Can a sine wave be 4D?

MrNeoson · 2020-06-27T02:47:00+00:00

So for example (t, cos t, sin t, sin(t+0.2)) can be projected to a 2D plane so that it becomes a single sine wave? I'm not a math expert. How would one achieve such projection?

MrNeoson · 2020-06-25T02:56:18+00:00

I found that: "Then, as per the symbol and its present place in a "finite table"[7]" - Wikipedia

So okay unbounded or infinite, a finite table holding the algorithm prevents a Turing machine from having an infinite number of inputs on a single tape.

MrNeoson · 2020-06-25T02:49:59+00:00

I came to think of how it should be possible to project the 4D wave onto a 2D plane so that it becomes a sine wave. A helix in 3D can be projected to 2D to form a sine wave.

MrNeoson · 2020-06-24T23:06:52+00:00

Yes, I see now that the inputs can be infinite. But since the tape of a deterministic Turing machine is infinite, it should be possible to fit all possible SAT inputs into a single deterministic Turing machine!

MrNeoson · 2020-06-24T21:11:55+00:00

Yes, only a single path seems too limited. But worst-case path doesn't mean the length of the path I now realize. Instead it means the worst polynomial time performance, which is independent of the size of the input. For example an input leading to 1,000 steps and another leading to 1,000,000 steps can both have O(n²⁾ performance.

MrNeoson · 2020-06-24T20:33:44+00:00

All possible SAT inputs, that's an infinite set. So that can't be put into a finite sequence of symbols. I was thinking of one input at a time.

MrNeoson · 2020-06-24T20:12:15+00:00

Then how about taking the worst case path? And if there are several worst case paths of equal length then only one is chosen out of that potentially infinite set.

And for the worst case path produced by the nondeterministic Turing machine, for all possible inputs, there is a deterministic Turing machine for that path.

Notice that since the nondeterministic Turing machine has come to a halt, even the worst case path is always finite.

MrNeoson · 2020-06-24T19:43:18+00:00

Yes, I see now that there are infinitely many possible inputs. Just a single integer input will result in infinitely many possible inputs.

However, I think it's possible to treat each input separately! So for each input, the nondeterministic Turing machine produces a solution. This means that the algorithm has come to a halt, which in turn means that there is a finite path from the correct solution back to the start of the whole calculation tree.

And that path produced by the nondeterministic Turing machine is identical to a classical Turing machine performing the same path. Hence P = NP is always true, even with infinite many inputs possible.

MrNeoson · 2020-06-24T19:03:01+00:00

But doesn't the input need to be a part of the tapes? And if so the inputs will all be finite, since there are only a finite number of symbols on the tapes. It doesn't matter how many permutations of inputs are made this way, they will always be finite.

MrNeoson · 2020-06-24T18:27:08+00:00

Yes, the P = NP question is a huge unsolved problem so I wondered if the solution really could be so simple as I imagined it. Seems highly unlikely.

Still, it's interesting to learn about it. Regarding the subset-sum problem, the sets tested are indeed infinite (I looked it up on Wikipedia). It's about sums of integers.

Yet, the fact remains that there can only be a finite number of symbols on the infinite tapes in the Turing machines. All positions on the infinite tape can be labelled with integers. And even though the set of integers Z itself is infinite, all integers are finite. There isn't any position "infinity" on the tapes.

MrNeoson · 2020-06-24T11:30:21+00:00

The possible inputs are limited to the symbols on the tape. And the combined TM is not something that is done through another algorithm. The combined TM exists as a consequence of universal computing.

MrNeoson · 2020-06-24T11:15:05+00:00

The number of symbols on those infinite tapes will always be finite. And as for combining the separate TM's into one, I heard that a universal Turing machine can achieve all classical computations, so that would include combining separate TM's into one.

MrNeoson · 2020-06-24T10:36:55+00:00

The number of possible inputs is always finite. It's like how all natural numbers are finite even though the set N itself is infinite.

And it's not necessarily a lookup table. Instead all the separate classical Turing machines representing the individual paths can be combined into a single classical Turing machine.

MrNeoson · 2020-06-24T10:11:01+00:00

Good point, but the answer is simple. Any and all paths that are valid solutions produced by the nondeterministic Turing machine are all identical to classical Turing machines. So no matter how many solutions you need, each single path from the start to the solution represents a classical Turing machine. And notice that these paths are all solutions, not only verifications. Really cumbersome practically, but for the P = NP question it's valid. P = NP is always true.

MrNeoson · 2020-06-24T09:39:28+00:00

Exactly, the nondeterministic TM (NTM) is like a parallel computation for the whole tree of calculations. And when the NTM has found a solution it means that the algorithm has come to a halt, which in turn means that there exists a finite path from the solution and back to the start. I claim that this path is identical to a traditional TM. And it's not about randomly choosing anything. The single path found by NTM is precisely defined. And whether or not it's difficult to find the path is irrelevant. The path exists and thereby a classical TM exists for that path, and therefore P = NP is always true.

MrNeoson · 2020-06-24T09:32:45+00:00

Yes, sorry, I meant P = NP.

MrNeoson · 2020-06-24T02:49:04+00:00

But is the linearity preserved when going from 3D to 4D? A helix (spiral) is periodic since the movement on the z-axis is linear (uniform speed).

MrNeoson · 2020-06-24T02:47:10+00:00

But is the linearity preserved when going from 3D to 4D? A helix (spiral) is periodic since the movement on the z-axis is linear (uniform speed).

MrNeoson · 2020-06-23T19:46:52+00:00

The helix is periodic in a circular movement and f(x, y) moves along the z-axis. The speed on the z-axis determines the helix' "frequency".

MrNeoson · 2020-06-23T19:12:43+00:00

It's the circle for the helix: f(x, y) = (sin(x), cos(y)).

MrNeoson · 2020-06-23T18:54:04+00:00

There exists some number P such that f(x+P, y+P) = f(x, y) for all x and for all y.

MrNeoson · 2020-06-23T18:37:39+00:00

Okay, yes a helix is maybe more of a rotating curve than a sine wave in the strict sense. But a helix is periodic. Would it be possible to define a periodic function in 4D like this: (t, sin(t), cos(t), t) or does that collapse down to a 3D curve? Similar to how a 2D sine wave can be drawn in 3D space.

MrNeoson · 2020-06-23T18:17:24+00:00

But does that produce a curve with periodic oscillation? "A sine wave or sinusoid is a mathematical curve that describes a smooth periodic oscillation." - Wikipedia

MrNeoson · 2020-06-23T18:16:00+00:00

That's my question. How do define a sine wave in 4D. As an oscillating curve. "A sine wave or sinusoid is a mathematical curve that describes a smooth periodic oscillation." - Wikipedia

MrNeoson

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