Good examples of “two is easy, three is hard” in computational sciences
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I recently encountered a formulation of the meta-phenomenon: "two is easy, three is hard" (phrased this way by Federico Poloni), which can be described, as follows:
When a certain problem is formulated for two entities, it is relatively easy to solve; however, an algorithm for a three-entities-formulation increases in the difficulty tremendously, possibly even rendering the solution not-feasible or not-achievable.
(I welcome suggestions to make the phrasing more beautiful, concise, and accurate.)
What good examples in various areas of computational sciences (starting from pure linear algebra and ending with a blanket-term computational physics) do you know?
linear-algebra computational-physics computational-geometry computational-chemistry
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show 5 more comments
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I recently encountered a formulation of the meta-phenomenon: "two is easy, three is hard" (phrased this way by Federico Poloni), which can be described, as follows:
When a certain problem is formulated for two entities, it is relatively easy to solve; however, an algorithm for a three-entities-formulation increases in the difficulty tremendously, possibly even rendering the solution not-feasible or not-achievable.
(I welcome suggestions to make the phrasing more beautiful, concise, and accurate.)
What good examples in various areas of computational sciences (starting from pure linear algebra and ending with a blanket-term computational physics) do you know?
linear-algebra computational-physics computational-geometry computational-chemistry
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The curse of dimensionality comes to mind.
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– Paul♦
May 16 at 0:13
4
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graph 2-coloring (easy) versus 3-coloring (NP-hard), see here
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– GoHokies
May 16 at 5:34
5
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@GoHokies Please don't post answers as comments.
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– David Richerby
May 16 at 11:01
4
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From foundation of math or recursion background, you might come across TREE function, where TREE(2)=3, and TREE(3) is ... quite large. (not being familiar with computational sciences, I'm not sure this is really an answer you are looking for, but it seems similar enough to leave a comment about)
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– BurnsBA
May 16 at 14:01
2
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A counterexample: "Never go to sea with two chronometers; take one or three." That said, there are so many good examples that there is no right answer. This question should be community wiki.
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– David Hammen
May 17 at 0:24
|
show 5 more comments
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I recently encountered a formulation of the meta-phenomenon: "two is easy, three is hard" (phrased this way by Federico Poloni), which can be described, as follows:
When a certain problem is formulated for two entities, it is relatively easy to solve; however, an algorithm for a three-entities-formulation increases in the difficulty tremendously, possibly even rendering the solution not-feasible or not-achievable.
(I welcome suggestions to make the phrasing more beautiful, concise, and accurate.)
What good examples in various areas of computational sciences (starting from pure linear algebra and ending with a blanket-term computational physics) do you know?
linear-algebra computational-physics computational-geometry computational-chemistry
$endgroup$
I recently encountered a formulation of the meta-phenomenon: "two is easy, three is hard" (phrased this way by Federico Poloni), which can be described, as follows:
When a certain problem is formulated for two entities, it is relatively easy to solve; however, an algorithm for a three-entities-formulation increases in the difficulty tremendously, possibly even rendering the solution not-feasible or not-achievable.
(I welcome suggestions to make the phrasing more beautiful, concise, and accurate.)
What good examples in various areas of computational sciences (starting from pure linear algebra and ending with a blanket-term computational physics) do you know?
linear-algebra computational-physics computational-geometry computational-chemistry
linear-algebra computational-physics computational-geometry computational-chemistry
edited May 16 at 0:15
Anton Menshov
asked May 16 at 0:08
Anton Menshov♦Anton Menshov
4,84422076
4,84422076
2
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The curse of dimensionality comes to mind.
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– Paul♦
May 16 at 0:13
4
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graph 2-coloring (easy) versus 3-coloring (NP-hard), see here
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– GoHokies
May 16 at 5:34
5
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@GoHokies Please don't post answers as comments.
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– David Richerby
May 16 at 11:01
4
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From foundation of math or recursion background, you might come across TREE function, where TREE(2)=3, and TREE(3) is ... quite large. (not being familiar with computational sciences, I'm not sure this is really an answer you are looking for, but it seems similar enough to leave a comment about)
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– BurnsBA
May 16 at 14:01
2
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A counterexample: "Never go to sea with two chronometers; take one or three." That said, there are so many good examples that there is no right answer. This question should be community wiki.
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– David Hammen
May 17 at 0:24
|
show 5 more comments
2
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The curse of dimensionality comes to mind.
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– Paul♦
May 16 at 0:13
4
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graph 2-coloring (easy) versus 3-coloring (NP-hard), see here
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– GoHokies
May 16 at 5:34
5
$begingroup$
@GoHokies Please don't post answers as comments.
$endgroup$
– David Richerby
May 16 at 11:01
4
$begingroup$
From foundation of math or recursion background, you might come across TREE function, where TREE(2)=3, and TREE(3) is ... quite large. (not being familiar with computational sciences, I'm not sure this is really an answer you are looking for, but it seems similar enough to leave a comment about)
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– BurnsBA
May 16 at 14:01
2
$begingroup$
A counterexample: "Never go to sea with two chronometers; take one or three." That said, there are so many good examples that there is no right answer. This question should be community wiki.
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– David Hammen
May 17 at 0:24
2
2
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The curse of dimensionality comes to mind.
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– Paul♦
May 16 at 0:13
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The curse of dimensionality comes to mind.
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– Paul♦
May 16 at 0:13
4
4
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graph 2-coloring (easy) versus 3-coloring (NP-hard), see here
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– GoHokies
May 16 at 5:34
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graph 2-coloring (easy) versus 3-coloring (NP-hard), see here
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– GoHokies
May 16 at 5:34
5
5
$begingroup$
@GoHokies Please don't post answers as comments.
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– David Richerby
May 16 at 11:01
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@GoHokies Please don't post answers as comments.
$endgroup$
– David Richerby
May 16 at 11:01
4
4
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From foundation of math or recursion background, you might come across TREE function, where TREE(2)=3, and TREE(3) is ... quite large. (not being familiar with computational sciences, I'm not sure this is really an answer you are looking for, but it seems similar enough to leave a comment about)
$endgroup$
– BurnsBA
May 16 at 14:01
$begingroup$
From foundation of math or recursion background, you might come across TREE function, where TREE(2)=3, and TREE(3) is ... quite large. (not being familiar with computational sciences, I'm not sure this is really an answer you are looking for, but it seems similar enough to leave a comment about)
$endgroup$
– BurnsBA
May 16 at 14:01
2
2
$begingroup$
A counterexample: "Never go to sea with two chronometers; take one or three." That said, there are so many good examples that there is no right answer. This question should be community wiki.
$endgroup$
– David Hammen
May 17 at 0:24
$begingroup$
A counterexample: "Never go to sea with two chronometers; take one or three." That said, there are so many good examples that there is no right answer. This question should be community wiki.
$endgroup$
– David Hammen
May 17 at 0:24
|
show 5 more comments
15 Answers
15
active
oldest
votes
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One example that appears in many areas of physics, and in particular classical mechanics and quantum physics, is the two-body problem. The two-body problem here means the task of calculating the dynamics of two interacting particles which, for example, interact by gravitational or Coulomb forces. The solution to this problem can often be found in closed form by a variable transformation into center-of-mass and relative coordinates.
However, as soon as you consider three particles, in general no closed-form solutions exist.
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3
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Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
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– llama
May 16 at 17:27
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good nitpick, thanks, "in general" is missing here.
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– davidhigh
May 16 at 18:08
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Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
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– WorldSEnder
May 16 at 19:53
add a comment |
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A famous example is the boolean satisfiability problem (SAT). 2-SAT is not complicated to solve in polynomial time, but 3-SAT is NP-complete.
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3
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3-SAT can be reduced to graph 3-coloring, or vice-versa
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– GoHokies
May 16 at 9:38
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@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
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– findusl
May 16 at 16:47
1
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@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
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– GoHokies
May 20 at 3:00
add a comment |
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In one and two dimensions, all roads lead to Rome, but not in three dimensions.
Specifically, given a random walk (equally likely to move in any direction) on the integers in one or two dimensions, then no matter the starting point, with probability one (a.k.a. almost surely), the random walk will eventually get to a specific designated point ("Rome").
However, for three or more dimensions, the probability of getting to "Rome" is less than one; with the probability decreasing as the number of dimensions increases.
So for instance, if conducting a stochastic (Monte Carlo) simulation of a random walk starting at "Rome", which will stop when Rome is returned to, then in one and two dimensions, you can be assured of eventually making it back to Rome and stopping the simulation - so easy. In three dimensions, you may never make it back, so hard.
https://en.wikipedia.org/wiki/Random_walk#Higher_dimensions
To visualize the two-dimensional case, one can imagine a person
walking randomly around a city. The city is effectively infinite and
arranged in a square grid of sidewalks. At every intersection, the
person randomly chooses one of the four possible routes (including the
one originally travelled from). Formally, this is a random walk on the
set of all points in the plane with integer coordinates.
Will the person ever get back to the original starting point of the
walk? This is the 2-dimensional equivalent of the level crossing
problem discussed above. In 1921 George Pólya proved that the person
almost surely would in a 2-dimensional random walk, but for 3
dimensions or higher, the probability of returning to the origin
decreases as the number of dimensions increases. In 3 dimensions, the
probability decreases to roughly 34%
See http://mathworld.wolfram.com/PolyasRandomWalkConstants.html for numerical values.
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Here's one close to the hearts of the contributors at SciComp.SE:
The Navier–Stokes existence and smoothness problem
The three-dimensional version is of course a famous open problem and the subject of a million-dollar Clay Millenium Prize. But the two-dimensional version has already been resolved a long time ago, with an affirmative answer. Terry Tao notes that the solution dates essentially back to Leray’s thesis in 1933!
Why is the three-dimensional problem so much harder to solve? The standard, hand-wavy response is that turbulence becomes significantly more unstable in three dimensions than in two. For a more mathematically rigorous answer, check out Charles Fefferman's official problem statement at the Clay Institute or Terry Tao's nice exposition on the possible proof strategies.
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In social choice theory, designing an election scheme with two candidates is easy (majority rules), but designing an election scheme with three or more candidates necessarily involves making trade-offs between various reasonable-sounding conditions. (Arrow's impossibility theorem).
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Simultaneous diagonalization of two matrices $A_1$ and $A_2$:
$$
U_1^T A_1 V = Sigma_1,quad U_2^TA_2V=Sigma_2
$$
is covered by existing generalized singular value decomposition.
However, when the simultaneous reduction of three matrices to a canonical form (weaker condition compared to the above) is required:
$$
Q^T A_1 Z = tilde{A_1},quad Q^T A_2 Z = tilde{A_2},quad Q^T A_3 Z = tilde{A_3}
$$
no direct methods exist. Therefor, one has to opt for more complicated routes using approximate SVDs, tensor decompositions, etc.
A practical application would be a solution for a quadratic eigenvalue problem:
$$
(A_1+lambda A_2+lambda^2 A_3)x=0
$$
Source: C. F. van Loan, "Lecture 6: The higher-order generalized singular value decomposition," CIME-EMS Summer School, Cetraro, Italy, June 2015.
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Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
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– Rosie F
May 17 at 10:27
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@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
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– Anton Menshov♦
May 17 at 21:11
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There are plenty of examples in quantum computing, although I've been out of this for a while and so don't remember many. One major one is that bipartite entanglement (entanglement between two systems) is relatively easy whereas entanglement among three or more systems is an unsolved mess with probably a hundred papers written on the topic.
The root of this is that rank-2 tensors (i.e. matrices) can be analyzed via singular value decomposition. Nothing similar exists for tensors of rank 3 or higher. In fact, even something as simple as $maxleft(u_a v_b w_c T^{abc} / leftlVert u rightrVert leftlVert v rightrVert leftlVert w rightrVert right)$ (with sub/superscripts denoting Einstein summation) is, IIRC, not believed to be efficiently solvable.
This paper seems relevant, although I haven't read it: Most tensor problems are NP-hard
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I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
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– Richard Zhang
May 17 at 16:36
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That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
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– Dan Stahlke
May 18 at 17:48
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Angle bisection with straightedge and compass is simple, angle trisection is in general impossible.
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Type inference for Rank-n types. Type inference for Rank-2 is not especially difficult, but type inference for Rank-3 or above is undecidable.
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Folding a piece of paper in half without tools is easy. Folding it into thirds is hard.
Factoring a polynomial with two roots is easy. Factoring a polynomial with three roots is significantly more complicated.
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Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
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– Bill K
May 17 at 23:24
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A smooth curve of degree 2 (i.e. given as the solution of $f(x,y) = 0$ where $f$ is a polynomial of degree 2) with a given point is rational, meaning that it can be parameterized by quotients of polynomials, of degree 3 it isn't. The former are considered well understood, the latter, called elliptic curves when a base point, i.e. a specific solution, is singled out, are the object of intense research.
This difference has several implications:
- In degree 2 there are algorithms to find all rational points (solutions in rational numbers), in degree 3 no such algorithm is known.
- Integrals involving $sqrt{f(x)}$ with $f$ of degree 1 or 2 have solutions in elementary functions, but not for $f$ of degree 3 or higher.
- The discrete logarithm problem is tractable on curves of degree 2, hence not suitable for cryptographic applications, whereas the assumed hardness of the same problem on elliptic curves is at the basis of some of the most popular public key cryptosystems.
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Here's a neat one from optimization: the Alternating Direction Method of Multipliers (ADMM) algorithm.
Given an uncoupled and convex objective function of two variables (the variables themselves could be vectors) and a linear constraint coupling the two variables:
$$min f_1(x_1) + f_2(x_2) $$
$$ s.t. ; A_1 x_1 + A_2 x_2 = b $$
The Augmented Lagrangian function for this optimization problem would then be
$$ L_{rho}(x_1, x_2, lambda) = f_1(x_1) + f_2(x_2) + lambda^T (A_1 x_1 + A_2 x_2 - b) + frac{rho}{2}|| A_1 x_1 + A_2 x_2 - b ||_2^2 $$
The ADMM algorithm roughly works by performing a "Gauss-Seidel" splitting on the augmented Lagrangian function for this optimization problem by minimizing $L_{rho}(x_1, x_2, lambda)$ first with respect to $x_1$ (while $x_2, lambda$ remain fixed), then by minimizing $L_{rho}(x_1, x_2, lambda)$ with respect to $x_2$ (while $x_1, lambda$ remain fixed), then by updating $lambda$. This cycle goes on until a stopping criterion is reached.
(Note: some researchers such as Eckstein discard the Gauss-Siedel splitting view in favor of proximal operators, for example see http://rutcor.rutgers.edu/pub/rrr/reports2012/32_2012.pdf )
For convex problems, this algorithm has been proven to converge - for two sets of variables. This is not the case for three variables. For example, the optimization problem
$$min f_1(x_1) + f_2(x_2) + f_3(x_3)$$
$$ s.t. ; A_1 x_1 + A_2 x_2 + A_3x_3 = b $$
Even if all the $f$ are convex, the ADMM-like approach (minimizing the Augmented Lagrangian with respect to each variable $x_i$, then updating the dual variable $lambda$) is NOT guaranteed to converge, as was shown in this paper.
https://web.stanford.edu/~yyye/ADMM-final.pdf
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The TREE
function.
We can calculate TREE(2) = 3
, but TREE(3)
is not calculable in the universe lifetime, we only know that it is finite.
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TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.
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– Solomonoff's Secret
May 16 at 22:39
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Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
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– justhalf
May 17 at 19:20
1
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Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
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– Novice C
Jun 4 at 9:26
add a comment |
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In a two-dimensional space, you can introduce complex structure, which can be used to elegantly solve many problems (e.g. potential flow problems), but no analogue exists in 3 dimensions.
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Real world:
Automation % - e.g It is easy to automate something in 30% or 50% or 80% meanwhile it is difficult to go e.g above 95% and incredibly difficult or even almost impossible to reach 100%.
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Can you provide references for your claims?
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– nicoguaro♦
May 18 at 14:33
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I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
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– Joelty
May 20 at 6:48
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Then, I think that your question is not appropriate for this site.
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– nicoguaro♦
May 20 at 14:14
add a comment |
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15 Answers
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15 Answers
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One example that appears in many areas of physics, and in particular classical mechanics and quantum physics, is the two-body problem. The two-body problem here means the task of calculating the dynamics of two interacting particles which, for example, interact by gravitational or Coulomb forces. The solution to this problem can often be found in closed form by a variable transformation into center-of-mass and relative coordinates.
However, as soon as you consider three particles, in general no closed-form solutions exist.
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3
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Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
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– llama
May 16 at 17:27
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good nitpick, thanks, "in general" is missing here.
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– davidhigh
May 16 at 18:08
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Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
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– WorldSEnder
May 16 at 19:53
add a comment |
$begingroup$
One example that appears in many areas of physics, and in particular classical mechanics and quantum physics, is the two-body problem. The two-body problem here means the task of calculating the dynamics of two interacting particles which, for example, interact by gravitational or Coulomb forces. The solution to this problem can often be found in closed form by a variable transformation into center-of-mass and relative coordinates.
However, as soon as you consider three particles, in general no closed-form solutions exist.
$endgroup$
3
$begingroup$
Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
$endgroup$
– llama
May 16 at 17:27
$begingroup$
good nitpick, thanks, "in general" is missing here.
$endgroup$
– davidhigh
May 16 at 18:08
$begingroup$
Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
$endgroup$
– WorldSEnder
May 16 at 19:53
add a comment |
$begingroup$
One example that appears in many areas of physics, and in particular classical mechanics and quantum physics, is the two-body problem. The two-body problem here means the task of calculating the dynamics of two interacting particles which, for example, interact by gravitational or Coulomb forces. The solution to this problem can often be found in closed form by a variable transformation into center-of-mass and relative coordinates.
However, as soon as you consider three particles, in general no closed-form solutions exist.
$endgroup$
One example that appears in many areas of physics, and in particular classical mechanics and quantum physics, is the two-body problem. The two-body problem here means the task of calculating the dynamics of two interacting particles which, for example, interact by gravitational or Coulomb forces. The solution to this problem can often be found in closed form by a variable transformation into center-of-mass and relative coordinates.
However, as soon as you consider three particles, in general no closed-form solutions exist.
edited May 17 at 11:06
Glorfindel
155119
155119
answered May 16 at 5:36
davidhighdavidhigh
1,086612
1,086612
3
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Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
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– llama
May 16 at 17:27
$begingroup$
good nitpick, thanks, "in general" is missing here.
$endgroup$
– davidhigh
May 16 at 18:08
$begingroup$
Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
$endgroup$
– WorldSEnder
May 16 at 19:53
add a comment |
3
$begingroup$
Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
$endgroup$
– llama
May 16 at 17:27
$begingroup$
good nitpick, thanks, "in general" is missing here.
$endgroup$
– davidhigh
May 16 at 18:08
$begingroup$
Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
$endgroup$
– WorldSEnder
May 16 at 19:53
3
3
$begingroup$
Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
$endgroup$
– llama
May 16 at 17:27
$begingroup$
Nitpick that I'm sure you know, but your answer doesn't state: There are closed-form solutions to the 3-body problem, but only for a few special cases
$endgroup$
– llama
May 16 at 17:27
$begingroup$
good nitpick, thanks, "in general" is missing here.
$endgroup$
– davidhigh
May 16 at 18:08
$begingroup$
good nitpick, thanks, "in general" is missing here.
$endgroup$
– davidhigh
May 16 at 18:08
$begingroup$
Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
$endgroup$
– WorldSEnder
May 16 at 19:53
$begingroup$
Do note that the 3-body problem does have a (very slowly converging) series solution found by Sundman in the early 20th century and a weaker version (that ignores a singularities where bodies collide) was found for the n-body problem in 1990.
$endgroup$
– WorldSEnder
May 16 at 19:53
add a comment |
$begingroup$
A famous example is the boolean satisfiability problem (SAT). 2-SAT is not complicated to solve in polynomial time, but 3-SAT is NP-complete.
$endgroup$
3
$begingroup$
3-SAT can be reduced to graph 3-coloring, or vice-versa
$endgroup$
– GoHokies
May 16 at 9:38
8
$begingroup$
@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
$endgroup$
– findusl
May 16 at 16:47
1
$begingroup$
@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
$endgroup$
– GoHokies
May 20 at 3:00
add a comment |
$begingroup$
A famous example is the boolean satisfiability problem (SAT). 2-SAT is not complicated to solve in polynomial time, but 3-SAT is NP-complete.
$endgroup$
3
$begingroup$
3-SAT can be reduced to graph 3-coloring, or vice-versa
$endgroup$
– GoHokies
May 16 at 9:38
8
$begingroup$
@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
$endgroup$
– findusl
May 16 at 16:47
1
$begingroup$
@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
$endgroup$
– GoHokies
May 20 at 3:00
add a comment |
$begingroup$
A famous example is the boolean satisfiability problem (SAT). 2-SAT is not complicated to solve in polynomial time, but 3-SAT is NP-complete.
$endgroup$
A famous example is the boolean satisfiability problem (SAT). 2-SAT is not complicated to solve in polynomial time, but 3-SAT is NP-complete.
answered May 16 at 9:20
Federico PoloniFederico Poloni
3,6031327
3,6031327
3
$begingroup$
3-SAT can be reduced to graph 3-coloring, or vice-versa
$endgroup$
– GoHokies
May 16 at 9:38
8
$begingroup$
@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
$endgroup$
– findusl
May 16 at 16:47
1
$begingroup$
@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
$endgroup$
– GoHokies
May 20 at 3:00
add a comment |
3
$begingroup$
3-SAT can be reduced to graph 3-coloring, or vice-versa
$endgroup$
– GoHokies
May 16 at 9:38
8
$begingroup$
@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
$endgroup$
– findusl
May 16 at 16:47
1
$begingroup$
@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
$endgroup$
– GoHokies
May 20 at 3:00
3
3
$begingroup$
3-SAT can be reduced to graph 3-coloring, or vice-versa
$endgroup$
– GoHokies
May 16 at 9:38
$begingroup$
3-SAT can be reduced to graph 3-coloring, or vice-versa
$endgroup$
– GoHokies
May 16 at 9:38
8
8
$begingroup$
@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
$endgroup$
– findusl
May 16 at 16:47
$begingroup$
@GoHokies I thought that is true for every np-complete problem? Or is something especially noteworthy about these two? Sry if this is a stupid question, my knowledge on this area is basic. But this is how I understand cooks theorem
$endgroup$
– findusl
May 16 at 16:47
1
1
$begingroup$
@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
$endgroup$
– GoHokies
May 20 at 3:00
$begingroup$
@findusl You’re perfectly right. What makes 3-SAT and 3-coloring “special” is the 2-vs-3 dichotomy of the OP.
$endgroup$
– GoHokies
May 20 at 3:00
add a comment |
$begingroup$
In one and two dimensions, all roads lead to Rome, but not in three dimensions.
Specifically, given a random walk (equally likely to move in any direction) on the integers in one or two dimensions, then no matter the starting point, with probability one (a.k.a. almost surely), the random walk will eventually get to a specific designated point ("Rome").
However, for three or more dimensions, the probability of getting to "Rome" is less than one; with the probability decreasing as the number of dimensions increases.
So for instance, if conducting a stochastic (Monte Carlo) simulation of a random walk starting at "Rome", which will stop when Rome is returned to, then in one and two dimensions, you can be assured of eventually making it back to Rome and stopping the simulation - so easy. In three dimensions, you may never make it back, so hard.
https://en.wikipedia.org/wiki/Random_walk#Higher_dimensions
To visualize the two-dimensional case, one can imagine a person
walking randomly around a city. The city is effectively infinite and
arranged in a square grid of sidewalks. At every intersection, the
person randomly chooses one of the four possible routes (including the
one originally travelled from). Formally, this is a random walk on the
set of all points in the plane with integer coordinates.
Will the person ever get back to the original starting point of the
walk? This is the 2-dimensional equivalent of the level crossing
problem discussed above. In 1921 George Pólya proved that the person
almost surely would in a 2-dimensional random walk, but for 3
dimensions or higher, the probability of returning to the origin
decreases as the number of dimensions increases. In 3 dimensions, the
probability decreases to roughly 34%
See http://mathworld.wolfram.com/PolyasRandomWalkConstants.html for numerical values.
$endgroup$
add a comment |
$begingroup$
In one and two dimensions, all roads lead to Rome, but not in three dimensions.
Specifically, given a random walk (equally likely to move in any direction) on the integers in one or two dimensions, then no matter the starting point, with probability one (a.k.a. almost surely), the random walk will eventually get to a specific designated point ("Rome").
However, for three or more dimensions, the probability of getting to "Rome" is less than one; with the probability decreasing as the number of dimensions increases.
So for instance, if conducting a stochastic (Monte Carlo) simulation of a random walk starting at "Rome", which will stop when Rome is returned to, then in one and two dimensions, you can be assured of eventually making it back to Rome and stopping the simulation - so easy. In three dimensions, you may never make it back, so hard.
https://en.wikipedia.org/wiki/Random_walk#Higher_dimensions
To visualize the two-dimensional case, one can imagine a person
walking randomly around a city. The city is effectively infinite and
arranged in a square grid of sidewalks. At every intersection, the
person randomly chooses one of the four possible routes (including the
one originally travelled from). Formally, this is a random walk on the
set of all points in the plane with integer coordinates.
Will the person ever get back to the original starting point of the
walk? This is the 2-dimensional equivalent of the level crossing
problem discussed above. In 1921 George Pólya proved that the person
almost surely would in a 2-dimensional random walk, but for 3
dimensions or higher, the probability of returning to the origin
decreases as the number of dimensions increases. In 3 dimensions, the
probability decreases to roughly 34%
See http://mathworld.wolfram.com/PolyasRandomWalkConstants.html for numerical values.
$endgroup$
add a comment |
$begingroup$
In one and two dimensions, all roads lead to Rome, but not in three dimensions.
Specifically, given a random walk (equally likely to move in any direction) on the integers in one or two dimensions, then no matter the starting point, with probability one (a.k.a. almost surely), the random walk will eventually get to a specific designated point ("Rome").
However, for three or more dimensions, the probability of getting to "Rome" is less than one; with the probability decreasing as the number of dimensions increases.
So for instance, if conducting a stochastic (Monte Carlo) simulation of a random walk starting at "Rome", which will stop when Rome is returned to, then in one and two dimensions, you can be assured of eventually making it back to Rome and stopping the simulation - so easy. In three dimensions, you may never make it back, so hard.
https://en.wikipedia.org/wiki/Random_walk#Higher_dimensions
To visualize the two-dimensional case, one can imagine a person
walking randomly around a city. The city is effectively infinite and
arranged in a square grid of sidewalks. At every intersection, the
person randomly chooses one of the four possible routes (including the
one originally travelled from). Formally, this is a random walk on the
set of all points in the plane with integer coordinates.
Will the person ever get back to the original starting point of the
walk? This is the 2-dimensional equivalent of the level crossing
problem discussed above. In 1921 George Pólya proved that the person
almost surely would in a 2-dimensional random walk, but for 3
dimensions or higher, the probability of returning to the origin
decreases as the number of dimensions increases. In 3 dimensions, the
probability decreases to roughly 34%
See http://mathworld.wolfram.com/PolyasRandomWalkConstants.html for numerical values.
$endgroup$
In one and two dimensions, all roads lead to Rome, but not in three dimensions.
Specifically, given a random walk (equally likely to move in any direction) on the integers in one or two dimensions, then no matter the starting point, with probability one (a.k.a. almost surely), the random walk will eventually get to a specific designated point ("Rome").
However, for three or more dimensions, the probability of getting to "Rome" is less than one; with the probability decreasing as the number of dimensions increases.
So for instance, if conducting a stochastic (Monte Carlo) simulation of a random walk starting at "Rome", which will stop when Rome is returned to, then in one and two dimensions, you can be assured of eventually making it back to Rome and stopping the simulation - so easy. In three dimensions, you may never make it back, so hard.
https://en.wikipedia.org/wiki/Random_walk#Higher_dimensions
To visualize the two-dimensional case, one can imagine a person
walking randomly around a city. The city is effectively infinite and
arranged in a square grid of sidewalks. At every intersection, the
person randomly chooses one of the four possible routes (including the
one originally travelled from). Formally, this is a random walk on the
set of all points in the plane with integer coordinates.
Will the person ever get back to the original starting point of the
walk? This is the 2-dimensional equivalent of the level crossing
problem discussed above. In 1921 George Pólya proved that the person
almost surely would in a 2-dimensional random walk, but for 3
dimensions or higher, the probability of returning to the origin
decreases as the number of dimensions increases. In 3 dimensions, the
probability decreases to roughly 34%
See http://mathworld.wolfram.com/PolyasRandomWalkConstants.html for numerical values.
answered May 16 at 23:13
Mark L. StoneMark L. Stone
1,227513
1,227513
add a comment |
add a comment |
$begingroup$
Here's one close to the hearts of the contributors at SciComp.SE:
The Navier–Stokes existence and smoothness problem
The three-dimensional version is of course a famous open problem and the subject of a million-dollar Clay Millenium Prize. But the two-dimensional version has already been resolved a long time ago, with an affirmative answer. Terry Tao notes that the solution dates essentially back to Leray’s thesis in 1933!
Why is the three-dimensional problem so much harder to solve? The standard, hand-wavy response is that turbulence becomes significantly more unstable in three dimensions than in two. For a more mathematically rigorous answer, check out Charles Fefferman's official problem statement at the Clay Institute or Terry Tao's nice exposition on the possible proof strategies.
$endgroup$
add a comment |
$begingroup$
Here's one close to the hearts of the contributors at SciComp.SE:
The Navier–Stokes existence and smoothness problem
The three-dimensional version is of course a famous open problem and the subject of a million-dollar Clay Millenium Prize. But the two-dimensional version has already been resolved a long time ago, with an affirmative answer. Terry Tao notes that the solution dates essentially back to Leray’s thesis in 1933!
Why is the three-dimensional problem so much harder to solve? The standard, hand-wavy response is that turbulence becomes significantly more unstable in three dimensions than in two. For a more mathematically rigorous answer, check out Charles Fefferman's official problem statement at the Clay Institute or Terry Tao's nice exposition on the possible proof strategies.
$endgroup$
add a comment |
$begingroup$
Here's one close to the hearts of the contributors at SciComp.SE:
The Navier–Stokes existence and smoothness problem
The three-dimensional version is of course a famous open problem and the subject of a million-dollar Clay Millenium Prize. But the two-dimensional version has already been resolved a long time ago, with an affirmative answer. Terry Tao notes that the solution dates essentially back to Leray’s thesis in 1933!
Why is the three-dimensional problem so much harder to solve? The standard, hand-wavy response is that turbulence becomes significantly more unstable in three dimensions than in two. For a more mathematically rigorous answer, check out Charles Fefferman's official problem statement at the Clay Institute or Terry Tao's nice exposition on the possible proof strategies.
$endgroup$
Here's one close to the hearts of the contributors at SciComp.SE:
The Navier–Stokes existence and smoothness problem
The three-dimensional version is of course a famous open problem and the subject of a million-dollar Clay Millenium Prize. But the two-dimensional version has already been resolved a long time ago, with an affirmative answer. Terry Tao notes that the solution dates essentially back to Leray’s thesis in 1933!
Why is the three-dimensional problem so much harder to solve? The standard, hand-wavy response is that turbulence becomes significantly more unstable in three dimensions than in two. For a more mathematically rigorous answer, check out Charles Fefferman's official problem statement at the Clay Institute or Terry Tao's nice exposition on the possible proof strategies.
answered May 17 at 0:17
Richard ZhangRichard Zhang
2,165826
2,165826
add a comment |
add a comment |
$begingroup$
In social choice theory, designing an election scheme with two candidates is easy (majority rules), but designing an election scheme with three or more candidates necessarily involves making trade-offs between various reasonable-sounding conditions. (Arrow's impossibility theorem).
$endgroup$
add a comment |
$begingroup$
In social choice theory, designing an election scheme with two candidates is easy (majority rules), but designing an election scheme with three or more candidates necessarily involves making trade-offs between various reasonable-sounding conditions. (Arrow's impossibility theorem).
$endgroup$
add a comment |
$begingroup$
In social choice theory, designing an election scheme with two candidates is easy (majority rules), but designing an election scheme with three or more candidates necessarily involves making trade-offs between various reasonable-sounding conditions. (Arrow's impossibility theorem).
$endgroup$
In social choice theory, designing an election scheme with two candidates is easy (majority rules), but designing an election scheme with three or more candidates necessarily involves making trade-offs between various reasonable-sounding conditions. (Arrow's impossibility theorem).
answered May 17 at 4:41
ajdajd
30114
30114
add a comment |
add a comment |
$begingroup$
Simultaneous diagonalization of two matrices $A_1$ and $A_2$:
$$
U_1^T A_1 V = Sigma_1,quad U_2^TA_2V=Sigma_2
$$
is covered by existing generalized singular value decomposition.
However, when the simultaneous reduction of three matrices to a canonical form (weaker condition compared to the above) is required:
$$
Q^T A_1 Z = tilde{A_1},quad Q^T A_2 Z = tilde{A_2},quad Q^T A_3 Z = tilde{A_3}
$$
no direct methods exist. Therefor, one has to opt for more complicated routes using approximate SVDs, tensor decompositions, etc.
A practical application would be a solution for a quadratic eigenvalue problem:
$$
(A_1+lambda A_2+lambda^2 A_3)x=0
$$
Source: C. F. van Loan, "Lecture 6: The higher-order generalized singular value decomposition," CIME-EMS Summer School, Cetraro, Italy, June 2015.
$endgroup$
$begingroup$
Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
$endgroup$
– Rosie F
May 17 at 10:27
1
$begingroup$
@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
$endgroup$
– Anton Menshov♦
May 17 at 21:11
add a comment |
$begingroup$
Simultaneous diagonalization of two matrices $A_1$ and $A_2$:
$$
U_1^T A_1 V = Sigma_1,quad U_2^TA_2V=Sigma_2
$$
is covered by existing generalized singular value decomposition.
However, when the simultaneous reduction of three matrices to a canonical form (weaker condition compared to the above) is required:
$$
Q^T A_1 Z = tilde{A_1},quad Q^T A_2 Z = tilde{A_2},quad Q^T A_3 Z = tilde{A_3}
$$
no direct methods exist. Therefor, one has to opt for more complicated routes using approximate SVDs, tensor decompositions, etc.
A practical application would be a solution for a quadratic eigenvalue problem:
$$
(A_1+lambda A_2+lambda^2 A_3)x=0
$$
Source: C. F. van Loan, "Lecture 6: The higher-order generalized singular value decomposition," CIME-EMS Summer School, Cetraro, Italy, June 2015.
$endgroup$
$begingroup$
Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
$endgroup$
– Rosie F
May 17 at 10:27
1
$begingroup$
@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
$endgroup$
– Anton Menshov♦
May 17 at 21:11
add a comment |
$begingroup$
Simultaneous diagonalization of two matrices $A_1$ and $A_2$:
$$
U_1^T A_1 V = Sigma_1,quad U_2^TA_2V=Sigma_2
$$
is covered by existing generalized singular value decomposition.
However, when the simultaneous reduction of three matrices to a canonical form (weaker condition compared to the above) is required:
$$
Q^T A_1 Z = tilde{A_1},quad Q^T A_2 Z = tilde{A_2},quad Q^T A_3 Z = tilde{A_3}
$$
no direct methods exist. Therefor, one has to opt for more complicated routes using approximate SVDs, tensor decompositions, etc.
A practical application would be a solution for a quadratic eigenvalue problem:
$$
(A_1+lambda A_2+lambda^2 A_3)x=0
$$
Source: C. F. van Loan, "Lecture 6: The higher-order generalized singular value decomposition," CIME-EMS Summer School, Cetraro, Italy, June 2015.
$endgroup$
Simultaneous diagonalization of two matrices $A_1$ and $A_2$:
$$
U_1^T A_1 V = Sigma_1,quad U_2^TA_2V=Sigma_2
$$
is covered by existing generalized singular value decomposition.
However, when the simultaneous reduction of three matrices to a canonical form (weaker condition compared to the above) is required:
$$
Q^T A_1 Z = tilde{A_1},quad Q^T A_2 Z = tilde{A_2},quad Q^T A_3 Z = tilde{A_3}
$$
no direct methods exist. Therefor, one has to opt for more complicated routes using approximate SVDs, tensor decompositions, etc.
A practical application would be a solution for a quadratic eigenvalue problem:
$$
(A_1+lambda A_2+lambda^2 A_3)x=0
$$
Source: C. F. van Loan, "Lecture 6: The higher-order generalized singular value decomposition," CIME-EMS Summer School, Cetraro, Italy, June 2015.
answered May 16 at 18:20
Anton Menshov♦Anton Menshov
4,84422076
4,84422076
$begingroup$
Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
$endgroup$
– Rosie F
May 17 at 10:27
1
$begingroup$
@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
$endgroup$
– Anton Menshov♦
May 17 at 21:11
add a comment |
$begingroup$
Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
$endgroup$
– Rosie F
May 17 at 10:27
1
$begingroup$
@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
$endgroup$
– Anton Menshov♦
May 17 at 21:11
$begingroup$
Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
$endgroup$
– Rosie F
May 17 at 10:27
$begingroup$
Should $U_1^T$ and $U_2^T$ both be $V^{-1}$? Here they're not even required to be equal.
$endgroup$
– Rosie F
May 17 at 10:27
1
1
$begingroup$
@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
$endgroup$
– Anton Menshov♦
May 17 at 21:11
$begingroup$
@RosieF not for (generalized) SVD. See first equations here, which are just not expressing $Sigma$'s.
$endgroup$
– Anton Menshov♦
May 17 at 21:11
add a comment |
$begingroup$
There are plenty of examples in quantum computing, although I've been out of this for a while and so don't remember many. One major one is that bipartite entanglement (entanglement between two systems) is relatively easy whereas entanglement among three or more systems is an unsolved mess with probably a hundred papers written on the topic.
The root of this is that rank-2 tensors (i.e. matrices) can be analyzed via singular value decomposition. Nothing similar exists for tensors of rank 3 or higher. In fact, even something as simple as $maxleft(u_a v_b w_c T^{abc} / leftlVert u rightrVert leftlVert v rightrVert leftlVert w rightrVert right)$ (with sub/superscripts denoting Einstein summation) is, IIRC, not believed to be efficiently solvable.
This paper seems relevant, although I haven't read it: Most tensor problems are NP-hard
$endgroup$
2
$begingroup$
I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
$endgroup$
– Richard Zhang
May 17 at 16:36
$begingroup$
That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
$endgroup$
– Dan Stahlke
May 18 at 17:48
add a comment |
$begingroup$
There are plenty of examples in quantum computing, although I've been out of this for a while and so don't remember many. One major one is that bipartite entanglement (entanglement between two systems) is relatively easy whereas entanglement among three or more systems is an unsolved mess with probably a hundred papers written on the topic.
The root of this is that rank-2 tensors (i.e. matrices) can be analyzed via singular value decomposition. Nothing similar exists for tensors of rank 3 or higher. In fact, even something as simple as $maxleft(u_a v_b w_c T^{abc} / leftlVert u rightrVert leftlVert v rightrVert leftlVert w rightrVert right)$ (with sub/superscripts denoting Einstein summation) is, IIRC, not believed to be efficiently solvable.
This paper seems relevant, although I haven't read it: Most tensor problems are NP-hard
$endgroup$
2
$begingroup$
I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
$endgroup$
– Richard Zhang
May 17 at 16:36
$begingroup$
That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
$endgroup$
– Dan Stahlke
May 18 at 17:48
add a comment |
$begingroup$
There are plenty of examples in quantum computing, although I've been out of this for a while and so don't remember many. One major one is that bipartite entanglement (entanglement between two systems) is relatively easy whereas entanglement among three or more systems is an unsolved mess with probably a hundred papers written on the topic.
The root of this is that rank-2 tensors (i.e. matrices) can be analyzed via singular value decomposition. Nothing similar exists for tensors of rank 3 or higher. In fact, even something as simple as $maxleft(u_a v_b w_c T^{abc} / leftlVert u rightrVert leftlVert v rightrVert leftlVert w rightrVert right)$ (with sub/superscripts denoting Einstein summation) is, IIRC, not believed to be efficiently solvable.
This paper seems relevant, although I haven't read it: Most tensor problems are NP-hard
$endgroup$
There are plenty of examples in quantum computing, although I've been out of this for a while and so don't remember many. One major one is that bipartite entanglement (entanglement between two systems) is relatively easy whereas entanglement among three or more systems is an unsolved mess with probably a hundred papers written on the topic.
The root of this is that rank-2 tensors (i.e. matrices) can be analyzed via singular value decomposition. Nothing similar exists for tensors of rank 3 or higher. In fact, even something as simple as $maxleft(u_a v_b w_c T^{abc} / leftlVert u rightrVert leftlVert v rightrVert leftlVert w rightrVert right)$ (with sub/superscripts denoting Einstein summation) is, IIRC, not believed to be efficiently solvable.
This paper seems relevant, although I haven't read it: Most tensor problems are NP-hard
answered May 17 at 3:49
Dan StahlkeDan Stahlke
1912
1912
2
$begingroup$
I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
$endgroup$
– Richard Zhang
May 17 at 16:36
$begingroup$
That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
$endgroup$
– Dan Stahlke
May 18 at 17:48
add a comment |
2
$begingroup$
I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
$endgroup$
– Richard Zhang
May 17 at 16:36
$begingroup$
That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
$endgroup$
– Dan Stahlke
May 18 at 17:48
2
2
$begingroup$
I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
$endgroup$
– Richard Zhang
May 17 at 16:36
$begingroup$
I feel like the real issue you're getting at is that the tensor rank decomposition is easy for order-1 tensors (vectors) and order-2 tensors (matrices), but NP-hard for the rest
$endgroup$
– Richard Zhang
May 17 at 16:36
$begingroup$
That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
$endgroup$
– Dan Stahlke
May 18 at 17:48
$begingroup$
That's a part of it, but even if you had a way of decomposing them there is still the issue of categorizing/classifying. For entanglement local unitaries don't matter, so all that's left in the order-2 case is a list of singular values (SVD is called Schmidt decomposition in this context). For higher orders there is a whole zoo of possibilities. Questions such as which states can be transformed into other states via local operations end up being very difficult (from a theoretical viewpoint, not necessarily computational).
$endgroup$
– Dan Stahlke
May 18 at 17:48
add a comment |
$begingroup$
Angle bisection with straightedge and compass is simple, angle trisection is in general impossible.
$endgroup$
add a comment |
$begingroup$
Angle bisection with straightedge and compass is simple, angle trisection is in general impossible.
$endgroup$
add a comment |
$begingroup$
Angle bisection with straightedge and compass is simple, angle trisection is in general impossible.
$endgroup$
Angle bisection with straightedge and compass is simple, angle trisection is in general impossible.
answered May 17 at 19:02
davidhighdavidhigh
1,086612
1,086612
add a comment |
add a comment |
$begingroup$
Type inference for Rank-n types. Type inference for Rank-2 is not especially difficult, but type inference for Rank-3 or above is undecidable.
$endgroup$
add a comment |
$begingroup$
Type inference for Rank-n types. Type inference for Rank-2 is not especially difficult, but type inference for Rank-3 or above is undecidable.
$endgroup$
add a comment |
$begingroup$
Type inference for Rank-n types. Type inference for Rank-2 is not especially difficult, but type inference for Rank-3 or above is undecidable.
$endgroup$
Type inference for Rank-n types. Type inference for Rank-2 is not especially difficult, but type inference for Rank-3 or above is undecidable.
answered May 16 at 20:06
André PopovitchAndré Popovitch
311
311
add a comment |
add a comment |
$begingroup$
Folding a piece of paper in half without tools is easy. Folding it into thirds is hard.
Factoring a polynomial with two roots is easy. Factoring a polynomial with three roots is significantly more complicated.
$endgroup$
2
$begingroup$
Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
$endgroup$
– Bill K
May 17 at 23:24
add a comment |
$begingroup$
Folding a piece of paper in half without tools is easy. Folding it into thirds is hard.
Factoring a polynomial with two roots is easy. Factoring a polynomial with three roots is significantly more complicated.
$endgroup$
2
$begingroup$
Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
$endgroup$
– Bill K
May 17 at 23:24
add a comment |
$begingroup$
Folding a piece of paper in half without tools is easy. Folding it into thirds is hard.
Factoring a polynomial with two roots is easy. Factoring a polynomial with three roots is significantly more complicated.
$endgroup$
Folding a piece of paper in half without tools is easy. Folding it into thirds is hard.
Factoring a polynomial with two roots is easy. Factoring a polynomial with three roots is significantly more complicated.
answered May 17 at 19:00
Arcanist LupusArcanist Lupus
1311
1311
2
$begingroup$
Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
$endgroup$
– Bill K
May 17 at 23:24
add a comment |
2
$begingroup$
Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
$endgroup$
– Bill K
May 17 at 23:24
2
2
$begingroup$
Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
$endgroup$
– Bill K
May 17 at 23:24
$begingroup$
Your first example doesn't fit the spirit of the quote. The idea is that as it gets higher past two it's more difficult, however with folding a paper, 4ths is just about as easy as half. The quote here would be "Even is easier than odd" I think the second one is good though--and 'grats on trying to hyper-simplify it with the paper!
$endgroup$
– Bill K
May 17 at 23:24
add a comment |
$begingroup$
A smooth curve of degree 2 (i.e. given as the solution of $f(x,y) = 0$ where $f$ is a polynomial of degree 2) with a given point is rational, meaning that it can be parameterized by quotients of polynomials, of degree 3 it isn't. The former are considered well understood, the latter, called elliptic curves when a base point, i.e. a specific solution, is singled out, are the object of intense research.
This difference has several implications:
- In degree 2 there are algorithms to find all rational points (solutions in rational numbers), in degree 3 no such algorithm is known.
- Integrals involving $sqrt{f(x)}$ with $f$ of degree 1 or 2 have solutions in elementary functions, but not for $f$ of degree 3 or higher.
- The discrete logarithm problem is tractable on curves of degree 2, hence not suitable for cryptographic applications, whereas the assumed hardness of the same problem on elliptic curves is at the basis of some of the most popular public key cryptosystems.
$endgroup$
add a comment |
$begingroup$
A smooth curve of degree 2 (i.e. given as the solution of $f(x,y) = 0$ where $f$ is a polynomial of degree 2) with a given point is rational, meaning that it can be parameterized by quotients of polynomials, of degree 3 it isn't. The former are considered well understood, the latter, called elliptic curves when a base point, i.e. a specific solution, is singled out, are the object of intense research.
This difference has several implications:
- In degree 2 there are algorithms to find all rational points (solutions in rational numbers), in degree 3 no such algorithm is known.
- Integrals involving $sqrt{f(x)}$ with $f$ of degree 1 or 2 have solutions in elementary functions, but not for $f$ of degree 3 or higher.
- The discrete logarithm problem is tractable on curves of degree 2, hence not suitable for cryptographic applications, whereas the assumed hardness of the same problem on elliptic curves is at the basis of some of the most popular public key cryptosystems.
$endgroup$
add a comment |
$begingroup$
A smooth curve of degree 2 (i.e. given as the solution of $f(x,y) = 0$ where $f$ is a polynomial of degree 2) with a given point is rational, meaning that it can be parameterized by quotients of polynomials, of degree 3 it isn't. The former are considered well understood, the latter, called elliptic curves when a base point, i.e. a specific solution, is singled out, are the object of intense research.
This difference has several implications:
- In degree 2 there are algorithms to find all rational points (solutions in rational numbers), in degree 3 no such algorithm is known.
- Integrals involving $sqrt{f(x)}$ with $f$ of degree 1 or 2 have solutions in elementary functions, but not for $f$ of degree 3 or higher.
- The discrete logarithm problem is tractable on curves of degree 2, hence not suitable for cryptographic applications, whereas the assumed hardness of the same problem on elliptic curves is at the basis of some of the most popular public key cryptosystems.
$endgroup$
A smooth curve of degree 2 (i.e. given as the solution of $f(x,y) = 0$ where $f$ is a polynomial of degree 2) with a given point is rational, meaning that it can be parameterized by quotients of polynomials, of degree 3 it isn't. The former are considered well understood, the latter, called elliptic curves when a base point, i.e. a specific solution, is singled out, are the object of intense research.
This difference has several implications:
- In degree 2 there are algorithms to find all rational points (solutions in rational numbers), in degree 3 no such algorithm is known.
- Integrals involving $sqrt{f(x)}$ with $f$ of degree 1 or 2 have solutions in elementary functions, but not for $f$ of degree 3 or higher.
- The discrete logarithm problem is tractable on curves of degree 2, hence not suitable for cryptographic applications, whereas the assumed hardness of the same problem on elliptic curves is at the basis of some of the most popular public key cryptosystems.
edited Jun 9 at 11:47
answered Jun 7 at 12:23
doetoedoetoe
30119
30119
add a comment |
add a comment |
$begingroup$
Here's a neat one from optimization: the Alternating Direction Method of Multipliers (ADMM) algorithm.
Given an uncoupled and convex objective function of two variables (the variables themselves could be vectors) and a linear constraint coupling the two variables:
$$min f_1(x_1) + f_2(x_2) $$
$$ s.t. ; A_1 x_1 + A_2 x_2 = b $$
The Augmented Lagrangian function for this optimization problem would then be
$$ L_{rho}(x_1, x_2, lambda) = f_1(x_1) + f_2(x_2) + lambda^T (A_1 x_1 + A_2 x_2 - b) + frac{rho}{2}|| A_1 x_1 + A_2 x_2 - b ||_2^2 $$
The ADMM algorithm roughly works by performing a "Gauss-Seidel" splitting on the augmented Lagrangian function for this optimization problem by minimizing $L_{rho}(x_1, x_2, lambda)$ first with respect to $x_1$ (while $x_2, lambda$ remain fixed), then by minimizing $L_{rho}(x_1, x_2, lambda)$ with respect to $x_2$ (while $x_1, lambda$ remain fixed), then by updating $lambda$. This cycle goes on until a stopping criterion is reached.
(Note: some researchers such as Eckstein discard the Gauss-Siedel splitting view in favor of proximal operators, for example see http://rutcor.rutgers.edu/pub/rrr/reports2012/32_2012.pdf )
For convex problems, this algorithm has been proven to converge - for two sets of variables. This is not the case for three variables. For example, the optimization problem
$$min f_1(x_1) + f_2(x_2) + f_3(x_3)$$
$$ s.t. ; A_1 x_1 + A_2 x_2 + A_3x_3 = b $$
Even if all the $f$ are convex, the ADMM-like approach (minimizing the Augmented Lagrangian with respect to each variable $x_i$, then updating the dual variable $lambda$) is NOT guaranteed to converge, as was shown in this paper.
https://web.stanford.edu/~yyye/ADMM-final.pdf
$endgroup$
add a comment |
$begingroup$
Here's a neat one from optimization: the Alternating Direction Method of Multipliers (ADMM) algorithm.
Given an uncoupled and convex objective function of two variables (the variables themselves could be vectors) and a linear constraint coupling the two variables:
$$min f_1(x_1) + f_2(x_2) $$
$$ s.t. ; A_1 x_1 + A_2 x_2 = b $$
The Augmented Lagrangian function for this optimization problem would then be
$$ L_{rho}(x_1, x_2, lambda) = f_1(x_1) + f_2(x_2) + lambda^T (A_1 x_1 + A_2 x_2 - b) + frac{rho}{2}|| A_1 x_1 + A_2 x_2 - b ||_2^2 $$
The ADMM algorithm roughly works by performing a "Gauss-Seidel" splitting on the augmented Lagrangian function for this optimization problem by minimizing $L_{rho}(x_1, x_2, lambda)$ first with respect to $x_1$ (while $x_2, lambda$ remain fixed), then by minimizing $L_{rho}(x_1, x_2, lambda)$ with respect to $x_2$ (while $x_1, lambda$ remain fixed), then by updating $lambda$. This cycle goes on until a stopping criterion is reached.
(Note: some researchers such as Eckstein discard the Gauss-Siedel splitting view in favor of proximal operators, for example see http://rutcor.rutgers.edu/pub/rrr/reports2012/32_2012.pdf )
For convex problems, this algorithm has been proven to converge - for two sets of variables. This is not the case for three variables. For example, the optimization problem
$$min f_1(x_1) + f_2(x_2) + f_3(x_3)$$
$$ s.t. ; A_1 x_1 + A_2 x_2 + A_3x_3 = b $$
Even if all the $f$ are convex, the ADMM-like approach (minimizing the Augmented Lagrangian with respect to each variable $x_i$, then updating the dual variable $lambda$) is NOT guaranteed to converge, as was shown in this paper.
https://web.stanford.edu/~yyye/ADMM-final.pdf
$endgroup$
add a comment |
$begingroup$
Here's a neat one from optimization: the Alternating Direction Method of Multipliers (ADMM) algorithm.
Given an uncoupled and convex objective function of two variables (the variables themselves could be vectors) and a linear constraint coupling the two variables:
$$min f_1(x_1) + f_2(x_2) $$
$$ s.t. ; A_1 x_1 + A_2 x_2 = b $$
The Augmented Lagrangian function for this optimization problem would then be
$$ L_{rho}(x_1, x_2, lambda) = f_1(x_1) + f_2(x_2) + lambda^T (A_1 x_1 + A_2 x_2 - b) + frac{rho}{2}|| A_1 x_1 + A_2 x_2 - b ||_2^2 $$
The ADMM algorithm roughly works by performing a "Gauss-Seidel" splitting on the augmented Lagrangian function for this optimization problem by minimizing $L_{rho}(x_1, x_2, lambda)$ first with respect to $x_1$ (while $x_2, lambda$ remain fixed), then by minimizing $L_{rho}(x_1, x_2, lambda)$ with respect to $x_2$ (while $x_1, lambda$ remain fixed), then by updating $lambda$. This cycle goes on until a stopping criterion is reached.
(Note: some researchers such as Eckstein discard the Gauss-Siedel splitting view in favor of proximal operators, for example see http://rutcor.rutgers.edu/pub/rrr/reports2012/32_2012.pdf )
For convex problems, this algorithm has been proven to converge - for two sets of variables. This is not the case for three variables. For example, the optimization problem
$$min f_1(x_1) + f_2(x_2) + f_3(x_3)$$
$$ s.t. ; A_1 x_1 + A_2 x_2 + A_3x_3 = b $$
Even if all the $f$ are convex, the ADMM-like approach (minimizing the Augmented Lagrangian with respect to each variable $x_i$, then updating the dual variable $lambda$) is NOT guaranteed to converge, as was shown in this paper.
https://web.stanford.edu/~yyye/ADMM-final.pdf
$endgroup$
Here's a neat one from optimization: the Alternating Direction Method of Multipliers (ADMM) algorithm.
Given an uncoupled and convex objective function of two variables (the variables themselves could be vectors) and a linear constraint coupling the two variables:
$$min f_1(x_1) + f_2(x_2) $$
$$ s.t. ; A_1 x_1 + A_2 x_2 = b $$
The Augmented Lagrangian function for this optimization problem would then be
$$ L_{rho}(x_1, x_2, lambda) = f_1(x_1) + f_2(x_2) + lambda^T (A_1 x_1 + A_2 x_2 - b) + frac{rho}{2}|| A_1 x_1 + A_2 x_2 - b ||_2^2 $$
The ADMM algorithm roughly works by performing a "Gauss-Seidel" splitting on the augmented Lagrangian function for this optimization problem by minimizing $L_{rho}(x_1, x_2, lambda)$ first with respect to $x_1$ (while $x_2, lambda$ remain fixed), then by minimizing $L_{rho}(x_1, x_2, lambda)$ with respect to $x_2$ (while $x_1, lambda$ remain fixed), then by updating $lambda$. This cycle goes on until a stopping criterion is reached.
(Note: some researchers such as Eckstein discard the Gauss-Siedel splitting view in favor of proximal operators, for example see http://rutcor.rutgers.edu/pub/rrr/reports2012/32_2012.pdf )
For convex problems, this algorithm has been proven to converge - for two sets of variables. This is not the case for three variables. For example, the optimization problem
$$min f_1(x_1) + f_2(x_2) + f_3(x_3)$$
$$ s.t. ; A_1 x_1 + A_2 x_2 + A_3x_3 = b $$
Even if all the $f$ are convex, the ADMM-like approach (minimizing the Augmented Lagrangian with respect to each variable $x_i$, then updating the dual variable $lambda$) is NOT guaranteed to converge, as was shown in this paper.
https://web.stanford.edu/~yyye/ADMM-final.pdf
answered Jun 9 at 20:12
Nathaniel KroegerNathaniel Kroeger
827
827
add a comment |
add a comment |
$begingroup$
The TREE
function.
We can calculate TREE(2) = 3
, but TREE(3)
is not calculable in the universe lifetime, we only know that it is finite.
$endgroup$
$begingroup$
TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.
$endgroup$
– Solomonoff's Secret
May 16 at 22:39
$begingroup$
Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
$endgroup$
– justhalf
May 17 at 19:20
1
$begingroup$
Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
$endgroup$
– Novice C
Jun 4 at 9:26
add a comment |
$begingroup$
The TREE
function.
We can calculate TREE(2) = 3
, but TREE(3)
is not calculable in the universe lifetime, we only know that it is finite.
$endgroup$
$begingroup$
TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.
$endgroup$
– Solomonoff's Secret
May 16 at 22:39
$begingroup$
Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
$endgroup$
– justhalf
May 17 at 19:20
1
$begingroup$
Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
$endgroup$
– Novice C
Jun 4 at 9:26
add a comment |
$begingroup$
The TREE
function.
We can calculate TREE(2) = 3
, but TREE(3)
is not calculable in the universe lifetime, we only know that it is finite.
$endgroup$
The TREE
function.
We can calculate TREE(2) = 3
, but TREE(3)
is not calculable in the universe lifetime, we only know that it is finite.
edited May 17 at 19:21
answered May 16 at 21:46
justhalfjusthalf
1134
1134
$begingroup$
TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.
$endgroup$
– Solomonoff's Secret
May 16 at 22:39
$begingroup$
Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
$endgroup$
– justhalf
May 17 at 19:20
1
$begingroup$
Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
$endgroup$
– Novice C
Jun 4 at 9:26
add a comment |
$begingroup$
TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.
$endgroup$
– Solomonoff's Secret
May 16 at 22:39
$begingroup$
Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
$endgroup$
– justhalf
May 17 at 19:20
1
$begingroup$
Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
$endgroup$
– Novice C
Jun 4 at 9:26
$begingroup$
TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.$endgroup$
– Solomonoff's Secret
May 16 at 22:39
$begingroup$
TREE(3)
is "calculable" given enough time. For example, for each $n$ you could generate all colored trees of size $n$ and verify if each meets the necessary criteria until no such trees exist. But it would take an unimaginable amount of space and time.$endgroup$
– Solomonoff's Secret
May 16 at 22:39
$begingroup$
Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
$endgroup$
– justhalf
May 17 at 19:20
$begingroup$
Right, sorry for the mistake. Fixed my statement. Thanks Solomonoff!
$endgroup$
– justhalf
May 17 at 19:20
1
1
$begingroup$
Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
$endgroup$
– Novice C
Jun 4 at 9:26
$begingroup$
Related numberphile video about Tree(3): youtube.com/watch?v=3P6DWAwwViU
$endgroup$
– Novice C
Jun 4 at 9:26
add a comment |
$begingroup$
In a two-dimensional space, you can introduce complex structure, which can be used to elegantly solve many problems (e.g. potential flow problems), but no analogue exists in 3 dimensions.
$endgroup$
add a comment |
$begingroup$
In a two-dimensional space, you can introduce complex structure, which can be used to elegantly solve many problems (e.g. potential flow problems), but no analogue exists in 3 dimensions.
$endgroup$
add a comment |
$begingroup$
In a two-dimensional space, you can introduce complex structure, which can be used to elegantly solve many problems (e.g. potential flow problems), but no analogue exists in 3 dimensions.
$endgroup$
In a two-dimensional space, you can introduce complex structure, which can be used to elegantly solve many problems (e.g. potential flow problems), but no analogue exists in 3 dimensions.
answered Jun 7 at 8:43
Patrick SananPatrick Sanan
664415
664415
add a comment |
add a comment |
$begingroup$
Real world:
Automation % - e.g It is easy to automate something in 30% or 50% or 80% meanwhile it is difficult to go e.g above 95% and incredibly difficult or even almost impossible to reach 100%.
$endgroup$
2
$begingroup$
Can you provide references for your claims?
$endgroup$
– nicoguaro♦
May 18 at 14:33
$begingroup$
I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
$endgroup$
– Joelty
May 20 at 6:48
$begingroup$
Then, I think that your question is not appropriate for this site.
$endgroup$
– nicoguaro♦
May 20 at 14:14
add a comment |
$begingroup$
Real world:
Automation % - e.g It is easy to automate something in 30% or 50% or 80% meanwhile it is difficult to go e.g above 95% and incredibly difficult or even almost impossible to reach 100%.
$endgroup$
2
$begingroup$
Can you provide references for your claims?
$endgroup$
– nicoguaro♦
May 18 at 14:33
$begingroup$
I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
$endgroup$
– Joelty
May 20 at 6:48
$begingroup$
Then, I think that your question is not appropriate for this site.
$endgroup$
– nicoguaro♦
May 20 at 14:14
add a comment |
$begingroup$
Real world:
Automation % - e.g It is easy to automate something in 30% or 50% or 80% meanwhile it is difficult to go e.g above 95% and incredibly difficult or even almost impossible to reach 100%.
$endgroup$
Real world:
Automation % - e.g It is easy to automate something in 30% or 50% or 80% meanwhile it is difficult to go e.g above 95% and incredibly difficult or even almost impossible to reach 100%.
answered May 17 at 13:04
JoeltyJoelty
95
95
2
$begingroup$
Can you provide references for your claims?
$endgroup$
– nicoguaro♦
May 18 at 14:33
$begingroup$
I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
$endgroup$
– Joelty
May 20 at 6:48
$begingroup$
Then, I think that your question is not appropriate for this site.
$endgroup$
– nicoguaro♦
May 20 at 14:14
add a comment |
2
$begingroup$
Can you provide references for your claims?
$endgroup$
– nicoguaro♦
May 18 at 14:33
$begingroup$
I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
$endgroup$
– Joelty
May 20 at 6:48
$begingroup$
Then, I think that your question is not appropriate for this site.
$endgroup$
– nicoguaro♦
May 20 at 14:14
2
2
$begingroup$
Can you provide references for your claims?
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– nicoguaro♦
May 18 at 14:33
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Can you provide references for your claims?
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– nicoguaro♦
May 18 at 14:33
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I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
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– Joelty
May 20 at 6:48
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I can't, but take a look at e.g self driving cars. Learning car to drive straight and control speed is probably times easier than learning it to drive like a normal person. The more complex process is, then more border cases appears when you want to make it fully automated
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– Joelty
May 20 at 6:48
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Then, I think that your question is not appropriate for this site.
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– nicoguaro♦
May 20 at 14:14
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Then, I think that your question is not appropriate for this site.
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– nicoguaro♦
May 20 at 14:14
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2
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The curse of dimensionality comes to mind.
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– Paul♦
May 16 at 0:13
4
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graph 2-coloring (easy) versus 3-coloring (NP-hard), see here
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– GoHokies
May 16 at 5:34
5
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@GoHokies Please don't post answers as comments.
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– David Richerby
May 16 at 11:01
4
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From foundation of math or recursion background, you might come across TREE function, where TREE(2)=3, and TREE(3) is ... quite large. (not being familiar with computational sciences, I'm not sure this is really an answer you are looking for, but it seems similar enough to leave a comment about)
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– BurnsBA
May 16 at 14:01
2
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A counterexample: "Never go to sea with two chronometers; take one or three." That said, there are so many good examples that there is no right answer. This question should be community wiki.
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– David Hammen
May 17 at 0:24