P versus NP Problem: Is P = NP?

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Oops! The distance between nodes, 6 and 7, is 3 for the previous diagram. Please see the corrected diagram below.

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Remark: Bidirectional connections between nodes are redundancies, and they should be discarded when solving TSP. And additional connections between nodes may be necessary.

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Remark: Assuming our theory here is correct, we should be able to solve TSP in polynomial time. Theory and practice (application) should be mutually beneficial.

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For the diagram below, we compute n, e, [tex]e_{d > B_{avg } }[/tex], [tex]S_{d > B_{avg } }[/tex], [tex]B_{avg }[/tex], and [tex]H_{avg }[/tex].

n = 12, e = 23, [tex]e_{d > B_{avg } } =12[/tex], [tex]S_{d > B_{avg } } = 112[/tex],

[tex]B_{avg } = \frac{S}{e} = \frac{3+3+10+12+14+3+8+7+7+3+8+11+9+3+5+5+9+8+4+6+4+9+5}{23} = \frac{156}{23} \approx 6.78[/tex],

and [tex]H_{avg } = \frac{S_{d > B_{avg } }}{e_{d > B_{avg } }} = \frac{10+12+14+8+7+7+8+11+9+9+8+9}{12} = \frac{112}{12} \approx 9.33[/tex]

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What is the shortest simple (optimal) cycle for the digraph below?

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What is the shortest simple (optimal) cycle for the digraph below?

What is the shortest simple (optimal) cycle for the digraph below?

Optimal cycle for the digraph is

[tex]0 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow 5 \rightarrow 4 \rightarrow 11 \rightarrow 12 \rightarrow 10 \rightarrow 9 \rightarrow 8 \rightarrow 7 \rightarrow 6 \rightarrow 0[/tex]

where [tex]S_{m } = 3+7+7+3+8+9+5+4+6+8+3+12 = 75[/tex].

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Oops! [tex]S_{min } = 75[/tex].

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Remark: We assumed the distance between node 1 and node 2 is 7.

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Remark: We reduced the number of edges from 23 to 12 and reduced the average distance per node from 6.78 to 6.25. And of course, S = 156 is reduced to [tex]S_{min } = 75[/tex], a 51.9% reduction.

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Remark: We assumed the distance between node 1 and node 2 is 7.

Check 0: We must make sure we are computing a complete directed digraph.

We have updated the digraph to include our assumption.

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Remark: We reduced the number of edges from 23 to 12 and reduced the average distance between two nodes from 6.78 to 6.25. And of course, S = 156 is reduced to [tex]S_{min } = 75[/tex], a 51.9% reduction.

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Check 1: We must reduce all bidirectional connections between two nodes to one connection. We can reverse direction without added cost/travel/distance/weight...

Check 2: We must recognize all nodes connected to more than one node.

Observation: Two nodes with weight/distance/... much higher than [tex]H_{avg }[/tex] may appear in the optimal cycle.

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Rule 0: The optimal cycle is based on the directed path(s) of the given complete digraph (G). We are not allowed to deviate or make changes to G.

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Rule 0: The optimal cycle is based on the directed path(s) of the given complete digraph (G). We are not allowed to deviate or make changes to G.

Observation: Some directions between two nodes in G may be reversed in the optimal cycle for G.

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Let's write an algorithm that inputs a complete digraph (G) with any number of nodes (n > 20), a number of directed or undirected edges (e > n + 1), and with positive distances ([tex]d_{ij }[/tex]) between nodes, i and j.

Moreover, the algorithm also finds the optimal cycle for G in polynomial time.

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FYI: "Will the world be a better place if P = NP or P ! = NP? by John Smith,

A humble Quora user bidding for your brief attention to a matter of importanceUpdated August 2, 2017
No. The world will only be a better place if P=NP (which is true, see my answer on Does P = NP? ) and there exists an efficient, low coefficient polynomial time algorithm to solve a problem in NP. In that case, the world will (in theory) be an immeasurably better place, because we will be able to solve numerous problems we otherwise would not be able to.

(In practice, however, such an efficient solution could create a power struggle. It essentially levels the playing field, and that is bound to create wars and chaos during the transition period… so it is a double-edged sword.)

So yes, in theory, an efficient solution could be immensely beneficial if handled with extreme care.

Anybody that says otherwise is lying to you for their own benefit or agenda (less likely), or what they believe to be everyone’s benefit, by saving you the time of trying to solve an “impossible” or immensely difficult problem (more likely), because they believe that P=NP is false.

Lies are never justified in circumstances where a person is willing to honestly listen and try to understand a whole argument. Their end argument, whether they realize it or not, is that it is not good for “the people” if you try to solve P vs. NP, because of the cumulative time and effort lost. That argument fails on the individual. They have no right to dictate to you what is good for you, nor does their argument hold any water. You may fail… SO WHAT??? EVERYONE FAILS. We learn from our failures. In fact, if we never failed we would never learn, so their argument would be an argument against learning itself, which makes complete sense in a communist society, but not in a free society. So, their arguments and attempts at justifications fail.

Now… let the communists downvote my answer in spite! But if you live in a free country, please show your support and upvote my answer, or do me the courtesy of having a debate with me in the comments if you believe I am wrong."

Source Link: https://www.quora.com/search?q=p%20%3D%20np%3F.

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"Lord GOD makes all possible or real."

Last Post by 'so-called Anonymous Author':

P = NP! Thank Lord GOD! Praise Lord GOD! Amen!

May all beings live in peace and in contentment. And may Lord GOD heal the suffering of all beings. Thank you, Lord GOD! Amen!

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my question is as simple as others
if P = NP are crypto currencies dead ?
aka with these Quantum computers you never know if SHA 256, 512 1024 whatever more will secure block chain !
so please answer this
P = NP proof or quantum computers will affect SHA or not ?

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my question is as simple as others
if P = NP are crypto currencies dead ?
aka with these Quantum computers you never know if SHA 256, 512 1024 whatever more will secure block chain !
so please answer this
P = NP proof or quantum computers will affect SHA or not ?

Good question? I do not know the answer...

P = NP!

SHA ( https://brilliant.org/wiki/secure-hashing-algorithms/) is probably safe and secure. Good luck!

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Go Blue!

https://lsa.umich.edu/math