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# Simulation Triggers¶

It is often useful to globally coordinate debug and instrumentation features using specific target-events that may be distributed across the target design. For instance, you may wish to enable collection of synthesized prints and sampling of AutoCounters simulataenously when a specific instruction is committed on any core, or alternatively if the memory system sees a write to a particular memory address. Golden Gate’s trigger system enables this by aggregating annotated TriggerSources distributed throughout the design using a centralized credit-based system which then drives a single-bit level-sensitive enable to all TriggerSinks distributed throughout the design. This enable signal is asserted while the design remains in the region-of-interest (ROI). Sources signal the start of the ROI by granting a credit and signal the end of the ROI by asserting a debit. Since there can be multiple sources, each of which might grant credits, the trigger is only disabled when the system has been debited as exactly as many times as it has been credited (it has a balance of 0).

## Quick-Start Guide¶

### Level-Sensitive Trigger Source¶

Assert the trigger while some boolean enable is true.

  import midas.targetutils.TriggerSource
TriggerSource.levelSensitiveEnable(enable)

Caveats:
• The trigger visible at the sink is delayed. See Trigger Timing.
• Assumes this is the only source; the trigger is only cleared if no additional credit has been granted.

### Distributed, Edge-Sensitive Trigger Source¶

Assert trigger enable when some boolean start undergoes a positive transition, and clear the trigger when a second signal stop undergoes a positive transition.

  // Some arbitarily logic to drive the credit source and sink. Replace with your own!
val start = lfsr(1)
val stop = ShiftRegister(lfsr(0), 5)

// Now annotate the signals.
import midas.targetutils.TriggerSource
TriggerSource.credit(start)
TriggerSource.debit(stop)
// Note one could alternatively write: TriggerSource(start, stop)

Caveats:
• The trigger visible at the sink is delayed. See Trigger Timing.
• Assumes these are the only sources; the trigger is only cleared if no additional credit has been granted.

## Chisel API¶

Trigger sources and sinks are Boolean signals, synchronous to a particular clock domain, that have been annotated as such. The midas.targetutils package provides chisel-facing utilities for annotating these signals in your design. We describe these utilities below, the source for which can be found in sim/midas/targetutils/src/main/scala/annotations.scala.

### Trigger Sources¶

In order to permit distributing trigger sources across the whole design, you must annotate distinct boolean signals as credits and debits using methods provided by the TriggerSource object. We provide an example below (the distributed example from the quick-start guide).

  // Some arbitarily logic to drive the credit source and sink. Replace with your own!
val start = lfsr(1)
val stop = ShiftRegister(lfsr(0), 5)

// Now annotate the signals.
import midas.targetutils.TriggerSource
TriggerSource.credit(start)
TriggerSource.debit(stop)
// Note one could alternatively write: TriggerSource(start, stop)

Using the methods above, credits and debits issued while the design is under reset are not counted (the reset used is implicit reset of the Chisel Module in which you invoked the method). If the module provides no implicit reset or if you wish to credit or debit the trigger system while the local module’s implicit reset is asserted, use TriggerSource.{creditEvenUnderReset, debitEvenUnderReset} instead.

### Trigger Sinks¶

Like sources, trigger sinks are boolean signals that have been annotated alongside their associated clock. These signals will be driven by a Boolean value created by the trigger system. If trigger sources exist in your design, the generated trigger will override all assignments made in the chisel to the same signal, otherwise, it will take on a default value provided by the user. We provide an example of annotating a sink using the the TriggerSink object below.

  // Note: this can be any reference you wish to have driven by the trigger.
val sinkBool = WireDefault(true.B)

import midas.targetutils.TriggerSink
// Drives true.B if no TriggerSource credits exist in the design.
// Note: noSourceDefault defaults to true.B if unset, and can be omitted for brevity
TriggerSink(sinkBool, noSourceDefault = true.B)

## Trigger Timing¶

Golden Gate implements the trigger system by generating a target circuit that synchronizes all credit and debits into the base clock domain using a single register stage, before doing a global accounting. If the total number of credits exceeds debits the trigger is asserted. This trigger is then synchronized in each sink domain using a single register stage before driving the annotated sink. The circuit that implements this functionality is depicted below:

Trigger generation circuit. Not shown: a sub-circuit analagous to that which totalCredit’ is replicated to count debits. Similarly, the sub-circuit feeding the add-reduction is generated for each clock domain that contains at least one source annotation.

Given the present implementation, an enabled trigger becomes visible in a sink domain no sooner than one base-clock edge and one local-clock edge have elapsed, in that order, after the credit was asserted. This is depicted in the waveform below.

Trigger timing diagram.

Note that trigger sources and sinks that reside in the base clock domain still have the additional synchronization registers even though they are uneeded. Thus, a credit issued by a source in the base clock domain will be visible to a sink also in the base clock domain exactly 2 cycles after it was issued.

Bridges that use the default HostPort interface add an additional cycle of latency in the bridge’s local domain since their token channels model a single register stage to improve simulation FMR. Thus, without using a different HostPort implementation, trigger sources generated by a Bridge and trigger sinks that feed into a Bridge will each see one additional bridge-local cycle of latency. In constrast, synthesized printfs and assertions, and AutoCounters all use wire channels (since they are unidirectional interfaces, the extra register stage is not required to improve FMR) and will see no additional sink latency.

## Limitations & Pitfalls¶

• The system is limited to no more than one trigger signal. Presently, there is no means to generate unique triggers for distinct sets of sinks.
• Take care to never issue more debits than credits, as this may falsely enable the trigger under the current implementation.


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# Tuesday, Nov. 29th

In class today students got back their 2-variable statistics project. The main thing I noticed that students could have done better was to interpret the real-world meaning of their results in a more specific manner. Papers with statements like the following

that the slope of the median median line is negative which means that crime is decreasing.

could be greatly improved if they wrote about the specific numbers, including units, and then included generalizations.

the slope of your median median line is negative .0078 which means that for each additional person in the population we expect a decrease in the number of violent crimes of .0078. In other words for every additional thousand people in the population I would expect to see almost 8 fewer violent crimes. This is interesting because it means that our population is growing at the same time as the number of violent crimes is shrinking.

Students had time to work on Chapter 4 review problems, update their statistics write-up and work on the next chapter 5 assignment. I also gave everyone a study guide for the test on Thursday.

Assignment:

• 5.3 on p. 256: 1-5, 10-16
• 5.4 on p. 263: 1-10

Reminder: Chapter 4 test on Thursday.

# Tuesday, Nov. 29th

Progress Report grades were entered yesterday by all teachers. Many students have already logged into StudentVUE to check grades for updates. The district will email progress reports but they will show only summary judgments (overall scores for the 4 criteria A, B, C and D). To see how students are doing on individual assignments, Student/Parent VUE is the place to check.

8th Graders attending BHS next year have an informational assembly tomorrow morning at 9:05 in the large gym. Check in with your Advisory teacher and bring a pencil.

In class today we did a warm up, I stamped students homework if they had all of the components of the assignment from last week completed (mathcounts worksheet, 4.5 textbook problems 1-17, study ?'s and summary). We reviewed some of the ideas covered before break and worked on today's assignment (4.3 – 4.5 worksheet).

Assignment: "Problem Set 11/29" worksheet (mathcounts problems) and 4.3  – 4.5 worksheet – students may skip the constructions problems

Chapter 4 Test coming up Friday or early next week. To be decided by tomorrow

# Tuesday, Nov. 29th

Progress Report grades were entered yesterday. I had a meeting and raced to get all my marks entered before I had to leave CPMS for a meeting with other BSD teachers to discuss the implementation of the new AGS1 course. One thing I noticed while entering in some scores was that there were a decent number of students in the class who would benefit from some more practice and a chance to try the module 2 test again. So, there will be a re-take of the Module 2 test on Friday. I am asking all students to do the retake and will count only the better of their two scores (retake vs. original).

In class we worked on a new problem set and a module 2 review. We also shared some great Thanksgiving stories  🙂

HW: Mathcounts w/s and Module 2 review

Note: Module 2 retake test on Friday. Best score from original and retake counts for grade.

# Tuesday, Nov. 22nd

In class today we worked on a mathcounts problem set and continued to study the triangle congruence shortcuts. In 5th period students also looked at an example of how triangle congruence can be used to prove/discover new ideas in geometry. So, I helped students to explain why the opposite sides and angles have to be congruent in a parallelogram.

Assignment. For Tuesday, students should have completed

• 4.5 on p. 227: 1-17
• 2+ Study Questions
• Sentence form summary of the topic of the assignment

Question

• What are you learning in Geometry about triangle congruence?
• How is triangle congruence related to the properties of a parallelogram?

• We are studying relationships relating to triangles. We have learned that there are shortcuts for determining when two triangles are congruent – SSS, SAS, ASA, and AAS are the triangle congruence shortcuts we have discussed. Also, students should be able to explain why ASS and AAA are not shortcuts. Again, chuckles, but really there is a reason to write these acronyms this way. For example, SAA congruence is the idea that if two separate triangles can be shown to have one pair of corresponding sides and two pairs of corresponding angles that are congruent then the triangles are congruent. In this case it is important that the position of the sides and angles correspond exactly.
• Triangle congruence can be used to prove the properties of a parallelogram if we assume only the definition of a parallelogram (quadrilateral with 2 pairs of opposite, parallel sides). We might notice that two triangles can be formed if we draw one of the diagonals of the parallelogram. And by proving these two triangles to be congruent, we can explain why the opposite sides and angles are congruent.

Now, we with each pair of parallel sides we can find a set of congruent alternate interior angles

So we can see that in $\dpi{300}\inline \triangle HJO$ and $\dpi{300}\inline \triangle JHN$ the segment $\dpi{300}\inline \overline{HJ} \cong \overline{HJ}$. So we have $\dpi{300}\inline \triangle HJO \cong \triangle JHN$ by ASA.

Since these two triangles are congruent, we can see that $\dpi{300}\inline \overline{OH} \cong \overline{NJ}$ and $\dpi{300}\inline \overline{OJ} \cong \overline{NH}$ and therefore that the opposite sides of a parallelogram are congruent.

Also, we can see that $\dpi{300}\inline \angle O \cong \angle N$ and that $\dpi{300}\inline \angle NJO$ and $\dpi{300}\inline \angle OHN$ must be congruent because they are made from congruent parts. So, the opposite angles of a parallelogram are congruent.

So, we can use triangle congruence to learn about shapes beyond triangles. And students will find that much of our knowledge of Geometry is learned through the application of triangle relationships.

Want to re-read these emails all in one place? Each email gets posted automatically to the class update blog. The little ;start ;end tags help with blog post formatting. You can see all Geometry updates at

http://teacher.nickhershman.com/blog/category/geometry/

What better way to spend the evening and the next six days off than to read and re-read each of these updates together?! Have a good break, see everyone on Tuesday.

# Monday, Nov 21

In class I answered questions from the last two sections of chapter 4 and the chapter 4 review and students had time to work on the first two assignments from chapter 5. Since it seems like a bad plan to schedule the chapter 4 test on tuesday, the Chapter 4 test will be on Thursday when we return from (va/stay)cation.

Answer Keys: The answer keys to the practice problem packets for Chapter 4 are attached.

Assignment

• Study for the chapter 4 test
• 5.1 on p. 240: 1-4, 7,8
• 5.2 on p. 248: 1-5, 7, 10, 12, 13, 16

# Tuesday, Nov. 22nd

ws-equationsolving-fractions-2.pdf
ws-mathcounts-20161122-OPLET.docx
In class today students worked on equation solving and mathcounts problems. I notice that generally they are comfortable solving multi-step equations like (equation 1)

$\dpi{300}\inline 72 + 4(5 - 3x) - (x + 3) - (4 - x) - 2(3 - 5x)$

But many students were not sure how to solve (equation 2):

$\dpi{300}\inline \dfrac{3x - 4}{5} = \dfrac{2x - 5}{2}$

So we did this example in class. I showed students that division can be distributed just like multiplication so we can turn the equation into

$\dpi{300}\inline \dfrac{3}{5}x - \dfrac{4}{5} = x - \dfrac{5}{2}$

and then solve it. Or, we can notice that if two fractions are equal, then their cross products are equal. For example, $\dpi{300}\inline \dfrac{\color{red} 3}{\color{blue}4} = \dfrac{\color{blue} 9}{\color{red}12}$ it is also the case that $\dpi{300}\inline {\color{red} 3 \cdot 12} = {\color{blue} 4 \cdot 9}$. This is true for all equivalent fractions and so another way to approach solving equation 2 is to find the cross products and set them equal

$\dpi{300}\inline 2(3x - 4) = 5(2x - 5)$

And solve from there

$\dpi{300}\inline 6x - 8 = 10x - 25$

Subtract $\dpi{300}\inline 6x$ from both sides and add $\dpi{300}\inline 25$ to both sides to get

$\dpi{300}\inline 17 = 4x$

And finally,

$\dpi{300}\inline \dfrac{17}{4} = x$

Assignment [handouts attached]

• Equations worksheet
• Mathcounts worksheet

Looking for yet another way to read these emails?

You might notice a little ;end showing up at the end of these emails which helps with the auto-formatting on the blog. If you would like to see old posts all in one place, you can find each email at

http://teacher.nickhershman.com/blog/category/ags1/

I make no promises about formatting there, but if it is helpful, you can find all the old posts for Algebra there.

Have a good break!

format: plain


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# OpenGL 2D rotations

## Recommended Posts

Hi all. Just registered! I'm starting out with GLUT and OpenGL and have been playing around. Wrote my own little version of Pong and now I'm moving on to 'Asteroids' But I've got a problem. I'm using the 2D view (gluOrtho2D) but don't know how to rotate something (ie. the spaceship) in 2D. I've rotated things in 3D before whilst learning some of the stuff, but glRotate doesn't seem to work in 2D. Is there a specific command to do this? And if not, how can I go about it? Is there some forumla I can use to calculate new x,y co-ords of a shape based on how much I want to rotate it by? Thanks. [Edited by - BigDogUK on May 27, 2005 3:10:44 AM]

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You could certainly do the calculations yourself. However, AFAIK working with 2d graphics in OpenGL ortho mode is basically just like drawing graphics to the xy plane in 3d. All rotations take place about the z axis, so your rotate calls look like this:

glRotate(angle, 0, 0, 1);

I believe there's a glVertex2(), but no glTranslate2(). So your translation calls will just be:

glTranslate(x, y, 0);

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Ahh, ok. My ship rotates now, but I'm not sure how to set the origin of rotation. It seems to be rotating around 0,0 on my 2D view and fiddling with the:

glRotatef(angle, x, y, 1)

x and y doesn't seem to help. One question. How do the float values correspond to places on a say 800x600 2D view. I'm having trouble working out where things are going to appear which is why I switched to 2DOrtho since it seemed to let me use real co-ords within the window. Ie. 400, 300 is the centre.

Here's my draw code, bit of a mess at the moment since I don't fully understand the viewports, matrix modes etc. Any guidance would be welcome :)

void renderGame(void) {

glClear(GL_COLOR_BUFFER_BIT);
glMatrixMode(GL_PROJECTION);
glViewport(0, 0, 760, 480);
gluOrtho2D(0, 760, 0, 480);
glPushMatrix();
glColor3f(0.0,0.0,0.0); // set bg colour
glVertex2i(0, 0); // top l
glVertex2i(550, 0); // top r
glVertex2i(550, 400); // bottom r
glVertex2i(0, 400); // bottom l
glEnd();
glColor3f(1.0,1.0,1.0); // set spaceship colour
glRotatef(craftangle, 0, 0, 1);
glBegin(GL_LINE_LOOP); // spacecraft
glVertex2i(260+craftx, 200+crafty); // bottom l
glVertex2i(267+craftx, 224+crafty); // middle
glVertex2i(274+craftx, 200+crafty); // bottom r
glEnd();

glutSwapBuffers();
glPopMatrix();

}

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This:

glRotatef(angle, x, y, 1);

Isn't what you want, but you're doing it correctly in the code below. A couple of other things... You probably need to add glMatrixMode(MODELVIEW) somewhere. And you shouldn't be including your ship's position in the glVertex() calls - use glTranslate() instead. Also, your ship triangle (I assume that's what it is) is currently not centered at the origin. For both those reasons the ship will not behave onscreen as you expect.

Here's your code with a few edits:

glClear(GL_COLOR_BUFFER_BIT);
glMatrixMode(GL_PROJECTION);
glViewport(0, 0, 760, 480);
gluOrtho2D(0, 760, 0, 480);
glPushMatrix(); // Is this doing anything?
glColor3f(0.0,0.0,0.0); // set bg colour
glVertex2i(0, 0); // top l
glVertex2i(550, 0); // top r
glVertex2i(550, 400); // bottom r
glVertex2i(0, 400); // bottom l
glEnd();
glColor3f(1.0,1.0,1.0); // set spaceship colour
glRotatef(craftangle, 0, 0, 1);

// You can make this basic shape whatever you want - I
// made it a triangle pointing down the +x axis.
glBegin(GL_LINE_LOOP); // spacecraft
glVertex2i(200, 0); // bottom l
glVertex2i(-200, 200); // middle
glVertex2i(-200, -200); // bottom r
glEnd();

glutSwapBuffers();
glPopMatrix();

Anyway, you might give that a try and see if it behaves any differently.

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Thanks! You've been a great help. I altered my code to your suggestions and it now spins properly on the spot. But I've got one last problem.

At the moment, the spacecraft spins around fine on the spot at the centre where it starts. But when I accelerate it off in a direction and try and rotate it, it still uses the same 0,0 co-ords so the rotation is off again! It's as though its attached to a peice of string connected to 0,0.

Do I need to technically keep it at 0,0 and move view or something? Or is there someway to update the origin it rotates around? Here's my current code:

void renderGame(void) {

glClear(GL_COLOR_BUFFER_BIT);
glMatrixMode(GL_PROJECTION);
glViewport(0, 0, 550, 400);
gluOrtho2D(-225, 225, -200, 200);
glColor3f(0.0,0.0,0.0); // set bg colour
glVertex2i(0, 0); // top l
glVertex2i(550, 0); // top r
glVertex2i(550, 400); // bottom r
glVertex2i(0, 400); // bottom l
glEnd();
glColor3f(1.0,1.0,1.0); // set spaceship colour

glRotatef(craftangle, 0, 0, 1);

glBegin(GL_LINE_LOOP); // spacecraft
glVertex2i(7, -12); // bottom l
glVertex2i(0, 12); // middle
glVertex2i(-7, -12); // bottom r
glEnd();

glutSwapBuffers();
glPopMatrix();

}

Thanks again, Oh and just for the record. I'm not actually altering the craft-x. Does this matter? I'm just altering craft-y to accelerate the craft in the direction it is pointing.

[Edited by - BigDogUK on May 26, 2005 3:14:33 AM]

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Have to leave for work in a moment, but very quickly, you need to put the glTranslate() line before glRotate(), not after. Also, you need to adjust both your ship x and y coordinates, i.e. in world space. You probably need something like this:

ship.y += cos(angle) * speed;
ship.x += sin(angle) * speed; // Or maybe -sin(angle)

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Ok. Thanks!

Been playing around with that and can't seem to get it to move in any logical direction!!! I've tried a few things and I've locked it so the angle is always between 0-360 and the angle of the ship does correspond to the correct rotation.

But it's behaving REALLY erratically. For one, it seems to completetly change direction just by adding or taking one degree from the angle. And it sure as hell isn't moving in the right direction :(

Can anyone have a quick look at my source?

http://www.zen77439.zen.co.uk/other/CurrentAsteroids.cpp

There it is! I don't seem to be doing anything strange and I've tried cleaning up the mix of floats and ints but it doesn't seem to help...

Thanks

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Quote:
 But it's behaving REALLY erratically. For one, it seems to completetly change direction just by adding or taking one degree from the angle. And it sure as hell isn't moving in the right direction :(
C++ trig functions take their arguments in radians, not degrees - I suspect that's where your problem lies. You can convert your angles from degrees to radians before submitting them to trig functions by multiplying by pi/180.

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Ahh, I suspected something was off with the angles because I was checking the output of the formula through cout and they were really weird. Sin 90 giving 0.44545 or something. But my knowledge of Maths isn't quite at that level. Getting on to that at A-level hopefully.

Just reworked it and the spaceship moves perfectly! Thanks for all your help :)

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# Local Volume Mapper

The Local Volume Mapper (LVM) is an optical, integral-field spectroscopic survey that will target the Milky Way, Small and Large Magellanic Clouds, and other Local Volume galaxies. LVM will make use of new small telescopes and newly built spectrographs that cover a wavelength range of 3600-10000 Å, with spectral resolution R~4000 (based on the DESI spectrograph design). It will collect roughly 20 million contiguous spectra over 2,500 square degrees of sky.

Broad-band (optical) and narrow-band ([O III], H$\alpha$, [S II]) imaging of the LMC (Smith et al. 2000). The left-hand column demonstrates sampling at a spatial resolution of 100 pc, typically achieved by the best IFU data available today for external galaxies. The resolution increases to the right, ending with the ~1 pc resolution that LVM will achieve in the Milky Way. Note the qualitative change in appearance around 25 pc, when networks of shocks and ionization fronts become visible and then resolved at less than 10 pc.

## Science Goals

Galaxy formation can be understood as a self-regulating process, with energy exchange between stars and interstellar gas occurring at numerous points in space and time. LVM will map the interstellar gas emission of the Milky Way and Local Group galaxies with unprecedented spatial resolution: reaching sub-parsec resolution in the Milky Way, 10s of parsecs resolution in the Magellanic Clouds, and down to <100 parsec resolution out to distances of several Mpc. LVM is thus well-positioned to study star formation and the physics of the interstellar medium in great detail by distinguishing individual star formation knots and the filamentary structures and shock networks between them. LVM will be able to connect studies across the parsec (molecular cloud) and kiloparsec (galaxy-wide) scales; this connection is fundamental to understand the physics governing star formation, the structure of the interstellar medium, the baryon cycle, and ultimately, the evolution of galaxies.

More details on the Local Volume Mapper and an extended description of its science goals can be found on the Science of SDSS-V page and in the SDSS-V White Paper SDSS-V: Pioneering Panoptic Spectroscopy.

## Contact information

For more information, contact the SDSS-V Program Head for the Local Volume Mapper, Niv Drory (University of Texas at Austin).


text:
### - Art Gallery -

In mathematical logic and computer science, two-variable logic is the fragment of first-order logic where formulae can be written using only two different variables [1]. This fragment is usually studied without function symbols.

Decidability

Some important problems about two-variable logic, such as satisfiability and finite satisfiability, are decidable.[2] This result generalizes results about the decidability of fragments of two-variable logic, such as certain description logics; however, some fragments of two-variable logic enjoy a much lower computational complexity for their satisfiability problems.

By contrast, for the three-variable fragment of first-order logic without function symbols, satisfiability is undecidable.[3]
Counting quantifiers

The two-variable fragment of first-order logic with no function symbols is known to be decidable even with the addition of counting quantifiers[4], and thus of uniqueness quantification. This is a more powerful result, as counting quantifiers for high numerical values are not expressible in that logic.

Counting quantifiers actually improve the expressiveness of finite-variable logics as they allow to say that there is a node with n neighbors, namely $${\displaystyle \Phi =\exists x\exists ^{\geq n}yE(x,y)}$$. Without counting quantifiers n+1 variables are needed for the same formula.
Connection to the Weisfeiler-Leman algorithm

There is a strong connection between two-variable logic and the Weisfeiler-Leman (or color refinement) algorithm. Given two graphs, then any two nodes have the same stable color in color refinement if and only if they have the same $$C^{2}$$ type, that is, they satisfy the same formulas in two-variable logic with counting[5].
References

L. Henkin. Logical systems containing only a finite number of symbols, Report, Department of Mathematics, University of Montreal, 1967
E. Grädel, P.G. Kolaitis and M. Vardi, On the Decision Problem for Two-Variable First-Order Logic, The Bulletin of Symbolic Logic, Vol. 3, No. 1 (Mar., 1997), pp. 53-69.
A. S. Kahr, Edward F. Moore and Hao Wang. Entscheidungsproblem Reduced to the ∀ ∃ ∀ Case, 1962, noting that their ∀ ∃ ∀ formulas use only three variables.
E. Grädel, M. Otto and E. Rosen. Two-Variable Logic with Counting is Decidable., Proceedings of Twelfth Annual IEEE Symposium on Logic in Computer Science, 1997.
Grohe, Martin. "Finite variable logics in descriptive complexity theory." Bulletin of Symbolic Logic 4.4 (1998): 345-398.


text:
# Triangle-free and Triangle-saturated Graphs

MacDougall, James A. and Eggleton, Roger B. (1997) Triangle-free and Triangle-saturated Graphs. Journal of Combinatorial Mathematics and Combinatorial Computing, 25 . pp. 3-21.

 Preview
PDF - Accepted Version
A graph $G$ is {\em triangle-saturated} if every possible edge insertion creates at least one new triangle. Furthermore, if no proper spanning subgraph has this property, then $G$ is {\em minimally triangle-saturated}. (Minimally triangle-saturated graphs of order at least 3 are the diameter 2 critical graphs.) The maximally triangle-free graphs of order $n$ are a proper subset of the minimally triangle-saturated graphs of order $n$ when $n \geq 6$. All triangle-saturated graphs are easily derivable from the minimally triangle-saturated graphs which are {\em primitive}, that is, have no duplicate vertices. We determine the 23 minimally triangle-saturated graphs of orders $n \leq 7$ and identify the 6 primitive graphs among them.


text:
# Visualization of a time-dependent Potential

We have a quantum particle in a harmonic potential which we will charge using a cyclic unitary processes.

Cyclic in this context means that our Hamiltonian

$$H(t) = H_0 + V(t)$$

consists of a time independent component $$H_0$$ and an applied external time-dependent potential $$V(t)$$ which is only non-zero in a certain time interval. This means in total that

$$H(0) = H(\tau) = H_0$$

or in other words after the field $$V(t)$$ has been applied the Hamiltonian returns to its initial state.

Now I am giving a presentation about this topic and would like to visualize this process, but am somewhat unsure as to how to show $$V(t)$$ acting on the system.

The paper (I provided the link for above) tries to show this on page 8 but I am unsure as to how 'valid' this is.

Maybe to be more exact: Is it possible to show a particle in a harmonic potential, described as a wave-function in the ground state $$\mid \psi_0 \rangle$$ like in this picture, then show 'some external field acting on this particle' and then show the particle to be in a higher energy state?

Or does anyone else have an idea how to show such a cyclic process acting on a particle in a presentation? (Does not have to be 100% exact, the professor said he prefers clarity over being exact.)


text:
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# NCERT Solutions for Class 10 Maths Exercise 1.3 Chapter 1 Real Numbers

## NCERT Solutions for Class 10 Maths Chapter 1 Real Numbers (Ex 1.3) Exercise 1.3

Last updated date: 20th Mar 2023
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Mathematics is an essential subject of the Class 10 board exam. The syllabus of Class 10 Mathematics has 15 different chapters on various. The Central Board of Secondary Education (CBSE) has included many topics of Mathematics such as Algebra, Geometry, Trigonometry, Statistics, etc. Chapter 1 of Class 10 Mathematics is all about real numbers. This chapter is a vital part of Arithmetic. The students should study and practice this chapter sincerely. There are many exercises and a practice set of this chapter. Here, we are discussing NCERT Solution for Class 10 Maths chapter 1 exercise 1.3.  The students will be benefited by the questions of chapter 1 ex 1.3 Class 10 maths. You can also download NCERT Solutions for Class 10 Maths to help you revise the complete syllabus and score more marks in your examination. Also, NCERT Solution Class 10 Science is available for free on Vedantu.

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## Maths Syllabus of CBSE Class 10

1. Prove that $\sqrt{5}$ is irrational.

Ans: We have to prove that $\sqrt{5}$ is irrational.

We will use a contradiction method to prove it.

Let $\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\sqrt{5}=\dfrac{a}{b}$

Now, squaring both sides, we get

${{\left( \sqrt{5} \right)}^{2}}={{\left( \dfrac{a}{b} \right)}^{2}}$

$\Rightarrow 5=\dfrac{{{a}^{2}}}{{{b}^{2}}}$

$\Rightarrow {{a}^{2}}=5{{b}^{2}}$ …….(1)

If ${{a}^{2}}$ is divisible by $5$ then $a$ is also divisible by $5$.

Let $a=5k$, where, $k$ is any integer.

Again squaring both sides, we get

$\Rightarrow {{a}^{2}}={{\left( 5k \right)}^{2}}$

Substitute the value in eq. (1), we get

$\Rightarrow {{\left( 5k \right)}^{2}}=5{{b}^{2}}$

$\Rightarrow {{b}^{2}}=5{{k}^{2}}$ …..(2)

If ${{b}^{2}}$ is divisible by $5$ then $b$ is also divisible by $5$.

From, eq. (1) and (2), we can conclude that $a$ and $b$ have $5$ as a common factor.

Therefore, we can say that $\sqrt{5}$ is irrational.

Hence proved.

2. Prove that $3+2\sqrt{5}$ is irrational.

Ans: We have to prove that $3+2\sqrt{5}$ is irrational.

We will use a contradiction method to prove it.

Let $3+2\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $3+2\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow 2\sqrt{5}=\dfrac{a}{b}-3$

$\Rightarrow \sqrt{5}=\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ ……..(1)

From eq. (1) we can say that $\dfrac{1}{2}\left( \dfrac{a}{b}-3 \right)$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $3+2\sqrt{5}$ is irrational.

Hence proved.

3. Prove that following are irrationals:

(i) $\dfrac{1}{\sqrt{2}}$

Ans: We have to prove that $\dfrac{1}{\sqrt{2}}$ is irrational.

We will use a contradiction method to prove it.

Let $\dfrac{1}{\sqrt{2}}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $\dfrac{1}{\sqrt{2}}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{b}{a}$ ………..(1)

From eq. (1) we can say that $\dfrac{b}{a}$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $\dfrac{1}{\sqrt{2}}$ is irrational.

Hence proved.

(ii) $7\sqrt{5}$

Ans: We have to prove that $7\sqrt{5}$ is irrational.

We will use a contradiction method to prove it.

Let $7\sqrt{5}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $7\sqrt{5}=\dfrac{a}{b}$

$\Rightarrow \sqrt{5}=\dfrac{a}{7b}$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{7b}$ is rational so $\sqrt{5}$ must be rational.

But this contradicts the fact that $\sqrt{5}$ is irrational. Hence the assumption is false.

Therefore, we can say that $7\sqrt{5}$ is irrational.

Hence proved.

(iii) $6+\sqrt{2}$

Ans: We have to prove that $6+\sqrt{2}$ is irrational.

We will use a contradiction method to prove it.

Let $6+\sqrt{2}$ is a rational number of the form $\dfrac{a}{b}$, where $b\ne 0$ and $a$ and $b$ are coprime i.e. $a$ and $b$ have only $1$ as a common factor.

Let $6+\sqrt{2}=\dfrac{a}{b}$

$\Rightarrow \sqrt{2}=\dfrac{a}{b}-6$ ………..(1)

From eq. (1) we can say that $\dfrac{a}{b}-6$ is rational so $\sqrt{2}$ must be rational.

But this contradicts the fact that $\sqrt{2}$ is irrational. Hence the assumption is false.

Therefore, we can say that $6+\sqrt{2}$ is irrational.

Hence proved.

### Maths Syllabus of CBSE Class 10

The Mathematics syllabus of Class 10 contains some crucial part of Mathematics such as Geometry, Algebra, Arithmetic, Trigonometry, Statistics, etc. There are 15 chapters in Class 10 in Mathematics. The chapters are real numbers, polynomials, pair of linear equations in two variables, Arithmetic progressions, quadratic equations, triangles, introduction to Trigonometry, coordinate Geometry, applications of Trigonometry, constructions, circles, the area related to circles, surface areas and volumes, probability, Statistics. There are chapters from different parts of Mathematics.

### CBSE Class 10 Mathematics Chapter 1

Central Board of Secondary Education (CBSE) has included some vital concepts of Arithmetic in the Class 10 Mathematics syllabus. Chapter 1 of Class 10 Mathematics is one of them. This chapter contains all about real numbers. Real numbers are a strong conceptual topic of Mathematics. From this chapter, the students will get to know the definition, properties and classification of the real number. This chapter is also an essential part of the mathematical number system. In this chapter, some examples and property proof is included for the students. By solving those examples, the students can learn the real number system easily. The property proofs will be beneficial for them. The students should practise the exercises of the textbook. They will get a good topic insight from Class 10 maths ex 1.3 NCERT solutions and it will be efficient for your practice.

### NCERT Solutions For Class 10 Maths Chapter 1 Exercise 1.3

Class 10 maths ex 1.3 is an important practice set for the students. This exercise contains some testing problems of real numbers. By practising those problems, the students will get a clear understanding of the properties and classification of real numbers. The examples are much efficient to clear the conceptual part. The solutions of this exercise are provided by NCERT. Class 10 maths ex 1.3 NCERT solutions are available online for free. The students can download it for their convenience.

### Necessity of NCERT Solutions Class 10 Maths Ch 1 Ex 1.3

Practising exercises of individual Mathematics chapters is essential for the Class 10 board exam. The students should practise the exercises of specific chapters sincerely. If they get solutions for those exercises after solving, that will be beneficial for the students. That is why Class 10 maths ex 1.3 NCERT solutions are necessary. It increases the confidence of the students.

### Irrational Numbers

Any number which cannot be represented in the form of p/q (where p and q are integers and q≠0.) is an irrational number. Examples √2,π, e, and so on.

Interesting Results of Number Theory

• If a number p (a prime number) divides a2, then p divides a. Example: 3 divides 62 i.e 36, which implies that 3 divides 6.

• The addition or subtraction of a rational and an irrational number always results in an irrational number.

• The multiplication value of a non-zero rational number and an irrational number is always irrational.

• The quotient obtained when division operation is performed between a non-zero rational number and an irrational number is always irrational.

• √m is irrational when ‘m’ is a prime. For example, 11 is a prime number and √11 is irrational which can be proved by the method of “Proof by contradiction”.

In a contradiction method, we start with an assumption which is contrary to what we are required to prove. Using a series of logical deductions from this assumption of contradiction, we will reach a mathematical inconsistency (error) – which enables us to conclude that our assumption of contradiction was incorrect.

Let’s suppose, we are going to prove √2 is an irrational number through the method of contradiction. Firstly, we assume the contradiction that √2 is a rational number and So it can be written in the form of a/b, where a and b are two co-prime numbers and n ≠ 0. By observing, we will find that there exist no coprime integers a and b for √2, so our assumption was wrong.

### Solved Examples

1. Prove that, √5 is an international number.

Solution:

Suppose, √5 is rational. Therefore, there are two integers a and b, where a/b=√5.

Suppose, the common factor of a and b is other than 1. So, we can assume them as co-prime numbers by dividing by the common factor.

a = b√5

Or,  5b2 = a2

Therefore, a2 is divisible by 5 as well as a is divisible by 5.

Let, a=5k, where k is an integer

Now, (5k)2=5b2

Or, b2 = 5k2

So, b2 and b both are divisible by 5.

Hence, we can say that a and b have common factor 5, which contradicts the co-prime fact.

Therefore, √5 cannot be expressed in p/q form. √5 is an irrational number (proved).

2. 3+2√5 is irrational - prove this.

Solution:

Suppose, 3+2√5 is a rational number.

That means, there are two integers called a and b such that

a/b=3+2√5

Or, a/b - 3 = 2√5

Or, √5 = 1/2 (a/b-3)

As a and b are integers and 1/2 (a/b - 3) will also be rational. So, √5 is rational.

This fact contradicts that √5 is irrational.

Hence, our consumption is false and √5 is an irrational number.

## FAQs on NCERT Solutions for Class 10 Maths Exercise 1.3 Chapter 1 Real Numbers

1. Why Practising Chapter 1 Class 10 Maths Exercise 1.3 is Important?

Class 10 board exam is the first board exam of the students. The students read and work hard to score well in their first board exam. The Mathematics syllabus of Class 10 has diversity in the chapters. The students have to read all the different chapters and topics with the same importance. The students should practise the chapters after completing the entire syllabus. There are several exercises and practise sets for each chapter of Class 10 Mathematics. Chapters 1 Class 10 maths exercise 1.3 is one of them. Chapter 1 Class 10 maths exercise 1.3 help the students to get a strong concept of this chapter and in practising.

Mathematics is a practice-based subject. The more the students will practise, the more they will get strong on the topics. They should read and complete their Mathematics syllabus. After that, they should practise as much as they can. There are individual exercises and practise sets for all the chapters of Class 10 Mathematics. For chapter 1 the exercises are ex 1.1, ex 1.2 and ex 1.3 Class 10 maths. They can download all the exercises of chapter 1 from different educational websites. Chapter 1 ex 1.3 Class 10 maths pdf is available online for free. The students can download the pdf or click the direct download link and get it.

3. How many questions are there in Exercise 1.3 of 10th Maths?

Chapter 1 in the syllabus for Class 10 Mathematics is “Real Numbers”. The chapter consists of four exercises in total. Exercise 1.3 of this chapter contains three questions in total.  Although most of these problems are short, their solutions are long. The exercise includes questions that require students to prove the given roots as irrational. These questions and the examples related to them are important and are often asked in the exam as well.

4. What is the main concept to learn exercise 1.3 of 10th Maths?

The 1st chapter in Class 10 Maths includes everything about real numbers which is an important topic in the Class 10 syllabus. Exercise 1.3 of this chapter includes questions of the proof of the irrationality of roots. Examples 9, 10, and 11 are also based on questions in this exercise. Solving these questions might seem difficult to understand at first. Students can take the help of NCERT Solutions for Class 10 Maths Exercise 1.3 Chapter 1.

5. Do I need to practice all the questions provided in Class 10 Maths Chapter 1 NCERT Solutions?

Class 10 Board Question Papers are framed directly based on the Class 10 Maths NCERT. Hence, it is important to practice all the questions that are provided in the NCERT Solution for Class 10 Maths Chapter 1. Any of them can be asked in the exams and it is not possible to know which ones they might be. Ignoring any questions or examples can lead to a loss of marks.

6. Where can I find NCERT Solutions for Class 10 Maths Exercise 1.3?

Students who need help understanding the method of solving the questions given in Exercise 1.3 can take the help of the NCERT Solutions for Class 10 Maths available on Vedantu. These solutions have been framed by experts of the subject in a language easy for students to understand and be able to grasp all the concepts taught in this exercise. You can find these solutions on the page NCERT Solutions for Class 10 Maths Exercise 1.3 Chapter 1 at free of cost. The PDFs of these NCERT Solutions are available on the Vedantu app for free.

7. How many exercises and questions are covered in NCERT Solutions for Class 10 Maths Chapter 1?

Chapter 1 in Class 10 Maths is known as Real Numbers. There are a total of four exercises covered in the NCERT Solutions for this chapter. The following is a list of total questions provided in each exercise:

• Exercise 1.1 - Five Questions

• Exercise 1.2 - Seven Questions

• Exercise 1.3 - Three Questions

• Exercise 1.4 - Three Questions

Students can find step-by-step NCERT Solutions for all four exercises on the page NCERT Solutions for Class 10 Maths Exercise 1.3 Chapter 1.


text:
Lemma 52.10.9. Let $I \subset \mathfrak a \subset A$ be ideals of a Noetherian ring $A$ and let $M$ be a finite $A$-module. Let $s$ and $d$ be integers. Suppose that

1. $A, I, V(\mathfrak a), M$ satisfy the assumptions of Situation 52.10.1 for $s$ and $d$, and

2. $A, I, \mathfrak a, M$ satisfy the conditions of Lemma 52.8.5 for $s + 1$ and $d$ with $J = \mathfrak a$.

Then there exists an ideal $J_0 \subset \mathfrak a$ with $V(J_0) \cap V(I) = V(\mathfrak a)$ such that for any $J \subset J_0$ with $V(J) \cap V(I) = V(\mathfrak a)$ the map

$H^{s + 1}_ J(M) \longrightarrow \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM)$

is an isomorphism.

Proof. Namely, we have the existence of $J_0$ and the isomorphism $H^{s + 1}_ J(M) = H^{s + 1}(R\Gamma _\mathfrak a(M)^\wedge )$ by Lemma 52.8.5, we have a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM) \to H^{s + 1}(R\Gamma _\mathfrak a(M)^\wedge ) \to \mathop{\mathrm{lim}}\nolimits H^{s + 1}_\mathfrak a(M/I^ nM) \to 0$

by Dualizing Complexes, Lemma 47.12.4, and the module $R^1\mathop{\mathrm{lim}}\nolimits H^ s_\mathfrak a(M/I^ nM)$ is zero because $\{ H^ s_\mathfrak a(M/I^ nM)\} _{n \geq 0}$ has Mittag-Leffler by Theorem 52.10.8. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).


text:
# Recurrent Variational Network: A Deep Learning Inverse Problem Solver applied to the task of Accelerated MRI Reconstruction

Magnetic Resonance Imaging can produce detailed images of the anatomy and physiology of the human body that can assist doctors in diagnosing and treating pathologies such as tumours. However, MRI suffers from very long acquisition times that make it susceptible to patient motion artifacts and limit its potential to deliver dynamic treatments. Conventional approaches such as Parallel Imaging and Compressed Sensing allow for an increase in MRI acquisition speed by reconstructing MR images from sub-sampled MRI data acquired using multiple receiver coils. Recent advancements in Deep Learning combined with Parallel Imaging and Compressed Sensing techniques have the potential to produce high-fidelity reconstructions from highly accelerated MRI data. In this work we present a novel Deep Learning-based Inverse Problem solver applied to the task of Accelerated MRI Reconstruction, called the Recurrent Variational Network (RecurrentVarNet), by exploiting the properties of Convolutional Recurrent Neural Networks and unrolled algorithms for solving Inverse Problems. The RecurrentVarNet consists of multiple recurrent blocks, each responsible for one iteration of the unrolled variational optimization scheme for solving the inverse problem of multi-coil Accelerated MRI Reconstruction. Contrary to traditional approaches, the optimization steps are performed in the observation domain ($k$-space) instead of the image domain. Each block of the RecurrentVarNet refines the observed $k$-space and comprises a data consistency term and a recurrent unit which takes as input a learned hidden state and the prediction of the previous block. Our proposed method achieves new state of the art qualitative and quantitative reconstruction results on 5-fold and 10-fold accelerated data from a public multi-coil brain dataset, outperforming previous conventional and deep learning-based approaches.

PDF Abstract CVPR 2022 PDF CVPR 2022 Abstract


text:
# Rate of change: find rate length is changing at moment

1. Apr 25, 2013

### syeh

1. The problem statement, all variables and given/known data

An equilateral triangle with side length x sits atop a rectangle with a base length of x, with the base of the triangle coinciding with the top base of the rectangle. the height of the rectangle is 7 times the length of its base. If the combined area of the figures is increasing at a rat of 8 mm^2/sec, find the rate at which a side of the triangle is lengthening at the moment the base of the rectangle is 20 mm.

3. The attempt at a solution

so i started by making a sketch. (I attached a sketch of the figures). then, i found the Area and then dA/dt:

A=lw+1/2 bh
=(7x)(x)+ 1/2 (x)((x*sqrt3)/2)
=7x^2 + (x^2*sqrt3)/4

dA/dt= 14x dx/dt + (x*sqrt3)/2 dx/dt

then, plug in x=20 mm, and dA/dt=8 mm^2/min:

8= 40 dx/dt + 2.474 dx/dt
8= dx/dt (40+2.474)
8= dx/dt (42.474)
dx/dt = 0.188 mm^2/sec

#### Attached Files:

• ###### mathcalc.jpg
File size:
14.5 KB
Views:
53
2. Apr 26, 2013

### Simon Bridge

That's great - did you have a question?

Note - for isomorphic change, $$A\propto x^2 \Rightarrow \frac{dA}{dt} \propto 2x\frac{dx}{dt}$$ ... which could save you a lot of trouble.

Last edited: Apr 26, 2013


text:
# Two recent works on spatio-temporal modelling

Scholar picked up on two spatio-temporal papers this week, both published in Statistical Science, which I believe are worth mentioning. The first is by Wikle, Milliff, Herbei and Leeds whilst the second is by Diggle, Moraga, Rowlingson and Taylor. In this post I will say some comments on the first work, which talks about BHMs in environmental sciences. The second one is an excellent review on point processes but I will not review it in detail here.

Modern Statistical Methods in Oceanography: A hierarchical perspective, by CK Wikle, RF Milliff, R Herbei and WB Leeds

This article considers the application of Bayesian hierarchical models (BHMs) in the context of oceanography, a field in which the authors have extensive experience. This is meant to be a review article, so if you are already familiar with the authors’ works there will be nothing particularly new here. However this article summarises very effectively, in one place, the statistical developments in this field in the last 15-20 years.

The article starts off by describing why oceanography is such a complex problem, and I find that many problems described here pertain to most environmental applications. For example:

Although  it is the case that there are too few in situ observations of the ocean to characterize its evolution and its interaction with marine ecosystems, in an ironic twist the discipline also suffers from having an abundance of particular data types when one factors in the the satellite observations […]

This problem is also what is being faced in glaciology at the moment. Satellites can provide  a seeming endless stream of data, however all of the same type, and rarely provide the whole picture. There is a lesson to be learnt here, that satellites, for all their importance in today’s environmental research arena, is no substitute for in situ data which can provide some mixture diversity. This is the only way multi-variate processes, such as that being considered by the authors at present, can be resolved.

The article proceeds to outline the hierarchical standard modelling paradigm before discussing three key application areas.

The first, data assimilation and inverse modelling, is the process by which numerical model outputs are combined with data to provide a consensus estimate (I avoid the word optimal intentionally). The application here is that of modelling surface vector winds (SVW) and the second column on Pg 472 does a good job in highlighting an important point, that today we no longer work with data but with data products. This ongoing concern to the statistician, also highlighted by Peter Guttorp at the last CliMathNet Conference, is what I believe deserves more attention in ST-stats. Remote sensors rarely measure the quantity of interest directly, rather they are used as proxy measurements and several “corrections” are applied in order for the measurement to reflect the quantity of interest. With SVW, the correction procedure seems relatively straightforward, however this is not always the case, for example with geodetic satellites (see last paragraph). As statisticians we are generally provided with “adjusted data” with some measure uncertainty but it’s important to be mindful of where the data and the associated uncertainties actually came from. The application example stresses some important examples of sceince-based parameterisations, for which more information can be found in this review article by two of these authors. Also, later on (in Section 4) the authors stress that:

In general, to be effective in this context, dimension reduction should not be independent of the physical and biological environment under consideration.

The second example of data assimilation in this work focuses on oceanographic tracer data for modelling ocean circulation. Personally I wasn’t aware of this work and found it interesting. The key idea is that the tracer’s stationary pattern is a solution to the advection-diffusion equation

${\bf u} \cdot \nabla C = \nabla \cdot (K \nabla C) + Qc,~~~ (x,y) \in \Omega$

with some Dirichlet boundary conditions. One solves for this on a grid, repeatedly for different parameters (in this case the velocity ${\bf u}$, the diffusivity $K$, boundary terms and the sink constant $Q$), and “compare” the solution for C with observations at respective locations (using the likelihood function). MCMC is used in practice, although I imagine solving for $C$ for each sample is quite computationally intensive.

The second application area is long-lead forecasting for sea-surface temperatures. The authors claim that

such “long-lead” forecasts have shown useful skill and correspond to one of the few situations in ocean science where a purely statistical forecast methodology is competitive with, and in may cases better than, equivalent deterministic model forecasts

Indeed, this work for me is a flagship spatio-temporal model and the original work of Mark Berliner is definitely worth a read for those who haven’t done so yet. The basic idea is to split the spatio-temporal model into correlated large-scale components (which I believe where EOFs in this case), and small-scale temporally independent components, but also to extensively use scientific insight. For example, the propagation matrix was set to vary according to the “climate regime” of the temporal year in question, depending on whether it was cool, normal or warm. This, clearly, is an educated model choice based on underlying physical considerations, and such inter-disciplinary thinking goes a long way in improving predictive assessments.

The third application considered is  a relatively new one by William Leeds, who uses BHMs for studying marine ecosystems. The reader should read this paper and this paper as they are some of the few works which combines BHMs with emulators. Emulation is the process by which complex numerical models are simplified as stochastic models which can be evaluated (i.e. sampled from) very quickly. This allows us to find distributions over unknown parameters in the model, and perform calibrated predictions.

The approach Leeds takes in his first work on the subject is to use a non-standard observation model (in his case a truncated normal distribution) and define a first-order emulator in the process layer. The emulator is constructed by taking the dominating singular vectors from model outputs (over time), and fitting a model to the respective weights with respect to the model parameters.  Once this model is fit (which can be very general and non-linear), MCMC can be used to sample the parameters and process in the usual manner.  The spatial element comes in by assuming spatial correlation between the model parameters. In the second work, the approach is extended to a dynamic setting. Now EOFs (from the model output) are used for the large scale features and the emulator is used (indirectly) to provide information about the propagation matrices and other parameters in the model. This paper shows, in a very neat way, how only certain aspects (characterisations?) of numerical model output is used in our process model. This was also the subject of a recent work of mine here.

On the whole, this paper was a very pleasant read and contains a ton of useful references which will come in handy for anyone interested in spatio-temporal BHMs.