Claim of proof that P = NP. Prove that x ± y = b? For questions like “Prove that x ± y = b where b is the solution to any given number x and y for which x is number i or j at every given input number (n)?” If a valid input ‘n’ exists which has already exists, then it would follow that P = NP. Unfortunately, many important problems have been shown to be NP-complete, and as of 2013 not a single fast algorithm for any of them is known. Basically, what the “proof x ± y = b” does is how one number (n), input once and only once, can give you two equal and opposite results. No one has ever been able to solve, nor provide the formula, or the solution to the proof till date. I have the solution to the proof and also I have written a computer program, an application to test it validity. Our assertion for P vs NP: Does P = NP? The answer is yes. There exists a valid input ‘n’ that satisfies x ± y = b and the computer takes the solution to be verified as input. For example, let consider the result of a computer program that I have written which runs on a command prompt that requires a valid input ‘n’ to test the proof x ± y = b i.e. Input any integer number or half-integer once and only once. Let say the value entered is 7; the result would be -2 and 2. The difference in result is 0. Input any integer number or half-integer once and only once. Let say the value entered is 1439; the result would be -719 and 719. The difference in result is 0. Without changing the instance; Test for validity: Input any decimal number that is not half-integer once and only once. Let say the value entered is 2.8; the result would be -0.6000000000000001 and 0.5999999999999996.The difference in result is -4.440892098500626E-16. Input any decimal number that is not half-integer once and only once. Let say the value entered is 3.7; the result would be -6.15 and 6.149999999999999.The difference in result is -1.7763568394002505E-15 Why is the different result when a decimal value is input? Since we have the same result when an integer value is input, the result must be the same when a decimal value is input. Else, there is a difference somewhere… Reason: When a decimal value that is not half integer is input once and only once, the floating point error enables us to see that what is done to achieve one value is not conversely of what is done to achieve the other value. Otherwise, despite the floating point error the result would have been the same when a decimal value is entered. Thus, the floating point numbers exist in this case of study! Conclusion: This tells us that the plus and minus values emerge from two different sources at a given input (n). So, we are not doing the same thing but different things to achieve the plus and minus values. Therefore, the plus and minus values can coexist. The problem: Unify plus and minus values into one place. “Prove that x ± y = b?” where b is the solution to any given number x and y for which x is number i or j at every given input number ‘n’. Given that x ± y = b. Such that, x – b = ±y for which x is number i or j. So that, i – b = -y ——— Equation 1. Also that, j – b = y ——— Equation 2. If it turns out that a valid input ‘n’ exist once and only once into the proof x ± y = b. Then the solution to the problem “Prove that x ± y = b?” can be quickly solved and verified by a computer. Therefore P = NP is true. Equation 1 and 2 conform. Otherwise, P ≠ NP is true. What are the formulas having at most one variable in common to create equation 1 and 2 at a time? Mainly, this are “The Proof” for x ± y = b. Thus, the shortest route that leads to P and NP. fofallthings/?page_id=235
Posted on: Mon, 25 Nov 2013 10:52:12 +0000
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