P=NP

Is True

I can prove it, no bullshit

I hope to claim the 1,000,000 dollar prize for this

this time I am pretty sure I have it almost correct

It just would be a bitch to write it all out

Basically you have a graph with the top node at x1+x2/2+x3/2+x4/4....xN/2

where x1...xN are the integers of the set in ascending order

Then you have movement down the graph in either a right or left direction. With the first movment being + or - xN/2. The second movement x(N-1)/2 etc

But along the way you have to deal with the fact there will be intersections if

X1+XN < X1+X2..+x(N-1)

To do this you change the following paths by a constant factor based on intersections.

Basically this will allow you to see if X1 is in the subset that adds up to a number (could be 0).

You should be able to do this in polynomial time and solve the subset sum theory in polynomial time.

I know it works in the following case

the set 6,7,8,9,10


28
+-5 I
+-4.5 I
+-4
+-3.5

the I denotes it causes intersections

I in all of these cases is 1

28
+-5 I=1
+-4.5 I =1
+-4 I = 0
+-3.5 I = 0

Assume we move +5 to start (right)
We set FD = -1R

If we moved -5 to start (left)

we set FD = -1L

Then we move the opposite direction next, so RL, LR (right left, left right)

we would arrive at 22.5 or 23.5

We check what value was set in FD

If FD = -1R then move right, and set the left movement to -1 whatever the right value is

RLR = 27.5
RLL = 26.5

For purposes of figuring out which direction to go next set the point inbetween the two new points

27

Then say we move to 27.5

we check our FD and have -1R, and -1L

so we set the two movement values to be one less. aka +-2.5 for this case

ps. the above alg doesnt really work but it is going in the right direction.

>>3
Aside from the fact this is all obvious bullshit, showing the subset sum problem is P would show that all NP problems are P, because it's NP complete.