of the Cellular Automata a1,1 Boolean Algebra or Circuit 𝑎3,3 a1,2 a1,3 a1,4 a1,5
- 11. Simply Cellular AutomataExample ai,j ai,j+1 Ai,j+2 ai+1,j+1 𝑎𝑖+1,𝑗+1 = 𝑎𝑖,𝑗 𝑎𝑖,𝑗+1 𝑎𝑖,𝑗+2 + 𝑎𝑖,𝑗𝑎𝑖,𝑗+1 𝑎𝑖,𝑗+2+ 𝑎𝑖,𝑗 𝑎𝑖,𝑗+1 𝑎𝑖,𝑗+2 + 𝑎𝑖,𝑗 𝑎𝑖,𝑗+1 𝑎𝑖,𝑗+2 Bar above ai,j means “Not” operation on ai,j Addition means “Or” operation Multiplication means “And” operation The rule can be implemented using the Boolean algebra below
- 12. Simply Cellular AutomataExample a1,1 a1,2 a1,3 a1,4 a1,5 a2,2 a2,3 a2,4 a3,3 𝑎2,2 = 𝑎1,1 𝑎1,2 𝑎1,3 + 𝑎1,1𝑎1,2 𝑎1,3+ 𝑎1,1 𝑎1,2 𝑎1,3 + 𝑎1,1 𝑎1,2 𝑎1,3 𝑎2,3 = 𝑎1,2 𝑎1,3 𝑎1,4 + 𝑎1,2𝑎1,3 𝑎1,4+ 𝑎1,2 𝑎1,3 𝑎1,4 + 𝑎1,2 𝑎1,3 𝑎1,4 𝑎2,4 = 𝑎1,3 𝑎1,4 𝑎1,5 + 𝑎1,3𝑎1,4 𝑎1,5+ 𝑎1,3 𝑎1,4 𝑎1,5 + 𝑎1,3 𝑎1,4 𝑎1,5 𝑎3,3 = 𝑎2,2 𝑎2,3 𝑎2,4 + 𝑎2,2𝑎2,3 𝑎2,4+ 𝑎2,2 𝑎2,3 𝑎2,4 + 𝑎2,2 𝑎2,3 𝑎2,4 Represent the Cellular Automata by Boolean algebra,
- 13. Simply Cellular AutomataExample 𝑎2,2,0 = 𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 + 𝑎1,1,0𝑎1,2,0𝑎1,3,0 𝑎2,2,1 = 𝑎1,1,1𝑎1,2,0𝑎1,3,0 + 𝑎1,1,0𝑎1,2,1𝑎1,3,1+ 𝑎1,1,0𝑎1,2,1𝑎1,3,0 + 𝑎1,1,0𝑎1,2,0𝑎1,3,1 𝑎2,3,0 = 𝑎1,2,1𝑎1,3,1𝑎1,4,1 + 𝑎1,2,1𝑎1,3,1𝑎1,4,0+ 𝑎1,2,1𝑎1,3,0𝑎1,4,1 + 𝑎1,2,0𝑎1,3,0𝑎1,4,0 𝑎2,3,1 = 𝑎1,2,1𝑎1,3,0𝑎1,4,0 + 𝑎1,2,0𝑎1,3,1𝑎1,4,1+ 𝑎1,2,0𝑎1,3,1𝑎1,4,0 + 𝑎1,2,0𝑎1,3,0𝑎1,4,1 𝑎2,4,0 = 𝑎1,3,1𝑎1,4,1𝑎1,5,1 + 𝑎1,3,1𝑎1,4,1𝑎1,5,0+ 𝑎1,3,1𝑎1,4,0𝑎1,5,1 + 𝑎1,3,0𝑎1,4,0𝑎1,5,0 𝑎2,4,1 = 𝑎1,3,1𝑎1,4,0𝑎1,5,0 + 𝑎1,3,0𝑎1,4,1𝑎1,5,1+ 𝑎1,3,0𝑎1,4,1𝑎1,5,0 + 𝑎1,3,0𝑎1,4,0𝑎1,5,1 𝑎3,3,1 = (𝑎1,1,1𝑎1,2,0𝑎1,3,0 + 𝑎1,1,0𝑎1,2,1𝑎1,3,1+ 𝑎1,1,0𝑎1,2,1𝑎1,3,0 + 𝑎1,1,0𝑎1,2,0𝑎1,3,1)(𝑎1,2,1𝑎1,3,1𝑎1,4,1 + 𝑎1,2,1𝑎1,3,1𝑎1,4,0+ 𝑎1,2,1𝑎1,3,0𝑎1,4,1 + 𝑎1,2,0𝑎1,3,0𝑎1,4,0)(𝑎1,3,1𝑎1,4,1𝑎1,5,1 + 𝑎1,3,1𝑎1,4,1𝑎1,5,0+ 𝑎1,3,1𝑎1,4,0𝑎1,5,1 + 𝑎1,3,0𝑎1,4,0𝑎1,5,0) + (𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 + 𝑎1,1,0𝑎1,2,0𝑎1,3,0)(𝑎1,2,1𝑎1,3,0𝑎1,4,0 + 𝑎1,2,0𝑎1,3,1𝑎1,4,1+ 𝑎1,2,0𝑎1,3,1𝑎1,4,0 + 𝑎1,2,0𝑎1,3,0𝑎1,4,1)(𝑎1,3,1𝑎1,4,0𝑎1,5,0 + 𝑎1,3,0𝑎1,4,1𝑎1,5,1+ 𝑎1,3,0𝑎1,4,1𝑎1,5,0 + 𝑎1,3,0𝑎1,4,0𝑎1,5,1)+ (𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 + 𝑎1,1,0𝑎1,2,0𝑎1,3,0)(𝑎1,2,1𝑎1,3,0𝑎1,4,0 + 𝑎1,2,0𝑎1,3,1𝑎1,4,1+ 𝑎1,2,0𝑎1,3,1𝑎1,4,0 + 𝑎1,2,0𝑎1,3,0𝑎1,4,1)(𝑎1,3,1𝑎1,4,1𝑎1,5,1 + 𝑎1,3,1𝑎1,4,1𝑎1,5,0+ 𝑎1,3,1𝑎1,4,0𝑎1,5,1 + 𝑎1,3,0𝑎1,4,0𝑎1,5,0) + (𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 + 𝑎1,1,0𝑎1,2,0𝑎1,3,0)(𝑎1,2,1𝑎1,3,1𝑎1,4,1 + 𝑎1,2,1𝑎1,3,1𝑎1,4,0+ 𝑎1,2,1𝑎1,3,0𝑎1,4,1 + 𝑎1,2,0𝑎1,3,0𝑎1,4,0)(𝑎1,3,1𝑎1,4,0𝑎1,5,0 + 𝑎1,3,0𝑎1,4,1𝑎1,5,1+ 𝑎1,3,0𝑎1,4,1𝑎1,5,0 + 𝑎1,3,0𝑎1,4,0𝑎1,5,1) Use the representation below to simplify the equations, 𝑎𝑖,𝑗,0 = 𝑎𝑖,𝑗 𝑎𝑖,𝑗,1 = 𝑎𝑖,𝑗 Combine all equations into 1 equation Later can do expansion or factorization on this 1 equation 𝑎3,3,1 = 𝑎2,2,1𝑎2,3,0𝑎2,4,0 + 𝑎2,2,0𝑎2,3,1𝑎2,4,1+ 𝑎2,2,0𝑎2,3,1𝑎2,4,0 + 𝑎2,2,0𝑎2,3,0𝑎2,4,1
- 14. Simply Cellular AutomataExample a1,1 a1,2 a1,3 a1,4 a1,5 a2,2 a2,3 a2,4 a3,3 Find a3,3 in terms of a1,1, a1,2, a1,3, a1,4, a1,5 is a3,3=a1,1,0a1,2,0a1,3,0a1,4,0a1,5,1 + a1,1,0a1,2,0a1,3,0a1,4,1a1,5,0 + a1,1,0a1,2,0a1,3,0a1,4,1a1,5,1 + a1,1,0a1,2,1a1,3,1a1,4,0a1,5,1 + a1,1,0a1,2,1a1,3,1a1,4,1a1,5,0 + a1,1,0a1,2,1a1,3,1a1,4,1a1,5,1 + a1,1,1a1,2,0a1,3,0a1,4,0a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,0a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,0a1,5,1 + a1,1,1a1,2,0a1,3,1a1,4,1a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,1a1,5,1 + a1,1,1a1,2,1a1,3,0a1,4,0a1,5,0 + a1,1,1a1,2,1a1,3,0a1,4,0a1,5,1 + a1,1,1a1,2,1a1,3,0a1,4,1a1,5,0 + a1,1,1a1,2,1a1,3,0a1,4,1a1,5,1 + a1,1,1a1,2,1a1,3,1a1,4,0a1,5,0 Use the representation below to simplify the equations, 𝑎𝑖,𝑗,0 = 𝑎𝑖,𝑗 𝑎𝑖,𝑗,1 = 𝑎𝑖,𝑗
- 15. Simply Cellular AutomataExample a1,1,0 Use the representation below to simplify the equations, 𝑎𝑖,𝑗,0 = 𝑎𝑖,𝑗 𝑎𝑖,𝑗,1 = 𝑎𝑖,𝑗 a1,1,1 a1,2,0 a1,2,1 a1,3,0 a1,3,1 a1,4,0 a1,4,1 a1,5,0 a1,5,1 a2,2,0 a2,2,1 a2,3,0 a2,3,1 a2,4,0 a2,4,1 A3,3,0 A3,3,1 Represent the Cellular Automata by Boolean algebra, 𝑎2,2,0 = 𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 + 𝑎1,1,0𝑎1,2,0𝑎1,3,0 𝑎2,2,1 = 𝑎1,1,1𝑎1,2,0𝑎1,3,0 + 𝑎1,1,0𝑎1,2,1𝑎1,3,1+ 𝑎1,1,0𝑎1,2,1𝑎1,3,0 + 𝑎1,1,0𝑎1,2,0𝑎1,3,1 𝑎2,3,0 = 𝑎1,2,1𝑎1,3,1𝑎1,4,1 + 𝑎1,2,1𝑎1,3,1𝑎1,4,0+ 𝑎1,2,1𝑎1,3,0𝑎1,4,1 + 𝑎1,2,0𝑎1,3,0𝑎1,4,0 𝑎2,3,1 = 𝑎1,2,1𝑎1,3,0𝑎1,4,0 + 𝑎1,2,0𝑎1,3,1𝑎1,4,1+ 𝑎1,2,0𝑎1,3,1𝑎1,4,0 + 𝑎1,2,0𝑎1,3,0𝑎1,4,1 𝑎2,4,0 = 𝑎1,3,1𝑎1,4,1𝑎1,5,1 + 𝑎1,3,1𝑎1,4,1𝑎1,5,0+ 𝑎1,3,1𝑎1,4,0𝑎1,5,1 + 𝑎1,3,0𝑎1,4,0𝑎1,5,0 𝑎2,4,1 = 𝑎1,3,1𝑎1,4,0𝑎1,5,0 + 𝑎1,3,0𝑎1,4,1𝑎1,5,1+ 𝑎1,3,0𝑎1,4,1𝑎1,5,0 + 𝑎1,3,0𝑎1,4,0𝑎1,5,1 𝑎3,3,0 = 𝑎2,2,1𝑎2,3,1𝑎2,4,1 + 𝑎2,2,1𝑎2,3,1𝑎2,4,0+ 𝑎2,2,1𝑎2,3,0𝑎2,4,1 + 𝑎2,2,0𝑎2,3,0𝑎2,4,0 𝑎3,3,1 = 𝑎2,2,1𝑎2,3,0𝑎2,4,0 + 𝑎2,2,0𝑎2,3,1𝑎2,4,1+ 𝑎2,2,0𝑎2,3,1𝑎2,4,0 + 𝑎2,2,0𝑎2,3,0𝑎2,4,1 The use of double variables remove the need of “Not” operation in the Boolean algebra
- 16. Simply Cellular AutomataExample a1,1,0 Find a3,3,1 in terms of a1,1,0, a1,2,0, a1,3,0, a1,4,0, a1,5,0, a1,1,1, a1,2,1, a1,3,1, a1,4,1, a1,5,1 is a3,3,1=a1,1,0a1,2,0a1,3,0a1,4,0a1,5,1 + a1,1,0a1,2,0a1,3,0a1,4,1a1,5,0 + a1,1,0a1,2,0a1,3,0a1,4,1a1,5,1 + a1,1,0a1,2,1a1,3,1a1,4,0a1,5,1 + a1,1,0a1,2,1a1,3,1a1,4,1a1,5,0 + a1,1,0a1,2,1a1,3,1a1,4,1a1,5,1 + a1,1,1a1,2,0a1,3,0a1,4,0a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,0a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,0a1,5,1 + a1,1,1a1,2,0a1,3,1a1,4,1a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,1a1,5,1 + a1,1,1a1,2,1a1,3,0a1,4,0a1,5,0 + a1,1,1a1,2,1a1,3,0a1,4,0a1,5,1 + a1,1,1a1,2,1a1,3,0a1,4,1a1,5,0 + a1,1,1a1,2,1a1,3,0a1,4,1a1,5,1 + a1,1,1a1,2,1a1,3,1a1,4,0a1,5,0 Use the representation below to simplify the equations, 𝑎𝑖,𝑗,0 = 𝑎𝑖,𝑗 𝑎𝑖,𝑗,1 = 𝑎𝑖,𝑗 a1,1,1 a1,2,0 a1,2,1 a1,3,0 a1,3,1 a1,4,0 a1,4,1 a1,5,0 a1,5,1 a2,2,0 a2,2,1 a2,3,0 a2,3,1 a2,4,0 a2,4,1 A3,3,0 A3,3,1 The use of double variables remove the need of “Not” operation in the Boolean algebra
- 17. Other Models without“Not” operation • Creating a Boolean algebra without “Not” operation may seem weird in the first place • But there is no “Not” operation in • Fault tree analysis • AND OR graph (for solving e.g. TIC-TAC-TOE) • Bayesian network or Markov Random Field
- 18. Expansion of theBoolean Algebra a3,3=a1,1,0a1,2,0a1,3,0a1,4,0a1,5,1 + a1,1,0a1,2,0a1,3,0a1,4,1a1,5,0 + a1,1,0a1,2,0a1,3,0a1,4,1a1,5,1 + a1,1,0a1,2,1a1,3,1a1,4,0a1,5,1 + a1,1,0a1,2,1a1,3,1a1,4,1a1,5,0 + a1,1,0a1,2,1a1,3,1a1,4,1a1,5,1 + a1,1,1a1,2,0a1,3,0a1,4,0a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,0a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,0a1,5,1 + a1,1,1a1,2,0a1,3,1a1,4,1a1,5,0 + a1,1,1a1,2,0a1,3,1a1,4,1a1,5,1 + a1,1,1a1,2,1a1,3,0a1,4,0a1,5,0 + a1,1,1a1,2,1a1,3,0a1,4,0a1,5,1 + a1,1,1a1,2,1a1,3,0a1,4,1a1,5,0 + a1,1,1a1,2,1a1,3,0a1,4,1a1,5,1 + a1,1,1a1,2,1a1,3,1a1,4,0a1,5,0 Polynomial Time Complexity Boolean algebra Non-Deterministic Polynomial Time Complexity Boolean algebra The Boolean algebra can be expanded 𝑎2,2,0 = 𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 +𝑎1,1,0𝑎1,2,0𝑎1,3,0 𝑎2,2,1 = 𝑎1,1,1𝑎1,2,0𝑎1,3,0 + 𝑎1,1,0𝑎1,2,1𝑎1,3,1+ 𝑎1,1,0𝑎1,2,1𝑎1,3,0 +𝑎1,1,0𝑎1,2,0𝑎1,3,1 𝑎2,3,0 = 𝑎1,2,1𝑎1,3,1𝑎1,4,1 + 𝑎1,2,1𝑎1,3,1𝑎1,4,0+ 𝑎1,2,1𝑎1,3,0𝑎1,4,1 +𝑎1,2,0𝑎1,3,0𝑎1,4,0 𝑎2,3,1 = 𝑎1,2,1𝑎1,3,0𝑎1,4,0 + 𝑎1,2,0𝑎1,3,1𝑎1,4,1+ 𝑎1,2,0𝑎1,3,1𝑎1,4,0 +𝑎1,2,0𝑎1,3,0𝑎1,4,1 𝑎2,4,0 = 𝑎1,3,1𝑎1,4,1𝑎1,5,1 + 𝑎1,3,1𝑎1,4,1𝑎1,5,0+ 𝑎1,3,1𝑎1,4,0𝑎1,5,1 +𝑎1,3,0𝑎1,4,0𝑎1,5,0 𝑎2,4,1 = 𝑎1,3,1𝑎1,4,0𝑎1,5,0 + 𝑎1,3,0𝑎1,4,1𝑎1,5,1+ 𝑎1,3,0𝑎1,4,1𝑎1,5,0 +𝑎1,3,0𝑎1,4,0𝑎1,5,1 𝑎3,3,0 = 𝑎2,2,1𝑎2,3,1𝑎2,4,1 + 𝑎2,2,1𝑎2,3,1𝑎2,4,0+ 𝑎2,2,1𝑎2,3,0𝑎2,4,1 +𝑎2,2,0𝑎2,3,0𝑎2,4,0 𝑎3,3,1 = 𝑎2,2,1𝑎2,3,0𝑎2,4,0 + 𝑎2,2,0𝑎2,3,1𝑎2,4,1+ 𝑎2,2,0𝑎2,3,1𝑎2,4,0 +𝑎2,2,0𝑎2,3,0𝑎2,4,1 • Note that all Boolean algebras do not have “Not” operation • Expansion is simply multiply throughout
- 19. Factorization of theBoolean Algebra a3,3 =a1,1,0a1,2,0a1,3,0a1,4,0a1,5,1+a1,1,0a1,2,0a1,3,0a1,4,1a1,5,0+ a1,1,0a1,2,0a1,3,0a1,4,1a1,5,1+ a1,1,0a1,2,1a1,3,1a1,4,0a1,5,1+ a1,1,0a1,2,1a1,3,1a1,4,1a1,5,0+ a1,1,0a1,2,1a1,3,1a1,4,1a1,5,1+ a1,1,1a1,2,0a1,3,0a1,4,0a1,5,0+ a1,1,1a1,2,0a1,3,1a1,4,0a1,5,0+ a1,1,1a1,2,0a1,3,1a1,4,0a1,5,1+ a1,1,1a1,2,0a1,3,1a1,4,1a1,5,0+ a1,1,1a1,2,0a1,3,1a1,4,1a1,5,1+ a1,1,1a1,2,1a1,3,0a1,4,0a1,5,0+ a1,1,1a1,2,1a1,3,0a1,4,0a1,5,1+ a1,1,1a1,2,1a1,3,0a1,4,1a1,5,0+ a1,1,1a1,2,1a1,3,0a1,4,1a1,5,1+ a1,1,1a1,2,1a1,3,1a1,4,0a1,5,0 Polynomial Time Complexity Boolean algebra Non-Deterministic Polynomial Time Complexity Boolean algebra The Boolean algebra can be factorized into multi- layer Boolean algebra that is more efficient if a more efficient Boolean algebra exists Naïve problem definition that solves a problem using brute force 𝑎2,2,0 = 𝑎1,1,1𝑎1,2,1𝑎1,3,1 + 𝑎1,1,1𝑎1,2,1𝑎1,3,0+ 𝑎1,1,1𝑎1,2,0𝑎1,3,1 +𝑎1,1,0𝑎1,2,0𝑎1,3,0 𝑎2,2,1 = 𝑎1,1,1𝑎1,2,0𝑎1,3,0 + 𝑎1,1,0𝑎1,2,1𝑎1,3,1+ 𝑎1,1,0𝑎1,2,1𝑎1,3,0 +𝑎1,1,0𝑎1,2,0𝑎1,3,1 𝑎2,3,0 = 𝑎1,2,1𝑎1,3,1𝑎1,4,1 + 𝑎1,2,1𝑎1,3,1𝑎1,4,0+ 𝑎1,2,1𝑎1,3,0𝑎1,4,1 +𝑎1,2,0𝑎1,3,0𝑎1,4,0 𝑎2,3,1 = 𝑎1,2,1𝑎1,3,0𝑎1,4,0 + 𝑎1,2,0𝑎1,3,1𝑎1,4,1+ 𝑎1,2,0𝑎1,3,1𝑎1,4,0 +𝑎1,2,0𝑎1,3,0𝑎1,4,1 𝑎2,4,0 = 𝑎1,3,1𝑎1,4,1𝑎1,5,1 + 𝑎1,3,1𝑎1,4,1𝑎1,5,0+ 𝑎1,3,1𝑎1,4,0𝑎1,5,1 +𝑎1,3,0𝑎1,4,0𝑎1,5,0 𝑎2,4,1 = 𝑎1,3,1𝑎1,4,0𝑎1,5,0 + 𝑎1,3,0𝑎1,4,1𝑎1,5,1+ 𝑎1,3,0𝑎1,4,1𝑎1,5,0 +𝑎1,3,0𝑎1,4,0𝑎1,5,1 𝑎3,3,0 = 𝑎2,2,1𝑎2,3,1𝑎2,4,1 + 𝑎2,2,1𝑎2,3,1𝑎2,4,0+ 𝑎2,2,1𝑎2,3,0𝑎2,4,1 +𝑎2,2,0𝑎2,3,0𝑎2,4,0 𝑎3,3,1 = 𝑎2,2,1𝑎2,3,0𝑎2,4,0 + 𝑎2,2,0𝑎2,3,1𝑎2,4,1+ 𝑎2,2,0𝑎2,3,1𝑎2,4,0 +𝑎2,2,0𝑎2,3,0𝑎2,4,1 • Simplification is simply factoring as there is no “Not” operation • We can find optimal time efficient Boolean algebra by factoring
- 20. Example of anNon-Deterministic problem (NP) expressed in Markov Random Field representation a1 a2 a3 a4 𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 = 𝑎1,𝑎2,𝑎3,𝑎4 ℎ(𝑎1, 𝑎2)ℎ(𝑎1, 𝑎3)ℎ(𝑎1, 𝑎4)ℎ(𝑎2, 𝑎3)ℎ(𝑎2, 𝑎4)ℎ(𝑎3, 𝑎4) Example of an NP problem expressed as a Graphical model (or Markov Random Field) Each variable ai can take finite values, e.g. 𝑎𝑖 ∈ {0,1,2,3} ℎ 𝑎𝑖, 𝑎𝑗 = 0 𝑜𝑟 1 If the NP problem has a solution, 𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 >0 Addition is “Or” operation Multiplication is “And” Operation
- 21. Boolean Algebra ofNP Problem • Boolean algebra • 𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 = 𝑎1,𝑎2,𝑎3,𝑎4 ℎ(𝑎1, 𝑎2)ℎ(𝑎1, 𝑎3)ℎ(𝑎1, 𝑎4)ℎ(𝑎2, 𝑎3)ℎ(𝑎2, 𝑎4)ℎ(𝑎3, 𝑎4) • Each h term, h(ai,aj), is a Boolean input variable taking an input 0 or 1 • Addition is “Or” operation • Multiplication is “And” operation • We can try to simplify this Boolean algebra by factorization h(a1=0,a2=0) h(a1=1,a2=0) h(a1=0,a2=1) h(a1=0,a2=2) … h(a1=0,a3=0) h(a1=1,a3=0) h(a1=0,a3=1) h(a1=0,a3=2) … h(a3=0,a4=0) h(a3=1,a4=0) h(a3=0,a4=1) h(a3=0,a4=2) … Boolean Algebra or Circuit ⋮ ⋮
- 22. “Not” Operations onh() Terms can be Replaced • Do not have to consider ℎ(𝑎𝑖 = 𝑣𝑖, 𝑎𝑗 = 𝑣𝑗) because “Not” operation on input can be replaced by • 𝑣𝑘,𝑣𝑙 (𝑣𝑖,𝑣𝑗) ℎ(𝑎𝑖 = 𝑣𝑘, 𝑎𝑗 = 𝑣𝑙) • which means sum of all h(ai,aj) terms not including h(ai=vi,aj=vj) • After replacing the ℎ(𝑎𝑖 = 𝑣𝑖, 𝑎𝑗 = 𝑣𝑗) term, the Boolean algebra remains Polynomial Time complexity if it is originally Polynomial time • Because it is replaced by at most 4*4-1=15 terms, where n=4 is the number of variables • Assume that the number of values a variable can take is 4 • E.g. ℎ(𝑎𝑖 = 0, 𝑎𝑗 = 0) can be replaced by h(ai=0,aj=1)+h(ai=0,aj=2) +h(ai=0,aj=3)+h(ai=1,aj=0)+h(ai=1,aj=1)+h(ai=1,aj=2) +h(ai=1,aj=3)+h(ai=2,aj=0)+h(ai=2,aj=1)+h(ai=2,aj=2)+h(ai=2,aj=3)+h(ai=3,aj=0) +h(ai=3,aj=1)+h(ai=3,aj=2)+h(ai=3,aj=3)
- 23. NP problem cannotbe simplified into P 𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 = 𝑎1,𝑎2,𝑎3,𝑎4 ℎ(𝑎1, 𝑎2)ℎ(𝑎1, 𝑎3)ℎ(𝑎1, 𝑎4)ℎ(𝑎2, 𝑎3)ℎ(𝑎2, 𝑎4)ℎ(𝑎3, 𝑎4) 𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 = 𝑎1,𝑎2 ℎ(𝑎1, 𝑎2) 𝑎3 ℎ(𝑎1, 𝑎3)ℎ(𝑎2, 𝑎3) 𝑎4 ℎ(𝑎1, 𝑎4) ℎ(𝑎2, 𝑎4)ℎ(𝑎3, 𝑎4) One possible factorization is shown below. There are many possible different factorizations An NP (Non-Deterministic Polynomial) problem cannot be simplified into Polynomial time by factorization Because of these 3 terms, it takes NP time to evaluate a1 a2 a3 a4 Actual proof is much more complex than this For details, please read my paper (end of slides)
- 24. About Me • Myjob uses Machine Learning to solve problems • Like my posts or slides in LinkedIn, Twitter or Slideshare • Follow me on LinkedIn • https://www.linkedin.com/in/sing-kuang-tan-b189279/ • Follow me on Twitter • https://twitter.com/Tan_Sing_Kuang • Look at my Slideshare slides • https://www.slideshare.net/SingKuangTan • 8 trends in Hang Seng Index • 4 types of Mathematical Proofs • How I prove NP vs P • Follow me on Slideshare
- 25. Share my links •I am a Small Person with Big Dreams • Please help me to repost my links to other platforms so that I can spread my ideas to the rest of the world • 我人小，但因梦想而伟大。 • 请帮我的文件链接传发到其他平台，让我的思想能传遍天下。 • Comments? Send to [email protected] • Link to my paper NP vs P paper • https://www.slideshare.net/SingKuangTan/brief-np-vspexplain-249524831 • Prove Np not equal P using Markov Random Field and Boolean Algebra Simplification • https://vixra.org/abs/2105.0181 • Other link • https://www.slideshare.net/SingKuangTan