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# class vectors

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## Recommended Posts

I'm getting compiling errors on this code and I'm not sure why... What I am trying to do is create a vector of number of players (an array of class objects that make up the number of players) Then when the user inputs the number of players I can resize the vector to the correct size and then have an element in the vector for each player to hold his/her info...(name, cash on hand, etc...) in my main.cpp

#define WIN32_LEAN_AND_MEAN

#include <string>

#include <vector>

#include <windows.h>

#include "resource.h"

#include "main.h"

#include "General.h"

#include "Player.h"

std::string str;
using namespace std;
const char g_szClassName[] = "myWindowClass";
WNDPROC lpfnOldWndProc;

GENERAL General;
vector<PLAYER> vPlayers;
vPlayers.reserve(6);
//more code....

player.cpp

#include "Player.h"

PLAYER::PLAYER()
{

name.reserve(15);
cash = 500;
}
PLAYER::~PLAYER()

{
}

player.h

#ifndef PLAYER_H
#define PLAYER_H
#include <vector>
using namespace std;

class PLAYER
{
public:
PLAYER();
int cash;
vector<char> name;
~PLAYER();
private:
};
#endif

when I compile this code I get the following errors in the main.cpp on this line "vPlayers.reserve(6);" d:\GDev\Cards\Ride2\main.cpp(17) : error C2143: syntax error : missing ';' before '.' d:\GDev\Cards\Ride2\main.cpp(17) : error C2501: 'vPlayers' : missing storage-class or type specifiers d:\GDev\Cards\Ride2\main.cpp(17) : error C2371: 'vPlayers' : redefinition; different basic types Any reasons why? I'm sure it is a minor problem that I'm overlooking... TIA Triston All the world's a stage...and I seem to fall off quite a bit. [edited by - Triston on September 2, 2002 1:43:24 PM] [edited by - Triston on September 2, 2002 1:44:29 PM]

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You can't call functions like that (vPlayers.reserve(6)) outside of a function. You can either move the reserve call into a function (like in you actual main () function), or if the default PLAYER is all you want initially, you can call this form of vector::vector:

vector<PLAYERS> vPlayers (6); // initializes w/ 6 "blank" PLAYERS

Those 6 players will be constructed by the default PLAYER ctor.

[edited by - Stoffel on September 2, 2002 1:55:25 PM]

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Woo Hoo thanks...Works great..
I got all caught up in working with vectors I didn''t realize I was outside of a function...thanks

Triston

All the world''s a stage...and I seem to fall off quite a bit.

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# Archimedean and Catalan solids: picking favorites

All five Platonic solids are perfect in every way; I cannot justify putting any one of them above others. Tetrahedron has the virtue of being self-dual: taking the centers of its faces as new vertices, we get another tetrahedron. It’s also the simplest of them all, but perhaps too simple.

Archimedean solids are another matter. Their isometry groups act transitively on vertices but not on faces. Which means the faces can vary in size and shape, though they are all regular polygons. Out of 13 Archimedean solids two stand out on the ground of being 50-50 interpolants between dual Platonic solids (and, as a result, being edge-transitive in addition to vertex-transitive).

Cuboctahedron is a 50-50 hybrid of Cube and Octahedron. Take either of two solids and truncate all vertices just enough for the new faces to meet at new vertices. Alternatively, takes the intersection of cube and octahedron, appropriately scaled. In coordinates $(x_1,x_2,x_3)$ a cuboctahedron is determined by inequalities $\forall i \ |x_i|\le 1$, $\sum |x_i|\le 2$. I’m going to take cuboctahedron as my Archimedean solid #1. It combines 8 triangles, 6 squares, and 4 hexagons (central section) into a shape of remarkable simplicity.

Similarly, Icosidodecahedron is the 50-50 hybrid of Icosahedron and Dodecahedron. Theoretically, it should be just as nice as cuboctahedron. Yet, I don’t find it nearly as appealing…

It’s either the disparity of face sizes, or of their number of sides (3 vs 5). Either way, Icosidodecahedron loses #2 spot to Truncated Icosahedron aka soccer ball (old design) aka buckyball:

This is also a hybrid, but it’s 2/3 of icosahedron and 1/3 of dodecahedron. The vertices of Icosahedron are truncated just enough so that triangular faces become regular hexagon: that is, 1/3 of sidelength is cut. The resulting combination of nearly equal hexagons and pentagons is both aesthetically pleasing and mathematically natural. Think of the standard hexagonal grid on the plane being forced to take spherical shape: some hexagons will become pentagons, allowing the surface to curve. Hence the structure of $C_{60}$.

The other asymmetric hybrid, 2/3 of dodecahedron and 1/3 of icosahedron, is so ugly that I don’t even want to show it here.

Unlike Platonic solids, the duals of Archimedean solids do not belong to the same class. Indeed, they form a dual class of 13 polyhedra known as Catalan solids. Given the perfect duality of the two classes, the gap between the times of Archimedes (-287 – -212) and Catalan (1814 – 1894) may appear surprising. The probable reason for neglect of Catalan solids is that their faces are not regular polygons. Yet, they beat their Archimedean counterparts in “real-world application” category: since their isometry group is face-transitive, they are equally likely to land on any face and thus can be used as game dice.

This is the dual of cuboctahedron… unremarkable, in my opinion. It does get an honorable mention for being space-tiling.

Having picked truncated icosahedron over icosidodecahedron, I should also prefer its dual, Pentakis Dodecahedron, to Rhombic Triacontahedron. But it seems that duality does not quite work this way: in my opinion, the latter solid looks better.

It’s the shape of these triangles (almost regular but not quite) that makes the pentakis dodecahedron look like some kind of a fake, a wannabe Platonic solid. In contrast, the rhombic triacontahedron does not pretend to be anything it’s not:

If you are going to be a rhombus, let your ratio of diagonals be the golden ratio. This is exactly what we see in the rhombic triacontahedron. But what clinches its #1 place among Catalan solid is the following. There are five cubes that can be inscribed into dodecahedron using some 8-subsets of the 20 vertices: [video]. The intersection of these five cubes is a rhombic triacontahedron: each of the 30 faces of the latter is a part of a face of some cube ($30=5\cdot 6$).

format: plain


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# Org-Mode’s Two Most Useful Data Table Functions That You Probably Never Used

It might be hard to admit it for us Emacsers but most of us are pretty comfortable using Excel (or its clones) for data management. Maybe the fun stops there but it doesn’t have to.

Here are the two most important functions for getting the most out of working together with Org-Mode data tables and Excel (without having to use literate programming):

• Do your work in Excel
• Export it to a tab separated (TSV) file
• Import that into Org with org-table-import
• Do you work
• Work with it manually
• Do simple sorting with org-table-sort-lines
• Handle copy pasted tabular data from wherever with org-table-convert-region
• Export that back to a Excel TSV with org-table-export

It might seem pretty simple but this is a great foot in the door to start integrating Org-Mode into your data workflow.


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# Why are there multiple lines in the hydrogen line spectrum?

Hydrogen has only one electron, yet it exhibits multiple lines in a spectral series, why is this?

• Have a look at this link and if that doesn't answer your question, consider editing your question to make it a bit more specific. The short answer is that the one electron can transition to/from multiple energy levels. – bobthechemist Jul 3 '13 at 16:01
• Yeah, another man how didn't use google before asking a question here. Have your link en.wikipedia.org/wiki/File:Hydrogen_transitions.svg – permeakra Jul 5 '13 at 9:44
• – Jan Oct 22 '15 at 15:44
• This is a 6 year late reply: in response to @permeakra, a google search brought me here :) – ExtremeRaider May 15 '19 at 9:37

As the electrons fall from higher levels to lower levels, they release photons. Different "falls" create different colors of light. A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength).

The visible wavelengths are caused a by single electron making the different transitions shown below. There are even more transitions that release invisible wavelengths.

 Wavelength   Transition  Color
(nm)
---------------------------------------------
383.5384      9 -> 2    Ultra Violet
388.9049      8 -> 2    Ultra Violet
397.0072      7 -> 2    Ultra Violet
410.174       6 -> 2    Violet
434.047       5 -> 2    Violet
486.133       4 -> 2    Bluegreen (cyan)
656.272       3 -> 2    Red
656.2852      3 -> 2    Red

*Values taken from Hyperphysics: Hydrogen Energies and Spectrum

(Why are there two different 3-> 2 transitions? See here: Hydrogen Fine Structure )

Though a hydrogen atom has only one electron, it contains a large number of shells, so when this single electron jumps from one shell to another, a photon is emitted, and the energy difference of the shells causes different wavelengths to be released... hence, mono-electronic hydrogen has many spectral lines.

Although hydrogen has only one electron, it contains many energy levels. When its electron jumps from higher energy level to a lower one, it releases a photon. Those photons cause different colours of light of different wavelengths due to the different levels. Those photons appear as lines. For this reason, though hydrogen has only one electron, more than one emission line is observed in its spectrum.

This arises due to the electron making transitions between the various energy levels in an atom . The electrons fall from a higher energy state to a lower energy state or vice versa emitting or absorbing wavelengths of different magnitude which fall directly on the spectrum of hydrogen thereby releasing colours of various wavelengths and creating a continuous spectrum .

• Welcome to chemistry.se! If you have questions about how to beautify your posts, have a look at the help center. Do you want to know more about this site, please take the tour. || Referring to your answer: A line spectrum is of course not a continuous spectrum, as the title already says. – Martin - マーチン Jan 16 '15 at 7:17


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# What does it mean for a set to be bounded?

1. Oct 26, 2009

### royzizzle

what does it mean for a set to be bounded??

in the context of the hein-borel theorem

i mean the mathematically rigorous definition

2. Oct 27, 2009

### Office_Shredder

Staff Emeritus
Re: what does it mean for a set to be bounded??

S a subset of Rn is bounded if there exists M>0 so that for all x in S, |x|<M

3. Oct 27, 2009

### Moo Of Doom

Re: what does it mean for a set to be bounded??

Another equivalent definition is that it has finite diameter, where

$$\mathrm{diam}(S) = \sup_{x, y \in S}(\mathrm{dist}(x, y)).$$

This is applicable to any metric space (though the Heine-Borel theorem is not!).


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# Floyd–Steinberg dithering

(Redirected from Floyd Steinberg)
original picture
no dithering
Floyd–Steinberg dithering
A 1-bit image of the Statue of David, dithered with Floyd–Steinberg algorithm

Floyd–Steinberg dithering is an image dithering algorithm first published in 1976 by Robert W. Floyd and Louis Steinberg. It is commonly used by image manipulation software, for example when an image is converted into GIF format that is restricted to a maximum of 256 colors.

## Implementation

The algorithm achieves dithering using error diffusion, meaning it pushes (adds) the residual quantization error of a pixel onto its neighboring pixels, to be dealt with later. It spreads the debt out according to the distribution (shown as a map of the neighboring pixels):

${\displaystyle {\begin{bmatrix}&&*&{\frac {\displaystyle 7}{\displaystyle 16}}&\ldots \\\ldots &{\frac {\displaystyle 3}{\displaystyle 16}}&{\frac {\displaystyle 5}{\displaystyle 16}}&{\frac {\displaystyle 1}{\displaystyle 16}}&\ldots \\\end{bmatrix}}}$

The pixel indicated with a star (*) indicates the pixel currently being scanned, and the blank pixels are the previously-scanned pixels. The algorithm scans the image from left to right, top to bottom, quantizing pixel values one by one. Each time the quantization error is transferred to the neighboring pixels, while not affecting the pixels that already have been quantized. Hence, if a number of pixels have been rounded downwards, it becomes more likely that the next pixel is rounded upwards, such that on average, the quantization error is close to zero.

The diffusion coefficients have the property that if the original pixel values are exactly halfway in between the nearest available colors, the dithered result is a checkerboard pattern. For example, 50% grey data could be dithered as a black-and-white checkerboard pattern. For optimal dithering, the counting of quantization errors should be in sufficient accuracy to prevent rounding errors from affecting the result.

In some implementations, the horizontal direction of scan alternates between lines; this is called "serpentine scanning" or boustrophedon transform dithering.

In the following pseudocode we can see the algorithm described above. This works for any approximately linear encoding of pixel values, such as 8-bit integers, 16-bit integers or real numbers in the range [0,1].

for each y from top to bottom do
for each x from left to right do
oldpixel := pixels[x][y]
newpixel := find_closest_palette_color(oldpixel)
pixels[x][y] := newpixel
quant_error := oldpixel - newpixel
pixels[x + 1][y    ] := pixels[x + 1][y    ] + quant_error × 7 / 16
pixels[x - 1][y + 1] := pixels[x - 1][y + 1] + quant_error × 3 / 16
pixels[x    ][y + 1] := pixels[x    ][y + 1] + quant_error × 5 / 16
pixels[x + 1][y + 1] := pixels[x + 1][y + 1] + quant_error × 1 / 16

When converting greyscale pixel values from a high to a low bit depth (e.g. 8-bit greyscale to 1-bit black-and-white), find_closest_palette_color() may perform just a simple rounding, for example:

find_closest_palette_color(oldpixel) = round(oldpixel / 255)

The pseudocode can result in pixel values exceeding the valid values (such as greater than 255 in 8-bit greyscale images). Such values should ideally be clipped by the find_closest_palette_color() function, rather than clipping the intermediate values, since a subsequent error may bring the value back into range. However, if fixed-width integers are used, wrapping of intermediate values would cause inversion of black and white, and so should be avoided.

The find_closest_palette_color is nontrivial for a palette that is not evenly distributed. In such a case, a nearest neighbor search in 3D is required.

## References

• Floyd–Steinberg Dithering (Graphics course project, Visgraf lab, Brazil)
• R.W. Floyd, L. Steinberg, An adaptive algorithm for spatial grey scale. Proceedings of the Society of Information Display 17, 75–77 (1976).


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# Category Archives: Mathematik

## Every finite division ring is a field

It is well-known for undergraduate students who major in mathematics that every finite integral domain is a field. Theorem 1. Every finite integral domain is a field. Proof. Let $$0,~1,~a_1 ,~a_2 ,~ \cdots ,~ a_n$$ be all the elements of a finite integral domain $$D .$$ We need to show that for $$a \in D$$, where… Read More »

## Which is better, e or π?

Top ln(e10) reasons why e is better than pi : e is easier to spell than pi. pi ≒ 3.14 while e ≒ 2.718281828459045. The character for e can be found on a keyboard, but pi sure can't. Everybody fights for their piece of the pie. ln(pi1) is a really nasty number, but ln(e1) = 1. e is… Read More »

## Euler Phi-Function is a Multiplicative Function

Let $$n$$ be a positive integer. Recall that the Euler phi-function $$\phi(n)$$ is defined as the number of positive integers less than or equal to $$n$$ and relatively prime to $$n.$$ Note that $$\phi(1)=1.$$ We have seen that Euler used this function to generalize the Fermat's Little Theorem. It is sometimes needed to calculate the value $$\phi(n)$$ of… Read More »

## Euler’s Generalization of Fermat’s Little Theorem

Fermat's Little Theorem says: Theorem 1. (Fermat) If $$p$$ is a prime number and $$(a,~p)=1,$$ that is, if $$a$$ and $$p$$ are relatively primes, then $$a^{p-1}\equiv 1$$ $$({\rm mod}~ p).$$ Euler gave a generalization of Fermat's theorem. His generalization will follow at once from next theorem, which is proceed by counting, using essentially the same argument as in… Read More »

## Fermat’s Little Theorem and Finite Prime Fields

The fact that "$$\mathbb{Z}_p$$ is a field if and only if $$p$$ is a prime" can be derived from the fact that "every finite integral domain is a field." Also, it is easily derived that every field include a subfield which is isomorphic to one of $$\mathbb{Q}$$ or $$\mathbb{Z}_p$$ for a prime $$p$$. That's why the fields $$\mathbb{Z}_p$$… Read More »

## “Associativity, Left Identity, Left Inverse Elements” Makes a Group

In most undergraduate algebra textbooks, you can find the definition of a group as following: A nonempty set $$G$$ with a binary operation satisfying the following conditions is called a group: Associative rule: $$a(bc)=(ab)c$$ for all $$a,~b,~c \in G$$. Existence of an identity element: There exists $$e\in G$$ s.t. $$ea = ae = a$$ for all \(a \in… Read More »


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697 meters in miles

Conversion

697 meters is equivalent to 0.433095720989422 miles.[1]

Conversion formula How to convert 697 meters to miles?

We know (by definition) that: $1\mathrm{m}\approx 0.00062137119\mathrm{mile}$

We can set up a proportion to solve for the number of miles.

$1  m 697  m ≈ 0.00062137119  mile x  mile$

Now, we cross multiply to solve for our unknown $x$:

$x\mathrm{mile}\approx \frac{697\mathrm{m}}{1\mathrm{m}}*0.00062137119\mathrm{mile}\to x\mathrm{mile}\approx 0.43309571943\mathrm{mile}$

Conclusion: $697  m ≈ 0.43309571943  mile$

Conversion in the opposite direction

The inverse of the conversion factor is that 1 mile is equal to 2.30895839311334 times 697 meters.

It can also be expressed as: 697 meters is equal to $\frac{1}{\mathrm{2.30895839311334}}$ miles.

Approximation

An approximate numerical result would be: six hundred and ninety-seven meters is about zero point four three miles, or alternatively, a mile is about two point three one times six hundred and ninety-seven meters.

Footnotes

[1] The precision is 15 significant digits (fourteen digits to the right of the decimal point).

Results may contain small errors due to the use of floating point arithmetic.


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# An external pressure P is applied on a cube at $0^{\circ}C$ so that it is equally compressed from all sides. $K$ is the bulk modulus of the material of the cube and $\alpha$ is its coefficient of linear expansion. Suppose we want to bring the cube to its original size by heating. The temperature should be raised by :
( A ) $\frac{P}{ \alpha K}$
( B ) $3 PK \alpha$
( C ) $\frac{P}{3 \alpha K}$
( D ) $\frac{3 \alpha}{ P K}$


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Enter two or more vectors and click Calculate to find the dot product. Find a ⋅ b when a = <3, 5, 8> and b = <2, 7, 1>, a ⋅ b = (a1 * b1) + (a2 * b2) + (a3 * b3) We can calculate the dot product for any number of vectors, however all vectors must contain an equal number of terms. Sometimes it may seem like the cross product is being carried out on a vectors of dimension lower than 3, and even NumPy does not seem to have any problem processing it either. The Matrix, … To find the dot product from vector coordinates we can use its algebraic definition. Rather than manually computing the scalar product, you can simply input the required values (two or more vectors here) on this vector dot product calculator and it does the math for you to find out the dot (inner) product. You can express this with the following equation: If you aren’t sure about the magnitude of a vector or how to perform the calculation, you’d be better off using a dot product of two vectors calculator. Geometrically the dot product is defined as . This is the formula used by the calculator Using magnetic potential energy as a dot product of a magnetic field and magnetic moment. The formula for the dot product in terms of vector components would make it easier to calculate the dot product between two given vectors. A dot product calculator is a convenient tool for anyone who needs to solve multiplication problems involving vectors. For this, you need this formula: If you’re wondering how this dot product solver works, you need to follow some steps. Working as a dot product of displacement and force. Vector Calculator: add, subtract, find length, angle, dot and cross product of two vectors in 2D or 3D.

The calculator will find the angle (in radians and degrees) between the two vectors, and will show the work. Using magnetic or electric flux as a dot product of a magnetic or electric field along with the surface which it flows through. This formula gives a clear picture on the properties of the dot product. Exercises. Solve for the product of each vector’s third components. You see, in 2 dimensions, you only need one vector to yield a cross product (which is in this case referred to as the perpendicular operator.). Since these parts are parallel, the result you get is the product of the lengths of both parts. $\begingroup$ It is true, 2 vectors can only yield a unique cross product in 3 dimensions. Solve for the product of each vector’s middle components. Dot product of two vectors a and b is a scalar quantity equal to the product of magnitudes of vectors multiplied by the cosine of the angle between vectors: The dot product is also known as Scalar product. Related Symbolab blog posts. This Dot Product calculator calculates the dot product of two vectors based on the vector's position and length. Direction cosines of a vector, Online calculator. In such a case, you can write each of the vectors using 3 components:eval(ez_write_tag([[300,250],'calculators_io-box-4','ezslot_8',104,'0','0']));eval(ez_write_tag([[300,250],'calculators_io-box-4','ezslot_9',104,'0','1']));eval(ez_write_tag([[300,250],'calculators_io-box-4','ezslot_10',104,'0','2'])); Geometrically speaking, the dot product is the product of the magnitudes of vectors multiplied by the value of the cosine of the angle between the vectors. Vectors may contain integers and decimals, but not fractions, functions, or variables. Here are the steps to follow for this matrix dot product calculator:eval(ez_write_tag([[728,90],'calculators_io-medrectangle-3','ezslot_7',110,'0','0'])); Despite the convenience of the dot product calculator which is also known as a dot product of two vectors calculator or a matrix dot product calculator, you may want to perform the calculation by hand. Detailed expanation is provided for each operation. In other words, the dot product comes from the multiplication of the length of vectors projected in the direction of one of these vectors. When you draw a triangle using 3 vectors, you can write the formula as. In this case you don't really have a 4D vector, but a 3D vector with a texture shift value. To see what I mean, even if you input vectors $\mathbf{u}$ and $\mathbf{v}$ as follows After inputting all of these values, the dot product solver automatically generates the values for the Dot Product and the Angle Between Vectors for you. If we defined vector a as and vector b as we can find the dot product by multiplying the corresponding values in each vector and adding them together, or (a1 * b1) + (a2 * b2) + (a3 * b3) .... + (an * bn). As long as you have the required values, you can use it to make automatic calculations. Select the vectors dimension and the vectors form of representation; Press the button The Matrix… Symbolab Version. Entering data into the dot product calculator . Defining different kinds of physical quantities as dot products. By using this website, you agree to our Cookie Policy. thus, we can find the angle as. Using power as a dot product of velocity and force. Guide - Dot product calculator To find the dot product of two vectors: Select the vectors dimension and the vectors form of representation; Type the coordinates of the vectors; Press the button "=" and you will have a detailed step-by-step solution. Therefore, if you have a vector with 3 components, your dot product formula would be:eval(ez_write_tag([[970,90],'calculators_io-banner-1','ezslot_12',105,'0','0'])); In any space which have more than 3 dimensions, add more terms to your summation.


text:
Ask Your Question

Gram-Schmidt with arbitrary inner product?

asked 2017-12-12 03:48:50 -0500

This post is a wiki. Anyone with karma >750 is welcome to improve it.

Hi,

I'd like to find an orthonormal basis of polynomials using an arbitrary inner product. I know how to do this in Maxima:

ip(f, g) := integrate(f * g, x, -1, 1); /* for example */
y : gramschmidt([1, x, x^2], ip);

But is there a nice way to do this in sage?

Thanks! -Andrew

edit retag close merge delete

1 answer

Sort by » oldest newest most voted

In sage, there is a method gram_schnidt associated to a matrix.

After the matrix is declared, the method can be immediately called. The following defines a matrix A , computes the Gram-Schnidt matrices G.M as in the examples from

A.gram_schmidt?

and does a check for the relation $A=MG$, it is a possible starting point:

var( 'x' )
def ip(f, g):
return integrate( f * g, x, -1, 1 )

f_list = [1, x, x^2, x^3, x^4]
A = matrix( QQ, len(f_list),
[ ip( f, g ) for f in f_list for g in f_list ] )

print "A is\n%s\n" % A

G, M = A.gram_schmidt()

print "G is\n%s\n" % G
print "M is\n%s\n" % M

print "Is MG = A? %s" % bool( M*G == A )

Results:

A is
[  2   0 2/3   0 2/5]
[  0 2/3   0 2/5   0]
[2/3   0 2/5   0 2/7]
[  0 2/5   0 2/7   0]
[2/5   0 2/7   0 2/9]

G is
[         2          0        2/3          0        2/5]
[         0        2/3          0        2/5          0]
[ -424/5439          0  1376/9065          0   248/1813]
[         0    -12/595          0      4/119          0]
[ 64/112035          0 -128/22407          0    64/9603]

M is
[       1        0        0        0        0]
[       0        1        0        0        0]
[675/1813        0        1        0        0]
[       0   75/119        0        1        0]
[425/1813        0 950/1067        0        1]

Is MG = A? True

Later edit:

I try now hard to reproduce in sagemath the following result from maxima, obtained by copy+pasting into the maxima interpreter the code, thanks go to danielvolinski for this sample,

load(eigen);
ip(f,g) := integrate(f*g, u, 0, 1);
y : gramschmidt([1, u, u^2, u^3, u^4], ip);

which delivers

                       2                           2
2 u - 1  6 u  - 6 u + 1  (2 u - 1) (10 u  - 10 u + 1)
(%o14) [1, -------, --------------, ----------------------------,
2           6                      20
4        3       2
70 u  - 140 u  + 90 u  - 20 u + 1
---------------------------------]
70

The situation is different, because there is no a priori knowledge of an orthonormal basis of the vector space generated by $1,u,u^2,u^3,u^4$ with the given scalar product $\langle f,g\rangle =\int_0^1f\;g$ .

To get a similar result i prefer to program from the scratch, there is no implemented facility...:

var( 'u' )
def ip(f, g):
return integrate( f * g, u, 0, 1 )

n = 5
B = vector( [u^k for k in range(n)] )
T = identity_matrix( QQ, n )    # in T we will have finally the transfromation matrix

C = vector( [u^k for k in range(n)] )    # in C we will have finally the Gram-Schmidt basis
for k in range(n-1):
S = identity_matrix( QQ, n )
for j in range(k+1, n):
S[j, k] = -ip( C[j], C[k] ) / ip( C[k], C[k] )

T = S*T
C = S*C

print T
print C

This gives the transformation matrix, and the new basis

[    1     0     0     0     0]
[ -1/2     1     0     0     0]
[  1/6    -1     1     0     0]
[-1/20   3/5  -3/2     1     0]
[ 1/70  -2/7   9/7    -2     1]
(1, u - 1/2, u^2 - u + 1/6, u^3 - 3/2*u^2 + 3/5*u - 1/20, u^4 - 2*u^3 + 9/7*u^2 - 2/7*u + 1/70)

in concordance with the maxima result.

more

Comments

Where is the resulting orthonormal basis of polynomials the user requested?

( 2018-03-23 03:51:49 -0500 )edit

There is it in a simpler example:

sage: A = matrix( QQ, 2,2, [4,3,3,4] )
sage: G, M = A.gram_schmidt()
sage: A, G, M, M.inverse()
(
[4 3]  [     4      3]  [    1     0]  [     1      0]
[3 4], [-21/25  28/25], [24/25     1], [-24/25      1]
)
sage: -24/25*vector([4,3]) + vector([3,4])
(-21/25, 28/25)

Gram-Schmidt is applied on rows. The first row of A is copied in G. The lower triangular M gives the transformation in one direction. To obtain the second row in G, we use the linear combination from the M.inverse().

By the way, the user did not request any transformation, only a way to "do something similar", knowing that he knows how to do it in maxima. Doing it in maxima as he did, we also have to search for stuff, including transformation.

( 2018-03-24 10:10:37 -0500 )edit
1

Hi Dan,

I think Andrew was looking for something similar to the following result in Maxima

Evaluate the following code in SageMath:

%maxima load(eigen)
%maxima ip(f,g):=integrate(f*g,u,0,1)
%maxima y:gramschmidt([1,u,u^2,u^3,u^4],ip)

You will get five polynomials which are orthogonal (not orthonormal).

Is there something similar in SageMatt?

Thanks,

Daniel

( 2018-03-26 06:06:17 -0500 )edit

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Asked: 2017-12-12 03:48:50 -0500

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Last updated: Mar 27 '18


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22

Like this? \documentclass[tikz]{standalone} \usepackage{tikz} \begin{document} \foreach \angle in {0,10,...,360} { \begin{tikzpicture} % fill circle and plot \fill[blue!50] (-1,0) arc (0:\angle:1) -- (-2,0) -- cycle; \fill[blue!50] plot[smooth,domain=0:\angle] (pi/180*\x,{sin(\x)}) |- (0,0); % draw connection \draw (-2,0) +(\angle:1) ...

19

I couldn't resist, so here's a solution using pgfplots (and some tikz), plus arara for creating the .gif animation. Output Click for bigger size Code % arara: animate: {density: 160, delay: 8} \documentclass[tikz]{standalone} \usepackage{amsmath,amssymb} \usepackage{pgfplots} \pgfplotsset{compat=1.13} \usepgfplotslibrary{fillbetween} \begin{document} ...

14

My answer consists of three parts. Step 1: decorate segments First, we would like to replace every segment by a dumbbell: This could be done, as the title suggests, by defining a decoration and applying it. So we may write \pgfdeclaredecoration{dumbbells}{initial} { \state{initial}[width=0pt,next state=depict one]{ \egroup % set up ...

7

So the problem with the code is that \subnode is never defined. You can get it defined by loading the tikzmark library. However, \newcommand\tikzmark... will then fail as the library defines the standard \tikzmark command. This problem can be avoided by simply choosing a different macro name, such as \mytikzmark. You cannot, however, use \mytikzmark or ...

6

A simple foreach can do it: \documentclass[tikz,border=2mm]{standalone} \begin{document} \begin{tikzpicture} \draw[->] (0,-1)--++(90:4); \draw[->] (-1,0)--++(0:12); \coordinate (aux) at (0,0); \foreach \i in {1,-1,1,1,1,-1,1,-1,-1,1,-1} \draw[->] (aux)--++(1,\i) coordinate (aux); \end{tikzpicture} \end{document}

5

For commutative diagrams I always use tikz's \matrix command together with matrix of math nodes. This gives an easy way of placing the nodes of your diagram into an array and then drawing lines between the nodes. To use this you need to have \usetikzlibrary{matrix} in your preamble. The basic idea is something like this: \begin{tikzpicture}[auto] ...

5

An example with tikz-cd. Output Code \documentclass{article} \usepackage{tikz-cd} \begin{document} \[\begin{tikzcd}[row sep={6mm,between origins},column sep={1cm,between origins}] hom_{D}(a,a) \arrow[rrrr, "\tau_{a}"] \arrow[ddd, "{hom_{D}(a,f)}", swap] & & & & Ta \arrow[ddd, "Tf"] \\ & 1_{a} \arrow[rr, |->] \arrow[d, ...

5

I didn't so much make it thinner as make the triangle bigger. Note: without setting inner sep=0pt the node was actually wider than the strip, causing a slight gap at the bottom. \documentclass[tikz]{standalone} \usetikzlibrary{shapes.geometric} \tikzset{my polygon/.style={regular polygon,regular polygon sides=#1,minimum size=4cm}} ...

5

You're not passing a set of options but something that contains a set of options. The error message is a bit misleading, because the expansion takes place when issuing the error, so it appears you have passed the right options. No key-value based package performs expansion when absorbing key-value pairs, because it could be disastrous. So the solution is to ...

5

You mean something like this? The arrows.meta library is not strictly necessary unless you want to use one of the new arrow tips. In that case, replace -{Straight Barb[length=5pt,width=5pt]} with ->. Output Code \documentclass[margin=10pt]{standalone} \usepackage{tikz} \usetikzlibrary{arrows.meta} \begin{document} \begin{tikzpicture} % axes ...

5

In the second part to this answer a custom coordinate system was given. This can be used to plot the grids (albeit a bit slowly). The other requirements (not done here) involve re-orienting the x, y, and z vectors, and changing the content and positioning of the labels. \documentclass[tikz,border=5]{standalone} \usetikzlibrary{arrows} \tikzset{declare ...

5

This answer offers two solutions, one with pgfplots and one with tikz. pgfplots This would be much easier to do with pgfplots, you don't need to draw everything manually. Output Code \documentclass{amsart} \usepackage{amsmath} \usepackage{amsfonts} \usepackage{pgfplots} \pgfplotsset{compat=1.13} \begin{document} \begin{tikzpicture} \begin{axis}[ ...

5

5

The simplest way is write matrix as TikZ matrix and add desired column frame and note to it: \documentclass[11pt]{book} \usepackage{amsmath,amsfonts,amssymb,amsthm} \usepackage{tikz} \usetikzlibrary{arrows.meta,bending,matrix,positioning} \begin{document} \begin{tikzpicture}[ node distance=1mm and 0mm, baseline] ...

5

As you have named the coordinates, just loop over their names: \documentclass[margin=10pt]{standalone} \usepackage{tikz} \usetikzlibrary{arrows,calc} \begin{document} \begin{tikzpicture}[line cap=round,line join=round,>=triangle 45,x=1.0cm,y=1.0cm] \clip(-0.5,-1) rectangle (5.5,3.5); % defining coordinates \coordinate (1) at (0,0); \coordinate (2) at ...

4

Another interpretation of "half dashed": \documentclass[tikz,border=5]{standalone} \usetikzlibrary{decorations.pathreplacing,calc} \tikzset{draw half paths/.style 2 args={% decoration={show path construction, lineto code={ \draw [#1] (\tikzinputsegmentfirst) -- ($(\tikzinputsegmentfirst)!0.5!(\tikzinputsegmentlast)$); \draw [#2] ...

4

The lines do end at the nodes, you have to consider that nodes have some padding (inner sep, outer sep), and lines are drawn to the edge of the node, not the center. Add draw to the node options and you'll see this: \documentclass{standalone} \usepackage{tikz} \begin{document} \begin{tikzpicture} \draw (4,2) node(p1)[draw,label={[label ...

4

This answer will focus on the funnel object, at least for the moment. Changes: Fixed funnel shapes. Removed one \foreach statement and included it in the previous one. Better node positioning. Output Code \documentclass{article} \usepackage[margin=2cm]{geometry} \usepackage{pgfplots} \definecolor{myellow}{RGB}{228,212,0} ...

4

This is a simple typo \arrow{r}{\iota} Not \arrow[r]{\iota}

4

A straightforward solution is add two small nodes placed inside "domain1" and "domain2" node: For above picture I define subelement style for small nodes. Beside this, I move node distance to common parameter in tikzset, slightly change it (that result is more similar to shoved image in question), for all nodes with rectangle shape use style element. And ...

4

First \visible command is from beamer not tikz. you can do that with a third variable \anim inside \foreach loop to control apparence of edges. Code \documentclass{beamer} \usepackage{tikz} \begin{document} \begin{frame} \centering \begin{tikzpicture} [scale=.5,auto=left,minimum size=2mm,inner sep=0pt,outer sep=0pt,every ...

4

This kind of groups can be easily drawn with a matrix of nodes. Following code shows this option: \documentclass{standalone} \usepackage{tikz} \usetikzlibrary{arrows,matrix,positioning} \begin{document} \begin{tikzpicture}[ field/.style={draw, rounded corners, minimum height=8mm, anchor=center, align=center}, frame/.style={matrix of nodes, column ...

3

The main mistake is that you use axis y line=center which implies that the axis should go through 0, or in the case of a log axis through 1. When you change this to axis y line=left and at least the xmin value, you should get the desired result. But then the ylabel is positioned left in the middle (again). This can be modified using the ylabel style key. ...

3

Edit: This solution in comparison with your MWE differ in the following: in preamble is added \pgfplotsset{compat=1.13}, so in the notation of node and drawing coordinates can be omitted axis cs: added are TikZ libraries \usetikzlibrary{calc,positioning} for determination of nodes coordinates the ymax in both axes are changed from 500 to 200 and 8000 to ...

3

If you want to repeat some tikzpicture or part of it, pics are a good option. Following code shows how to use them. A pic is declared to draw your original graph, and this pic is used inside several tikzpictures distributed within a tabular where fisrt column is a multirow. As you can see, array pic only draws part of the graph which can be completed with ...

3

The problem is that the inner sep and line width do not change with the font. The solution is to define a length that does change proportionately with the font. I chose the height of \mathstrut, but almost anything will do. Also, by drawing the circle separately, you can adjust its size and the size of the contents ($r$) separately. ...

3

For the lines, you can use \pgfextra and draw lines using the node's anchors. For the label, you can use a label as an argument for the style. Output Code \documentclass[tikz]{standalone} \usetikzlibrary{calc,arrows,shapes.geometric,positioning} \tikzset{ element/.style={ draw, thick, node distance=1.5cm, minimum ...

3

Some notes: I ignored the warnings when possible because your document has a lot, especially underfull \hbox and overfull \hbox. The first one means that you have maybe inserted a line break where the line had no content, such as writing \\\\, the second one means that your content exceeds the page margins. I removed most of the content from your example ...

3

I am not very familiar with the datavisualization of tikz. So I recreated your picture with pgfplots. \documentclass[border=5mm]{standalone} \usepackage{pgfplots} \begin{document} \begin{tikzpicture} \begin{axis}[ height=6cm, width=13cm, axis lines=middle, grid=both, domain={-360:360}, ymin=-1.3, ...

3

Concerning your example, there are some issues: A \matrix already creates nodes, so I don't think that nesting nodes is advisable. You need to enclose each \item inside curly braces when inside a \matrix, like: \item{<content goes here>} For coloring a cell, use |[fill=<color>]| Output Code \documentclass{article} ...

Only top voted, non community-wiki answers of a minimum length are eligible


text:
## Courses

### Topics on Matroid Theory

#### Henry Crapo (Centre de Recherche Les Moutons Matheux, La Vacquerie-et-Saint-Martin-de-Castries, France)

The concept of matroid is a combinatorial generalization of those relations holding in any finite set of vectors in any vector space, or any finite set of points in a projective geometry. The theory was initially developed in the 1930's by Hassler Whitney, Saunders MacLane, and Garrett Birkhoff. Characterizations of matroids representable as structures related to graphs, those representable over the field of two elements, and similar problems, were solved by William Tutte in the 1960's.

For those who have not previously done any serious study of matroids, we will give a brief introduction to the basics, in a preliminary talk.

Our first talk deals with the interplay between algebra and geometry, as developed in the now classical works of Hermann Grassmann, Giuseppe Peano, Henry Baker, Luigi Cremona, and Emil Artin. Our primary goal is to explain the significance of projective conditions, which may or may not hold in a geometric figure, and which determine properties of the figure relative to rigidity, relative to the possibility of lifting to higher dimensions, and the like. Recent developments include the definition of the Whitney algebra of a matroid, in which such properties find an algebraic formulation.

The second talk looks more closely at two of these applications: the rigidity of graphs viewed as bar and joint structures in d-dimensional space, and the closely related topic of spatial interpretation of polyhedral figures drawn in the plane.

Research in matroid theory is a domain currently witnessing major developments, chiefly within what is known as the matroid minors project, headed up by Jim Geelen, Bert Gerards and Geoff Whittle. Their goal is to extend to matroids the recent theory of graph minors, due to Neil Robertson and Paul Seymour. The aim is to settle a number of conjectures put forward in the 1960's and 1970's by William Tutte and Gian Carlo Rota. Our final talk will provide a glimpse of these remarkable and contemporary theoretical advances.

Topics covered are:

A brief introduction to the basics of matroid theory.
The dialectics of algebra and geometry.
• Arithmetic on the circle.
• Theorem proving.
• The Whitney algebra of a matroid.
Matroids from mechanics and scene analysis.
• Isostatic structures.
• Lifting planar scenes.
• Pure conditions, special position.
A quick sketch of the new frontiers: The matroid minors project. Geelen, Gerards and Whittle.

### Triangulations of Polytopes

#### Francisco Santos (Universidad de Cantabria, Spain)

Triangulations of polytopes play an important role in several parts of mathematics, both pure and applied. This course is an introduction to them, focusing on their relation to oriented matroids and in the theory of regular triangulations and secondary polytopes developed in the last twenty years. We will follow parts of Chapters 2 and 4 of the recent book Triangulations: Structures for Algorithms and Applications (Algorithms and Computation in Mathematics, Vol 25), by J. A. de Loera, J. Rambau and F. Santos.

Topics covered are:

The oriented matroid of a point configuration.
Triangulations of point configurations. Regular triangulations.
Circuits and flips.
Gale transforms. Triangulations vs. Chambers
The secondary polytope of a point configuration.

There will be a problem session for this course.

## Contributed Talks

### Bipartite hypermaps

Abstract:

A hypermap is a cellular imbedding of a hypergraph on a surface. A hypermap is bipartite if we can divide its set of hypervertices in two parts in such a way that consecutive hypervertices around a hyperedge or a hyperface are contained in alternate parts.

In this talk we will see some operators from the category of hypermaps to the category of bipartite hypermaps, and their properties. We will also see the classification of bipartite hypermaps with large symmetry group on surfaces of small genus.

### Key polynomials and crystal graphs in type $$C$$

Abstract:

Key polynomials (in $$n$$ variables) were introduced by Demazure for all Weyl groups and were studied combinatorially in type $$A$$ by Lascoux and Schützenberger. In type $$A$$ they are a linear basis for the ring $$\Z[x_1, x_2, \ldots,x_n]$$ which contain the linear basis of Schur polynomials for the subring of symmetric polynomials under the action of the symmetric group, and, in type $$C$$, for the ring of Laurent polynomials in the variables $$(x_1,\ldots,x_n)$$ which contain the linear basis of symplectic Schur polynomials for the subring of symmetric Laurent polynomials under the action of the hyperoctahedral group. In type $$A$$, Lascoux and Schützenberger produced a crystal graph indexed by Young tableaux to describe a key polynomial and have characterized the tableaux indexing the vectors of the corresponding Demazure crystal.

A key polynomial of type $$C$$ may also be described by a subset of a crystal graph of type $$C$$ indexed by sympletic tableaux in the sense of De Concini or Kashiwara and Nakashima. This crystal structure is obtained by means of the crystal operations in type $$C$$ to give a combinatorial interpretation of a Demazure operator of type $$C$$. This is a joint work with Ricardo Mamede.

### A meeting point between matroid and representation theory

Abstract:

In this talk, we introduce a matroid invariant known as the rank partition which was defined by J. A. Dias da Silva in 1990. For matroids realizable over the complex numbers, following recent work of A. Berget, we explain how this concept relates matroid theory to the representation theory of the complex general linear group $${\GL}_m(\C)$$.

In addition, we present an elegant duality, usually referred to as Schur-Weyl duality, between the representations of $${\GL}_m(\C)$$ and those of an algebraic structure called the rook monoid.

The rook monoid is an extension of the symmetric group and thus its representation theory provides a natural context to generalize concepts such as symmetry classes of tensors and symmetrized tensors. We present such generalizations and how they may be applied to obtain new results in matroid theory involving the rank partition.

### A new encoding for Generation, Ranking and Unranking of $$(k,l)$$-trees and $$(k,l)$$-Catalan numbers

Abstract:

Let $$T$$ be a tree with root $$r$$. For each vertex $$v$$ of $$T$$, we say that $$v$$ is on level $$j$$ if the unique path from $$v$$ to $$r$$ is of length $$j$$, and the root is said to be on the level $$0$$. We can then define $$(k,m)$$-ary tree of order $$n$$ as follows: For $$k,m\ge 1$$ and $$n\ge 0$$, a $$(k,m)$$-ary tree of order $$n$$ is a tree that:

a) All vertices on even levels have degree $$k$$, and
b) All vertices on odd levels have degree $$m$$ or $$0$$, and there are exactly $$n$$ vertices of degree $$m$$ [1].
We use $$T_{k,m}(n)$$ to denote the set of $$(k,m)$$-ary trees of order $$n$$. A $$(k,m)$$-ary tree, which is also called a $$(k,m)$$-Catalan number is a generalization of $$k$$-ary tree [1].

The problem of generation is to propose a linear ordering among the studied trees or using an available one, in order to generate the trees with respect to this ordering. Two such natural orderings are the $$A$$-order and the $$B$$-order [2]. The problem of ranking consists of determining the order number of a given tree and conversely unranking means constructing the tree of a given rank [2].

Generation, ranking and unranking of different classes of trees were studied earlier in the literature, but there is no generation, ranking and unranking algorithms for $$T_{k,m}(n)$$. In this work, we use Zaks encoding presented in [3] for representing the $$T_{k,m}(n)$$ trees. We use this encoding for generating $$T_{k,m}(n)$$ trees in $$A$$-order. Generating algorithm has constant average time and $$O(n)$$ time complexity in the worst case. Also, new ranking and unranking algorithms are presented in time complexity of $$O(n)$$ and $$O(n \log n)$$,respectively.

### On algorithms for constructing $$(0,1)$$-matrices with prescribed row and column sum vectors

Abstract:

Given partitions $$R$$ and $$S$$ with the same weight and $$S\preceq R^*$$ let $${\cal A}(R,S)$$ be the class of the $$(0,1)$$-matrices with row sum $$R$$ and column sum $$S$$. These matrices play an active role in modern mathematics and the applications extend from their natural context (Matrix Theory, Combinatorics or Graph Theory) to many other areas of knowledge. The Robinson-Schensted-Knuth correspondence establishes a bijection between the class $${\cal A}(R,S)$$ and pairs of Young tableaux with conjugate shape $$\lambda$$ and $$\lambda^*$$ with $$S\preceq \lambda\preceq R^*$$. We give a canonical construction for matrices in $${\cal A}(R,S)$$ whose insertion tableau has a prescribed shape (\lambda\), with $$S\preceq \lambda\preceq R^*$$. This algorithm generalizes some recent constructions due to R. Brualdi for the extremal cases $$\lambda =S$$ and $$\lambda =R^*$$.

### Polytopes of high rank for symmetric and alternating groups

Abstract:

There exists only one polytope with automorphism group $${Sym}(n)$$ of rank $$n-1$$ for $$n>6$$, that is the symplicial complex. We show that when $$n>7$$ there is also a unique polytope of rank $$n-2$$, up to duality and isomorphism.

For the alternating group, we show that for $$n>11$$, $${Alt}(n)$$ acts on at least one abstract regular polytope of rank equal to the floor of $$(n-1)/2$$. We conjecture that this is the highest possible rank. In this talk we show the work done in order to prove this conjecture.

## Posters

### A new encoding for efficient Generation, Ranking and Unranking of trees with $$n$$ nodes and $$m$$ leaves

Abstract:

Trees with $$n$$ nodes and $$m$$ leaves represents the $$2^{nd}$$ structure of RNAs, which is an object of research in Bioinformatics and Combinatorics [1].

The problem of generation is proposing a linear ordering among the studied trees or using an available ordering; then generating the trees with respect of this ordering. Rank of a tree is its position in the exhaustive generated list. Unranking is, as usual, the inverse of ranking.

Generation, ranking and unranking of different classes of trees were studied earlier in the literature, but the problem of generation, ranking and unranking of trees with $$n$$ nodes and $$m$$ leaves has been studied only by Pallor in [2]. He generalized a coding of binary trees (introduced in a previous work) to regular $$k$$-ary trees. This coding is used in order to generate all the trees with $$n$$ nodes and $$m$$ leaves as a list, by embedding them in a set of regular $$m$$-ary trees. It is quite challenging to find an encoding without embedding a given tree to a regular $$m$$-ary tree to propose the efficient algorithms.

In this work, a new encoding is introduced for representing the trees with $$n$$ nodes and $$m$$ leaves. This encoding is used for generating this class of trees in $$A$$-order (an ordering based on global information concerning tree nodes and appear to be a natural ordering of trees) with constant average time and $$O(n)$$ time complexity in the worst case. Also new ranking and unranking algorithms are presented in time complexity of $$O(n \log n)$$.

References
1. R. Giegerich, B. Voss and M. Rehmsmeier, Abstract shapes of RNA, Nucleic Acids Research 32 (2004), 4843--4851.
2. J. Pallor, Generating trees with n nodes and m leaves, Intern. J. Comput. Math. I 21 (1987), 133--144.

### The coloring and clique cover problems for triangulations of a polygon and their products

Abstract:

It is well know that triangulations of a polygon can be vertex colored with three colors, however, there seems lack of such results for complements and strong product of triangulations of a polygon. It is interesting to note that such problems have certain applications in information theory, e.g. to determine index codes and Witsenhausen's rates of graphs, which are obtained from polygon triangulations. We present some new results for the above-mentioned problems, for example we give bounds on the chromatic number of the complements of triangulations.

### Estimating Betti numbers of cut ideals from path graphs

Abstract:

A cut ideal is an homogeneous ideal in a polynomial ring which can be associated to any simple graph. We study minimal free resolutions of such ideals for the class of path graphs. By employing basic methods from topological combinatorics, we obtain simple formulas on the number of vertices of a path graph, which bound the Betti numbers of its cut ideal from above. These formulas arise from the enumeration of induced subgraphs of certain incomparability graphs associated to the path graphs. Ongoing work with Samu Potka and Hakan Gueler.

### A symbolic umbral characterization of Riordan arrays

Abstract:

Riordan arrays are infinite lower triangular complex valued-matrices that have been applied to a wide range of subjects, from Computer Science to Combinatorial Physics, in connection with combinatorial identities, recurrence relations, walk problems, asymptotic approximation and the problem of normal ordering for boson strings, among other relevant topics. The usual way of approaching Riordan arrays is by means of generating functions. I will present a promising alternative characterization of Riordan arrays based on a symbolic renewed approach to umbral calculus. A deep generalization of an Abel's identity for polynomials is a key tool in our symbolic approach.

## Schedule

Here is the timetable for the workshop.

## Social Dinner

It will take place at restaurant, on Tuesday at 8pm. Here are the menu options you can choose from. You will be asked on Monday to sign up a form letting us know your preferences.

### Contact

Email: worklisb2012 [at] fc.ul.pt + (35) 1 21 294 83 88 + (35) 1 21 750 03 24 + (35) 1 21 790 47 08 + (35) 1 21 795 42 88


text:
# Htam::Diary

 Themes: { PHOTO, SURF, TT, VB, TACIT, CS, SCIENCE, SPORT, BIBLE, IM, HTAG, MATH, COMP, FAITH, SYSOP, TEST } =
Years: { 2004, 2005, 2006, 2007, 2008, 2009, 2010, 2011, 2012, 2013 }
Pages: { newest, 165, 164, 163, 162, 161, 160, 150, 140, 130, 120, 110, 100, 90, 80, 70, 60, 50, 40, 30, 25, 20, 15, 10, 5, 3, 2, 1 }
Total 3299 posts.
 htam@127.0.0.12004-07-28 22:42:39 [FAITH] West Coast Christian Conference 2004 0 comments West Coast Christian Conference 2004 was from July 21 to 28.Topic: Knowing the Holy Spirit.''Location: Pacific Lutheran University, Seattle, WA.The three main sub topics are* The Person of the Holy Spirit by Lance Lambert,* The Ministry of the Holy Spirit by Christian Chen, and* The Manifestation of the Holy Spirit by Stephen Kaung.The three short topics are* The Holy Spirt and the Bible by Kwok-Hin Wong,* The Holy Spirt and Spiritual Life by Godwin Sun, and* The Holy Spirt and the Church by Hoseah Wu.I taught the second/third graders during the first Christian Chen message, therefore I felt I should go to the message replay before I listen to any more of his messages. Because bro Chen goes very fast and his materials are always extremely dense. He ALWAYS goes over the time limit. He has a lot to say.So because of the message replay and the morning teaching, I was too tired to attend the first special topic by brother Wong. I attended all the other meetings.Some of this must be true:* the topic this year is comparably simpler, * this year the speakers all spoke very clearly, or* I matured a bit in the Spirt,I felt I understood more of the messages than the previous years.May I be able* to subjectify the objectiveness of his Spirit,* to experience him first hand,* to establish or strengthen my relationship with him, and* to grow in the knowledge of him and become mature in the Spirit. htam@127.0.0.12004-07-17 21:55:06 [MATH] Differential equations 0 comments Try to solve the second-order non-homogeneous ordinary linear differential equation with the method of variation of parametersx'' - x' - 2x = e^3tand you'll see that this equation is so humorous!! htam@127.0.0.12004-07-16 16:23:33 Placement tests 0 comments I finished the following Caltech placement tests:* Math diagnostic* Math 1a: Improper integrals, limits, series* Math 1b: Linear Algebra* Math 1c: Multi-variable/vector calculus* Chemistry placementThe math ones should be perfect/near perfect.The chemistry one, if I get 50% or above, I'll laugh.I took chemistry in Taiwan three years ago.I studied 8 hours yesterday and today.I still have:* Math 2b: Differential Equation* Physics placement* CS 1* CS 2I'll take the math one before 07.19 (deadline).The others will be after WCCC before 07.31 (deadline). htam@127.0.0.12004-07-13 23:03:47 [MATH] Plaque of unknown origin 2 comments I just received a plaque from USPS today, which reads:  ANNUAL HIGH SCHOOL    MATH CONTEST        2004Highest Regional Score      JED YANGbut I don't really know what this is all about...I better call the mailer and confirm what I did to deserve this. htam@127.0.0.12004-07-12 14:38:37 [SYSOP] LaTeX support 0 comments So I just installed* TeX,* LaTeX,* netpbm, and* textogif.When I have time to integrate it with php, everyone can post math equations like this:But for now, this process has not been automatede yet, and we'll need to wait.I envision also writing a semi-WYSIWYG LaTeX math online editor.Till then, refer to TeX manuals. htam@127.0.0.12004-07-10 12:06:56 [MATH] Linear algebra 1 comment Let {u_1, ..., u_p} be an orthogonal basis for a subspace W of R^n.For each y in W, the weights in the linear combination  y = c_1 u_1 + ... + c_p u_pare given by  c_j = (y.u_j)/(u_j.u_j)    (j = 1, ..., p).This decomposes each y in Span {u_1, ..., u_p} into the sum of pprojections onto one-dimensional subspaces that are mutually orthogonal.This is why orthogonal basis is so cool! htam@127.0.0.12004-07-09 18:46:40 Rubik's cube record 0 comments Yay, I broke my previous 71s record on rubik's cube today with a 69s one.More info at my rubik's cube page.Update: After Bible study, I got a 63s one.Update: Saturday I got a 62s one, yay. htam@127.0.0.12004-07-09 17:01:31 Health checkup 0 comments So I went to the doctor again, and beholdMy tuberculosis test is positive (duh...)So I had three X-ray's taken... and more blood drawn.I also need to pay a visit to the UW Medical Center sometime.Squashed with all the test I need to preprare for, I'm so busy. htam@127.0.0.12004-07-09 00:54:21 Digital camera 0 comments My digital camera was fixed and returned.I took a few pictures of my sister holding my first place plague.Maybe I'll take a few pictures of my auction items when I have time.Now I better go regenerate... It's a looong day in front of me:I got to study for Caltech advanced placement test.In 8 hrs:  Linear Algebra, here I come! htam@127.0.0.12004-07-09 00:49:21 [COMP] Software upgrades 1 comment So I updated/installed* Mozilla 1.7* Gaim 0.79, with addons  o libgpg-error,  o libgcrypt,  o libtasn1,  o GnuTLS,   o Auto-profile, and  o Extended Prefs* fortuneit feels like I'm on a new systemI* ./configure'd,* make'd,* su'ed, and* make install'ed so many times today... htam@127.0.0.12004-07-07 23:23:17 Health checkup 0 comments I went to have a check-up as required by Caltech.I was poked three times today.* draw blood (it's actually the least uncomfortable)* Tuberculosis test (which I am positive for sure)* Tetanus and Diphteria vaccine (Td)Td hurts so much, now my left arm just feels swollen and stiff. htam@127.0.0.12004-07-07 22:43:08 Math placement test 1 0 comments So I just finished my Caltech Math1a advanced placement exam on Calculus (limits, complex numbers, sequences, series, and Taylor series).It was a three hour test.It was one hour and I still only did three out of six questions. The other three I did not know how to do. So I skipped them. Really!But then I thought harder, prayed, ate, and walked around during the test... until I came up with ideas on how to tackle the other questions.The answers should all end up right actually, but the explanation might not be fully mathematically precise.I only have 8 more to go ;-)* Math1b Linear Algebra (UW 308)* Math1c Multi-variable and Vector Calculus (UW 324)* Math2a Probablity and Statistics* Math2b Differential Equations (UW 307)* Chemistry (Thanks Dennis for the textbook/review booklet)* Physics* CS1* CS2That's all.I finished the essay and received a 4 out of 6 (1 the worst), it's good to my standards.I also finished the eazy math placement (not the advanced placement) exam last month.So busy with school stuff, eh?But I still find time to read Bible, pack my room, and code. htam@127.0.0.12004-07-06 15:48:47 [SYSOP] Todo list 0 comments Some stuff to be worked on:* text MACRO beautifying project* math beautifying (LaTeX?)# a script to remove all BR's, since I am using PRE now.# stress test# wrap* more authoring tools (preview, math equation editor, etc)* user-friendlier* login and account support (while keeping guest posts privileges)and more to come. htam@127.0.0.12004-07-05 11:30:43 USA 0 comments Now I've been in the US three years straight now.I came on July the fourth, two thousand and one. htam@127.0.0.12004-07-05 00:10:28 [FAITH] JY sharing 0 comments I shared with the Junior Youth kids today (or yesterday, Sunday).Some said it was good.I hope they actually learned something and will read the Bible more.I urge all of them to find spiritual companions... I hope they remember.Anyways, I am really tired.The best thing about sharing with them is: I lost my rubik's cube a few days ago, but some of them found it.Now I have my rubik's cube back. :) htam@127.0.0.12004-07-03 11:44:54 [MATH] Infinite series 8 comments To see if the infinite series (n^2+1)/(n^4+1) converges,I used the limit comparison.The ratio of that previous thing and 1/n^2 goes to 1 as n goes to oo.So since 1/n^2 converges, the previous thing also converge.Correct? Yes. htam@127.0.0.12004-07-03 09:46:49 [FAITH] Bible reading 0 comments Yesterday I led Bible study (with Grace).Therefore, I only started reading my daily Bible reading after church.I spend so loooong that I actually broke my record of late sleep.I think I went to bed THIS MORNING around 12:30. Whoa!This is THE latest...I should start reconsidering when I read Bible. :) htam@127.0.0.12004-07-02 15:38:57 [FAITH] Service 0 comments I have been preparing my sharing with the Junior Youth group for this Sunday.I also will be leading Bible study tonight.Just so busy...I have a bunch of placement tests I need to complete in a short amount of time. htam@127.0.0.12004-07-01 20:33:06 [SYSOP] Initiation! 0 comments Finally. After a day's of work, this section is now at least usable.Fell free to flood me with comments regarding programming (or otherwise).I will be revising this area closely in the next few days.Consider it ALPHA testing stage :)So, don't complain, but be loud nevertheless.Jed htam@127.0.0.12004-07-01 20:10:24 [TEST] 27 comments final test run before publishing...


text:
# Defining a function to be the minimum of two other functions

I cannot find a way to define a function as the minimum of two other functions that each have the same two independent variables. I have f(x,y) and g(x,y), and I would like to find $h = \min\{f(x,y),g(x,y)\}$. $h$ is not a numerical answer, but a piecewise function; i.e., $h = f(x,y)$ for $f(x,y)<g(x,y)$, and $h=g(x,y)$ for $f(x,y)>=g(x,y)$.

Additionally, I would like to find the function that maximizes $h$ with respect to the variable $y$ and plot it in the xy-plane.

• Sounds like homework. – Alan May 1 '18 at 0:18
• What have you tried? Have you looked at Min in the documentation? – eyorble May 1 '18 at 0:22
• I am doing an analysis on a baseball bunting play where x is the distance the fielder is from the plate and y is the angle the ball is hit at. G and F represents the likelihood of an out given that a specific player fields the ball. Thus it is natural that the minimum of these functions would indicate who should field the ball. While the hitter maximizes this likelihood of a hit fcn, h, by choosing the angle y. – Alexander Pingry May 1 '18 at 1:10
• Under Min in documentation it is not used as function, rather an operation. – Alexander Pingry May 1 '18 at 1:11

Since you haven't said what f and g are, I've just picked something at random:

f[x_, y_] := x y;
g[x_, y_] := x^2 + y^2;
h[x_, y_] := Min[f[x, y], g[x, y]]

If you prefer, you can explicitly make h a piecewise function:

h2[x_, y_] := Piecewise[{{f[x, y], f[x, y] <= g[x, y]},
{g[x, y], f[x, y] > g[x, y]}}]

• For fun: h3 = Through@*Min[f, g]. – AccidentalFourierTransform May 1 '18 at 3:28
• h[x_, y_] = Min[f[x, y], g[x, y]] // PiecewiseExpand – Bob Hanlon May 1 '18 at 5:37


text:
This is an old revision of the document!

### Entering Angles

Angles are their own object type. An angle is entered by prefixing a number with the angle symbol (∡, AL-RS-6), followed by a number, then finally a suffix that indicates the system (° = degrees, r = radians, g = grads, or d = DMS [DDD.MMSS, where DDD=degrees, MM=minutes, SS=seconds])1). To enter an angle in the current angle mode of the calculator, the suffix can be left off. For example, the following all indicate an angle of 90.5 degrees:

∡90.5°

∡1.5795.r

∡100.5556g

90.3000d

and

∡90.5 with the angle mode set to degrees.

### Working with Angles

Trigonometric functions operate on angle objects as expected:

∡45° SIN yields the real number result 0.7071.

Basic arithmetic operations can also be applied to angle objects:

∡90° ∡50° + yields ∡135°

∡1.57r ∡50° - yields ∡0.6973r

∡90° ∡45° * yields 4050

∡90° ∡45° / yields 2

The result of an operation on angle objects has the following conventions:

• Addition and subtraction between angles returns an angle expressed in the system of the first argument,
• For all other operations, angles will be converted to the current angle mode and output a number.

### Commands for Angle Objects

1)
r, g, and d are case insensitive
• manual/chapter3/angles.1502500374.txt.gz


text:
# Find the last two digits of a sum of eighth powers of $100$ consecutive integers

Find the last two digits of a sum of eighth powers of $100$ consecutive integers.

Attempt:

Since $(100k+x)^8 \equiv x^8 \pmod{100}$, we solely need to find $S=1^8+2^8+\cdots+100^8 \pmod{100}$. By the Chinese Remainder Theorem, if we determine the sum modulo $4$ and modulo $25$, we will know it modulo $100$. Taking it modulo $4$ we have $S \equiv 1+0+1+\cdots+1+0 \equiv 50 \equiv 2 \pmod 4$.

Now, since the number of units modulo $25$ is $\phi(25)$ and $\gcd(8,\phi(25)) = 4$, does that mean we can replace the sum of the $8$-th powers mod $25$ by the sum of the $4$-th powers mod $25$?

• Hint: $a^8=(k-a)^8 mod k$ Now how can this be of use to calculate your congruence? – Bahbi May 15 '16 at 14:41

I don't know about your last argument, but the problem is not that complicated if you consider the following reductions:

• to compute $1^8+...+100^8$ modulo $25$ it suffices to find $1^8+...+25^8$ modulo $25$ and multiply the result by $4$.
• to compute $1^8+...+25^8$ modulo $25$ it suffices to find $1^8+...+12^8$ modulo $25$ and multiply the result by $2$ and this is not too long

I guess you may find a more elegant solution using rings of the form $(\Bbb{Z}/n\Bbb{Z})$.

• Cant we take mod 10 upto 10 and multiply by 10? – N.S.JOHN May 15 '16 at 17:15
• Yes, that might work also. :) – Beni Bogosel May 15 '16 at 18:57
• So wouldn't the answer be 0 as we are multiplying 10? – N.S.JOHN May 16 '16 at 2:55
• We want to find the last two digits, not only the last one. – Beni Bogosel May 16 '16 at 9:15
• @Beni Bogosel: $1+2^8+3^8+4^8+....+25^8=504 213 653 645=20 168 546 145\cdot 25+20$ and $20\cdot 4=80$. The problem is that $80\ne 30$. Regards. – Piquito May 16 '16 at 16:26

If $N=100M+ab$ then $N^8\equiv (ab)^8\pmod{100}$ so the question concerns the integers of the form $10a+b$ where $a$ and $b$ are digits.

Because of $(10a+b)^8\equiv 80ab^7+b^8 \pmod{100}$, we use this table modulo $100$ for calculations $$\begin{array}{|c|c|}\hline a & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 0\\\hline a^7 & 1 & 28 & 87 & 84 & 25 & 36&43&52&69&00\\\hline a^8&1&56&61&36&25&16&01&16&21&00\\\hline\end{array}$$ Take the sum of ten consecutive eight powers $$S_x=(10a+x)^8+(10a+x+1)^8+(10a+x+2)^8+(10a+x+3)^8+(10a+x+4)^8+(10a+x+5)^8+(10a+x+6)^8+(10a+x+7)^8+(10a+x+8)^8+(10a+x+9)^8$$ One has $$S_x\equiv {80a[x^7+(x+1)^7+...+(x+9)^7]+[x^8+(x+1)^8+…+(x+9)^8]}\pmod{100}$$ Calculation gives $$S_0\equiv {80a(25)+33}\equiv 33\pmod{100}\\S_1 \equiv {80a(25)+33}\equiv 33\pmod{100}\\S_2 \equiv {80a(95)+13}\equiv 13\pmod{100}\\S_3 \equiv {80a(75)+53}\equiv 53\pmod{100}\\S_4 \equiv {80a(5)+13}\equiv 13\pmod{100}\\S_5 \equiv {80a(25)+33}\equiv 33\pmod{100}\\S_6 \equiv {80a(75)+33}\equiv 33\pmod{100}\\S_7 \equiv {80a(95)+13}\equiv 13\pmod{100}\\S_8 \equiv {80a(25)+53}\equiv 53\pmod{100}\\S_9 \equiv {80a(5)+13}\equiv 13\pmod{100}$$ We see the sum of ten consecutive eight powers beginning by $(10a+x)^8$ is independent of $a$ modulo $100$. It follows $$\sum_{k=0}^{99}(10a+x+k)^8=\sum_{k=0}^9(10a+x+k)^8+ \sum_{k=10}^{19}(10a+x+k)^8+\sum_{k=20}^{29}(10a+x+k)^8+...+\sum_{k=90}^{99}(10a+x+k)^8\equiv \underbrace{(S_x+S_x+…+S_x)}_{10\text{ times}}\equiv 10S_x\pmod{100}$$ According to the calculation above we have $10S_x\in\{330,130,530\}$ Thus $$\sum_{k=0}^{99}(10a+x+k)^8\equiv \color{red}{30}\pmod {100}$$


text:
# String Bit Manipulation- There and Back Again

Your job is to encrypt a string using a number key.

Your program will receive 2 strings: one containing a sequence of characters to encrypt, and another one containing an even amount of digits (1-9, never 0).

The encrypting will work like this:

1. take the string to be encrypt and decode it to binary (ISO-8859-1)
2. for every pair of numbers x and y in the number string do: for every character in the string of zeroes and ones, starting from the first one, if the bit is 0 toggle the xth bit in front of the bit you are reading right now. If the bit is 1, toggle the yth bit in front of the bit you are reading right now. If the bit to be toggled is out of bounds, do not toggle anything.

Other than that:

This is a code golf challenge, shortest code wins. Standard Loopholes apply.

Here's a more graphic explanation of what happens:

inputs: "a", "23"
"a" =BIN> "01100001"
"01100001" //first bit is a 0, get the first number in the key, 2.
- ^       //toggle the bit 2 steps in front
"01000001" //next bit is 1, get second number in the key, 3.
-  ^     //toggle the bit 3 steps in front
"01001001" //0 bit, toggle the bit that is 2 steps in front.
- ^
"01000001" //etc
- ^
"01000101"
- ^
"01000111"
-  ^ //no change, out of bounds.
"01000111"
-  ^
"01000111"
-  ^
"01000111" =ASCII> "G"

So: a, 23 = G. Then, running G, 23 should be a because this process is perfectly reversible. To decrypt a string encrypted using key 12438135, reverse the ordering of the pairs, 35814312 and run the encrypted string through a similar program that performs the operations in reverse order (starting at the end, approaching the beginning) using that key. (thanks to Martin Büttner for pointing that out)

Good luck and have fun! (My first challenge, tell me if i missed anything :))

• I've reordered and made some minor changes to make this less unclear, but it's still not entirely clear to me what the process is. "Extract the bits of each character": in what encoding? I would guess ISO-8859-1, but you should make that explicit. "Form a long string of zeroes and ones, appending each character" using big-endian? "For every pair of numbers": that is, splitting the number string into two, not advancing for (i=0; i+1 < length; i++)? What does "in front of" mean? And why does the code block only use x and never y? – Peter Taylor Sep 17 '14 at 14:55
• I guess i wasn't quite clear with the question, I'll edit it now. – rodolphito Sep 17 '14 at 14:56
• I will create a base program now to illustrate what the program should accomplish – rodolphito Sep 17 '14 at 15:12
• Btw, I don't think this process is reversible, because some of the toggles depend on earlier toggles. So the first affected bit will return to what it was before. But once you get there, it will cause the toggling of a different bit than during the encoding. My program below yields n if I feed it G. – Martin Ender Sep 17 '14 at 15:38
• You are right. To reverse the process you have to run the program again, but instead of starting at the first bit, start at the last one and work backwards to cancel out the flipped bits. Good catch. I will edit the question when i get back from school. – rodolphito Sep 17 '14 at 16:17

# CJam, 53 bytes

SSl++256b2bl2/{Em<_,E-{_0=P=~_2$=!t1m<}*}fPE>2b256b:c  Reads two lines of input: the string, followed by the key. Try it online. For decryption, this code can be used: SSl++256b2bl2/W%{1m>_,E-{_0=P=~_2$=!t1m>}*Dm>}fPE>2b256b:c

$LANG=en_US cjam enc.cjam # First two lines are input. codegolf 345678 q¼M³)6¯J  ### How it works CJam's built-in base conversion discards leading zeros, so we start by prepending two spaces to the input string. Since a space character is [0 0 1 0 0 0 0 0], this prepends 14 zeros to the bit encoding for the input string. After each transformation, we rotate the array to the left to avoid adding indexes, skipping the extra 14 bytes from the spaces. This way, toggling bits "out of bounds" affects only those 14 bits, which are discarded after the last toggle. SSl++ " Read one line of input (plaintext) and prepend two spaces. "; 256b2b " String ↦ byte array ↦ bit array. "; l2/ " Read one line of input (key) and split it in chunks of length 2. "; { " For each digit pair P: "; Em< " Rotate the array 14 places to the left. "; _,E- " Push the arrays length minus 14. "; { " Do the following that many times: "; _0=P=~ " Push the digit of P that corresponds to the first bit of the array. "; _2$=    " Retrieve the corresponding bit from the array. ";
!t      " Compute its logical NOT and modify the array at the given position. ";
1m<     " Rotate the array one place to the left. ";
}*        " ";
}fP         " ";
E>          " Discard the first 14 bits of the array. ";
2b256b:c    " Bit array ↦ byte array ↦ string. ";

• Wow this is pretty cool. Padding the array and discarding out of bounds modifications is quite ingenious. It doesn't really have to be spaces though, right? Its discarded at the end anyways. – rodolphito Sep 17 '14 at 22:44
• Almost any other character would have worked (null bytes, e.g., would accomplish nothing), but S is a convenient shorthand for " ". The only thing that varies the the shift amount. – Dennis Sep 17 '14 at 22:49

## Ruby, 230191181178175173171 159 bytes

i,n=$* s=i.gsub(/./){|c|"%08b"%c.ord} n.scan(/(.)(.)/){s.size.times{|i|s[j=i+(s[i]==?0?$1:$2).to_i]&&s[j]=s[j]==?0??1:?0}}$><<s.gsub(/.{8}/){|c|c.to_i(2).chr}

It's just a very naive implementation of the spec. Input via command-line arguments, output to STDOUT.

# Read ARGV into i and n
i,n = $* # Turn each character into a string of eight 0s and 1s using a format string. s = i.gsub(/./){|c|"%08b"%c.ord} # For each pair of characters in n n.scan(/(.)(.)/){ # For each index into s s.size.times{|i| # Determine the index that is to be toggled.$1 and $2 refer to the captures # of /(.)(.)/. j=i+(s[i]==?0?$1:$2).to_i # Using short-circuiting, toggle s[j] if it isn't nil. s[j]&&s[j]=s[j]==?0??1:?0 } } # Convert each group of 8 characters back to a single character and print$><<s.gsub(/.{8}/){|c|c.to_i(2).chr}

# GolfScript, 45

~2/{\{8{(..2\?3$&!!4$=48-- 2\?@\^\.}do;}%\;}/

This takes arguments from stdin in the form of two strings (e.g. "a""23") and outputs to stdout.

Test it online here

• For "codegolf""345678", it produces an incorrect result. With the official interpreter, it just crashes (Rational can't be coerced into Rational for bitwise arithmetic). – Dennis Sep 21 '14 at 1:24

# JavaScript 243

function E(s,k,d){for (var o="",a=0,b=0;a<s.length;a++,b%=k.length){
var j,i=8,z,n=s.charCodeAt(a),x=parseInt(k.charAt(b++)),y=parseInt(k.charAt(b++))
for (;i--;)j=d?7-i:i,z=j-(n&(1<<j)?y:x),n=z>0?n^=1<<z:n
o+=String.fromCharCode(n)
}return o}

Example:

var s0 = "codegolf";
var k = "5556";
var s1 = E(s0,k);
var s2 = E(s1,k,true);
console.log(s0+"\n"+s1+"\n"+s2)

Output:

codegolf
ekbaakjb
codegolf


text:
## Simple Question = Silly Answer

For all your silly time-killing forum games.

Moderators: jestingrabbit, Moderators General, Prelates

### Re: Simple Question = Silly Answer

Somewhere near the arse of the alphabet.

How did we get here?
The image link changes whenever I find a new cool website.
Spoiler:

The Hyphenator

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### Re: Simple Question = Silly Answer

Well... you see... when a daddy and a mummy love each other very much, they sit down together and have a game croquet! And then, 9 months later, you came out!

And then mummy has to explain WHY you came out, because daddy was away on business most of the month.

The universe... why is it so big?
phlip wrote:sleepy, the only thing you're worse at is being not awesome.*
*All links to be treated as NFSW.

sleepygamer
More metal than thou

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### Re: Simple Question = Silly Answer

To fit your mother.

How do you solve a problem like maria?

.Taylor.

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### Re: Simple Question = Silly Answer

To fit my ego inside of it.

How much blood does a church-mouse have?
Wolydarg wrote:That was like a roller coaster of mathematical reasoning. Problems! Solutions! More problems!

****************Signature Dehosted, New Signature Under Construction************************

Antimony-120

Posts: 831
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### Re: Simple Question = Silly Answer

Well, that really depends on how you define churchmouse. If you're talking about a mouse that is in a church, then there is no way of directly telling because mice are invincible while on holy ground (it was part of that deal between Reepicheep and Aslan; the tail was just a bonus). You can logically infer from the size of the mouse, however, that it ranges from about five milliliters to about twenty, depending on age. If you're talking about a member or descendant of the Churchmouse clan from Redwall, however, that's a whole different story. That universe has a kind of inconsistent measuring system. On the one hand, fish, trees, and such are much larger than the mice; on the other hand, when they refer to "inches," it's to the mice approximately what an inch to humans would be. So that implies that the mice, and everything else, are friggin' huge, man! So for this breed of Churchmouse, you would either have a similar estimate, or one probably approximating the amount of blood a human has, i.e. approximately eight pints for an adult, about two for a wee baby.

Why didn't they just have the Haitian erase Sylar's memory when they captured him?
Love may be blind, but lust has x-ray vision.
Avatar shamelessly plucked from PMOG.

Also, it turns out I'm not dead--I'm just a right bastard who disappears from the internet for months at a time every so often.

Matthias

Posts: 275
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### Re: Simple Question = Silly Answer

//actually, I think the trees are proportionate to the mice, just that they live in a big forest. I mean, the highest trees stretch to the hight of the abby wall, which is built in proportion to the mice and also in proportion to the tree, apparently. it's the horses and badgers that confuse me. badger lords would make an AWESOME D&D class. they'd have to have like ecl-20.
It rains on the enemy too!

pkuky

Posts: 612
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### Re: Simple Question = Silly Answer

//You could probably get much the same effect by making a naturally lycanthropic werebadger, with levels in Barbarian. Although technically you might be able to take Frenzied Berserker from the get-go. I seem to recall making a rather silly twink like that, actually, though I was talking about a regular badger. They have a trait that allows them to go into a frenzy when attacked, which they will not come out of until they or their opponent is dead. Then you take ten levels of Frenzied Berserker and get deathless frenzy, and you cannot lose in single combat. Ridiculous.
Love may be blind, but lust has x-ray vision.
Avatar shamelessly plucked from PMOG.

Also, it turns out I'm not dead--I'm just a right bastard who disappears from the internet for months at a time every so often.

Matthias

Posts: 275
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Location: Out of context, probably.

### Re: Simple Question = Silly Answer

Matthias wrote:Why didn't they just have the Haitian erase Sylar's memory when they captured him?

Because there is no such thing as erasing peoples' memory, even on Haiti.

Why did you try to kill this thread with obscure references?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

Because only lame people never watched Heroes.

Why do they use cardinal directions to label the lanes on the I-285 Perimeter loop instead of simple terms like "Outer Loop" and "Inner Loop" or "Sunwise" and "Widdershins?"

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

Because they're Catholic.

(now groan)

Why is it,t that for the first 1hr i spent on irc (6 channels on 3 australian servers), no-one spoke a word?

abrenecki

Posts: 62
Joined: Mon Sep 10, 2007 8:39 am UTC
Location: Mid North SA, Australia

### Re: Simple Question = Silly Answer

Because you shouldn't talk if you have nothing to say.

Why do the Swiss laugh unexpectedly in church Sunday morning?

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

They're on drugs.

Is Randall plotting against the world-wide conspiracies?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

A better question would be: "Why does everyone use the fact that you're paranoid as an excuse to ignore the lethal threat that is about to destroy the earth?"

Why does everyone use the fact that I'm paranoid as an excuse to ignore the ignominious threat that is about to discombobulate the planet?

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

Because IT'S HAPPENED BEFORE! We're still discombobulated, too confused to know that it's happening again. To save the earth you must go back to the future!

Why is it life seems about to work out just before it smacks me upside the head?
Wolydarg wrote:That was like a roller coaster of mathematical reasoning. Problems! Solutions! More problems!

****************Signature Dehosted, New Signature Under Construction************************

Antimony-120

Posts: 831
Joined: Wed Apr 09, 2008 4:16 am UTC
Location: Wherever you can look - wherever there's a fight, so hungry people can eat.

### Re: Simple Question = Silly Answer

Typo in God's Divine Plan(TM).

Why does appending a trademark notice make everything funnier?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

A failure to be silly does not appropriately create a situation wherein a query about the humor value of a comedic device can be issued. Especially if said humor device is never truly humorous in and of itself.

Sex?

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

quintopia wrote:Sex?

Not for me, thanks, I'm married.

Is Santa real?
Some of us exist to find out what can and can't be done.

Others exist to hold the beer.
btilly

Posts: 1877
Joined: Tue Nov 06, 2007 7:08 pm UTC

### Re: Simple Question = Silly Answer

Well, it all depends on your perspective. In certain cultures, the character known as "Santa Claus" is real enough in the minds of children. As children are more prone to solipsism, Santa is very real from their perspective. The author of "Yes, Virgina, There Is A Santa Claus" can further elaborate. However, from a scientific perspective, this character would have to be so ridiculously advanced in terms of technology that we could only postulate his existence in terms of Clarke's Third Law. Furthermore, from a historical perspective, there was surely a real person named "Nicholas" who was sainted and whose persona was eventually exaggerated both by literature (see Clement Moore) and by clever marketing campaigns (see Coca-Cola). This evolution has led the Santa Claus character to be a very culturally real symbolic phenomenon of the spirit of Christmas, both as it was and is now. Lastly, if you read it in the Sun, then it is so.

Why is nudism taboo in this culture?

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

quintopia wrote:Why is nudism taboo in this culture?

'Cause waaaaay back in the olden days, a great, respected elder in the community declared nudity to be a sin so that he wouldn't have to show anyone his saggy butt.

How do you make pancakes?
videogamesizzle wrote:so, uh, seen any good arbitrary, high numbers lately?

Platypodes

Posts: 201
Joined: Mon Nov 19, 2007 8:42 am UTC
Location: On a knot on a log in a hole in the bottom of the sea

### Re: Simple Question = Silly Answer

Code: Select all
\$ make pancacesmake: *** No rule to make target pancaces'.  Stop.`

What do you want me to ask?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

mat-tina wrote:What do you want me to ask?

I'd like you to ask, "Would you like some chocolate-covered peanuts?" And then I'd like you to send them to me over the internet. My computer has a candy dispenser right next to the cup holder.

How many shopping days are left till Christmas?
videogamesizzle wrote:so, uh, seen any good arbitrary, high numbers lately?

Platypodes

Posts: 201
Joined: Mon Nov 19, 2007 8:42 am UTC
Location: On a knot on a log in a hole in the bottom of the sea

### Re: Simple Question = Silly Answer

How many shopping days are left till Christmas?

Thats the question to why Jesus died.

Am I going to become a teacher?
DrPhibes

Posts: 14
Joined: Sat Jul 26, 2008 12:34 pm UTC

### Re: Simple Question = Silly Answer

Yes, you're going to teach astrophysics to sentient marshmallows on a giant flying zamboni.

Who will be the next president of the United States?
videogamesizzle wrote:so, uh, seen any good arbitrary, high numbers lately?

Platypodes

Posts: 201
Joined: Mon Nov 19, 2007 8:42 am UTC
Location: On a knot on a log in a hole in the bottom of the sea

### Re: Simple Question = Silly Answer

Terry Pratchett.

Could you use a cylindrical white hole with an aperture on one end, containing an equally cylindrical but slightly smaller black hole, as a gravity laser?
Love may be blind, but lust has x-ray vision.
Avatar shamelessly plucked from PMOG.

Also, it turns out I'm not dead--I'm just a right bastard who disappears from the internet for months at a time every so often.

Matthias

Posts: 275
Joined: Wed Sep 12, 2007 4:55 am UTC
Location: Out of context, probably.

### Re: Simple Question = Silly Answer

if that was true, wouldn't we have one already?

how much wood can a woodchuck chuck?

falcon434

Posts: 0
Joined: Fri Apr 11, 2008 2:57 pm UTC

### Re: Simple Question = Silly Answer

The rate of a woodchucks woodchucking, V' m³/s, varies according to V' = k \cdot V_b \cdot f where V_b m³ is the volume of wood that each chuck chucks away when chucking an SI standard piece of wood, f Hz is the frequency of the chucking and k is a constant depending on the hardness of the wood.

Will you teach me The Way, oh Guru?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

I would, but I already did. It will take you a lifetime to understand.

Where's Nemo?
Some of us exist to find out what can and can't be done.

Others exist to hold the beer.
btilly

Posts: 1877
Joined: Tue Nov 06, 2007 7:08 pm UTC

### Re: Simple Question = Silly Answer

Have you tried calling the local hospitals?

Where are my glasses?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

You remember that *crunch* when you sat down? Well those were your glasses.

Who am I?
Some of us exist to find out what can and can't be done.

Others exist to hold the beer.
btilly

Posts: 1877
Joined: Tue Nov 06, 2007 7:08 pm UTC

### Re: Simple Question = Silly Answer

The source of words which have been transmitted to me via the Internet. I can truly know nothing else about you.

What happens if you ask a simple question?

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

You get a silly dingbat wallflower turnip goose android b'kaw b'kaw chirp chirp squirtle doorknob ding

Why was the last question so meta?

malec2b

Posts: 138
Joined: Thu Nov 22, 2007 5:40 am UTC

### Re: Simple Question = Silly Answer

Coz it was quintopia asking it... and let's face it, he's meta.

Why do I want a big bowl of ice-cream with milo on top right now?
Teacher: "Right, you are no longer allowed to let sock puppets take responsibility for any of your actions."

Damn. There goes my education.

dragongrrl

Posts: 7
Joined: Wed Jul 16, 2008 10:24 am UTC

### Re: Simple Question = Silly Answer

Because aspiring dragons prefer their ice cream with tabby kittens on .

What has that got to do with anything?

quintopia

Posts: 2790
Joined: Fri Nov 17, 2006 2:53 am UTC
Location: atlanta, ga

### Re: Simple Question = Silly Answer

She wanted to understand why she was contributing so much to the key leading economic indicator called "ice cream sales".

Why do people like Fridays so much?
Some of us exist to find out what can and can't be done.

Others exist to hold the beer.
btilly

Posts: 1877
Joined: Tue Nov 06, 2007 7:08 pm UTC

### Re: Simple Question = Silly Answer

They don't they just dislike all the other days.

Why does it always rain on me?
It's turtles all the way down
Panda
BHCM
protect the flag
snuggles
Jahoclave wrote:Tip: Shotgun, they could be a zombie. Remember, you're the first step in preventing the zombie apocalypse.

Grooving Tony

Posts: 100
Joined: Sun May 04, 2008 9:55 pm UTC

### Re: Simple Question = Silly Answer

you live in scotland man... it rains on everyone here... even indoors.

what makes hybernation possible?
Signature : 01000011 01101000 01110101 01100011 01101011
00100000 01001110 01101111 01110010 01110010 01101001 01110011 00100000

ShaKri

Posts: 155
Joined: Tue Jun 10, 2008 8:15 pm UTC
Location: Glasgow, Scotland

### Re: Simple Question = Silly Answer

Because Scotland is so boring all the animals just pass out.

Was that mean?

Jebobek

Posts: 2216
Joined: Tue Jun 17, 2008 7:19 pm UTC
Location: Harrisburg, Pennsylvania Geohash graticule

### Re: Simple Question = Silly Answer

no it was no the average of a number

How many epic fails does it take to equal one Bush fail?
Mindless
Tillian wrote:Yeah, but the polar bears get more territorial during the summer, so we have to stay indoors.

Asmodieus

Posts: 174
Joined: Fri Aug 01, 2008 5:16 pm UTC
Location: Teenage wasteland

### Re: Simple Question = Silly Answer

Fail is a verb. The noun is "failure".

Understood?
Felltir wrote:has no sig, and therefore something to hide
GENERATION n: The first time you see this, copy it into your sig on any forum. If n is an even number, divide it by 2. If it's odd, multiply it by 3 and add 1. Prove that this sequence converges to 1 for all n.

mat-tina

Posts: 331
Joined: Mon Jun 02, 2008 3:33 pm UTC

### Re: Simple Question = Silly Answer

No becasue this is the Internet, interwebs, interblag, interblog, and a meme cest pool

Now why does everyone avoid that question?
Mindless
Tillian wrote:Yeah, but the polar bears get more territorial during the summer, so we have to stay indoors.

Asmodieus

Posts: 174
Joined: Fri Aug 01, 2008 5:16 pm UTC
Location: Teenage wasteland

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