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# What are these “div-by-0-supporting product of elements in set”-beasts known as?

I remember asking about this in some Norwegian Usenet group many many years back, and as I recall (although I may be just imagining it) a mathematician at the University of Oslo gave me a good summary. But I have forgotten it all!

Background: during my student days at Heriot Watt U. (Edinburgh) I wrote dissertation about combination of uncertain evidence by means of a model called Dempster-Shafer. With this method one had to update products of the elements of multisets (by this I mean that an element could be present any number of times), where it was necessary to support both that an element was removed, and that it was added. Addition of x to a set meant just multiplying the product by x, but removal was trickier: when x was zero one could not just divide the product by x!

So, for each set I maintained a pair (P, Z) where P was the product of the elements if possible or else 1, and Z the count of zeroes in the set. It was practical to allow a negative number of occurrences of a number in a set. And then the computer implementation of updates of these pairs (when sets were updated) corresponded directly (identical to) the computer implementation of of multiplication and division of complex numbers in polar form, i.e. (P1,Z1) "times" (P2,Z2) = (P1*P2, Z1+Z2), where the main trick is that the set {0} has product (1,1).

I find it difficult to write in the math notation required by this site. To be concrete, trying to formalize things a little by defining count of an element in a multiset, and then product of set, I couldn't get nice formula result for x raised to expression with multi-character name function. So I gave up on that, and tried simple textual formulas.

But even those proved too challenging for me here (and now I can appreciate the formula editor in Word!), I'm sorry. Also, sorry about the tags. I tried "multi-set" and "pair" but they don't exist, and I don't have rating to create new tags. I was unable to find good synonyms -- so, formulas lacking and tags lacking!

But hopefully the above description gives the gist of it: updates of sets (element added or removed) corresponded to updates of pairs, and the product, if defined, is trivially deduced from the pair. I wonder what those pair beasts are called and what the common notation is. And e.g. if there is any way to cut down on the complexity of results of addition and subtraction (I can't see any way, apparently addition and subtraction produce ugly ugly things not even deserving of name "beast"...).

?

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I don't really understand what you mean by "product." Could you give a simple example with explicit sets? –  Qiaochu Yuan Jul 13 '11 at 3:14
I am not 100% sure that [set-theory] is relevant. –  Asaf Karagila Jul 13 '11 at 4:27
@Asaf: Agreed. It looks like probability and statistics to me... –  Zhen Lin Jul 13 '11 at 6:38
Qiaochu: I'll have to invent a notation for the sets, first. Let e.g. S1={3,3,5} denote a multiset where 3 occurs twice and 5 occurs once. The product is 3*3*5=45. When the set is updated by adding 0 forming S2={3,3,5,0} the product is 0. The pair for this set is (45,1), which implies product 0. Update yet again to remove one count of 0, yielding S3={3,3,5}, corresponds to updating the pair to (45,0), which yields product 45. Doing that again yields a set which I don't know how to express, but with pair (45,-1) and undefined product. Now remove the sets and the (P,Z) pairs live on their own. –  Alf P. Steinbach Jul 13 '11 at 11:28
@Asaf: "set-theory" was tag I found closest in meaning to "multiset". –  Alf P. Steinbach Jul 13 '11 at 11:38

Do you have other arithmetic operations defined on the pairs $(P,Z)$? As far as I can tell, what you have defined so far is just a pair $(P,Z)$ where $P$ takes values in the non-zero reals ($P\in \mathbb{R}\setminus{0}$) and $Z$ takes values in the reals (or the integers?) with a binary operation $(P_1,Z_1) \cdot (P_2,Z_2) = (P_1 \times P_2, Z_1 + Z_2)$.

In terms of abstract algebra, you have that $P$ is in the multiplicative group of non-zero reals, and $Z$ is in the additive group of either real numbers or integers. And what you have constructed, as the pair $(P,Z)$, is the group direct product of the two groups.

To define the process of evaluating $(P,Z)$ to the "product", assuming that $Z$ takes value in the integers, you can actually embed $(P,Z)$ into the formal Laurent series over the reals, by sending $(P,Z)$ to a monomial

$$(P,Z)\mapsto P x^Z$$

which you can then formally evaluate at $z = 0$. With this you can get an analogy for $Z \neq 0, 1$ to the zeroes of higher degree, and the poles, of complex analysis.

Now let's think about the underlying multiset and how it corresponds to the pair $(P,Z)$. Your collection of multisets actually form a groupoid. The elements are given by the multisets and their formal inverses, and the partial function is given by "set addition and subtraction" in the following way.

Each element of your groupoid shall be either a multiset $m$ consisting of a collection of real numbers (or integers), or a formal inverse of a multiset (which we can denote as $m^{-1}$). The partial function is defined as follows: given $m$ and $n$ multisets, $m\times n$ is the multiset union of $m$ and $n$. Given $m^{-1}$ and $n^{-1}$ inverse multisets, $m^{-1}\times n^{-1} = (m\times n)^{-1}$ is the inverse of the multiset union of their corresponding multisets. If $m$ is a multiset and $n^{-1}$ is a multiset inverse, $m\times n^{-1}$ is only defined if $n$ is a multiset subset of $m$, and the output is $m$ with the elements of $n$ removed. Similarly we can defined $m^{-1}\times n$.

The mapping from the multiset notation to the $(P,Z)$ notation then corresponds to a groupoid homomorphism.

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Of course, the mapping from the multiset notation to the $(P,Z)$ notation is not a monomorphism (i.e. you lose some information). The key is that if the information you only care about is the "product", that operation can be factored through this homomorphism, and so it is enough to keep track of the "pair" notation and not the full multiset notation. –  Willie Wong Jul 13 '11 at 14:03
Well, the practically useful operations are multiplication and division. In the question I'm asking about e.g. addition, can it be usefully defined? The pairs would be meaningless if they didn't have connection to sets (where set addition corresponds directly to pair multiplication), and mapping to real numbers. I don't have the background to understand either the words or the notation in your mapping pair -> number, but the idea is simple enough and I wish I'd thought of that! I think mapping numbers to pairs can't be simplified that much, since 0 maps to (1,1) as opposed to general (x,0)? –  Alf P. Steinbach Jul 13 '11 at 14:09
Yes, addition can be usefully defined, but not in the form you have written as pairs. You can extend $(P,Z)$ to the formal Laurent series I wrote about above. And as the formal Laurent series form a ring, you can do all the usual arithmetic operations, except for "division by zero"... but that is not a problem, since the "zero" element in the Laurent series expression corresponds to the $(0,0)$ element of the $(P,Z)$ notation, which you expressedly disallow. –  Willie Wong Jul 13 '11 at 14:19
"m×n−1 is only defined if n is a multiset subset of m, and the output is m with the elements of n removed" I think this must be wrong, or else you're talking about multisets where an element can't occur a negative number of times (have negative "multiplicity", oh how I love these words! :-) )? –  Alf P. Steinbach Jul 13 '11 at 14:19
Lifting back to the $(P,Z)$ notation, what you have is that you have addition while keeping track of the number of zeros. That is, if $Z_1 = Z_2$, then $(P_1,Z_1)+(P_2,Z_2) = (P_1 + P_2, Z_1)$. But if $Z_1 \neq Z_2$, you just leave it as the symbol $(P_1,Z_1)+(P_2,Z_2)$ without simplifying. Kinda like how you can add an subtract polynomials. (Indeed, you should think of this whole exercise as treating $0$ as the symbol $x$ in the usual notation for a polynomials....) –  Willie Wong Jul 13 '11 at 14:22

Generally, suppose you are working with multisets over a commutative monoid, i.e. a set with a commutative and associative product operation and identity element $1$. Apparently you desire to annotate each multiset with its product, so that, after insertions or deletions, the product can be efficiently updated by a single operation (vs. multiplying together a possibly very large number of elements contained in the multiset). For insertions, to update the product one simply multiplies the product by the inserted element. But this is a noninvertible operation for noncancellable elements (e.g. a $0$ such that $\rm\:0\cdot x = 0)\:.\:$ Thus, to efficiently support deletions, such noncancellable elements must be maintained separately from the product, in their own distinguished multiset.

Thus one may simply collapse all of the cancellable elements of a multiset into their product, and leave the noncancellable elements in multiset representation. So, generally, it suffices to represent the multiset by a triple containing (1) the submultiset of noncancellable elements; (2) the submultiset of cancellable elements; and (3) the product of the submultiset of cancellable elements.

To compute the product of all elements of the multiset one multiplies the cached product of all cancellable elements by the on-the-fly computed product of all of the noncancellable elements. This is a significant optimization when cancellable elements occur much more frequently than noncancellable elements (such as your example when $0$ is the only noncancellable element). One could also cache the entire product and invalidate it when a deletion occurs.

Note that one cannot eliminate the submultiset of cancellable elements from the triple if one desires it to be easily computable, since generally it cannot "decoded" from their product, or it may be computationally expensive to do so (e.g. factorization of integers).

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# Entangled or Not?

Is the following state $$|\Psi\rangle$$ entangled?

$|\Psi\rangle = \frac{1}{\sqrt{2}} (|\downarrow\downarrow\rangle + |\downarrow\uparrow\rangle).$

Note: The notation $$|\uparrow\downarrow\rangle = |\uparrow\rangle \otimes |\downarrow\rangle$$ is a common shorthand for tensor products of spin states.

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3,120 views

Let $P$ be a Mealy machine that has $N$ states and $M$ outputs. Let $Z$ be the number of states of the corresponding Moore machine $Q$ which is equivalent to $P$. Which of the following conditions always holds?

1. $Z<M+N$
2. $Z=M*N$
3. $Z=P*M+Q*N$
4. $Z \leq M*N$

if we convert melay to moore then no of state remain same or may increase so first 3 options are ruled out

@Nilabja Sarkar how can you elemenate first 3?

RHS is given as function of both states and output ,where as LHS is only states of MOORE machine

P->Melay m/c now corresponding  moore m/c means we are covering melay to moore . so the states will remain same otherwise increase let take a case of 2s complement no melay take 2 state   2 o/p moore take 3 state . its not perfect  proof i am thinking that way

In Moore to Mealy conversion, the number of states are same.

In Mealy to Moore conversion, the number of states in worst case can be N*M.

i.e the number of states in Moore are equal or more than its equivalent Mealy machine.

### 1 comment

baburavula says

If a Mealy machine has "n" states and "m" output then corresponding Moore

machine may have "m*n" states.

In a Moore machine as we know each state has a fixed output. Hence suppose if we make a transition from one state whose output is 1 to another state, it will have a definite output defined for that state. In Mealy machine that is not the case the Automata accept an input symbol, generates an output symbol and if designed may or may not change states. Perhaps the diagram below will make it clear.

So for each state N in Mealy machine, we have Z number of states is Moore machine where Z=N*(number of output symbols). or Z=M*N

Z<N *(number of output symbol) or Z<N*M only when we define redundant states in Mealy Machine that is when further reduction is possible. Thus, a general answer $Z\leqslant M*N$

If a Mealy machine has "n" states and "m" output then corresponding Moore

machine may have "m*n" states.

D is correct
by


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## Thinking Mathematically (6th Edition)

Let's suppose we want to convert 24 inches from units of inches to units of feet. In this case, we need to introduce the units of feet. $24~in = \frac{24~in}{1}\times \frac{1~ft}{12~in} = 2~ft$ In this example, we needed to put the units of feet in the numerator and not in the denominator. In some cases, we need to put the units to be introduced in the numerator and in other cases, we need to put the units to be introduced in the denominator. Therefore, although the statement would be correct in some specific cases, the statement does not make sense in general.


text:
11 questions linked to/from Is HMAC needed for a SHA-3 based MAC?
4k views

### Attacks of the MAC construction $\mathcal{H}(m\mathbin\|k)$ for common hashes $\mathcal{H}$?

Consider a common practically-collision-resistant hash function $\mathcal{H}$ (e.g. SHA-1, SHA-256, SHA-512, RIPEMD-160), perhaps based on the Merkle–Damgård construction as are the first three. We ...
3k views

### HMAC-SHA-256 vs SHA-512/256 vs truncated SHA-512 for use as a MAC

I'm looking at three options for shared-secret message authentication codes: HMAC-SHA-256(secret, payload) SHA-512/256(secret+payload) SHA-512(secret+payload) truncated to 256 bits Option 1 is the ...
217 views

### What is the collision resistance of a 2048-bit hash based on SHA-3?

Let $M$ denote the message. Let $B_i$ denote eight different bitstrings (chosen arbitrarily or randomly) such that $$0 \le \text{length}(B_i) \le 12345678.$$ Consider the following function (it ...
431 views

### Can Skein be used as a secure MAC in format H(k || m)?

Can Skein be used as a secure MAC in format H(k || m)? I understand it can work with HMAC, but it is fine as a one-pass MAC in the format above and resists length extension attacks. Is this ...
3k views

It would be possible to implement the HMAC construction with (draft) SHA-3, leading to HMAC-SHA3-224, HMAC-SHA3-256, HMAC-SHA3-384, HMAC-SHA3-512 (the last 3 digits are the output size $\ell$, where $\... 1answer 69 views ### Security implications of using hash as a MAC What are the security risks, if I use a hash function$H$such that$Y=H(K||M||N)$where$K$is a secret key,$M$is a message,$N$is a nonce and$||$is concatenation operation (where the length of$...
633 views

According to the pseudocode, keys longer than the block-size of the underlying hash function are shortened by hashing the key. Furthermore, keys shorter than the blocksize are zero-padded up the ...
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### Do we still need HMAC now that SHA3 is here? [duplicate]

In Another look at HMAC from Koblitz and Menezes, they state that HMAC came along because doing h(secret || message) was vulnerable to length extension attacks. But as far as I know, these kind of ...
1k views

### Is the HMAC construction really neccessary for a fixed length message?

The HMAC Construction is defined as: Hash( (Key XOR opad) || Hash((Key XOR ipad) || Message) ) This was designed because a simpler construction such as: ...


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Browse Questions

# Explain:(a)The basis of similarities and difference between metallic and ionic crystals.(b) Unit cell is not simply a cube of $4Na^+$ ions and $4Cl^−$ ions.(c) Can a cube consisting of $Na^+$ and $Cl^−$ ions at alternate corners serve as satisfactory unit cell for the sodium chloride lattice?(d) Ionic solids are hard and brittle.

(a) Similarities:
(i) Both involve electrostatic forces of attraction
(ii) Both are non-directional.
Differences: Ionic bond is a strong bond due to electrostatic forces of attraction while metallic bond may be weak or strong depending upon the kernels.
(b) Unit cell of NaCl has fcc arrangement of $Cl^−$ ions and $Na^+$ ions are present at the edge centres and one at the body-centre.
Thus there are $14Cl^−$ ions and $13Na^+$ ions in the unit cell. However their net contribution towards the unit cell is $4Na^+$ and $4Cl^−$ ions.
(c)Yes because its repetition in different directions produces the complete space lattice.
(d) Ionic solids are hard because there are strong electrostatic forces of attraction. However they are brittle because the bond is non-directional.


text: Let $X$ be a compact Riemann surface of genus $g \geq 2$ and let $G$ be a semisimple simply connected algebraic group. We introduce the notion of a {\em parahoric} $G$--bundle or equivalently a torsor under a suitable Bruhat-Tits group scheme. We also construct the moduli space of semistable parahoric $G$--bundles and identify the underlying topological space of this moduli space with certain spaces of homomorphisms of Fuchsian groups into maximal compact subgroup of $G$. These results generalize the earlier results of Mehta and Seshadri on parabolic vector bundles. (joint work with C.S Seshadri).


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# How do find the quotient of (-4sqrt3 )/( 3sqrt2)?

Nov 8, 2015

• Rewrite the square roots to exponents
• A little bit of algebra
.. and you will get the answer $- \frac{\sqrt{8}}{\sqrt{3}}$

#### Explanation:

Sometimes square roots can be painful to work with, so we have some rules that we can use to rewrite them. One of them is:
$\sqrt{a} = {a}^{\frac{1}{2}}$
And this is the one we will use for this problem.

We also need to know how to deal with eventual negative exponents. Let's have an example:
${a}^{- n} = \frac{1}{{a}^{n}}$
In this problem, we will use this rule the other way around as well. So let's start!

To make it easier for us, we can split up the expression into parts; one for the base of 2, and one of the base of 3:

$\frac{- 4 \sqrt{3}}{3 \sqrt{2}} = \frac{- 4}{\sqrt{2}} \cdot \frac{\sqrt{3}}{3}$

Let's tweek on the expression so we can see the exponents

$- {2}^{2} / {2}^{\frac{1}{2}} \cdot {3}^{\frac{1}{2}} / 3$

Now, let's move the factors to the same place, using the formula ${a}^{- n} = \frac{1}{{a}^{n}}$:

$\frac{- {2}^{2} \cdot {2}^{- \frac{1}{2}}}{1} \cdot \frac{1}{{3}^{1} \cdot {3}^{- \frac{1}{2}}}$
= -2^(2/2-1/2)/1 * 1/(3^(2/2 - 1/2)
$= - {2}^{\frac{3}{2}} / 1 \cdot \frac{1}{3} ^ \left(\frac{1}{2}\right)$
$= - \frac{\sqrt{{2}^{3}}}{\sqrt{3}}$
$= - \frac{\sqrt{8}}{\sqrt{3}}$


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Browse Questions

Suppose you have three variable resistors each of range 15 ohms to 30 ohms. Which of the following values of equivalent resistance cannot be obtained using these 3 variable resistors?

$\begin {array} {1 1} (A)\;10\Omega & \quad (B)\;45\Omega \\ (C)\;61\Omega & \quad (D)\;100\Omega \end {array}$

The range of equivalent resistances possible to generate using these is
$\large\frac{15}{3}$$ohms$ to $30 \times 3\: ohms.$ Since variable resistors have a continuous range,
equivalent resistors have a continuous range. Hence, one can obtain all values
between $5\: ohms\: to\: 90\: ohms.$
Ans : (D)
edited Mar 14, 2014


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# The connection between compressed sensing and sparse representation

If I understand correctly, Compressed Sensing as an application of Sparse Representation is defined as:

To find linear compression schemes for huge input signals that are known to have a sparse representation, so that the input signal can be recovered efficiently from the compression (the "sketch").(ref: this question)

So based on my understanding, compressed sensing is like a precoding before analog to digital data conversion.
And my take on sparse representation is:
A signal can be represented as a linear combination of basis functions where the set of basis functions is called dictionary and data samples are much more than their features. Mathematically, in the system of linear equations $Y=DX$ where $Y \in \mathbb{R}^{n \times N} (n \ll N)$ we seek a dictionary that results in sparse representation of $Y$.

So as far as I've understood, the output of sparse representation is used in compressed sensing. In my thesis on dictionary learning, after an introduction about compressed sensing, I'd like to talk a bit about the connection between compressed sensing and sparse representation before getting into the details of sparse representation. What I fail to do is to explain the connection between the two even though I am aware of the concepts. I'd like to have a good understanding of the connection between compressed sensing and sparse representation. Any explanation would be greatly appreciated!


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# Spahn categorical logic and type theory

• Bart Jacobs, categorical logic and type theory

## 11.5 Full higher order dependent type theory (Fho DTT)

(p.684)

Definition (dependency): (…) $s\succ t$ reads “$s$ depends on $p$”.

In Fho DTT we have: $Type\succ Kind$, $Kind\succ Type$, $Type\succ Type$, and $Kind\succ Kind$. It is a main difference to polymorphic dependent type theory (PDTT) that there we do not have $Kind\succ Type$.

For sorts $s_1, s_2$ such that $s_1\succ s_2$ the $(s_1,s_2)$-product resp. -sum resp. equality rules are defined by

$\frac{\Gamma \vdash C:s_1\;\;\;\Gamma,x:C\vdash D:s_2}{\Gamma\vdash \Pi x :C.D:s_2}$
$\frac{\Gamma \vdash C:s_1\;\;\;\Gamma,x:C\vdash D:s_2}{\Gamma\vdash \Sigma x :C.D:s_2}$
$\frac{\Gamma\vdash C:s_1}{\Gamma, x,x^\prime:C\vdash Eq_C(x,x^\prime):s_2}$

The weak elimination rule for $s_1,s_2)$-sums is

$\frac{\Gamma\vdash B:s_2\;\;\;\Gamma,x:C,y:D\vdash Q:B}{\Gamma,z:\Sigma x:C.D\vdash Q:B}$

(…)

## 11.6 Fho DTT categorically

(p.692)

First we import a definition from §9:

Definition (p.539): Let $p:E\to B$ be a fibration. A functor $P:E\to B$ is called a comprehension category on $p$ if we have:

(1) $cod P=p$. (2) For each cartesian map $f$ in$E$, the induced square $P(f)$ in $B$ is a pullback.

Definition (p.694): A weak Fho DTT-structure is defined to be a diagram

$\array{ D&\stackrel{\stackrel{R}{\leftarrow}}{\underset{I}{\hookrightarrow}}&E&\stackrel{P}{\to}&Arr(B)\\ &\searrow^q&\downarrow&\swarrow^{cod}\\ &&B }$

where

• $P:E\to Arr(B)$ is a closed comprehension category of kinds.
• $q=cod P I$ is a fibration (of types), and $D\stackrel{\stackrel{R}{\leftarrow}}{\underset{I}{\hookrightarrow}}E$ is a fibered reflection.
• There is an object $\Omega\in E$ over $\{\omega\}=dom P(\Omega)\in B$.

In this case the composite $Q:=PI:D\to Arr(B)$ is a comprehension category which has a unit $R 1:B\to D$ where $1:B\to E$ is the unit of $P$.

The following example of an $Fho DTT$ is a degenerate one since the subobject fibration for types is a poset.

Example (p.695): Let $E$ be a topos. Then $B$ is a regular category, such that there is a fibered reflection $Sub(B)\stackrel{\leftarrow}{\hookrightarrow} Arr(B)$ as in Theorem 4.4.4. where the reflector sends a morphism to its monic part (using the epi-mono factorization in a topos). Since monos are closed under composition, $Sub(B)\to Arr(B)$ is a closed comprehension category. So we obtain a Fho DTT-structure

$\array{ Sub(B)&\stackrel{\stackrel{R}{\leftarrow}}{\underset{I}{\hookrightarrow}}&Arr(B)&\stackrel{id}{\to}&Arr(B)\\ &\searrow^q&\downarrow&\swarrow^{cod}\\ &&B }$

Last revised on February 19, 2013 at 03:25:07. See the history of this page for a list of all contributions to it.


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# Algorithm for optimizing decision trees

### Background

A binary decision tree $T$ is a rooted tree where each internal node (and root) is labeled by an index $j \in \{1,..., n\}$ such that no path from root to leaf repeats an index, the leafs are labeled by outputs in $\{A,B\}$, and each edge is labeled by $0$ for the left child and $1$ for the right child. To apply a tree to an input $x$:

1. Start at the root
2. if you are at leaf, you output the leaf label $A$ or $B$ and terminate
3. Read the label $j$ of your current node, if $x_j = 0$ then move to the left child and if $x_j = 1$ then move to the right child.

The tree is used as a way to evaluate a functions, in particular we say a tree $T$ represents a total function $f$ if for each $x \in \{0,1\}^n$ we have $T(x) = f(x)$. The query complexity of a tree is its depth, and the query complexity of a function is the depth of the smallest tree that represents it.

### Problem

Given a binary decision tree T output a binary decision tree T' of minimal depth such that T and T' represent the same function.

### Question

What is the best known algorithm for this? Are any lower bounds known? What if we know that the $\text{depth}(T') = O(\log \text{depth}(T))$? What about if we only require $T'$ to be of approximately minimal depth?

### Naive approach

The naive approach is given $d = \text{depth}(T)$ to recursively enumerate all binary decision trees of depth $d - 1$ while testing if they evaluate to the same thing as $T$. This seems to require $O(\frac{d 2^n n!}{(n - d)!})$ steps (assuming that it takes $d$ steps to check what $T(x)$ evaluates to for an arbitrary $x$). Is there a better approach?

### Motivation

This question is motivated by a previous question on the trade off between query complexity and time complexity. In particular, the goal is to bound the time separation for total functions. We can make a tree $T$ from a time optimal algorithm with runtime $t$, and then we would like to convert it to a tree $T'$ for a query optimal algorithm. Unfortunately, if $t \in O(n!/(n - d)!)$ (and often $d \in \Theta(n)$) the bottleneck is the conversion. It would be nice if we could replace $n!/(n - d)!$ by something like $2^d$.

• Finding the optimal decision tree is NP-complete. I was taught that in Decision theory and Data mining classes, however those were based on notes and I am not aware of the original paper that introduced the result. – chazisop Jul 31 '11 at 13:09
• @chazisop cool, thanks. It is not obvious to me that finding the optimal decision tree is in NP, but I will think about it/search for it some more. Sometimes knowing the theorem statement is halfway to proving it :D. – Artem Kaznatcheev Jul 31 '11 at 16:42
• I think the earliest reference for this is: Lower Bounds on Learning Decision Lists and Trees. (Hancock et al. 1994) cs.uwaterloo.ca/~mli/dl.ps – Lev Reyzin Aug 25 '11 at 1:37
• The proof that finding the optimal decision tree is a NP-complete problem was given by Laurent Hyafil and Ronald L. Rivest in Constructing optimal binary decision trees is NP-complete (1976). reference: here – antoine Nov 5 '13 at 20:47

I have 3 answers, all giving somewhat different hardness results.

Let $f: \{0,1\}^n \rightarrow \{0,1\}$ be some function.

Given a decision tree $T$ computing $f$ and a number, it is NP-hard to tell if there exists a decision tree $T'$ computing $f$ of size at most that number. (Zantema and Bodlaender '00)

Given a decision tree $T$ computing $f$, it is NP hard to approximate the smallest decision tree computing $f$ to any constant factor. (Sieling '08)

Let $s$ be the size of the smallest decision tree computing $f$. Given a decision tree $T$ computing $f$, assuming $NP \subsetneq DTIME(2^{n^\epsilon})$ for some $\epsilon < 1$, one cannot find an equivalent decision tree $T'$ of size $s^k$ for any $k \ge 0$.

I think that this stronger answer (relying on a weaker assumption) can be made from known results in the learning theory of Occam algorithms for decision trees, via the following argument:

1. Is it possible to find a decision tree on $n$ variables in time $n^{\log s}$, where $s$ is the smallest decision tree consistent with examples coming from a distribution (PAC model). (Blum '92)
2. Assuming $NP \subsetneq DTIME(2^{n^\epsilon})$ for some $\epsilon < 1$, we cannot PAC learn size $s$ decision trees by size $s^k$ decision trees for any $k \ge 0$. (Alekhnovich et al. '07)

These two results seem to imply a hardness result for your problem. On the one hand (1), we can find a large decision tree; on the other hand (2), we shouldn't be able to minimize it to get an equivalent "small" one, of size $s^k$, even when one exists of size $s$.

• (I found your answer from this answer, which was posted less than an hour ago.) $\:$ It looks like "$\epsilon < 1$" can be replaced with "positive $\epsilon$, since decreasing $\epsilon$ makes the containment's right-hand-side smaller. $\:$ Also, where in that paper is 2. shown? $\;\;\;\;$ – user6973 Jul 7 '15 at 0:51
• See bullet point #2 in the abstract here: researcher.watson.ibm.com/researcher/files/us-vitaly/… – Lev Reyzin Jul 7 '15 at 11:46
• (coming from the same answer as Ricky Demer) could you detail a bit more how do you get "answer 3" from points 1. and 2.? I am not very familiar with learning theory and have a hard time connecting the parts... – Marc Jul 8 '15 at 17:03
• This consistency problem and learnability are closely related via Occam's razor. The idea is that if you can find a consistent function from a small set, you can succeed in PAC learning. Therefore a hardness of learning result implies a "hardness of consistency" result. I'm not sure how much more I can explain in a comment... – Lev Reyzin Jul 8 '15 at 18:24
• As far as I understand it, the algorithm evoked for 1. does not run in time $Poly(n,s)$ which would be necessary to get a contradiction with 2. (the precise result in the article if I got it correctly says that there is no polytime learning algorithm for decision trees). So there might be a problem with your argumentation. – Marc Jul 10 '15 at 9:09


text:
# And let GxH = {(g,h):ge Gand he H} Suppose (G;) and (Hp) are groups GxH is a under the

###### Question:

and let GxH = {(g,h):ge Gand he H} Suppose (G;) and (Hp) are groups GxH is a under the operation, &,given by Prove group (g1,ha) @ (gz,hz) = (91*g2,hzshz) Prove that if G and Hare both commutative then GxH is commutative:

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### What’s a Good GRE Score?

Any answer to the “what’s a good GRE score?” question should come with a disclaimer. Do you want a good GRE score for a PhD program in medieval history, or for an EMBA at your state university? Do you have weak college grades but ten years of work experience, or are you a current student with a 4.0 GPA? The real question to ask is “what’s a good GRE score for me?”—and the simple answer is, a good score is whatever score will impress the programs you’re applying to!

But how good does your score need to be in order to do that? Most of my students are taking the GRE in order to get into MBA programs. Assuming that you’re one of those students, you’re probably reading this article because:

• you want to know what GRE score to target as part of your MBA application.
• you’re wondering whether to take the GMAT or GRE.
• you’re curious about how good a certain GRE score is, relative to the average.

#### What Makes Up a Good GRE Score?

Your GRE score isn’t just one score, it’s three. Schools can see your Quant score, your Verbal score, and your Analytical Writing score. You could have a good Quant score and a bad Verbal score, or vice versa—even on the same test!

By the way, you can’t submit only your Quant score, or only your Verbal score, to a school. You have to send all of the scores from a certain GRE together. You can submit results from multiple GREs, but schools generally won’t pick and choose your highest sections—they’ll look at each test as a whole.

Your GRE Quant score will be a number from 130 to 170. About 4% of test takers earn a perfect Quant score. At the opposite end, very low Quant scores are rare: only about 8% of test takers score in the bottom quarter of the range (140 or below).

A perfect GRE Verbal score is much rarer than a perfect GRE Quant score. The Verbal scale also goes from 130 to 170, but only 1% of test takers score a 169 or 170, and only 3% earn a 166 or higher. Almost all of us will score between 135 and 165 on Verbal.

Finally, your writing score will be a number from 0.0 to 6.0. Very high and very low GRE writing scores are uncommon. A 4.0, 4.5 or 5.0 is a good GRE writing score, and a 5.5 or 6.0 is exceptionally good. Even a slightly lower score might not seriously affect your applications, as long as you can demonstrate your writing skills in another way.

#### How Common is a Good GRE Score?

The ETS provides data about how common or rare each GRE score is. This can give you a broad sense of what counts as a good score. Here’s a summary of the data for GRE Quant, based on scores from 2014 to 2017.

So, 166 is certainly a good GRE Quant score: only one test taker out of every ten has a score that high. For comparison, in 2017, one out of every ten people in the United States earned more than $108,000. Here’s the data for GRE Verbal: Only one out of every hundred test takers scores a 169+ on Verbal, which is as rare as an individual in the US earning$300,000 a year.

The ETS doesn’t publish data on combined, or overall, GRE scores. However, schools that publish their average GRE scores will typically publish the Quant and Verbal scores separately, so you can compare them to your own scores.

#### Do You Need a Good GRE Score for Top MBA Programs?

A 166 on GRE Quant is a one-in-ten score, while a 160 is one-in-four. Those are both good GRE scores, since they compare well against the scores that most people earn. But in order to get into a great MBA program, do you need to be one out of ten, or is one out of four good enough?

As the GRE becomes more popular for MBA applicants, a number of top programs have started publishing their average scores online. This article from Poets & Quants analyzes the data. Here’s a sample of the average GRE scores earned by successful applicants to top MBA programs.

On average, successful applicants to top-10 MBA programs have both GRE scores in the high 150s to mid 160s. For most top-50 schools, a good GRE score is in the mid-to-high 150s on both sections. Based on this data, an overall GRE score of 310 to 315 is a good GRE score for MBA applicants, while a score of 325 is excellent.

#### What’s Better: A Good GRE Score, or a Good GMAT Score?

There are a couple of different ways to compare a GRE score to a GMAT score. For instance, you could look at the percentile that corresponds to your GRE score and find the GMAT score that’s at the same percentile. A 153 on GRE Quant is a 50th-percentile score; a 50th-percentile score on GMAT Quant, correspondingly, is a 44.

However, that would give you the wrong impression of your results. Percentiles compare you to everyone who takes a certain test. That’s fine when you’re taking the GMAT, since most people who take the GMAT are MBA applicants. But MBA applicants are only a minority of GRE test takers. Comparing your score to everyone else’s is like comparing apples to oranges.

One alternative is to use this tool, created by the ETS, to compare GRE and GMAT scores. It isn’t perfect, but it does suggest something interesting: schools seem to be accepting applicants with “lower” GRE scores, compared to their GMAT counterparts. Here’s an article that analyzes the GMAT or GRE scoring question in greater depth.

Does that mean that weaker test takers can use the GRE to get admitted to top programs? Possibly, but you should be skeptical.

First, the statistics can’t tell you whether you would get a good score on the GRE or GMAT. Across a large population, some people will perform better on the GMAT, and others will perform better on the GRE. If you’re stronger at the GMAT, your “GMAT advantage” will more than cancel out any “GRE advantage” in applications.

It’s also possible, without more data, that the GRE applicants were better than the GMAT applicants in other ways—aside from their test scores. The number of people submitting GRE scores to most top schools is relatively small, and we don’t know exactly what those applicants look like outside of their scores.

Your best bet is actually to decide which test to take independently: the GMAT or the GRE? Take your target schools’ average scores for both tests into consideration, but don’t assume automatically that you’ll have better chances with the GRE.

Once you’ve picked your test, research the score data for the schools you’re applying to. Start with the average score for admitted students. Then, set a target right around that average—or a few points higher or lower, depending on the rest of your application. For instance, if you’ve gotten weak grades in Quant-related classes in the past, you might aim for a slightly-higher-than-average GRE Quant score to show that you’ve mastered the subject.

What if your situation is more complicated, and you really don’t know what a good GRE score is for you? Consider contacting an admissions consulting service, such as mbaMission—they’ll even do a free half-hour phone call with applicants. Take a look at our guide to starting your GRE studies to learn what’s involved in taking the GRE, and consider signing up for a free trial session of any Manhattan Prep GRE course. ?

See that “SUBSCRIBE” button in the top right corner? Click on it to receive all our GRE blog updates straight to your inbox!

Chelsey Cooley is a Manhattan Prep instructor based in Seattle, Washington. Chelsey always followed her heart when it came to her education. Luckily, her heart led her straight to the perfect background for GMAT and GRE teaching: she has undergraduate degrees in mathematics and history, a master’s degree in linguistics, a 790 on the GMAT, and a perfect 170Q/170V on the GRE. Check out Chelsey’s upcoming GRE prep offerings here.


text:
1. ## stationary point

hey having a problem with the following qus.

investigate the function f(x) = 2x - 6/(5-3x) for the nature of the stationary points (if any) and the value of the function at these points.

i found dy/dx = 2 + 18(5-3x)^-2 make it equal to zero

can't seem the solve for x?

hey having a problem with the following qus.

investigate the function f(x) = 2x - 6/(5-3x) for the nature of the stationary points (if any) and the value of the function at these points.
i found dy/dx = 2 + 18(5-3x)^-2 make it equal to zero
can't seem the solve for x?
Hello,

you've made a minor mistake because you forgot to use the chain rule:

$\displaystyle \frac{dy}{dx} = 2 -6 \cdot (-1) \cdot (5-3x)^{-2} \cdot (-3)=2-\frac{18}{(5-3x)^2}$

Set this term equal to zero.

Multiplying by the denominator and dividing by 2 will give:

$\displaystyle (5-3x)^2=9$

I'll leave this to you.

Greetings

EB

3. thanks agian earboth


text:
# Thread: Finding unknown variable in a Limit problem

1. ## Finding unknown variable in a Limit problem

Thanks.

2. Originally Posted by moremathlearner
Thanks.

Note that the as x → 1, the denominator → 0.

3. Yeah. That is the obvious problem.

Then how to find the possible values of a. Is it what I indicated i.e. a>1.

4. Please think about the limit existing at all. Factor the denominator.

5. Originally Posted by tkhunny
Factor the denominator.

If you click on an image thumbnail, you'll get an enlarged pop-up display of the image. If you click on the pop-up (once or twice), you'll get further enlargements. If you don't see any enlargements, then the original image was already displayed at full size (i.e., user uploaded small image). In this case, try your browser's zoom controls.

6. Thanks. Browser wasn't working at my other installation.

moremathlearner... You did NOT do what I suggested. You only factored the denominator. How about the other part?

With a factor of (x-1) in the denominator, this limit certainly fails to exist unless it has that same factor in the numerator. Demonstrate that, first.

Note: When your problem is quadratic, why in the name of reason would you make it quartic? Don't do that.

7. Originally Posted by moremathlearner
For interested readers, the text in the images appears to be as follows:

4. For what value(s) of a is the following limit positive?

. . . . .$\displaystyle \lim_{x \rightarrow 1}\, \dfrac{x^2\, -\, 2x\, +\, a}{2x^2\, -\, x\, -\, 1}$

$\displaystyle \lim_{x \rightarrow 1}\, \dfrac{x^2\, -\, 2x\, +\, a}{2x^2\, -\, x\, -\, 1}\, >\, 0$

$\mbox{Factorize den }\, \longrightarrow\, (2x\, +\, 1)(x\, -\, 1)$

$(x^2\, -\, 2x\, +\, a)\, (2x\, +\, 1)\, (x\, -\, 1)\, >\, 0$
________________________________

. . . . . . .$[(4x^2\, -\, 1)\, (x^2\, -\, 1)]\, \nearrow$

$(x^2\, -\, 2x\, +\, a)\, (2x\, +\, 1)\, (x\, -\, 1)\, >\, 0$
. . . . . . . . . . . . . . .$\searrow\, \swarrow$
. . . . . . . . . . . . .$\mbox{both}\, >\, 0$

$\therefore (x^2\, -\, 2x\, +\, a)\, >\, 0$

$\mbox{sub lim }\, x\, \rightarrow\, 1$

$1\, -\, 2\, +\, a\, >\, 0$

$-1\, +\, a\, >\, 0$

$\underline{ a\, >\, 1 }$

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text:
# NMSA407 Linear Regression: Tutorial

Matrix algebra background of linear regression

## Dataset

Everything in this tutorial will be exemplified on pseudodata that we now create.

y <- c(-9, -11, 1, 19)
x <- c(-3, -1, 1, 3)
x2 <- c(9, 1, 1, 9)
x2 <- x^2                     ### the same
X <- cbind(1, x, x2)
colnames(X) <- c("intcpt", "x", "x^2")
print(X)

##      intcpt  x x^2
## [1,]      1 -3   9
## [2,]      1 -1   1
## [3,]      1  1   1
## [4,]      1  3   9

Data <- data.frame(y = y, x = x, x2 = x2)
print(Data)

##     y  x x2
## 1  -9 -3  9
## 2 -11 -1  1
## 3   1  1  1
## 4  19  3  9

## Linear model

We will be fitting the following linear model: $\mathsf{E}Y = \beta_0 + \beta_1\,x + \beta_2\,x^2.$

## Raw solution to normal equations

First, we calculate solution to normal equations directly from theoretically derived expressions, i.e., as $$\widehat{\beta} = \bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1}\mathbf{X}^\top\mathbf{Y}$$.

### Matrix $$\mathbf{X}^\top\mathbf{X}$$

XtX <- t(X) %*% X
XtX <- crossprod(X)    ## the same
print(XtX)

##        intcpt  x x^2
## intcpt      4  0  20
## x           0 20   0
## x^2        20  0 164

### Matrix $$\bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1}$$

invXtX <- solve(XtX)
print(invXtX)

##           intcpt    x       x^2
## intcpt  0.640625 0.00 -0.078125
## x       0.000000 0.05  0.000000
## x^2    -0.078125 0.00  0.015625

### Solution to normal equations using already calculated inverse

b <- invXtX %*% t(X) %*% y
print(b)

##         [,1]
## intcpt -6.25
## x       4.80
## x^2     1.25

### Solution to normal equations calculated directly as a solution to a linear system

solve(XtX, t(X) %*% y)

##         [,1]
## intcpt -6.25
## x       4.80
## x^2     1.25

solve(crossprod(X), crossprod(X, y))   ## the same

##         [,1]
## intcpt -6.25
## x       4.80
## x^2     1.25

## Linear algebra and solution to normal equations

Calculation of the solution to normal equations using the above “raw” approach is numerically not too optimal. In practice, numerically more stable and much better methods are used which are based on decompositions of the model matrix $$\mathbf{X}$$ known from linear algebra. The advantage of these methods is also the fact that once the decomposition of the model matrix $$\mathbf{X}$$ is available, calculation of all the rest (fitted values, residuals, solution to normal equations, …) is trivial (and also computationally fast). This is in particular useful in situations when the least squares solutions are needed for a series of “response vectors” $$\mathbf{Y}$$ while keeping the model matrix $$\mathbf{X}$$ fixed.

The most often used decompositions are

• QR decomposition (used also by the lm function in R);
• singular value decomposition (SVD) which is also useful for calculation of the Moore-Penrose pseudoinverse matrix.

In the following, the above two decompositions will be explored in more detail. Nevertheless first, we explore other two decompositions of a symmetric positive semidefinite matrix $$\mathbf{X}^\top\mathbf{X}$$ that can also be used to solve the normal equations $$\mathbf{X}^\top\mathbf{X}\,\mathbf{b}= \mathbf{X}^\top\mathbf{Y}$$ and possibly for some other tasks.

These will be

• spectral decomposition;
• Cholesky (square root) decomposition.

Both the spectral decomposition and the Cholesky decomposition should be known from the bachelor courses Linear algebra and Fundamentals of Numerical Mathematics, Cholesky decomposition perhaps in a more general form of the LU decomposition.

## Spectral decomposition of a symmetric positive semidefinite matrix $$\mathbf{X}^\top\mathbf{X}$$

$\mathbf{X}^\top\mathbf{X} = \mathbf{V}\,\Lambda\,\mathbf{V}^\top = \sum_{j=1}^p \lambda_j\,\mathbf{v}_j\,\mathbf{v}_j^\top, \qquad \lambda_j \geq 0,\;\;j=1,\ldots,p.$

• Columns of matrix $$\mathbf{V}$$: orthonormal eigenvectors $$\mathbf{v}_1,\ldots,\mathbf{v}_p$$ of a matrix $$\mathbf{X}^\top\mathbf{X}$$.
• Matrix $$\Lambda$$: diagonal matrix with eigenvalues $$\lambda_1, \ldots, \lambda_p$$ on a diagonal.

It is additionally satisfied: $\mathbf{V}\,\mathbf{V}^\top = \sum_{j=1}^p \mathbf{v}_j\,\mathbf{v}_j^\top = \mathbf{I}_p,$ $\mathbf{V}^\top\,\mathbf{V} = \bigl(\mathbf{v}_j^\top\,\mathbf{v}_l\bigr) = \mathbf{I}_p.$ That is, $\mathbf{V}^{-1} = \mathbf{V}^\top.$ If $$\lambda_j > 0$$ for all $$j=1,\ldots,p$$ then $\bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1} = \sum_{j=1}^p \lambda_j^{-1}\,\mathbf{v}_j\,\mathbf{v}_j^\top = \mathbf{V}\,\Lambda^{-1}\,\mathbf{V}^\top.$

### Spectral decomposition of $$\mathbf{X}^\top\mathbf{X}$$

eigen(XtX)

## $values ## [1] 166.462113 20.000000 1.537887 ## ##$vectors
##            [,1] [,2]       [,3]
## [1,] -0.1221833    0  0.9925076
## [2,]  0.0000000   -1  0.0000000
## [3,] -0.9925076    0 -0.1221833

### Elements of an object returned by the eigen function

e <- eigen(XtX)
names(e)

## [1] "values"  "vectors"

#### Does it satisfy what it should satisfy?

V <- e$vectors Lambda <- diag(e$values)

V %*% Lambda %*% t(V)

##      [,1] [,2] [,3]
## [1,]    4    0   20
## [2,]    0   20    0
## [3,]   20    0  164

round(V %*% t(V), 13)             ## only numerically identity matrix

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

round(t(V) %*% V, 13)             ## only numerically identity matrix

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

round(XtX %*% V - V %*% Lambda, 13)    ## X'X * V = V * Lambda

##        [,1] [,2] [,3]
## intcpt    0    0    0
## x         0    0    0
## x^2       0    0    0

invLambda <- diag(1 / e$values) V %*% invLambda %*% t(V)  ## [,1] [,2] [,3] ## [1,] 0.640625 0.00 -0.078125 ## [2,] 0.000000 0.05 0.000000 ## [3,] -0.078125 0.00 0.015625  round(V %*% invLambda %*% t(V) - invXtX, 13) ## V * Lambda^(-1) * V' = (X'X)^(-1)  ## intcpt x x^2 ## intcpt 0 0 0 ## x 0 0 0 ## x^2 0 0 0  ## Cholesky (square root) decomposition of a symmetric positive semidefinite matrix $$\mathbf{X}^\top\mathbf{X}$$ $\mathbf{X}^\top\mathbf{X} = \mathbf{L}\,\mathbf{L}^\top,$ where $$\mathbf{L}$$ is a lower triangular matrix. This is a special case of so called LU decomposition where in the case of a symmetric positive semidefinite matrix $$\mathbf{U} = \mathbf{L}^\top$$ (or $$\mathbf{L} = \mathbf{U}^\top$$). In statistics, symmetric positive semidefinite matrices are quite often faced with as all covariance matrices are of this type. The Cholesky decomposition is in fact a generalization of a square root to matrices (as will immediately be shown). The Cholesky decomposition of a covariance matrix is then in fact a multivariate version of a standard deviation. Its calculation then simplifies many problems. If $$\mathbf{D}$$ is a symmetric positive semidefinite matrix (e.g., a covariance matrix or a matrix $$\mathbf{X}^\top\mathbf{X}$$ in a regression context) then it can be decomposed as $$\mathbf{D} = \mathbf{L}\,\mathbf{L}^\top$$, where $$\mathbf{L}$$ is lower triangular matrix. Among other things, the following tasks are practically trivial when $$\mathbf{L}$$ is already available. • Linear systems $$\mathbf{D}\,\mathbf{b} = \mathbf{c}$$ are easily solved by backward and forward substitution: • $$\mathbf{L}\,\mathbf{L}^\top\,\mathbf{b} = \mathbf{c}$$, • first solve $$\mathbf{L}\,\mathbf{v} = \mathbf{c}$$ (forward substitution), • then solve $$\mathbf{L}^\top\,\mathbf{b} = \mathbf{v}$$ (backward substitution). This is especially useful when the linear system $$\mathbf{D}\,\mathbf{b} = \mathbf{c}$$ is to be solved for (many) different right-hand-side vectors $$\mathbf{c}$$. • $$\mathsf{det}(\mathbf{D})$$ $$= \mathsf{det}\bigl(\mathbf{L}\,\mathbf{L}^\top\bigr)$$ $$= \mathsf{det}(\mathbf{L})\,\mathsf{det}(\mathbf{L})$$ $$= (\mbox{product of diagonal elements of }\mathbf{L})^2$$. • $$\mathbf{D}^{-1}$$ $$= \bigl(\mathbf{L}\,\mathbf{L}^\top\bigr)^{-1}$$ $$= \bigl(\mathbf{L}^{-1}\bigr)^\top\,\mathbf{L}^{-1}$$. That is, to calculate $$\mathbf{D}^{-1}$$, it is only needed to invert a triangular matrix (which is easy). • If a random vector $$\mathbf{Z}$$ has a distribution with a unit covariance matrix then $$\mathbf{L}\,\mathbf{Z}$$ has a distribution with a covariance matrix $$\mathbf{D}$$. This is useful for random numbers generation. ### Cholesky decomposition of a matrix $$\mathbf{X}^\top\mathbf{X}$$ R returns matrix $$\mathbf{U} = \mathbf{L}^\top$$. chol(XtX)  ## intcpt x x^2 ## intcpt 2 0.000000 10 ## x 0 4.472136 0 ## x^2 0 0.000000 8  #### Does it satisfy what it should satisfy? U <- chol(XtX) t(U) %*% U  ## intcpt x x^2 ## intcpt 4 0 20 ## x 0 20 0 ## x^2 20 0 164  ### Calculate $$\bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1}$$ from the Cholesky decomposition chol2inv(U)  ## [,1] [,2] [,3] ## [1,] 0.640625 0.00 -0.078125 ## [2,] 0.000000 0.05 0.000000 ## [3,] -0.078125 0.00 0.015625  invXtX2 <- chol2inv(U) round(invXtX2 - invXtX, 13) ## compare with previously calculated (X'X)^(-1), numerically the same  ## intcpt x x^2 ## intcpt 0 0 0 ## x 0 0 0 ## x^2 0 0 0  ### Determinant of $$\mathbf{X}^\top\mathbf{X}$$ prod(diag(U))^2  ## [1] 5120  det(XtX) ## R function to calculate determinant  ## [1] 5120  ### Solution to normal equations via backward and forward substitution using the Cholesky decomposition of a matrix $$\mathbf{X}^\top\mathbf{X}$$ We need to solve $$\mathbf{X}^\top\mathbf{X}\,\mathbf{b} = \mathbf{X}^\top\mathbf{Y}$$. Since $$\mathbf{X}^\top\mathbf{X} = \mathbf{U}^\top\mathbf{U}$$, we have to solve $$\mathbf{U}^\top\mathbf{U}\,\mathbf{b} = \mathbf{X}^\top\mathbf{Y}$$. (v <- forwardsolve(t(U), crossprod(X, y))) ## solution to U' * v = X'Y  ## [,1] ## [1,] 0.00000 ## [2,] 21.46625 ## [3,] 10.00000  (bchol <- backsolve(U, v)) ## solution to U * b = v  ## [,1] ## [1,] -6.25 ## [2,] 4.80 ## [3,] 1.25  print(b) ## compare to previously calculated solution  ## [,1] ## intcpt -6.25 ## x 4.80 ## x^2 1.25  ## Singular value decomposition (SVD) of the model matrix $$\mathbf{X}$$ Suppose that $$n = \mathrm{nrow}(\mathbf{X})$$, $$k = \mathrm{ncol}(\mathbf{X})$$, $$r = \mathrm{rank}(\mathbf{X}) \leq k \leq n$$. The singular value decomposition (SVD) of the model matrix $$\mathbf{X}$$ takes the following form: $\mathbf{X} = \mathbf{U}\,\mathbf{D}\,\mathbf{V}^\top,$ where • Columns of a matrix $$\mathbf{U}_{n\times k}$$: $$\mathbf{u}_1,\ldots,\mathbf{u}_k$$ $$=$$ the first $$k$$ orthonormal eigenvectors of a matrix $$\mathbf{X}\,\mathbf{X}^\top$$; • Columns of a matrix $$\mathbf{V}_{k\times k}$$: $$\mathbf{v}_1,\ldots,\mathbf{v}_k$$ $$=$$ (all) orthonormal eigenvectors of a matrix $$\mathbf{X}^\top\,\mathbf{X}$$; • Matrix $$\mathbf{D}_{k\times k}$$: diagonal matrix with $$d_1, \ldots, d_k$$ on a diagonal • $$d_j^2$$, $$j=1,\ldots,k$$: the first $$k$$ eigenvalues of a matrix $$\mathbf{X}\,\mathbf{X}^\top$$, which are the same as (all) eigenvalues of a matrix $$\mathbf{X}^\top\,\mathbf{X}$$. Numbers $$d_1,\ldots,d_k$$ are called singular values of the matrix $$\mathbf{X}$$. It holds: • $$r = k$$: $$d_1 \geq d_2 \geq \cdots \geq d_k > 0$$, • $$r < k$$: $$d_1 \geq d_2 \geq \cdots \geq d_r > d_{r+1} = \cdots = d_k = 0$$. Further, • $$\mathbf{U}^\top\,\mathbf{U} = \bigl(\mathbf{u}_i^\top\mathbf{u}_j\bigr) = \mathbf{I}_{k\times k}$$; • $$\mathbf{V}^\top\,\mathbf{V} = \bigl(\mathbf{v}_i^\top\mathbf{v}_j\bigr) = \mathbf{I}_{k\times k}$$; • $$\mathbf{V}\,\mathbf{V}^\top = \sum_{j=1}^k\mathbf{v}_j\,\mathbf{v}_j^\top = \mathbf{I}_{k\times k}$$. • $$\mathbf{X} = \mathbf{U}\,\mathbf{D}\,\mathbf{V}^\top = \sum_{j=1}^k d_j\,\mathbf{u}_j\,\mathbf{v}_j^\top$$; • $$\mathbf{X}^\top\,\mathbf{X} = \mathbf{U}\,\mathbf{D}\,\mathbf{D}\,\mathbf{V}^\top = \sum_{j=1}^k d_j^2\,\mathbf{u}_j\,\mathbf{v}_j^\top$$. If for all $$j=1,\ldots,k$$, $$d_j > 0$$, matrix $$\mathbf{X}^\top\,\mathbf{X}$$ is invertible and $\bigl(\mathbf{X}^\top\,\mathbf{X}\bigr)^{-1} = \mathbf{U}\,\mathbf{D}^{-2}\,\mathbf{V}^\top = \sum_{j=1}^k \frac{1}{d_j^2}\,\mathbf{u}_j\,\mathbf{v}_j^\top,$ where $$\mathbf{D}^{-2}$$ is a diagonal matrix with $$d_1^{-2},\ldots,d_k^{-2}$$ on a diagonal. Since matrix $$\mathbf{U}$$ contains an orthonormal basis of $$M(\mathbf{X})$$ in its columns, the projection matrix into $$M(\mathbf{X})$$ is $\mathbf{H} = \mathbf{U}\,\mathbf{U}^\top.$ The SVD decomposition can also be used to calculate the Moore-Penrose pseudoinverse matrix to $$\mathbf{X}$$: $\mathbf{X}^- = \tilde{\mathbf{V}}\,{\tilde{\mathbf{D}}}^{-1}\,{\tilde{\mathbf{U}}}^\top,$ where matrices $$\tilde{\mathbf{V}}_{k\times r},\,\tilde{\mathbf{D}}_{r\times r},\,\tilde{\mathbf{U}}_{n\times r}$$ contains/corresponds only to positive singular values. ### Singular value decomposition of a matrix $$\mathbf{X}$$ svd(X)  ##$d
## [1] 12.902020  4.472136  1.240116
##
## $u ## [,1] [,2] [,3] ## [1,] 0.70180882 0.6708204 -0.08639661 ## [2,] 0.08639661 0.2236068 0.70180882 ## [3,] 0.08639661 -0.2236068 0.70180882 ## [4,] 0.70180882 -0.6708204 -0.08639661 ## ##$v
##           [,1] [,2]       [,3]
## [1,] 0.1221833    0  0.9925076
## [2,] 0.0000000   -1  0.0000000
## [3,] 0.9925076    0 -0.1221833

### Content of an object returned by the svd function

svdX <- svd(X)
names(svdX)

## [1] "d" "u" "v"

(U <- svdX$u)  ## [,1] [,2] [,3] ## [1,] 0.70180882 0.6708204 -0.08639661 ## [2,] 0.08639661 0.2236068 0.70180882 ## [3,] 0.08639661 -0.2236068 0.70180882 ## [4,] 0.70180882 -0.6708204 -0.08639661  (D <- diag(svdX$d))

##          [,1]     [,2]     [,3]
## [1,] 12.90202 0.000000 0.000000
## [2,]  0.00000 4.472136 0.000000
## [3,]  0.00000 0.000000 1.240116

(V <- svdX$v)  ## [,1] [,2] [,3] ## [1,] 0.1221833 0 0.9925076 ## [2,] 0.0000000 -1 0.0000000 ## [3,] 0.9925076 0 -0.1221833  ### Matrices $$\mathbf{D}^2,\,\mathbf{D}^{-1},\,\mathbf{D}^{-2}$$ (D2 <- diag(svdX$d^2))       ## = D * D

##          [,1] [,2]     [,3]
## [1,] 166.4621    0 0.000000
## [2,]   0.0000   20 0.000000
## [3,]   0.0000    0 1.537887

(iD  <- diag(1 / svdX$d)) ## = D^(-1)  ## [,1] [,2] [,3] ## [1,] 0.07750724 0.0000000 0.0000000 ## [2,] 0.00000000 0.2236068 0.0000000 ## [3,] 0.00000000 0.0000000 0.8063762  (iD2 <- diag(1 / svdX$d^2))   ## = (D * D)^(-1)

##             [,1] [,2]      [,3]
## [1,] 0.006007373 0.00 0.0000000
## [2,] 0.000000000 0.05 0.0000000
## [3,] 0.000000000 0.00 0.6502426

#### Do the above matrices satisfy what they should satisfy?

U %*% D %*% t(V)              ### --> X

##      [,1] [,2] [,3]
## [1,]    1   -3    9
## [2,]    1   -1    1
## [3,]    1    1    1
## [4,]    1    3    9

V %*% D %*% t(D) %*% t(V)     ### --> X'X

##      [,1] [,2] [,3]
## [1,]    4    0   20
## [2,]    0   20    0
## [3,]   20    0  164

round(t(U) %*% U, 13)         ### --> I (numerically)

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

round(t(V) %*% V, 13)         ### --> I (numerically)

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

round(V %*% V, 13)            ### --> I (numerically)

##      [,1] [,2] [,3]
## [1,]    1    0    0
## [2,]    0    1    0
## [3,]    0    0    1

Xmin <- V %*% diag(1/svdX$d) %*% t(U) print(Xmin)  ## [,1] [,2] [,3] [,4] ## [1,] -0.0625 0.5625 0.5625 -0.0625 ## [2,] -0.1500 -0.0500 0.0500 0.1500 ## [3,] 0.0625 -0.0625 -0.0625 0.0625  ### Properties of the Moore-Penrose pseudoinverse round(X %*% Xmin %*% X - X, 13) ## X * X^- * X = X  ## intcpt x x^2 ## [1,] 0 0 0 ## [2,] 0 0 0 ## [3,] 0 0 0 ## [4,] 0 0 0  round(Xmin %*% X %*% Xmin - Xmin, 13) ## X^- * X * X^- = X^-  ## [,1] [,2] [,3] [,4] ## [1,] 0 0 0 0 ## [2,] 0 0 0 0 ## [3,] 0 0 0 0  round(t(X %*% Xmin) - X %*% Xmin, 13) ## (X * X^-)' = X * X^-  ## [,1] [,2] [,3] [,4] ## [1,] 0 0 0 0 ## [2,] 0 0 0 0 ## [3,] 0 0 0 0 ## [4,] 0 0 0 0  round(t(Xmin %*% X) - Xmin %*% X, 13) ## (X^- * X)' = X^- * X  ## [,1] [,2] [,3] ## intcpt 0 0 0 ## x 0 0 0 ## x^2 0 0 0  ### $$\mathbf{H}$$ matrix to project into $$M(\mathbf{X})$$ (regression space) (H <- U %*% t(U))  ## [,1] [,2] [,3] [,4] ## [1,] 0.95 0.15 -0.15 0.05 ## [2,] 0.15 0.55 0.45 -0.15 ## [3,] -0.15 0.45 0.55 0.15 ## [4,] 0.05 -0.15 0.15 0.95  ### $$\mathbf{M}$$ matrix to project into the orthogonal complement of $$M(\mathbf{X})$$ (residual space) (M <- diag(nrow(X)) - H)  ## [,1] [,2] [,3] [,4] ## [1,] 0.05 -0.15 0.15 -0.05 ## [2,] -0.15 0.45 -0.45 0.15 ## [3,] 0.15 -0.45 0.45 -0.15 ## [4,] -0.05 0.15 -0.15 0.05  ### Fitted values and residuals (yhat <- H %*% y)  ## [,1] ## [1,] -9.4 ## [2,] -9.8 ## [3,] -0.2 ## [4,] 19.4  (u <- M %*% y)  ## [,1] ## [1,] 0.4 ## [2,] -1.2 ## [3,] 1.2 ## [4,] -0.4  ### Fitted values and residuals using the solution to normal equations (to compare) (yhat_from_b <- X %*% b)  ## [,1] ## [1,] -9.4 ## [2,] -9.8 ## [3,] -0.2 ## [4,] 19.4  (u_from_b <- y - yhat_from_b)  ## [,1] ## [1,] 0.4 ## [2,] -1.2 ## [3,] 1.2 ## [4,] -0.4  ### Matrix $$\mathbf{X}^\top\mathbf{X}$$ V %*% D2 %*% t(V) ## X'X  ## [,1] [,2] [,3] ## [1,] 4 0 20 ## [2,] 0 20 0 ## [3,] 20 0 164  crossprod(D %*% t(V)) ## X'X calculated even in a better way  ## [,1] [,2] [,3] ## [1,] 4 0 20 ## [2,] 0 20 0 ## [3,] 20 0 164  print(XtX) ## previously calculated X'X  ## intcpt x x^2 ## intcpt 4 0 20 ## x 0 20 0 ## x^2 20 0 164  ### Matrix $$\bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1}$$ V %*% iD2 %*% t(V) ## (X'X)^(-1)  ## [,1] [,2] [,3] ## [1,] 0.640625 0.00 -0.078125 ## [2,] 0.000000 0.05 0.000000 ## [3,] -0.078125 0.00 0.015625  crossprod(iD %*% t(V)) ## (X'X)^(-1) calculated even in a better way  ## [,1] [,2] [,3] ## [1,] 0.640625 0.00 -0.078125 ## [2,] 0.000000 0.05 0.000000 ## [3,] -0.078125 0.00 0.015625  print(invXtX) ## previously calculated X'X  ## intcpt x x^2 ## intcpt 0.640625 0.00 -0.078125 ## x 0.000000 0.05 0.000000 ## x^2 -0.078125 0.00 0.015625  ## QR decomposition of the model matrix $$\mathbf{X}$$ Still suppose that $$n = \mathrm{nrow}(\mathbf{X})$$, $$k = \mathrm{ncol}(\mathbf{X})$$, $$r = \mathrm{rank}(\mathbf{X}) \leq k \leq n$$. The QR decomposition of the model matrix $$\mathbf{X}$$ takes the following form: $\mathbf{X} = \mathbf{Q}\,\mathbf{R},$ where • Columns of a matrix $$\mathbf{Q}_{n\times k}$$: $$\mathbf{q}_1,\ldots,\mathbf{q}_k$$; • the first $$r$$ of these columns: orthonormal basis of a linear span of columns of $$\mathbf{X}$$ (regression space $$M(\mathbf{X})$$); • Matrix $$\mathbf{R}_{k\times k}$$: upper triangular matrix. Further, $\mathbf{X}^\top\mathbf{X} = \mathbf{R}^\top\,\mathbf{Q}^\top\mathbf{Q}\,\mathbf{R} = \mathbf{R}^\top\mathbf{R},$ since the columns of the matrix $$\mathbf{Q}$$ form an orthonormal basis and hence $$\mathbf{Q}^\top\mathbf{Q} = \mathbf{I}_k$$. System of normal equations is then easily solvable by one backward substition since: $\mathbf{X}^\top\mathbf{X}\,\mathbf{b} = \mathbf{X}^\top\mathbf{Y}$ $\mathbf{R}^\top\mathbf{R}\,\mathbf{b} = \mathbf{R}^\top\mathbf{Q}^\top\,\mathbf{Y}$ $\mathbf{R}\,\mathbf{b} = \mathbf{Q}^\top\,\mathbf{Y}.$ That is, it is only requested to solve one linear system with an upper triangular matrix. If $$r = k$$ then $$\mathbf{X}^\top\mathbf{X}$$ is invertible and $\bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1} = \bigl(\mathbf{R}^\top\mathbf{R}\bigr)^{-1} = \mathbf{R}^{-1}\bigl(\mathbf{R}^{-1}\bigr)^\top.$ That is, it is only needed to invert a triangular matrix (which is easy). ### QR decomposition of a matrix $$\mathbf{X}$$ qr(X)  ##$qr
##      intcpt          x         x^2
## [1,]   -2.0  0.0000000 -10.0000000
## [2,]    0.5 -4.4721360   0.0000000
## [3,]    0.5  0.4472136   8.0000000
## [4,]    0.5  0.8944272  -0.9296181
##
## $rank ## [1] 3 ## ##$qraux
## [1] 1.500000 1.000000 1.368524
##
## $pivot ## [1] 1 2 3 ## ## attr(,"class") ## [1] "qr"  Note that results are stored and shown in a compact form. ### Content of an object returned by the qr function qrX <- qr(X) names(qrX)  ## [1] "qr" "rank" "qraux" "pivot"  ### Rank of the model matrix qrX$rank

## [1] 3

r <- qrX\$rank
n <- nrow(X)

### $$\mathbf{Q}$$ matrix

In our case, $$r = k = 3$$, all columns of $$\mathbf{Q}$$ form an orthonormal basis of $$M(\mathbf{X})$$.

qr.Q(qrX)

##      [,1]       [,2] [,3]
## [1,] -0.5  0.6708204  0.5
## [2,] -0.5  0.2236068 -0.5
## [3,] -0.5 -0.2236068 -0.5
## [4,] -0.5 -0.6708204  0.5

Q <- qr.Q(qrX)

### $$\mathbf{R}$$ matrix

qr.R(qrX)

##      intcpt         x x^2
## [1,]     -2  0.000000 -10
## [2,]      0 -4.472136   0
## [3,]      0  0.000000   8

R <- qr.R(qrX)

### Is it really $$\mathbf{X} = \mathbf{Q}\,\mathbf{R}$$?

Q %*% R - X                ## not really zeros

##      intcpt x          x^2
## [1,]      0 0 0.000000e+00
## [2,]      0 0 0.000000e+00
## [3,]      0 0 8.881784e-16
## [4,]      0 0 0.000000e+00

round(Q %*% R - X, 14)     ## numerically yes

##      intcpt x x^2
## [1,]      0 0   0
## [2,]      0 0   0
## [3,]      0 0   0
## [4,]      0 0   0

### $$\mathbf{Q}$$ matrix complemented by $$n - r$$ columns to form an orthonormal basis of $$R^n$$ (in our case $$R^4$$)

The orthonormal basis of $$R^n$$ will be stored in a matrix $$\mathbf{P}$$.

(P <- qr.Q(qrX, complete = TRUE))

##      [,1]       [,2] [,3]       [,4]
## [1,] -0.5  0.6708204  0.5  0.2236068
## [2,] -0.5  0.2236068 -0.5 -0.6708204
## [3,] -0.5 -0.2236068 -0.5  0.6708204
## [4,] -0.5 -0.6708204  0.5 -0.2236068

### Orthonormal basis of the residual space (orthogonal complement to $$M(\mathbf{X})$$)

The last $$n-r$$ columns of the matrix $$\mathbf{P}$$ form an orthonormal basis of the residual space. We store this basis in a matrix $$\mathbf{N}$$.

(N <- P[, (r + 1):n])

## [1]  0.2236068 -0.6708204  0.6708204 -0.2236068

### Projection matrix $$\mathbf{H}$$ into the regression space $$M(\mathbf{X})$$

(H <- Q %*% t(Q))

##       [,1]  [,2]  [,3]  [,4]
## [1,]  0.95  0.15 -0.15  0.05
## [2,]  0.15  0.55  0.45 -0.15
## [3,] -0.15  0.45  0.55  0.15
## [4,]  0.05 -0.15  0.15  0.95

### Projection matrix $$\mathbf{M}$$ into the residual space (orthogonal complement to $$M(\mathbf{X})$$)

(M <- N %*% t(N))

##       [,1]  [,2]  [,3]  [,4]
## [1,]  0.05 -0.15  0.15 -0.05
## [2,] -0.15  0.45 -0.45  0.15
## [3,]  0.15 -0.45  0.45 -0.15
## [4,] -0.05  0.15 -0.15  0.05

### Fitted values and residuals

(yhat <- H %*% y)

##      [,1]
## [1,] -9.4
## [2,] -9.8
## [3,] -0.2
## [4,] 19.4

(u <- M %*% y)

##      [,1]
## [1,]  0.4
## [2,] -1.2
## [3,]  1.2
## [4,] -0.4

### Fitted values and residuals calculated using the solution to normal equations

(yhat_from_b <- X %*% b)

##      [,1]
## [1,] -9.4
## [2,] -9.8
## [3,] -0.2
## [4,] 19.4

(u_from_b <- y - yhat_from_b)

##      [,1]
## [1,]  0.4
## [2,] -1.2
## [3,]  1.2
## [4,] -0.4

### Matrix $$\mathbf{X}^\top\mathbf{X}$$ easily calculated from the $$\mathbf{R}$$ matrix

crossprod(R)      ## = R'R = X'X

##        intcpt  x x^2
## intcpt      4  0  20
## x           0 20   0
## x^2        20  0 164

print(XtX)        ## previously calculated X'X

##        intcpt  x x^2
## intcpt      4  0  20
## x           0 20   0
## x^2        20  0 164

### Matrix $$\bigl(\mathbf{X}^\top\mathbf{X}\bigr)^{-1}$$ calculated from its Cholesky decomposition $$\mathbf{R}^\top\mathbf{R}$$

chol2inv(R)       ## inverse of X'X calculated from its Cholesky factor R (programme R uses the information that R matrix is upper triangular)

##           [,1] [,2]      [,3]
## [1,]  0.640625 0.00 -0.078125
## [2,]  0.000000 0.05  0.000000
## [3,] -0.078125 0.00  0.015625

print(invXtX)     ## compare to previously calculated (X'X)^(-1)

##           intcpt    x       x^2
## intcpt  0.640625 0.00 -0.078125
## x       0.000000 0.05  0.000000
## x^2    -0.078125 0.00  0.015625

### Solution to normal equations by solving $$\mathbf{R}\,\mathbf{b} = \mathbf{Q}^\top\mathbf{Y}$$

(b_from_QR <- backsolve(R, crossprod(Q, y)))

##       [,1]
## [1,] -6.25
## [2,]  4.80
## [3,]  1.25

print(b)          ## previously calculated b for comparison

##         [,1]
## intcpt -6.25
## x       4.80
## x^2     1.25

### Residual sum of squares

The residual sum of squares is also easily calculated, using the $$\mathbf{N}$$ matrix (orthonormal basis of the residual space) since $\mathrm{SS}_e = \bigl(\mathbf{M}\mathbf{Y}\bigr)^\top\,\mathbf{M}\mathbf{Y} = \mathbf{Y}^\top\,\mathbf{M}^\top\mathbf{M}\,\mathbf{Y} = \mathbf{Y}^\top\,\mathbf{N}\mathbf{N}^\top\mathbf{N}\mathbf{N}^\top\,\mathbf{Y} = \mathbf{Y}^\top\,\mathbf{N}\mathbf{N}^\top\,\mathbf{Y} = \bigl\|\mathbf{N}^\top\,\mathbf{Y}\bigr\|^2.$

(NtY <- crossprod(N, y))

##          [,1]
## [1,] 1.788854

(SSe <- as.numeric(crossprod(NtY)))

## [1] 3.2

## Conclusions

It should now be clear that when dealing with the least squares, it is not necessary to calculate the matrix $$\mathbf{X}^\top\mathbf{X}$$, or even its inverse directly from $$\mathbf{X}^\top\mathbf{X}$$. Everything can be obtained from a suitable decomposition of the model matrix $$\mathbf{X}$$, and once this decomposition is available, all the rest is easy (also with respect to computational demands).


text:
Definition 29.49.4. Let $X$ be a scheme. The ring of rational functions on $X$ is the ring $R(X)$ whose elements are rational functions with addition and multiplication as just described.

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text:
The Dry: A Novel
by Jane Harper
Hardcover- \$14.01

INSTANT NEW YORK TIMES BESTSELLER
“A breathless page-turner, driven by the many revelations Ms. Harper dreams up…You’ll love [her] ...

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A very graphic and disturbing start to the novel leads to a story of murder committed long ago and at present. Setting of drought in Australia and its effect on the characters is central to the story. Told in flashbacks and true to most mysteries there are a lot of potential perpetrators. Many twists and turns.

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Its title refers to a drought. The setting of THE DRY is a small farming town in Australia, and everyone is suffering because of the drought. God-awful things are going on in this small town, and often the drought is being blamed.

But Federal Agent Aaron Falk knows better. He and his father left there when he was 16 years old because it was such a nasty place, with some really bad people making them miserable.

So why is he back now, 20 years later? It appears that his best friend there has shot himself along with his wife and five-year-old son. But Falk and the town sheriff, Raco, are finding that all is not as it appears.

So that is that mystery. But there is another, the reason Falk and his father left the small town all those years ago.

THE DRY is unputdownable. The mysteries have twists and turns on every page. But you also want to watch every page for clues. Believe it, this is a book you will be sorry to see end.

"" by kate.spil[email protected] (see profile) 02/12/20

This book started off well, but it was quite slow in the middle. The ending was completely random, which I didn’t like!

"" by [email protected] (see profile) 02/20/20

Good start and ending, a bit boring in the middle. Unpredictable but I didn’t love the ending.

"" by nanawild (see profile) 06/18/20

"" by [email protected] (see profile) 11/11/20

"" by [email protected] (see profile) 08/24/21


text:
# Basic Finite Element Method (FEM) question: assembly and re-assembly

I'm reading up on the Finite Element Method (Zienkiewicz's Book), so I understand better what I'm doing in FEniCS and COMSOL. Currently, I'm wondering about this:

• Using FEM to solve fluid flow problems, do I have to re-assemble the entire system in each iteration of the solution process?

I would assume not since re-assembly only needs to be done if the coordinates of the nodes of the elements change. I can see how this is important in structural analysis, however, this is not the case in fluid dynamics (unless you work with moving meshes). So I assume in a common CFD computation it should be safe to only assemble the system of equations once and carry that through the entire solution process?

I'm asking because, if I understand correctly, not reassembling the system in each iteration of the solution should (in my current understanding) significantly reduce the computation time.

• "If you are doing Picard iteration rather than Netwon-Raphson, then you should only have to reform the right-hand-side vector." No. For secant method, or "modified Newton", you don't have to recompute every step. Picard is generally formulated for semi-linear equations as $A(u^n) u^{n+1} = f(u^n)$, and clearly needs reassembly. Furthermore, matrix-free finite differencing is not available for Picard, in contrast to Newton, for which you can ues JFNK to lag the preconditioner while still solving the Newton step. – Jed Brown Nov 8 '13 at 5:01


text:
In this work, higher-order moving average polynomials are defined by straightforward generalization of the standard moving average. The self-similarity of the polynomials is analyzed for fractional Brownian series and quantified in terms of the Hurst exponent H by using the detrending moving average method. We prove that the exponentH of the fractional Brownian series and of the detrending moving average variance asymptotically agree for the first-order polynomial. Such asymptotic values are compared with the results obtained by the simulations. The higher-order polynomials correspond to trend estimates at shorter time scales as the degree of the polynomial increases. Importantly, the increase of polynomial degree does not require to change the moving average window. Thus trends at different time scales can be obtained on data sets with the same size. These polynomials could be interesting for those applications relying on trend estimates over different time horizons (financial markets) or on filtering at different frequencies (image analysis).

Self-similarity of higher order moving averages / Arianos, Sergio; Carbone, ANNA FILOMENA; Turk, Christian. - In: PHYSICAL REVIEW E, STATISTICAL, NONLINEAR, AND SOFT MATTER PHYSICS. - ISSN 1539-3755. - STAMPA. - 84:(2011), pp. 046113-1-046113-9. [10.1103/PhysRevE.84.046113]

### Self-similarity of higher order moving averages

#### Abstract

In this work, higher-order moving average polynomials are defined by straightforward generalization of the standard moving average. The self-similarity of the polynomials is analyzed for fractional Brownian series and quantified in terms of the Hurst exponent H by using the detrending moving average method. We prove that the exponentH of the fractional Brownian series and of the detrending moving average variance asymptotically agree for the first-order polynomial. Such asymptotic values are compared with the results obtained by the simulations. The higher-order polynomials correspond to trend estimates at shorter time scales as the degree of the polynomial increases. Importantly, the increase of polynomial degree does not require to change the moving average window. Thus trends at different time scales can be obtained on data sets with the same size. These polynomials could be interesting for those applications relying on trend estimates over different time horizons (financial markets) or on filtering at different frequencies (image analysis).
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text:
## Operations

Now that we have the basic shapes, we need to combine them to create full scenes.

To do so, two basic shapes can be combined with an operation. Two operations can be again combined with an operation etc. This way, a tree builds up representing the final scene.

In this chapter, we just care about the operations and cover union, intersection, difference, negation and scale. They are shown in figure 1, except the scale.

Figure 1: CSG Operations: Union, intersection, difference, negation

The frayed edges of the difference shows that this implementation of dual marching cubes doesn’t preserve sharp edges.

Wang and Kaufman show the operations union, intersection, difference and negation on the density values in those formulas [WK94, S. 4 – 5]:

$f_{a \cup b}(x, y, z) = max(f_a(x, y, z), f_b(x, y, z))$
$f_{a \cap b}(x, y, z) = min(f_a(x, y, z), f_b(x, y, z))$
$f_{a - b}(x, y, z) = min(f_a(x, y, z), 1 - f_b(x, y, z))$
$f_{\bar{a}} = 1 - f(x, y, z)$

It doesn’t seem logical, that the difference is calculated by $1 - f_b(x, y, z)$ and the negation via $1 - f(x, y, z)$ because the resulting density value is changed by 1. That’s why those formulas are changed to this:

$f_{a - b}(x, y, z) = min(f_a(x, y, z), -f_b(x, y, z))$
$f_{\bar{a}} = -f(x, y, z)$

Beside the density value, the new gradient has to be calculated in a CSG operation. It is the one of the choosen density function by applying $min()$ or $max()$. For example if $f_a(x, y, z)$ has the greater density value in the union $f_{a \cup b}(x, y, z)$, its gradient is choosen. For the difference with $-f_b(x, y, z)$ being the smaller value, the gradient of $f_b(x, y, z)$ multiplied by -1 is taken.

The last operator is another unary one (taking only one input, beside the union for example taking two inputs), the scaling. It scales its input (whether its a basic shape or a whole CSG tree) by a numerical factor $sf$. First we have to bring the coordinate into the space of the original density function of the input via a division by $sf$. The resulting value is multiplied by $sf$:

$f_{s}(x, y, z) = f(\frac{x}{sf}, \frac{y}{sf}, \frac{z}{sf}) \cdot sf$

If you like this work and want to support it, feel free to donate something or click on any Flattr button!


text:
# Reference request: group theory

Currently I'm studying differential geometry and PDEs - so I often meet the use of groups. I also studied symmetries methods for solutions of differential equations but the connection between Lie groups and Lie algebras is still not implicit for me.

I am looking for a literature which gives a nice description of group theory (I am especially interested in continuous groups) and necessary covers Lie groups and Lie algebras. Thanks in advance.

-
I don't much about this subject: But a friend a mine recently attended a $\text{Lie Algebra}$ workshop and was told to refer: $\text{Introduction to Lie algebras and representation theory}$ by Humphreys. – user9413 Aug 2 '11 at 16:35
When most people say "group theory" they generally mean the study of abstract groups, not of topological etc. groups. It sounds like you really want a book on Lie theory (that maybe covers applications to differential geometry and PDE). – Qiaochu Yuan Aug 2 '11 at 17:11
@Qiaochu: maybe you're right. Which book can you recommend on Lie theory? – Ilya Aug 2 '11 at 18:30
I found Howe's short article Very basic Lie theory brilliant. – t.b. Aug 2 '11 at 20:06


text:
# Rails refactoring has_many

I have a RoR 4 app, and its model have these associations:

has_many :accreditations
has_one :legal, -> { where(kind_cd: Accreditation.legal) }, class_name: 'Accreditation'
has_many :departments, -> { where(kind_cd: Accreditation.department) }, class_name: 'Accreditation'

So, you see that these associations similar, but legal & departments have more conditions than accreditations. Can I replace class_name: 'Accreditation' with something like using :accreditations?

I don't see what you will gain with your hypothetical using syntax, let's say there is one... your code will look like this:

has_many :accreditations
has_one :legal, -> { where(kind_cd: Accreditation.legal) }, using: :accreditations
has_many :departments, -> { where(kind_cd: Accreditation.department) }, using: :accreditations

How is it better than your current code? It is not more expressive, nor is it more succinct, nor DRY.

Also, these associations are similar to some extent, but they look as DRY as you can make them - one is has_one, while the others are has_many, one uses the default idiom, while the others have names different than the associated class, conditions are different (and though one might argue you can predict the filter by the association name, one is singular, and the other is plural...)

In short, I think your current code is good enough - any change will only harm readability.

• Uri, but if I make two queries like @accreditations = @contractor.accreditations;@legal = @contractor.accreditations I'll get second query faster because it will use cache from first query, isn't it? – asiniy Mar 27 '14 at 5:31

There are two ways to tackle this situation. One of them is what you did: follow the Law of Demeter; I don't have any advice to improve your code. The second way is move the logic to the other model. In that case you'd write:

class Company
has_many :accreditations
end

class Accreditation
scope :departments, where(kind_cd: department)

def self.legal
where(kind_cd: legal).first
end
end

Both are ok even though I prefer the second because the logic of departments/legal is within the accreditation model.


text:
# A rural county is seeing more and more of its agricultural land being converted to residential...

###### Question:

A rural county is seeing more and more of its agricultural land being converted to residential property. It is estimated that the increase I in the county's tax base, in hundreds of dollars, due to the zoning change, after t years can be described by the function, . At the end of 15 years, what is the annual rate of change in the county's tax base? Please explain step by step.

The tax base is increasing by $334,066 per year. The tax base is increasing by$3305 per year.

The tax base is increasing by $72 per year. The tax base is decreasing by$3708 per year.

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