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# Is the Square a Secure Polygon?

In this week's episode of PBS Infinite Series, I shared the following puzzle:

Consider a square in the xy-plane, and let A (an "assassin") and T (a "target") be two arbitrary-but-fixed points within the square. Suppose that the square behaves like a billiard table, so that any ray (a.k.a "shot") from the assassin will bounce off the sides of the square, with the angle of incidence equaling the angle of reflection. Puzzle: Is it possible to block any possible shot from A to T by placing a finite number of points in the square?

As I mention in the video, this is one of a number of billiard problems that folks studying dynamical systems have asked. It was mentioned in a 2014 lecture given by Maryam Mirzakhani, which you can watch in the video on the left! Maryam describes the puzzle but leaves the audience to think about the solution. And in that audience was category theorist Emily Riehl, who did indeed think about a solution on and off for a few weeks.

(You can read Emily's reflection on her experience in this AMS tribute to Maryam.)

Fast forward to last fall, when I ran into Emily at a Women in Topology workshop at MSRI. During a bus ride up the hill to the MSRI campus, Emily suggested that the puzzle might be a good fit for Infinite Series. I couldn't agree more! The solution she shared with me is delightfully simple for what seems to be an incredibly complicated puzzle. It's the solution she believes Maryam intended the audience to find, and it's what I want to share with you today! In the video above, I set up some of the mathematics needed to work through the solution, so if you haven't watched it already, now would be a good time. In any case, I'll assume you're comfortable with the notions of flat tori, a tiling of the xy-plane with flat tori, and a lattice in the xy-plane.

Alright, let's get to it!

My sincerest thanks go to Emily Riehl for today's episode and blog post!

The answer is YES. It is possible to place a finite number of points in the square that will block any possible shot from $A$ to $T$. Moreover, 16 points will suffice!

## The Solution

### Step #1: Transfer the problem to the xy-plane.

Billiard trajectories (i.e. zig-zags on a square) can be tricky to analyze, so to make things easier, we want to find a way to replace any biliard trajectory by a line in the plane. (Why? Because zig-zags are hard. Lines are easy!) Now anytime a mathematician mentions the words "square" and "trajectory" and "line" in the same sentence, there's a 99.2% chance that flat torus is involved! So the goal for the first step in the solution is to somehow (we'll see how in a bit) use a flat torus to transfer the puzzle from the billiard table to the $xy$-plane.

To start, I'll pick up where we left off in the video — let's draw the billiard table as a square whose edges are each a different color. This will remind us that unlike a flat torus, none of the sides of this square are identified. I'll also pick two arbitrary-but-fixed points in the square, labeled $A$ (assassin) and $T$ (target). To distinguish the points visually, I've drawn the assassin as a white dot and the target as a black dot.

Now even though this square billiard table is not a flat torus, we can make a flat torus by sticking four of these billiard tables together! How? First notice that the square can be reflected in four different ways (including the "empty reflection").

Square #1 represents the original square (with no reflection), and I've labeled its edges by north, south, east, and west. Square #2 is obtained by reflecting square #1 across a horizontal axis, so that the north and south sides get swapped. Square #3 is obtained by reflecting square #1 across a vertical axis, so that the east and west sides get swapped. We can also perform these two reflections in succession to swap both north/south and east/west. That's square #4. To get a torus, simply assemble these four squares as shown below!

Now we can tile the plane with that torus. (In math lingo one says, "We can pass to the universal cover of the torus.")

In what follows, I'll refer to that yellow square that you see (labeled with 1) as the "original square." We'll think of it as the original billiard table, while the other squares are the copies/reflected copies.

Okay, we’ve transferred our target/assassin problem from a square to the plane. So any trajectory on the billiard table will correspond to a line segment in the plane! For example, the trajectory from the assassin to the target on the left-hand side below IS the line segment on the right-hand side! (Verify this for a good mental check!)

So we've successfully moved the puzzle from a square to the plane. That was the first step.

### Step #2: Observe that the problem involves lattices!

Until now I've omitted a detail, but now's a good time to point it out. In the very first graphic above, you see that I plopped two points $A$ and $T$ (white and black) on the original square. Since those points lie on the original square, they must lie on the reflected copies, too!

Therefore they also appear in the tiling of the plane!

While this may look pretty, it'll be more helpful if we focus on the assassin and the target in their respective squares, rather than all jumbled together. I've drawn this below.

For the moment, let's only focus on what's happening with the copies of square #1. We've got a bunch of repeating black and white dots! The technical word for  repeating dots is a lattice.

##### More formally: Given a point $(x,y)\in\mathbb{R}\times \mathbb{R}$, the lattice generated by $(x,y)$ is the set of points $\{(x,y)+(n,m)\;\vert\;n,m\in\mathbb{Z}\}$. In other words, start at $(x,y)$ and move $m$ units left/right and $n$ units up/down, where $m$ and $n$ range over all possible integers. The resulting collection of points is called a lattice, which is really just a translated copy of $\mathbb{Z}\times\mathbb{Z}$. (You can think of $\mathbb{Z}\times\mathbb{Z}$ as the lattice generated by $(0,0)$.)

In the previous graphic, we observed that the target in the original square generates a lattice, which we'll call Lattice #1. In the graphic below, I've done a little rescaling so that the set of black dots really is like $\mathbb{Z}\times\mathbb{Z}$, rather than $2\mathbb{Z}\times2\mathbb{Z}$. (i.e. so that the black dots really are spaced out by a factor of one, rather than a factor of two. This is what I'm referring to in the "small technicality alert" in the graphic below.)

Of course, a similar story holds for the assassin $A$ — it generates its own lattice, though I won't draw it because we won't need it! It'll suffice to consider the original assassin (the assassin in the original square) and not its copies. Also remember — we've only focused on square #1! We actually get more lattices because of the three other squares, and we'll focus on those generated by the target. So in addition to Lattice #1, let's call Lattice #2 the lattice generated by the target in square #2; Lattice #3 the lattice generated by the target in square #3; Lattice #4 the lattice generated by the target in square #4. Here are the pictures.

Woo! That's a lot of dots. At this point, you might wonder why we're not worrying about the lattices generated by the assassin. Why represent the assassin as a single dot rather than as lattices, as we did with the target? The answer lies in the following observation: The original problem consists of two points $A$ and $T$ on a square billiard table,  and so when we tile the plane with those cleverly-crafted flat tori,

This is the key! In fact, it bears repeating:

##### Any shot from the assassin to the target in the original billiard table can be represented by a line segment in the plane that goes from the assassin in the original square to a copy of the target in one of the four lattices.

So to solve the puzzle, let's make the following strategy.

### Strategy: To block any shot from the assassin to the target, we need to block any shot that can occur in each of the four lattices.

Let's call this process protecting the lattice. To solve the puzzle, then, we first need to protect Lattice #1, then we need to protect Lattice #2, then we need to protect Lattice #3, and finally we need to protect Lattice #4. Fortunately, this isn't as complicated as it seems! As we'll see below, once we figure out a way to protect Lattice #1, we'll be able to apply similar arguments to protect the remaining three lattices. In other words, if we can protect Lattice #1 using $k$ "security agents," then we may conclude that a total of $4k$ security agents are needed protect the full plane. That is, it will take a total of $4k$ security agents to block any shot from the assassin to the target! So what is this special integer $k$? Let's find out in the next steps.

### Step 3: Protect Lattice #1

Here's a picture of Lattice #1 again. I've plotted the original assassin in the original square, circled below.

To make things easier to visualize, let's introduce a new coordinate plane - one where the original assassin lies at the origin, like this:

I've drawn the new axes in dark blue to emphasize that this really is a new grid! Now suppose that the original target has coordinates $(a,b)$ in this plane. (I realize it looks like $(a,b)=(\frac{1}{2},\frac{1}{2})$, but that's only due to poor artistic planning. The target needn't be in the center of the blue squares!) Then the coordinates of the target in the other three squares nearest the origin are as shown below.

And now we're getting really close! Remember, our goal is to block any shot from the assassin (the origin) to the target (the dots). So naturally, we want block those four closest shots, right? So let’s place a "security agent" — indicated by a red dot and labled with $B_1, B_2, B_3,$ and $B_4$ — in between the assassin and the target in each of the four closest squares. Where exactly? How about the midpoint! It's the easiest thing we can try. Plus, we can write down exact coordinates for the agents!

Moreover, each point $B_1,B_2,B_3,B_4$ generates its OWN lattice! And the idea is that $B_1$ (and its copies), for instance, could potentially block other shots, too — those that come from all angles that might have bounced off the walls several times, but haven’t yet hit the target. I won't draw the pictures for these lattices, but the idea comes in handy soon, so be sure to keep it in mind!

In any case, we've certainly blocked the four closest shots from the assassin to the target. But what about ANY shot? How many more agents do we need? I claim the answer is zero. ZERO! In other words — and this is really amazing — the claim is that $B_1,B_2,B_3,B_4$ will block any shot from the assassin to the target, not just the four closest shots! It only takes FOUR agents to protect Lattice #1!

Observing why this is true is the final step.

### Step #4: Make the final observation!

The goal in this step is to show that ANY trajectory from the assassin (origin) to the target (one of the  dots) in Lattice #1 must intersect one of the points in the lattice generated by either $B_1$, $B_2$, $B_3$, or $B_4$.

Here's the proof of that claim.

Consider an arbitrary shot from the assassin to the target. On the plane, this corresponds to a straight line segment from the origin to  a point of the form $(a+m,b+n)$ for some integers $m$ and $n$. To block this shot, let's place a new point $B$ at the midpoint of that line segment. Thus $B$ has coordinates $(\frac{a+m}{2} , \frac{b+n}{2})$.

Now here's the punchline! Depending on the parity of $m$ and $n$, the point $B$ is REALLY a copy of either $B_1$ or $B_2$ or $B_3$ or $B_4$!! In other words, this new security agent $B$ isn't so new after all! It lies in the lattice generated by one of the four blockers already chosen! Indeed:

For example, $m=1$ and $n=2$ in the illustration above. This puts us in case 4, so that $B=(\frac{a-1}{2},\frac{b+2}{2})$ is a point in the lattice generated by $B_2$. And this agent blocks that particular trajectory that we saw earlier! Indeed, notice that in the middle graphic below, the target is in one of the squares labeled 1, which guarantees it's a point in Lattice #1. Moreover, we can get to that square by starting at the original square, then moving up by two squares labeled 1 (this corresponds to $n=2$) and then moving right by one square labeled 1 (this corresponds to $m=1$).

BINGO! Once we place the first four security agents down, we automatically block any other shot in the first lattice. It takes just four agents to protect Lattice #1! (So that special integer $k$ we were looking for earlier is $k=4$.) And what's great is that the argument we used to protect Lattice #1 works for any lattice! In particular, this means we can protect Lattice #2, Lattice #3, and Lattice #4 using just four agents each! Therefore we need a total of $$4\times 4 = 16$$ agents to secure the square!

Conclusion? The square is a secure polygon!

## Food for thought

I didn't draw the lattices generated by the agents, but you might have a bit of fun doing this — see where they lie on the original plane (the one with all the colors) and convince yourself that every shot the assassin takes really is blocked! Also, we showed that 16 points will suffice, but could fewer than 16 work, depending on where the assassin and target are located? (For example, if the target is in the center of the billiard table, then you only get one lattice rather than four.)

You might also wonder about other (regular) polygons. Today we saw that the square is secure. But what about triangles? Pentagons? Hexagons? For which $n$ is an $n$-gon a secure polygon? I'll leave you to ponder the answers.

Until next time!

## An interactive demo! (added 7/23/18)

As if this puzzle weren't cool enough, Jeremy Kun of Math ∩ Programming recently made an interactive demo to go along with it! The gray dot represents the assassin, the green dot represents the target, and the 16 red dots are the security agents. If you hover your mouse over the grid, you'll see the trajectory of the assassin's shot. Below is a GIF of the demo. Check out the real thing here!

Notice how the trajectory never reaches the target? EVER. How cool (and totally unintuitive) is that? Even better, this holds no matter where the assassin and target are located. To see this, just refresh your browser window! The assassin and target will randomly change positions, and the guards will adjust accordingly. (You can also click and drag the green dot around. The agents shift position!) And they still block every possible shot, just as we predicted in our proof above.

Thanks so much for coding this, Jeremy!! It's really neat to see the dynamics in action.

If you're interested in Jeremy's code, check out his blog post, Visualizing an Assassin Puzzle!

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# Valuing urban green spaces using DCEs: Replicating Fruth et al. (2019)

In this blog post, I will replicate the results from Fruth et al (2019), a recently published choice experiment study.

Fruth et al. (2019) investigate preferences towards urban green features that have received little to no research focus: green facades, street greening or green initiatives to teach local communities. Urban green spaces improve the provision levels of urban ecosystem services, such as recreation and asthetics, climate regulation, food provision and freshwater cycling.

They conduct a choice experiment survey in Berlin, Germany, in a single neighborhood. The sample includes 128 respondents, which is unfortunately not representative of the neighborhood’s population.

The survey describes three attributes: the number of green building facades, street green in the form of trees or planter boxes, and green initiatives, such as eco-events and education programs. Each attribute had three levels, representing either the status quo or possible improvements (increases in greening). Moreover, the authors included a cost attribute described as a mandatory annual payment to an “urban green fund”, which would be managed by the local government. The cost attribute had six levels ranging from 12 € to 360 € (and zero for the status quo). In the choice card, respondents selected their preferred alternative, given the description in terms of attributes and cost. The picture below illustrates the choice card (as copied from the original paper):

The authors conducted an internet-based survey in 2018 and collected 128 completed questionnaires. The authors estimate a conditional logit model and a latent class logit model with 2 classes. Generally, the authors find that respondents are willing to pay for urban greening, thus planners should fund these green features. They also show that preferences are heterogeneous and the two classes have distinct preferences.

The original article webpage gives access to the original data. The original data is an RDS file, which means I have to have the readRDS function.

urbangreen <- readRDS(file = "DIRECTORY/gmnldata.rds")

I use the gmnl package to estimate a conditional logit model:

install.packages("gmnl")
library(gmnl)

The authors estimate the following utility function:

$U_{in} = ASC0 + \beta_1*Fac2_n + \beta_2*Fac3_n + \beta_3*Str2_n + \beta_4*Str3_n + \beta_5*Ini2_n + \beta_6*Ini3_n + \beta_7*Ini2_n*Age_i + \beta_8*Ini3_n*Age_i + \beta_9*Price_n + \beta_{10}*Price_n*Income_i + e_{in}$

In order to estimate the regression above, we need to create an interaction between the price variable and individual’s income (all other interactions are already included in the original data):

urbangreen$PRICEXINC <- urbangreen$PRICE * urbangreen$INCOME  Finally, we can run a conditional logit model by using the gmnl function: model_cl <- gmnl(pref1 ~ ASC + D.FACAD.2 + D.FACAD.3 + D.STREET.2 + D.STREET.3 + D.INITI.2 + D.INITI.3 + D.INITI.2xAGE + D.INITI.3xAGE + PRICE + PRICEXINC | 0, data = urbangreen, model = "mnl") summary(model_cl) The results are identical to those published in the paper: > summary(model_cl) Coefficients: Estimate Std. Error z-value Pr(>|z|) ASC 0.4876727 0.1583468 3.0798 0.0020716 ** D.FACAD.2 0.5514503 0.1151176 4.7903 1.665e-06 *** D.FACAD.3 0.7617965 0.1141175 6.6755 2.463e-11 *** D.STREET.2 0.8160700 0.1153793 7.0729 1.517e-12 *** D.STREET.3 0.8293647 0.1168767 7.0961 1.284e-12 *** D.INITI.2 0.2432129 0.1112497 2.1862 0.0288018 * D.INITI.3 0.1428693 0.1119393 1.2763 0.2018458 D.INITI.2xAGE -0.1791532 0.0588993 -3.0417 0.0023526 ** D.INITI.3xAGE -0.2153725 0.0590365 -3.6481 0.0002642 *** PRICE -0.0694812 0.0051119 -13.5920 < 2.2e-16 *** PRICEXINC 0.0063972 0.0024031 2.6621 0.0077653 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Optimization of log-likelihood by Newton-Raphson maximisation Log Likelihood: -1085.1 Number of observations: 1152 Number of iterations: 5 Exit of MLE: gradient close to zero Finally, we can obtain average willingness to pay for various attributes by using the wtp.gmnl function: > wtp.gmnl(model_cl , wrt = "PRICE") Willigness-to-pay respect to: PRICE Estimate Std. Error t-value Pr(>|t|) ASC -7.018777 2.467293 -2.8447 0.0044449 ** D.FACAD.2 -7.936689 1.719150 -4.6166 3.900e-06 *** D.FACAD.3 -10.964073 1.770363 -6.1931 5.898e-10 *** D.STREET.2 -11.745198 1.815254 -6.4703 9.782e-11 *** D.STREET.3 -11.936542 1.808887 -6.5988 4.144e-11 *** D.INITI.2 -3.500415 1.611808 -2.1717 0.0298759 * D.INITI.3 -2.056231 1.615658 -1.2727 0.2031282 D.INITI.2xAGE 2.578443 0.854826 3.0163 0.0025585 ** D.INITI.3xAGE 3.099726 0.864337 3.5862 0.0003355 *** PRICEXINC -0.092071 0.034314 -2.6832 0.0072925 ** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 The price variable has been scaled down, so the WTP estimates need to be multiplied by 10 and by -1. > as.data.frame(wtp.gmnl(model_cl , wrt = "PRICE"))$Estimate*(-10)
[1]  70.1877659  79.3668862 109.6407321 117.4519836 119.3654206  35.0041482  20.5623150 -25.7844318 -30.9972599   0.9207083

We conclude that on average respondents are willing to pay 79 euro to increase green facades by 1 out of 10 buildings with green facades, and 110 euro to have 2 out of 10 buildings with green facades. All of the estimates are statistically significant (except Green Initiatives 3, which represents increasing the frequency of eco-events).

To uncover patterns of preference heterogeneity, the authors also estimate a latent class model with 2 classes. again, I will use the gmnl package. The command is similar to the previous one, but the model has to be specified as “lc”, panel sets to true (T) and the number of classes to 2 (Q=2).

latent_model <- gmnl(
pref1 ~ ASC
+ D.FACAD.2 + D.FACAD.3
+ D.STREET.2 + D.STREET.3
+ D.INITI.2 + D.INITI.3 + D.INITI.2xAGE + D.INITI.3xAGE
+ PRICE + PRICEXINC
| 0 | 0 | 0 |  RESIDENT + CAR,
data = urbangreen,
model = "lc",
panel = T,
Q = 2
)

The regression output is identical to the study’s.

> summary(latent_model)

Estimate Std. Error z-value  Pr(>|z|)
class.1.ASC            0.5611493  0.2976105  1.8855 0.0593602 .
class.1.D.FACAD.2      0.4225581  0.2458068  1.7191 0.0856023 .
class.1.D.FACAD.3      0.6559344  0.2453457  2.6735 0.0075062 **
class.1.D.STREET.2     0.7881459  0.2451663  3.2147 0.0013056 **
class.1.D.STREET.3     0.4923308  0.2537271  1.9404 0.0523317 .
class.1.D.INITI.2      0.1219113  0.2349468  0.5189 0.6038383
class.1.D.INITI.3     -0.4667315  0.2763268 -1.6891 0.0912087 .
class.1.D.INITI.2xAGE -0.4304923  0.1163966 -3.6985 0.0002169 ***
class.1.D.INITI.3xAGE -0.6721978  0.1351556 -4.9735 6.575e-07 ***
class.1.PRICE         -0.2302094  0.0262598 -8.7666 < 2.2e-16 ***
class.1.PRICEXINC      0.0141420  0.0135507  1.0436 0.2966548
class.2.ASC           -0.8895071  0.2468243 -3.6038 0.0003136 ***
class.2.D.FACAD.2      0.7075329  0.1496369  4.7283 2.264e-06 ***
class.2.D.FACAD.3      1.0257079  0.1543898  6.6436 3.061e-11 ***
class.2.D.STREET.2     0.9235234  0.1485329  6.2176 5.047e-10 ***
class.2.D.STREET.3     1.1218419  0.1525922  7.3519 1.954e-13 ***
class.2.D.INITI.2      0.3133084  0.1441028  2.1742 0.0296900 *
class.2.D.INITI.3      0.4670330  0.1486033  3.1428 0.0016733 **
class.2.D.INITI.2xAGE -0.1616624  0.0900741 -1.7948 0.0726900 .
class.2.D.INITI.3xAGE -0.1797597  0.0920344 -1.9532 0.0507982 .
class.2.PRICE         -0.0647417  0.0076188 -8.4977 < 2.2e-16 ***
class.2.PRICEXINC      0.0029378  0.0040429  0.7267 0.4674337
(class)2               0.1998110  0.1956655  1.0212 0.3071660
RESIDENT:class2       -0.0694162  0.1255315 -0.5530 0.5802783
class2:CAR             0.6049938  0.1078917  5.6074 2.054e-08 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Optimization of log-likelihood by BFGS maximization
Log Likelihood: -861.45
Number of observations: 1152
Number of iterations: 473
Exit of MLE: successful convergence

The function to estimate WTP of each class for attributes is identical to before:

> as.data.frame(wtp.gmnl(latent_model , wrt = "class.1.PRICE")[1:9,])$Estimate*(-10) [1] 24.375603 18.355381 28.492947 34.236044 21.386217 5.295668 -20.274218 -18.700032 -29.199407 > as.data.frame(wtp.gmnl(latent_model , wrt = "class.2.PRICE")[12:20,])$Estimate*(-10)
[1] -137.39318  109.28546  158.43075  142.64734  173.27959   48.39359   72.13788  -24.97036  -27.76567

As seen on Table 5 of the paper, these WTP are identical to those reported in the paper. Individuals belonging to class 1 are willing to pay 34 euros to have mostly trees (every 10 meters), while individuals belonging to class 2 are willing to 143 euros for the same change.

The paper concludes that there are two distinct consumer profiles. While individuals in class 1 are not willing to pay much for street greening, individuals in class 2 are willing to pay significantly more for urban greening measures. This is one way to uncover preference heterogeneity. Increasing the number of classes (Q=) allows for more flexible patterns of preference heterogeneity (while it complicates convergence of the model).

References:

Fruth, E., Kvistad, M., Marshall, J., Pfeifer, L., Rau, L., Sagebiel, J., … & Winiarski, B. (2019). Economic valuation of street-level urban greening: A case study from an evolving mixed-use area in Berlin. Land Use Policy89, 104237.

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## Thursday, 28 January 2010

### Review

This week has been one of a glowing in-box, lots of tasty job offers, good news and interesting opportunities. What a joy.

The most recent being a review of my luxury this-a-job-really residency in the south of France, with French House Party.

Oh, the heat.

## Tuesday, 26 January 2010

### Unhappiest Day of the Year

Obviously the day itself isn't unhappy, but apparently yesterday (the Monday closest to 23rd January) - Blue Monday is the day when we of the northern hemisphere feel most unhappy.
And here's an equation to prove it

$\frac{[W + (D-d)] \times T^Q}{M \times N_a}$

Created by an unnamed mental health charity, where W = weather, d= debt, T= time since christmas and Q = our failing new year resolutions. Not too sure what all the other letters signify. But if you were unhappy yesterdau, this may go part way to explaiing why ...

Although for me, yesterday was pretty good, and I do struggle with this new year greyness.

Why pretty good? Because yesterday (parp of trumpets) was the day I spotted the first snowdrops in my garden.

## Sunday, 24 January 2010

### The Garden of Cosmic Speculation

I went to a talk by Charles Jencks yesterday, hosted by the Wordsworth Trust as part of their arts and books weekend. He's an architect, but I discovered him through his Garden of Cosmic Speculation, which is a garden inspired by mathematics, science, astronomy.

He was an absolutely wonderful speaker: funny, current, sharp and extremely stimulating.

When I say I discovered him through the garden, I should say I discovered the garden through a book. I haven't yet been to the garden. It's only open for half a day a year. I have the date in my diary for this year. And will make sure I spend as long as possible there.

It touches on so many themes that I'm finding myself increasingly interested in: macrocosms and microcosms, wave patterns, and man's ambigious relationship with nature: being both connected to and disconnected from it - in almost all we do. To experience those ideas in a consciously physical way, to walk through the manifestation of someone else's interpretation of the universe, is an opportunity to untangle some of my own thinking; to experience my half-baked ideas, and other theories in a visceral sense, absorb them through my pores. This is always my favourite way of experience, hence the sailing.

How much can poetry do this? I constantly aim for visceral language, but language by its essence forces a basic separation from physical to intellectual.

It's one of the reasons why I enjoy ee cummings - he presents emotional realities by bypassing the intellectual interpretation. He manages to set words down so they communicate emotionally and short-circuit analysis.

## Tuesday, 19 January 2010

### The Work Play Conundrum

Having seemingly survived half a foot of snow and three inch thick ice (snow and ice remain resolutely imperial in my mind), the good yacht Sunshine was subject to human overwintering at the weekend.

Fortunately we had already filled the engine cooling system with anti-freeze a week before the snowfall, tucking it up with a duvet as an extra preventative.

But she's an old boat and the snowmelt had found all the routes inwards, down the mast and under the cockpit seats being two preferred pathways. We'd removed the cushions too so she was looking an unappealing and unlikely pleasurecraft on Sunday.

But owning a boat (even only as a third share) is a good lesson in taking the good with the bad, the rough with the smooth, and work with play. Although I'm not quite sure when the play begins when sailing - certainly not at the chart table, the sail changes, the weather forecasting, or skippering inexperienced crew.

It's a bit like being a poet. When are you not a poet? Certainly not when you're asleep (that's often when I'm busiest), or watching telly (although I've given up on most of the ideas that floated by as I was watching the British Museum's History of the World), or being with friends (too many of them have become sounding boards or inspirations).

And there was me, in my twenties, vowing not to follow in my father's footsteps of living and breathing work. The Protestant Work Ethic is not just an ethic, it is a gene.

But then it isn't really work in that sense of management, salary and pension. It's work in the sense of a romanticised freedom; as in I'm on a boat and can go anywhere (the wind and sea permit); as in I'm a writer I can do anything (I'm capable of expressing). What is real freedom?

## Wednesday, 13 January 2010

### First Draft Drafted

Yey! I've drafted the first draft of the John Denver play. Draft being the operative word. It is very sketchy. And as usual I have a sheet of questions to consider when I read it through in a week or so's time.

I like this switch into editor. It really takes the pressure off me as a writer. I get to pose all sorts of dilemmas and theories to the text which I then (as writer) have to tackle - but not yet.

So the key dilemmas now (before reading) are: how fleshed or cliched are the characters? How obvious is their development? How interesting/illumintaing is what they say?

Pretty big, really. But not in my mind - in that I have the sketch. The next step is the oil and colour - the texture.

And, best of all, I'm ahead of schedule, so have time to cook the material. What I give to Suzy, the director, won't be a hastily written first draft, but hopefully one with a little more flesh she can then add her own layer and opinions to. There's no point passing work on in the state of flimsy two-dimensions for feedback when I can build it up and, hopefully, upping the ante for feedback. Suzy'll only respond to the material I give her, so the better the material, the better the response and there, with luck, the better the final script.

My ideal is that it becomes a play that people who aren't necessarily fans of John Denver will be interested in coming along to see as much as fans of John Denver.

So any suggestions for appeal in this field, as ever, welcome ...

## Monday, 11 January 2010

### New Discovery

It's not often that I pick up a poetry book and at almost every page turn think oh wow, oh, yes, yes, and then sob deeply at the next page, and then continue reading the whole book in my first sitting and then hold it in my hands, stroke the cover, admire that too, and open randomly and reread a poem and sigh.

Perhaps that's a good job. Life would be pretty exhausting if I did.

This weekend, instead of going up to Edinburgh (warned off because of the snow) I picked up Grain by John Glenday. I'd ordered it because I'd read a review and a poem from it in the Poetry Book Society, a great way of being introduced to a whole bunch of new books every quarter that I can either buy or not buy.

So, what got me about this? Well, it's always high anticipation at opening a book of someone I've never read before (apart from the good feeling generated from the PBS Bulletin). So it's already a little like opening a treasure box (especially given the lush cover, which this picture does not fully
reveal).

What really hit me in the reading was how John Glenday repeatedly stripped away not just the layers of our world, but of our time. Narratives ran backwards, images unpeeled shade, found the dark under light, the rhythms echoed what was behind words, and he was forever asking questions, doubting, uncertain as he did so.

I've been thinking a lot lately about poetry's contribution to our current society, not too enthralled by the few war poems I've read. Eco-poetry (despite its handy label), for me, displays an effective reflection of now, a banner for what is now (middle-class) mainstream concerns. But this book does something so much bigger, more consuming. It reflects the need for us to stop and reassess everything we know, strip it back to we can, individually measure its worth. And, more beautifully, it does it in a way that's so quiet, so tentative, that I just want to crawl inside each poem, alongside each word sound, sit there and listen with my eyes closed.

Yes, there is a predominance of nature poems. There is also writing on relationships, myth, religion and work. And all of them show a care in handling the language, in the respect of words and communication. This collection, of 43 poems, reminds me of my father's adage: only speak if you have something worth saying. John Glenday isn't as much speaking as sharing what he has witnessed and is trying to connect together.

A generous book. A great start to my year of reading.

## Saturday, 2 January 2010

### Another Murmuration

Just over a year ago, one of my first posts on this blog was of a visit to Leighton Moss where I watched thousands of starlings roost.

It seemed like a good thing to do on New Year's Day.

The pools and reedbeds were still frozen so walking around the reserve was a surreal spectacular entrance into an almost monochrome, almost sepia world. Very few birds about. Sensible really, it was very very cold.

We got to the public hide just as the sun was turning the sky an apricot mousse in the west, settled ourselves and opened the little window facing the woods in the east. Then I saw a smudge. And another. They swarmed and grew for the next five or ten minutes in the distance over the woods, with clear parameters, never swerving too far from a centre. All the time more fingerprints of birds adding to the mass.

Then they started to fly closer to us, spinning out, stretching the shimmering wider, thinner and then looping back into each other. Thousands of them. Back to the woods. Back to picking up more groups.

The turning, the there-are-thousands-suddenly-cutting-into-transparency movement of this huge group was hypnotising. Not so much that I couldn't leap up and out of the hide as they turned and flew towards us, then over us.

The spray of their wings, an airy cascade. The scattered plop of shit. The squeaking shrillness. A sky darkening. They swooped into the seaward reedbed and caught wave upon wave of themselves, calling, whistling as only they can, as they settled in. Wave upon wave. Low down, nearer to us, the numbers magnified. The reedbed still and tall, as they skittered through it.

It seemed a far larger display than last year. I felt closer (physically not emotionally) to them. There was more variation in their groupings and swoopings. Or maybe I was was more alert to it all.

Whichever, it has sent me into 2010 with a renewed sense of purpose about my work - to have patience, be focused on each project, enjoy current and seek new collaborations, listen intently and watch out for the bird shit.


text:
# Using built-in type Decimal or Entity type Advance, to hold AdvanceAmount

In a wage processing system which employees MVP pattern We have Model classes WageInfo, EarningInfo, DeductionInfo and AdvanceInfo and entity class Advance. (With several other classes)

EarningInfo and DeductionInfo has a composition relationship with WageInfo (contained in WageInfo) and also since Advance amount is a part of wages (paid in advance) I've modeled the relationship of Advance with DeductionInfo as aggregation because finally it becomes a deduction to the wages (advance amount should be deducted from wages).

public class DeductionInfo
{
public int EmployeeID { get; set; }
public string Name { get; set; }
public decimal LoanInstalmentAmount { get; set; }
public decimal UniformInstalmentAmount { get; set; }
public decimal InsuranceInstalmentAmount { get; set; }

}

{
public int AdvanceID { get; set; }
public decimal Amount { get; set; }
public DateTime EnteredDate { get; set; }
}

So when I create a DeductionInfo object I simply assign a value to AdvanceAmount property and I'm not initializing other properties of Advance class in this instance.

My question are...

1. Is the way I've modeled these class relationships logical and correct?
2. Whether I've properly implemented aggregation in DeductionInfo?
3. Since the advance amount is just a amount is it worth to use decimal type instead?

This question is very, very similar to another question you recently asked, so this answer will basically (hopefully!) clarify my other answer.

### KISS: Keep It Simple, Stupid.

These classes represent database records. With Entity Framework you would define the relationships like this (I'd drop the Info suffix):

public class Deduction
{
// primary key:
public int Id { get; set; }

public string Name { get; set; }
public decimal LoanInstalmentAmount { get; set; }
public decimal UniformInstalmentAmount { get; set; }
public decimal InsuranceInstalmentAmount { get; set; }

// foreign keys:
public int AdvanceId { get; set; }
public int EmployeeId { get; set; }

public virtual Employee Employee { get; set; }
}

{
// primary key:
public int Id { get; set; }

public decimal Amount { get; set; }
public DateTime EnteredDate { get; set; }

public virtual ICollection<Deduction> Deductions { get; set; }
}

EF works with conventions - it determines that a property called Id is mapped to your primary key, and that a property called [AnotherEntityTypeName]Id is a foreign key property when there's a virtual navigation property of that other entity type - just with the types involved and the names of the properties, EF is "smart" enough to determine that there's a one-to-many relationship between Deduction and Advance. Now as @Malachi pointed out, the question is is this really what you want?

Only you can answer that. We review code, not database schemas.

Per your other question I see that you're not using Entity Framework. I don't see how that would impact the design of your POCO classes though; at the end of the day they're just classes that convey data between the database and your app, and if you're using ADO.NET then you have full control over how the classes get instantiated and loaded with data, so the classes you're using can be anything your sweet heart desires. Ok anything your business logic needs.

For example, the below code returns an IEnumerable<Deduction>, where Deduction is defined as above, eager-loading the related Advance when it exists:

public IEnumerable<Deduction> GetAllDeductions()
{
var sql = "SELECT * FROM dbo.Deductions d LEFT JOIN dbo.Advance a ON d.AdvanceId = a.Id"
using (var connection = new SqlConnection(_connectionString)
{
connection.Open();
using (var command = new SqlCommand(sql, connection))
{
{
var deduction = new Deduction();

// assuming column names... this is just an example
{
}

yield return deduction;
}
}
}
}

This boilerplate-level gruntwork could all be automated with EF and look like this instead:

public IEnumerable<Deduction> GetAllDeductions()
{
using (var context = new MyDbContext(_connectionString))
{
return context.Deductions.Select(e => e).Include(e => e.Advance).ToList();
}
}

### ViewModels

What you seem to want to display is different. It's a ViewModel, something that does not represent database records, but something meant to be displayed.

Then you create another class to represent just that. It's possible that such a class looks very similar to the entity types; it's also possible that it contains properties from two or more entity types:

public class DeductionInfo // or DeductionViewModel
{
public int Id { get; set; } // only if the View needs it

public string Name { get; set; }
public decimal LoanInstalmentAmount { get; set; }
public decimal UniformInstalmentAmount { get; set; }
public decimal InsuranceInstalmentAmount { get; set; }
public decimal AdvanceAmount { get; set; }
}

You'll have a service whose job is to get the data, prepare your ViewModels and spit out these DeductionInfo instances so that your View can display them.

Bottom line: the View doesn't care about the database schema, and doesn't need to know about your entity types. It works with a ViewModel that only contains the data that it needs.

An alternative could be to create a complex type (basically, the DeductionInfo class above), and to have the service fetch the data directly into that type, and return that type for the view to consume: you only select the data you need, and you consume it directly. Works just as well. Which is best? You decide!

1. No. You have one class that needs another classes Properties for it's own Properties, Why aren't they the same Class/Object? take parameters from a constructor for filling in this information.

2. Why Aggregate DeductionInfo as a Class? The Name DeductionInfo doesn't describe the parts well. By that I mean that the things inside this class don't say to me that this is Deduction information.

3. Yes. That looks correct to me.

# Clarification on Question #1

An Advance and a Deduction are two different things one adds and one subtracts, and Advance should go in Earnings and not Deductions. it is an advance of earnings, not a deduction from earnings.

# Rants

Do you really want Advance to be public? Do you want just anything to change the values in that class? Is an Advance a Deduction?

It doesn't look like a good idea to me to have one classes properties be a reflection of another classes properties, why aren't they the same class?

I also don't think that you want to create a new Advance object inside of DeductionInfo all of the information will be blank or default values.

I don't see a reason why you would want your DeductionInfo setting your Advance.Amount it should not be set there.

public decimal AdvanceAmount
{

• So is it ok if I use a decimal type property in DeductionInfo instead of Advance object while having a separate class called advance? If its said that the relation ship is aggregation then how to model that in code please? Making advance public was a mistake really sorry for that. – CAD May 28 '14 at 16:11


text:
Sylow subgroups of projective general linear groups - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:21:05Z http://mathoverflow.net/feeds/question/88017 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups Sylow subgroups of projective general linear groups Alex Bartel 2012-02-09T17:40:47Z 2012-02-10T13:25:13Z
What is known about the Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$, where $q$ is any prime power? For example, according to Theorem 7.9 in Isaacs's Character Theory, these Sylow 2-subgroups cannot be generalised quaternion. Is there a classification of all these Sylows available?

Concretely, let me make the following semi-conjecture, which, if true, would make me very happy:

The irreducible complex representations of Sylow 2-subgroups of $\rm{PGL}_n(\mathbb{F}_q)$ have trivial Schur index.

Is this known? Any references to literature that discusses such questions would be very welcome.

If anyone has information on Sylow $l$-subgroups for arbitrary $l$, I would also be very interested.
 http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups/88030#88030 Answer by Jim Humphreys for Sylow subgroups of projective general linear groups Jim Humphreys 2012-02-09T19:56:01Z 2012-02-09T19:56:01Z
To reinforce Geoff Robinson's cautious comment, I'd encourage you to dig into a volume of the ongoing treatise by Gorenstein-Lyons-Solomon The Classification of the Finite Simple Groups, Number 3 (Amer. Math. Soc., 1998): Section 3.3 on "Equicharacteristic Sylow Structure" and Section 4.10 on "Cross-characteristic Sylow Structure", supplemented by references to the literature. The question raised concerns just one of the adjoint-type groups of Lie type, but even here the difficulty is apparent in the treatment by G-L-S of standard structural matters. As their section headings indicate, there are fundamental differences depending on whether or not the prime r (such as 2) involved in the Sylow subgroup is the natural/defining prime for the Lie-type group (having q as a power). But in either case it's already nontrivial to compute the r-rank and some of the commutator structure.

The references (as of 1998) seem to cover the essential literature, including the work of Aschbacher and the paper by Carter and Fong mentioned by Geoff. All of this is above my pay grade, as they say, so I can only wish you luck.
 http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups/88046#88046 Answer by Derek Holt for Sylow subgroups of projective general linear groups Derek Holt 2012-02-09T22:38:09Z 2012-02-09T23:58:09Z
For $q$ odd, it is not difficult to describe the structure of a Sylow 2-subgroup of $G={\rm GL}(n,q)$, so let me do that.

If $q \equiv 1 \bmod 4$, then the subgroup of monomial matrices of $G$ contains a Sylow $2$-subgroup of $G$, which is a wreath product of a cyclic group of order $t$ (where $t$ is the 2-part of $q-1$), and a Sylow 2-subgroup of the symmetric group $S_n$. The Sylow subgroups of $S_n$ are of course themselves built up as wreath products of cyclic groups.

It is a little more complicated when $q \equiv 3 \bmod 4$. In that case, let $t$ be the 2-part of $q^2-1$. Then, if $n$ is even, a Sylow 2-subgroup of $G$ is a wreath product of a Sylow 2-subgroup of ${\rm GL(2,q)}$, which is semidihedral of order $2t$, with a Sylow 2-subgroup of $S_{n/2}$. If $n$ is odd, then it is a direct product of a Sylow 2-subgroup of ${\rm GL}(n-1,q)$ with a cyclic group of order 2.

To get a Sylow 2-subgroup of ${\rm PGL}(n,q)$, you have to factor our the cyclic scalar subgroup, which is a diagonal of the base group of the wreath product.

For $q$ even, the upper unitriangular matrices form a Sylow 2-subgroup of ${\rm PGL}(n,q)$.

I don't know much about Schur indices but I did some quick computations in Magma, and I found that for Sylow 2-subgroups of ${\rm GL}(n,q)$ for small $n,q$, going up $(n,q)=(6,5)$ and $(8,3)$, the Schur indices of all irreducible representations are indeed 1, which seems to provide good experimental evidence for your conjecture. You might find it easier to try and prove it for ${\rm GL}(n,q)$, since the structure of the Sylow 2-subgroups can be described so explicitly, at least for odd $q$.

Added later. For odd prime $r$ not dividing $q$, let $d$ be minimal such that $r$ divides $q^d-1$, let $t$ be the $r$-part of $q^d-1$, and let $m = \lfloor n/d \rfloor$. Then a Sylow $r$-subgroup of ${\rm GL}(n,q)$ is a wreath product of a cyclic group of order $t$ with a Sylow $r$-subgroup of $S_m$.

For all of the classical groups, Sylow subgroups in coprime characteristic arise in a similar way, as wreath products of a group coming from a base case with a Sylow subgroup of a symmetric group. But of course there are lots of little complications for the individual types. I had a student a few years ago write Magma code to construct all of these, so I can produce very explicit descriptions!
 http://mathoverflow.net/questions/88017/sylow-subgroups-of-projective-general-linear-groups/88098#88098 Answer by F. Ladisch for Sylow subgroups of projective general linear groups F. Ladisch 2012-02-10T12:55:43Z 2012-02-10T13:25:13Z
The irreducibles of the Sylow 2-subgroups of ${\rm GL}(n,q)$, $q$ odd, have indeed trivial Schur indices: Let $S$ be such a Sylow 2-subgroup. First, observe that the natural module $(\mathbb{F}_q)^n$ splits into a sum of simple modules $U_1\oplus \dotsb \oplus U_l$, and the dimension of each simple module is a power of $2$. This shows that $S \cong S_1 \times \dotsb \times S_l$, where the $S_i$'s are Sylow 2-subgroups of a ${\rm GL}(2^k, q)$. Thus, w.l.o.g. we may assume that $n=2^k$. Then I use induction on $k$. First note that if $S$ is a Sylow 2-subgroup of ${\rm GL}(2^k, q)$, then $$T=\lbrace \begin{pmatrix} s & \\ & t \end{pmatrix} \mid s, t\in S \rbrace \cup \lbrace \begin{pmatrix} & s \\ t & \end{pmatrix} \mid s,t\in S\rbrace \cong S\wr C_2$$ is a Sylow 2-subgroup of ${\rm GL}(2^{k+1}, q)$, except when $k=0$ and $q\equiv 3\mod 4$. This follows from the description in Derek Holt's answer, but it can also be seen directly by observing that $T$ has the right order.
 Write $N=S\times S$, so that $T= C_2 N$, and let $\chi\in {\rm Irr} (T)$. Three cases have to be considered:
 1. We have $\chi_N \in {\rm Irr} (N)$. By induction, $\chi_N$ has trivial Schur index. By Lemma~10.4 in Isaacs' character theory book, for example, it follows that $\chi$ has trivial Schur index.
 2. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are not Galois conjugate. Then $\chi = \theta^T$ and $\theta$ have the same field of values and the same Schur index (again, see Lemma~10.4 in Isaacs).
 3. $\chi_N = \theta + \theta^g$, where $\theta$ and $\theta^g$ are Galois conjugate. This means that $\chi=\theta^T$, but $|\mathbb{Q}(\theta):\mathbb{Q}(\chi)|=2$. Now a representation $R\colon N\to {\rm GL}(\chi(1), \mathbb{Q}(\chi) )$ affording the character $\theta+\theta^g$ can be extended to a representation over the same field, since $N$ has a complement (of order 2) in $T$.
 To get the induction going when $q\equiv 3\mod 4$, one has to check that the Sylow 2-subgroup of ${\rm GL}(2,q)$ has trivial Schur indices, but that is clear since it's a semidihedral group.


text:
# Speiser's criterion

A. Speiser (Math Annahlen 110 (1934) 514-521) proved that the Riemann Hypothesis is equivalent to the non-vanishing of the derivative in the left-half of the critical strip Levinson and Montgomery [MR 54 #5135] gave an alternative, quantitative version of this result which led to Levinson's [MR 58 #27837] discovery of his method for counting zeros on the critical line. He then proceeded to prove that at least 1/3 of the zeros of are on the critical line.

Back to the main index for The Riemann Hypothesis.


text:
American Institute of Mathematical Sciences

November  2010, 14(4): 1511-1535. doi: 10.3934/dcdsb.2010.14.1511

Uniform stabilization in weighted Sobolev spaces for the KdV equation posed on the half-line

 1 Instituto de Matemática, Universidade Federal do Rio de Janeiro, P.O. Box 68530, CEP 21941-909, Rio de Janeiro, RJ, Brazil 2 Institut Élie Cartan, UMR 7502 UHP/CNRS/INRIA, B.P. 239, F-54506 Vandoeuvre-lès-Nancy Cedex

Received  August 2009 Revised  December 2009 Published  August 2010

Studied here is the large-time behavior of solutions of the Korteweg-de Vries equation posed on the right half-line under the effect of a localized damping. Assuming as in [19] that the damping is active on a set $(a_0,+\infty)$ with $a_0>0$, we establish the exponential decay of the solutions in the weighted spaces $L^2((x+1)^mdx)$ for $m\in$N* and $L^2(e^{2bx}dx)$ for $b>0$ by a Lyapunov approach. The decay of the spatial derivatives of the solution is also derived.
Citation: Ademir Fernando Pazoto, Lionel Rosier. Uniform stabilization in weighted Sobolev spaces for the KdV equation posed on the half-line. Discrete & Continuous Dynamical Systems - B, 2010, 14 (4) : 1511-1535. doi: 10.3934/dcdsb.2010.14.1511
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2019 Impact Factor: 1.27


text:
In [5]:
import numpy as np
import matplotlib.pyplot as plt
%matplotlib inline

We're going to start with generating a random asset price:

In [9]:
def generate_brownian_asset_price(S, sigma, r, T):
"""
S     : current stock
sigma : volatility
r     : rate
T     : time
"""
ret = S * np.exp((r - 0.5 * sigma**2) * T + sigma * np.sqrt(T) * np.random.randn())
return ret

Now set some values for the current stock price S, the volatility sigma, the interest rate r, and the expiration date T (in terms of days away from the current date)

In [10]:
import datetime
S = 857.29
sigma = 0.2076
r = 0.0014
T = (datetime.date(2013,9,21) - datetime.date(2013,9,3)).days / 365.0

Is it anywhere close? yep:

In [13]:
generate_brownian_asset_price(S, sigma, r, T)

Out[13]:
833.91332634495018

Next, let's make the call payout: $$C_t = e^{-r(T-t)} E[max(0, S_T - K)]$$ But we'll adjust for the risk-free interest rate (or discount rate) at the end.

In [1]:
def call_payout(S_T, K):
return max(0, S_T - K)

In [2]:
call_payout(863.91, 860)

Out[2]:
3.909999999999968


text:
# Gold reserves in central banks – 2020 survey results

Reserve managers share their views on future gold holdings, target allocations, purchasing and storage approaches, the use of ETFs, and the impact of Covid-19, in the results of a new joint Central Banking-Invesco survey. By Nick Carver and Robert Pringle

This article reports the findings of a joint survey of central banks carried out by Central Banking, with the support of Invesco, in August 2020. Those that took part did so on the condition that neither their names nor those of their central banks would be mentioned.

### Executive summary

• The Covid‑19 pandemic has not, in the main, changed the view of central banks on gold, although almost one-quarter of respondents said they view gold as a more attractive asset.
• Central bankers typically expect central bank gold holdings to increase over the next 12 months; no respondent expected a decrease.
• When determining a central bank’s gold holding, the benefits of diversification stand out as the most relevant factor for reserve managers.
• One in three central banks said they maintain a target allocation for gold: this rose to 39% of respondents when only those holding gold were considered.
• Purchases in the global market are by far the most popular means of buying and selling gold, with derivatives second.
• Overseas storage at a central bank is the preferred way to store gold: more than 80% of respondents said they did this.
• Central banks are positively disposed to gold exchange-traded funds (ETFs), but active interest in investing is the preserve of a minority.
• A combination of gold’s quality as a hedge against the US dollar and ETFs’ cost-effectiveness are the main benefits for holding ETFs.
• Liquidity risk is the chief concern associated with holding gold ETFs, although larger holders say safety regarding physical gold is more of an issue.

### Profile of respondents

The Gold for Central Banking 2020 survey questionnaire was distributed in August 2020 and had received replies from 26 central banks by the first week of September. The reserve managers that replied did so on condition of anonymity. The 26 central banks are, as a group, responsible for just over $2.6 trillion in reserves as of June 2020.1 The average holding was just over$100 billion. Collectively, the banks are responsible for $344 billion worth of gold, with an average holding of$13 billion. The group included six central banks with no gold reserves. A breakdown of respondents by national income classification and geography is set out below.

1. International Monetary Fund International Financial Statistics, with gold valued at market prices.
Percentages in some tables may not total 100 due to rounding.

The Covid‑19 outbreak has not changed how central banks see gold. This was the view of just over three-quarters of respondents, responsible for $2 trillion in reserves. Several respondents from this group noted that gold was an elemental part of their reserve management framework and as such was not impacted by market moves. A reserve manager from Asia explained: “Gold has been an integral part of our reserves regardless of market condition, and it may likely continue to be so.” A respondent from central Europe said their central bank remained committed: “We were very positive on gold in terms of the role it plays in any foreign reserve portfolio, even before Covid‑19, and that has not changed.” Another reserve manager from Europe had seen their view of gold’s strength in a crisis confirmed: “Gold is seen as a safe-haven asset in times of financial and political stress. Current market conditions have no impact on its strategic role in foreign reserves, other than to reaffirm the validity of the initial view.” A contrarian view was offered by a reserve manager from Africa, whose central bank had yet to be persuaded on the benefits of holding gold: “In our view, gold does not provide a regular income, has high volatility and, with the Covid‑19 outbreak, its price has risen substantially, making it unattractive to buy for our portfolio.” Six respondents did however say their central bank’s view had changed. The reserve holdings of this group ranged from$3 billion to more than $300 billion and they were mostly drawn from middle-income countries. In comments, two reserve managers, from Africa and Europe respectively, noted that an increase in the price of gold had strengthened its safe-haven status. ### What issues are most relevant in determining your institution’s gold holdings? The benefits from diversification stand out as the most relevant factor for reserve managers in determining their institution’s holdings of gold. Just under two-thirds of respondents – a group of central banks responsible for$1.9 trillion, and with an average holding of $134 billion – said this was “very relevant”. This group, which included several large holders, was dominated by central banks from middle-income countries. It included over half the respondents from the Americas. Low real interest rates also figure prominently in decision-making: just over 40% said this was very relevant. This group was also dominated by central banks from middle-income countries, and there was significant overlap with views on diversification benefits: all but one of the nine also said diversification benefits were very relevant. The nine were smaller reserve holders, however, holding$85 billion on average. Eight respondents reported that gold’s contribution to the risk of a portfolio was very relevant for their central bank, a group that contained several very large holders and was responsible for $1.4 billion. All eight also said that diversification benefits were very relevant. A similar pattern was observed when looking at the central banks with gold holdings. Of the 18 that answered this question, 61% placed diversification benefits first, followed by low real interest rates (50%) and then contribution to portfolio risk (39%). Viewing the results through the prism of national income classification revealed a variation in views, however. Diversification benefits scored highest for reserve managers from high- and middle-income countries. Immediately below this, those from high-income countries placed more emphasis on the contribution to the risk of the portfolio, but for middle-income countries, low real interest rates matter more. Just over half of respondents said “historical performance” was relevant, and several mentioned legacy or legal requirements in their comments. “Decline in oil prices” and “domestic political challenges” were not, however, seen as relevant for this group, and there was only marginal support for “weakening of the US dollar”, “Covid-19 pandemic”, “declining global growth” and “inflation risk”. ### How will global central bank gold reserves change over the next 12 months? There is little prospect of central banks reducing their gold holdings over the next 12 months, with expectations centred on stock increasing or remaining flat. Just over 60% of respondents – a group of central banks responsible for$1.7 trillion in reserves – said they expect gold holdings to increase over the next 12 months. The percentage was slightly higher among gold holders at 70%. The group of 16 was dominated by central banks from high-income countries (10), with the remainder from lower middle-income (4) and upper middle-income countries (2).

A reserve manager from Europe, whose central bank has significant gold reserves, explained what they saw as the two driving forces behind this: “The growing demand from emerging-market central banks reflecting current geopolitical, political and economic conditions, as well as structural changes in the global economy, will probably last in the future.” Six reserve managers, of which two-thirds were from upper middle-income countries, said they thought holdings would remain the same. Indeed, among upper middle-income countries, this was the majority view: only one in three among this group expect gold holdings to increase. In contrast, all four reserve managers from lower middle-income countries expect central banks to hold more gold in 12 months’ time. Three of the four reserve managers who were “unsure” were from high-income countries.

### What are the main benefits for central banks that hold ETFs?

A combination of gold’s quality as a hedge against the US dollar and cost-effectiveness are considered the main benefits for holding ETFs. Over half the survey respondents selected these two options, and three-quarters of the sample of 15 chose at least one of them. This group of 12 was responsible on average for reserves with $46 billion and gold worth$19 billion. It is notable that those mentioning the hedge against the dollar tended to be smaller holders on average of both reserves and gold than those that viewed cost-effectiveness as a main benefit. Liquidity, cited by six respondents, was typically a benefit for larger holders: three of these six were responsible for reserves worth more than $100 billion. In contrast, all but one of the six that saw reduction of the risk against the gold price as a benefit were responsible for less than$10 billion. Among those that said their institution had a positive view on gold ETFs, there was marginally more support for the hedge against the US dollar than cost-effectiveness, but the pattern broadly followed that in the previous table (What are the main benefits for central banks that hold ETFs?).

### What is the greatest concern for your central bank regarding holding gold ETFs?

Liquidity risk is the chief concern associated with holding gold ETFs: over half of respondents ranked this either first or second. These six were on average smaller holders, at $15 billion in reserves and$270 million in gold, than the survey average. The second greatest concern was that gold was not as safe as physical gold, and, indeed more respondents (five) ranked this first than did liquidity risk (three). These five were on average much larger holders – three held reserves worth more than $100 billion – and were all from Europe. Counterparty risk was third overall and ranked first or second by five respondents. Four of these five were middle-income countries and, as a group, tended to be smaller holders – on average responsible for$30 billion in reserves and $7.5 billion in gold. No single central bank saw the lack of rights of redemption as the primary concern, with only two ranking it second. There was some variation across geography and income groups. Central banks in Europe were unanimous in their concern that ETFs were not as safe as gold: all five European respondents ranked this first, with trust in the custodian the second greatest concern. In contrast, respondents from Asia were more concerned about counterparty risk and as concerned about liquidity risk as they were safety. Large holders, with more than$10 billion, were most concerned about safety compared to physical gold, whereas for smaller holders liquidity was the main concern.

This feature forms part of the Central Banking focus report, Gold for central banking 2020

Only users who have a paid subscription or are part of a corporate subscription are able to print or copy content.

To access these options, along with all other subscription benefits, please contact info@centralbanking.com or view our subscription options here: http://subscriptions.centralbanking.com/subscribe

#### More on Gold

##### Sanctions fear drives surge in central bank gold purchases

China, Turkey and India are largest declared buyers, while Russia probably bought big quantities

##### PBoC reports rise in gold reserves for first time since 2019

Analysts say the move is part of China’s ongoing efforts to diversify its vast dollar-denominated assets

##### The canary in the goldmine

Gold accumulation may herald broader concerns about dollar holdings, writes Jennifer Johnson-Calari

##### Central banks’ gold demand hits record level

Purchases in 2022 reach 673 tonnes, higher than in any full year since 1967

##### Bolivian ex-governor calls opposition audit request ‘nonsense’

Most central bank gold reserves are held outside country, Morales says

##### Bolivian legislators ask to inspect gold reserves

Opposition politicians claim government wants to monetise and spend gold

##### Will the dollar remain the world’s reserve currency?

Bank of Russia sanctions raise questions about the US dollar’s central role

##### Bolivian central bank wants to become sole buyer of gold

Proposal would allow BCB to use gold reserves as collateral or swap them for foreign exchange


text:
Count ways to increase LCS length of two strings by one
Medium Accuracy: 21.6% Submissions: 162 Points: 4

Given two strings S1 and S2 of lower alphabet characters of length N1 and N2, we need to find the number of ways to insert a character in the first string S1 such that length of LCS of both strings increases by one.

Example 1:

Input:
N1 = 4
S1 = abab
N2 = 3
S2 = abc
Output:
3
Explanation:
LCS length of given two
strings is 2. There are 3
ways of insertion in str1,to
increase the LCS length by
one which are enumerated below,
str1 = “abcab” str2 = “abc” LCS length = 3
str1 = “abacb” str2 = “abc” LCS length = 3
str1 = “ababc” str2 = “abc” LCS length = 3

Example 2:

Input:
N1 = 6
S1 = abcabc
N2 = 4
S2 = abcd
Output:
4
Explanation:
LCS length of given two
strings is 3. There are 4
ways of insertion in str1,to
increase the LCS length by
one which are enumerated below,
str1 = “abcdabc” str2 = “abcd” LCS length = 4
str1 = “abcadcb” str2 = “abcd” LCS length = 4
str1 = “abcabdc” str2 = “abcd” LCS length = 4
str1 = “abcabcd” str2 = “abcd” LCS length = 4

You don't need to read input or print anything. Your task is to complete the function waysToIncreaseLCSBy1() which take string S1 and string S2 of length N1 and N2 respectively as input parameters and returns the number of ways to insert a character in the first string S1 such that length of LCS of both strings increases by one.

Expected Time Complexity: O(N1 * N2)
Expected Space Complexity: O(N1 * N2)

Constraints:
1<= N1, N2 <=100
S1 and S2 contains lower case English character

We are replacing the old Disqus forum with the new Discussions section given below.


text:
Adversarial Combinatorial Bandits with General Non-linear Reward Functions

Yanjun Han · Yining Wang · Xi Chen

Keywords: [ Applications -> Computer Vision ] [ Applications -> Computational Photography ] [ Theory -> Online Learning Theory ]

Abstract: In this paper we study the adversarial combinatorial bandit with a known non-linear reward function, extending existing work on adversarial linear combinatorial bandit. {The adversarial combinatorial bandit with general non-linear reward is an important open problem in bandit literature, and it is still unclear whether there is a significant gap from the case of linear reward, stochastic bandit, or semi-bandit feedback.} We show that, with $N$ arms and subsets of $K$ arms being chosen at each of $T$ time periods, the minimax optimal regret is $\widetilde\Theta_{d}(\sqrt{N^d T})$ if the reward function is a $d$-degree polynomial with $d< K$, and $\Theta_K(\sqrt{N^K T})$ if the reward function is not a low-degree polynomial. {Both bounds are significantly different from the bound $O(\sqrt{\mathrm{poly}(N,K)T})$ for the linear case, which suggests that there is a fundamental gap between the linear and non-linear reward structures.} Our result also finds applications to adversarial assortment optimization problem in online recommendation. We show that in the worst-case of adversarial assortment problem, the optimal algorithm must treat each individual $\binom{N}{K}$ assortment as independent.


text:
Tag Info

27

Disclaimer: I'm not an expert in number theory. Short answer: If you're willing to assume "reasonable number-theoretic conjectures", then we can tell whether there is a prime in the interval $[n, n+\Delta]$ in time $\mathrm{polylog}(n)$. If you're not willing to make such an assumption, then there is a beautiful algorithm due to Odlyzko that achieves $n^{1/... 21 ($=$is a logical symbol, hence I will not write it as part of the signature.) The satisfiability problem is decidable, as$\gcd$has both a universal and an existential definition in terms of$|$,$+$, and$\le$:$\begin{align*} \gcd(a,b)=c&\iff c\ge0\land c\mid a\land c\mid b\land\forall d\:(d\mid a\land d\mid b\to d\mid c)\\ &\iff c\ge0\land c\... 18 This problem has a variation with a single integer input: Doesn$have a divisor strictly in between its two largest prime factors? The idea is to use the same randomized reduction from subset sum described in the top answer to the linked question, but with the target range encoded as the largest two primes instead of given separately. The definition ... 18 First, the name of the conjecture is "Hartmanis-Stearns", not "Hartmanis-Stearn". Second, the Hartmanis-Stearns conjecture concerns those real numbers computable by a multi-tape Turing machine in real time; in other words, the TM must compute the n'th digit in n time. Third, the result of Adamczewski et al. is only about finite automata and deterministic ... 17 One interesting example from number theory is expressing a positive integer as a sum of four squares. This can be done relatively easily in random polynomial time (see my 1986 article with Rabin at https://dx.doi.org/10.1002%2Fcpa.3160390713), and if I remember correctly, there is now even a deterministic polynomial-time solution. But counting the number ... 17 A very nice and simple example from Graph Theory is counting the number of Eularian circuits in an undirected graph. The decision version is easy (... and the Seven Bridges of Königsberg problem has no solution :-) The counting version is #P-hard: Graham R. Brightwell, Peter Winkler: Counting Eulerian Circuits is #P-Complete. ALENEX/ANALCO 2005: 259-262 14 First of all, there is a formal definition of "quantum-NC", see QNC on the zoo. GCD is indeed a good candidate for a problem that could be shown to be in QNC, but it's not known to be in NC. However, finding a QNC algorithm for GCD is still an open problem. The feeling for which this is believed to be true comes from the fact that the Quantum Fourier ... 11 We can show that if all$\alpha_i$are different, then square removal and factoring of$n$are equally hard. It is obvious, that if we can factor$n$, we can also compute square removal of$n$. The other direction is a bit more tricky. First compute the square removal of$n$and let's call this$m$. From the definition it follows that$m$divides$n$. ... 10 First note that this algorithm only computes$\lceil \log_2 v \rceil$, and as the code is written, it works only for$v$that fit in a$32$-bit word. The sequence of shifts and or-s that appears first has the function of propagating the leading 1-bit of$v$all the way down to the least significant bit. Numerically, this gives you$2^{\lceil \log_2 v \rceil}...

10

I believe no polynomial algorithm is known. According to a paper this is used in at least one cryptosystem: Abstract. We propose a cryptosystem modulo $p^k q$ based on the RSA cryptosystem. We choose an appropriate modulus $p^ k q$ which resists two of the fastest factoring algorithms, namely the number field sieve and the elliptic curve method. ...

10

Of course $k \geq 2$ here. There once was a manuscript by Horváth that claimed to solve the problem, but it was unclear in several places and to my knowledge was never published. As far as I know, the problem is still open. One direction of the implication is easy, of course.

9

I'm not sure this is a statement about primes so much as it is a statement about secret key generation: if the method is deterministic (e.g. take the smallest prime larger than 10^20), then your adversary can simply reproduce the computation to find your secret key.

9

See my paper with Eric Bach, "Factoring with cyclotomic polynomials", where we show that if the cyclotomic polynomial $\Phi_k(p)$ is $B$-smooth for any $p$ dividing $N$, then we can factor $N$ in time polynomial in $\log N$ and $k$ and $B$. In particular this gives a $(p+1)$-method (see the earlier work of Williams) and $(p^2+1)$ method. http://www.ams.org/...

9

There are essentially only two algorithms that I'm aware of: Use repeated-squaring, along the lines you mentioned. Factor $n$ using a state-of-the-art algorithm, then use the Chinese remainder thoerem. If $p$ is prime, you can compute $a^{b^c} \bmod p$ efficiently by computing $b^c \bmod p-1$ using fast exponentiation, call the result $d$, then computing $... 9 A something that might be too long for a comment, based on the previous answer by Emil. In the case you are interested in the complexity of such a logic, consider reading LICS'2015 paper by Joël Ouaknine, Antonia Lechner and Ben Worrell. A preprint is available here: https://www.cs.ox.ac.uk/people/james.worrell/LICS-main.pdf According to the authors, the ... 8 TL;DR The decimal expansion of a fixed rational number is not pseudorandom in the cryptographic sense, but irrational numbers (are conjectured to) exhibit some weaker but interesting forms of pseudorandom behavior. Roughly speaking, a sequence$s \in \{0, \ldots, B\}^n$is pseudorandom with respect to distinguishers$\cal A$, if it cannot be distinguished (... 8 The textbook of Computational Complexity: Modern Approach, by Arora and Barak gives such example: They define the decision problem Integer Factoring on input of three positive integers,$\text{Integer Factoring} = \{\langle L, U, N \rangle \;|\; (\exists \text{ a prime } p \in \{L, \ldots, U\})[p | N]\}$. They state that Alon and Kilian showed that if ... 8 Here's a truly excellent example (I may be biased). Given a partially ordered set: a) does it have a linear extension (i.e., a total order compatible with the partial order)? Trivial: All posets have at least one linear extension b) How many does it have? #P-complete to determine this (Brightwell and Winkler, Counting Linear Extensions, Order, 1991) c) ... 7 Some comments (not really an answer). Let's classify 32-bit integers$c$as follows: Type X:$c$(as a binary string) is De Bruijn sequence (for all rotations, bits [27,31] are distinct). An example: 11111011100110101100010100100000 Type Y: bits [27,31] of$2^i \cdot c$are distinct for$i = 0, 1, ..., 31$. This is what Leiserson et al. uses. Examples: ... 7 Your problem seems a special case of the turnpike reconstruction problem (for which no polynomial time algorithm is known). See for example: Shiteng Chen, Zhiyi Huang, and Sampath Kannan, "Reconstructing Numbers from Pairwise Function Values". Abstract: The turnpike problem is one of the few natural problems that are neither known to be NP-complete nor ... 7 Somewhat late in the day but the following paper by Allender, Saks, Shparlinski proves that (among other lower bounds) that GCD is not in$\mathsf{AC}^0$or$\mathsf{AC}^0[p]$for any prime$p$. 7 [Certainly not a complete answer, but too long for a comment] Testing whether a given DFA accepts the base-2 representation of at least one prime number is not known to be computable. If it were uncomputable, that's some kind of weak evidence that there's no "regular-ish" formula for primality. (I mean, we know the set of primes itself is not regular, but ... 6 Concerning your second question, problems such as Monotone-2-SAT (deciding of the satisfiability of a CNF-formula having at most 2 positive literals by clause) is completely trivial (you just have to check if your formula is empty or not) but the counting problem is #P-hard. Even approximating the number of satisfying assignments of such formula is hard (see ... 6 (I understand the description of the problem so that the input numbers are bounded by a constant, so I will not track dependence on the bound.) The problem is solvable in linear time and logarithmic space using sums of logarithms. In more detail, the algorithm is as follows: Using binary counters, count the numbers of occurrences of each possible input ... 6 This is impossible. No finite number of bits of$f(a_0,\dots,a_n)$suffices to determine any of$a_0,\dots,a_n$; in fact, any nondegenerate real interval contains the values$f(a_0,\dots,a_n)$for infinitely many vectors$(a_0,\dots,a_n)\in\mathbb Z^{n+1}$, and this holds even you fix all but two of the$a_0,\dots,a_n$in advance. 5 Update: The description below is for a different problem (in which you have all pairwise distances in a set rather than pairwise distances between two distinct sets). I'll leave it up anyway since it is closely related. This problem is called the beltway problem, and is a special case of the general$d$-torus embedding problem. It is also closely related to ... 5 There have been interesting developments on this problem, however Replacing$AC^0$with ACC(2) (Namely allowing mod 2 gates as well) is still well out-of-reach. Some progress beyond Ben Green's theorem can be found in this MO question https://mathoverflow.net/questions/57543/walsh-fourier-transform-of-the-mobius-function as well as this one https://... 5 Consider the function$T: \mathbb N \rightarrow \mathbb N$, where$T(n)=n/2$when$n$is even and$T(n)=n+1$when$n$is odd. Then it is known that for any$n \in \mathbb N$, there exists a$k \in \mathbb N$such that$T^{(k)}(n)=1$. If instead of$T(n)=n+1$when$n$is odd, we had defined$T(n)=3n+1$when$n\$ is odd, we would have the Collatz Conjecture, ...

5


text:
#### Cannot Register Document

Windows XP Home.  Dell Inspiron Laptop.  MS Office 2003.

I have just finished a recovery installation on my daughters laptop.
Everything worked fine - including reactivating Office by telephone -
however when I open any document in Word 2003 (not Excel) I get the
message...:

"This document could not be registered. It will not be possible to create
links from other documents to this document."

I have checked in Options etc but I cannot find any reference to
registering. I do not know what this is.

Can anyone help me with this issue?

TIA for any help.

Ed

 0
Ed
1/8/2010 10:01:01 AM
word.newusers 801 articles. 0 followers.

0 Replies
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Can anyone help me figure out how to print the MS Money register? I want to print all the information (ie - date, amount, payee, category) just like it appears on the register. I don't need a report just a print out of the register so I can compare with my bank statement. Thanks. Rick While viewing the register, you can click File and Print, but if you do this, what you get is what you get. If you use an Account Transactions report, you can customize it to show the date range, account, category, etc. that you choose. "Rick Vila" <rick@rickvila.com> wrote in message ...

Duplicate Entries in Register
I dint know what i did, but i now have duplicate entris for all my transactions in my register. This is very frustrating !! I do not use MSN and only downlaod directly from my bank - HELP ...

opening document has gridlines how do I remove them
When I opend Microsoft word, new document page isn't blank, how do I return it to a blank page. It has little gridlines all over it. Vickey, On the drawing toolbar, click Draw button, choose grid, remove the check next to show gridlines. Luc Sanders MVP - PowerPoint "Vickey" wrote: > When I opend Microsoft word, new document page isn't blank, how do I return > it to a blank page. It has little gridlines all over it. For Microsoft Word 2007 the gridlines checkbox is in the View tab in the Show/Hide box. Charlie ...

PM_Mass_Voucher_Save_Button trigger is not registered
I just updated to V10, SP3 and now when I enter Dynamics GP, I get the above message. I click OK and Dynamics seems to be running fine. Any idea how to correct the above message? thanks! -- Doug Do you have PSTL (Professional Services Tools Library ) loaded? If so, had you enabled the PM Minimum Check amount feature? If so, you need to get a revised version of the PSTL dictionary. This is not incorporated within the DynamicsGP10 SP3 download. https://mbs.microsoft.com/customersource/serviceplans/professionalservices/noam_pstl.htm?printpage=false ------ Robert "Doug" <...

Using 2007 Word to open 2003 Word document
I just installed Office 2007. When I opened any Word 2003 document, the formatting seems to be all skewered. For example, a page breaks off in the middle, and a five page document becomes five and half. Will someone please tell how to us Word 2007 in opening a Word 2003 document without the former formatting? The difference is almost certainly caused by differences in the Printer Driver. (It also points to some probably irrational formatting in the original document). -- Hope this helps. Please reply to the newsgroup unless you wish to avail yourself of my services o...

MS Outlook Registered User
I've changed the registered user info in the registry before, but have fogotten where it is !! Can anyone remind me which key ? This is the name that shows when starting up. This Product is registered to - I want to change it back to Authorised User and not an individuals name. The registry key uses Hex for setting this info. Ta ...

Item transaction entry
We are on GP 9.0, and have several users that are getting the same message on the Item Transaction Entry screen. Whenever they pull it up and try to change the Document Type, they get the message "You don't have access to this Document type." We've checked Advanced Security but haven't found anything that jumps out at us that might cause the message. Has anyone seen this or do you know how to correct it? thanks, Hi Nick, Do you know whether you have customisations or third party module integrated with your Inventory Control Management. Customisatio...

Open any document type from Access 2K2
My databases can contain attachments such as PDF, doc, xls, jpg. At the moment I just open these within Internet Explorer but I would really prefer to open them in the associated application directly. I could have a big IF block to create an instance of the application with VBA but surely there is a better way. I'm comfortable with API calls, etc so any method is appreciated. Thankyou, Luke To launch any document from ms-access, simply go: application.FollowHyperlink "path name to document" -- Albert D. Kallal (Access MVP) Edmonton, Alberta Canada pleaseNOOSpamKallal@ms...

possible to have different page colors in the same document?
For MS07...Is it possilbe to have different page colors in the same document?? For other tasks (margins, page size etc.) there are options like 'selected section only' or 'whole document' but not for page color...or am i missing something??? SOS...thanks The page color is primarily for web pages and you can have only one such color per document. -- Stefan Blom Microsoft Word MVP "DCB" <DCB@discussions.microsoft.com> wrote in message news:E7B82DC8-ABF8-4055-A78E-4B1E542F5E71@microsoft.com... > For MS07...Is it possilbe to have dif...

Scan multiple pages into one Word 2010 document
How do I scan multiple pages into one Word 2010 document? Thanks Partha One at a time - unless you are using OCR software that has a multi-page facility. -- <>>< ><<> ><<> <>>< ><<> <>>< <>><<> Graham Mayor - Word MVP My web site www.gmayor.com Word MVP web site http://word.mvps.org <>>< ><<> ><<> <>>< ><<> <>>< <>><<> "mcp111" <mcp111@discussions.microsoft.com> wrote in mes...

How do I eliminate an unwanted second page of a document in word.
Let say I type a simple letter document in word 2003. My letter goes on to a second page, but after editing and changes I really have just a one page letter. However my document now has two pages, but my second page is blank. How do I get rid of that second blank page? Turn on Show Non-Printing Characters (Ctrl-Shift-8), and delete all the extra paragraph marks at the end. There are other ways to get blank pages, but extra stuff at the bottom is the most likely with a simple letter. On Dec 10, 10:32=A0am, Paper Man II <Paper Man I...@discussions.microsoft.com> wrote...