Monotone Circuit can implement an algorithm to run Non-Deterministic Polynomial time complexity (NP) problem in Polynomial time complexity (P). I developed a method to implement all algorithms without "Not" operations. Using this information, I manage to prove that NP is not equal to P. Kung Fu Computer Science, Geometric complexity theory

1. A Weird Soviet Method to
Partially Solve the Perebor
Problem
How is this method related to my NP vs P solution
Sing Kuang Tan
singkuangtan@gmail.com
17 August 2021

2. Perebor Problems
• Perebor problems 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 (from Soviet 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-clique in 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
• What is 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 are unnecessary
• 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 can Emulate 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 Automata Example
5 bit binary input
1 bit binary output

9. Simple Cellular Automata Example
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 Automata Example
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 Automata Example
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 Automata Example
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 Automata Example
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 Automata Example
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 Automata Example
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 Automata Example
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 the Boolean 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 the Boolean 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 an Non-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 of NP 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 on h() 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 cannot be 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)

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