P vs NP solution

P vs NP solution: Can be solved by splitting into conditions can make it easy to solve in same as check P=NP. Solution: splitting into max conditions If to find the sum of two numbers must be equal to 150 from 100 numbers. Sort by odd and odd or even and even Sort by Case 1: First-digit sum = 14, second-digit sum = 10. Case 2: First-digit sum = 15, second-digit sum = 0. Normal method: Finding the list of possibilities of sum of two numbers which equal to 150 and matches with that list. Basic conditions check: Condition 1: a=±(10x+y) b=±(10z+a) c=±(10b+c) a+b=c ±(10x+y)+±(10z+a)=±(10b+c) ±(10x+y+10z+a)=±(10b+c) 10(x+z)+(a+y)=10b+c a+y can be greater or smaller than 10. Sub condition 1: a+y greater than 10. Something = (a+y)-10 10(x+z)+10+something=10b+c 10(x+z+1)+something=10b+c b=x+z+1 x+z=b-1 Something = c a+y=c+10 Subcondition 2: x+z=b a+y=c Condition 2: A=±(10x+y) B=-+(10z+a) C=±(10b+c) A+b=c ±(10x+y)+-+(10z+a)=±(10b+c) ±(10x+y-10z-a)=±(10b+c) 10(x-z)+(a-y)=10b+c A-y can be greater or smaller than 0. Sub condition 1: A-y smaller than 0. Something = (a-y)a-y+10 10(x-z)-10+something=10b+c 10(x-z-1)+something=10b+c B=x-z-1 X-z=b+1 Something = c A-y=c-10 Subcondition 2: X-z=b A-y=c