The document proposes a proof that NP ≠ P using Markov random fields and Boolean algebra simplification. It represents non-deterministic polynomial problems as graphical models and shows that while linear chains and trees can be solved efficiently in polynomial time, fully connected graphs cannot due to remaining complex terms. The author believes propagating "not" operations in Boolean algebras will preserve polynomial time for polynomial problems but not for NP-complete problems. They provide examples and ask others to share links to their paper proposing this proof.

1. Brief explanation of how I
prove NP≠P
Sing Kuang Tan
singkuangtan@gmail.com
Link to my paper
Prove Np not equal P using Markov Random Field and
Boolean Algebra Simplification
https://vixra.org/abs/2105.0181

2. 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

3. Linear Chain Graphical Model
a1 a2 a3
a4
𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 =
𝑎1,𝑎2,𝑎3,𝑎4
ℎ(𝑎1, 𝑎2)ℎ(𝑎2, 𝑎3)ℎ(𝑎3, 𝑎4)
𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4 =
𝑎1 𝑎2
ℎ(𝑎1, 𝑎2)
𝑎3
ℎ(𝑎2, 𝑎3)
𝑎4
ℎ(𝑎3, 𝑎4)
It is well known in the literature that linear chain graphical model can be solved in Polynomial time
This expression needs Non-Deterministic
Polynomial time to evaluate
Can be factorized into
This expression needs Polynomial time to
evaluate

4. Tree Structure Graphical Model
a1
a2 a3
𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4, 𝑎5, 𝑎6, 𝑎7 =
𝑎1,𝑎2,𝑎3,𝑎4,𝑎5,𝑎6,𝑎7
ℎ(𝑎1, 𝑎2)ℎ(𝑎1, 𝑎3)ℎ(𝑎2, 𝑎4)ℎ(𝑎2, 𝑎5)ℎ(𝑎3, 𝑎6)ℎ(𝑎3, 𝑎7)
It is well known in the literature that tree
structure graphical model can be solved in
Polynomial time
This expression
needs Non-
Deterministic
Polynomial time
to evaluate
Can be factorized into
This expression
needs Polynomial
time to evaluate
a5
a4 a7
a6
𝑓 𝑎1, 𝑎2, 𝑎3, 𝑎4, 𝑎5, 𝑎6, 𝑎7
=
𝑎1 𝑎2
ℎ(𝑎1, 𝑎2)
𝑎4
ℎ(𝑎2, 𝑎4)
𝑎5
ℎ(𝑎2, 𝑎5)
𝑎3
ℎ(𝑎1, 𝑎3)
𝑎6
ℎ(𝑎3, 𝑎6)
𝑎7
ℎ(𝑎3, 𝑎7)

5. Transform any Boolean algebra so that only
the first layer has ‘Not’ operations
a1
a2
a3
a4
a6
a5
𝐻
Output
Not operation
at input or
output
Or operation
And operation
Propagate the ‘Not’ operations to the previous
layer using DE Morgan’s theorem

6. a1
a2
a3
a4
a6
a5
𝐻
Output
Not operation
at input or
output
Or operation
And operation
Propagate the ‘Not’ operations to the previous
layer using DE Morgan’s theorem

7. a1
a2
a3
a4
a6
a5
𝐻
Output
Not operation
at input or
output
Or operation
And operation
Propagate the ‘Not’ operations to the previous
layer using DE Morgan’s theorem

8. a1
a2
a3
a4
a6
a5
𝐻
Output
Not operation
at input or
output
Or operation
And operation
Propagate the ‘Not’ operations to the previous
layer using DE Morgan’s theorem

9. a1
a2
a3
a4
a6
a5
𝐻
Output
Not operation
at input or
output
Or operation
And operation
Propagate the ‘Not’ operations to the previous
layer using DE Morgan’s theorem

10. a1
a2
a3
a4
a6
a5
𝐻
Output
Not operation
at input or
output
Or operation
And operation
Until only the first layer has ‘Not’ operations
For details, please read my paper
Note that the number
of gates is at most 2
times the original
Boolean algebra after
propagation.
Polynomial time
Boolean algebra
remains Polynomial
time after
propagation.

11. 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

12. Why NP problem cannot be simplified
a1 a2
a3
a4
a1 a2
a3
a4
a1 a2
a3
a4
If we remove constraints
such that the Graphical
model becomes a linear
chain, it can be solved
efficiently in Polynomial
time
If we remove constraints
such that the Graphical
model becomes a tree, it
can be solved efficiently
in Polynomial time
However if no constraint
is removed, it can never
be solved efficiently in
Polynomial time

13. Boolean algebra representation of NP
problem
The NP Markov Random Field can be expressed as a 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)
=∨𝑎1,𝑎2,𝑎3,𝑎4
(ℎ 𝑎1, 𝑎2 ∧ℎ 𝑎1, 𝑎3 ∧ℎ 𝑎1, 𝑎4 ∧ℎ 𝑎2, 𝑎3 ∧ℎ 𝑎2, 𝑎4 ∧ℎ 𝑎3, 𝑎4 )
From the previous example that every Boolean algebra can be expressed with only first layer contains ‘Not’ operations, and
this NP Markov Random Field Boolean algebra has no ‘Not’ operation, the simplification of this NP Boolean algebra can be
done without any ‘Not’ operation. (For details, please read my paper)

14. 2sat problem
• There is a special case for the fully connected Graphical
model
• If each variable ai can only takes 2 values (0 or 1)
• Then it is a 2sat problem
• The Graphical Model can be solved in Polynomial time
• It can be solved by dynamic programming
• 𝐻0 𝑎𝑖, 𝑎𝑗 = 𝐻 𝑎𝑖, 𝑎𝑗
• 𝐻𝑙
𝑎𝑖, 𝑎𝑗 = 𝐻𝑙−1
(𝑎𝑖, 𝑎𝑗) 𝑎𝑘
𝐻𝑙−1
𝑎𝑖, 𝑎𝑘 𝐻𝑙−1
(𝑎𝑗, 𝑎𝑘)
• where 1 ≤ 𝑙 ≤ 𝑛
• n is the number of variables 𝑎1, 𝑎2, … , 𝑎𝑛
• For details, please read my paper
a1 a2
a3
a4
Note that for Hl() and Hl-1(),
the superscripts are indexes
and do not mean H raised to
the power of l

15. 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 singkuangtan@gmail.com
• Link to my paper
• Prove Np not equal P using Markov Random Field and Boolean Algebra
Simplification
• https://vixra.org/abs/2105.0181