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A binary tree is a type of tree data structure in which each parent node has no more than two children. A binary tree node is made up of three components:
• Data item
• Address of left child
• Adress of right child
Snippet from Wikipedia: Heap (data structure)
In computer science, a heap is a tree-based data structure that satisfies the heap property: In a max heap, for any given node C, if P is a parent node of C, then the key (the value) of P is
greater than or equal to the key of C. In a min heap, the key of P is less than or equal to the key of C. The node at the "top" of the heap (with no parents) is called the root node.
The heap is one maximally efficient implementation of an abstract data type called a priority queue, and in fact, priority queues are often referred to as "heaps", regardless of how they may
be implemented. In a heap, the highest (or lowest) priority element is always stored at the root. However, a heap is not a sorted structure; it can be regarded as being partially ordered. A
heap is a useful data structure when it is necessary to repeatedly remove the object with the highest (or lowest) priority, or when insertions need to be interspersed with removals of the
root node.
A common implementation of a heap is the binary heap, in which the tree is a complete binary tree (see figure). The heap data structure, specifically the binary heap, was introduced by J. W.
J. Williams in 1964, as a data structure for the heapsort sorting algorithm. Heaps are also crucial in several efficient graph algorithms such as Dijkstra's algorithm. When a heap is a
complete binary tree, it has the smallest possible height—a heap with N nodes and a branches for each node always has log[a] N height.
Note that, as shown in the graphic, there is no implied ordering between siblings or cousins and no implied sequence for an in-order traversal (as there would be in, e.g., a binary search
tree). The heap relation mentioned above applies only between nodes and their parents, grandparents. The maximum number of children each node can have depends on the type of heap.
Heaps are typically constructed in-place in the same array where the elements are stored, with their structure being implicit in the access pattern of the operations. Heaps differ in this way
from other data structures with similar or in some cases better theoretic bounds such as Radix trees in that they require no additional memory beyond that used for storing the keys.
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EDL's Fib Sequence and Hebrew EAST
First line of Magnetic Current is special to me.. as I was told this same thing in a Near Death Experience .. "Look to the East, Look to the east, Look to the east"... this happened in 1996 and I
first heard of Edward Leedskalnin on the Transit of Venus June 8, 2004. So... what does it mean? Amazingly, I find EDL seems to point to math.. and in part, the Fibonacci Sequence and Hebrew
Gematria. I will share some thoughts.. though there is MUCH MORE to it most of which I can't figure out yet.
This is page 3 of Magnetic Current (MC):
This is page 6 of Magnetic Current (MC):
And here is the 4 Hebrew Directions.. aligned with the numbers 1 to 22.. or alef to tav of Hebrew alphabet,
also the same thing as 22 squared:
Fib Sequence and running sum:
Now.. EDL mentioned the last 4 of his 5 works are meant to count together. If you count the lines in all 5 works you get 2024 lines. If each line was counted as a spherical cannonball and stacked in
a tetrahedron shape 22 triangular layers high. The top 15 layers would equal 680 cannonballs...and EDL's first work is 680+1 lines. The last 4 works would be 1343 lines.. which is the bottom 7 of 22
triangular layers.. 1344-1 lines. This is taking the 22 squared.. which can be also shown as sum 1+2+3... to 22 and back to 1 again. In other words.. The 4 Hebrew directions are 2 of the 4 faces of
the 22 layered cannonball tetrahedron.
The 2nd work of EDL is Magnetic Base Sound Base and contains 153 lines. This is equivalent to EAST I believe. But the numbers breathe from 1 to 10.. the monad to decad. .. the Tetractys...as shown on
back of US one dollar bill that EDL points out at very beginning of this work.. with a dot.
Now.. the next work EDL mentions is MC.. and thus there is 7 lines on cover of MC, 2 lines on copyright page, then the 10th line is the title of page 3.. 10 lines total. We have here also the 1,9,10
relationship that is found in EAST... 1 to 1,2,3,4... tip of pyramid shining bright with the EYE OF HORUS. Now.. after this "MAGNETIC CUPRENT" as EDM calls it.. is the word "EAST" in the next line
reading left to right and also top to bottom.. down the next 21 lines. And the first 3 paragraphs start acrostically .. TFF.. subtract 1 in a caesar cipher.. SEE. Now.. is it a coincidence that the
153rd line is the first one in MC to have an ending line touch both the right and left margins? And the line after that is the 144th line not counting the first 10 lines. In Hebrew gematria QEDEM is
144.. which is also happened to be a fibonacci number... along with SOUTH which is 55.. the other fib direction.
I ask anyone with a partially open mind that even dares to strain their brain to comprehend what I present here.. to ask themselves is all these fib numbers a pure coincidence? The right hand margin
of page 3.. counting lines per paragraph.. that touch the right margin.. 2,3,5,8.. this is the end of the 21 line EAST. and the first 3 paragraphs that start MC that start with TFF .. 2,1,3,1,5,1...
13.. which in Hebrew Gematria is the same as ECHAD or ONE. But also a fib number. So, one one end of 144 lines starting on page 3 MC.. Hebrew EAST.. we have 1,1,1,2,3,5,8..13...21...and the other end
on page 6 MC..
we have 34 lines in the 3 paragraphs starting with "BNN".. which also is a caesar cipher shift of "SEE". Actually, it is a -17 caesar cipher.. as 19 S - 2 B = 17.
I also ask you to ask yourself if 144 being a fib number and also the Hebrew Gematria for EAST.. and the fact that he says..
"This writing is lined up so when you read it you look East,".. and from there down.. at the capital "E".. 21 lines down span spells "EAIST"... with an "I" in the center.. which is 9th letter of
English alphabet.. which is also the same as the 2+3+4.. that breathes or twinkles in the 4 Hebrew directions.. 153,144,143.... when the monad becomes the decad.
Here is another reason to show a reason that the 1 becomes the 10 in EDL's works. MC I believe is a mathematical representation of the EARTH's polar diameter in English miles... where each line is 10
English miles. .. minus the 17 page numbers in this case. .. and overlapping the equator.. as being 395 lines from each end. .. the earth being 3950 miles polar radius.. thus each line is 10 English
miles. We have ONE line is TEN English miles. Thus.. another breathing of 1 to 10. Right? So, could EDL somehow be pointing to a mathematical pattern of EARTH involving the Fibonacci sequence that
also somehow relates to EAST...and more specifically the Hebrew "QEDEM" with its corresponding gematria? Could it also maybe have something to do with the US DOLLAR BILL and the 22 layered
Tetrahedron? Does the EAST "breathe" or "twinkle" from 1 to 10.. 153 to 144 to 143?
I say ED's hints all suggest this to me but suspect there is way more to the story that I still as yet have figured out.
April 2016
poughkeepsieblue said:
based on what my friend Dave experienced with you in the past.
allrighty then ' ----- looking at your writing style , I know which Dave are you referring to
Just remember --- people won't just hand over their knowledge they've spent years of countless hours (right or wrong) to anyone on the net , just for asking.
I'm curious Charlie.. is it "(right or wrong)" in your opinion?
Here is my answer (962 or 269).
Keep wondering since we have addition of 1 to 22 to 1 and it involves fib sequence.. if Pascal's triangle could be involved here somehow in EDL's EAST:
This is Pascal's Triangle in more readable format showing the fib sequence.
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Given AB of length 7.3 cm and CD of length 3.4 cm, construct a line segment XY such that the length of XY is equal to the difference between the lengths of AB and CD
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
A day full of math games & activities. Find one near you.
Given AB of length 7.3 cm and CD of length 3.4 cm, construct a line segment XY such that the length of XY is equal to the difference between the lengths of AB and CD. Verify by measurement.
With the help of a ruler and a compass, we can draw a line segment by following steps:
Given: length of AB = 7.3 cm and length of CD = 3.4 cm
Steps of construction:
Step 1: Construct a line AB = 7.3 cm and CD = 3.4 cm.
Step 2: Open the compasses up to the length of CD and draw an arc from point A that cuts AB at point P.
Now, open the compasses up to the length of PB.
Step 3: Draw a line l and mark X point on a line l.
Step 4: Place the compass pointer on X and draw an arc to cut the line l at Y.
Hence, XY is the required line segment
Now, XY = AB - CD = 7.3 – 3.4 = 3.9 cm
With the help of a ruler, we can measure the length of XY. It will definitely measure a length of 3.9 cm.
NCERT Solutions for Class 6 Maths Chapter 14 Exercise 14.2 Question 5
Given AB of length 7.3 cm and CD of length 3.4 cm, construct a line segment XY such that the length of XY is equal to the difference between the lengths of AB and CD. Verify by measurement.
Given AB of length 7.3 cm and CD of length 3.4 cm, construct a line segment XY such that the length of XY is equal to the difference between the lengths of AB and CD. The measurement is verified by
using steps of construction.
☛ Related Questions:
Math worksheets and
visual curriculum
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The Transportation Problem in AMPL
AMPL Formulation
The formulation of the transportation problem is AMPL is a straightforward translation of the mathematical programme for
the transportation problem
. We will build the file
. The sets
set SUPPLY_NODES;
set DEMAND_NODES;
The supply
param Supply {SUPPLY_NODES} >= 0, integer;
param Demand {DEMAND_NODES} >= 0, integer;
The cost
param Cost {SUPPLY_NODES, DEMAND_NODES};
Now, the mathematical programme follows directly:
var Flow {SUPPLY_NODES, DEMAND_NODES} >= 0, integer;
minimize TotalCost:
sum {i in SUPPLY_NODES, j in DEMAND_NODES} Cost[i, j] * Flow[i, j];
subject to UseSupply {i in SUPPLY_NODES}:
sum {j in DEMAND_NODES} Flow[i, j] = Supply[i];
subject to MeetDemand {j in DEMAND_NODES}:
sum {i in SUPPLY_NODES} Flow[i, j] = Demand[j];
Note that we assume the transportation is balanced.
Adding Bounds
In the main discussion of
transportation problems
, we saw that adding bounds to the flow variables allowed us to easily either bound the transportation of good from a supply node to a demand node or remove an arc from the problem altogether. We can
add bounds to our AMPL formulation by
declaring 2 new parameters with defaults
param Lower {SUPPLY_NODES, DEMAND_NODES} integer default 0;
param Upper {SUPPLY_NODES, DEMAND_NODES} integer default Infinity;
and adding them to the
variable declaration:
var Flow {i in SUPPLY_NODES, j in DEMAND_NODES}
>= Lower[i, j], <= Upper[i, j], integer;
Also, since some arcs no longer exist you should set a default of 0 for
(thus you don't have to define costs for non-existent arcs).
param Cost {SUPPLY_NODES, DEMAND_NODES} default 0;
Balancing Transportation Problems
Balanced transportation models
are preferred as there is no confusion about the relational operators for the supply and demand constraints. We can use a script file (
) to balance any transportation problem automatically:
model transportation.mod;
param costFromDummy {DEMAND_NODES} default 0;
param costToDummy {SUPPLY_NODES} default 0;
param difference;
# Add the problem date file here
# e.g., data brewery.dat;
let difference := (sum {s in SUPPLY_NODES} Supply[s])
- (sum {d in DEMAND_NODES} Demand[d]);
if difference > 0 then
let DEMAND_NODES := DEMAND_NODES union {'Dummy'};
let Demand['Dummy'] := difference;
let {s in SUPPLY_NODES} Cost[s, 'Dummy'] := costToDummy[s];
else if difference < 0 then
let SUPPLY_NODES := SUPPLY_NODES union {'Dummy'};
let Supply['Dummy'] := - difference;
let {d in DEMAND_NODES} Cost['Dummy', d] := costFromDummy[d];
}; # else the problem is balanced
# Make sure the problem is balanced
check : sum {s in SUPPLY_NODES} Supply[s] = sum {d in DEMAND_NODES} Demand[d];
option solver cplex;
display Flow;
Note the
check statement
to ensure that the balancing has been done properly before solving. Also, note that
allow for the definition of costs on any flow from/to a dummy node in the data file. --
- 02 Apr 2008
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How many calories does walking for 8 hours burn?
How many calories does walking for 8 hours burn?
An hour walk burns between 210 and 360 calories for most people. At a casual pace you will cover 3 miles in an hour walk. Doing an hour walk 5 days of the week will burn an extra 1,050 to 1,800
calories. If your diet remains the same, this increased exercise could lead to ⅓ to ½ a pound of fat loss a week.
How many calories does walking for 3 hours burn?
A 140 pound (63.5kg) person will burn 1,000 calories in 5 hours walking on level, firm surface at 2.5mph (4kph). On the same surface at 4mph (6.4kph), they will burn 1,000 calories in 3 hours.
How many calories do I burn walking for 2 hours?
Tip. Depending on your weight and how fast you walk, you can burn approximately 480 to 888 calories speed walking for two hours.
How many calories burned walking 10?
Overall, you can expect to burn 700–1,200 calories when walking 10 miles, depending on several factors. Your pace will matter less in regards to the total number of calories burned. However, walking
faster will help you hit the 10-mile mark much sooner.
How long do I have to walk to burn 1000 calories?
Walk on a treadmill at an incline for an hour. I am 6′ and 200 lbs, and when I walk at 4 mph and a 6% incline, I burn about 1,000 calories an hour. So one way to reach your goal is to do this for 5
hours (adjusting for your calorie burn based on your own research).
How many calories do you burn in an hour of walking?
To sum it up, it is tough to estimate the number of calories that you burn in an hour without any additional data. However, we can say that for an average person, the number of calories burned in an
hour or walking ranges from 200-500 kcal.
How many calories does it take to walk 10 miles?
Walking 10 miles on a flat surface is relatively easy compared to a route that has you trudging up steep hills. Walking at a 5 percent incline, in fact, burns as many as five extra calories per
minute, which over the course of 10 miles could amount to an increase of as much as 750 calories burned.
How many calories does it take to walk 10000 steps?
Assuming: 1) 10,000 steps is 5 miles, and 2) an average pace of 2.5mph (total of 2 hours walking), a 140 pound (63.5kg) person will burn 401 calories and a 200 pound (90.7kg) person will burn 573
calories in this time. How long do I have to walk to burn 1,000 calories?
How long does it take to walk 10 miles a day?
A typical walking pace is 15–20 minutes per mile. To go any faster results in you essentially jogging or running, which does have other benefits and downsides compared with walking. At the typical
walking rate, it’ll take you 2–3 hours to get to 10 miles. If you have the time for this, great.
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A consumer spends Rs.800 on the consumption of a particular commodity of price Rs.10 per unit. When the price increases to Rs.20 per unit, he spends Rs.960 on it. Calculate the price elasticity of demand.
A consumer spends Rs.800 on the consumption of a particular commodity of price Rs.10 per unit. When the price increases to Rs.20 per unit, he spends Rs.960 on it. Calculate the pri... A consumer
spends Rs.800 on the consumption of a particular commodity of price Rs.10 per unit. When the price increases to Rs.20 per unit, he spends Rs.960 on it. Calculate the price elasticity of demand.
Understand the Problem
The question is asking us to calculate the price elasticity of demand based on the changes in price and consumer expenditure on a particular commodity. We will use the formula for price elasticity of
demand to solve it.
The calculated price elasticity of demand will depend on the specific values used in the formulas provided.
Answer for screen readers
The final answer will depend on the specific values of the old and new quantities and prices provided in the problem.
Steps to Solve
1. Identify the Formula for Price Elasticity of Demand
The price elasticity of demand (PED) can be calculated using the following formula:
$$ PED = \frac{% \text{ change in quantity demanded}}{% \text{ change in price}} $$
2. Calculate the Percentage Change in Quantity Demanded
To find the percentage change in quantity demanded, use the formula:
$$ % \text{ change in quantity demanded} = \frac{\text{New Quantity} - \text{Old Quantity}}{\text{Old Quantity}} \times 100 $$
3. Calculate the Percentage Change in Price
Similarly, calculate the percentage change in price using:
$$ % \text{ change in price} = \frac{\text{New Price} - \text{Old Price}}{\text{Old Price}} \times 100 $$
4. Substituting Values into the PED Formula
Once you have both percentage changes, substitute them into the PED formula to calculate the price elasticity of demand.
5. Interpret the Result
A calculated PED value can indicate the responsiveness of consumers to price changes:
• If $|PED| > 1$, the demand is elastic.
• If $|PED| < 1$, the demand is inelastic.
• If $|PED| = 1$, the demand is unitary elastic.
The final answer will depend on the specific values of the old and new quantities and prices provided in the problem.
More Information
Understanding price elasticity can help businesses set prices effectively and understand consumer behavior regarding their products. A higher elasticity means that a small change in price leads to a
larger change in quantity demanded.
• Failing to convert the changes into percentages correctly.
• Taking the change in price or quantity in the wrong direction (i.e., not following the correct order of old and new).
• Misunderstanding what elastic, inelastic, and unitary elastic demand means.
AI-generated content may contain errors. Please verify critical information
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14,776 research outputs found
This article is based on a talk given at the ``Strings'97'' conference. It discusses the search for the universality class of confining strings. The key ingredients include the loop equations, the
zigzag symmetry, the non-linear renormalization group. Some new tests for the equivalence between gauge fields and strings are proposed.Comment: 13 pages, latex; talk at STRINGS'9
The leading twist parametrization of dipion mass distribution in hard exclusive reactions is proposed. Its parameters are related to quark distributions (usual and skewed) in the pion and to
distributions amplitudes of mesons ($\pi$, $\rho$, etc.). We show that measurements of the shape of dipion mass distribution in hard exclusie reactions can give important information about partonic
structure of the pion. The expression for the amplitude of the reaction $\gamma^*\gamma\to \pi\pi$ near the threshold in terms of singlet quark distribution in the pion is presented.Comment: Talk
given at 7th International Workshop on Deep Inelastic Scattering and QCD (DIS 99), Zeuthen, Germany, 19-23 Apr 1999. Minor typos are correcte
We discuss several closely related concepts in the NSR formulation of superstring theory. We demonstrated that recently proposed NSR model for superstrings on $AdS_5 \times S^5$ is described by the
world-sheet logarithmic conformal field theory (LCFT). The origin of LCFT on a world-sheet is closely connected to the matter-ghost mixing in the structure of a brane-like vortex operators. We
suggest a dynamical origin of M theory as a string theory with an extra dimension given by bosonised superconformal ghosts.Comment: 20 pages, no figures, harvmac, corrected some typo
We use the homological algebra context to give a more rigorous proof of Polyakov's basic variational formula for loop spaces.Comment: Latex, 17 pages, no figure
The effects bringing about by the finiteness of the photon mass due to the Debye screening in the monopole gas in three-dimensional compact QED are studied. In this respect, a representation of the
partition function of this theory as an integral over monopole densities is derived. Dual formulation of the Wilson loop yields a new theory of confining strings, which in the low-energy limit almost
coincides with the one corresponding to the case when the photon is considered to be massless, whereas in the high-energy limit these two theories are quite different from each other. The confining
string mass operator in the low-energy limit is also found, and its dependence on the volume of observation is studied.Comment: 7 pages, LaTeX, no figures, 1 reference is added, final version to
appear in Phys. Lett.
The Polyakov's "soldering procedure" which shows how two-dimensional diffeomorphisms can be obtained from SL(2,R) gauge transformations is discussed using the free-field representation of SL(2,R)
current algebra. Using this formalism, the relation of Polyakov's method to that of the Hamiltonian reduction becomes transparent. This discussion is then generalised to N=1 superdiffeomorphisms
which can be obtained from N=1 super Osp(1,2) gauge transformations. It is also demonstrated that the phase space of the Osp(2,2) supercurrent algebra represented by free superfields is connected to
the classical phase space of N=2 superconformal algebra via Hamiltonian reduction.}Comment: 21 pages, revised version contains minor grammatical changes and an update of refernce
We discuss the NSR formulation of the superstring action on AdS_5 X S^5 proposed recently by Kallosh and Tseytlin in the Green-Schwarz formalism.We show that the stress-energy tensor corresponding to
the NSR action for AdS superstring contains the branelike terms, corresponding to exotic massless vertex operators (refered to as the branelike vertices). The corresponding sigma-model action has the
manifest SO(1,3) X SO(6) invariance of superstring theory on AdS_5 X S^5. We argue that adding the branelike terms is equivalent to curving the space-time to obtain the AdS_5 X S^5 background. We
commence the study of the proposed NSR sigma-model by analyzing the scattering amplitudes involving the branelike vertex operators.The analysis shows quite an unusual momentum dependence of these
scattering amplitudes.Comment: 15 pages more corrections and references adde
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Calculate Felt Weight at Acceleration
Calculate Felt Weight at Acceleration
Calculator for the weight or mass that is sensed when accelerating or braking. Acceleration or braking is measured in g-force, which is the unit of the gravitational acceleration. The calculation of
this can be done with one of the previous calculators. In resting state on Earth, the force is one g, the own weight is felt. When accelerating with 2 g, or braking with -2 g, twice the own weight is
felt additionally. The direction of the gravitational acceleration is down, that of the additional acceleration can be any. Both forces add according to their angle.
The formula for the addition of two forces is F = √ F[1]² + F[2]² + 2 F[1] F[2] cos(α)
Example: a person with 70 kilograms, accelerating forward with 1.5 g, perpendicular (90 degrees) to the gravity, experiences a force of 1238 newtons, which feels like about 126 kilograms.
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Understanding the Equation: Why 1 + 2 + 3 ... = -1/12 in Physics
Written on
The Paradox of Infinite Sums
The equation 1 + 2 + 3 ... = -1/12 is often met with skepticism due to its perplexing nature. How can an infinite sum of positive integers yield a negative result? Let's explore the underlying
reasons behind this surprising conclusion.
Despite appearing nonsensical, the equation has significant applications in physics. Understanding its relevance requires a closer look at the concept of vacuum energy.
Vacuum Energy in Physics
When does this peculiar sum emerge in the realm of physics? It primarily arises in the calculation of vacuum energy, particularly within string theory. In quantum physics, energy is derived from the
vibrations of various fields, such as electromagnetic fields. The vacuum energy correlates with the frequencies of these modes, leading to equations like this:
However, physics is grounded in measurable quantities. Since we cannot quantify infinity, the computed vacuum energy cannot logically be infinite either, creating a paradox. Thus, we must reconcile
this contradiction.
A Leap of Faith
To progress, we must entertain the possibility that our equation is incomplete. This could indicate that our models are not fully developed. While they may provide results, those results might only
be applicable within a specific context.
This situation is akin to a calculator that functions well with small numbers but produces erroneous outputs when faced with very large values. To make sense of the infinite sum, we must account for
our lack of understanding—a process termed regularization.
Regularization: Taming the Infinite
To address the discrepancies in the formula, we need to parameterize our inaccuracies. This parameterization must possess certain characteristics:
1. It should treat each term uniformly, honoring the symmetries present in the physical world.
2. It must taper off for large values of n to appropriately manage the infinite aspects.
3. It needs to be adjustable, allowing us to observe varying effects.
In mathematical terms, we introduce a regularizer ( R(n, N) ) that encapsulates our uncertainties in physics. This regularizer should range between 0 and 1, diminishing as n increases. Additionally,
N governs the extent of our ignorance. Thus, we can reformulate our sum as follows:
To clarify this concept, let's examine an explicit example of exponential regularization.
Exponential Regularization Explained
The regulator ( e^{-n/N} ) is effective because, with a large N, it minimally affects the terms until we reach substantial n values. The beauty of this regularization lies in its ability to yield
exact computations!
As we let N approach infinity, and by ignoring the infinite ( N^2 ) terms, we discover that ( 1 + 2 + 3 ... = -1/12 ) holds true, albeit under specific conditions.
Addressing Concerns of Ambiguity
This leads to a critical question: how can we disregard infinity without it seeming like a fallacy? The answer lies in the fact that the outcome must be finite, as it represents an observable
quantity. The appearance of infinity suggests our theories are incomplete or that our mathematical techniques require refinement. By integrating corrections into our theoretical framework, we can
align our models with natural phenomena.
Mathematical Uniqueness
From a mathematical perspective, we can ensure the uniqueness of our answer, provided we adhere to the requirements established by physics-driven regularization. Prof. Terry Tao's blog offers a
rigorous proof supporting this assertion.
While theoretical justifications may appear tenuous, experimental validation is the ultimate arbiter. It is feasible to measure the attractive forces resulting from the vacuum energy calculations, as
demonstrated in this experimental paper — the results affirm the validity of our approach (albeit in a different but analogous context).
Conclusion: Embracing the Infinite
In conclusion, to comprehend the infinite, we must acknowledge our limitations and amend our results. Fortunately, a consistent and rational method exists to rectify our misunderstandings and yield
finite outcomes. Perhaps nature imparts a valuable lesson: humility and groundedness can transform the seemingly impossible into the comprehensible.
Appendix: Riemann Zeta Function Insights
Some articles may hastily link the infinite sum to the Riemann zeta function ( zeta(s) ), defined as follows:
While this substitution yields the correct result, it can lead to confusion, as the definition is valid only for ( s > 1 ). A more robust analysis would necessitate complex analysis techniques. For
instance, employing the Abel formula allows us to convert the summation into an integral, where regularization acts as a cap. This intricate manipulation yields results comparable to those obtained
through exponential regularization, linking finite terms to ( zeta(2) ), which is intertwined with ( zeta(-1) ) via the zeta reflection identity.
In the first video, "Numberphile v. Math: the truth about 1+2+3+...=-1/12," the complexities of this equation are explored in detail, shedding light on its mathematical implications.
The second video, "ASTOUNDING: 1 + 2 + 3 + 4 + 5 + -", further delves into the astounding nature of this equation, revealing its intriguing aspects and significance.
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TS Inter 1st Year Maths 1A Study Material Pdf Download | TS Intermediate Maths 1A Solutions
TS Inter 1st Year Maths 1A Textbook Solutions Pdf Download | TS Inter Maths 1A Study Material Pdf
TS Inter 1st Year Maths 1A Functions Solutions
TS Inter 1st Year Maths 1A Mathematical Induction Solutions
TS Inter 1st Year Maths 1A Matrices Solutions
TS Inter 1st Year Maths 1A Addition of Vectors Solutions
TS Inter 1st Year Maths 1A Products of Vectors Solutions
TS Inter 1st Year Maths 1A Trigonometric Ratios upto Transformations Solutions
TS Inter 1st Year Maths 1A Trigonometric Equations Solutions
TS Inter 1st Year Maths 1A Inverse Trigonometric Functions Solutions
TS Inter 1st Year Maths 1A Hyperbolic Functions Solutions
TS Inter 1st Year Maths 1A Properties of Triangles Solutions
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projection matrix
A new family of limited-memory variable metric or quasi-Newton methods for unconstrained minimization is given. The methods are based on a positive definite inverse Hessian approximation in the form
of the sum of identity matrix and two low rank matrices, obtained by the standard scaled Broyden class update. To reduce the rank of matrices, various … Read more
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The Set
A TriAx piece consists of three dicubes (1*1*2), one in a north-south direction, one in an east-west direction and one in an up-down direction. The set of 28 pieces consists of all different ways a
piece can be formed this way. Two pieces are considered to be equal if by spatial rotations they can exactly be mapped onto each other, with the dicubes in corresponding positions.
Compare pieces 27 and 28. These are:
(I) unequal if we take the relative positions of the composing dicubes into account.
(II) equal if we only consider the spaced occupied by these pieces.
Since each piece consists of 6 unit-cubes, the set comprises 168 cubes. Thus, spacial blocks with a volume of 168 cubes may be filled with it. Because a TriAx-piece is never flat, edges of such
blocks will measure at least 2 unit-edges.
Block solutions
The prime factors of 168 are 2 * 2 * 2 * 3 * 7. Thus the following blocks may be tried:
a) 2*12*7 b) 2*6*14 c) 4*6*7 d) 8*3*7
e) 4*3*14 f) 2*2*42 g) 2*4*21 h) 2*3*28
or combinations of smaller blocks, e.g.:
i) 2 blocks of 2*6*7
j) 2 blocks of 3*4*7
k) 4 blocks resp. 2*3*4 + 3*3*4 + 4*3*4 + 5*3*4
l) 4 blocks resp. 2 of 3*3*4 + 2 of 4*3*4
Solutions of k) render, by repositioning, solutions of c), d), e) en j). A solution of i) likewise renders solutions of a) and b).
With the exception of f) and g), solutions for these possibilities have been found. Some 'block-solutions' are impossible because of the following theorem, the proof of which is omitted here:
A block of arbitrary size with edges x, y and z units, cannot be filled with TriAx-pieces if x*y*z is not divisible by 12.
This means that there are no solutions for, for instance, 3 blocks of 2*4*7 or for a block of 2*4*10 and a block of 2*4*11.
A single dicube in particular cannot be triplicated (3*3*6), thus the 'universal' triplication of all piece, by using three triplicated dicubes, does not exist.
Although no solutions have been found for the cases f) and g), theorem 1 does not exclude these. Nevertheless f) seems, by intuition, a hard case.
A 4-boxes solution
Obviously these four boxes, 2*3*4, 3*3*4, 4*3*4 and 5*3*4 can be rearranged to one in several ways: 14*3*4, 7*6*4 and 7*3*8.
A triplication of #9
An interesting property of the set is that it allows triplications of any piece with the 27 pieces that are left. This renders essentially 15 new puzzles because 2 of the 28 pieces are unchanched
by reflection. The remaining 26 pieces come in pairs and of course solving one of a pair means solving both.
Triplications have been found for all pieces!
As a consequence of theorem 1, it is not possible to triplicate a di-cube. If it were, all triplications would be solved at once, but a di-cube's volume, 3*3*6, is not divisible by 12.
Also, it is not possible to obtain a triplication by dividing and rearranging another triplication: Each triplication is a puzzle in itself.
This property appears to be unique for the TriAx.
TriAx ©
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Sequential properties of function spaces with the compact-open topology
The main results of the paper are:. (1)If X is metrizable but not locally compact topological space, then Ck(X) contains a closed copy of S[2], and hence does not have the property AP;(2)For any
zero-dimensional Polish X, the space C[k](X,2) is sequential if and only if X is either locally compact or the derived set X×, is compact; and(3)All spaces of the form C[k](X,2), where X is a
non-locally compact Polish space whose derived set is compact, are homeomorphic, and have the topology determined by an increasing sequence of Cantor subspaces, the nth one nowhere dense in the (n+1)
Bibliographical note
Funding Information:
✩ The third named author acknowledges the support of the FWF grant P19898-N18. We thank the referees for comments and suggestions.
✩ The third named author acknowledges the support of the FWF grant P19898-N18. We thank the referees for comments and suggestions.
Funders Funder number
Austrian Science Fund P19898-N18
• AP
• Arens space
• Compact-open topology
• Fréchet-Urysohn
• Polish space
• Pytkeev property
• Sequential
• Sequential fan
• WAP
Dive into the research topics of 'Sequential properties of function spaces with the compact-open topology'. Together they form a unique fingerprint.
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Graph Theory Notes - EduTechLearners
Graph Theory Notes
Graph theory has many practical applications in various disciplines including, to name a few, biology, computer science, economics, engineering, informatics,linguistics, mathematics, medicine, and
social science, graphs are excellent modelling tools, etc.
Now the question arises that What is a Graph ? right…
A linear graph (or simply a graph) G = (V,E) consists of a set of objects V = {v1, v2,…..} called vertices, and another set E = {e1, e2,…..} whose elements are called edges, such that each edge ek
is identified with an unordered pair (vi , vj) of vertices. So graph theory is simply a overall picture of graphs and it’s application.
So, Start learning such awesome in handwritten notes specially designed for the students under teacher’s surveillance. These notes are handwritten Notes of the Computer Subject “Graph Theory” in Pdf
format.These notes enables students to understand every concept of the the term “Graph Theory”. Graph Theory Notes can be easily download from EduTechLearners without signup or login.
Unit wise notes consists of following topics:-
Unit 1 – Introduction
Unit 2 – Planar Graphs
Unit 3 – trees
Unit 4 – Optimization and Matching
Unit 5 – Fundamental Principles of Counting
Unit 6 – The Principle of Inclusion and Exclusion
Unit 7 – Generating Functions
Unit 8 – Recurrence Relations
For more clarification on the graph Theory please refer the below book which explains all the concepts of Graph Theory in full details.
Graph Theory with Applications to Engineering By NARSINGH DEO
For more Queries related to above notes comment below…!!!!
If you wants some more notes on any of the topics please mail to us or comment below. We will provide you as soon as possible and if you want your’s notes to be published on our site then feel free
to contribute on EduTechLearners or mail your content to contribute@edutechlearners.com ( The contents will be published by your Name).
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// In secondary classrooms
The NAD Secondary Mathematics Committee (NADSMC), which met in June 2017, was composed of five math teachers and two union secondary leaders. They reviewed textbooks from five publishers. Materials
for consideration were presented by Houghton Mifflin Harcourt (HMH), Kendall Hunt, Pearson, and McGraw Hill/Glencoe. The NAD Curriculum Committee has accepted the recommendation and publishes the
following list of approved mathematics textbooks. Mathematics textbooks were previously reviewed in 2012 and adopted in 2013.
Algebra I
Algebra 1 is the critical element in secondary mathematics education. Topics introduced in Algebra 1 provide the foundation students require for future success in high school mathematics, critical
thinking, and problem solving. The primary goal in Algebra 1 is to help students transfer their concrete mathematical knowledge to more abstract algebraic generalizations.
Algebra II
Mathematics provides the conceptual basis for the structure of many things around us. This course is an extension of the Algebra 1 curriculum. Topics that were first introduced in Algebra 1 will be
built upon and applied to problems that require higher order thinking skills. Additional topics will also be introduced in a variety of methods, including self-discovery activities, group project and
presentations, and teacher led class discussions. Algebra 2 builds a foundation of mathematics for those students going on to PreCalculus and/or students who are college bound. Along with many
colleges, a majority of careers require a successful completion of an Algebra 2 course.
Due to the variation in curricular offerings across the NAD, the committee did not review Calculus textbooks. For further information and assistance, please consult your union conference office of
Geometry is a critical component of a mathematics education because students are required to relate concepts from Algebra I and Algebra II to geometric phenomena. This course requires students to
focus on logical proof and critical thinking when solving problems or evaluating arguments. Much of the course is focused on preparation for Pre-Calculus, and thus several concepts and activities
preview topics from these higher-level mathematics courses and analytic geometry. Most post-secondary institutions require students to take a geometry course in high school because this subject
provides the necessary mathematical tools for complex reasoning and solving problems in the sciences, technology, engineering, and many skilled trades and professions.
This course is designed to cover topics in Algebra ranging from polynomial, rational, and exponential functions to conic sections. Trigonometry concepts such as Law of Sines and Cosines will be
introduced. Students will then begin analytic geometry and calculus concepts such as limits, derivatives, and integrals. This class is important for any student planning to take a college algebra or
college pre-calculus class.
Consumer Math
Due to the variation in curricular offerings across the NAD, the committee did not review Consumer Math textbooks. For further information and assistance, please consult your union conference office
of education.
Due to the variation in curricular offerings across the NAD, the committee did not review Calculus textbooks. For further information and assistance, please consult your union conference office of
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How to Save Trained Model In Tensorflow?
In order to save a trained model in TensorFlow, you can make use of the tf.train.Saver() class. This class provides methods to save and restore variables from checkpoints.
First, you need to create an instance of the tf.train.Saver() class, typically at the end of your model's code. You can specify which variables you want to save by passing them as a list or a
dictionary to the var_list parameter of the Saver() constructor.
After creating the Saver object, you can save the model by calling the save() method and passing the TensorFlow session object (sess) along with the path where you want to save the model checkpoint.
The path can be the file name of the checkpoint or a directory path. TensorFlow will create multiple files with the same prefix but different suffixes for the checkpoint.
Here's an example:
1 import tensorflow as tf
3 # Define your model here...
4 # ...
6 # Create a Saver instance
7 saver = tf.train.Saver()
9 # Training code...
10 # ...
12 # Save the model
13 saver.save(sess, "/path/to/save/model.ckpt")
This will save the model in the specified directory or file path. The saved model will include the variables you have specified in the Saver constructor.
To restore the saved model, you can use the tf.train.Saver() class again. You need to define the same variables as before and call the restore() method, passing the session and the path of the saved
model checkpoint.
1 # Create a Saver instance
2 saver = tf.train.Saver()
4 # Your model definition...
5 # ...
7 # Restore the saved variables
8 saver.restore(sess, "/path/to/save/model.ckpt")
By doing this, the model will be restored with the saved variable values, and you can use it for further operations or predictions.
How to save a TensorFlow model using the tf.train.Saver() API?
To save a TensorFlow model using the tf.train.Saver() API, you can follow the steps below:
1. Define your computation graph and train your model.
2. Create a tf.train.Saver() object within your TensorFlow session.
3. Specify which variables you want to save by passing them to the var_list parameter of the Saver() constructor. If left empty, it saves all trainable variables by default.
4. Call the saver.save() method with the session and the path where you want to save the model.
5. The saver.save() method returns the path to the checkpoint file, which will have a .ckpt extension.
Here's an example code to save a TensorFlow model:
1 import tensorflow as tf
3 # Build and train your model
4 # ...
6 # Create a Saver object
7 saver = tf.train.Saver(var_list=None) # Saves all trainable variables by default
9 # Start a session
10 with tf.Session() as sess:
11 # Run your model and training steps
12 # ...
14 # Save the model
15 save_path = saver.save(sess, "path/to/save/model.ckpt")
16 print("Model saved in path: %s" % save_path)
In the above example, the model is saved in the "path/to/save/" directory with the name "model.ckpt". The checkpoint file is saved with a .ckpt extension.
What is the benefit of saving a TensorFlow model at specific checkpoints?
Saving a TensorFlow model at specific checkpoints has several benefits:
1. Resume training: Checkpoints allow you to save the current state of your model during training. If the training process is interrupted, you can resume from the last saved checkpoint and continue
training from where you left off. This helps to ensure that you don't lose progress and allows for better utilization of computing resources.
2. Avoid overfitting: While training deep learning models, there is always a risk of overfitting, where the model becomes overly specialized to the training data and performs poorly on unseen data.
Saving checkpoints helps to monitor the model's performance during training. You can periodically evaluate the model on a validation dataset using the saved checkpoints and choose the one with
the best performance, preventing overfitting.
3. Model deployment: Saved checkpoints serve as snapshots that capture the weights and parameters of the model at a given point in time. These checkpoints can be used for deploying the trained model
into production environments without the need for retraining. It allows you to restore the model and use it for inference tasks, making predictions on new unseen data.
4. Model evaluation: Saving checkpoints enables you to evaluate and compare different model architectures or hyperparameters. By training multiple models with different configurations and saving
checkpoints, you can assess their performance on a common evaluation metric and select the best model checkpoint based on those metrics.
Overall, saving TensorFlow model checkpoints provides flexibility in terms of training interruption, overfitting prevention, model deployment, and model evaluation, facilitating the development and
deployment of effective machine learning models.
How do I save a TensorFlow model for future use?
To save a TensorFlow model for future use, you can follow these steps:
1. Import the required libraries:
1 import tensorflow as tf
1. After training your model, save the model using tf.saved_model.save(). This function saves the model in the SavedModel format, which is a flexible and efficient serialization format for
TensorFlow models.
1 model = ... # define and train your model
3 # Provide a directory path where you want to save the model
4 path_to_save = 'path/to/save/model'
6 # Save the model
7 tf.saved_model.save(model, path_to_save)
1. The saved model will create a folder with several subdirectories within the specified path. These subdirectories contain the model's variables, assets, and the TensorFlow runtime.
2. Now, you can load the saved model using tf.saved_model.load(). This function loads the model from the SavedModel directory and returns a TensorFlow SavedModel object.
1 # Load the saved model
2 loaded_model = tf.saved_model.load(path_to_save)
1. To use the loaded model, you need to retrieve the model's inference function. The inference function can vary depending on your model's specific structure. For example, if your model has a
default serving signature function called serving_default, you can use it as follows:
1 inference_fn = loaded_model.signatures["serving_default"]
1. Now you can use the inference_fn to make predictions or perform inference on new data as required.
Note: TensorFlow also provides options to save and load models in various other formats like TensorFlow SavedModel, TensorFlow Hub, Keras HDF5 format, etc. The aforementioned method showed how to
save and load models using TensorFlow's SavedModel format.
What is the difference between saving and exporting a TensorFlow model?
Saving a TensorFlow model and exporting a TensorFlow model serve different purposes.
Saving a TensorFlow model: Saving a TensorFlow model refers to the process of serializing the model's architecture and trained weights to disk. It allows you to save the entire state of the model,
including the model's structure, weights, optimizer's state, and the configuration. Saved models can be easily loaded and used for further training or inference in TensorFlow or converted to other
formats such as TensorFlow Lite or TensorFlow.js.
Exporting a TensorFlow model: Exporting a TensorFlow model refers to the process of converting a trained model into a format that is compatible with serving or deploying the model for production use
or inference. The exported model typically includes the model's architecture and weights, but not necessarily other components like the optimizer's state or training configuration. Exported models
are intended for deployment purposes and can be used for serving predictions, integrating with other systems, or running on different platforms or devices.
In summary, saving a TensorFlow model ensures you can load and continue training or performing inference on it within the TensorFlow ecosystem. On the other hand, exporting a TensorFlow model
prepares the model for deployment by converting it into a format that can be used by other systems or platforms.
How to save a TensorFlow model in the SavedModel format?
To save a TensorFlow model in the SavedModel format, you can follow these steps:
1. Train and build your model using TensorFlow.
2. After training, create a directory where you want to save the model.
3. Call the tf.saved_model.save() function to save the model in the SavedModel format. Pass the model and the directory path as arguments. Here's an example:
1 import tensorflow as tf
3 # Build and train your model
5 # Create a directory to save the model
6 export_dir = '/path/to/save/model'
8 # Save the model
9 tf.saved_model.save(model, export_dir)
The model argument is your trained TensorFlow model, and export_dir is the path to the directory where you want to save the model.
1. TensorFlow will save the model in the SavedModel format, which includes the model architecture, variables, and any other associated assets.
2. Inside the export directory, TensorFlow will create a subdirectory named with a timestamp (e.g., YYYYMMDD-HHMMSS). This subdirectory will contain the saved model. The saved model can be easily
loaded in TensorFlow for inference or retraining using tf.saved_model.load().
What is the recommended way to save model weights in TensorFlow?
In TensorFlow, there are multiple ways to save and load model weights:
1. tf.keras API: If you are using the tf.keras API, you can save and load model weights using the model.save() and tf.keras.models.load_model() methods. model.save() saves the entire model,
including the architecture, optimizer, and weights, in a single directory. You can later load the saved model using tf.keras.models.load_model().
1 # Saving model weights
2 model.save('model_directory')
4 # Loading model weights
5 loaded_model = tf.keras.models.load_model('model_directory')
1. Checkpoints: TensorFlow provides the tf.train.Checkpoint mechanism for saving and restoring model weights without saving the entire model. You can manually manage checkpoint objects and save
specific variables or layers of the model.
1 # Create checkpoint object
2 checkpoint = tf.train.Checkpoint(model=model)
4 # Save model weights
5 checkpoint.save('model_weights.ckpt')
7 # Restore model weights
8 checkpoint.restore('model_weights.ckpt')
1. tf.train.Saver: If you are using the lower-level TensorFlow API, you can use tf.train.Saver to save and restore model weights. It allows you to save specific variables or the entire graph.
1 # Create a saver object
2 saver = tf.train.Saver()
4 # Save model weights
5 saver.save(sess, 'model_weights.ckpt')
7 # Restore model weights
8 saver.restore(sess, 'model_weights.ckpt')
It is important to note that when saving and loading model weights, ensure that the model architecture and the order of the layers/variables remain the same. Otherwise, loading the weights might lead
to incorrect results or errors.
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Theme: Verified Artificial Intelligence and Machine Learning
If you are interested in any of the topics listed below or in the general area, please contact me (Nils Jansen). My webpage can be found here. I'm always happy to discuss directions and topics, and
this webpage may often be outdated due to recent research developments.
We conduct the supervision of Bachelor and Master theses very interactively. Often, one or more PhD student is involved, and in regular meetings with all students there is the opportunity to give
short presentations and discuss vividly. Some of the topics may be co-supervised with Jurriaan Rot or Sebastian Junges.
With the rapidly growing application of artificial intelligence and machine learning in social life, considering for instance autonomous systems such as self-driving cars, the need for verified or
dependable guarantees against potentially fatal accidents is self-evident. An important question is then, how systems can be made safe. A very nice article discussing challenges and open problems can
be found here. If one of the open problems is of interest, let me know.
Generally, we offer topics within the following exciting directions. In most of the directions, we have both Bachelor and Master topics available, and there are deep theoretical challenges as well as
rather practical topics that mostly involve an implementation.
1) Neural Network Robustness
We are interested in the correctness and robustness of neural networks. Join us to write your thesis on a very recent and interesting topic in collaboration with TNO or the University of Antwerp.
All topics would involve at least a lightweight implementation.
2) Decision-Making under Uncertainty
The real world is inherently uncertain. We look into formal models for agents that have to operate under various sources of such uncertainty, for instance in the form of Markov decision processes.
Topics can be very theoretical or very practical. We use techniques from robust optimization, sampling, abstraction, and many more.
3) Industrial Theses
We offer multiple theses as part of industry projects, for instance with the companies Alfalval, Rolsch, Nexperia, or Canon. As an example for a project, within PrimaVera you can work on the
predictive maintenance of high-tech systems.
If you have a company and are looking for a supervisor, feel free to contact me and we can see if that fits.
4) Human-Robot Interaction
In recent years, the interaction between humans and robots has become more and more relevant to our daily life. The task here is to formalize correctness or confidence measures in the safe
interaction between humans and robots.
The project builds on prior work and may be performed with experts from other universities. See this overview article to get an idea.
5) Model Learning
In order to verify an AI system, a model is needed. However, in most cases, such a model is not directly available. This project will investigate learning techniques for deriving a model that is
amenable to formal verification techniques.
A particular focus can be learning state representations from Atari games and verifying that a player will always make a safe choice. To learn more, have a look at https://arxiv.org/pdf/
1906.08226.pdf and https://arxiv.org/pdf/1806.01363.pdf, or Frits Vaandrager's famous paper on model learning.
6) Robot Control Systems
Real-world robotic systems carry problems that go beyond the current capabilities of AI. We aim to enable machine learning methods for continuous and high-dimensional state spaces for systems that
exhibit various types of noisy sensors. As an example, see this or this paper.
7) Green IT
I am an ambassador of Radboud's Green IT center. Within this program, my personal goal is to derive IT solutions that help towards more sustainability, for instance via improved ressource use or
optimal power management for renewable energies.
8) Case Studies and Examples
An important part of our research is to work with exciting examples, case studies, and driver cases. If you are interested in finding new cool examples with us, there are always topics available.
The task could for instance be to think of a nice problem, and then model it in a formal way so it can be analyzed with our methods.
See below a few examples of our case studies: satellite motion planning, aircraft collision avoidance, and PacMan respectively.
Ongoing and Finished theses
Have a look at the topics that we did so far!
Bachelor theses
• Marnix Suilen (2018, Convex optimization for uncertain Markov decision processes)
• Manuela Bergau (2019, Human-in-the-loop strategy synthesis: PAC-MAN verified)
• Sjoerd Hemels (2019, A comparison of model checking tools for synchronization problems)
• Laura Philipse (2019, Routing Algorithms for Autonomous Agricultural Vehicles, with company Phact)
• Johan Sijtsma (2020, Creating a formal model of the game 2048)
• Niels van Welzen (2020, An iterative version of Tarjan's algorithm)
• Tom Smitjes (2020, Connecting Mixed-Integer Linear Programming and Finite-memory POMDP Strategies)
• Koen Verdenius (2021, A case study for predictive maintenance, ongoing, co-supervised with Marnix Suilen)
Master theses
• Marnix Suilen (2020, Entropy-guided decision making in multiple-environment Markov decision processes, co-supervised with Sebastian Junges)
• Serena Rietbergen (2021, Reward machines for POMDPs, ongoing)
• Okan Ok (2021, Safe Reinforcement Learning With Quasi-Convex Optimisation-Based Model Repair, ongoing)
• Toon Lenaerts (2021, Adversarial POMDPs, ongoing, co-supervised with Marnix Suilen)
• Jeremy Guijt (2021, Explainability for recurrent neural networks, ongoing, with TNO)
• David Kerkkamp (2021, Deep Learning for water pipe networks, with company Rolsch, ongoing)
• Ilse Pool (2021, An MDP model for Scrubber Systems, with company Alfalaval, ongoing, co-supervised with Thom Badings)
• Reinier Joosse (2021, Evaluating Adversarial Attack Detectors using Formal Verification Methods, ongoing, with company Info Support)
• Marck van der Vegt (2021, Model Learning for Probabilistic Systems, ongoing, co-supervised with Jurriaan Rot)
• Pleun Koldewijn (2021, Optimal Execution Problem for FX trading, ongoing, with company ING)
Master internships
• Jeremy Guijt (2019, Shielding POMDPs)
• Marnix Suilen (2019, Robust Policy Synthesis for Uncertain POMDPs via Convex Optimization)
• Reinier Joosse (2020, Applying Machine Learning to create Control Software for a Model Factory, with company ICT group)
• David Kerkkamp (2020, Learning State Representations for Formal Verification using Atari 2600 Games
• Serena Rietbergen (2020, Optimization of Maintenance Intervention)
• Pleun Koldewijn (2021, Finite-state controller for POMDPs, ongoing, co-supervised with Sebastian Junges)
• Marck van der Vegt (2021, Processing and Generating Observations for Uncertain MDPs, co-supervised with Marnix Suilen)
• Anass Fakir (2021, Creating a toolchain to automate policy calculations for POMDPs, co-supervised with Marnix Suilen)
The concrete topics below may be partially outdated, but feel free to have a look or contact me about them!
Thesis Topics
a) Data-consistent Machine Learning with Formal Guarantees
In machine learning algorithms, a system model is learned based on observation of the real world. Upon further observation, this model may be subject to change. The problem of applying formal
verification to such a model is that it is not a well-defined and fixed model of the system at hand. This project proposes to robustify a current learned model against further changes based on future
observations. If it can be verified, that given system specifications are satisfied for this robust model, all future observations that are consistent with the intervals will not change this fact.
The project will involve the development of underlying formalisms and a small prototype implementation using Python libraries for reinforcement learning and model checking using the probabilistic
model checker Storm.
b) Human-in-the-loop Verification of Partially Observable Environments
Take a problem that is sufficiently modeled by a POMDP, for instance, an agent navigating in a partially unknown environment. Intuitively, the agent does not really know, in which state the system
is, in fact, there is a probability distribution over possible states, called the belief state. This yields an infinite number of possibilities.
The task is to make use of human capabilities in inferring from partial observation to resolve or reduce the belief space of the POMDP. We propose to immerse a human either actively or as an observer
into a verification scenario. Possibilities range from a sophisticated virtual reality immersion to a plain 2-D representation of the scenario. Check this article on gathering human feedback from the
OpenAI blog.
c) Adversarial Examples to Guide Learning
Reinforcement learning largely suffers from the problem, that it is basically uninformed exploration of the state space of a system. We want to study the exploitation of counterexamples to identify
corner cases and critical parts of state spaces to guide learning. For applications in Deep Neural Networks, check this article.
d) Permissive Scheduling for Partially Observable Environments
Verification for POMDPs is not feasible in general. We propose to use so-called permissive schedulers computed on the observable part of a POMDP to allow for as much freedom in decision making as
In a POMDP exploration scenario such as typical motion planning for robots, this will reduce the number of situations, where the exact system state needs to be computed: Consider a situation where
the belief state is a distribution over a certain number of states. In other words, there are many possibilities of states, the system can be in. Usually, in order to make a decision, this needs to
be based on all possibilities. If now each state has a precomputed permissive choice that agrees with all states, no further computation is necessary.
e) Robotics Case Studies for Formal Verification
This very broad and practical project aims at finding and modeling typical robotics case studies that can be formally verified. Here, we are not fixed on probabilistic systems but also care for
hybrid systems or even plain transitions systems. Many case studies may for instance be found in this book about probabilistic robotics by Sebastian Thrun.
f) System Design under Sensor Imprecision Probabilities
Many applications are structured in the sense that the underlying systems exploit a certain structured in the sense that there exist multiple but finitely many pre-specified options for the system
Imagine a system incorporating surveillance of the environment using a number of sensors. Each of these sensors has a fixed probability for false alarms or misclassification, depending on the quality
of the sensor. Increased quality may come at an increased acquisition or maintenance cost. We propose to model such a given structured scenario as a parametric Markov decision process, where the
several fixed options influence both the cost and the system dynamics.
The goal is to synthesize a system configuration that adheres to given cost and safety specifications. The thesis topic involves work on an existing and practical case study and exploitation of
current theoretical achievements.
g) Permissive Scheduling for Uncertain Probabilistic Systems
Traditional schedulers (also referred to as strategies or adversaries) resolve all non-determinism in a system such as Markov decision process (MDP). When probabilities are not fixed but given by an
interval range of probabilities, synthesis methods require permissiveness for a scheduler in the sense, that only critical behavior (that is, non-deterministic choices) is excluded.
The goal is to work on a case study focused on autonomous systems such as self-driving cars, where schedulers need to be synthesized that guarantee safe behavior while allowing for as much freedom as
possible. Prior work can be found here.
h) Convex Optimization for Parametric Markov Decision Processes
Many verification problems for parametric probabilistic systems can be cast as a nonlinear program. Solving such programs is in general infeasible. However, for a certain subclass, so-called convex
programs, efficient methods exist. In many applications such as networks, planning problems, and control theory in general, it is even stated that only convex problems are well stated, consider this
great book by Stephen Boyd.
Some old topics where we could still find relevant thesis topics are below.
1) Model checking and machine learning for the 2048 game.
The famous game 2048 exhibits probabilistic behavior. The task is to use machine learning and verification methods to compute an optimal AI for the game, or to prevent humans from making disastrous
moves. See here for instance. Topic can be rather applied or very theoretical.
2) Memoryless (positional) strategies (policies) for partially observable Markov decision processes (POMDPs).
POMDPs are extremely important in many robotics and AI applications. As in the standard AI textbook by Stuart Russell and Peter Norvig: Partially observable MDPs ... are usually viewed as much more
difficult than ordinary MDPs. We cannot avoid POMDPs, however, because the real world is one.
The task is to work on methods to compute strategies with finite or no memory, which is (contrary to infinite memory) still feasible but not investigated very much. Approaches can range from
dedicated branch and bound methods (algorithmic), mixed-integer linear programming (optimization), or SMT solving (satisfiability). All approaches that lead to convincing results are expected to be
able to lead to research publications. Topic is rather theoretical with some practical aspects.
3) Model checking and machine learning for PAC-MAN.
We seek methods that help a machine learning algorithm to solve PAC-MAN. See for instance the videos uploaded here for a new result we recently obtained. The paper is available here. Approaches range
from model checking, machine learning to SMT solving. Topic can be rather applied or very theoretical.
Building on prior work, this project concerns the application of several concepts from convex optimization to well-known verification problems for parametric probabilistic systems that cannot be
solved efficiently.
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Monte Carlo simulation of state-space models
[Y,X] = simulate(Mdl,numObs) simulates one sample path of observations (Y) and states (X) from a fully specified, state-space model (Mdl). The software simulates numObs observations and states per
sample path.
[Y,X] = simulate(Mdl,numObs,Name,Value) returns simulated responses and states with additional options specified by one or more Name,Value pair arguments.
For example, specify the number of paths or model parameter values.
[Y,X,U,E] = simulate(___) additionally simulate state disturbances (U) and observation innovations (E) using any of the input arguments in the previous syntaxes.
Simulate States and Observations of Time-Invariant State-Space Model
Suppose that a latent process is an AR(1) model. The state equation is
where ${u}_{t}$ is Gaussian with mean 0 and standard deviation 1.
Generate a random series of 100 observations from ${x}_{t}$, assuming that the series starts at 1.5.
T = 100;
ARMdl = arima('AR',0.5,'Constant',0,'Variance',1);
x0 = 1.5;
rng(1); % For reproducibility
x = simulate(ARMdl,T,'Y0',x0);
Suppose further that the latent process is subject to additive measurement error. The observation equation is
${y}_{t}={x}_{t}+{\epsilon }_{t},$
where ${\epsilon }_{t}$ is Gaussian with mean 0 and standard deviation 0.75. Together, the latent process and observation equations compose a state-space model.
Use the random latent state process (x) and the observation equation to generate observations.
Specify the four coefficient matrices.
A = 0.5;
B = 1;
C = 1;
D = 0.75;
Specify the state-space model using the coefficient matrices.
Mdl =
State-space model type: ssm
State vector length: 1
Observation vector length: 1
State disturbance vector length: 1
Observation innovation vector length: 1
Sample size supported by model: Unlimited
State variables: x1, x2,...
State disturbances: u1, u2,...
Observation series: y1, y2,...
Observation innovations: e1, e2,...
State equation:
x1(t) = (0.50)x1(t-1) + u1(t)
Observation equation:
y1(t) = x1(t) + (0.75)e1(t)
Initial state distribution:
Initial state means
Initial state covariance matrix
x1 1.33
State types
Mdl is an ssm model. Verify that the model is correctly specified using the display in the Command Window. The software infers that the state process is stationary. Subsequently, the software sets
the initial state mean and covariance to the mean and variance of the stationary distribution of an AR(1) model.
Simulate one path each of states and observations. Specify that the paths span 100 periods.
[simY,simX] = simulate(Mdl,100);
simY is a 100-by-1 vector of simulated responses. simX is a 100-by-1 vector of simulated states.
Plot the true state values with the simulated states. Also, plot the observed responses with the simulated responses.
title({'True State Values and Simulated States'})
legend({'True state values','Simulated state values'})
title({'Observed Responses and Simulated responses'})
legend({'Observed responses','Simulated responses'})
By default, simulate simulates one path for each state and observation in the state-space model. To conduct a Monte Carlo study, specify to simulate a large number of paths.
Simulate State-Space Models Containing Unknown Parameters
To generate variates from a state-space model, specify values for all unknown parameters.
Explicitly create this state-space model.
$\begin{array}{c}{x}_{t}=\varphi {x}_{t-1}+{\sigma }_{1}{u}_{t}\\ {y}_{t}={x}_{t}+{\sigma }_{2}{\epsilon }_{t}\end{array}$
where ${u}_{t}$ and ${\epsilon }_{t}$ are independent Gaussian random variables with mean 0 and variance 1. Suppose that the initial state mean and variance are 1, and that the state is a stationary
A = NaN;
B = NaN;
C = 1;
D = NaN;
mean0 = 1;
cov0 = 1;
stateType = 0;
Mdl = ssm(A,B,C,D,'Mean0',mean0,'Cov0',cov0,'StateType',stateType);
Simulate 100 responses from Mdl. Specify that the autoregressive coefficient is 0.75, the state disturbance standard deviation is 0.5, and the observation innovation standard deviation is 0.25.
params = [0.75 0.5 0.25];
y = simulate(Mdl,100,'Params',params);
title 'Simulated Responses';
xlabel 'Period';
The software searches for NaN values column-wise following the order A, B, C, D, Mean0, and Cov0. The order of the elements in params should correspond to this search.
Estimate Monte-Carlo Forecasts of State-Space Model
Suppose that the relationship between the change in the unemployment rate (${x}_{1,t}$) and the nominal gross national product (nGNP) growth rate (${x}_{3,t}$) can be expressed in the following,
state-space model form.
$\left[\begin{array}{c}{x}_{1,t}\\ {x}_{2,t}\\ {x}_{3,t}\\ {x}_{4,t}\end{array}\right]=\left[\begin{array}{cccc}{\varphi }_{1}& {\theta }_{1}& {\gamma }_{1}& 0\\ 0& 0& 0& 0\\ {\gamma }_{2}& 0& {\
varphi }_{2}& {\theta }_{2}\\ 0& 0& 0& 0\end{array}\right]\left[\begin{array}{c}{x}_{1,t-1}\\ {x}_{2,t-1}\\ {x}_{3,t-1}\\ {x}_{4,t-1}\end{array}\right]+\left[\begin{array}{cc}1& 0\\ 1& 0\\ 0& 1\\ 0&
1\end{array}\right]\left[\begin{array}{c}{u}_{1,t}\\ {u}_{2,t}\end{array}\right]$
$\left[\begin{array}{c}{y}_{1,t}\\ {y}_{2,t}\end{array}\right]=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 1& 0\end{array}\right]\left[\begin{array}{c}{x}_{1,t}\\ {x}_{2,t}\\ {x}_{3,t}\\ {x}_{4,t}\
end{array}\right]+\left[\begin{array}{cc}{\sigma }_{1}& 0\\ 0& {\sigma }_{2}\end{array}\right]\left[\begin{array}{c}{\epsilon }_{1,t}\\ {\epsilon }_{2,t}\end{array}\right],$
• ${x}_{1,t}$ is the change in the unemployment rate at time t.
• ${x}_{2,t}$ is a dummy state for the MA(1) effect on ${x}_{1,t}$.
• ${x}_{3,t}$ is the nGNP growth rate at time t.
• ${x}_{4,t}$ is a dummy state for the MA(1) effect on ${x}_{3,t}$.
• ${y}_{1,t}$ is the observed change in the unemployment rate.
• ${y}_{2,t}$ is the observed nGNP growth rate.
• ${u}_{1,t}$ and ${u}_{2,t}$ are Gaussian series of state disturbances having mean 0 and standard deviation 1.
• ${\epsilon }_{1,t}$ is the Gaussian series of observation innovations having mean 0 and standard deviation ${\sigma }_{1}$.
• ${\epsilon }_{2,t}$ is the Gaussian series of observation innovations having mean 0 and standard deviation ${\sigma }_{2}$.
Load the Nelson-Plosser data set, which contains the unemployment rate and nGNP series, among other things.
Preprocess the data by taking the natural logarithm of the nGNP series, and the first difference of each. Also, remove the starting NaN values from each series.
isNaN = any(ismissing(DataTable),2); % Flag periods containing NaNs
gnpn = DataTable.GNPN(~isNaN);
u = DataTable.UR(~isNaN);
T = size(gnpn,1); % Sample size
y = zeros(T-1,2); % Preallocate
y(:,1) = diff(u);
y(:,2) = diff(log(gnpn));
This example proceeds using series without NaN values. However, using the Kalman filter framework, the software can accommodate series containing missing values.
To determine how well the model forecasts observations, remove the last 10 observations for comparison.
numPeriods = 10; % Forecast horizon
isY = y(1:end-numPeriods,:); % In-sample observations
oosY = y(end-numPeriods+1:end,:); % Out-of-sample observations
Specify the coefficient matrices.
A = [NaN NaN NaN 0; 0 0 0 0; NaN 0 NaN NaN; 0 0 0 0];
B = [1 0;1 0 ; 0 1; 0 1];
C = [1 0 0 0; 0 0 1 0];
D = [NaN 0; 0 NaN];
Specify the state-space model using ssm. Verify that the model specification is consistent with the state-space model.
Mdl =
State-space model type: ssm
State vector length: 4
Observation vector length: 2
State disturbance vector length: 2
Observation innovation vector length: 2
Sample size supported by model: Unlimited
Unknown parameters for estimation: 8
State variables: x1, x2,...
State disturbances: u1, u2,...
Observation series: y1, y2,...
Observation innovations: e1, e2,...
Unknown parameters: c1, c2,...
State equations:
x1(t) = (c1)x1(t-1) + (c3)x2(t-1) + (c4)x3(t-1) + u1(t)
x2(t) = u1(t)
x3(t) = (c2)x1(t-1) + (c5)x3(t-1) + (c6)x4(t-1) + u2(t)
x4(t) = u2(t)
Observation equations:
y1(t) = x1(t) + (c7)e1(t)
y2(t) = x3(t) + (c8)e2(t)
Initial state distribution:
Initial state means are not specified.
Initial state covariance matrix is not specified.
State types are not specified.
Estimate the model parameters, and use a random set of initial parameter values for optimization. Restrict the estimate of ${\sigma }_{1}$ and ${\sigma }_{2}$ to all positive, real numbers using the
'lb' name-value pair argument. For numerical stability, specify the Hessian when the software computes the parameter covariance matrix, using the 'CovMethod' name-value pair argument.
params0 = rand(8,1);
[EstMdl,estParams] = estimate(Mdl,isY,params0,...
'lb',[-Inf -Inf -Inf -Inf -Inf -Inf 0 0],'CovMethod','hessian');
Method: Maximum likelihood (fmincon)
Sample size: 51
Logarithmic likelihood: -170.92
Akaike info criterion: 357.84
Bayesian info criterion: 373.295
| Coeff Std Err t Stat Prob
c(1) | 0.06750 0.16548 0.40791 0.68334
c(2) | -0.01372 0.05887 -0.23302 0.81575
c(3) | 2.71201 0.27039 10.03006 0
c(4) | 0.83816 2.84586 0.29452 0.76836
c(5) | 0.06274 2.83470 0.02213 0.98234
c(6) | 0.05196 2.56873 0.02023 0.98386
c(7) | 0.00272 2.40771 0.00113 0.99910
c(8) | 0.00016 0.13942 0.00113 0.99910
| Final State Std Dev t Stat Prob
x(1) | -0.00000 0.00272 -0.00033 0.99973
x(2) | 0.12237 0.92954 0.13164 0.89527
x(3) | 0.04049 0.00016 256.77783 0
x(4) | 0.01183 0.00016 72.52162 0
EstMdl is an ssm model, and you can access its properties using dot notation.
Filter the estimated, state-space model, and extract the filtered states and their variances from the final period.
[~,~,Output] = filter(EstMdl,isY);
Modify the estimated, state-space model so that the initial state means and covariances are the filtered states and their covariances of the final period. This sets up simulation over the forecast
EstMdl1 = EstMdl;
EstMdl1.Mean0 = Output(end).FilteredStates;
EstMdl1.Cov0 = Output(end).FilteredStatesCov;
Simulate 5e5 paths of observations from the fitted, state-space model EstMdl. Specify to simulate observations for each period.
numPaths = 5e5;
SimY = simulate(EstMdl1,10,'NumPaths',numPaths);
SimY is a 10-by- 2-by- numPaths array containing the simulated observations. The rows of SimY correspond to periods, the columns correspond to an observation in the model, and the pages correspond to
Estimate the forecasted observations and their 95% confidence intervals in the forecast horizon.
MCFY = mean(SimY,3);
CIFY = quantile(SimY,[0.025 0.975],3);
Estimate the theoretical forecast bands.
[Y,YMSE] = forecast(EstMdl,10,isY);
Lb = Y - sqrt(YMSE)*1.96;
Ub = Y + sqrt(YMSE)*1.96;
Plot the forecasted observations with their true values and the forecast intervals.
h = plot(dates(end-numPeriods-9:end),[isY(end-9:end,1);oosY(:,1)],'-k',...
ylabel('Change in the unemployment rate')
legend(h([1,2,4:6]),{'Observations','MC forecasts',...
'95% forecast intervals','Theoretical forecasts',...
'95% theoretical intervals'},'Location','Best')
title('Observed and Forecasted Changes in the Unemployment Rate')
h = plot(dates(end-numPeriods-9:end),[isY(end-9:end,2);oosY(:,2)],'-k',...
ylabel('nGNP growth rate')
legend(h([1,2,4:6]),{'Observations','MC forecasts',...
'95% MC intervals','Theoretical forecasts','95% theoretical intervals'},...
title('Observed and Forecasted nGNP Growth Rates')
Input Arguments
numObs — Number of periods per path to simulate
positive integer
Number of periods per path to generate variants, specified as a positive integer.
If Mdl is a time-varying model (see Decide on Model Structure), then the length of the cell vector corresponding to the coefficient matrices must be at least numObs.
If numObs is fewer than the number of periods that Mdl can support, then the software only uses the matrices in the first numObs cells of the cell vectors corresponding to the coefficient matrices.
Data Types: double
Name-Value Arguments
Specify optional pairs of arguments as Name1=Value1,...,NameN=ValueN, where Name is the argument name and Value is the corresponding value. Name-value arguments must appear after other arguments, but
the order of the pairs does not matter.
Before R2021a, use commas to separate each name and value, and enclose Name in quotes.
Example: [Y,X] = simulate(Mdl,numObs,'NumPaths',100)
Output Arguments
Y — Simulated observations
matrix | cell matrix of numeric vectors
Simulated observations, returned as a matrix or cell matrix of numeric vectors.
If Mdl is a time-invariant model with respect to the observations (see Decide on Model Structure), then Y is a numObs-by-n-by-numPaths array. That is, each row corresponds to a period, each column
corresponds to an observation in the model, and each page corresponds to a sample path. The last row corresponds to the latest simulated observations.
If Mdl is a time-varying model with respect to the observations, then Y is a numObs-by-numPaths cell matrix of vectors. Y{t,j} contains a vector of length n[t] of simulated observations for period t
of sample path j. The last row of Y contains the latest set of simulated observations.
Data Types: cell | double
U — Simulated state disturbances
matrix | cell matrix of numeric vectors
Simulated state disturbances, returned as a matrix or cell matrix of vectors.
If Mdl is a time-invariant model with respect to the state disturbances, then U is a numObs-by-h-by-numPaths array. That is, each row corresponds to a period, each column corresponds to a state
disturbance in the model, and each page corresponds to a sample path. The last row corresponds to the latest simulated state disturbances.
If Mdl is a time-varying model with respect to the state disturbances, then U is a numObs-by-numPaths cell matrix of vectors. U{t,j} contains a vector of length h[t] of simulated state disturbances
for period t of sample path j. The last row of U contains the latest set of simulated state disturbances.
Data Types: cell | double
E — Simulated observation innovations
matrix | cell matrix of numeric vectors
Simulated observation innovations, returned as a matrix or cell matrix of numeric vectors.
If Mdl is a time-invariant model with respect to the observation innovations, then E is a numObs-by-h-by-numPaths array. That is, each row corresponds to a period, each column corresponds to an
observation innovation in the model, and each page corresponds to a sample path. The last row corresponds to the latest simulated observation innovations.
If Mdl is a time-varying model with respect to the observation innovations, then E is a numObs-by-numPaths cell matrix of vectors. E{t,j} contains a vector of length h[t] of simulated observation
innovations for period t of sample path j. The last row of E contains the latest set of simulated observations.
Data Types: cell | double
Simulate states from their joint conditional posterior distribution given the responses by using simsmooth.
[1] Durbin J., and S. J. Koopman. Time Series Analysis by State Space Methods. 2nd ed. Oxford: Oxford University Press, 2012.
Version History
Introduced in R2014a
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Algebra _ Remainder theorem, factor theorem and synthetic division - SACE Mathematics
Algebra _ Remainder theorem, factor theorem and synthetic division
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Algebra _ Remainder theorem, factor theorem and synthetic division
The Remainder Theorem states that the remainder that we end up with when synthetic division is applied actually gives us the functional value. Another use is finding factors and zeros. The Factor
Theorem states that if the functional value is 0 at some value c, then x - c is a factor and c is a zero.
Added March 2022
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The Diffusion Equation (Part 1)
May 17th, 2015
The Diffusion Equation (Part 1)
Diffusion is the complicated process by which particles in two or more different substances mix together as a result of random motion. The particles move around, bump into each other, and
redistribute themselves in a random way over time. Amazingly, after enough time has passed, the substances will be completely mixed together with no help from anybody stirring the mixture. We see
diffusion in many areas of physics and chemistry - mixing solutions, Brownian motion, osmosis, etc.
A particle undergoing diffusion will experience huge numbers of collisions every second from all of the other particles randomly moving about and bumping into each other. Classical mechanics is not
equipped to handle this problem because of the huge number of particles and collisions. Instead, we will turn to statistical physics to tell us something about the system.
Deriving the Diffusion Equation
We can’t possibly keep track of every single particle that diffuses through a mixture. Instead, we will keep track of the number density of particles in every unit of space at every unit of time. We
will call this number density \(\phi(x,t)\). Notice that we are only looking at particle density along a single direction (x). We are only going to derive the diffusion equation in one dimension for
now, just to get a good feel for what it entails. I might derive the three-dimensional diffusion equation at a later date (although it is pretty straightforward to do).
Let’s start our analysis with the continuity equation. The continuity equation states that the amount of “stuff” that goes into some volume of space minus the amount of “stuff” that leaves that
volume of space, must equal the change in the amount of “stuff” in that volume of space. We can write that as:
\[$$\frac{\partial \phi}{\partial t} + abla \cdot \vec{j} = 0$$\]
In the equation above, \(\phi\) is the number density of particles at some point in space and time while \(\vec{j}\) is the flux of particles entering or leaving that point in space at that time. The
continuity equation is a broad expression for any conserved quantity. We can combine the continuity equation with Fick’s first law in order to derive the diffusion equation.
\[$$\vec{j} = -D abla \phi$$\] \[$$\frac{\partial \phi}{\partial t} + abla \cdot \left [ -D abla \phi \right ] = 0$$\]
Let’s assume that the diffusivity is a constant. In other words, \(D\) does not depend on position (\(x\)), time (\(t\)), or number density (\(\phi\)). Then:
\[ \frac{\partial \phi}{\partial t} + \nabla \cdot \left [ -D \nabla \phi \right ] &= 0\\[0.1cm] \frac{\partial \phi}{\partial t} - D \nabla \cdot \left [ \nabla \phi \right ] &= 0\\[0.1cm] \frac{\
partial \phi}{\partial t} - D \nabla^2 \phi &= 0 \]
We are only going to analyze the diffusion equation in one dimension (along the \(x\)-axis), so the expression becomes:
\[$$\frac{\partial \phi}{\partial t} - D \frac{\partial^2 \phi}{\partial x^2} = 0$$\] \[$$\frac{\partial \phi}{\partial t} = D \frac{\partial^2 \phi}{\partial x^2}$$\]
Solving the Diffusion Equation
Oftentimes when you solve the diffusion equation, you focus on the technique of separation of variables. Paul’s Online Notes is a great resource if you are not sure how to go through this procedure.
I personally am not a huge fan of separation of variables for this formula because your final answer is an infinite series of sines and exponentials. It is a technically correct answer, but it is
also difficult to draw real and useful conclusions from it.
I want to tackle this problem in a different way using the Laplace Transform. Yes, this is slightly more advanced math, but it will yield a pretty cool answer that actually makes intuitive sense.
Let’s begin by taking the Laplace transform of each side of the diffusion equation. I have included the arguments \((x,t)\) for the number density, just to keep everything clear. I write the Laplace
transform of the number density as \(\mathscr{L} \phi(x,t) = \Phi(x,s)\). Otherwise, I simply used a table of Laplace transforms and some math:
\[$$\frac{\partial \phi(x,t)}{\partial t} = D \frac{\partial^2 \phi(x,t)}{\partial x^2}$$\] \[ \mathscr{L} \left [ \frac{\partial \phi(x,t)}{\partial t} \right ] &= \mathscr{L} \left [ D \frac{\
partial^2 \phi(x,t)}{\partial x^2} \right ]\\[0.1cm] s\Phi(x,s) - \phi(x,0) &= D \frac{\partial^2 \Phi(x,s)}{\partial x^2}\\[0.1cm] D\frac{\partial^2 \Phi(x,s)}{\partial x^2} - s\Phi(x,s) &= -\phi
(x,0) \]
We have converted a differential equation for \(\phi\) in terms of \(x\) and \(t\) to a new differential equation for \(\Phi\) in terms of \(x\) only. The one for \(\Phi\) is much easier to solve
than the one for \(\phi\). In fact, if you have taken a course in differential equations, you might be able to solve it by inspection alone. Let’s work it out in detail. The total solution to the
differential equation is the sum of the homogeneous and particular solutions:
\[$$\Phi = \Phi_h + \Phi_p$$\]
First, we need to solve for the homogeneous solution. Let \(\Phi_h = e^{\lambda x}\). Remember, this differential equation is only in terms of \(x\), so variable \(s\) is treated as a constant. Then:
\[ D\frac{\partial^2 \Phi_h(x,s)}{\partial x^2} - s\Phi_h(x,s) &= 0\\[0.1cm] D\frac{\partial^2 e^{\lambda x}}{\partial x^2} - se^{\lambda x} &= 0\\[0.1cm] D \lambda^2 e^{\lambda x} - se^{\lambda x} &
= 0\\[0.1cm] D \lambda^2 - s &= 0\\[0.1cm] \lambda &= \pm \sqrt{\frac{s}{D}} \]
The homogeneous solution is:
\[$$\Phi_h = C_1 e^{x \sqrt{\frac{s}{D}}} + C_2 e^{-x \sqrt{\frac{s}{D}}}$$\]
Now we can solve for the particular solution. We do not know what \(\phi(x,0)\) is, but we can still use the method of variation of parameters to get a genral solution. We start off with the
\[$$W = \begin{vmatrix} e^{x \sqrt{\frac{s}{D}}} & e^{-x \sqrt{\frac{s}{D}}}\\[0.1cm] \sqrt{\frac{s}{D}} e^{x \sqrt{\frac{s}{D}}} & -\sqrt{\frac{s}{D}} e^{-x \sqrt{\frac{s}{D}}} \end{vmatrix}$$\] \[
W &= -\sqrt{\frac{s}{D}} e^{x \sqrt{\frac{s}{D}}} e^{-x \sqrt{\frac{s}{D}}} - \sqrt{\frac{s}{D}} e^{x \sqrt{\frac{s}{D}}} e^{-x \sqrt{\frac{s}{D}}}\\[0.1cm] &= -\sqrt{\frac{s}{D}} - \sqrt{\frac{s}
{D}}\\[0.1cm] &= -2 \sqrt{\frac{s}{D}} \]
And then write out the integral for the particular solution:
\[ \Phi_p &= e^{x \sqrt{\frac{s}{D}}} \int_0^x \frac{e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0)}{W}d\alpha - e^{-x \sqrt{\frac{s}{D}}} \int_0^x \frac{e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha,
0)}{W}d\alpha\\[0.1cm] &= \frac{e^{x \sqrt{\frac{s}{D}}}}{W} \int_0^x e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha - \frac{e^{-x \sqrt{\frac{s}{D}}}}{W} \int_0^x e^{\alpha \sqrt{\frac{s}
{D}}} \phi(\alpha, 0) d\alpha\\[0.1cm] &= -\frac{e^{x \sqrt{\frac{s}{D}}}}{2} \sqrt{\frac{D}{s}} \int_0^x e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha + \frac{e^{-x \sqrt{\frac{s}{D}}}}{2}
\sqrt{\frac{D}{s}} \int_0^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha\\[0.1cm] \]
The general solution then becomes:
\[$$\Phi = C_1 e^{x \sqrt{\frac{s}{D}}} + C_2 e^{-x \sqrt{\frac{s}{D}}} -\frac{e^{x \sqrt{\frac{s}{D}}}}{2} \sqrt{\frac{D}{s}} \int_0^x e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha + \frac
{e^{-x \sqrt{\frac{s}{D}}}}{2} \sqrt{\frac{D}{s}} \int_0^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha\\[0.1cm]$$\] \[$$\Phi = \left [ C_1 - \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{-\
alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right ]e^{x \sqrt{\frac{s}{D}}} + \left [ C_2 + \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right
] e^{-x \sqrt{\frac{s}{D}}}$$\]
We can simplify the expression further by adding the condition that \(\phi(x,t)\) be bounded (and then, by extension, \(\Phi(x,s)\) also be bounded). In other words, the limit of \(\Phi\) as variable
\(x\) approaches infinity must equal zero.
Then the first part of the expression yields:
\[ \lim_{x \rightarrow \infty} \left [ C_1 - \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right ]e^{x \sqrt{\frac{s}{D}}} &= 0\\[0.1cm] \lim_{x \
rightarrow \infty} \left [ C_1 - \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right ] &= 0\\[0.1cm] C_1 - \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^\
infty e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha &= 0\\[0.1cm] \] \[$$C_1 = \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^\infty e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha$$\]
And the second part of the expression yields:
\[ \lim_{x \rightarrow -\infty} \left [ C_2 + \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right ] e^{-x \sqrt{\frac{s}{D}}} &= 0\\[0.1cm] \lim_{x \
rightarrow -\infty} \left [ C_2 + \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right ] &= 0\\[0.1cm] C_2 + \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^{-\
infty} e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha &= 0 \] \[ C_2 &= - \frac{1}{2} \sqrt{\frac{D}{s}} \int_0^{-\infty} e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha\\[0.1cm] &= \
frac{1}{2} \sqrt{\frac{D}{s}} \int_{-\infty}^0 e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \]
Alright. Take \(C_1\) and \(C_2\). Plug them back into the expression for \(\Phi\). Do some cancelling. I swear we are in the home stretch..
\[ \Phi &= e^{x \sqrt{\frac{s}{D}}}\frac{1}{2} \sqrt{\frac{D}{s}} \int_0^\infty e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha - e^{x \sqrt{\frac{s}{D}}} \frac{1}{2} \sqrt{\frac{D}{s}} \int_0
^x e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha\\[0.1cm] &+ e^{-x \sqrt{\frac{s}{D}}} \frac{1}{2} \sqrt{\frac{D}{s}} \int_{-\infty}^0 e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha +
e^{-x \sqrt{\frac{s}{D}}}\frac{1}{2} \sqrt{\frac{D}{s}} \int_0^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \] \[$$\Phi = e^{x \sqrt{\frac{s}{D}}}\frac{1}{2} \sqrt{\frac{D}{s}} \int_x^\
infty e^{-\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha + e^{-x \sqrt{\frac{s}{D}}} \frac{1}{2} \sqrt{\frac{D}{s}} \int_{-\infty}^x e^{\alpha \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha$$\] \[$$
\Phi = \frac{1}{2} \sqrt{\frac{D}{s}} \int_{-\infty}^\infty e^{-|x - \alpha| \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha$$\]
Ok, assuming that I did not make any stupid math mistakes, we have gotten to a nice, simple(?), expression for \(\Phi\). To finish this problem, I am going to take the inverse Laplace transform (once
again using a table of Laplace transforms):
\[ \phi &= \mathscr{L}^{-1} \left [ \frac{1}{2} \sqrt{\frac{D}{s}} \int_{-\infty}^\infty e^{-|x - \alpha| \sqrt{\frac{s}{D}}} \phi(\alpha, 0) d\alpha \right ]\\[0.1cm] &= \frac{1}{2} \int_{-\infty}^\
infty \mathscr{L}^{-1} \left [ \sqrt{\frac{D}{s}} e^{-|x - \alpha| \sqrt{\frac{s}{D}}} \right ] \phi(\alpha, 0) d\alpha\\[0.1cm] &= \frac{1}{\sqrt{4\pi Dt}} \int_{-\infty}^\infty e^{-|x - \alpha|^2/
(4Dt)} \phi(\alpha, 0) d\alpha \]
The exponent inside the integral sign is called the heat kernel. We will definitely revisit it at a later date - there is lots of interesting structure in that expression. For now, let’s finish up
the problem. Suppose that at time \(t=0\), all of the particles are at the origin. In other words, we can represent the total initial number density as a Dirac delta function -> \(\phi(\alpha,0) = \
delta(\alpha)\). The final expression becomes:
\[ \phi(x,t) &= \frac{1}{\sqrt{4\pi Dt}} \int_{-\infty}^\infty e^{-|x - \alpha|^2/(4Dt)} \delta(\alpha) d\alpha\\[0.1cm] &= \frac{1}{\sqrt{4\pi Dt}} e^{-x^2/(4Dt)} \]
Whew, that was a lot of math today. But we did get an answer for the diffusion equation that does not involve a bunch of sines and exponentials! Instead, it is a strange Gaussian-looking equation
that evolves with time.
Next week, we will visualize this answer and take a look at it on a deeper level.
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The Upper Confidence Bound Algorithm
We now describe the celebrated Upper Confidence Bound (UCB) algorithm that overcomes all of the limitations of strategies based on exploration followed by commitment, including the need to know the
horizon and sub-optimality gaps. The algorithm has many different forms, depending on the distributional assumptions on the noise.
The algorithm is based on the principle of optimism in the face of uncertainty, which is to choose your actions as if the environment (in this case bandit) is as nice as is plausibly possible. By
this we mean that the unknown mean payoffs of each arm is as large as plausibly possible based on the data that has been observed (unfounded optimism will not work — see the illustration on the
right!). The intuitive reason that this works is that when acting optimistically one of two things happens. Either the optimism was justified, in which case the learner is acting optimally, or the
optimism was not justified. In the latter case the agent takes some action that they believed might give a large reward when in fact it does not. If this happens sufficiently often, then the learner
will learn what is the true payoff of this action and not choose it in the future. The careful reader may notice that this explains why this rule will eventually get things right (it will be
“consistent” in some sense), but the argument does not quite explain why an optimistic algorithm should actually be a good algorithm among all consistent ones. However, before getting to this, let us
clarify what we mean by plausible.
Recall that if $X_1, X_2,\ldots, X_n$ are independent and $1$-subgaussian (which means that $\E[X_i] = 0$) and $\hat \mu = \sum_{t=1}^n X_t / n$, then
\Prob{\hat \mu \geq \epsilon} \leq \exp\left(-n\epsilon^2 / 2\right)\,.
Equating the right-hand side with $\delta$ and solving for $\epsilon$ leads to
\Prob{\hat \mu \geq \sqrt{\frac{2}{n} \log\left(\frac{1}{\delta}\right)}} \leq \delta\,.
This analysis immediately suggests a definition of “as large as plausibly possible”. Using the notation of the previous post, we can say that when the learner is deciding what to do in round $t$ it
has observed $T_i(t-1)$ samples from arm $i$ and observed rewards with an empirical mean of $\hat \mu_i(t-1)$ for it. Then a good candidate for the largest plausible estimate of the mean for arm $i$
\hat \mu_i(t-1) + \sqrt{\frac{2}{T_i(t-1)} \log\left(\frac{1}{\delta}\right)}\,.
Then the algorithm chooses the action $i$ that maximizes the above quantity. If $\delta$ is chosen very small, then the algorithm will be more optimistic and if $\delta$ is large, then the optimism
is less certain. We have to be very careful when comparing the above display to \eqref{eq:simple-conc} because in one the number of samples is the constant $n$ and in the other it is a random
variable $T_i(t-1)$. Nevertheless, this is in some sense a technical issue (that needs to be taken care of properly, of course) and the intuition remains that $\delta$ is approximately an upper bound
on the probability of the event that the above quantity is an underestimate of the true mean.
The value of $1-\delta$ is called the confidence level and different choices lead to different algorithms, each with their pros and cons, and sometimes different analysis. For now we will choose $1/\
delta = f(t)= 1 + t \log^2(t)$, $t=1,2,\dots$. That is, $\delta$ is time-dependent, and is decreasing to zero slightly faster than $1/t$. Readers are not (yet) expected to understand this choice
whose pros and cons we will discuss later. In summary, in round $t$ the UCB algorithm will choose arm $A_t$ given by
A_t = \begin{cases}
\argmax_i \left(\hat \mu_i(t-1) + \sqrt{\frac{2 \log f(t)}{T_i(t-1)}}\right)\,, & \text{if } t > K\,; \\
t\,, & \text{otherwise}\,.
The reason for the cases is that the term inside the square root is undefined if $T_i(t-1) = 0$ (as it is when $t = 1$), so we will simply have the algorithm spend the first $K$ rounds choosing each
arm once. The value inside the argmax is called the index of arm $i$. Generally speaking, an index algorithm chooses the arm in each round that maximizes some value (the index), which usually only
depends on current time-step and the samples from that arm. In the case of UCB, the index is the sum of the empirical mean of rewards experienced and the so-called exploration bonus, also known as
the confidence width.
Besides the slightly vague “optimism guarantees optimality or learning” intuition we gave before, it is worth exploring other intuitions for this choice of index. At a very basic level, we should
explore arms more often if they are (a) promising (in that $\hat \mu_i(t-1)$ is large) or (b) not well explored ($T_i(t-1)$ is small). As one can plainly see from the definition, the UCB index above
exhibits this behaviour. This explanation is unsatisfying because it does not explain why the form of the functions is just so.
An alternative explanation comes from thinking of what we expect from any reasonable algorithm. Suppose in some round we have played some arm (let’s say arm $1$) much more frequently than the others.
If we did a good job designing our algorithm we would hope this is the optimal arm. Since we played it so much we can expect that $\hat \mu_1(t-1) \approx \mu_1$. To confirm the hypothesis that arm
$1$ is indeed optimal the algorithm better be highly confident about that other arms are indeed worse. This leads very naturally to confidence intervals and the requirement that $T_i(t-1)$ for other
arms $i\ne 1$ better be so large that
\hat \mu_i(t-1) + \sqrt{\frac{2}{T_i(t-1)} \log\left(\frac{1}{\delta}\right)} \leq \mu_1\,,
because, at a confidence level of $1-\delta$ this guarantees that $\mu_i$ is smaller than $\mu_1$ and if the above inequality did not hold, the algorithm would not be justified in choosing arm $1$
much more often than arm $i$. Then, planning for $\eqref{eq:ucbconstraint}$ to hold makes it reasonable to follow the UCB rule as this will eventually guarantee that this inequality holds when arm
$1$ is indeed optimal and arm $i$ is suboptimal. But how to choose $\delta$? If the confidence interval fails, by which we mean, if actually it turns out that arm $i$ is optimal and by unlucky chance
it holds that
\hat \mu_i(t-1) + \sqrt{\frac{2}{T_i(t-1)} \log\left(\frac{1}{\delta}\right)} \leq \mu_i\,,
then arm $i$ can be disregarded even though it is optimal. In this case the algorithm may pay linear regret (in $n$), so it better be the case that the failure occurs with about $1/n$ probability to
fix the upper bound on the expected regret to be constant for the case when the confidence interval fails. Approximating $n \approx t$ leads then (after a few technicalities) to the choice of $f(t)$
in the definition of UCB given in \eqref{eq:ucb}. With this much introduction, we state the main result of this post:
Theorem (UCB Regret): The regret of UCB is bounded by
\begin{align} \label{eq:ucbbound}
R_n \leq \sum_{i:\Delta_i > 0} \inf_{\epsilon \in (0, \Delta_i)} \Delta_i\left(1 + \frac{5}{\epsilon^2} + \frac{2}{(\Delta_i – \epsilon)^2} \left( \log f(n) + \sqrt{\pi \log f(n)} + 1\right)\
\begin{align} \label{eq:asucbbound}
\displaystyle \limsup_{n\to\infty} R_n / \log(n) \leq \sum_{i:\Delta_i > 0} \frac{2}{\Delta_i}\,.
Note that in the first display, $\log f(n) \approx \log(n) + 2\log\log(n)$. We thus see that this bound scales logarithmically with the length of the horizon and is able to essentially reproduce the
bound that we obtained for the unfeasible version of ETC with $K=2$ (when we tuned the exploration time based on the knowledge of $\Delta_2$). We shall discuss further properties of this bound later,
but now let us present a simpler version of the above bound, avoiding all these epsilons and infimums that make for a confusing theorem statement. By choosing $\epsilon = \Delta_i/2$ inside the sum
leads to the following corollary:
Corollary (UCB Simplified Regret): The regret of UCB is bounded by
R_n \leq \sum_{i:\Delta_i > 0} \left(\Delta_i + \frac{1}{\Delta_i}\left(8 \log f(n) + 8\sqrt{\pi \log f(n)} + 28\right)\right)\,.
and in particular there exists some universal constant $C>0$ such that for all $n\ge 2$, $R_n \le \sum_{i:\Delta_i>0} \left(\Delta_i + \frac{C \log n}{\Delta_i}\right)$.
Note that taking the limit of the ratio of the bound above and $\log(n)$ does not result in the same rate as in the theorem, which is the main justification for introducing the epsilons in the first
place. In fact, as we shall see the asymptotic bound on the regret given in \eqref{eq:asucbbound}, which is derived from~\eqref{eq:ucbbound} by choosing $\epsilon = \log^{-1/4}(n)$, is unimprovable
in a strong sense.
The proof of the theorem relies on the basic regret decomposition identity that expresses the expected regret as the weighted sum of the expected number of times the suboptimal actions are chosen. So
why will $\EE{T_i(n)}$ be small for a suboptimal action $i$? This is based on a couple of simple observations: First, (disregarding the initial period when all arms are chosen once) the suboptimal
action $i$ can only be chosen if its UCB index is higher than that of an optimal arm. Now, this can only happen if the UCB index of action $i$ is “too high”, i.e., higher than $\mu^*-\epsilon>\mu_i$
or the UCB index of that optimal arm is “too low”, i.e., if it is below $\mu^*-\epsilon<\mu^*$. Since the UCB index of any arm is with reasonably high probability an upper bound on the arm’s mean, we
don’t expect the index of any arm to be below its mean. Hence, the total number of times when the optimal arm’s index is “too low” (as defined above) is expected to be negligibly small. Furthermore,
if the sub-optimal arm $i$ is played sufficiently often, then its exploration bonus becomes small and simultaneously the empirical estimate of its mean converges to the true value, making the
expected total number of times when its index stays above $\mu^*-\epsilon$ small.
We start with a useful lemma that will help us quantify the last argument.
Lemma Let $X_1,X_2,\ldots$ be a sequence of independent $1$-subgaussian random variables, $\hat \mu_t = \sum_{s=1}^t X_s / t$, $\epsilon > 0$ and
\kappa = \sum_{t=1}^n \one{\hat \mu_t + \sqrt{\frac{2a}{t}} \geq \epsilon}\,.
Then, $\displaystyle \E[\kappa] \leq 1 + \frac{2}{\epsilon^2} (a + \sqrt{\pi a} + 1)$.
Because the $X_i$ are $1$-subgaussian and independent we have $\E[\hat \mu_t] = 0$, so we cannot expect $\hat \mu_t + \sqrt{2a/t}$ to be smaller than $\epsilon$ until $t$ is at least $2a/\epsilon^2$.
The lemma confirms that this is indeed of the right order as an estimate for $\EE{\kappa}$.
Let $u = 2a \epsilon^{-2}$. Then, by the concentration theorem for subgaussian variables,
&\leq u + \sum_{t=\ceil{u}}^n \Prob{\hat \mu_t + \sqrt{\frac{2a}{t}} \geq \epsilon} \\
&\leq u + \sum_{t=\ceil{u}}^n \exp\left(-\frac{t\left(\epsilon – \sqrt{\frac{2a}{t}}\right)^2}{2}\right) \\
&\leq 1 + u + \int^\infty_u \exp\left(-\frac{t\left(\epsilon – \sqrt{\frac{2a}{t}}\right)^2}{2}\right) dt \\
&= 1 + \frac{2}{\epsilon^2}(a + \sqrt{\pi a} + 1)\,.
Before the proof of the UCB regret theorem we need a brief diversion back to the bandit model. We have defined $\hat \mu_i(t)$ as the empirical mean of the $i$th arm after the $t$th round, which
served us well enough for the analysis of the explore-then-commit strategy where the actions were chosen following a deterministic rule. For UCB it is very useful also to have $\hat \mu_{i,s}$, the
empirical average of the $i$th arm after $s$ observations from that arm, which occurs at a random time (or maybe not at all). To define $\hat \mu_{i,s}$ rigorously, we argue that without the loss of
generality one may assume that the reward $X_t$ received in round $t$ comes from choosing the $T_i(t)$th element from the reward sequence $(Z_{i,s})_{1\le s \le n}$ associated with arm $i$, where $
(Z_{i,s})_s$ is an i.i.d. sequence with $Z_{i,s}\sim P_i$. Formally,
X_t = Z_{A_t,T_{A_t}(t)}\,.
The advantage of introducing $(Z_{i,s})_s$ is that it allows a clean definition (without $Z_{i,s}$, how does one even define $\hat \mu_{i,s}$ if $T_i(n) \leq s$?). In particular, we let
\hat \mu_{i,s} &= \frac{1}{s} \sum_{u=1}^s Z_{i,u}\,.
Note that $\hat \mu_{i,s} = \hat \mu_i(t)$ when $T_i(t)=s$ (formally: $\hat \mu_{i,T_i(t)} = \hat \mu_i(t)$).
Proof of Theorem
As in the analysis of the explore-then-commit strategy we start by writing the regret decomposition.
R_n = \sum_{i:\Delta_i > 0} \Delta_i \E[T_i(n)]\,.
The rest of the proof revolves around bounding $\E[T_i(n)]$. Let $i$ be some sub-optimal arm (so that $\Delta_i > 0$). Following the suggested intuition we decompose $T_i(n)$ into two terms. The
first measures the number of times the index of the optimal arm is less than $\mu_1 – \epsilon$. The second term measures the number of times that $A_t = i$ and its index is larger than $\mu_1 – \
&= \sum_{t=1}^n \one{A_t = i} \nonumber \\
&\leq \sum_{t=1}^n \one{\hat \mu_1(t-1) + \sqrt{\frac{2\log f(t)}{T_1(t-1)}} \leq \mu_1 – \epsilon} + \nonumber \\
&\qquad \sum_{t=1}^n \one{\hat \mu_i(t-1) + \sqrt{\frac{2 \log f(t)}{T_i(t-1)}} \geq \mu_1 – \epsilon \text{ and } A_t = i}\,. \label{eq:ucb1}
The proof of the first part of the theorem is completed by bounding the expectation of each of these two sums. Starting with the first, we again use the concentration guarantee.
\EE{\sum_{t=1}^n \one{\hat \mu_1(t-1) + \sqrt{\frac{2 \log f(t)}{T_1(t-1)}} \leq \mu_1 – \epsilon}}
&= \sum_{t=1}^n \Prob{\hat \mu_1(t-1) + \sqrt{\frac{2 \log f(t)}{T_1(t-1)}} \leq \mu_1 – \epsilon} \\
&\leq \sum_{t=1}^n \sum_{s=1}^n \Prob{\hat \mu_{1,s} + \sqrt{\frac{2 \log f(t)}{s}} \leq \mu_1 – \epsilon} \\
&\leq \sum_{t=1}^n \sum_{s=1}^n \exp\left(-\frac{s\left(\sqrt{\frac{2 \log f(t)}{s}} + \epsilon\right)^2}{2}\right) \\
&\leq \sum_{t=1}^n \frac{1}{f(t)} \sum_{s=1}^n \exp\left(-\frac{s\epsilon^2}{2}\right) \\
&\leq \frac{5}{\epsilon^2}\,.
The first inequality follows from the union bound over all possible values of $T_1(t-1)$. This is an important point. The concentration guarantee cannot be applied directly because $T_1(t-1)$ is a
random variable and not a constant. The last inequality is an algebraic exercise. The function $f(t)$ was chosen precisely so this bound would hold. If $f(t) = t$ instead, then the sum would diverge.
Since $f(n)$ appears in the numerator below we would like $f$ to be large enough that its reciprocal is summable and otherwise as small as possible. For the second term in \eqref{eq:ucb1} we use the
previous lemma.
&\EE{\sum_{t=1}^n \one{\hat \mu_i(t-1) + \sqrt{\frac{2 \log f(t)}{T_i(t-1)}} \geq \mu_1 – \epsilon \text{ and } A_t = i}} \\
&\qquad\leq \EE{\sum_{t=1}^n \one{\hat \mu_i(t-1) + \sqrt{\frac{2 \log f(n)}{T_i(t-1)}} \geq \mu_1 – \epsilon \text{ and } A_t = i}} \\
&\qquad\leq \EE{\sum_{s=1}^n \one{\hat \mu_{i,s} + \sqrt{\frac{2 \log f(n)}{s}} \geq \mu_1 – \epsilon}} \\
&\qquad= \EE{\sum_{s=1}^n \one{\hat \mu_{i,s} – \mu_i + \sqrt{\frac{2 \log f(n)}{s}} \geq \Delta_i – \epsilon}} \\
&\qquad\leq 1 + \frac{2}{(\Delta_i – \epsilon)^2} \left(\log f(n) + \sqrt{\pi \log f(n)} + 1\right)\,.
The first part of the theorem follows by substituting the results of the previous two displays into \eqref{eq:ucb1}. The second part follows by choosing $\epsilon = \log^{-1/4}(n)$ and taking the
limit as $n$ tends to infinity.
Next week we will see that UCB is close to optimal in several ways. As with the explore-then-commit strategy, the bound given in the previous theorem is not meaningful when the gaps $\Delta_i$ are
small. Like that algorithm it is possible to prove a distribution-free bound for UCB by treating the arms $i$ with small $\Delta_i$ differently. Fix $\Delta>0$ to be chosen later. Then, from the
proof of the bound on the regret of UCB we can derive that $\EE{T_i(n)} \le \frac{C \log(n)}{\Delta_i^2}$ holds for all $n\ge 2$ with some universal constant $C>0$. Hence, the regret can be bounded
without dependence on the sub-optimality gaps by
&= \sum_{i:\Delta_i > 0} \Delta_i \E[T_i(n)]
= \sum_{i:\Delta_i < \Delta} \Delta_i \E[T_i(n)] + \sum_{i:\Delta_i \geq \Delta} \Delta_i \E[T_i(n)] \\
&< n \Delta + \sum_{i:\Delta_i \geq \Delta} \Delta_i \E[T_i(n)]
\leq n \Delta + \sum_{i:\Delta_i \geq \Delta} \frac{C \log n}{\Delta_i} \\
&\leq n \Delta + K\frac{C \log n}{\Delta}
= \sqrt{C K n \log(n)}\,,
where in the last step we chose $\Delta = \sqrt{K C \log(n) / n}$, which optimizes the upper bound.
There are many directions to improve or generalize this result. For example, if more is known about the noise model besides that it is subgaussian, then this can often be exploited to improve the
regret. The main example is the Bernoulli case, where one should make use of the fact that the variance is small when the mean is close to zero or one. Another direction is improving the worst-case
regret to match the lower bound of $\Omega(\sqrt{Kn})$ that we will see next week. This requires a modification of the confidence level and a more complicated analysis.
Note 1: Here we argue that there is no loss in generality in assuming that the rewards experienced satisfy $\eqref{eq:rewardindepmodel}$. Indeed, let $T’ = (A’_1,X’_1,\dots,A’_n,X’_n)$ be any
sequence of random variables satisfying that $A_t’ = f_t(A’_1,X’_1,\dots,A’_{t-1},X’_{t-1})$ and that for any $U\subset \R$ open interval
\Prob{X_t’\in U\,|\,A’_1,X’_1,\dots,A’_{t-1},X’_{t-1},A’_t} = P_{A’_t}(U)\,,
where $1\le t\le n$. Then, choosing $(Z_{i,s})_s$ as described in the paragraph before $\eqref{eq:rewardindepmodel}$, we let $T=(A_1,X_1,\dots,A_n,X_n)$ be such that $A_t = f_t(A_1,X_1,\dots,A_
{t-1},X_{t-1})$ and $X_t$ be so that it satisfies $\eqref{eq:rewardindepmodel}$. It is not hard to see then that the distributions of $T$ and $T’$ agree. Hence, there is indeed no loss of generality
by assuming that the rewards are indeed generated by $\eqref{eq:rewardindepmodel}$.
Note 2: The view that $n$ rewards are generated ahead of time for each arm and the algorithm consumes these rewards as it chooses an action was helpful in the proof as it reduced the argument to the
study of averages of independent random variables. The analysis could also have been done directly without relying on the “virtual” rewards $(Z_{i,s})_s$ with the help of martingales, which we will
meet later.
A third model of how $X_t$ is generated could have been that $X_t = Z_{A_t,t}$. We will meet this “skipping model” later when studying adversarial bandits. For the stochastic bandit models we study
here, all these models coincide (they are indistinguishable in the sense described in the first note above).
Note 3: So is the optimism principle universal? Does it always give good algorithms, even in more complicated settings? Unfortunately, the answer is no. The optimism principle leads to reasonable
algorithms when using an action gives feedback that informs the learner about how much the action is worth. If this is not true (i.e., in models where you have to choose action $B$ to learn about the
rewards of action $A$, and choosing action $A$ would not give you information about the reward of action $A$), the principle fails! (Why?) Furthermore, even if all actions give information about
their own value, the optimistic principle may give rise to algorithms whose regret is overly large compared to what could be achieved with more clever algorithms. Thus, in a way, finite-armed
stochastic bandits is a perfect fit for optimistic algorithms. While the more complex feedback models may not make much sense at the moment, we will talk about them later.
The idea of using upper confidence bounds appeared in ’85 in the landmark paper of Lai and Robbins. In this paper they introduced a strategy which plays the leader of the “often sampled” actions
except that for any action $j$ in every $K$th round the strategy is checking whether the UCB index of arm $j$ is higher than the estimated reward of the leader. They proved that this strategy, when
appropriately tuned, is asymptotically unimprovable the same way UCB as we defined it is asymptotically unimprovable (we still owe the definition of this and a proof, which will come soon). The
cleaner UCB idea must have been ready to be found in ’95 because Agrawal and Katehakis & Robbins discovered this idea independently in that year. Auer et al. later modified the strategy slightly and
proved a finite-time analysis.
41 thoughts on “The Upper Confidence Bound Algorithm”
1. Sorry that I just can’t understand
why at a confidence level of 1−δ this guarantees that μi is smaller than μ1 and if the above inequality did not hold (after inequality (3))
2. The statement is slightly informal, but roughly $\hat \mu_i(t-1)$ is an empirical estimate of $\mu_i$ based on $T_i(t-1)$ samples. Since we assumed that the rewards are $1$-subgaussian we know
that for $T_i(t-1) = u$ that $\Prob{\hat \mu_i + \sqrt{\frac{2}{u} \log\left(\frac{1}{\delta}\right)} \leq \mu_i} \leq \delta$. The informality comes from the fact that $T_i(t-1)$ is usually also
a random variable, which makes the analysis a little trickier, but does not change much the intuition.
Note that we treat the concentration of subgaussian random variables in a previous post (https://banditalgs.com/2016/09/14/first-steps-explore-then-commit/)
1. Hi! Thanks for this amazing series of blogs.
I have a small question. I think it might be better to say that the value of 1-δ is called the confidence level (instead of saying δ is the confidence level). δ is sort of like the upper
bound on the probability of error that we allow. Also, using 1-δ will possibly make it more consistent with confidence interval terminology used in statistics. Please correct me if I am
wrong. Thanks! 🙂
1. You are right! I’ll fix this:) Thanks for the comment!
3. Hey Tor, do you mind expanding on inequality (7)? I sort of understand that the first and second terms in the inequality represent the events that the UCB index for the optimal arm is “too low”
and that the UCB index for the sub-optimal arm is “too high”, respectively. However, I’m confused as to how either of these events imply that the UCB index for the sub-optimal arm is less than
the UCB index for the optimal arm in a given round.
1. Hi,
I’ll dub as Tor. The implication is easiest to see by inverting things. We want to see that $A_t=i$ implies that $\UCB_i$ is high, or $\UCB_1$ is low. Well, if $\UCB_i$ was low and $\UCB_1$
was high, then arm $1$ would have been preferred to arm $i$, so it must be that if arm $i$ is selected then either $\UCB_i$ is high, or $\UCB_1$ is low. Does this make sense?
PS: Sorry for the slow reply.
1. I am confused too about this inequality
In the first case ($UCB_1$ is low) the left hand side of the inequality is the index of arm 1, right? Shall we not use $T_1(t-1)$ in the denominator of the square root?
1. And why we compare the indices to $\mu_1 – \epsilon$? Why not simply $\mu_1?
1. Another good point. With $\epsilon=0$, what we would need to bound is the probability of the index of the optimal arm smaller than the optimal mean. The way the index is defined,
if the optimal arm is pulled a fixed, say, $s$ number of times, this probability happens to be constant. This is too large; it would render the bound vacuous. I hope this helps.
And sorry for the slow response; somehow I did not get a notification of the comment, or I just missed it.
2. Hmm, I have not caught this before. True. We should have had $T_1(t-1)$ in the denominator. I have corrected this now, thanks!
1. Hi Csaba, above you mentioned “The way the index is defined, if the optimal arm is pulled a fixed, say, s number of times, this probability happens to be constant.”
I am not really seeing how this is happening and how that would make it ‘too large’. Could you please explain this?
4. Hi, in the UCB Simplified Regret, does the universal constant C rely on suboptimality gaps? I suspect that if the suboptimality gaps are not bounded, we cannot find such a constant C.
Also, in the UCB regret (3), should the last constant be ‘1’, but not ‘3’?
Thank you!
1. Hi Xiang. You’re right on all counts. See the updated theorem for the (hopefully) correct statement of this kind of result.
Thanks for pointing out the bugs!
1. Great thanks for the quick response!
However, after the correction, there still exists a flaw in the final distribution-free bound for UCB. This bound also requires the suboptimality gaps be bounded, right?
1. Hi Xiang! Sorry for the slow response. Where is the bug? The universal constant just relies on bounding constant + $\log \log n$ by $C \log n$, it seems to me.
5. Hi, professor, when reasoning arm 1 is optimal and not the arm j (j \neq 1), we say that we have a 1-\delta level of confidence. But, should we also say arm 1 is optimal compared to all the other
K-1 arms, so the confidence level would be (1-\delta)^{K-1}?
1. Where is this? The trick we use is that we bound the *expected* number of pulls of suboptimal arms. Hence, each suboptimal arm is compared to the optimal arm, one by one, separately, avoiding
the need to argue about multiple suboptimal arms at the same time. I hope this clarifies things.
6. Hey!
It would be a big help for me if you could explain where the infimum condition in the equation for the UCB regret (Eq. (4)) comes from. It is comprehensible in the sense that one wants to keep
the upper bound as small as possible, but why is the range of epsilon chosen like this?
By the way, thanks for creating this blog – I think this is a really nice medium to get into this topic!
7. Greetings,
You define 1/δ=f(t)=1+t*log²(t).
Most authors seem to define 1/δ=f(t)=t (see for example (1) in https://agrawal.wikischolars.columbia.edu/file/view/Lecture+3+part+1.pdf)
You can see that 1/δ in your (3) was replaced by t in his (1).
You do say “If f(t)=t instead, then the sum would diverge.” I am confused because everywhere I say UCB explained they use 1/δ=t…
I am implementing an UCB-based solution for a problem I have… Is there maybe a list of pros-and-cons of several f(t) functions? I could not find any, and I have tried to iterate the literature as
much as I could…
Sorry if I am not making sense. I am a bandit newbie. 🙂
1. Hi Ricardo
The choice of f(t) is very delicate and one has to be careful about comparisons when the underlying variance is different. When the noise is Gaussian with variance V you generally want the
confidence interval to look like
sqrt(2V / T_k log f(t))
where f(t) = t + O(log^p(t)) for as small p as your analysis will allow. Actually for the Gaussian case you may use f(t) = t, but even for subgaussian we do not know if this works. Roughly
speaking things are easy if sum_t 1/f(t) converges. Now in Shipra’s notes the rewards are bounded in [0,1]. Of course they cannot be Gaussian, but the maximum variance (or subgaussian
constant) of [0,1] bounded rewards is 1/4. If we substitute this into the formula above you have a confidence interval of
sqrt(2(1/4) / T_k log f(t)) = sqrt(1/(2T_k) log f(t))
Choosing f(t) = t^2 yields the choice in those notes and this is definitely summable. In general, if what you care about is expected regret, then you want f(t) to be “as small as possible”,
but you will pay a price for this in terms of the variance of the regret, so take care. Finally, there are lots of more sophisticated choices. For example the arm-dependent confidence
sqrt(2V/T_k log(t/T_k)) will give you a big boost. I recently wrote a big literature review on all these choices. Feel free to email me if you want a copy.
8. I am suddenly a bit confused by the analysis for bounding the second term in eq (7). Why is the second inequality true. I understand that removing the intersection with $A_t = i$ gives an upper
bound. But how do we deal with the change to $\hat{\mu}_{T_i(t-1)}$ to $\hat{\mu}_{i,s}$? Thanks for the nice write-up.
9. In the analysis after Eq. (7), why sum on s? Instead, you could write
… \le sum_{t=1}^n exp(-T_1(t-1)(epsilon + sqrt(2 lot f(t)/T_1(t-1)))^2)
\le sum_{t=1}^n exp(-T_1 esilon^2)/f(t)
\le sum_{t=1}^n 1/f(t)
This would be a stronger bound, although ultimately it wouldn’t change the scaling of regret with n. But am I missing something?
P.S. I love your blog!
1. The problem is that T_1(t-1) is a random variable, but we only proved the concentration guarantees hold for a fixed sample size. Actually the bounds do not hold more generally without
slightly increasing the logarithmic term, which you can prove using the law of the iterated logarithm.
A little more directly, it’s always worth checking types. A (unconditional) probability should be a number so I better never write
P(predicate(X, Y)) <= f(X, Y)
because the left-hand side is a number and the right-hand side is a random quantity!
And thanks for the kind work. We hope you like the book even more!
1. I’ll admit — I didn’t totally understand that. Probably because I’m basically a physicist, so I’m sloppy when it comes to probabilities. 😉 But let me attempt to be more rigorous. Dropping
the dependence on t-1, and using d1 as shorthand for \hat{\mu}_1(t-1) – \mu_1, we have
sum_t P(d1 + sqrt(2 log f(t)/T1) < -epsilon)
\le sum_t P(d1 + sqrt(2 log f(t)/T1) < 0)
= sum_t sum_T1 P(d1 + sqrt(2 log f(t)/T1) < 0|T1) P(T1)
\le sum_t (1/f(t)) sum_T1 P(T1)
= sum_t 1/f(t)
Is that any more correct?
1. Now the problem appears in the second inequality. When you condition on the number of samples T1 you change the underlying measure of the samples observed. Under this measure they are
not independent so again the concentration bound cannot be applied. This is a bit counter intuitive, but hopefully the following example sheds some light on the issue.
Consider the case where the rewards are {0,1} valued and we are using a silly algorithm that plays arm 1 until it observes a zero and then plays other arms. When you condition on
having played arm 1 some number of times (say, 100 times), then the empirical mean can only be 1 or 99/100 by the definition of the algorithm. But maybe the true mean is 1/2, which is
not close at all to 1 or 99/100.
Of course the probability that this occurs is very small (because you are unlikely to ever observe 99 heads in a row. And this silly algorithm is not UCB. The point is just that by
conditioning on the sample counts in problems involving sequential design is a risky business. Hence the naive union bound.
1. Thanks. That is indeed subtle.
10. Hi, thanks for posting theses. I have a small question about the sub-gaussian assumption(which leads to equation 1), do they have to have zero means? Or the equation should actually consider it’s
mean and have a confidence band of the difference between empirical mean and the latent mean?
1. The usual definition is that X is 1-subgaussian if E[exp(lambda X)] <= exp(lambda^2/2) for all lambda. This definition implies that X has zero mean (proving this is a good exercise). In this
post we are assuming that Z_{i,s} - mu_i is 1-subgaussian for all i and s.
1. I see, thank you for replying. Before, I thought the reward is assumed as a zero-mean subgaussian.
11. Now I’m confused about the union bound (going from the first to the second line in the equation after Eq. 7). Written in its most general form (suppressing the sum on n), it looks like
P(\hat{mu}(t-1) – mu_1 < g(n,T(t-1))) \le sum_s P(\hat{mu}(s) < g(n,s))
Presumably this should hold for arbitrary g(n,s). But I've been trying to convince myself of that for days, with no luck. Is there an easy way to see (or prove) it?
1. Hi
Let A_s be the event {T(t-1) = s and ^mu_s – mu <= g(s)}. Then the event F = {^mu_{T(t-1)} - mu <= g(T(t-1))} is a subset of the union of A_1,...,A_n since T(t-1) must be between 1 and n.
Then the union bound says that Prob(F) <= Prob(union A_t) <= sum_t Prob(A_t)
1. That’s kind of what I thought. But then I started worrying: For the event F, ^mu_{T(t-1)} and T(t-1) are not independent, whereas for the event A_s, ^mu_s and s are independent. (OK, not
quite independent, since s determines the variance of ^mu_s, but more independent than for F.)
1. I’m not sure what you’re getting at here. The independence of quantities that define F is not being used here. The union bound holds regardless of any independence.
The core is that F is indeed a subset of the union of all A_s and so the probability of F is less than the probability of the union. The second important part is that concentration
analysis can bound the probability of A_s.
By the way, a slightly more straightforward analysis of a simpler algorithm is given in Chapter 7 of the book.
1. Let’s take a specific example, based on your example from an earlier thread:
x_i \in {0,1}; for definiteness let’s say P(x_i=1) = 1/2, so mu_i = 1/2.
strategy: choose arm i until x_i=1; then stop. under this strategy, if T_i = 3 then ^mu_i = 1/3.
let g(s) = 1/2 if s=3 and -1 if s \ne 3.
with this setup,
P(^mu(T_i=3) – mu_i < g(3)) = 1,
sum_s P(^mu_{i,s} – mu_i < g(s)) = P(^mu_{i,3} – mu_i < 1/2) < 1.
Thus, P(^mu(T_i=3) – mu_i < g(3)) is not less than sum_s P(^mu_{i,s} – mu_i < g(s)).
I'm guessing I have a serious misconception, but I can't figure out where.
P.S. The first time I left a comment, it took me forever to get past the robot with the missing eye, because its right eye is missing, and it asks to add the left eye. Is that
part of the test? 😉
2. Hi
Probably the misconception is coming from your definition of P(^mu(T_i=3) – mu_i < g(3)). My interpretation of your argument is that this is meant to be a conditional probability.
These are very difficult to handle in sequential settings and we avoid them. Our argument comes from the following view. To each arm associate a big stack of rewards, which are
sampled independently at the beginning of the game and not observed. Each time the learner chooses an action, it gets the top reward on the stack corresponding to that arm. Then ^
mu_{1,s} is the mean of the first s rewards in the stack corresponding to arm 1. Since these really are independent we can apply our concentration bound to show that with high
probability ^mu_{1,s} + sqrt(2/s log f(t)) is never much smaller than mu_1 - epsilon for any s. Now the event F that ^mu_1(t-1) + sqrt(2/T_1(t-1) log f(t)) < mu_1 - epsilon is
definitely a subset of the union of F_s with s in [n] where F_s is the event that ^mu_{1,s} + sqrt(2/s log f(t)) < mu_1 - epsilon. This is true because T_1(t-1) must be in [n] =
{1,2,...,n}. Hence Prob(^mu_1(t-1) + sqrt(2/T_1(t-1) log f(t)) < mu_1 - epsilon) <= sum_{s=1}^n Prob(^mu_{1,s} + sqrt(2/s log f(t)) < mu_1 - epsilon)) Notice that in the left-hand
side probability there is no conditioning. By the way, in the pdf book we have two versions of UCB, the first of which has slightly worse bounds than what we present here, but an
easier proof (see Chapters 7 and 8).
12. I don’t really understand your argument, in the sense that it implies (to me) that P(^mu_i(t-1) – mu_i < g(T_i(t-1)) = P(^mu_{i,s} – mu_i < g(s)) and we know that's wrong.
But you are right that I'm interpreting the above probability as a conditional. If I instead thought of it as P((^mu_i(t-1) – mu_i < g(T_i(t-1)) & (T_i(t-1) was reached)), it makes much more
sense. Under this interpretation, if I take the strategy of stopping when x_i=1, then (using my previous example) when T_i(t-1)=3:
P(^mu_i(t-1) – mu_i < g(3)) = P(observing 0, 0, 1 from arm i).
And in general, P(^mu_i(t-1) – mu_i < g(T_i(t-1)) is the probability of observing a particular set of data. Given that, the union bound makes perfect sense.
Is that a reasonably way of thinking about this?
I know you keep mentioning a simpler bound with an easier proof, but it's driving me nuts that I don't understand this one.
1. I think you’re on the right track. So indeed none of the probabilities in this post are conditional. $\Prob{A}$ is the probability that $A$ happens and in this case $A$ is the event that $\
hat \mu_1(t-1) + \sqrt{2/T_1(t-1) \log f(t)} \le \mu_1 – \epsilon$.
Now we can write this event as a union of other events:
$A = \cup_s A_s$ where $A_s$ is the event that $\hat \mu_1(t-1) + \sqrt{2/T_1(t-1) \log f(t)} \le \mu_1 – \epsilon \text{ and } T_1(t-1) = s$.
On the event $A_s$ we have $T_1(t-1) = s$, so $A_s$ can also be written as the event that $\hat \mu_{1,s} + \sqrt{2/s \log f(t)} \le \mu_1 – \epsilon \text{ and } T_1(t-1) = s$. Now this is a
subset of the event $B_s$ defined to occur when $\hat \mu_{1,s} + \sqrt{2/s \log f(t)} \le \mu_1 – \epsilon$ (clearly if $A_s$ happens, then $B_s$ also happens). So now we have $\Prob{A} = \
Prob{\cup_s A_s} \le \sum_s \Prob{A_s} \le \sum_s \Prob{B_s}$.
First inequality is the union bound. Second inequality because $B_s$ is a subset of $A_s$. The price we pay for the fact that $T_1(t-1)$ is a random quantity is that we must sum over all its
possible values when applying our concentration inequality.
13. Hi Csaba, above you mentioned “The way the index is defined, if the optimal arm is pulled a fixed, say, s number of times, this probability happens to be constant.”
I am not really seeing how this is happening and how that would make it ‘too large’. Could you please explain this?
1. Simply, plugging in $\epsilon=0$, we have $\Prob{\hat \mu_{1,s} + \sqrt{\frac{2 \log f(t)}{s}} \leq \mu_1}\le \exp\left(-\frac{s\left(\sqrt{\frac{2 \log f(t)}{s}} \right)^2}{2}\right)=\frac
{1}{f(t)}$, which is too large for our purposes. I hope this makes sense.
14. I am wondering if we can also utilize the upper confidence bound for making life decisions?
Usually, the same life decision could not be made twice. That is, even if we are able to reduce uncertainty by making a specific choice, we would never be able to choose among the same
alternative choices again. Is UCB still applicable in such situations?
1. UCB is not applicable in this case, for exactly the reason you point out. Life is a reinforcement learning problem, and a very difficult one.
Nevertheless, ideas based on optimism can work in reinforcement learning with some modification and assumptions. Somehow you need to construct confidence sets about what the world is like and
then act as if the world is as nice as plausibly possible. You’ll have to make assumptions to do this.
Another big caveat. Optimism works well in bandits because you can never suffer too much regret with one wrong decision. This is obviously not true more generally. So caution is advised!
Tali Sharot’s book “Optimism Bias” is a nice exploration of optimistic behavior in humans, which maybe would interest you.
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Program to find Pivot Element of a Sorted and Rotated Array
• Write a program in C to find pivot element of a sorted and rotated array using binary search.
Given an sorted integer array of size N which is also rotated by an unknown position. Input Array is not monotonically increasing as it is rotated at some unknown pivot element. We have to find the
pivot element of array.
Pivot element is the only element in input array which is smaller than it's previous element. A pivot element divided a sorted rotated array into two monotonically increasing array.
For Example :
Sorted Rotated Array : 4 5 6 7 8 1 2 3
1 is the Pivot Element
Let inputArray be a sorted and rotated integer array of size N and we want to find the pivot element(minimum element).
By linearly searching input Array
• In sorted and rotated array, pivot element(minimum element) is the only element which is smaller than it's previous element.
• Traverse inputArray from index 0 to N-1 and search for an element inputArray[i] which is smaller than previous element inputArray[i-1].
By using modified binary search
Algorithm to find pivot element of a rotated array.
• Initialize leftIndex and rightIndex to 0 and N-1 respectively.
• If leftIndex == rightIndex(size of the array is 1), return leftIndex.
• Find the middle index as (leftIndex + rightIndex)/2. Let middle index be mid.
• Check if inputArray[mid] is a pivot element. If (inputArray[mid-1] > inputArray[mid] < inputArray[mid+1]) is true then inputArray[mid] is pivot element.
• If inputArray[leftINdex] >= inputArray[mid] then recursively search on sub left sub array from index leftIndex to mid-1.
• Else recursively search on sub array from index mid+1 to rightIndex.
Time Complexity : O(Logn)
C program to find pivot element of rotated array
#include <stdio.h>
int getPivotElement(int *array, int left, int right){
if (right < left) /* Array not rotated */
return -1;
/* Only element in sub array */
if (right == left)
return left;
/* Find the mid element */
int middle = (left + right)/2;
/* Only the pivot element will be
more than it's next element */
if (middle < right && array[middle] > array[middle + 1])
return middle;
if (middle > left && array[middle] < array[middle - 1])
return middle-1;
if (array[left] >= array[middle]){
/* Pivot element is between left and mid index */
return getPivotElement(array, left, middle-1);
} else {
/* Pivot element is between mid and right index */
return getPivotElement(array, middle + 1, right);
int main(){
int array[11] = {16, 18, 22, 25, 1, 3, 5, 6, 7, 10, 14};
printf("Pivot Element : %d \n", array[getPivotElement(array, 0, 10) + 1]);
return 0;
Pivot Element : 1
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4.04 Lengths of similar figures
A stick of height $1.1$1.1 m casts a shadow of length $2.2$2.2 m. At the same time, a tree casts a shadow of $6.2$6.2 m.
If the tree has a height of $h$h metres, solve for $h$h.
A $4.9$4.9 m flagpole casts a shadow of $8.6$8.6 m. Amelia casts a shadow of $2.5$2.5 m.
If Amelia is $h$h metres tall, solve for $h$h correct to one decimal place.
A school building reaching $h$h metres high casts a shadow of $30$30 m while a $3$3 m high tree casts a shadow of $6$6 m. Solve for $h$h.
Get full access to our content with a Mathspace account
|
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Cutting Length Calculator - CivilGang
What is a Cutting Length Calculator?
A cutting length calculator is a web-based tool that helps users calculate the total cutting length of cylindrical pieces (e.g., bars or pipes) based on their diameter and the number of pieces.
This is commonly used in manufacturing, construction, and various industries.
Why Use a Cutting Length Calculator?
1. Material Estimation: Ensures the accurate estimation of cutting lengths required for manufacturing or construction projects.
2. Cost Estimation: Helps in budgeting by providing precise information on material needs and costs.
3. Project Efficiency: Aids in planning and executing projects that involve cutting cylindrical materials.
4. Resource Management: Efficiently manages material resources, reducing waste and costs.
Cutting Length Calculator
Enter the diameter and the number of pieces to calculate the total cutting length:
Total Cutting Length: 0 inches
|
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1. Introduction
Artificial Neural Networks arise from the interpretation of the functioning of a human brain. Although the first to relate computing to the human brain was Alan Turing in 1936, it was Warren
McCulloch and Walter Pitts who created the theory of how a neuron works^10.
Joel W. Johnson in 2012 presents the main factors affecting vote inequality among incumbent cohorts (members of the same party and district), indicating the strong influence of vote splitting
incentives on electoral environments focused on the candidates^11.
The study developed by Ching-Hsing Wang in 2014 indicates that awareness and emotional stability can significantly increase female participation in electoral votes, but have no effect on male
participation^18. Furthermore, openness to experience has opposite effects on male and female participation. As openness to experience increases, men are more likely to vote, while women are less
likely to cast ballots. However, extraversion and agreeableness are not associated with participation, regardless of gender^18. Orlando D'Adamo in July 2015, makes a study of the usefulness and scope
of the use of social networks during electoral campaigns. The authors in ^5 present results of an investigation that analyzes the use of social networks made by candidates for deputies and senators
for the city of Buenos Aires in the legislative elections.
In 2017, Dimitrios Xefteris studied several factors influencing electoral voting, including religion, race, culture, and others, making use of optimized data access for the Data Warehouse maximizes
differentiated voting participation^8. Artificial Neural Networks or ANN for its acronym in English, have numerous publications made each year. Several researchers have done studies on the different
neural networks, the simplest being the multilayer perceptron which has a pattern recognition architecture, which means that its neurons are only connected from one layer to the next^6^,^13.
Artificial Neural Networks can be used to predict the difficulties of the electoral process, since they have been used for different complex problems with adaptive and cognitive mechanisms of human
learning. The literature tells us that training a neural network is an NP-hard optimization problem that represents several theoretical and computational limitations^7. Dat Thanh Tran in January
2019, proposes a library to avoid the bottleneck in Machine Learning using the perceptron^15, Recurrent Neural Networks for Sequential Data Modeling have also been published, for voice recognition
considering the morphology of words, the syntax and semantics^16. The literature shows that the information that influences the electoral choice is uncertainty. it is clear that various factors and
attributes influence to know the winner of the elections, it should be noted that doing this manually represents a very laborious workload and knowing the main attributes that influence is always of
great importance for both voters and the nominated candidates.
2. Problem statement
A database with artificial data is randomly generated with which the Artificial Neural Network can evaluate the attributes that affect a vote. These must be safe to avoid inconsistencies since the
data must be managed by software which, according to the restrictions of the problem and the input information obtained from the database, will show the efficiency of the proposed algorithm. This
software is an algorithm which will give a solution to find the attribute more efficiently with a back propagation method. The database will have the necessary initial information that is required
for the Artificial Neural Network. These represent the different possibilities to make a choice when voting, among which are the economy, socio-cultural movements, and work. Taking that into account,
the input data to the Artificial Neural Network are the following:
And the output variables are the political parties to take into consideration, in this case we will use the most common political ideology patterns among which are usually classified as Left, Right
and agnostic or central. With which we will create three fictitious political parties called:
To understand the way in which supporters of each party are classified, Fig. 1 shows the schematic distribution of the political spectrum.
This political distribution has its origin in 1789 in France during the National Constituent Assembly in which the revocation of political power to the monarchy was discussed. Those who were against
it were on the right and those who promoted a change, seeking national sovereignty, were on the left. Péronnet in 1985 Said distribution was modified while preserving the same political bases since
at the beginning of the 19th century the aristocracy was supplanted by the bourgeoisie as the predominant class^12. With which we can say that liberal or left politics seeks political equality, for
the progress of the people. Without imposing on the most the law of the least. Rightwing or conservative politics represents maintaining the current political order, in which it represents those who
possess power and wealth seeking the individual good without taking into account all social classes.
Moderate politics has gained popularity in recent years because it represents the union between liberal politics and conservative politics trying to get the best of both parties. The idea of joining
and discerning the vote on the scale from Left to Right entails acceptance to the way that group of people work, taking into account the way in which they deal with problems, that is, the means used
to resolve conflicts. Thus, generating the right with the other part of the scale, attributing pejorative con- notations to the identity of the opposition and vice versa^3. Recently there is a
tension due to not so clear circumstances between political geometries, leaving out the debate by differentiating political thought, focusing on the discrediting of the adversary. These acts cause
confusion in the voters, frustrating the reasoning behind their vote, this means that the electoral decision in many cases is overwhelmed by the economy, socio-cultural movements and work, when
making a decision. The disturbance in our representation Politics is represented by sociocultural movements and other events not initially considered as natural phenomena, electoral fraud, among
3. Backpropagation neural network
An Artificial Neural Network is a complex mathematical function inspired by the operation of its biological namesake. But it's also the interaction of many simpler parts called neurons, working
together, which have numerical inputs and outputs. And its goal is to solve problems in a way similar to the human brain. The neural network is the integration of many neurons in which each neuron
performs a weighted sum whose weighting is given by the weight assigned to each of the connections of entry. This means that each connection that reaches the neuron will have an associated value that
will define the intensity with which the input variable will affect the neuron and therefore will influence the result that the output layer throws^2. Backpropagation networks use feedback as a
supervised method, consisting of three layers: input, hidden, and output. Having better precision because the error propagates inversely, that is, it starts from the output layer passing between the
hidden layer to reach the input layer^14.
As shown in Fig. 2, the X variables from X [ 1 ] to X [ n ] represent the inputs to the network, and the Y variable from Y [ 1 ] to Y [ n ] represents the result obtained from the neurons in the
output layer.
4. Kohonen neural network
Unlike the backward propagation Neural Network, the Kohonen Neural Network, as shown in Fig. 3, is simpler as it has only one layer, which uses an unsupervised method, therefore it does not have a
specific vector to be trained, among others. reasons that affect the veracity of the output result^1.
5. Model of an artificial neuron
It is the base unit of a neural network, basically it is an elementary processor which processes a vector X and produces an output resulting from the weighted sum^9. The model of an artificial neuron
is an imitation of the process of a biological neuron, as seen in Fig. 4.
Where the X [ 1 ] are the inputs (through the dendrites) to the neuron. These undergo a multiplying effect on the weight W [ 1 ] , due to their communication to the nucleus of the neuron, and b is
bias. Thus, obtaining the Eq. (1) as can be seen in Fig. 4.
Z=activationfunction∑weight*input+bias (1)
The basic characteristic of a neural network is that it is composed of three layers. The input layer is in charge of receiving the input values and sending these values to the second layer called the
hidden layer and these carry out their process and transfer the information to the output layer, the network can contain more layers if required, they can be modified by adding or removing input or
output variables or by changing the learning or training process^4. A conventional neural network is composed of three characteristics:
• The interconnection model between the different layers of the network.
• Development of learning in the variation of weights between the interconnections.
• The activation function modifies the weighted result format of the network in the output activation value.
In this case, a Trained Neural Network will be used with a backward propagation algorithm oriented towards gradient descent and the use of the chain-rule.
6. Activation function
The activation function is used to modify the data and enter it within a shorter range to make a simpler calculation. Next, we have the activation function that we will use in our backpropagation
algorithm^19. This function modifies the input values. where the high values are close to 1 and the very low values are close to 0 and is represented in Eq. (2), called Sigmoid function.
f(x)=11-e-x (2)
6.1. Backpropagation
Backpropagation is an algorithm widely used in the training of forward neural networks for supervised learning. It works by calculating the gradient of the loss function with respect to each weight
by the chain rule, iterating backwards one layer at a time from the last layer to avoid redundant calculations of middle terms in the chain rule, and is based on partial derivatives of calculus. Each
weight and bias value have an associated partial derivative. You can think of a partial derivative as a value that contains information about how much and in which direction a weight value should be
adjusted to reduce error. The collection of all partial derivatives is called a gradient. However, for simplicity, each partial derivative is commonly called a gradient^17.
6.2. Chain rule
If a variable y depends on a second variable u, which in turn depends on a third variable x, then the rate of change of y with respect to x can be calculated as the product of the rate of change of y
with respect to u multiplied by the rate of change of u with respect to x. If g(x) is differentiable at the point x and f(x) is differentiable at the point g(x), then f is differentiable in x. Also,
let y=f(g(x)) y u=g(x), then Eq. (3) is obtained. What is the chain rule.
dydx=dydu∙dudx (3)
6.3. Cost function
The cost function tries to determine the error between the estimated value and the real value, in order to optimize the parameters of the neural network. In this case we will use the root mean square
error. In regression analysis, Mean Square Error refers to the mean of the squared deviations of the predictions from the true values, over one space outside the test sample, generated by a model
estimated over one sample space particular. Its formula is shown in the Eq. (4) root mean square error.
c(ajl)=12∑j(yj-ajl)2 (4)
6.4. Mathematical model of our artificial neural network
Initially the network values are randomly generated. With this, it is very probable that the error obtained is very high, for which reason the network must be trained to obtain the minimum possible
error. We will begin by calculating the derivative of the parameters in the last layer, in which the result obtained from the weighted sum that is show below in Eq. (5), z ^ 1 is the result of the
weighted sum, is the weight 1 is the representation of the bias.
zl=wlx+bl (5)
Subsequently, the activation and cost function are added to this result, resulting in the error obtained from the network represented in the Eq. (5) where Result of the weighted sum activation
function F. cost. What we will look for will be the partial derivative of the cost with respect to the weight and bias parameters, with which we will have to calculate two derivatives. As we have
said, we are going to start working from back to front, therefore we begin to calculate the derivative of the parameters of the last layer. The number of the layer to which the parameter belongs if
our neural network has layers. To calculate this derivative it is important to analyze which is the path that connects the value of the parameter and the final cost in the last layer of this path is
not very long, although it still has several steps, previously we saw that in the operation of the neuron the parameter participated in a weighted sum now which we will refer to as which would then
be passed by the activation function represented in the Eq. (2) and the result of the activations of the neuron in the last layer would conform to the result of the network that would later be
evaluated by the cost function Having to determine the network error. With this, a composition of functions is formed and we will use a mathematical calculation tool called the Chain Rule represented
in the Eq. (6) To calculate the derivative of composition of functions, what it tells us is that to calculate the derivative of a composition of functions we simply have to multiply each of the
intermediate derivatives. Considering the Eqs. (7) and (10) represent the weighted sum of the last layer.
zl=wlal-1+bl (6)
c(a(zl))=error (7)
We will obtain the derivative of the weight with respect to the cost in Eq. (8) and the derivative of the bias with respect to the cost in Eq. (9) Derivative of weight with respect to cost.
∂c∂wl=∂c∂al∙∂al∂zl∙∂zl∂wl (8)
∂c∂bl=∂c∂al∙∂al∂zl∙∂zl∂bl (9)
Thus, obtaining three partial derivatives where the derivative of the activation with respect to the cost Eq. (9) to obtain the cost variation of the network when the output of the activation of the
neurons in the last layer is varied, that is to say that a derived from the cost function with respect to the output of the neural network. The cost function that we will use in this case will be the
root mean square error described in Eq. (4) with the parameters of our network Eq. (10) partial derivative of activation with respect to cost.
∂c∂al (10)
Thus, the derivative of the function with respect to the output of the network, Eq. (11), would be represented. Eq. (12) indicates the derivative of the activation function with respect to the output
of the network.
c(ajl)=12∑j(yj-ajl)2 (11)
∂c∂ajl=(ajl-yi) (12)
al(zl)=11+e-zl (13)
∂al∂zl=alzl1-alzl (14)
zl=∑iail-1wil+bl (15)
∂zl∂bl=1 (16)
∂zl∂wl=ail-1 (17)
∂c∂bl=∂c∂al∙∂al∂zl∙∂zl∂bl (18)
∂c∂zl=δl (19)
We continue with the activation function of the weighted sum Eq. (13) Activation function of the weighted sum of and its Derivative of the activation function with respect to the weighted sum in Eq.
(14). This reveals the output variation of the neuron when the weighted sum of the neuron is varied, thus calculating the activation function. This derivative is calculated depending on the type of
activation function, in this case we will use the sigmoid function Eq. (15) with all that we would only be missing two partial derivatives with respect to bias Eq. (16). Partial derivative with
respect to bias and weight Eq. (17). Partial derivative with respect to weight. These are obtained by deriving the weighted sum of the neuron as shown below in Eq. (15). Derivation of the weighted
sum of the neuron. By applying the chain rule and using two partial derivatives in the derivative of the bias with respect to cost Eq. (18), the error is obtained as a function of the value of
represented in Eq. (19). What is the weighted sum calculated within the neuron, that is, what this derivative tells us is to what degree the cost error is modified when there is a small change in the
sum of the neuron if this derivative is large, before a small change in the value of the neuron this will be reflected in the final result and on the contrary if the derivative is small it does not
matter how we vary the value of the sum since this will not affect the error of the network, that is, the derivative from here is the one that will tell us what responsibility the neuron has in the
final result and therefore in the error, this is what we said before, if the neuron is a part responsible for the final error then we should use this information to extract part of that mistake for
this one. We will need the derivation of the bias with respect to the cost with error imputed to neuron 1 is represented by Eq. (20) which will be the error imputed to the neuron or derivation of the
bias with respect to the cost with error imputed to neuron 2 shown in the Eq. (21), which is calculated in Eq. (22), which is the error imputed to the neuron.
∂c∂bl=δl∙∂zl∂bl (20)
∂c∂bl=δl∙1 (21)
∂c∂bl=δl (22)
Later we will do the same, but with the partial derivative of the weight with respect to the cost of the error imputed to neuron 1, Eq. (23), which reduces to Eq. (24). Derivative of the weight with
respect to the cost with error imputed to neuron 2.
∂c∂wl=δl∙∂zl∂wl (23)
∂c∂wl=δl∙ail-1 (24)
We have deduced three different expressions that allow us to obtain the partial derivatives that we are looking for the last layer, one that tells us how to calculate the error of the neurons in the
last layer, and another for each of the partial derivatives and thus we obtain the result of the last layer. To obtain the result of the previous layer, we apply the Chain Rule again to the following
composition. Eq. (25) Indicates the Error in the penultimate layer that with the chain rule generates two derivatives; derivative of weight with respect to cost in the penultimate layer Eq. (26) and
derivative of bias with respect to cost in the penultimate layer, Eq. (27).
c(al(wlal-1(wlal-2+bl-1)+bl)) (25)
∂c∂wl-1=∂c∂al∙∂al∂zl∙∂zl∂al-1∙∂al-1∂zl-1∙∂zl-1∂wl-1 (26)
∂c∂bl-1=∂c∂al∙∂al∂zl∙∂zl∂al-1∙∂al-1∂zl-1∙∂zl-1∂bl-1 (27)
Calculated that is the error of the layer and these derivatives are operated the same as before minus 1 and the activation of the previous layer and this here is the derivative of the function of the
activation in this expression. The only thing that would need to be calculated would be this derivative that tells us about how the weighted sum of a layer varies when the output of a neuron in the
previous layer is varied. This derivative is also simple to calculate and is basically the parameter matrix, Eq. (28), which connects both layers, what it does is move the error from one layer to the
previous layer, distributing the error based on the weights of the connections, with this we would again have an expression from which to obtain the derivatives partials we are looking for. Again,
the block highlighted in Eq. (29). It becomes this derivative, Eq. (30). Which again represents the error of the neurons in this layer.
∂c∂zl-1=δl-1 (28)
The effectiveness of the back propagation algorithm lies in the fact that what we have done in this layer is already extensible to the rest of the layers of the network, applying the same logic, we
take the error from the previous layer, we multiply it by the weight matrix in a transformation that comes to represent the back propagation of the errors, Eq. (28), and we calculated the partial
derivatives with respect to the parameters and so on, going through all the layers of the network until the end, with a single step we had calculated all the errors and the partial derivatives of our
network using only four expressions.
With what in the end we will obtain four expressions to calculate error starting with the last layer, Eq. (30), that computationally indicates the error of the last layer and later perform a retro
propagation of the error of the previous layer, Eq. (31), to do the retro propagation from the error to the previous layer and the calculation of the derivatives of each layer, Eq. (32), the
derivative bias of the layer is developed using the error with the Eq. (28) of our network the weight of the house is calculated using the error.
δl=∂c∂al∙∂al∂zl (29)
δl-1=wlδl∙∂al-1∂zl-1 (30)
∂c∂bl-1=δl-1 (31)
∂c∂wl-1=δl-1al-2 (32)
To calculate the partial derivatives that we are looking for in the Eq. (32) and Eq. (31) expressions are quite intuitive because we are simply telling ourselves how we have to use the error from the
previous layer to calculate the error in this layer there are two different cases. one is in the last layer where the error already belongs to the cost function, Eq. (30), and others are the rest of
the layers of our network that depend on another layer, Eq. (31), and of course once we have these two expressions that tell us how we can calculate the error in the current layer with respect to the
previous one.
7. Implementation of the artificial neural network
The program randomly creates an artificial database, it starts by randomly generating 1,000 synthetic data items.
Each data item has four input values and three output values as can be seen in Fig. 5. The four input values are all between -10.0 and +10.0 and correspond to predicted values that have been
normalized, so that values below zero are less than the average, and values above zero are greater than the average. The three output values correspond to a variable to predict that can take one of
the three categorical values. To predict a person's political inclination: conservative, moderate or liberal.
The program randomly divides the data into a training set of 800 items and a test set of 200 items Fig. 6. The training set is used to create the neural network model, and the test set is used to
estimate the accuracy of the model. After the data is partitioned, the program creates an instance of a neural network with n hidden nodes. The number of hidden nodes is arbitrary and must be
determined by trial and error. To finish the program, it generates a final neural network with the optimal weights and bias with values generated during the previous training of the network to obtain
the final result.
8. Algorithm
In Fig. 7, we present a flowchart illustrating the application of the backpropagation algorithm within an artificial neural network. The primary goal is to develop an algorithm capable of predicting
electoral votes. The process commences by defining the parameters of the network, followed by the generation of a randomized database, as depicted in Fig. 5. Subsequently, an artificial neural
network is created.
The input data extracted from the randomized database is then introduced into the input layer of the neural network. After traversing hidden layers, the processed data ultimately reaches the output
layer. The ensuing step involves a comparison between the neural network's output and the anticipated results derived from the training data. This assessment yields an error or loss metric, which
quantities the network's performance.
If the calculated error falls within an acceptable range, the results are displayed, as illustrated in Fig. 8, and the process concludes. However, if the error remains outside this acceptable
threshold, the data is looped back to the input layer, where it undergoes further processing iterations. This iterative procedure continues until the error converges to an acceptable value.
9. Results
These results are tuned in order to know if the dependence of the parameters of the Artificial Neural Network finds the normalized average value. And know if the number of hidden Layers affects the
results of the voting. We can see in Fig. 8 the optimal number of hidden layers (NumHidden), obtained when there is a hidden layer, because it is the smallest value with a mean of 0. 0667240612695.
We also see the worst result when having 8 hidden layers in the network, with an average of 0.116324855868.
10. Conclusions
In conclusion, the parameters with the best performance in our Network Artificial Neural are those expressed below in Fig. 9.
It only remains for us to take the approach based on electoral votes, with experimental data, and the disturbance as a determining event in the alteration, or not, of the final result. We have three
possible outcomes which are on a scale of 0 to 10. If you are representing a person who is younger than average, has a much lower income than average, is somewhat more educated than average, and has
more debt than average. The person has a liberal political view. Likewise, if another person on the same scale is older than average, has a higher-than-average income, is slightly less than, equal
to, or slightly more educated than average, and has less than average debt, that person has a conservative political vision. But when a person is within the average or a little below or a little
above in all the parameters, he is in a moderate ideology.
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Gini coefficient measurement | Rated Documentation
The Gini coefficient (Gini) is a measure of inequality across a certain set of values. In this case, we're measuring the Gini of the validator market share of entities we have identified.
Entity groupings
We have taken the commonality by highest order such that grouping by pools takes precedence over node operators, which take precedence over deposit addresses.
Interpreting the Gini
A Gini coefficient of 0 reflects perfect equality, where all income or wealth values are the same, while a Gini coefficient of 1 (or 100%) reflects maximal inequality among values.
Gini calculation
Basing our calculation of the work by Evgeny Medvedev here, we first take the validator_count of each of the entities on the latest day and rank them in descending order accordingly. We then use the
1-2B formula for measuring the Gini, where B is the area under the Lorenz curve such that:
Gini == 1 - 2 * SUM((validator_count * (rank - 1) + validator_count / 2)) / COUNT(entities) / SUM(validator_count)
Function components
validator_count * (rank — 1)
is the area of the rectangular horizontal slice under the Lorenz curve.
is the area of the triangle on the left of the rectangular slice.
SUM((validator_count * (rank — 1) + validator_count / 2))
is the sum of all the slices
normalizes the x axis to the range 0 to 1
normalizes the y axis to the range 0 to 1
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Quantum gravity and higher curvature actions
Effective equations are often useful to extract physical information from quantum theories without having to face all technical and conceptual difficulties. One can then describe aspects of the
quantum system by equations of classical type, which correct the classical equations by modified coefficients and higher derivative terms. In gravity, for instance, one expects terms with higher
powers of curvature. Such higher derivative formulations are discussed here with an emphasis on the role of degrees of freedom and on differences between Lagrangian and Hamiltonian treatments. A
general scheme is then provided which allows one to compute effective equations perturbatively in a Hamiltonian formalism. Here, one can expand effective equations around any quantum state and not
just a perturbative vacuum. This is particularly useful in situations of quantum gravity or cosmology where perturbations only around vacuum states would be too restrictive. The discussion also
demonstrates the number of free parameters expected in effective equations, used to determine the physical situation being approximated, as well as the role of classical symmetries such as Lorentz
transformation properties in effective equations. An appendix collects information on effective correction terms expected from loop quantum gravity and string theory.
All Science Journal Classification (ASJC) codes
• Physics and Astronomy (miscellaneous)
Dive into the research topics of 'Quantum gravity and higher curvature actions'. Together they form a unique fingerprint.
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Types Of Numbers Worksheets
Types Of Numbers Worksheets act as fundamental devices in the world of maths, offering an organized yet functional platform for learners to discover and master numerical ideas. These worksheets offer
a structured method to understanding numbers, supporting a solid foundation whereupon mathematical efficiency thrives. From the easiest counting workouts to the intricacies of innovative
calculations, Types Of Numbers Worksheets deal with learners of varied ages and ability degrees.
Unveiling the Essence of Types Of Numbers Worksheets
Types Of Numbers Worksheets
Types Of Numbers Worksheets -
Types of Numbers We ve got your number Whether it s positive or negative rational or irrational prime or composite even odd decimal whole mixed or fractional an absolute value or even a variable
Identify Types of Whole Numbers Natural Numbers and Integers Liveworksheets transforms your traditional printable worksheets into self correcting interactive exercises that the students can do online
and send to the teacher
At their core, Types Of Numbers Worksheets are lorries for conceptual understanding. They envelop a myriad of mathematical principles, guiding learners with the labyrinth of numbers with a collection
of appealing and purposeful exercises. These worksheets transcend the boundaries of conventional rote learning, urging energetic involvement and fostering an intuitive understanding of numerical
Nurturing Number Sense and Reasoning
Types Of Numbers GCSE Maths Steps Examples Worksheet
Types Of Numbers GCSE Maths Steps Examples Worksheet
These learning numbers worksheets provide practice in recognizing and printing numbers including numbers written as words and ordinal numbers Even and odd numbers are also introduced Free math
worksheets from K5 Learning no login required
Help your students to master different types of numbers in this versatile worksheet Primed for Years 7 8 learning the types of numbers worksheet support classroom work or independent study thanks to
its supportive structure An opening section briefs the learner on factors multiples primes square numbers and cube numbers before challenging
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From Theory to Real-World Application
Write Number Names For Given Numbers Math Worksheets MathsDiary
Write Number Names For Given Numbers Math Worksheets MathsDiary
It is a comprehensive resource for visualizing the application of the base ten number system The simple and interesting format of these worksheets is highly efficient for kids to comfortably begin
their learning Download Identifying Numbers Worksheet PDFs These math worksheets should be practiced regularly and are free to download in PDF
A worksheet with questions on all types of number The worksheet does get quite challenging towards the end and consolidates multiples factors primes triangle numbers squares cubes and square roots
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In a progressively varied world, Types Of Numbers Worksheets accept inclusivity. They go beyond cultural boundaries, integrating examples and problems that reverberate with students from varied
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Types Of Numbers Worksheets chart a program towards mathematical fluency. They infuse determination, essential reasoning, and problem-solving abilities, vital qualities not only in maths but in
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Number Types Worksheet EdPlace
Types Of Numbers Worksheet
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Types Of Numbers Worksheet Grade 7 Worksheet
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Number Types Worksheet EdPlace
Types Of Numbers Teaching Mathematics Mathematics Teaching
Identifying Types Of Numbers Worksheet Live Worksheets
Identify Types of Whole Numbers Natural Numbers and Integers Liveworksheets transforms your traditional printable worksheets into self correcting interactive exercises that the students can do online
and send to the teacher
Classify Numbers Algebra practice Khan Academy
8th grade Course 8th grade Unit 1 Lesson 3 Irrational numbers Intro to rational irrational numbers Classifying numbers rational irrational Classify numbers rational irrational Classifying numbers
Classify numbers Classifying numbers review Worked example classifying numbers Math Numbers and operations Classify numbers
Identify Types of Whole Numbers Natural Numbers and Integers Liveworksheets transforms your traditional printable worksheets into self correcting interactive exercises that the students can do online
and send to the teacher
8th grade Course 8th grade Unit 1 Lesson 3 Irrational numbers Intro to rational irrational numbers Classifying numbers rational irrational Classify numbers rational irrational Classifying numbers
Classify numbers Classifying numbers review Worked example classifying numbers Math Numbers and operations Classify numbers
Types Of Numbers Worksheets Worksheet
Types Of Numbers Worksheet
Number Types Worksheet EdPlace
Types Of Numbers Teaching Mathematics Mathematics Teaching
Classifying Real Numbers Worksheet Doc Worksheet
Types Of Numbers GCSE Maths Steps Examples Worksheet
Types Of Numbers GCSE Maths Steps Examples Worksheet
Types Of Numbers GCSE Maths Steps Examples Worksheet
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The Value of a Peer-Reviewed Activity
This week, we have been talking about proof writing in the discrete mathematics course I'm teaching. Yesterday, I started class by having students answer how confident they are about their proof
writing skills on a scale of 1 to 10, 1 being "clueless and not sure where to start", 5 being "Okay and ready for homework," and 10 being "I can do most any proof you throw at me with ease." I then
had them individually complete three proofs in 15 minutes.
Many students struggled with Problem 2 because they are not comfortable with sequence notation. Some misunderstood Problem 3 and tried proving something completely different than intended. After 15
minutes, they traded papers (with anonymous codes instead of their names so there was no embarrassment) and reviewed each someone else's papers as we went over them in class. They wrote some lovely,
encouraging comments to each other. For example, one student had no idea of what to do on the second problem so they left it blank and their reviewer wrote, "Bet you can do it now! :D." Others noted
where the proof failed and wrote a comment that they too had that difficulty. It's refreshing to see such kindness. Finally, they took the same survey about their proof writing again and there were
dramatic changes in confidence.
Many more students were confident in their abilities. I'm not sure who the one who reported 1 after the exercise is. I hope they come to office hours.
This is in no way a rigorous test, but students expressed they learned more from this exercise because it forced them to think about the material instead of just going through the proof together on
the board. I imagine there is a psychological benefit of seeing how someone else is doing too and being kind to them in written comments. It was also suggested that I do a problem example in class
without proving it before class so students could hear my raw thought process first hand. I'll have to think about that. I like being prepared, but I'm sure I could find some way to do that.
import matplotlib.pyplot as plt
import numpy as np
# set up the data
rank = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
pre = [0, 6, 25, 25, 13, 13, 19, 0, 0, 0] # pre exercise confidence
post = [7, 0, 7, 13, 33, 13, 7, 7, 13, 0] # post exercise confidence
# weighted average of confidence
pre_mean = np.sum(np.array(rank) * (np.array(pre)/100))
post_mean = np.sum(np.array(rank) * (np.array(post)/100))
# figure generation
fig, axs = plt.subplots(ncols=2, sharey=True)
axs[0].bar(rank, pre, align='center')
axs[1].bar(rank, post, align='center')
axs[0].set_title("Pre-exercise (mean={:0.1f})".format(pre_mean))
axs[1].set_title("Post-exercise (mean={:0.1f})".format(post_mean))
for ax in axs:
ax.set_xlabel("Student confidence")
axs[0].set_ylabel("Percentage of students answering")
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On lattices, Euclid, and zeta functions
Speaker: Valeriia Starichkova
Affiliation: University of New South Wales
I would like to talk about a project in progress that we started this year. Let K denote a number field and R its ring of integers. We can think of K as a vector space over rationals Q with some
finite dimension n (i.e. K is Q^n) and of R as a lattice over integers Z (i.e. R is Z^n).
In 1977, Lenstra constructed a criterion for R to be Euclidean. This criterion connects the discriminant of R, the “packing” information about R, and some more specific information related to the
units of R. In particular, this criterion implies a bound for the discriminant of K, which, assuming the Generalised Riemann Hypothesis, holds only if the degree of K (number n) is not too large.
This fact attracted our attention at the very beginning of the project, by making a link between Lenstra's criterion and the analytic number theory.
I will introduce the required theory on my way, explain Lenstra’s criterion and talk about tools from different maths areas which are all combined in this criterion; namely, we will cover some
properties of number fields, a little bit of analytic number theory, and a tiny bit of packing theory.
About Pure mathematics seminars
We present regular seminars on a range of pure mathematics interests. Students, staff and visitors to UQ are welcome to attend, and to suggest speakers and topics.
Seminars are usually held on Tuesdays from 2 to 3pm.
Talks comprise 45 minutes of speaking time plus five minutes for questions and discussion.
Information for speakers
Researchers in all pure mathematics fields attend our seminars, so please aim your presentation at a general mathematical audience.
Contact us
To volunteer to talk or to suggest a speaker, email Ole Warnaar or Ramiro Lafuente.
Priestley Building (67)
Room: 443
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What are the 4 rules of concave mirror?
Image Formation By Concave Mirror At the infinity. Beyond the centre of curvature. At the centre of curvature. Between the centre of curvature and principal focus.
What are 3 examples of concave mirrors?
• Shaving mirrors.
• Head mirrors.
• Ophthalmoscope.
• Headlights.
• Solar furnaces.
What is the formula of a concave mirror?
1/f= 1/u + 1/v. This equation is referred to as the mirror formula. The formula holds for both concave and convex mirrors.
How can we solve the mirror problem?
1. Step 1: Make a list of the known quantities given in the problem.
2. Step 2: Determine if the unknown quantities require you to use the mirror equation, the magnification equation, or both.
3. Step 3: Solve the needed equation(s) symbolically for the unknown quantities.
Which type of image is formed by concave mirror?
Concave mirrors form both real and virtual images. When the concave mirror is placed very close to the object, a virtual and magnified image is obtained and if we increase the distance between the
object and the mirror, the size of the image reduces and real images are formed.
What is the formula of mirror?
Let’s explore the mirror formula (1/f = 1/v+1/u) and see how to locate images without drawing any ray diagrams.
What is the distance of a concave mirror?
At what distance from a concave mirror focal length 10 cm should an object 2 cm long be placed in order to get an erect image 6 cm tall? To find: Distance of the object from the mirror, u . Thus, the
distance of the object from the mirror u is -6.67 cm.
What is the image distance in concave mirror?
According to sign conventions, for a concave mirror, the focal length is negative, the object distance is negative, the image distance is positive for a virtual image and negative for a real image.
How do you find the location of an image in a concave mirror?
When the object is located at a location beyond the focal point, the image will always be located somewhere on the opposite side of the mirror. Regardless of exactly where in front of F the object is
located, the image will always be located behind the mirror.
How do you calculate the focal length of a concave mirror?
1. 1f.
2. 1f.
3. Or 1f.
4. Or f=uvuโ v.
5. F=R2.
How do you derive a mirror formula?
The distance of the principal focus from the pole is called the focal length (f). There is a relationship between these three quantities given by the mirror formula which is expressed as 1v+1u=1f.
This formula is valid in all situations for all spherical mirrors for all positions of the object.
Do concave mirrors produce real images?
Concave mirrors, on the other hand, can have real images. If the object is further away from the mirror than the focal point, the image will be upside-down and real—meaning that the image appears on
the same side of the mirror as the object.
Which image Cannot be obtained by a concave mirror?
the type of image that cannot be obtained by a concave mirror is virtual image.
What are 10 uses of convex mirror?
• Convex Mirrors in ATMs. Convex mirrors are usually placed on top of the ATMs.
• Convex Mirrors as Rear-View Mirrors.
• Convex Mirrors in Parking Lots.
• Convex Mirrors for Security Purposes.
• Convex Mirrors inside Buildings & Offices.
Which mirror gives real image?
Solution: Concave mirrors can form real images. Convex and plane mirrors always form virtual images.
What are the properties of concave mirror?
Concave mirrors will form both real and virtual images. When the object is closer to the mirror, a virtual and magnified image is formed. When the object is further away from the mirror, a real and
diminished image is formed.
Which mirror used in cars?
Convex mirror The side mirrors of the car and the rear view mirror of a car are made up of convex mirrors. This is because the image formed by a convex mirror is diminished and erect image, thus it
provides a larger field of view.
What is V in concave mirror?
The distance between the image and the pole of the mirror is called Image distance(v).
How is image distance calculated?
What is the distance of the image from the mirror?
The image distance always equals the object distance. The size of the image is the same as the object (the mirror does not magnify the image).
When an object is placed 20 cm from a concave mirror?
When an object is placed 20 cm from a concave mirror, a real image magnified three times is formed.
What is the distance of a concave mirror of focal length 10cm?
Answer. So the object should be placed at a distance of 6.7 cm from the pole of the concave mirror (on the left, in front of the mirror).
What is the distance from a concave mirror to focal length 25?
Answer : 50 cm away.
What is the formula of height of image in mirror?
Magnification Equation: The magnification equation for mirrors relates the magnification index to the heights of the object and the image as well as to the distance between the mirror and the object
and image. The magnification equation states that m=hiho=โ dido m = h i h o = โ d i d o .
Is height of image in concave lens positive or negative?
The real image is always formed as inverted, therefore, for real image height is taken as negative.
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Convex hull of zeros
There’s a well-known theorem in complex analysis that says that if p is a polynomial, then the zeros of its derivative p′ lie inside the convex hull of the zeros of p. The convex hull of a set of
points is the smallest convex set containing those points.
This post gives a brief illustration of the theorem. I created a polynomial with roots at 0, i, 2 + i, 3-i, and 1.5+0.5i. The convex hull of these points is the quadrilateral with corners at the
first four roots; the fifth root is inside the convex hull of the other roots.
The roots are plotted with blue dots. The roots of the derivative are plotted with orange ×’s.
In the special case of cubic polynomials, we can say a lot more about where the roots of the derivative lie. That is the topic of the next post.
More complex analysis posts
One thought on “Convex hull of zeros”
1. Check out this cute proof using electrostatics:
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33. Suppose your group would like to do a research about the
quality of sleeping on...
33. Suppose your group would like to do a research about the quality of sleeping on...
33. Suppose your group would like to do a research about the quality of sleeping on doing exercises,
and you believe that the quality of sleeping will significantly be improved if a person has regular
exercises habit. A study is conducted of a random sample of 16 participants who has not been
exercised for more than 40 days. The duration for the study is 25 days, and each participant needs
to do exercise 30 minutes every day. During the 36 days, the average total sleeping hours increased
10 minutes, with the standard deviation is 8. Could we conclude that the quality of sleeping is
significantly affected by doing exercises?
a) Parametric or nonparametric hypotheses?
b) Z distribution, t distribution, chi-square test or hypothesis test of a proportion
c) Please indicate the null hypotheses.
d) Please indicate the alternative hypotheses
e) Please calculate the stand error? (If it is a chi-square test, type NA for this question)
f) If z table will be used, type NA for this question. If t table will be used, indicate the degree of
g) Please calculate the statistical value.
h) Hypothesis is supported or no supported, and what is your conclusion?
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SCIENTIFIC ABSTRACT KUNICKI-GOLDFINGER, WL. - KUNIN, I. K.
SCIENTIFIC ABSTRACT KUNICKI-GOLDFINGER, WL. - KUNIN, I. K.
Document Number (FOIA) /ESDN (CREST):
----------- --------- ---- 0 Poland Aicrobiology. General Microbiology. F-1 Abs Jour: Referat.Zh.-Biol., No. 9, 1957, 3548o Author : Kunicki-Goldfinger, Wl. Title : Changeability of Bacteria Orig
Pub: Acta microbiol. polon., 1954, 3, No. 3, 199-347 Abstract: A c!ritical survey. The essence of the problem of changeability of bacteria is discussed In the 1st chapter. In the 2nd chapter the idea
of the individual and specie is determined. The 3rd ebapter is devoted to a classification of the types of changeability which the author divides into changeability of development, mod- ification
changeability, mutilization and hybrid- ization, Variability of develop-,nent-eytomorphosis Card 1/3 Poland /Mcrobiology. General Microbiology, F-1 Abs Jour: Referat.Zh.-Biol., No. q.. 1957, 3548o
and heteromorphosis are analyzed in chapters 4 and 5. Under the latter, the author enumerates dissociation, reactive forms, and filtering and L-forms; he criticizes the opinions of Brown of
dissociation as a manifestation of spontaneous mutations. In chapters 6-9, modification change- abill.ty Is described-biochemical changeability, the formation of phago- and drug-resistant forms,
variability of antigens, mutilations, The author considers all these forms adaptations. Also given is a criticism of the work of Louis, Ryan, Demerec, Luria, and Delbrook. In chapter 10 the problem
of sex, transformation and hybridization in bac- teria is discussed. Chapter 9 is devoted to the process of specie formation, in which, in the opinion of the author, modification changeability, C ard
2/3 Poland Aicrobiology. General Microbiology. F-1 Abs Jour: Referat,,Zh.-Biol., No. 9, 1957, 3548o hybridization and selection plays a fundamental role. Included are 18 drawings and 21 photo-
graphs. A bibliography of 784 titles. Card 3/~ C4 f /The bemolya~n of R-scht-ithia j was rw-UL'l irL C-Wtu.- 'I cc 7 hem I tulatr variwus c and it) i 1 dn 74 hr'l TI'I'm S!',nra ~4 I ru -,U -ldfinul
Runs,ki-CODI 2nd A. Miirz- Acto mirr.4i--4. P'L. L 127-J29- '111. ~Ifect of cultu:, fill: itrm; cos&"ging tox,iL, 000misum WtIckil, 0 bolujimum typ. i .-- 1 4. and ( .,) r b.wlzribm diphlbriur 'pi
parxlmt~Aum a-r- t-b~d :,::~ I &t%= tb. il-~ 11-tixxi of 101`i~ Ik, lun!w pTopcrim to toxln,-OnL,~ ,43 flui&. a. VLNUY. KWIATKOWSKI, Z.; KEMCKI". LDFINGM, W.; IORKIF.WICZ, Z. Certain physiological
properties of Proteus vulgaris L form. Acts. microb. polon 5 no.1-2:15-19 1956. 1. Z Zakladu Mlkrobiologii Ogolnej UMCS w Lublinie. (PROTEUS VULGARIS, L form, physiol. (Pol)) LD I 1111R 11.; DTGDALA,
K.; LAGOWSKA, M.; VIERCIENSKA, D. Rffect of lithium chloride on Ischeriebla coli and on otber bacteria; preliminary communication. Act& microb. polon 5 ao.1-2:31-40 1956. 1. Z Zakladu Mikrobiologii
Ogolnej UMCS w Lublinie. (LITHIUM, effects, chlorides, on 3. coli, Bacillus subtilis & Proteus (Pol)) (CHLORIDES, effects '' 'lithium chloride. on Bacillus subtilis. 3. coli & Proteus (POW (BACILLUS
SUBTILIS. effect of drugs on, lithium chloride (Pol)) (ESCHERICHIA COLI, effects of drugs on, same) (Pr OTXJS. effect of drugs on, same)) POLAND/Microbiolocw - General Microbioloao F-1 Abs J:)ur :
Rof Zhur - Biol., No 11, 1958, 47854 Author ; KSi:i a ki -Go ~-4f in ger, 'flala, K.) LaGowska, M., DyG Wiercienska, Do Inst Title GoniOial Bacteria. orig Pub Acta Microbiol Polon, 5, no 1-2~ 41-43
(1956) (in Polish witi an English sur=x5) Abstract G,.,nidial bacteria were isolated from the intestinal contents of small rodents and insectivora and cultured by the mathod Df 00.urer LTN-. spellin
[; uncertain_7 (Ann Inst 1?astc=, 2~1 395 (1954)~. Thaae bacteria forn microcolonies on a[pr consisting of elementary Wics 0-2-0-3.,-- in diam. and in broth rive a li&A opalescence. The addition of
blood, serum, of yeast and liver extracts, and of intestinal con- tents extract from roOwrta O.U. not change the character Card 1/2 POLAIM/Microblolor ~;y - General Microbiolo(Z F-1 Abs jour Ref
21iur - Diol,j No 11, 1958, 47854 Of L-XOvth- On further transplantations the elementary bodies transform into 0-ijAltheraids 0-5-1.5,,o- in size. On aGar the latter fori'.i cOlonica resembling
streptococci colonies and in broth they produce turKdity and a residue. The reverse transforimtion Of diplithercids into tjjc Goni- aial forms could not be observcd. The UOnidial bacteria Jescribed
are sensitive to penicillin, are very stable on storar;e, and retain their viability on dehydration or in broth for two years. Card 2/2 KTJNICKI,GOLDFINGBR, We, ROVIVSKI, Se Some studies on the
structure of bacterial colony. Acta Microbe polono 6 no.4021-330 1957 1. Z' Zakladu Hikrobiologii Palveraytetu Wroclawskiego i Zaklndu Mikrobiologii Ogolney Instytutu Immunologii i Ternpii
Doewiadotalnej im. L# Hirsafelda, we Wroclawtu Wplynelo dnia I wrzeenia 1957 r. (BACILLUS. culture growth & colony form (Pol)) KUNICII-GOIX&INGF-R, W. ', DROZANSKI, W. -, BLASZCZAK, D. ;KAZIM, J. ;
SKIBINSIA, J. Bacteria as tood for soil amoebae. Acts Hicrob. polon. 6 no-4:331-344 1957- 1. Z Zakladu Mikrobiologii Uniwerartatu. Wroclnwakiego we Wroclnwiu i Zakladu Mtkrobiologii Ogolnej
Uniwaroytetu Marii-Curia-Sklodowskiej w Lublinie Wplynelo dnia 20 wrzeanin 1957 r (AMOBBA, metabolism soil bact. qa food source, growth & develop (Pol)) (SOIL, mibrobiology bact, nn food source for
amooba, growth & develop. (Pol)) 91: KUNICKI-GOLDFINGER. W. Bronislaw fliklewski (1"-1961) as a zicrobiologist, Acta microbiol. pol. 10 no.2:123-127 161. (MICROBIOLOGY hist) (BIOGWRIES)
KUNICKI-GOLDFINGER, Vladyslaw J. H. Adaptive enzywo in the pathway of tr7ptophane synthesis in Escherichia coli. (preliminary note). Aota microbiol. pol. 10 no.2:129-133 '61. 1. From the Department
of Microbiology, The University, Wroclaw. (ESCHMCHIA COLI metab) (TRYPTOPHAN metab) (MYVM metab) SKURSKIp Adam; SLOPEK, Stefan; KUNICKI-GOLDFINGER, Wladyslaw; MICHALSKA, Eugenia. Studies on the
mechanism of the phagocytic reaction. VII. Phagocytosis and S - R dissociation of Brucella bacilli. Arcfi. immun.ter.dosw. 8 no-3:389-394 160. 1. Department of )tycology, Department of Bacteriology
and Department of Microbial Geneticsp Institute of Immunology and Experimental Therapy, Polish Acadewj of Sciences, Wroclaw. (PHAGOCYTOSIS) (BRUQELLA izaranol) kV)UCKI-GOLDFINGER, Wladyslaw:
KUNICKA-GOLDFINGtRj Wladyslawa; My wspolpracy Intestinal microflora of Sorax amnius araneus L. and C16thrionomys glareolus glareolus Schreb. in natural conditions. I. Quantitative and qualitative
characteristics of the intestinal microflo'r'a. Acta-microbiol. Pol. 11 no.1/2%43-75 162. 1. Z Katedry Mikrobiologii Univeraytetu WarszawBkiego w Wars2AWie i Zakladu Badania Ssakow PAN w Bialowiezy.
(INTESTINES microbio-1) (INSECTIVORA microbiol) (RODENTS microbiol) KUNICKI-,GOLDFINGER., Wladyslaw; KUNICKA-GOLDFINGEH, Wladyslawa Intestinal microflora of Sorex araneus araneus L. and Clothriononys
glareolus glareolus Schreb..in natural conditions. II. Gawral characteristics of separate strains. Acts. microbiol. Pol. 11 no.1/2: 77-91 '62. 1. Z Katedry Mikrobiologii Uniwersytetu Warsaawskiego w
Warazawie. (INTESTINES microbiol) (INSECTIVORA microbiol) (RODENTS microbiol) KUNICKI-GOLDFINGER., Wladyslaw; KUNICKA-GOLDFINGM, Wladyslawa Intestinal microflora of Borax aranaus araneus L. and
ClethrionoWn glareolus glareolus Schreb. in natural conditions. III. Seasonal variations. Acts. microbiol. Pol. 12 no.1/2:95-110 '62. 1. Z Katedr7 Mikrobiologii Uniwersytetu Warazavskiego w Warazavie
i Zakladu Badania Ssakow PAN w Bialoviazy. (INTESTINES microbiol) (RODENTS microbiol) UNSECTIVORA microbiol) (WEATHER) KUNICKI-WLDFINGER, Wladyslaw J.H.; CZMUINSKA, Katarwjma The environmental
control of the conjugation in Rooherichia coli K-12. II. ThA offoot of temperaL-ure on effective pairs formation and on chromosomal transfer. Acts. microbiol. Pol, 13 no.1:13-21 164 1. From the
Department of Microbiology, Warsaw-University, Warsaw and the-Microbic-ory Department, Wroclaw University, Wroclaw. Pasteurella-like microorganism in ~Jriall rodents. Acta mic-o- biol. Pol. 13
no-4341-347 26.1, ly~, tInc. War-sssaw 'Unilversity, 1. From the Department of Ificrobiolog W-arsav, Poland. HERDA, M., inz. We.; GESAK, K., inz.; WFBER, B., inz.; VYHNANEEKJ, V.,inz.;
RP49KY,-L.,,,inz.; SDEK, J.,, inz.; PROSTREDNIK, K., inz. Maps for area planning and records of the built constructions. Good kart obzor 10 no.9/10j232-235 0 '64 ~TJNIGKY., 4dialar., Inz, Aerial
photogrametry and railroads. Iatecky obzor 8 no.39 70-71 W064 - KUNIGKY, Ladislav, inz.; VYHANANEK, Vlastimil, inz. Use of ground photogranmetry for technical documentation. Geod kart obzor 9
no.8:210-213 Ag 163. 1. Ceskosloslovenske statni drahy. I KUNIh`V, S. Preparation of machine-tractor stations for autumn ad winter repairs of a~.ricidtural machirei7~. P. 16 Vol. 6. no. 10) Oct- 1955
IMSHINIZIF,kNO ZE~,IEDELIE Oofiya, Bulgaria Sol Eastern Luropean Accession Vol. 5 No. 1 Jan. 1956 KUNIVICZ, Helena; BROAKAN . Jadviga; JOKAJTIS, Merin. Significance of hmato-carebrospinal orgar index
in tuberculous meningitis and encephalitis. Grazlica 23 no-10:701-706 Oct 55. 1. Z I Kliniki Chorob Dzieciecyok A.M. w Gdansku. Kierownik: prof. dr. nod. H.Brokman. Gdansk, I Kliui*a Pediatryesna
A.M. ul. Debinki 7a. (TUMCULOSIS, NUINGIAI~. metabolism, carbob ydrates, hemato-encephalic passage) (RUATO-21CWHALIC BARRM, . permeability of sugar in tuberc. meningitis) (CARBORYDRAM, metabolism,
hemat o- ene ophalic.,V&13 sage in tubere. meningitis) KMTIAIICZ, Helena Intoxication with antictine in a 3-year-old child. Pediat.polika 30 no.6:575-576 Je '55. 1. Z KlIniki Chorob Dzieclecych A.M.
w Gdansku. Kierovnik: prof, dr med. H. Brokman Gdansk, Debinki 7a. (AYTIRISTAMINICS, injurious 01!ects, antasoline, in child) KUNIEWICZ, lielena; SKARZYNSKA, Halina; ZYCHOWICZ) Czealaw Primary
pneumonia In the course of varicella in children. Polski tygod. lek. 16 no.28:1071e-1076 lo J1 161. i. Z I Kliniki Chorob Dzieci AMG w Gdanoku; kierownik: prof. dr Med. K. Erecinski. (CHICKENPOX
compl) (PNEUMONIA in inf & child) KUNIEVIICZ '..qel-e-na; USIEWSKA, Jadwlg~q ZYCHOWIC3, Gzaglaw Inflamnation of the'larynx qrA lower respiratory tract U measles in cUldren. Podiat, Pol. 37
noll2tl289-1296 D 162. l. Z I Kliniki Chorob Dzieci MI w Gdansku Kiorownik: prof. dr med. K. Erecinski. (MEASLES) (LARYNGITIS) (TRACHEITIS) (BRONCHITIS) C:Z He I 'I a-5Zl'A!'O*,,SKk-D/dlr,0j 'Aar,
da; SZICZUROW; 14a- I FST 7117 ~ j? f-TLIK A , A 11 f!,(, a ~ KTIE-7 G 4Y, I la 1 in a A.--,:uf,e ciltarrhen-I syndrome with alcorative and rhan-es --n infants. Padi-at. Po2. 39 Z 1. K L In ik.' "ho
rob d~,A roc i a-adie.-ri I! Mled,,,c 7,rv~i -w (Atur, I-.-, I -1 (RierovnAk., pf:)fa dro IIM7.,d. K. PalolngIcne-* kkadenil a Worownilk~ prof. dr. ned. Q . , . . I -- . - , 9 Ftfv 2 621.31121 621
31N f,71.392.2 919--l KunlewO.1 11 The Mittt of P4-wrr Transformers an 11, F. C"Trent Vlow lAn;-:L ,,Wplyv, tran7zfurtnutor6w mocy nt, rozPly%, p-;iJ6v' W101ktel thwDkI w Dn~ach prz--rriy-0ov,,ych .;
rys~kiogo n-jpvQcIa-. (Prate Przem !iiar- Tele~tvrn No. 13-14). Worrza-wa. 1054. PINT, 13 pp., 21 figg.. I taL The rvsult2 ,f the rnrazqurements, in a case of .9 sing Ir- -cond uc t;,r V'UPTIng
6y,~',P'Tj. of real -ckd 1-iaginary rnrhpon(-nL% (4 the on Ignte of vAr)(,~A P,,,,-r .--nmf~rmrr ', in !he frequency range from 2V kc"3 tip to 300 krIs. Chararterislics of p-ow-,r trnn5foi-mers were
imnl,ysed ir, conne-c- ,Ion with ffivse of tYpIcal double-frequency blocking chokeq with an In- cluctnripe of 0 15 mli There 15 also a dlsru3sioti of the attenomion iritr-y- dured at the ?nrl 0 in
if. F. line section by po-xrr tr-nniiformerr withnut hlr~lking vhDke5 A ne-. method of using H. F b~orking devices Li ex- p1nined In ronclu=n, the-paper gives the results of the meaziurements of
attenuation caused hy power translctrmers, Inserted between different i,ectiowk -if H F ty-nnsmIsMon lines, The lowest attenu;itllan vnlue-- In the r.-ingf, 20 . 3W kcis are recorded for both star
and delta connection. KUNIEWSKI, H. Short-range unbalanced telemetric systems with self-inductive electric power. p. 202 Vol. 28, no. 6, June 1955 PRZEGIAD TEL--PKOMUNIKACYJNY Warszawa SO: Yonthly
List of East European Accessions (EEAI,), W. Vol. 5, no. 2 Feb. 1956 KLINImIIKTP H. Self-controlled talemetric systems. (To be contd.) p.272 PRFOLAD TELEKOMIKAMNY. (Stowarzyszenie Blektrykow
Polakich. Sekcja Telekomunikacyjna) Warszawa,, Poland Vol.28, no.8, Aug. 1955 Monthly list of East European Acceasione (EEAI) LC, Vol.9, no.1, Jan. 1960 Uncl. KUNIEWSKI, H. Telemetric
self-controlling systems. (Conclusion) p. A Vol. 28, no. 9,1 Sept. 1955 PRZEGLAD TELEKOMUNIKACYJNY. Warszawa* SOURCE: East European Accessions Idst 'FEAL), LC, Vol. 5, no. 3, March 1956 7.04 AUTHOR:
Kuniewski, H., Docent 22847 P/02 2 6o/ooo/olO/012/012 A222/A126 TITLE: Selective ringing equipment, ITR system PERIODICALs Przegl~d telekomunikacyjny, no. 10, 1960, 326-328 TEXTs The Instytut Tele-i
Radiotechniczny (Institute of Telecommu- nication and Radio Engineering) designed a radio intercommunication system with frequency-selective ringing, in which the master station uses 13 audio
frequencies and the subordinate stations 2 audio frequencies each. The variation results in 78 combinations; thus, the system comprises 78 remote stations which, in turn, are set up into 13 groups of
6 stations each. Such arrangement permits to call each group of 6 stations by means of only 4 frequencies transmitted at the same time. The 13 frequencies were alloca- ted within the range of
420-3,000 cps with an irregular spacing. A block diagram of the master station comprises: a) 5 variable-tuning, RC audio generators (4basic and I stand-by generator); b) output amplifier; c) cyclic
ringing assembly; d) control and test board; e) power supply. It Card 1/4 22847 P/022160/000/010/012/012 Selective ringing equipment, ITR system A222/A126 has 78 push buttons for individual ringing
of subordinate stations. The block diagram of a subordinate station "is shown in Fig.3. The ready sub ordinate receiver is shown in Fig.4. Technical data of the transmitter: frequency range 420-3,000
cps; alignment accuracy � 1 cps; frequency sta- bility at a feed-voltage variation of + 10% and temperature variation of + 100C and tube replacement is as good as + 1 cps below 1,180 cps or 2% ';bove
1,400 cps; amplitude stability unde'~ above conditions + 3%; linear distortion under unfavorable conditions is lower than 2cv; output resistance 600+ 10%; maximum power drain 130 va. Technical data
of the receiver: ban7dwith is + 20 cps at 420 cps and + 75 cps at 3 000 cps; maximum toler- able tranami-asion level variation fr-om about -0.8; (neper) to about +0.8N as measured against the 1,000
cps level; inpat resistance higher than 0.25 megohms; stand-by power drain 15 ma, on power drain 15 to 20 ma; dimensions 125 x 105 x 75 mm; weight 1-5 kg. There are 4 figures. ASSOCIATION: Instytut
Tele-i Radiotechniczny (Institute of Telecommunica- tion and Radio Engineering) Card 2/4 !6,6 0 P/022/61/000/00,3/002/002 A076/AI26 AUTHOR: Kuniewski, Henryk, Docent a TITU- Transmitting sets of
non-periodic impulse systems in long-distanco telemetry PERIOMCCAL: Przegl#d Telekomunikacyjny, no. 3, 1961, 78 - 84 TEXT; After generally describing the main characteristics of non-periodi., impulse
systems, the author describes transmitting and receiving sets produced by the A.T.M. Strowger; the Bristol; the AEG; the L.M. Ericeson and the Landis Gyr Firms. The MiZ transmitter, produced in the
USSR, is also sbown and its main parts described. There are 18 figures. Card 1/1 KUNIEWSKI, Henryk., doe. Instruments for personal paging. Prace Inst teletechn 6 no,3185-90 t62, 1. Instytut Tele-i
Radiotechniczny, Warszawa. KUNIEWSKI, _~enryk, _ doe. "Garrier-frequency teletransmission of information over high-voltage networks" by II.K, Podazook. Reviewed by ifenryk Kuniewski. Przegl
elektrotech V no.11:478 162. I ~ KUNIEWSKI, Henryk, doe. A set of dquipment for selective calling installations. Prace Inst taletechn 4 no.1:90-93 160. -.1- e~~I ~,e . -.- Use of curved screens in
coal preparation plants. Sbor. inform po obog, i brike ugl, no.4:61-63 '57. (miu 11.-6) (Mining)) preparation) (Screens (Ooal KUNIK, V.P., inzh. improving coal properties for briquetting purposes at
the Rhine I*iquet plant in Germany. Obog. i brik. ugl. no.6:63-66 '58. (MIRA 12:7) (Germany, West-Briquets (Fuel)) KMTIK, V.P.. inzh. Increasing the efficiency of tubular steam driers by means of
preliminary partial drying. Obog. i brik. ugl. n0-7:74-76 '58. (MIRA 12:7) (Coal--Drying) (Drying apparatus) KURKIN, Yn.F.. inzh.; KUNIK, V.P., inzh. Graphic method of determining the results of coal
crashing. Obog.i brik.ugI. no.12:48-50 '59. (MM 13:6) (Coal preparation) ISAYEV, Ivan Ilikolayevich; KW,IK, V.P oty. red.; LOMILA.'A, L.L., �--*-s tekbra. red.; S1fKLYAR-,-S-.Yi-.- to-k-b-n-. rod.
[Concentrating table ti I Hontse ntratsionnye ctoly. Moukvn, Goo- torgizdato 1962. 100 1.. (MIRA 15:10) (Ore drossing-Equipment and supplien) KLIMANOV, Aleksey Dmitriyovich, kand. tekhr... ns%ik,
dots.; HUDYNKO, Konstantin Gerasimovich., kand. tekbn. nauk, dots.; KARPUKHDI, V.D.0 dots., retsenzent; OGLOBLIH, N.D., inzh.p retsenzent; DREMLO, P.G... inzh., retsenzent; KUNIK, V.P., otv. red.;
BOLVYREVA, Z.A., tekhn. red. [Safety techniques and fire prevention in ore dreusing and briquetting plants]Tokhnika bazopasnosti i protivopo7liarnaia teklmika na obogatitollrorkh i briketnykh
fabrikakh. Moskva, Goagortekhizdnt, 1962, 362 pe (IAIRA 15:10) (Coal preparation planta-Fire, and fire prevention) (Ore dressing-Safety measures) KUNIK, Ya., kand. yurid. nauk A firm asks for the
floor. Sov. torg. 37 no.10:16-20 0 163. (MIRA 17:1) KWIIK, -~ , kand. yarid. muk. Accounting by means of checks. Sov. torg. no.3:54-56 Ur 158. (Accounting) (Chooko) (MIRA Ilt2) ONIK, Yakov
Abramovich: STARGIMOVA, I.I., red.; BABICHKTA, V.V., ~1:1:111--E--- eV, "" [Legal forms for Intraoity accounts in Soviet state trade) Pravovye formy vnutrigorodskikh rasebstov v 'sovetskoi
goBudarBtvannoi torgovle. Moskva, Gos.i2d~-vo torg.lit-r7, 1959. 61 p. (MIRA 12:6) (Banks and banking) K M K, Ya., kand.yurid.nauk Delivery of goods and payment methods. Sov. torg. 33 no. 9:20- 23 S
160. (MRA, 14:2) (Delivery of goods) (Payment) --- KUNIK,,U, lAt's inculcate progresaive fonro of payments. Sov.torg. 35 no./+:26-28 AP 162, WITIA 15 - 40 (Russia-Commerce) (Payment) ANTIMONOV, V.S.,
prof.; VEDENIN, N.N., kand. yurid. nauk; GENKIN, D. M., prof.; GRAVE,. K.A., prof.; YEPANESFINIKOV, N.V., dot.4%; ZHUKOVA, L.F., dots.; KUNI~, Ya.A., dots.; LIVOVICH, Yu.Ya.; MARGOLIN, M.Z.;
111011OVSKAYA, T.A., dots.; POLENINA, S.V., kand. yurid. nauk; SADIKOV, I.N.; FIALKOV, M,A., kand. yurid. nauk; YAZEV, V.A., knnd. yurid. nauk; YAKHNINA, N.A., kand. yurid. nauk; KIRAKOZOVA, N,Sh.,
red.; ELIKINA, E.M., tekhn. red. (Government trade regulation] Regulirovanie gosudarstvennoi torgovli. Moskva, Gostorgizdat, 1963. 339 p. (MIRA 16:7) (Commercial law) TYPOVSKY, K., As. Dr; FARGAS,
Xd., Dr; KUHIK, Z., MUC Largical treatment of Intra-articular fractures of the condyle of the tibia with the aid of a clip. Acta chir orthop Cz 21.no.l: 8-14 F '54. (M" 3 '- 8) 1. Z chirargicke
kliniky PU v Olomouci. Prednosta prof. MUDr Vlad, Bapant. (TIBIA. fractures, *intra-articular fract. of condyle, ourg. reduction with metal clip) OMACTUR33, *tibia, intra-articular fract. of cond7le.
aurg. reduction with metal.clip) BELIKOVICH, V.V.; KUNIWV, M.V. Mnthod for the quadratic trrinnformation of' PJ.gnal amplitudon. Prib. t tokh. oknp. 9 no.1:115-116 Ja-F 164. (MIRA 17:4) 1.
Nauchno-issledovatollskiy radlofizicheskiy In-3titut Gorlkovskogo gosudarst-rennogo universitetn. /0 W(d)1Pa(l)1FS(Y)-3,1?8S-2 TT/AST/GW ACCESSION NRt AP5=55 Ult/0293/65/003/004/0610/062
629.195.21621.39 W, 1 no AMORS t p., Kdj~ 1yko___1_ Ve M~ Halik Ich. V. VIA Bnkhnin.' V. HN Kanto L. - YaVWiK0=Q'-bh"kaL! A. Gjtge-r-ep'MVME~~* Wm~_%_A ir 3 V T 4q 6,qq, TITM The results of an
experiment on radio co icatiopR via lEcho 2* and the moon at a frequency of 162.4 megacycles between the observatories of Jodrell Bank. and Zimenki SOURCM Koamichankiye iaeledovaniyal v. 3# no. 41
19651 618-4629 TOPIC TAGSs moon, eltell 'Yiadio telescope, radio tranamiesiono _k&j_qmmicati pj satellite trackingp scientific research coordination / Jodrell Bank radio tele- scope, Zimen1d
obaervator7 radio telescope, BESH 2 electronic coWuter ABSTRACM During February-4ktrch 1964 the Acadew of Sciences of the SSSR, NASA of the USAj and the General post office Department of Great
Britain conducted an experiment to establish or*-mq radio communication at 162.4 megao7oles via the passive satellite nftho-2v and the moon. Echo-2 van used for 34 oomounication cam ]A L 65295-65
ACCESSION VRt AP5021255 r) toots of 10-15 minutes (the time i6terval permitted by Echols orbit), and the moon woo used for 15 test runs between tho Echo toots. The transmitting equip- ment at Jodrell
Bank and the receiving unit of the Zimenki Observatory are do- scribed In detail. Echo orbit information furnished by NASA, visual observations, and radio trucking data from fixed stations were fed
to a BESR-2 electronic cal- Puintor which providod prograrmod tracking control. The receivel signal exhibitod strong fluctuations separable into two periodat 1) a 1-2 minuto fluctuation as- vociatod
vith Echo-2 distortion from a sphoro and with tracking errora; 2) a 3-10 second period associated with small surface irregularities. The rapid fluctua- tiono varied with each teste Voice signals,
slowed by a factor of 8, were barely intelligible. Telegraph, tolotM, one rbotofaceimile transmission, in general, were unsatisfactory, but in periods of high signal-to-noise ration intelligible
moonagoo were received. The moon transmissions wore not an clear but did furnish scientific Information. Unexpected transmission losses included 3-5 db for polar- ization losses and 1-2 db for
unknown causes. The international cooperation was excellent, with the Soviet submitting a.couplete report. rOffers for further co- operation have been extended. Orig. art. hast 3 tables, Tfigures,
and 1. for- mulas. ASSDCIATIONj none SUBHrrMDs 1SApK-65 MICLs 00 SUB ODIXt AA, EG NO MV MY v OOD 1 002 Cc!dj2/2- BDGATYREV, A.S., konstruktor zavoda, g.Irkutsk; MIKHALICHINED, V.; TSUKASOV, 1.
(pos.Ili, Alma-Atinskov obl.); KRYLOV, ff.; SKRYABIN, A.;_~~IIWV- 3U_T, K., (Leningrad, Siuopskaya nab., 66, kv-5) Advertisement board. Isobr. i rate. no-11:521-53 v 16o. (KIRA 13:10) 1.
Leznikovskoye karlyerotipravleniye, Zhitomirskoy obl, (for Hikhall- chenko). 2. P~redsedatell pervichnoy organizatsii Voesoyusnogo obshchestva izobretateley i ratsionalizatorov, g.Ivanovo (for
Skryabin). (Technological innovations) Ili-A t ------- "O1& 04 00 00 It -1 post .1,10 jilt I-Iliv "11113111 1111.1 00 .1 v Pus, -pil, py IwI!fIq I I fixHl P-'P.!o w IWO 411PIM 1 00 it J0111,4111111
Im"1011'.1mil -lip" Jill 01111 milla It twill J~.vmb)%% I mitill v 11114A 11411sl 'Mint ILIP Aill 01141 %III its .1"Alix &tit UI Aspos, .1141 its pU3 all$ ^Imu 00 in 00 All 41tisulp Suisvisid "Ill
"jill in aw 041111 o f1lir w.111 P0 JO *1 ot-I 01 *ucmpAquInb 1. 4., I I I lism .111 tll~ -jisr ul to -.%3 MI ,v `~ ppv-,(IR .!.F ;N '0 x .11'0176 0! -Lv 111 4 it -Rum S140=031 000 49RU41611,11 10
4WPw oql Iscs wpollIssilmlea dF 0 lols;)q .. 00 I-wicv -A 00 n as ILI 44 W- 7 -1 ' Ir" VT_A~ _R 5 4 It w .7 - & Ma Riabtsev, N. 1. General fuel technology; a textbook Moskva, Goo. nauch.-tekho.
izd-vo neftianoi i gorno-toplivnoi lit-ry, 1949. 326. p. (50-15032) TP318.R5 h-I r - . -N-%'i, A. ~-.. GOYEMM, I.M.; KUNIN, A.M. [Semicoking of coal) Polukoksovanio uglia. Moskva, Goo. nauchno-
tekhn. izd-vo neftianoi i gorno-toplivnoi lit-ry, 1953. 193 P. (Coke industry) (HLRA 7:8) I I I i X- t&' -"~ Y/ li~ TURSKIY, Yu.I.; BRIX, A.N.; VNIN, A.M.; GALIPERN, Ye.M. Determination of small
quantities of butYl naltate in water. Gas.prom. no.9:11-13 S '57. (MIRA 10:10) (Acetates--Analysis) (Water-Analyois) .A2) PHASE I BOOK EXPLOITATION SOV/334o Kunin, Aleksandr Maksimovich, and Mark
1khelevich DerbaremdIker Tekh-no-khimicheakiy kontroll gazovogo proizvodstva (Technical and Chemical Control of Gas Production) Moscow, Gostoptekhizdat,, 1958. 331 P. 3,000 copies printed. Executive
Ed.: Ye.S. Lozbyakova, Engineer; Tech..,Ed.: A.S. Polosina. PURPOSE: The book is intended for laboratory personnel in g'as,works and gas,-generating plants. COVERAGE: The book is an attempt at a
systematized presentation of the problem of quality control in the produr-tion of gas. 9~he_ following steps of the production process-are treated: control of the quality of coal used for
gasificafion; quality control in the processes of.p?~oduction, dehydration and purification of gas from tars and hydrogen sulfide; and control In the dephenolization and repurification of waste
waters. D.A. Muravlev collaborated .with the authors in writing Chapter 5. Chapter 4 was written Card-=1/1-3--- Technical and Chemical Control (Cont.) SOV/334o jointly by S.M. Golyand, T.K. Krapivina
and M.M. Kuzmak. There are 46 references: 45 Soviet and 1 German. TABLE OF CONTENTS: Foreword Ch. 1. Controlling the Quality of Coal Used for Gasification Coal as an industrial raw material for
gasification 5 Methods of analyzing solid fuel 11 Composition of solid fuel 11 An average fuel test sample 11 Sampling and separating a coal test sample 13 Separating initial samples in the
laboratory 13 Preparation of analytical samples for general analysis 16 Determining moisture content 17 Determining moisture content (Wa) in an analytical sample for general analysis 19 Rapid methods
for determining moisture content in solid fuel 19 Determining ash content in solid fuel 22 Determining the specific gravity Qf solid fuel 25 Card,194,3,- RAKOVSKIY, V.Ya., dol-tor tekhn.nauk;
RIVKIIIA, Kh.l., kand.tokhn.muk; K0111, A.M., kand.tekhn.nauk; RAYRUBERG, M.M., inzh. -tiz Peat bakeliten In tho manufacture of sawdust boards. Torf. prom, 36 no.8:8-12 '59. (MIRA UO) 1. Kalininskly
torfyanoy inatitut (for Mayzenberg). (Peat) (Phenol condensation products) NUZIMMOV, L.N.; KOIN, A.M. Removal of water from peat and shale tare by the action of ultra- sonic waves. Torf.prome P
no*7:19-22 060. (MIRA 13:11) 1. Uningradakiy metrologicheakiy inBtitUt imeal D.I.Mendeleyeva (for luz8memkov). 2. Kalininskiy torfyanoy institut (for Kunin). (Post-Drying) (Ultrasonic
waves-Industrial applications) MMNDERG,, M.M.,, inzh.; RAKOVSKIY,, V.Ye.,, doktor tekhn.nauk; RIVKINA, Kh.I,, kand.tokhn.nauk; KU14IN A.M. kand.tekhn.nauk Synthesis of resol resin by the condensation
of peat phenols with formaldehyde in an oil medium, Torf. prom, 38 no.8:24,- 25 161, (14IRA 14:12) 1. Kalininskiy torfyanoy institut (for K~zain). (Phenol condensation products) (peat) FEDOROV,
N.A.;,BELYANOVA Ye.M.; GRIDNEVA) K.I.; RAKOVSKIY, V.Ye.; KUNIN AM.; YAKdA, 1". S. Composition and ways of using the liquiq products .of under- I ground gwrification, of coals. Nauch. trjdft
VNIIPodzemgaza no.8:95-103 162. (MIRA 16:6) 1. Voesoyuznyy nauchno-iseledevatellokiy inatitut podzemnoy gazifikataii ugley, Kalininskiy torfyanoy inatitut i Vasoyuznyy nauchno-ionlodovntal'okiy
institut udobrady i Agropoahvo- vedaniya. (Coal gasification., Underground--By-products) --UUU14 A. V. Favorable conditions of production guarantee success* Tranop* strol. 10 n0-5:6-7 Vq 160. (MIRA
13:7) 1. GlavMy inzhener Kontrollno-proverochnogo punkta stroitelistva Pemetroy-put' (for Kunin). (Reinforced concrete) MMIN, B.A. Ixtensivo resection of the humerus with fibular substitution. Ortop.
travm. i protez. 20 to.2:59 P 159. (MIRA 12:12) 1. Iz Tul'skogo garnizonnogo voyennogo goapitalya. (HUMERUS, surg, extensive resection, fibular substitution (Ran)) (FIBULA, trarispl. in extensive
resection of humerus (Rue)) W WNIN, B.A.v polkovnikmea. eluihby Diagnosis, treatment, and late results of injuries to the meniscus of the Icuee joint. Voeu.-med.zhur. n0-208-40 P 160. (MIRA 13:5) ( 1
11 wds. & inj.) I P TF, 'Y, A ZEXLYANOY, N. G. KOSTOGRY-'/Oi"l .7.S., hand. tehhn. ra,,ik,- ICROF31INICHFIM". ),,.V.~ MwJind for objective control of thc- intensity of carhon dicx`.de em-'s-4ion from
a tub. A-vtom. I pi-lb. no.1;9-12 1-1,; 165. (MIRA 1M) KUNIN, B.Z., inzb. Designing walls and slabs fixed on three sides only with the fourth unsupported. Promestroi- 38 no.3:6o-62 16o. (Walls)
(Goncrete slabs) (MIRA INO h'J?~'!Nj D.; ANTUNOITA, T. N.; RAKOVSKIY, V. Ye. I - --- "Chemical and heat processing of peat." Reoort submitted for the 2nd International Peat Congress, Leningrad 15-22
Aug 63. PZTROVSKIY, V., in-rh.; KUITIH, F. Improving the filter centrifuge for the removal of fat from a protein-water-fat mixture. Mias. ind. SSSR. 30 no-4:37-38 '59. 1 (MR& 12:12) l.VeasoyuzW
nauchno-iseledovatellakly inatitut myasnoy promVshlen- nooti (for Petrovskiy). (Poltava-Packing houses--Equipment and supplies) (Oils and fate) KUNIN, G.L.;_.IJGWV, P.A., takhnik Measurement of
capacities b7 means of the MVU-49 bridge. Avtom. telem. i sviaz' 3 no.8:24-25 Ag '59. (MIRA 13.-2) l.S-arshi7 inzhener Isboratorii signalizataii i evyazi Kuybyshevskoy dorogi (for Kunin).
2.Isboratori7a signalizateii i evyazi Kuyby9hevekoy dorogi (for Uglov). (Electric measurements) (Bridge circuits) SOV/124-58-7-7725 Translation from: Referativnyy zhurnal, Mekhanika, 1958, Nr 7, p 58
(USSR) AUTHOR: Kunin, I.A. TITLE: __'__contribution to the Hydrodynamic Theory of the Lubrication of a Thrust-Bearing (K gidrodinamicheskoy teorii smazki pod- pyatnika) PERIODICAL: Izv. %,ost. fil.
AN SSSR, 1957, Nr 4 - 5, pp 128-137 ABSTRACT: The solution of the problem of the th ree -dimensional flow of a lubricant with varying viscosity in a thrust bearing is de- scribed concisely. The
Reynolds equation and the approximated heat-balance equation are discussed, wherein the heat transfer through the walls of the thrust-plate and the thrust-bearing seg- ment is accounted for
approximately by a coefficient. In solving the Reynolds equation the author assumes the viscosity of the lubricant to be dependent upon the flow angle in the direction of the segment rotation. In
this case there are two possible meth- ods of solving the Reynolds equation. The first method consists in changing over to new variables, in which the equation does not change_ but the viscosity is
little dependent on the angle. By Card 1/2 treating the viscosity as constant, the Poisson equation is SOV/124-58-7-7725 Contribution to the Hydrodynamic Theory (cont.) obtained, the solution of
which does not present any difficulties. The newly obtained expression for the pressure distribution is substituted in the heat- balance equation, which serves to determine the value of the parameter
entering into the relationship between the viscosity and the angle. The second method assumes that the relationship between the viscosity and the angle is expressed by means of a harmonic function.
In this case the pro- duct of this function by the pressure also produces the Poisson equation. This method of solution is simpler (but less general) as compared to the first, and it is recommended
for the calculation of the thru st.-bea rings. A description of a calculation method is given with pertinent nomograms for a case when the ratio of the outer and the inner diameters of the
thrust--bear- ing is 1.57. A.I. Golubev 1. Thrust bearings--Lubrication 2. Thrust bearings--Hydrodynamic characteristics 3. Harmonic functions--Applications 4. Mathematics--Applications Card 2/2
AUTHOR: Kunin, I. A. (Novosibirsk) 24-10-23/26 TITLE: Solution of the Reynolds equation of the hydrodynamic theory of lubrication in the case of variable viscosity. (Resheniye uravneniya Reynolldsa
gidrodinwaicheskoy teorii smazki pri peremennoy vjazkosti). PERIODICAL: Izvestiya Akademii Nauk SSSR, Otdeleniye Tekhnicheskikh Nauk-, 1957, No.109 pp. 109-110 (USSR) ABSTRACT: A method is described
of solving the basicequation of the hydrodynamic theory of lubrication (Reynolds equation) for the case of variable viscosity, which is based on the following idea: the viscosity is approximated by
an appropriate coordinate function which depends also on non-determined parameters, which have to be determined from the thermal balance equation, whereby the approximate function is so chosen that
the Reynolds equation can be easily solved. The case of a thrust bearing is considered; the solution will be similar for a radial bearing. There are 2 figures and 1 Slavic reference. SUBMITTED: May
9, 1957. AVAILABLE: Library of Congress. Card 1/1 KUNINI-I...A., Cand phys-Math Sel -- (diss) "Hydrodynamic theory r ~ of lubrication of 00 footstep bearing$." Fvovosibirsk'-', 10158. L_ J 12 pp,
(Len Polytechnic Inst im M. I. Kalinin, Acad Sci USSR* West-Siberian Affiliate), 110 copies M, 18-58, 95) 1 -7- KUNIN, I.A. - ~Solvi~ne some classes of problems by analogy in an electrol7tic tank.
Izv. Sib. otd. AN. SSSR n0-7:53-61 '58. (MM 11:9) 1.Zapadno-Sibirs iV filial AN SSSR. tilectromechunical analogies) 3OV/24-58-10-29/34 AUTHOR: Kunin, I. A. (Novosibirsk) - - J_ TITLE: An Approximate
Method for the Solution of Boundary Problems for Some Equations of Elliptical Type (Priblizhennyy metod resheniya granichnykh zadach dlya nekotorykh uravneniy ellipticheskogo tipa) PERIODICAL%
Izvestiye, Akademii nauk SSSR., Otdeleniye tekhnicheskikh nauk, 1958, 11r 1-0, p]? 146-J50 (U&")*R) ABSTRACT: An aceount is given of an approximate method of solving boundary problems for equations
of elliptical type to which many field problems may be reduced. Their solution is divided into two stages. In the first stagel, the o-riginal equation with variable coefficients is reduced, using
partial solut- ions of a homogeneous equation, to an equation with almost constant coefficients. In the second sta-e the latter equat- ion is solved approximately by solving the corresponding equa-
tion with constant coefficients. As an example, the problem of lubrication of a bearing in the form of a sectcr of a circle is considered, the viscosity being variable and obeying a linear Card 1/2
SOV/24-58-10-29/34 An Approximate Method for the Solution of Boundary Problems for SDme Equations of Elliptical Type law. The solution obtained is in agreement with that obtained by Mitchel (Ref,l)
in a special case. There are 3 figures and 2 Soviet references. SUBMITTED: June 3., 1957, Card 2/2 SOV/179-59-2-10/40 AUTHOR. Kunin, I, A. (Novosibirsk) TITLE: ~fi~*_etilodynariiic Theory of Flat
Film Lubrication with Res- pect to Viscosity and Temperature (Ploskaya zadacha gidrodinam- icheskoy teorii smazki pri ucliete zavisimosti vyazkosti ot temperatury) PERIODICAL: Izvestiya Akadeinii
nauk SSSH OTN.. Mekhunika i mashino- stroyeniye, 1959, Nr 2, pp ?0-?4 031"~R) ABSTRACT: In this article 11--b--ication of the bearings of hydro- generators and ships' turbines is considered. The
problem is illustrated in Fig 1, where ab - a segment resting on a po-inL 0 P load, ed - resisting surface moving with velocitty U 0 The hydrod namic equaition is given as Eq (1.1) for the conditions
U.N The equation of thermal equilibrium, in the rarige of temperatures between 30 to 700C, is given as Eq (1-3), where It- - viscosity at the initial .L temperature, t -- inere'ase of
talliperat'llre, ID -- tomperat- ure characterizing the relationship of It and t . Assum- ing that most of the heat is taken with the grease, the above eauation becomes Eq (1.4) where y specific
weight of grease, c - heat conductivity, m the coefficient !~;0.9. Card 1/4 SOV/1 -0/40 On 'the Hydrodynamic Theory of Flat F3_1r-, Lubrination When no t is considered the Eq (1.115.) can be applied.
If the expressica cf velocity is substituted in the third equat- ion of the expression (1.1) and in the thermal. equation (1.5), the Eqs (1.6), (1.?) and (1.8) are obtained, from which Eq (1.9) can
be found. As Tj is not known, E- (l 6) can be found as follows. The. function ~iQ, a, J3 f~r a con- (0) (4) stant & and -~L and' 11 are defined, then in the region of parameters oL and 4Y the
fun;.~tions ~L(~ ) increase from th~_~ value It-0 '- -~).Ta.-nEq (2.1) can be defined. Fig 2 represents ji(o) and ji (4) for 1 and t,~ = 3 which shows that p is not affected by a The function is also
given. The viscosity can be calculated from the approximate Eq (2.2) (dotted line) which gives an accuracy of 3%. The characteristic ooetfieient of the minimum film L. tliicl~ness is defined as EqB
(2-3) and (2-4), and the eccen- tricity is given by Eq (2.5). The increase of temperature ilt can be found from Eq -(2.6)~ In general, th~e problem Card 2/4 is solved when the relations n , 112 ~ 0 ,
s , a and ,r are iit jOV1V 9- ~59-2 On the Hydrodynamic Theory of Flat Film Lubrication determined. This can be done, for examnle, as follows. The following axe given: dimension of the segjilentl
velocilty, initial. tam'oerature and type of grease; the following are found,., relaiioji of film thickness and increase of tempera- ture at; various loarls and the eccentri,:~ity for their max mum
values. Thiis L-9 T and ., are known a_nd p h~_ 2 M 0 and At are proportional to 1-1 if and 6 Therefore, it is sufficient to determine R 2 a-rid J . This is illus- trated in Figs 3 a-rLd 4, where Q
cx = 0 corresponds to the limit of possible value. The oui~v,- a -- const in Fig 3 is shown as a dotted line. The relation of H 2 and 0 to n for E - 61 can be determined from Eq (3.1). Similarly, the
loss of power due to fric~tion N can be determined from Card 3/4 On the Hydrodynamic Theory of Flat 10i lia, Lubrication Eq (3.2). The effect of grease on the charac-teristics of .U the bearings
(with given -o. , *r 1 110and iniuial tempera- ture of the grease) for k ^1 Iii ~ F1 /\./ T.-I can be shown as Eq (3.3) and the initial tampo-rarLire of the grease, with other parameters constant,
can be detoi' rqj.ned from Eqs (3.4) or (3.5). The relationship of the characteristic of the bearings to the velooity is def4ned as Eq (3.6). Fig 5 shows the function n (a, 0) for v =-l and /J- = 3
defined by the method of linear visoosity (a), mean viscosity (c) and from the results of this work (b). It shows that the least error is produced by the method des- cribed in this work. There are 5
figurcs and 2 references, of which 1 is Sovieb and I English. SUBMITTED: July 21, 1956. Card 4/4 67590 SOV/179-59-5-9/41 AUTHOR: Kunin, I.A. (Novosibirsk) TITLE: Contribution to the Theory of the
PlanetarX-Vibrator in an Infinite Fluid Medium PERIODICAL:Izvestiya Akademii nauk SSSR, Otdeleniye tekhnicheskikh nauk, Mekhanika i mashinostroyeniye, 1959, Nr 5, pp 48-52 (USSR) ABSTRACT: High
frequency mechanical vibrators of the planetary type without bearings are finding increasing favour in Russia. The design of certain types of such vibrators is described by L.P.Petrunlkin (Ref 1).
The elementary theory of this type of vibrator for compacting a concrete mixture has been given in the sr~me paper. The problem of the generation by the planetary vibrator of sonic waves in an
infinite fluid medium is considered by the present author. The mechanical model investigated has a roller rotating under a constant external torque at a constant angular velocity. Simultaneously, the
roller rolls without sliding along the internal surface of a hollow cylinder. The latter is so placed in an infinite, viscous, compressible fluid that it can take part in translational Card 1/3
motion in a plane at right angles to the cylinder axis. 67590 SOV/179-59-5-9/41 Contribution to the Theory of tile Planetary Vibrator in an Infinite Fluid Medium In the steady state, the centres of
gravity of the roller and the cylinder rotate at a certain angular velocity about a certain fixed point. This, in general, lies outside the straight line joining the roller and cylinder centres.
Hence the oscillations of the roller and cylinder will have a phase difference other than direct opposition. The forces oxerted by the fluid on the cylinder are first found, treating the plane
problem only. Under certain conditions, defined by relations between the dimensions of the vibrating bodies, the frequency, speed of sound in the fluid and its kinematic viscosity (conditions which
are fulfilled in all cases of practical interests), the fluid outside the vibrating body can be divided into two regions: (a) a thin layer containing vorticity, where the viscous forces are
significant and (b) the region of sound waves. In the latter region, a velocity potential exists which satisfies the wave equation. To find this potential, the conditi-~ns of emission at infinity and
the equality of the normal Card 2/3 velocities of the fluid and the body at their boundary must be satisfied. In the boundary 67590 SOV/179-59-5-9/fil Contribution to the Theory of the Planotary
Vibrator in an Infinite Fluid Medium layer, the tangential component of velocity satisfies an equation of the parabolic type and decays exponentially across the thickness of the layer. Its boundary
condition is determined by the step of the tangential compononts of the velocity of Clio potontial t'low. The potential flow is found first. The velocity distribution and the friction force in the
boundary layer are then determined. It is noted that the resistance caused by sound radiation is predominant in the range of medium frequency, where the losses caused by friction in the boundary
layer are negligible. The power absorption of the vibrator is computed and the conditions for rolling without sliding of the roller in the cylinder are stated. There are 2 figures and 3 Soviet
references. SUBMITTED: May 11, 1959 Card 3/3 DYKIM, A.M. ; KUNIN, I.A. Determining the surface area of a convex body from its projections. Izv. Sib. otd. AV SSSR n0-8:3-12 '59. (MIRA 13:2) 1.1natitut
radiofiziki t alaktroniki, Institut gornogo dela Sibirskogo otdoleniya AN SSSR. (Surfaces) KUIIIH, I.A. Absolute minimum of one functional. Izv.Sib.olvd.AN SSSR no.11:90-91 159. WIRA 13:4) 1.
Institut rnogo dola Sibirakogo otdelenlya AN SSSIR. rFunctional analysis) .'i MME :~Boo.K npwiunon sov/469o Minin, Isaak Abramovich G Ii dro&inamIcheskaya teoriya smazki upornykh podshipnikov (The
Hydrodynamic*Theory. of Lubrication of Thrust Bearin6s) Novosibirsk, Izd-vo Sibirskogo otd-niya AN S*3SP), 1960. 129 p. Trrata slip insirtel. 1,000 copies printed. Sp~woring Agency- Akademiya nauk
SSSR. Sibirskoye Gtdeleniye. Resp. Ed.: B*Vi. Sudnisbuikc;v, Candidate of Technical Sciences; Ed.: G.L. Ivanova; Tech. Ed.: A.F. Mazourovs,- PURPOSE: This book is intended,for teebideal peraminel of
the machine-building industry and vorkeers of scientific research institutes. C'OVERAGE: The bock develops the hydrodynamic theory of lubrication of slider Iih-rLwt bea---ngs for steady operating
ecaditions. F14wic equations of this theory are analyzed, and new methoftvhich give qikrtial tcaldetation to the dependence ~d vittt:osity on temperature, are developed for solving these equations.
Special att*ention is given to an investigation of the drsy-andence of bearing character- --Lgtics ofi their design parameters.' The auggestea enItulating method makes T-I~e Hy(In.-dynamic Theory
(Cont.) SOV/4690 posgible the choice of optimum design parmaters. Some methods for improving bearing characteristics are'elaborated and M&Y be u-seii ir. the develo]pment of the t22zust bearing
theory.* It is mentioned in the forevord that hydrogeneXitor, thrust bearings for very high loads are cowtru,~ted by the "Elektrosila" and ""ira:Lelektro,apparat" plants. The book was prepared at the
request of the NTGZ (Nc7oeibirsk Turbogenerator Plant),, and computations and graphs necessary for the deteminaticn of characteristic coefficients wre made in the calculation cffice of thia'plaut by
E.Gj KaluzbskV&. There axe 74 references: 36 Soviet, 27 English, 10 German and 1 French. T ABTZ OF CONrMM: 3 Ch. 1. Bsaic Problems of the Hyd2vdynsmic Theory of Thrmat Bearing Lubrication 5 1.
Generul picture of phenomena occurring in the .1abricant film 5 2. Brief descriptim of thruBt bearing design 6 3. Stating the ptoblein of the hydrodynamic tbeory of lubrication 7 4. Three ccmposite
parts of the theory, 10 KUNIN, I.A.; (Novosibirsk); KHON, V.G. (Novosibirsk) Interaction of a vibrator and a bounded liquid medium. PM77 no.2:144-146 Jl-Ag 60. (MIRA 1416) 1. Institut gornogo dela
Sibirskogo otdeleniya AN SSSR i Novosibirskiy clektrotekhnicbeskiy institut. (Vibrators) (Hydrodynamics) KUNIN# I.A.; KHONv V*F* Theory of interaction of a vibrator with the absorbing fluid. Izv.
Sib. otd, AN SSSR no. ll:X36-139 160. (MIRA 14:1) 1. Instiut gornogo dela SibirskQgo otdeleniya AN SSSR i Novosibirskiy elaktrotekhnichesl~iy i4stitut. (Vibrationo) KUNIN,-.!-.A.; MBKO, V.D. Pendulum
apparatus for determining the coefficient of rolling friction. Izv.Sib.otd.AN SSSR m.8:116-119 t61. O-MIA 14'. 8) 1. Institut, gornogo dela Sibirskogo otdeleniya AN SSSH, Novosibirsk. (Friction.)
(Pendulum) ALABUZHEV, P.M.; KUNIN, I.A.; FETREYV1j A.M.; KHONy V.F. Interaction of a submerged vibrator with an unlimited medium. Izv. Sib. otd. AN SSSR no-3:25-29 162. (MIRA 17:7) 1. Novosibirskiy
elektrotekhnicheskiy institut i Institut gornogo dela Sibirskogo otdeleniya AN SSSR, Novosibirsk. jr, Le mirid st r--~ I MF~L+ KUNIN, J-.A. Green's tensor for an anisotropic elastic 1110(ill"D with
sources of internal stress. Dokl. AN SSIIJR 157 no.6:1319-1320 Ag 164. (MIRA 17:9) 1. Irintitut teplofiziki Siblrakogo otdoDinlyu lki; !Tod:,tavlono akudemikoin NJ.'. liaboLnovyin. _ -JLLV -ULIII,
S-*Ic,,., hFirld. teklin. n-lik.. KVJ~~JjlT _ 1&7a~ Kopelovich; 111K 1. . I retson'zent [Ore drawing and haulago in undorground ,rdning] Vypusk i dostavl(a rudy pri poduinmoi dol~vcho. Moskva,
Rodl-lit l9u~. 196 p. (tilftA 17:9)
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11. Prove that the parallelogram circumscribing a circle is a r... | Filo
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11. Prove that the parallelogram circumscribing a circle is a rhombus.
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Question Text 11. Prove that the parallelogram circumscribing a circle is a rhombus.
Updated On Feb 8, 2023
Topic All topics
Subject Mathematics
Class Class 11
Answer Type Video solution: 1
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Avg. Video Duration 7 min
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Level Order Traversal, Print each level in separate line
This post is completed by 1 user
• 0 Add to List
39. Level Order Traversal, Print each level in separate line
Objective: Given a Binary tree, Print each level of a tree in a separate line.
NOTE: This problem is very similar to Create Linked Lists of all the nodes at each depth
Naive Approach:
1. Get the height of the tree.
2. Put a for loop for each level in the tree.
3. for each level in step 2, do pre-order traversal and print only when the height matches the level.
4. Look at the code for a better explanation
Time Complexity: O(N^2) - because at each level we are traversing the entire tree.
Better Solution: Time Complexity - O(N)
1. Do the level order traversal using the queue(Breadth First Search). Click here to learn about how to level order traversal.
2. For getting all the nodes at each level, before you take out a node from the queue, store the size of the queue in a variable, say you call it levelNodes.
3. Now while levelNodes>0, take out the nodes print it, and add their children into the queue.
4. After this while loop put a line break.
levelNodes = q.size();
Node n = (Node)q.remove();
System.out.print(" " + n.data);
if(n.left!=null) q.add(n.left);
if(n.right!=null) q.add(n.right);
• Since we had taken the queue size before we added new nodes, we will get the count at each level and after printing this count, put a line break, see the example below
Output by Naive Approach :
Output by Better Approach :
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Transactions Online
Ienari IGUCHI, Takuya IMAIZUMI, Tomoyuki KAWAI, Yukio TANAKA, Satoshi KASHIWAYA, "Josephson and Quasiparticle Tunneling in Anisotropic High-Tc d-Wave Superconductors" in IEICE TRANSACTIONS on
Electronics, vol. E85-C, no. 3, pp. 789-796, March 2002, doi: .
Abstract: We report the measurements on the ramp-edge type Josephson and quasiparticle tunnel junctions with the different interface angle geometry using high-Tc YBa[2]Cu[3]O[7-y] (YBCO) electrodes.
The YBCO/I/Ag tunnel junctions with different crystal-interface boundary angles are fabricated for the investigation of zero bias conductance peak. The angle dependent zero bias conductance peak
typical to a d[x^2-y^2]-wave superconductor is observable. For Josephson junctions, YBCO ramp-edge junctions with different ab-plane electrodes relatively rotated by 45are fabricated using a CeO[2]
seed-layer technique. The temperature dependence of the maximum Josephson current for YBCO/PBCO/YBCO junctions (PBCO: PrBa[2]Cu[3]O[7-y]) exhibits angle-dependent behavior, qualitatively different
from the Ambegaokar-Baratoff prediction. Under microwave irradiation of 9 GHz, the Shapiro steps appear at integer and/or half integer multiples of the voltage satisfying Josephson voltage-frequency
relation, whose behavior depends on the sample angle geometry. The results are reasonably interpreted by the d[x^2-y^2]-wave theory by taking the zero energy state into account.
URL: https://global.ieice.org/en_transactions/electronics/10.1587/e85-c_3_789/_p
author={Ienari IGUCHI, Takuya IMAIZUMI, Tomoyuki KAWAI, Yukio TANAKA, Satoshi KASHIWAYA, },
journal={IEICE TRANSACTIONS on Electronics},
title={Josephson and Quasiparticle Tunneling in Anisotropic High-Tc d-Wave Superconductors},
abstract={We report the measurements on the ramp-edge type Josephson and quasiparticle tunnel junctions with the different interface angle geometry using high-Tc YBa[2]Cu[3]O[7-y] (YBCO) electrodes.
The YBCO/I/Ag tunnel junctions with different crystal-interface boundary angles are fabricated for the investigation of zero bias conductance peak. The angle dependent zero bias conductance peak
typical to a d[x^2-y^2]-wave superconductor is observable. For Josephson junctions, YBCO ramp-edge junctions with different ab-plane electrodes relatively rotated by 45are fabricated using a CeO[2]
seed-layer technique. The temperature dependence of the maximum Josephson current for YBCO/PBCO/YBCO junctions (PBCO: PrBa[2]Cu[3]O[7-y]) exhibits angle-dependent behavior, qualitatively different
from the Ambegaokar-Baratoff prediction. Under microwave irradiation of 9 GHz, the Shapiro steps appear at integer and/or half integer multiples of the voltage satisfying Josephson voltage-frequency
relation, whose behavior depends on the sample angle geometry. The results are reasonably interpreted by the d[x^2-y^2]-wave theory by taking the zero energy state into account.},
TY - JOUR
TI - Josephson and Quasiparticle Tunneling in Anisotropic High-Tc d-Wave Superconductors
T2 - IEICE TRANSACTIONS on Electronics
SP - 789
EP - 796
AU - Ienari IGUCHI
AU - Takuya IMAIZUMI
AU - Tomoyuki KAWAI
AU - Yukio TANAKA
AU - Satoshi KASHIWAYA
PY - 2002
DO -
JO - IEICE TRANSACTIONS on Electronics
SN -
VL - E85-C
IS - 3
JA - IEICE TRANSACTIONS on Electronics
Y1 - March 2002
AB - We report the measurements on the ramp-edge type Josephson and quasiparticle tunnel junctions with the different interface angle geometry using high-Tc YBa[2]Cu[3]O[7-y] (YBCO) electrodes. The
YBCO/I/Ag tunnel junctions with different crystal-interface boundary angles are fabricated for the investigation of zero bias conductance peak. The angle dependent zero bias conductance peak typical
to a d[x^2-y^2]-wave superconductor is observable. For Josephson junctions, YBCO ramp-edge junctions with different ab-plane electrodes relatively rotated by 45are fabricated using a CeO[2]
seed-layer technique. The temperature dependence of the maximum Josephson current for YBCO/PBCO/YBCO junctions (PBCO: PrBa[2]Cu[3]O[7-y]) exhibits angle-dependent behavior, qualitatively different
from the Ambegaokar-Baratoff prediction. Under microwave irradiation of 9 GHz, the Shapiro steps appear at integer and/or half integer multiples of the voltage satisfying Josephson voltage-frequency
relation, whose behavior depends on the sample angle geometry. The results are reasonably interpreted by the d[x^2-y^2]-wave theory by taking the zero energy state into account.
ER -
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[Julia] Spin gap behavior and boundary effects
Hello ITensors Team ,
Ok I decide ask a new question and close the last one.
i'm trying to investigate the spin Gap behavior of Heisenberg spin-1/2 J1-J2(J1 and J2 >0, non-frustrated) Isotropic model with interations between first pair of spins with exchange paramenter J1 and
the following pair of spins with interations mediated by J2 an so on. This simple model presents an topological phase transitions at J1=J2 between "Haldane-Even" and "Haldane-Odd" phases.
I wrote an simple code for this model considering open boundary conditions as follows :
using ITensors
using DelimitedFiles
int = range(0.95, step=0.5, stop=1.5)
Data = []
Data_TW = []
for i in int
N = 120
sites = siteinds("S=1/2", N; conserve_qns=true)
J_1 = 1.0
J_2 = i
ampo = AutoMPO()
for j=1:2:N
ampo += 0.5*J_1,"S+",j,"S-",j+1
ampo += 0.5*J_1,"S-",j,"S+",j+1
ampo += J_1,"Sz",j,"Sz",j+1
for j=2:2:N-2
ampo += 0.5*J_2,"S+",j,"S-",j+1
ampo += 0.5*J_2,"S-",j,"S+",j+1
ampo += J_2,"Sz",j,"Sz",j+1
H = MPO(ampo,sites)
state1 = []
push!(state1, "Up", "Up")
for n=2:N/2
push!(state1, "Up", "Dn")
init_state = [isodd(n) ? 1 : 2 for n = 1:N]
psi0 = randomMPS(sites,init_state,32)
psi1 = randomMPS(sites,state1,32)
@show flux(psi0)
@show flux(psi1)
sweeps0 = Sweeps(10)
maxdim!(sweeps0, 160,320,600,800,1000)
cutoff!(sweeps0, 1e-11)
noise!(sweeps0, 1e-8,1e-9, 1e-10,0.0)
energy0,psi0 = dmrg(H,psi0,sweeps0)
H02 = inner(H,psi0,H,psi0)
E0 = inner(psi0,H,psi0)
var0 = H02 - E0^2
@show var0
sweeps1 = Sweeps(18)
maxdim!(sweeps1, 320,600,800,1000,1200)
cutoff!(sweeps1, 1e-11)
noise!(sweeps1, 1e-8, 1e-9,1e-10,0.0)
energy1,psi1 = dmrg(H,psi1,sweeps1)
H12 = inner(H,psi1,H,psi1)
E1 = inner(psi1,H,psi1)
var1 = H12 - E1^2
@show var1
gap = energy1-energy0
sgap = N*gap
println("The Gap is =",gap)
println("The Scaled gap is=", sgap)
println("The J_2 value is :", J_2)
push!(Data, ["$J_2" "$energy0" "$gap" "$sgap"])
open("Data(test).dat", "w") do f
writedlm(f, Data, '\t')
In my code the first loop concerning the J1 interacting spin terms i sum up to N for each 2 points and the second loop for the J2 interacting terms i sum up to N-2 terms(OBC).
The strange fact is, i cannot reproduce the literature results because my spin gap behaves like an Gapped-Gapless transition. The curious fact is, when i change the first loop summation(Up to N-2), i
reproduce the correct behavior with an clear quantum phase transition at J = J2/J1 = 1.0 , but with the penalty of the missing interaction at last pair of spins(which shows up when i compute the
local magnetization).
I already tried several initial different states , productMps vs randomMPS wavefunctions and different values of noise term .
Probably i'm missing something, i'm grateful with any help ! Thanks !
Hi, thanks for the question. The j=1:2:N loop makes me a bit nervous since j+1 might be greater than N. But I guess not as long as N is an even number, since the last value there will be j=N-1.
But the key issue here, I would bet, is that you are just not doing enough DMRG sweeps. Especially when obtaining excited states with DMRG, it’s important to check that enough sweeps have been done,
because for excited states convergence can be rather slow.
Relatedly, I don’t recommend putting a sequence of DMRG calculations into one big loop (the loop over i here). It’s tempting, but almost every time I’ve seen users do it, the resulting problem is
that they are studying different points in their phase diagram using the same DMRG settings, whereas near phase transitions or at various other harder points, they need to do a more careful study of
DMRG convergence in both number of sweeps and other parameters such as maxdim and cutoff.
Please also check that at various points your results are converged in those parameters (maxdim and cutoff) too, since if you don’t have convergence in those, it could also strongly affect what gaps
you find.
Finally, these kinds of phases (Haldane) phase can have subtle zero-energy edge states which can complicate how one calculates and defines the gap. Please check that the state you are treating as the
first excited state is not actually a second, third, or even fourth degenerate ground state.
Hope that helps!
Hello miles, Thank you for the reply
Concerning the convergence, i tried to chose an number of parameters(Sweeps, maxdim and cutoff) with returns an variation V = <H^2> - <H>^2 proportional to E-8 in mostly points(i'm using maxdim up to
1200, cutoff e-12 for small chains). So its possible the system still stuck in an local minima even using the noise term and sweeps numbers sufficiently large to given an variation V proportional to
Concerning the degenerated ground state, indeed it seems me code target wrongly the ground state of the system after the Quantum phase transition, where the system presents an degenerated non-trivial
topological phase with virtual spins at edges. So, just giving as i did an state1 array with total number QN(sz = 1) could not represent the first excited state ? Maybe i had to search excited states
at the same sector QN(sz=0) ? can you clarify this to me please ? Thank you !
Yes, good question. So if the physics after the phase transition is similar to that of the Haldane phase (the one that is in the same phase as the S=1 Heisenberg chain or the AKLT model), then with
open boundary conditions there are *four* degenerate ground states. You can think of this as if the edges each had an S=1/2 edge state and then the four states are the four combinations of two spin 1
/2's (singlet, and 3 triplets).
So if you only obtained the lowest-energy Sz=0 and Sz=+1 states, those would have nearly the same energy (up to exponentially small corrections in the system size). To obtain the bulk gap, you'll
either need to obtain the lowest 5 eigenstates or else, as you mentioned **obtain the lowest excited state in the Sz=0 sector**.
So yes, I'd recommend looking just at Sz=0, but remember there is a second ground state (Sz=0 triplet) there, so you will need the second excited state.
Another idea would be to obtain the ground state and first excited state in the Sz=+1 (or +2 in "ITensor units") sector.
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Coordinate systems and projections for beginners - Resource Centre | Esri UK
This blog post gives a basic introduction to coordinate systems and projections, with a focus on UK data. To seasoned geographers, I apologise for all the things I’ve simplified or simply left out!
My intention is to provide GIS novices who are a bit confused by the topic with just enough about various different coordinate systems to get started working with them in ArcGIS. Anyone looking for
an excellent comprehensive introduction should refer to Ordnance Survey’s guide.
A coordinate system lets us define where a location is in space. In GIS, there are many types of coordinate systems, of which the two most used are geographic (3D) and projected (2D). The difference
is shown below:
A geographic coordinate system (GCS) uses a grid on the surface of a 3D globe (the technical term for this grid is graticule; the North/South lines are lines of longitude and the East/West lines are
lines of latitude). Graticule lines are not parallel to each other because they are defined using angles (e.g. degrees) from the centre of the globe, not linear units (e.g. metres) on a flat surface.
A projected coordinate system (PCS) is a flattened version of a 3D coordinate system. Grid lines are parallel, and coordinates are given in “flat” units like metres or feet. Although the Earth isn’t
flat, we often use flat surfaces to represent it (paper surfaces, computer screens, etc), so we frequently need to convert 3D coordinates to 2D coordinates. However, there’s a problem: imagine
peeling the skin off an orange and trying to make it sit flat on a table. It’s not possible to do this without ripping or stretching the peel! It’s the same with a 3D coordinate system. Whenever we
flatten, or “project”, a 3D coordinate grid to a 2D coordinate grid, we have to distort the grid’s proportions somehow. Distances, shapes, areas, and angles, or some combination of all four, are
If you work with UK data, there are two geographic and two projected coordinate systems that you should know about. They are shown in the diagram below, along with the numeric WKID (Well-Known ID) of
that GCS/PCS, as well as their relationships with one another and example coordinates in each system showing the location of Esri UK’s head office in Aylesbury.
World Geodetic System (WGS-84) is familiar to many non-geographers because it is used by GPS devices to describe locations all over the Earth. A different GCS, called OSGB-36, which is more accurate
for describing locations in Britain but not as good for other countries, is used specifically for British data. Web Mercator is a PCS based on WGS-84 used for global maps, and British National Grid
is a PCS based on OSGB-36 used for British maps.
For each GCS, there are many different ways of converting, or “projecting”, 3D coordinates to 2D coordinates. Web Mercator and British National Grid are the most important projections for their
respective geographic coordinate systems. There are alternative projections, each with its own pros and cons.
Converting between coordinate systems that are based on the same GCS is relatively straightforward, but when converting, for example, GPS (WGS-84) coordinates to BNG eastings and northings, a
mathematical transformation is required. The “Petroleum” transformation is an accurate transformation from WGS-84 to OSGB-36 (and vice-versa) included with ArcGIS.
To convert data between WGS-84 and BNG in ArcGIS Desktop, you should use the Project tool. Make sure you select the “Petroleum” transformation otherwise your results will not be accurate.
Select the optional Petroleum transformation for the best results
For GB data, you only really need to know about WGS-84 and BNG, and that you should use the Petroleum transformation to get an accurate conversion between them. You will also need to use “Petroleum”
to convert between BNG/OSGB-36 and Web Mercator (or WGS 1984 Web Mercator Auxiliary Sphere, to give it the name used by Esri). Web Mercator has become the standard projection for international
consumer web maps, such as Google Maps, Bing Maps, and OpenStreetMap. All ArcGIS Online basemaps are also in this projection.
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:: Lecture 10
STAM101 :: Lecture 10 :: T-test
Definition – Assumptions – Test for equality of two means-independent and paired t test
Student’s t test
When the sample size is smaller, the ratio
This fact was brought out by Sir William Gossest and Prof. R.A Fisher. Sir William Gossest published his discovery in 1905 under the pen name Student and later on developed and extended by Prof. R.A
Fisher. He gave a test known as t-test.
Inference About Two Means
Applications (or) uses
• To test the single mean in single sample case.
• To test the equality of two means in double sample case.
• Independent samples(Independent t test)
(ii) Dependent samples (Paired t test)
• To test the significance of observed correlation coefficient.
• To test the significance of observed partial correlation coefficient.
• To test the significance of observed regression coefficient.
Test for single Mean
Ho: µ=µo
(i.e) There is no significance difference between the sample mean and the population mean
• Form the Alternate hypothesis
H1: µ≠µo (or µ>µo or µ<µo)
ie., There is significance difference between the sample mean and the population mean
3. Level of Significance
The level may be fixed at either 5% or 1%
4. Test statistic
• Find the table value of t corresponding to (n-1) d.f. and the specified level of significance.
• Inference
If t < ttab we accept the null hypothesis H0. We conclude that there is no significant difference sample mean and population mean
(or) if t > ttab we reject the null hypothesis H0. (ie) we accept the alternative hypothesis and conclude that there is significant difference between the sample mean and the population mean.
2-Sample t-Test Using Minitab Student-t-Test
Example 1
Based on field experiments, a new variety of green gram is expected to given a yield of 12.0 quintals per hectare. The variety was tested on 10 randomly selected farmer’s fields. The yield
(quintals/hectare) were recorded as 14.3,12.6,13.7,10.9,13.7,12.0,11.4,12.0,12.6,13.1. Do the results conform to the expectation?
Null hypothesis H0: m=12.0
(i.e) the average yield of the new variety of green gram is 12.0 quintals/hectare.
Alternative Hypothesis: H1:m≠ 12.0
(i.e) the average yield is not 12.0 quintals/hectare, it may be less or more than 12 quintals / hectare
Level of significance: 5 %
Test statistic:
From the given data
= 1.0853
Table value for t corresponding to 5% level of significance and 9 d.f. is 2.262 (two tailed test)
t < ttab
We accept the null hypothesis H0
We conclude that the new variety of green gram will give an average yield of 12 quintals/hectare.
Before applying t test in case of two samples the equality of their variances has to be tested by using F-test
is the variance of the second sample whose size is n2.
It may be noted that the numerator is always the greater variance. The critical value for F is read from the F table corresponding to a specified d.f. and level of significance
F <Ftab
We accept the null hypothesis H0.(i.e) the variances are equal otherwise the variances are unequal.
Test for equality of two Means (Independent Samples)
Given two sets of sample observation x11,x12,x13…x1n , and x21,x22,x23…x2n of sizes n1 and n2 respectively from the normal population.
• Using F-Test , test their variances
• Variances are Equal
Ho:., µ1=µ2
H1 µ1≠µ2 (or µ1<µ2 or µ1>µ2)
Test statistic
where the combined variance
The test statistic t follows a t distribution with (n1+n2-2) d.f.
• Variances are unequal and n1=n2
It follows a t distribution with
• Variances are unequal and n1≠n2
This statistic follows neither t nor normal distribution but it follows Behrens-Fisher d distribution. The Behrens – Fisher test is laborious one. An alternative simple method has been suggested by
Cochran & Cox. In this method the critical value of t is altered as tw (i.e) weighted t
where t1is the critical value for t with (n1-1) d.f. at a dspecified level of significance and
t2 is the critical value for t with (n2-1) d.f. at a dspecified level of significance and
Example 2
In a fertilizer trial the grain yield of paddy (Kg/plot) was observed as follows
Under ammonium chloride 42,39,38,60 &41 kgs
Under urea 38, 42, 56, 64, 68, 69,& 62 kgs.
Find whether there is any difference between the sources of nitrogen?
Ho: µ1=µ2 (i.e) there is no significant difference in effect between the sources of nitrogen.
H1: µ1≠µ2 (i.e) there is a significant difference between the two sources
Level of significance = 5%
Before we go to test the means first we have to test their variances by using F-test.
Ho:., s12=s22
H1:., s12≠s22
Ftab(6,4) d.f. = 6.16
Þ F < Ftab
We accept the null hypothesis H0. (i.e) the variances are equal.
Use the test statistic
The degrees of freedom is 5+7-2= 10. For 5 % level of significance, table value of t is 2.228
t <ttab
We accept the null hypothesis H0
We conclude that the two sources of nitrogen do not differ significantly with regard to the grain yield of paddy.
Example 3
The summary of the results of an yield trial on onion with two methods of propagation is given below. Determine whether the methods differ with regard to onion yield. The onion yield is given in Kg/
Method II
Method I
n1=12 n2=12
SS1=186.25 SS2=737.6667
Ho:., µ1=µ2 (i.e) the two propagation methods do not differ with regard to onion yield.
H1 µ1≠µ2 (i.e) the two propagation methods differ with regard to onion yield.
Level of significance = 5%
Before we go to test the means first we have to test their variability using F-test.
Ho: s12=s22
H1: s12≠s22
Ftab(11,11) d.f. = 2.82
Þ F > Ftab
We reject the null hypothesis H0.we conclude that the variances are unequal.
Here the variances are unequal with equal sample size then the test statistic is
t =1.353
The table value for
We accept the null hypothesis H0
We conclude that the two propagation methods do not differ with regard to onion yield.
Example 4
The following data relate the rubber yield of two types of rubber plants, where the sample have been drawn independently. Test whether the two types of rubber plants differ in their yield.
│ │6.21│5.70│6.04│4.47│5.22│4.45│4.84│5.84│5.88│5.82│6.09│5.59│
│Type I ├────┼────┼────┼────┼────┼────┼────┼────┼────┼────┼────┼────┤
│ │6.06│5.59│6.74│5.55│ │ │ │ │ │ │ │ │
│Type II │4.28│7.71│6.48│7.71│7.37│7.20│7.06│6.40│8.93│5.91│5.51│6.36│
Ho:., µ1=µ2 (i.e) there is no significant difference between the two rubber plants.
H1 µ1≠µ2 (i.e) there is a significant difference between the two rubber plants.
Level of significance = 5%
Before we go to test the means first we have to test their variability using F-test.
Ho:., s12=s22
H1:., s12≠s22
\ if
Ftab(11,15) d.f.=2.51
Þ F > Ftab
We reject the null hypothesis H0. Hence, the variances are unequal.
Here the variances are unequal with unequal sample size then the test statistic is
t1=t(16-1) d.f.=2.131
t2=t(12-1) d.f .=2.201
We reject the null hypothesis H0. We conclude that the second type of rubber plant yields more rubber than that of first type.
Equality of two means (Dependant samples)
Paired t test
In the t-test for difference between two means, the two samples were independent of each other. Let us now take particular situations where the samples are not independent.
In agricultural experiments it may not be possible to get required number of homogeneous experimental units. For example, required number of plots which are similar in all; characteristics may not be
available. In such cases each plot may be divided into two equal parts and one treatment is applied to one part and second treatment to another part of the plot. The results of the experiment will
result in two correlated samples. In some other situations two observations may be taken on the same experimental unit. For example, the soil properties before and after the application of industrial
effluents may be observed on number of plots. This will result in paired observation. In such situations we apply paired t test.
Suppose the observation before treatment is denoted by x and the observation after treatment is denoted by y. for each experimental unit we get a pair of observation(x,y). In case of n experimental
units we get n pairs of observations : (x1,y1), (x2,y2)…(xn,yn). In order to apply the paired t test we find out the differences (x1- y1), (x2-y2),..,(xn-yn) and denote them as d1, d2,…,dn. Now d1,
d2…form a sample . we apply the t test procedure for one sample (i.e)
the mean
Example 5
In an experiment the plots where divided into two equal parts. One part received soil treatment A and the second part received soil treatment B. each plot was planted with sorghum. The sorghum yield
(kg/plot) was absorbed. The results are given below. Test the effectiveness of soil treatments on sorghum yield.
│ │49│53│51│52│47│50│52│53│
│Soil treatment A │ │ │ │ │ │ │ │ │
│Soil treatment B │52│55│52│53│50│54│54│53│
H0: m1 = m2 , there is no significant difference between the effects of the two soil treatments
H1: m1 ¹ m2, there is significant difference between the effects of the two soil treatments
Level of significance = 5%
Test statistic
│ │y │d=x-y │d2│
│x │ │ │ │
│49│52│ -3 │9 │
│53│55│ -2 │4 │
│51│52│ -1 │1 │
│51│52│ -1 │1 │
│47│50│ -3 │16│
│50│54│ -4 │16│
│52│54│ -2 │4 │
│53│53│ 0 │0 │
│Total│ -16 │44│
Table value of t for 7 d.f. at 5% l.o.s is 2.365
We reject the null hypothesis H0. We conclude that the is significant difference between the two soil treatments between A and B. Soil treatment B increases the yield of sorghum significantly,
Download this lecture as PDF here
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Fluid friction
The pressures in a flowing fluid are calculated assuming the value of the Fanning friction factor is known. Determination of the Fanning friction factor, in fact, may be the most difficult step in
this calculation. Fluid friction is studied by the science of rheology.
Fluid rheology
The science of rheology is concerned with the deformation of all forms of matter, but has had its greatest development in the study of the flow behavior of suspensions in pipes and other conduits.
The rheologist is interested primarily in the relationship between flow pressure and flow rate, and in the influence thereon of the flow characteristics of the fluid. There are two fundamentally
different relationships:
1. The laminar flow regime prevails at low flow velocities. Flow is orderly, and the pressure-velocity relationship is a function of the viscous properties of the fluid.
2. The turbulent flow regime prevails at high velocities. Flow is disorderly and is governed primarily by the inertial properties of the fluid in motion. Flow equations are empirical.
The laminar flow equations relating flow behavior to the flow characteristics of the fluid are based on certain flow models, namely:
• Newtonian
• Bingham plastic
• Pseudoplastic
• Yield power-law
• Dilatant
Only the first four are of interest in drilling-fluid technology. Most drilling fluids do not conform exactly to any of these models, but drilling-fluid behavior can be predicted with sufficient
accuracy by one or more of them. Flow models are usually visualized by means of consistency curves, which are plots either of flow pressure vs. flow rate or of shear stress vs. shear rate.
Shear stress is force per unit area and is expressed as a function of the velocity gradient of the fluid as
where μ is the fluid viscosity and dv/dr is the velocity gradient. The negative sign is used in Eq. 1 because momentum flux flows in the direction of negative velocity gradient. That is, the momentum
tends to go in the direction of decreasing velocity The absolute value of velocity gradient is called the shear rate and is defined as
Then, Eq. 1 can be written as
Viscosity is the resistance offered by a fluid to deformation when it is subjected to a shear stress. If the viscosity is independent of the shear rate, the fluid is called a Newtonian fluid. Water,
brines, and gases are examples of Newtonian fluid. The shear stress is linear with the shear rate for a Newtonian fluid and is illustrated by Curve A in Fig. 1. The symbol μ without any subscript is
used to refer to the viscosity of Newtonian fluid. Most of the fluids used in drilling and cementing operations are not Newtonian, and their behavior is discussed next.
If the viscosity of a fluid is a function of shear stress (or, equivalently, of shear rate), such a fluid is called non-Newtonian fluid. Non-Newtonian fluids can be classified into three general
1. Fluid properties are independent of duration of shear.
2. Fluid properties are dependent on duration of shear.
3. Fluid exhibits many properties that are characteristics of solids.
Time independent
The following three types of materials are in this class.
Bingham plastic
These fluids require a finite shear stress, τ[y]; below that, they will not flow. Above this finite shear stress, referred to as yield point, the shear rate is linear with shear stress, just like a
Newtonian fluid. Bingham fluids behave like a solid until the applied pressure is high enough to break the sheer stress, like getting catsup out of a bottle. The fluid is illustrated by Curve B in
Fig. 1. The shear stress can be written as
where τ[y] is called the yield point (YP), and μ[p] is referred to as the plastic viscosity (PV) of the fluid. Some water-based slurries and sewage sludge are examples of Bingham plastic fluid. Most
of the water-based cement slurries and water-based drilling fluids exhibit Bingham plastic behavior. Drilling muds are often characterized with YP and PV values, but this is for historical reasons
and does not necessarily imply that the Bingham fluid model is the best model for all muds.
These fluids exhibit a linear relationship between shear stress and shear rate when plotted on a log-log paper. This is illustrated by Curve C in Fig. 1. This fluid is also commonly referred to as a
power-law non-Newtonian fluid. The shear stress can be written as
where K is the consistency index, and n is the exponent, referred to as power-law index. A term μ[a] is defined that is called the apparent viscosity and is
Note that apparent viscosity and effective viscosity as defined by different authors are not always defined in the sense used here, so read with caution. The apparent viscosity decreases as the shear
rate increases for power-law fluids. For this reason, another term commonly used for pseudoplastic fluids is "shear thinning." Polymeric solutions and melts are examples of power-law fluid. Some
drilling fluids and cement slurries, depending on their formulation, may exhibit power-law behavior.
Yield power law
Also known as Herschel-Bulkley fluids, these fluids require a finite shear stress, τ[y], below which they will not flow. Above this finite shear stress, referred to as yield point, the shear rate is
related to the shear stress through a power-law type relationship. The shear stress can be written as
where τ[y] is called the yield point, K is consistency index, and m is the exponent, referred to as power-law index.
These fluids also exhibit a linear relationship between shear stress and shear rate when plotted on a log-log paper and are illustrated as Curve D in Fig. 1. The shear stress expression for dilatant
fluid is similar to power-law fluid, but the exponent n is greater than 1. The apparent viscosity for these fluids increases as shear rate increases. For this reason, dilatant fluids are often called
Quicksand is an example of dilatant fluid. In cementing operations, it would be disadvantageous if fluids increased in viscosity as shear stress increased.
Time dependent
These fluids exhibit a change in shear stress with the duration of shear. This does not include changes because of reaction, mechanical effects, etc. Cement slurries and drilling fluids usually do
not exhibit time-dependent behavior. However, with the introduction of new chemicals on a regular basis, one should test and verify the behavior.
Solids characteristic
These fluids exhibit elastic recovery from deformation that occurs during flow and are called viscoelastic. Most of the cement slurries and drilling fluids do not exhibit this behavior. However, as
mentioned earlier, new polymers are being introduced on a regular basis, and tests should be conducted to verify the behavior.
The unit of viscosity is Pascal-second (Pa-s) in the SI system and lbf/(ft-s) in oilfield units. One Pa-s equals 10 poise (P), 1,000 centipoise (cp), or 0.672 lbf/(ft-s). The exponent n is
dimensionless, and consistency index, K, has the units of Pa-s^n in the SI system and lbf/(sec^n-ft^–2) in oilfield units. One Pa-s^n equals 208.86 lbf/(sec^n.ft^–2). The yield point for Bingham
fluids is often characterized in units of lbf/(1,00ft^2), and plastic viscosity is usually given in centipoise.
The rheology parameters of the fluids, μ and μ[p] , and τ[o] , K, and n, are determined by conducting tests in a concentric viscometer. This consists of concentric cylinders with one of them
rotating, usually the outer one. A sample of fluid is placed between the cylinders, and the torque on the inner cylinder is measured. Assuming an incompressible fluid, with flow in the laminar flow
regime, the equations of motion can be solved for τ to give
τ = shear stress, Pa;
M[T] = torque, N-m;
L = length, m;
r = radius, m.
In a concentric viscometer, torque, M[T], is measured at a different rotational speed of the outer cylinder. Shear stress is then calculated from Eq. 8 , and shear rate is given by
R[b] = radius of inner cylinder (bob), m;
R[c] = radius of outer cylinder (cup), m;
κ = ratio of radius of inner cylinder to outer cylinder;
Ω[0] = angular velocity of outer cylinder.
Shear stress and shear rate are then analyzed to determine the rheology model.
A number of commercially available concentric cylinder rotary viscometers are suitable for use with drilling muds. They are similar in principle to the viscometer already discussed. All are based on
a design by Savins and Roper, which enables the plastic viscosity and yield point to be calculated very simply from two dial readings, at 600 and 300 rpm, respectively.^[1] They are referred to in
the industry as the direct-indicating viscometer and typically are called Fann viscometers.
The underlying theory is as follows: Eqs. 8 and 9 are combined to give
where a[vs] and b[vs] are constants that include the instrument dimensions, the spring constant, and all conversion factors; ω is the rotor speed in revolutions per minute (rpm).
where θ[1] and θ[2] are dial readings taken at ω[1] and ω[2] rpm, respectively. PV is the conventional oilfield term for plastic viscosity, thus measured. Then, the yield point is determined.
YP is the conventional oilfield term for yield point, thus measured. The numerical values of a[vs] , b[vs] , ω[1] , and ω[2] were chosen so that
Apparent viscosity μ[a] may be calculated from the Savins-Roper viscometer reading as
where θ is the dial reading at ω rpm. Typical viscometer results are shown in Fig. 2.^[2] Notice that real fluids are not ideally any of the models shown, but generally are pretty close to one model
or another. The selection of the model may be motivated by a particular fluid velocity of interest. For instance, fluid 6 in Fig. 2 would be modeled well by a yield-power law for rpm below about 100.
• Fig. 2—Typical drilling fluid consistency curves.^[2] (Reprinted from Composition and Properties of Oil Well Drilling Fluids, G.R. Gray and H.C.H. Darley, fourth edition, © 1980, with permission
from Elsevier.)
Fanning friction factor correlations
Flow in pipes and annuli are typically characterized as laminar or turbulent flow. Laminar flow often can be solved analytically. Correlation for turbulent flow is usually developed empirically by
conducting experiments in a flow loop. Typical data will look like those that are shown in Fig. 3. Experimental data are usually analyzed and correlated through the use of two dimensionless numbers:
f, the Fanning friction factor, and Re, the Reynolds number. The relationship between the friction factor, f, and Reynolds number for Newtonian fluids is given in Fig. 4,^[3] with the pipe roughness
given in Fig. 5. This figure is based on the experimental results of Colebrook.^[4] The relationship between friction factor f vs. Re for pseudoplastic fluids is shown in Fig. 6. This figure is based
on the experimental results of Dodge and Metzner.^[5] Here non-Newtonian fluids usually assume this pseudoplastic friction factor for turbulent flow.
The pressure drop per unit length for flow through a duct is given by
where f is Fanning friction factor, Δz is the length, v is the velocity, ρ is the density, D[hyd] is a characteristic "diameter," and ΔP is the pressure drop. The friction factor depends on Reynolds
number, Re, and the roughness of the pipe. The Reynolds number, Re, is defined as
where ρ is the density of the fluid, v is the average velocity, D is a characteristic length (e.g., pipe diameter), and μ is a characteristic viscosity. Correlations for friction factor, f, in both
laminar and turbulent flow regime and for critical Reynolds number are available for a number of fluids and geometries. However, in critical situations, it is recommended that flow-loop tests be
conducted and data compared with calculations that are based on fundamental equations for flow. For example, experimental data in laminar flow should be compared with estimated values from
correlation such as Eq. 20. However, some solid-laden polymers are known to exhibit what is known as shear-induced diffusion, in which solids migrate away from the walls to the center of the pipe.
These fluids show deviation in calculated and experimental values in laminar flow. Correlations should be modified as needed to reflect this behavior. Several polymers are known to exhibit drag
reduction in turbulent flow. Theoretical prediction of polymer-flow behavior is not yet good enough, so flow-loop data are almost always needed.
Commonly used Fanning friction correlations are summarized in the next section. Correlations are provided for three geometric configurations: pipe flow, concentric annular flow, and slit flow. For
each case, the ΔP and Re are defined for the specific geometry and flow model. The laminar flow equations for annular flow are approximate for Newtonian and power-law flow in annuli with
low-clearance but reasonably accurate and much simpler than the exact solutions. Note that for low clearance annuli, the slit flow model provides almost as accurate a result as the concentric annular
model but can also be modified to account for eccentric annuli.
Rheological model 1: Newtonian fluids
Pipe flow
Frictional pressure drop:
Reynolds number:
where D[i] is the pipe inside diameter (ID).
Laminar flow:
for Re < 2,100. Turbulent flow:
for Re > 3,000, and k is the absolute pipe roughness in the same units as D.
Annular flow
Frictional pressure drop:
Reynolds number:
where D[o] is the annulus outside diameter (OD), and D[i] is the ID.
Laminar flow:
for Re < 2,100. Turbulent flow:
for Re > 3,000, and k is the absolute pipe roughness in the same units as D.
Rheological model 2: Bingham plastic fluids
Pipe flow
Frictional pressure drop:
Reynolds number:
where D[i] is the pipe ID, and μ[p] is the plastic viscosity.
Laminar flow:
Turbulent flow:
for Re > Re[BP2] , where: For He < = 0.75 × 10^5 , A = 0.20656, and B = 0.3780. For 0.75 × 10 5 < He< = 1.575 × 10 5 , A = 0.26365, and B = 0.38931. For He > 0.75 × 10 5 , A = 0.20521, B = 0.35579,
and He = τ[o]ρD^2/μ[p]^2 .
Annular flow
Frictional pressure drop:
Reynolds number:
where D[o] is the annulus OD; D[i] is the ID; and μ[p] is the plastic viscosity.
Laminar flow:
Turbulent flow:
for Re > Re[BP2], where: For He < = 0.75 × 10^5 , A = 0.20656, and B = 0.3780. For 0.75 × 10^5 < He< = 1.575 × 10^5 , A = 0.26365, and B = 0.38931. For He > 0.75 × 10^5 , A = 0.20521, B = 0.35579,
and He = τ[o]ρ(D[o]^2 – D[i]^2)/ μ[p]^2 .
Slit flow
Frictional pressure drop:
Reynolds number:
where D[o] is the annulus OD; D i is the ID; and μ[p] is the plastic viscosity.
Laminar flow:
Turbulent flow:
for Re > Re[BP2] , where: For He < = 0.75 × 10 5, A = 0.20656, and B = 0.3780. For 0.75 × 10^5 < He< = 1.575 × 10^5 , A = 0.26365, and B = 0.38931. For He > 0.75 × 10^5 , A = 0.20521, B = 0.35579,
and He = τ[o]ρ(D[o]^2 – D[i]^2)/(1.5μ[p])^2 .
Rheological model 3: power law fluids
Pipe flow
Frictional pressure drop:
Reynolds number:
where D[i] is the pipe ID. Laminar flow:
for Re ≤ 3,250 – 1,150n .
Turbulent flow:
for Re ≥ 4,150 – 1,150n.^[5]
Annular flow
Frictional pressure drop:
Reynolds number:
where D[o] is the annulus OD, and D[i] is the ID.
Laminar flow:
for Re ≤ 3,250 – 1,150n.
Turbulent flow:
for Re ≥ 4,150 – 1,150n.
Slit flow
Frictional pressure drop:
Reynolds number:
Laminar flow:
for Re ≤ 3,250 – 1,150n.
Turbulent flow:
for Re ≥ 4,150 – 1,150n.
Rheological model 4: yield power law (YPL) fluids
Pipe flow
Fictional pressure drop:
Reynolds number:
Laminar flow:
for Re ≥3,250 – 1,150n.
Turbulent flow:
for Re ≥ 4,150 – 1,150n.
Slit flow
Frictional pressure drop:
Reynolds number:
Laminar flow:
for Re ≤ 3,250 – 1150n .
Turbulent flow:
for Re ≥ 4,150 – 1,150n .
Frictional pressure drop in eccentric annulus
The frictional pressure drop in an eccentric annulus is known to be less than the frictional pressure drop in a concentric annulus. For laminar flow of Newtonian fluids, the pressure drop in a fully
eccentric annulus is half the pressure drop in a concentric annulus. For turbulent flow, the difference is about 10%. For non-Newtonian fluids, the effect is less but still significant. In deviated
wells, the drillpipe should be fully eccentric over much of the deviated wellbore, resulting in reduced fluid friction.
Define the correction factor for eccentricity.
where subscript e denotes eccentric, and subscript c denotes concentric.
C[e] for laminar flow is determined based on the methods used by Uner et al.^[6] The flow rate through a concentric annulus is given by
where R[r] = r[i]/r[o] . The flow rate through an eccentric annulus was determined to be
where δ[r] is the distance between centers of the inside and outside pipes (e.g., δ[r] = 0 for concentric pipes). The geometry of the eccentric annulus is illustrated in Fig. 7.
The function E may be evaluated using a six-coefficient approximation. The function F must be evaluated using numerical methods (e.g., a seven-point Newton-Cotes numerical integration formula).
Setting q[a] and q[e] equal, then
Because C[e] depends only on f, n, and R[r], C[e] need be calculated only once, then used for all future frictional pressure drop calculations, as long as the property n does not vary.
C[e] for turbulent flow is determined by applying the same techniques to the turbulent velocity profile determined by Dodge and Metzner.^[5]
The volume flow rate through the concentric annulus is given by
h = r[o] – r[i] ,
w = π(r[o] + r[i] ). Integrating Eq. 36 gives
where A is the flow area.
The equivalent integral to Eq. 71 for eccentric flow is given by
The integral in Eq. 73 must be evaluated numerically (e.g., by a seven-point Newton-Cotes numerical integration). C[e] can be determined by setting Eq. 72 equal to Eq. 73 and noting that
where v[c]* is determined from the concentric solution given by Dodge and Metzner.^[5] The resulting nonlinear equation must be solved for C[e] numerically (e.g., by using Newton ’ s method). Because
C[e] depends only on f[1], f[2] , f[3] , n, and R[r] , C[e] need be calculated only once, then used for all future frictional pressure-drop calculations, as long as the properties ρ,K, and n do not
Sample calculations
The most important consideration in making hydraulic calculations is the use of consistent units. Unfortunately, oilfield units are rarely consistent; in some cases they are unique to the industry.
The universal set of consistent units is the SI Metric System of Units. The Society of Petroleum Engineers (SPE) has available a publication: "The SI Metric System of Units and SPE Metric Standard"
that contains every conversion factor necessary. Whenever there is a question of units, the safest solution is to convert all units to SI units, solve the problem, and then convert the answer back to
the common engineering units.
Sample problem
A deviated well kicks off at 3,000 ft and is drilled to total depth (TD) at an angle of 30° to the vertical. The well ’ s total measured depth is 11,000 ft. The well is cased with 72-ppf 13 ^3/[8]
-in. casing (13.375 × 12.347 in.) set at 3,000 ft. The drillstring consists of 900 ft of 8-in. 147-ppf drill collars (8 × 3 in.), 19 ½-ppf drillpipe (5 × 4.206 in.), a 9 ^5/[8] -in. bit with 3× ^13/
[22]-in. nozzles. The undisturbed temperature is 70°F at the surface with a 1.4°F/100-ft gradient. We will neglect the build section and assume the well trajectory is vertical to 3,000 ft measured
depth, and deviated at 30° to the vertical from 3,000 ft measured depth to 11,000 ft measured depth. We will assume the open hole is gauge (9.625 in.).
True vertical depth
For measured depth < 3,000 ft, Z = z. For measured depth > 3,000 ft, Z = 3,000 ft + (z-3,000)sin(30)≅ 402 + 0.866z, where z is measured depth in feet, Z is true vertical depth in feet.
Hydrostatic pressure
1. Assume the wellbore is filled with 8.34 lbm/gal fluid (fresh water). What is the pressure at TD? True vertical depth at TD is 402 + 0.866 × 11,000 = 9,928 ft. Using (Eq. 10 from Static wellbore
pressure solutions) and converting to SI units:
This pressure is gauge pressure at TD. For absolute pressure, add atmospheric pressure, 14.7 psi:
2. Assume a layered wellbore with 14 lbm/gal mud from surface to 5,000 ft (measured depth) and 9 lbm/gal mud from 5,000 ft to TD. What is the pressure at TD?
For layer 1:
For layer 2:
3. Assume the wellbore is filled with nitrogen with a surface pressure of 2,000 psi. What is the pressure at TD? This problem is much more difficult because the gas density and temperature vary over
the length of the wellbore. The pressure change is given by
The temperature distribution is given by T(Z) = 70 + .014 Z, where Z is true vertical depth. Because we need absolute temperature, in Kelvin: (T °F+ 459.67)/1.8 = T°K,
The integral of 1/T with respect to Z is
Frictional pressure loss
4. Assume fresh water is being circulated at 600 gal/min. What is the pressure change inside a single vertical 30-ft joint of drillpipe? Assume the density is 8.34 lbm/gal and the viscosity is 1 cp.
This Reynolds number indicates turbulent flow. To determine the friction factor, first determine the relative roughness k/D. From Fig. 5, the relative roughness is about .0004 for commercial steel.
The friction factor is about .011 from Fig. 4. Friction pressure drop is given by
The hydrostatic pressure change per foot is
Total pressure change per length of pipe for flow downward is
The total pressure change in a 30-ft pipe joint is 0.166 × 30 = 4.98 psi.
5. Assume a 10-lbm/gal mud is being circulated at 100 gal/min. What is the frictional pressure change in the annulus outside a single 30-ft joint of drillpipe? Use the Bingham plastic model and
assume the plastic viscosity is 40 cp and the YP is 15 lbf/100 ft ^2.
6. Repeat Calculation 5, but assume the fluid is a power-law fluid. Remember that PV and YP were determined from the 300-rpm and 600-rpm readings of the Fann viscometer. The equivalent shear stresses
7. For a flow rate of 600 gal/min, what is the fluid pressure in the bit nozzles? The mud density is 12 lbm/gal. What is the pressure recovery in the annulus?
a = acoustic velocity, m/s
α[vs] , b[vs] = constants that include the viscometer dimensions, the spring constant, and all conversion factors
A = flow area (see subscripts), m^2
c = average concentration of cuttings overall
c[a] = cuttings concentration in annular region
c[o] = feed concentration of cuttings
c[p] = cuttings concentration in plug region
C = compressibility
C[d] = discharge coefficients for the flow through an area change, dimensionless
C[D] = drag coefficient, dimensionless
C[e] = pressure drop correction factor for pipe eccentricity, dimensionless
C[p] = heat capacity at constant pressure, J/kg-K
C[v] = heat capacity at constant volume, J/kg-K
d[s] = particle diameter, m
dv/dr = velocity gradient, s ^–1
dv/dt = total derivative of velocity with respect to time, Pa/s
D = characteristic length in Reynolds number, m
D[e] = special equivalent diameter for yield power law fluid, m
D[eq] = equivalent diameter, m
D[hyd] = hydraulic diameter, m
D[h] = wellbore diameter, m
D[i] = inside diameter, m
D[o] = outside diameter, m
D[p] = drillpipe outside diameter, m
D[plug] = plug diameter, m
E[f] =
s modulus for the formation, Pa
E(k) = complete elliptic integral of the second kind, parameter k
ƒ = Fanning friction factor, dimensionless
ƒ[1] , ƒ[2] , ƒ[3] = turbulent flow velocity profile parameters, dimensionless
ƒ[i] = the fraction of flow in the pipe, ith iteration
ƒ[m] = the fraction of flow in the pipe, lower annular pressure
ƒ[p] = the fraction of flow in the pipe, higher annular pressure
F(ƒ,n, R[r]) = eccentric flow function
F[d] = total viscous drag force on the particle, N
g = acceleration of gravity, m/s^2
G = mass flow rate density of mixture, kg/m^3–s
G[s] = mass flow rate density of solids, kg/m^3–s
h = specific enthalpy, J/kg
h = total friction pressure drop, Pa/m
He = Hedstrom number
H[s] = holdup of solid particles, volume fraction of solids
k = absolute pipe roughness, m
k = c[p]/c[v]
K = consistency index for pseudoplastic fluid, Pa-s^n
K[b] = elastic bulk modulus, Pa
L = length of viscometer bob, m
m = power-law exponent for Herschel-Bulkley fluids
ṁ = mass flow rate, kg/s
ṁ[s] = mass flow rate of solid, kg/s
M[T] = torque measured by viscometer, N-m
n = power law exponent for pseudoplastic fluids
p[n] = pressure in bit nozzle, Pa
p[r] = pressure in bit annular area, Pa
P = pressure, Pa
p[atm] = atmospheric pressure, Pa
= aerodynamic force exerted on the cuttings by the air, N
q[a] = total volumetric flow rate, m^3/s
q[c] = volumetric flow rate through concentric annulus, m^3/s
q[e] = volumetric flow rate through eccentric annulus, m^3/s
Q = heat transferred into volume, W
Q[c] = volumetric flow rate of the cuttings, m^3/s
Q[m] = volumetric flow rate of the mud, m^3/s
r[i] = inside radius of annulus, m
r[o] = outside radius of annulus, m
R = ideal gas constant, m^3 Pa/kg-K
R[b] = radius of inner cylinder (bob) of viscometer, m
R[c] = radius of outer cylinder (cup) of viscometer, m
Re = Reynolds number
Re[p] = particle Reynolds number
R[r] = r[i]/r[o]
S = entropy, J/K
t = time, s
T = absolute temperature, °K
u = radial displacement, m
v* = characteristic velocity for turbulent flow calculations, m/s
v = average velocity, m/s
v[a] = average annulus velocity, m/s
v[mix] = mixture velocity, m/s
v[n] = velocity in bit nozzle, m/s
v[p] = plug velocity, m/s
v[r] = velocity in bit annular area, m/s
v[s] = average settling velocity
v[sa] = average cuttings velocity in annular region, m/s
v[sl] = particle slip velocity, m/s
v[sp] = average cuttings velocity in plug, m/s
W = buoyant weight or particle, N
W[s] = buoyant weight of the cuttings, N
x = parameter in settling velocity equation
y = parameter in settling velocity equation
Y[a] = parameter in settling velocity equation
z = measure depth, ft
Z = true vertical depth, ft
α[c] = parameter in Bingham fluid friction factor
β = = coefficient of thermal expansion, 1/K
γ = shear stress, s^–1
γ[e] = equivalent shear stress, s^–1
δ = ratio of the average particle volume to its cross-sectional area
δ[r] = the distance between centers of the inside and outside pipes, m
ΔP = pressure drop, Pa
Δt = time increment, s
Δv = change in velocity, m/s
Δz = length of flow increment, m
ε = internal energy, J/kg
ζ = measured depth integration variable, m
θ = viscometer reading, degrees
ϑ = integration variable
κ = ratio of radius of inner cylinder to outer cylinder
λ[P] = D[plug]/D[eq] , the plug diameter ratio
μ = Newtonian viscosity of the fluid, Pa-s
μ[a] = apparent viscosity, Pa-s
μ[p] = plastic viscosity, centipoise
ξ = integration variable corresponding to depth z, m
ρ = fluid density, kg/m^3
= fluid in-mixture density, kg/m^3
ρ[ƒ] = fluid density in solid/fluid mixture, kg/m^3
ρ[s] = solid density in solid/fluid mixture, kg/m^3
= solid in-mixture density, kg/m^3
τ = shear stress, Pa
τ[w] = wall shear stress, Pa
τ[y] = yield point, Pa
υ[ƒ] =
s ratio for the formation
Φ = angle of inclination from the vertical
Φ = viscous dissipation, W
Ψ = sphericity
ω = rotor speed, rev/min
Ω[o] = angular velocity of outer cylinder
1 = properties inside pipe, surge calculations
2 = properties inside annulus, surge calculations
3 = properties of moving pipe, surge calculation
c = concentric
e = eccentric
n = properties in bit nozzle, surge calculations
o = upstream, initial, or inlet
r = properties in annulus outside bit, surge calculations
- = upstream properties
Noteworthy papers in OnePetro
Quigley, M.C., Mobil R and D Corp.: Advanced Technology for Laboratory Measurements of Drilling Fluid Friction Coefficient, 19537-MS, http://dx.doi.org/10.2118/19537-MS
E. Kaarstad, SPE, B.S. Aadnoy, SPE, and T. Fjelde, SPE, University of Stavanger: A Study of Temperature Dependent Friction in Wellbore Fluids, 119768-MS, http://dx.doi.org/10.2118/119768-MS
External links
See also
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Piezo buzzer resonance with a square wave generator
A square wave is used to drive a piezo buzzer, and we attempt to identify which harmonic of the input frequency is used. The results also confirm that only F,3F, 5F... frequencies are available in a
square wave.
With the advanced data logger of expeyes, one can vary one output parameter, and study the effect on some other aspect of the experiment.
We will study the resonance of a piezo buzzer using a square wave, and observe that the buzzer automatically resonates(becomes loud) as long as the natural frequency matches with the input frequency
or its odd harmonics. This shows that a square wave is composed of a fundamental sine wave and its odd harmonics with decreasing amplitude.
A settling delay of 100mS is set to allow the piezo buzzer to settle into a new frequency before measuring the stable amplitude.
500 datapoints were acquired, and the piezo buzzer was found to have two resonant frequencies with distinct shapes at 3110 and 3740Hz.
At the lower end of the frequency range, multiple peaks were observed resembling the major peaks. So a smaller frequency range up to 2000Hz was chosen to identify these more clearly
A square wave generator is composed of sine waves, and the series expansion looks something like Asin(fx) + Asin(3fx)/3 + Asin(5fx)/5 + …
Sure enough, the first peak of the buzzer was found to appear when driven with frequencies 1036 (3 x 1036 is close to 3110), 625 (5 x 625 close to 3110), 445(7 x f), or 341( 9 x f )
The second peak shape appeared at 1250 (3 times this frequency is close to 3750), 752 (5 x 752 is close to 3740) , 536 (7 x f), or 417 (9 x f)…
Following this, we recorded the dominant frequency emitted by the buzzer against the input frequency. The buzzer would automatically choose one of the component frequencies in the square wave
(F,3F,5F,7F …) depending on its natural frequency.
It was confirmed, that the emitted frequency was always an odd multiple of the input frequency
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|
Ebook Representation Theory Of Finite Groups And Finite Dimensional Algebras: Proceedings Of The Conference At The University Of Bielefeld From May 15–17, 1991, And 7 Survey Articles On Topics Of Representation Theory 1991
Ebook Representation Theory Of Finite Groups And Finite Dimensional Algebras: Proceedings Of The Conference At The University Of Bielefeld From May 15–17, 1991, And 7 Survey Articles On Topics Of
Representation Theory 1991
Ebook Representation Theory Of Finite Groups And Finite Dimensional Algebras: Proceedings Of The Conference At The University Of Bielefeld From May 15–17, 1991, And 7 Survey Articles On Topics Of
Representation Theory 1991
by Natalia 3.4
If you are at an ebook Representation Theory of Finite Groups and Finite Dimensional Algebras: Proceedings or early war, you can be the town verification to assign a hand across the school doing for
lead or PART Facts. What are Saint Louis University's small bridge lasers and GPA?
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10. How many words can be formed with the letters of the word '... | Filo
Question asked by Filo student
10. How many words can be formed with the letters of the word 'DELHI' if and never occur together? 11. How many words can be formed with the letters of the word 'GANESHPURI' in which vowels occupy
odd positions?
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10. How many words can be formed with the letters of the word 'DELHI' if and never occur together? 11. How many words can be formed with the letters of the word 'GANESHPURI' in which
Question Text vowels occupy odd positions?
Updated On Oct 14, 2022
Topic Calculus
Subject Mathematics
Class Class 11
Answer Type Video solution: 1
Upvotes 119
Avg. Video 6 min
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Power Analysis for Two-group Independent sample t-test | Stata Data Analysis Examples
Example 1. A clinical dietician wants to compare two different diets, A and B, for diabetic patients. She hypothesizes that diet A (Group 1) will be better than diet B (Group 2), in terms of lower
blood glucose. She plans to get a random sample of diabetic patients and randomly assign them to one of the two diets. At the end of the experiment, which lasts 6 weeks, a fasting blood glucose test
will be conducted on each patient. She also expects that the average difference in blood glucose measure between the two group will be about 10 mg/dl. Furthermore, she also assumes the standard
deviation of blood glucose distribution for diet A to be 15 and the standard deviation for diet B to be 17. The dietician wants to know the number of subjects needed in each group assuming equal
sized groups.
Example 2. An audiologist wanted to study the effect of gender on the response time to a certain sound frequency. He suspected that men were better at detecting this type of sound then were women. He
took a random sample of 20 male and 20 female subjects for this experiment. Each subject was be given a button to press when he/she heard the sound. The audiologist then measured the response time –
the time between the sound was emitted and the time the button was pressed. Now, he wants to know what the statistical power is based on his total of 40 subjects to detect the gender difference.
Prelude to The Power Analysis
There are two different aspects of power analysis. One is to calculate the necessary sample size for a specified power as in Example 1. The other aspect is to calculate the power when given a
specific sample size as in Example 2. Technically, power is the probability of rejecting the null hypothesis when the specific alternative hypothesis is true.
For the power analyses below, we are going to focus on Example 1, calculating the sample size for a given statistical power of testing the difference in the effect of diet A and diet B. Notice the
assumptions that the dietician has made in order to perform the power analysis. Here is the information we have to know or have to assume in order to perform the power analysis:
• The expected difference in the average blood glucose; in this case it is set to 10.
• The standard deviations of blood glucose for Group 1 and Group 2; in this case, they are set to 15 and 17 respectively.
• The alpha level, or the Type I error rate, which is the probability of rejecting the null hypothesis when it is actually true. A common practice is to set it at the .05 level.
• The pre-specified level of statistical power for calculating the sample size; this will be set to .8.
• The pre-specified number of subjects for calculating the statistical power; this is the situation for Example 2.
Notice that in the first example, the dietician didn’t specify the mean for each group, instead she only specified the difference of the two means. This is because that she is only interested in the
difference, and it does not matter what the means are as long as the difference is the same.
Power Analysis
In Stata, it is fairly straightforward to perform power analysis for comparing means. For example, we can use Stata’s power command for our calculation as shown below. We first specify that we have
two means. Next, we specify the two means, the mean for Group 1 (diet A) and the mean for Group 2 (diet B). Since what really matters is the difference, instead of means for each group, we can enter
a mean of zero for Group 1 and 10 for the mean of Group 2, so that the difference in means will be 10. Next, we specify the standard deviation for the first population and standard deviation for the
second population. The default significance level (alpha level) is .05. For this example we will set the power to be at .8, which is the default value.
power twomeans 0 10, sd1(15) sd2(17)
Performing iteration ...
Estimated sample sizes for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1 versus Ha: m2 != m1
Study parameters:
alpha = 0.0500
power = 0.8000
delta = 10.0000
m1 = 0.0000
m2 = 10.0000
sd1 = 15.0000
sd2 = 17.0000
Estimated sample sizes:
N = 84
N per group = 42
The calculation results indicate that we need 42 subjects for diet A and another 42 subject for diet B in our sample in order the effect. Now, let’s use another pair of means with the same
difference. As we have discussed earlier, the results should be the same, and they are.
power twomeans 5 15, sd1(15) sd2(17)
Performing iteration ...
Estimated sample sizes for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1 versus Ha: m2 != m1
Study parameters:
alpha = 0.0500
power = 0.8000
delta = 10.0000
m1 = 5.0000
m2 = 15.0000
sd1 = 15.0000
sd2 = 17.0000
Estimated sample sizes:
N = 84
N per group = 42
Now the dietician may feel that a total sample size of 84 subjects is beyond her budget. One way of reducing the sample size is to increase the Type I error rate, or the alpha level. Let’s say
instead of using alpha level of .05 we will use .07. Then our sample size will reduce by 4 for each group as shown below.
power twomeans 5 15, sd1(15) sd2(17) alpha(.07)
Performing iteration ...
Estimated sample sizes for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1 versus Ha: m2 != m1
Study parameters:
alpha = 0.0700
power = 0.8000
delta = 10.0000
m1 = 5.0000
m2 = 15.0000
sd1 = 15.0000
sd2 = 17.0000
Estimated sample sizes:
N = 76
N per group = 38
Now suppose the dietician can only collect data on 60 subjects with 30 in each group. What will the statistical power for her t-test be with respect to alpha level of .05?
power twomeans 0 10, sd1(15) sd2(17) n(60)
Estimated power for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1 versus Ha: m2 != m1
Study parameters:
alpha = 0.0500
N = 60
N per group = 30
delta = 10.0000
m1 = 0.0000
m2 = 10.0000
sd1 = 15.0000
sd2 = 17.0000
Estimated power:
power = 0.6610
What if she actually collected her data on 60 subjects but with 40 on diet A and 20 on diet B instead of equal sample sizes in the groups?
power twomeans 0 10, sd1(15) sd2(17) n(60) nratio(2)
Estimated power for a two-sample means test
Satterthwaite's t test assuming unequal variances
H0: m2 = m1 versus Ha: m2 != m1
Study parameters:
alpha = 0.0500
N = 60
N1 = 20
N2 = 40
N2/N1 = 2.0000
delta = 10.0000
m1 = 0.0000
m2 = 10.0000
sd1 = 15.0000
sd2 = 17.0000
Estimated power:
power = 0.6232
As you can see the power goes down from .66 to .62 even though the total number of subjects is the same. This is why we always say that a balanced design is more efficient.
An important technical assumption is the normality assumption. If the distribution is skewed, then a small sample size may not have the power shown in the results, because the value in the results is
calculated using the method based on the normality assumption. We have seen that in order to compute the power or the sample size, we have to make a number of assumptions. These assumptions are used
not only for the purpose of calculation, but are also used in the actual t-test itself. So one important side benefit of performing power analysis is to help us to better understand our designs and
our hypotheses.
We have seen in the power calculation process that what matters in the two-independent sample t-test is the difference in the means and the standard deviations for the two groups. This leads to the
concept of effect size. In this case, the effect size will be the difference in means over the pooled standard deviation. The larger the effect size, the larger the power for a given sample size. Or,
the larger the effect size, the smaller sample size needed to achieve the same power. So, a good estimate of effect size is the key to a good power analysis. But it is not always an easy task to
determine the effect size. Good estimates of effect size come from the existing literature or from pilot studies. One may also want to consider using the minimum effect size of interest.
See Also
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782 Arcmin/Square Hour to Turn/Square Millisecond
Arcmin/Square Hour [arcmin/h2] Output
782 arcmin/square hour in degree/square second is equal to 0.000001005658436214
782 arcmin/square hour in degree/square millisecond is equal to 1.005658436214e-12
782 arcmin/square hour in degree/square microsecond is equal to 1.005658436214e-18
782 arcmin/square hour in degree/square nanosecond is equal to 1.005658436214e-24
782 arcmin/square hour in degree/square minute is equal to 0.0036203703703704
782 arcmin/square hour in degree/square hour is equal to 13.03
782 arcmin/square hour in degree/square day is equal to 7507.2
782 arcmin/square hour in degree/square week is equal to 367852.8
782 arcmin/square hour in degree/square month is equal to 6954980.92
782 arcmin/square hour in degree/square year is equal to 1001517253.2
782 arcmin/square hour in radian/square second is equal to 1.7552050862392e-8
782 arcmin/square hour in radian/square millisecond is equal to 1.7552050862392e-14
782 arcmin/square hour in radian/square microsecond is equal to 1.7552050862392e-20
782 arcmin/square hour in radian/square nanosecond is equal to 1.7552050862392e-26
782 arcmin/square hour in radian/square minute is equal to 0.00006318738310461
782 arcmin/square hour in radian/square hour is equal to 0.22747457917659
782 arcmin/square hour in radian/square day is equal to 131.03
782 arcmin/square hour in radian/square week is equal to 6420.24
782 arcmin/square hour in radian/square month is equal to 121387.32
782 arcmin/square hour in radian/square year is equal to 17479773.58
782 arcmin/square hour in gradian/square second is equal to 0.00000111739826246
782 arcmin/square hour in gradian/square millisecond is equal to 1.11739826246e-12
782 arcmin/square hour in gradian/square microsecond is equal to 1.11739826246e-18
782 arcmin/square hour in gradian/square nanosecond is equal to 1.11739826246e-24
782 arcmin/square hour in gradian/square minute is equal to 0.004022633744856
782 arcmin/square hour in gradian/square hour is equal to 14.48
782 arcmin/square hour in gradian/square day is equal to 8341.33
782 arcmin/square hour in gradian/square week is equal to 408725.33
782 arcmin/square hour in gradian/square month is equal to 7727756.58
782 arcmin/square hour in gradian/square year is equal to 1112796948
782 arcmin/square hour in arcmin/square second is equal to 0.00006033950617284
782 arcmin/square hour in arcmin/square millisecond is equal to 6.033950617284e-11
782 arcmin/square hour in arcmin/square microsecond is equal to 6.033950617284e-17
782 arcmin/square hour in arcmin/square nanosecond is equal to 6.033950617284e-23
782 arcmin/square hour in arcmin/square minute is equal to 0.21722222222222
782 arcmin/square hour in arcmin/square day is equal to 450432
782 arcmin/square hour in arcmin/square week is equal to 22071168
782 arcmin/square hour in arcmin/square month is equal to 417298855.5
782 arcmin/square hour in arcmin/square year is equal to 60091035192
782 arcmin/square hour in arcsec/square second is equal to 0.0036203703703704
782 arcmin/square hour in arcsec/square millisecond is equal to 3.6203703703704e-9
782 arcmin/square hour in arcsec/square microsecond is equal to 3.6203703703704e-15
782 arcmin/square hour in arcsec/square nanosecond is equal to 3.6203703703704e-21
782 arcmin/square hour in arcsec/square minute is equal to 13.03
782 arcmin/square hour in arcsec/square hour is equal to 46920
782 arcmin/square hour in arcsec/square day is equal to 27025920
782 arcmin/square hour in arcsec/square week is equal to 1324270080
782 arcmin/square hour in arcsec/square month is equal to 25037931330
782 arcmin/square hour in arcsec/square year is equal to 3605462111520
782 arcmin/square hour in sign/square second is equal to 3.35219478738e-8
782 arcmin/square hour in sign/square millisecond is equal to 3.35219478738e-14
782 arcmin/square hour in sign/square microsecond is equal to 3.35219478738e-20
782 arcmin/square hour in sign/square nanosecond is equal to 3.35219478738e-26
782 arcmin/square hour in sign/square minute is equal to 0.00012067901234568
782 arcmin/square hour in sign/square hour is equal to 0.43444444444444
782 arcmin/square hour in sign/square day is equal to 250.24
782 arcmin/square hour in sign/square week is equal to 12261.76
782 arcmin/square hour in sign/square month is equal to 231832.7
782 arcmin/square hour in sign/square year is equal to 33383908.44
782 arcmin/square hour in turn/square second is equal to 2.79349565615e-9
782 arcmin/square hour in turn/square millisecond is equal to 2.79349565615e-15
782 arcmin/square hour in turn/square microsecond is equal to 2.79349565615e-21
782 arcmin/square hour in turn/square nanosecond is equal to 2.79349565615e-27
782 arcmin/square hour in turn/square minute is equal to 0.00001005658436214
782 arcmin/square hour in turn/square hour is equal to 0.036203703703704
782 arcmin/square hour in turn/square day is equal to 20.85
782 arcmin/square hour in turn/square week is equal to 1021.81
782 arcmin/square hour in turn/square month is equal to 19319.39
782 arcmin/square hour in turn/square year is equal to 2781992.37
782 arcmin/square hour in circle/square second is equal to 2.79349565615e-9
782 arcmin/square hour in circle/square millisecond is equal to 2.79349565615e-15
782 arcmin/square hour in circle/square microsecond is equal to 2.79349565615e-21
782 arcmin/square hour in circle/square nanosecond is equal to 2.79349565615e-27
782 arcmin/square hour in circle/square minute is equal to 0.00001005658436214
782 arcmin/square hour in circle/square hour is equal to 0.036203703703704
782 arcmin/square hour in circle/square day is equal to 20.85
782 arcmin/square hour in circle/square week is equal to 1021.81
782 arcmin/square hour in circle/square month is equal to 19319.39
782 arcmin/square hour in circle/square year is equal to 2781992.37
782 arcmin/square hour in mil/square second is equal to 0.00001787837219936
782 arcmin/square hour in mil/square millisecond is equal to 1.787837219936e-11
782 arcmin/square hour in mil/square microsecond is equal to 1.787837219936e-17
782 arcmin/square hour in mil/square nanosecond is equal to 1.787837219936e-23
782 arcmin/square hour in mil/square minute is equal to 0.064362139917695
782 arcmin/square hour in mil/square hour is equal to 231.7
782 arcmin/square hour in mil/square day is equal to 133461.33
782 arcmin/square hour in mil/square week is equal to 6539605.33
782 arcmin/square hour in mil/square month is equal to 123644105.33
782 arcmin/square hour in mil/square year is equal to 17804751168
782 arcmin/square hour in revolution/square second is equal to 2.79349565615e-9
782 arcmin/square hour in revolution/square millisecond is equal to 2.79349565615e-15
782 arcmin/square hour in revolution/square microsecond is equal to 2.79349565615e-21
782 arcmin/square hour in revolution/square nanosecond is equal to 2.79349565615e-27
782 arcmin/square hour in revolution/square minute is equal to 0.00001005658436214
782 arcmin/square hour in revolution/square hour is equal to 0.036203703703704
782 arcmin/square hour in revolution/square day is equal to 20.85
782 arcmin/square hour in revolution/square week is equal to 1021.81
782 arcmin/square hour in revolution/square month is equal to 19319.39
782 arcmin/square hour in revolution/square year is equal to 2781992.37
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CBSE Class 8 Maths Sample Paper Set 9 - 2022
CBSE Sample Question Paper for Class 8 Maths
Maths for Class 8 is considered to be one of the most important and immensely scoring subjects. And the best way to prepare apart from completing NCERT and reference books is solving CBSE Sample
Papers. Here on Ribblu one can get immense collection of Sample Question Papers for Class 8 Maths in PDF format for free.
Class 8 Maths Marks Distribution
Units Marks
Number Systems 08
Algebra 17
Coordinate Geometry 04
Geometry 28
Mensuration 13
Statistics & Probability 10
Total 80
Internal Assessment 20
Grand Total 100
Maths Topics to be covered for Class 9
• Real Numbers
• Polynomials
• Coordinate Geometry
• Linear Equations in Two Variables
• Introduction to Euclid's Geometry
• Lines and Angles
• Triangles
• Quadrilaterals
• Area
• Circles
• Constructions
• Areas
• Surface Areas and Volumes
• Statistics
• Probability
Structure of CBSE Maths Sample Paper for Class 9 is
Type of Question Marks per Question Total No. of Questions Total Marks
Objective Type Questions 1 20 20
Short Answer Type Questions - I 2 6 12
Short Answer Type Questions - II 3 8 24
Long Answer Type Questions 4 6 24
Total 40 80
For Preparation of board exams students can also check out other resource material
CBSE Class 8 Maths Question Papers
Important Questions for Class 8 Maths Chapter Wise
Maths Revision Notes for class 8
Sample Papers of Other Subjects of Class 8
CBSE Sample Papers of Class 8 Science CBSE Sample Papers of Class 8 English CBSE Sample Papers of Class 8 Social Science CBSE Sample Papers of Class 8 Computer Science CBSE Sample Papers of Class 8
Hindi CBSE Sample Papers of Class 8 Sanskrit
What are CBSE Sample Papers?
Sample papers are mainly a kind of mock tests or model test papers which are prepared in accordance with the latest syllabus and guidelines that are issued by the central board. These examination
test papers are designed as the replica of the actual papers that are asked in final examinations. All the marking schemes , number of questions , types of questions asked are followed as per board
scheme and are issued to students two or three months before the examinations so that students get enough time to practice.
What is the importance of Sample Papers for Students?
In order to access the level of preparation done by any particular student he or she needs to solve CBSE Sample Papers. These papers are the perfect way to practise for the final board exam. If one
wants to have a clear idea of how the final exam papers would be in terms of level of difficulty, time and other aspects then , all students must make sure that they do sample papers once their
course revision is finished.
Few benefits of solving CBSE sample papers are given below:
• Gauging Self Performance: Understanding and revising the subject is very good, but unless one attempts the sample paper in the lookalike environment as in board exam, seldom can the student
identify and check whether the understanding of all concepts of the subject are complete or not. Once students try the question paper in the same time frame he or she is able to judge the
capability of solving the paper in the stipulated time frame. It highlights the weak areas if any and gives students ample amount of time to work on those areas and be better prepared before
• Testing Time Criticality: Knowing is not everything as far as board papers are concerned. Sometimes it happens that in spite of knowing everything a student falls short of time to complete the
entire paper and thus loses marks. Generally CBSE sample papers are generally of 3 hour duration. So while practicing sample papers it is imperative to create a board like environment at home and
ensure that sample paper is attempted only in 3 hours and then check whether it was possible to complete the paper in the desired amount of time. Often at first students take longer than
expected, and thus they get early warning to practice more and increase the speed.
• Exam Anxiety: Sometimes sensitive students feel anxious to sit in the examination hall with a 3 hour length paper and for them it becomes a more intense requirement to practice prior to main
exams and get rid of any kind of fear. Since they do not know what questions will be asked in the CBSE board they create panic in their mind due to this fear of the unknown and get scared with
the idea that they might not be able to do well in exams. Thus such students needs to complete at-least 7-10 sample papers prior to the exams, to gain confidence and get into better frame of
Best Time to Practice Sample Papers
This mainly varies from student to student but in general students should start dealing with sample papers as soon as their book revisions are over. Infact along with sample papers students should
also look for model test papers of various publishers and attempt them to get the idea of level of preparation. As the dates of exam are approaching near by that time only samples papers should be
mainly considered for practice and in case of any shortcomings those should be thoroughly discussed with teachers freids and other concerned person so that one has clarity before attempting the final
Sample Papers of Other Classes
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MIP Relative Optimality Tolerance | AIMMS Community
In the Options Tree, I have been changing the MIP Relative Optimality Tolerance to 0.5%, 1.0%, 5.0% and 10.0% to analyse the performance of the solution, gap and solving time.
I wish to change the gap to 0.5%, 1.0%, 5.0% and 10.0% to analyse whether the solving time can be reduce by changing the gap. However, the CPLEX always stop at either 0.04%, 0.12% or 0.00% of gap. Is
there any way to stop the CPLEX at 10% of gap?
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Handling Mathematical Operations in Ruby
When working with Ruby, you may often need to perform various mathematical operations such as addition, subtraction, multiplication, and division. In this article, we will explore how to handle these
operations in Ruby and provide examples to help you better understand the concepts.
Basic Arithmetic Operations
To perform addition in Ruby, you can simply use the + operator. Here is an example:
num1 = 10
num2 = 5
result = num1 + num2
puts result
In this example, the variables num1 and num2 are added together, and the result is stored in the result variable. The output will be 15.
Subtraction in Ruby is done using the - operator. Here is an example:
num1 = 10
num2 = 5
result = num1 - num2
puts result
In this example, the value of num2 is subtracted from num1, and the result is stored in the result variable. The output will be 5.
For multiplication in Ruby, you can use the * operator. Here is an example:
num1 = 10
num2 = 5
result = num1 * num2
puts result
In this example, the variables num1 and num2 are multiplied together, and the result is stored in the result variable. The output will be 50.
Division in Ruby is performed using the / operator. Here is an example:
num1 = 10
num2 = 5
result = num1 / num2
puts result
In this example, the value of num1 is divided by the value of num2, and the result is stored in the result variable. The output will be 2.
Advanced Mathematical Operations
To calculate the exponentiation of a number in Ruby, you can use the ** operator. Here is an example:
num = 2
exponent = 3
result = num ** exponent
puts result
In this example, the value of num raised to the power of exponent is calculated, and the result is stored in the result variable. The output will be 8.
The modulo operation in Ruby is performed using the % operator. It returns the remainder of the division of two numbers. Here is an example:
num1 = 10
num2 = 3
result = num1 % num2
puts result
In this example, the remainder of dividing num1 by num2 is calculated, and the result is stored in the result variable. The output will be 1.
In this article, we have covered the basic and advanced mathematical operations in Ruby. By understanding how to handle these operations, you can perform various calculations in your Ruby programs
with ease. Practice using these operators in your code to become more familiar with them and improve your programming skills.
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Variant surveillance example
In this vignette, we use the phylosamp package to prepare for the emergence of a new variant of a SARS-CoV-2-like pathogen into a fictional population, based on current whole genome sequencing
capacity and experience with previous variants of concern.
There are three steps in applying our method:
1. Determine the population of interest (Figure 1). In this example, we’ll assume we are interested in tracking variants of Pathogen X in a small country with a well-defined population.
2. Identify the key question we are trying to answer with our surveillance scheme. In this example, we will assume we are interested in calculating the sample size needed to detect the emergence of
a new variant of Pathogen X by the time it reaches a frequency of 1% across all infected individuals in our country. In other words, we will focus on variant detection.
3. Identify the sampling frequency we have the capacity to maintain. In our case, we’ll assume we want to develop a weekly sampling scheme, in which pathogen samples collected over a 7-day period
are sequenced in weekly batches (i.e., periodic surveillance).
Now that we’ve identified our surveillance goals, we need to estimate some basic parameters for our population of interest, such as the pathogen testing rate, the sensitivity of the tests used, etc.
However, since these values may vary by pathogen variant, we need to explore and estimate these parameters in a variant-specific context. In the current implementation of the sample size calculation
methodology described herein, the specific parameters we will need to consider are (see Table 1): the variant-specific asymptomatic rate, the asymptomatic and symptomatic testing rates, the
variant-specific testing sensitivity using currently available technologies, the variant-specific sampling success rate (i.e., the expected number of samples of high enough quality for variant
characterization by whole genome sequencing), and the sequencing success rate. Let’s consider each of these parameters in turn (see also: Estimating bias in observed variant prevalence).
Guidance for determining variant-specific parameters
The asymptomatic rate (\(\psi\)).
• So far, epidemiologists have determined that the asymptomatic rate of Pathogen X has ranged from 30-40%. The currently circulating variant has an asymptomatic rate of 30%. Given this, we want to
plan for variants that could have asymptomatic rates ranging from 25-45%. A lower asymptomatic rate causes enrichment of the variant in sampled infections; conversely, a higher asymptomatic rate
would artificially deplete samples belonging to the variant of interest in the pool of detected infections. Therefore, an asymptomatic rate of 25% represents the least conservative scenario
(since enrichment of a variant of interest would mean fewer sequences are required to detect it) while an asymptomatic rate of 45% represents the most conservative scenario.
The testing rate (\(\tau\)).
• Given the widespread availability of rapid antigen tests for Pathogen X, we assume that only 50% of symptomatic infections (of any variant) are tested, and only 10% of asymptomatic infections are
detected and samples sent to national public health laboratories. We anticipate that testing rates could drop as low as 40% (symptomatic) / 5% (asymptomatic) as the population becomes
increasingly desensitized to disease spread. Because of the complex relationship between testing rates and sampling bias, we’ll explore these two scenarios independently when performing sample
size calculations.
The testing sensitivity (\(\phi\)).
• The current gold-standard PCR test for Pathogen X has a sensitivity of 95% for the currently circulating variant. Historical data shows this rate has changed very little between variants.
However, to account for the possibility that a future variant may significantly change the viral load present in patient samples or mutate in such a way that tests temporarily become less
effective (until an updated PCR target can be developed), we will perform sample size calculations assuming no change in sensitivity (least conservative scenario) as well as a drop in sensitivity
down to 90% (most conservative scenario).
The sampling success rate (\(\gamma\)).
• In many laboratory settings, viral load is measured for each sample by qPCR prior to sequencing, and the results are used to select samples for sequencing. Only sequencing the highest quality
samples ensures the sequencing process is maximally cost effective. For the sake of example, we will assume that the sampling success rate is expected to be the same across all potential
variants. However, we can imagine that a variant with a lower sampling success rate would require additional sampling for accurate detection.
The sequencing success rate (\(\omega\)).
• Not all samples selected for sequencing will produce high quality genomes that can be used for variant characterization. We assume that sequencing success is fixed across all variants, as the
factors that affect sequencing success are not independent of those affecting infection detection and sample quality. In the national laboratory of our country of interest, the sequencing success
rate is 80%.
The coefficient of detection ratio
Once we have estimated the parameter ranges of interest, we can calculate the coefficient of detection in the most and least conservative scenarios. We can do this using the vartrack_cod_ratio()
function as shown below (for more details, see Estimating bias in observed variant prevalence).
When calculating the coefficient of detection, keep in mind that the \(\gamma\) parameter can be left out for both the variant of interest and general population parameters, since (as discussed
above) we are assuming that this parameter does not change between variants. In the least conservative scenario as described above, the testing sensitivity \(\phi\) also does not differ between
potential new variants and the currently circulating pathogen population.
We can provide the remaining parameters as follows. (Note that \(V_1\) represents the future variant we want to capture and \(V_2\) parameters correspond to the general pathogen population.)
# Least conservative scenario assuming higher testing rate:
vartrack_cod_ratio(psi_v1=0.25, psi_v2=0.3, tau_a=0.1, tau_s=0.5)
## [1] 1.052632
# Least conservative scenario assuming lower testing rate:
vartrack_cod_ratio(psi_v1=0.25, psi_v2=0.3, tau_a=0.05, tau_s=0.4)
## [1] 1.059322
# Most conservative scenario assuming higher testing rate:
vartrack_cod_ratio(psi_v1=0.45, psi_v2=0.3, phi_v1=0.9, phi_v2=0.95,
tau_a=0.1, tau_s=0.5)
## [1] 0.7977839
# Most conservative scenario assuming lower testing rate:
vartrack_cod_ratio(psi_v1=0.45, psi_v2=0.3, phi_v1=0.9, phi_v2=0.95,
tau_a=0.05, tau_s=0.4)
## [1] 0.778769
Given these results, we can move forward to sample size calculations with two values of the coefficient of detection ratio to test: 0.779 (most conservative scenario) and 1.059 (least conservative
Sample size calculations
Once we have determined the range of scenarios we’d like to explore, we can perform sample size calculations. As our aim is to ensure variant detection using a periodic sampling strategy, we need to
use the sampling_freq = "cont" option of the vartrack_samplesize_detect() function of the phylosamp R package (see Estimating the sample size needed for variant monitoring: periodic sampling for more
To do this, there are a few more parameters we need to estimate:
The desired probability of detection (\(prob\)).
• We can again select a parameter range to explore for our sample size calculations. In our case, we want to ensure a good chance of detecting a new variant of Pathogen X when it enters our
country, so we will explore probabilities of detection between 75% (least conservative) to 95% (most conservative).
The desired variant prevalence (\(p_{\text{detect}}\)).
• As stated above, we want to ensure we catch any variant by the time it has reached 1% prevalence in the population of infected individuals.
Initial variant prevalence (\(p_0\)).
• The method we will use for sample size calculations assumes logistic growth of any new variants of concern, with a starting prevalence and growth rate that can be specified. This initial
prevalence depends on the number of simultaneous variant introductions into our country of interest as well as the total infected population size. Over the last year, we have observed a total of
between 5,000 and 10,000 total cases of Pathogen X in our country at any given time, and we expect the number of cases to be similar if a new variant is introduced. Because of the complex
relationship between initial prevalence and the shape of the logistic growth curve, we will estimate the required sample size in two scenarios: (1) if a new variant could be introduced via a
single index case, at a time when nearly 10,000 people are infected (initial prevalence \(=1/10000\)); and (2) if 5 different travelers are be infected by a new variant and bring it into the
country in the same week, at a time when only 5,000 individuals are infected (initial prevalence \(=5/5000=1/1000\)).
Logistic growth rate (\(r\)).
• We also need to estimate a variant growth rate over time. Based on historical data of Pathogen X, we know that a recently introduced variant may grow as slowly as 0.1x/day (least conservative, as
fewer samples are needed to ensure we catch the variant before it goes above 1% frequency) or as quickly as 0.2x/day (most conservative).
We now have all of the values we need to estimate the sample size needed for detecting a variant by the time it reaches 1% in the population, assuming weekly periodic sampling. Additionally, it is
important to remember that the number of required sequences is not the same as the number of required samples, because of the sequencing success rate (\(\omega\)) discussed above. The phylosamp
functions output the number of samples required, taking into account that not all samples selected for sequencing will result in high quality samples suitable for variant characterization:
# Least conservative scenario with low initial prevalence:
vartrack_samplesize_detect(prob=0.75, p_v1=0.01, p0_v1=1/10000, r_v1=0.1,
omega=0.8, c_ratio=1.059, sampling_freq="cont")
## Calculating sample size for variant detection assuming periodic sampling
## [1] 15.106
# Least conservative scenario with high initial prevalence:
vartrack_samplesize_detect(prob=0.75, p_v1=0.01, p0_v1=1/1000, r_v1=0.1,
omega=0.8, c_ratio=1.059, sampling_freq="cont")
## Calculating sample size for variant detection assuming periodic sampling
## [1] 16.41287
# Most conservative scenario with low initial prevalence:
vartrack_samplesize_detect(prob=0.95, p_v1=0.01, p0_v1=1/10000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating sample size for variant detection assuming periodic sampling
## [1] 80.14972
# Most conservative scenario with high initial prevalence:
vartrack_samplesize_detect(prob=0.95, p_v1=0.01, p0_v1=1/1000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating sample size for variant detection assuming periodic sampling
## [1] 96.2708
Based on these calculations, we need to be sequencing between 16 and 97 samples per day (or 112 and 679 samples per week) in order to detect a new variant by the time it reaches 1% in the population.
As this is a rather wide range, we can use the reverse functionality of the sample size calculation method to determine the probability of detecting a variant given a fixed number of samples and most
conservative parameter values.
Estimating the probability of detection
Given the recommendation of 112-679 samples per week, the government of our country of interest has decided that funding will be allocated to support sequencing of 200 Pathogen X samples per week.
Given our most conservative scenario of a coefficient of detection of 0.779 and a growth rate of 0.2, we can use the vartrack_prob_detect() function to calculate the probability of detecting a
variant before it crosses the 1% prevalence threshold in the population (see Estimating the probability of detecting a variant: periodic sampling).
# Most conservative scenario with low initial prevalence
vartrack_prob_detect(n=28, p_v1=0.01, p0_v1=1/10000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating probability of detection assuming periodic sampling
## [1] 0.6488521
# Most conservative scenario with high initial prevalence
vartrack_prob_detect(n=28, p_v1=0.01, p0_v1=1/1000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating probability of detection assuming periodic sampling
## [1] 0.5815917
In both the high and low initial prevalence scenarios, the probability of detection (assuming roughly 28 samples selected per day, to be sequenced in weekly batches) remains above 58% even using the
most conservative parameters. Furthermore, the probability of detecting a new variant by the time it reaches 2% in the population is approximately 85% in both scenarios (as shown below), with numbers
approaching 99% chance of detection before the variant hits 5% prevalence. These values may be sufficient for country officials to feel confident in their ability to detect a variant soon after it is
introduced regardless of its biological properties; if it is not, the calculations can simply be repeated with a higher number of weekly samples.
## DETECTION BEFORE REACHING 2% PREVALENCE
# Most conservative scenario with low initial prevalence
vartrack_prob_detect(n=28, p_v1=0.02, p0_v1=1/10000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating probability of detection assuming periodic sampling
## [1] 0.8514334
# Most conservative scenario with high initial prevalence
vartrack_prob_detect(n=28, p_v1=0.02, p0_v1=1/1000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating probability of detection assuming periodic sampling
## [1] 0.8693221
## DETECTION BEFORE REACHING 5% PREVALENCE
# Most conservative scenario with low initial prevalence
vartrack_prob_detect(n=28, p_v1=0.05, p0_v1=1/10000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating probability of detection assuming periodic sampling
## [1] 0.9940441
# Most conservative scenario with high initial prevalence
vartrack_prob_detect(n=28, p_v1=0.05, p0_v1=1/1000, r_v1=0.2,
omega=0.8, c_ratio=0.779, sampling_freq="cont")
## Calculating probability of detection assuming periodic sampling
## [1] 0.989769
Of course, there are many assumptions that underlie these calculations, the most obvious being that the weekly batch of samples for sequencing are assumed to be well-distributed across the days of
the week, and that they capture all regions or ports of entry into the country. Even so, this method provides sampling guideposts that can be applied in a variety of settings. For example, it is
clear from the simple calculations above that 100 samples per week would be unlikely to be particularly informative for detecting new variants early and with high confidence.
Although the example provided here focuses on the question of detection with periodic sampling, the same principles (though different functions/spreadsheet tabs) can be applied to a cross-sectional
sampling scheme (see Estimating the sample size needed for variant monitoring: cross-sectional and Estimating the probability of detecting a variant: cross-sectional) and/or estimating variant
prevalence. The section on the coefficient of detection remains identical, and only the sampling calculations need to be updated to suit the surveillance goals.
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Meta-mathematical reasoning
It's common in programming languages to write proofs about proofs. For example, when using a Gentzen-style deduction system for specifying a type system or a big-step operational semantics, we often
write a proof by induction on the size of the proof tree of a derivation.
When proving type soundness, you can use a number of different dynamic semantics. It's sanest to attempt this with an operational semantics, but there are many flavors to choose from: big-step
operational semantics, small-step semantics with compatibility rules for lifting the basic reductions into arbitrary evaluation contexts, small-step semantics with Gentzen-style rules for creating an
inductive proof tree that lifts the reductions into evaluation contexts, or Felleisen-style small-step semantics where the evaluation contexts are reified as plain old mathematical objects.
The benefit of the latter is that it requires no meta-mathematical reasoning--this is not to say it's impossible to use any of the others to prove soundness, but it perhaps requires less subtlety to
write a proof about ordinary objects than one about proofs.
It reminds me of what
Andrew Pitts
has been saying for a while about his and Gabbay's denotational framework for theories of binding. They originally formulated their theory by working in a different set theory (one invented in the
early 20th century by Fraenkel and Mostowski), but discovered that this required a certain "meta-logical sophistication" that made their work less accessible:
[Pitts and Gabbay [2002]] expresses its results in terms of an axiomatic set theory, based on the classical Fraenkel-Mostowski permutation model of set theory. In my experience, this formalism
impedes the take up within computer science of the new ideas contained in that article. There is an essentially equivalent, but more concrete description of the model as standard sets equipped
with some simple extra structure. These so-called nominal sets are introduced by Pitts [2003], and I will use them here to express α-structural recursion and induction within "ordinary
From "
Alpha-structural recursion and induction
", p. 462
Again, I think the moral of the story is that when you place all the objects you want to reason about at the object level, you avoid the need for meta-reasoning, which can make your proofs simpler
and more accessible.
No comments:
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Total Surface Area of Icosahedron given Midsphere Radius Calculator | Calculate Total Surface Area of Icosahedron given Midsphere Radius
What is an Icosahedron?
An Icosahedron is a symmetric and closed three dimensional shape with 20 identical equilateral triangular faces. It is a Platonic solid, which has 20 faces, 12 vertices and 30 edges. At each vertex,
five equilateral triangular faces meet and at each edge, two equilateral triangular faces meet.
What are Platonic Solids?
In three-dimensional space, a Platonic solid is a regular, convex polyhedron. It is constructed by congruent (identical in shape and size), regular (all angles equal and all sides equal), polygonal
faces with the same number of faces meeting at each vertex. Five solids who meet this criteria are Tetrahedron {3,3} , Cube {4,3} , Octahedron {3,4} , Dodecahedron {5,3} , Icosahedron {3,5} ; where
in {p, q}, p represents the number of edges in a face and q represents the number of edges meeting at a vertex; {p, q} is the Schläfli symbol.
How to Calculate Total Surface Area of Icosahedron given Midsphere Radius?
Total Surface Area of Icosahedron given Midsphere Radius calculator uses Total Surface Area of Icosahedron = 5*sqrt(3)*((4*Midsphere Radius of Icosahedron)/(1+sqrt(5)))^2 to calculate the Total
Surface Area of Icosahedron, The Total Surface Area of Icosahedron given Midsphere Radius formula is defined as the total quantity of plane enclosed by the entire surface of the Icosahedron and
calculated using the midsphere radius of the Icosahedron. Total Surface Area of Icosahedron is denoted by TSA symbol.
How to calculate Total Surface Area of Icosahedron given Midsphere Radius using this online calculator? To use this online calculator for Total Surface Area of Icosahedron given Midsphere Radius,
enter Midsphere Radius of Icosahedron (r[m]) and hit the calculate button. Here is how the Total Surface Area of Icosahedron given Midsphere Radius calculation can be explained with given input
values -> 846.8282 = 5*sqrt(3)*((4*8)/(1+sqrt(5)))^2.
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The following numbers are obviously not perfect - WorkSheets Buddy
The following numbers are obviously not perfect
The following numbers are obviously not perfect. Give reason.
(i) 567
(ii) 2453
(iii) 5298
(iv) 46292
(v) 74000
The square of which of the following numbers would be an odd number or an even number? Why?
(i) 573
(ii) 4096
(iii) 8267
(iv) 37916
More Solutions:
Leave a Comment
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Nim | Brilliant Math & Science Wiki
Nim is a combinatorial game, where two players alternately take turns in taking objects from several heaps. The only rule is that each player must take at least one object on their turn, but they may
take more than one object in a single turn, as long as they all come from the same heap.
Nim is the most well-known example of an impartial game, a game where both players have the same moves all the time, and thus the only distinguishing feature is that one player goes first. It is also
completely solved, in the sense that the exact strategy has been found for any starting configuration.
The basic Nim begins with two players and several heaps, each containing several objects. Occasionally, heaps are also called piles, and the objects are called stones.
Each player, in turn, must take at least one stone, but they may take more than one stone as long as they all come from the same pile. It's allowed to make a pile empty, effectively removing the pile
out of the game. When a player is unable to move, the game ends. Naturally, as long as there is a stone, either player can take that stone, and thus can move. So the ending condition can be
rephrased, where the game ends if there is no stone left.
In normal Nim, the loser is the player unable to move. It is called normal condition by convention from combinatorial game theory, where a normal game gives the win to the last player making a move.
In misère Nim, the player unable to move wins instead; this is equivalent to the player taking the last stone losing.
Consider the following example of the game. There are three piles, initially having \(3, 4, 5\) stones respectively. Alice and Bob are playing, with Alice starting.
Pile 1 Pile 2 Pile 3 Move
\(3\) \(4\) \(5\) Starting position
\(1\) \(4\) \(5\) 1. Alice takes \(2\) stones from Pile 1
\(1\) \(4\) \(3\) 2. Bob takes \(2\) stones from Pile 3
\(1\) \(2\) \(3\) 3. Alice takes \(2\) stones from Pile 2
\(0\) \(2\) \(3\) 4. Bob takes \(1\) stone from Pile 1
\(0\) \(2\) \(2\) 5. Alice takes \(1\) stone from Pile 3
\(0\) \(1\) \(2\) 6. Bob takes \(1\) stone from Pile 2
\(0\) \(1\) \(1\) 7. Alice takes \(1\) stone from Pile 3
\(0\) \(1\) \(0\) 8. Bob takes \(1\) stone from Pile 3
\(0\) \(0\) \(0\) 9. Alice takes \(1\) stone from Pile 2
In normal play, Alice wins, as Alice has taken the last stone and thus leaving Bob with no move. In misère play, Alice would lose instead, but in this case Alice would have taken \(2\) stones from
Pile 3 on move 7, leaving Bob with pile sizes \(0, 1, 0\) and thus forcing Bob to take the last object.
In the above game, Alice has played a perfect game, never letting Bob to be able to snatch the win. This can be generalized into a general strategy.
The nim-sum \(a \oplus b\) of two non-negative integers \(a\) and \(b\) is defined as follows. Represent \(a\) and \(b\) as sums of distinct powers of two. Cancel powers of two appearing twice, and
add up the remaining numbers. Nim-sum is also known as XOR by computer scientists.
For example, \(3 \oplus 5\) can be computed as follows. We have \(3 = 2^1 + 2^0\) and \(5 = 2^2 + 2^0\). We cancel \(2^0\) for appearing twice, and add up the remaining \(2^1 + 2^2\), giving \(3 \
oplus 5 = 6\).
It can be shown that \(\oplus\) is associative, thus making the nim-sum of several numbers \(a_1 \oplus a_2 \oplus \ldots \oplus a_n\) defined.
Now, given a normal Nim position (the sizes of the piles) \(a_1, a_2, \ldots, a_n\), the player to move wins if \(a_1 \oplus a_2 \oplus \ldots \oplus a_n \neq 0\); a winning move can be found by
determining a pile \(i \in \{1, 2, \ldots, n\}\) and a number \(b_i \in \{0, 1, \ldots, a_i - 1\}\) such that \(a_1 \oplus a_2 \oplus \ldots \oplus a_{i-1} \oplus b_i \oplus a_{i+1} \oplus a_{i+2} \
oplus \ldots \oplus a_n = 0\), and taking some stones from pile \(i\) such that \(b_i\) stones are left. If \(a_1 \oplus a_2 \oplus \ldots \oplus a_n = 0\), then the player to move loses.
For example, in the above game with \(3, 4, 5\), we see that \(3 \oplus 4 \oplus 5 = 7 \oplus 5 = 2\), and \(1 \oplus 4 \oplus 5 = 5 \oplus 5 = 0\), so the player to move (in this case Alice) wins by
reducing the first pile from \(3\) stones to \(1\) stone. After that, \(1 \oplus 4 \oplus 5 = 0\), so theoretically the player to move (in this case Bob) loses, as long as the opponent plays
In misère Nim, the strategy is almost identical. As long as after the suggested move, there will be at least one pile of size \(2\) or larger, follow the strategy of the normal Nim. However, if after
the suggested move, there will be no pile of size \(2\) or larger, play a different move:
• If the suggested move makes the pile have \(1\) stone left, make it have \(0\) stones instead, or
• If the suggested move makes the pile have \(0\) stones left, make it have \(1\) stone instead.
In other words, the correct move is to leave an odd number of piles of size \(1\). (In normal play, there should be an even number of piles of size \(1\) instead, to make the nim-sum zero.)
Proof of Winning Strategy
The following proof is for normal Nim's strategy, given by C. Bouton.
The moving player wins in normal Nim if and only if the nim-sum of the pile sizes is not zero.
We will begin with the easy base case: if the pile sizes are all zero, then the moving player loses and the nim-sum is zero. From now on, assume that not all pile sizes are zero.
First, we observe that the nim-sum obeys the following several properties for all non-negative integers \(a,b,c:\)
□ Associativity: \((a \oplus b) \oplus c = a \oplus (b \oplus c)\)
□ Commutativity: \(a \oplus b = b \oplus a\)
□ Identity: \(0 \oplus a = a\)
□ Self-inverse: \(a \oplus a = 0\)
□ Computing nim-sum of multiple numbers at once is possible: write all numbers as sums of distinct powers of 2, find all powers of 2 that appear an odd number of times, and sum up one
occurrence of each power of 2. For example, \(1 \oplus 3 \oplus 7 = \big(2^0\big) \oplus \big(2^0 + 2^1\big) \oplus \big(2^0 + 2^1 + 2^2\big) = 2^0 + 2^2 = 5\).
Suppose the pile sizes are \(a_1, a_2, \ldots, a_n\) before a move, and \(b_1, b_2, \ldots, b_n\) after a move. Suppose that the move is on pile \(k\); then for all \(i \neq k\), \(a_i = b_i\).
Let \(s = a_1 \oplus a_2 \oplus \ldots \oplus a_n\) and \(t = b_1 \oplus b_2 \oplus \ldots \oplus b_n\). We have
\[\begin{align*} t &= 0 \oplus t \\ &= (s \oplus s) \oplus t \\ &= s \oplus (s \oplus t) \\ &= s \oplus \big((a_1 \oplus a_2 \oplus \ldots \oplus a_n) \oplus (b_1 \oplus b_2 \oplus \ldots \oplus
b_n)\big) \\ &= s \oplus \big((a_1 \oplus b_1) \oplus (a_2 \oplus b_2) \oplus \ldots \oplus (a_n \oplus b_n)\big) \\ &= s \oplus \big(0 \oplus 0 \oplus \ldots \oplus 0 \oplus (a_k\oplus b_k) \
oplus 0 \oplus \ldots \oplus 0\big) \\ &= s \oplus (a_k \oplus b_k). \end{align*}\]
Now we will prove two results.
Result 1: If \(s = 0\), then \(t \neq 0\). If the nim-sum of the original sizes is zero, then the moving player is losing (they must make the nim-sum nonzero).
We claim that \(a_k\oplus b_k \neq 0\). Indeed, suppose it is, then
\[\begin{align*} a_k &= a_k \oplus 0 \\ &= a_k \oplus (a_k \oplus b_k) \\ &= (a_k \oplus a_k) \oplus b_k \\ &= b_k. \end{align*}\]
Thus \(a_k = b_k\). But this contradicts the fact that the moving player moved on pile \(b_k\), and thus must make the size different.
Thus, since \(a_k \oplus b_k \neq 0\), we have
\[\begin{align*} t &= s \oplus (a_k \oplus b_k) \\ &= 0 \oplus (a_k \oplus b_k) \\ &= a_k \oplus b_k \\ &\neq 0. \end{align*}\]
Result 2: If \(s \neq 0\), it's possible to make \(t = 0\). If the nim-sum of the original sizes is not zero, the moving player is winning (they can make the nim-sum zero).
Consider the largest power of 2, \(2^k\), not greater than \(s\). There must be at least one \(a_i\) such that it also contains \(2^k\), otherwise \(2^k\) cannot appear in \(s\). Now, take \(b_i
= s \oplus a_i\). The value \(b_i\) decreases by \(2^k\), and increases by at most \(2^{k-1} + 2^{k-2} + \cdots + 2^0 = 2^k - 1\) (each remaining powers of 2 making up \(s\) adds to the value;
for example \(s = 2^2 + 2^1 + 2^0\) and \(a_i = 2^3 + 2^2\) gives \(b_i = 2^3 + 2^1 + 2^0\)), so \(b_i < a_i\). Moreover,
\[\begin{align*} t &= s \oplus (a_i \oplus b_i) \\ &= s \oplus \big(a_i \oplus (s \oplus a_i)\big) \\ &= (s \oplus s) \oplus (a_i \oplus a_i) \\ &= 0. \end{align*}\]
This proves the theorem. \(_\square\)
The strategy given above for misère Nim is correct: follow normal Nim strategy, except that when the moving player is going to make all pile sizes less than \(2\) stones, the moving player makes
the number of piles of \(1\) stone odd instead of even.
The only change is when the moving player needs to reduce the single pile having size \(2\) or more to less than \(2\) \((\)other piles have sizes no more than \(1).\)
Suppose the piles are \(a_1, a_2, \ldots, a_n\), where \(a_1 \ge 2\) and \(a_2, a_3, \ldots, a_n \le 1\). Then \(a_2 \oplus a_3 \oplus \ldots \oplus a_n = 0 \vee 1\). If \(a_1 \oplus a_2 \oplus \
ldots \oplus a_n = 0\), this would imply \(a_1 = 0 \vee 1\), contradiction. So the moving player is winning. It's also clear that the moving player can make the first pile to have only zero or
one stone, since it begins with at least \(2\) stones.
Once the moving player decides the move, the rest of the game is forced; with an odd number of \(1\)-sized piles and the opponent starting, the opponent will also take the last stone, thus
This proves the correctness. \(_\square\)
There are many variations of Nim. The most well-known ones are normal Nim and misère Nim; in several textbooks, normal Nim is considered to be the standard game and misère Nim is the variant, while
in others, it's the other way around. Due to Nim being a very basic game, there are so many possible variants obtained by simply adding one extra rule; this list doesn't intend to cover all
possibilities, only the well-known ones.
A variant is known as subtraction game. This is equivalent to Nim, but where the number of stones taken is limited to some set of positive integers: for example, the first \(k\) numbers or the square
numbers. Generally, this is played with one pile only.
Usually, Nim is played with the stones clumped together in heaps. A variant places the stones in rows so that taking stones in the middle of a row breaks the row into two. In other words, a player's
move is to take at least one stone from a pile and optionally split the pile into two piles. Kayles is played with a single pile, allowing splitting, but a player may take at most \(2\) stones at a
time. Dawson's chess is another variant of Kayles, where a player may take at most \(3\) stones at a time; however, taking one stone is only allowed if it's the only stone in a pile, and taking two
stones doesn't allow splitting. Circular Nim is played with stones initially arranged in a circle, so the first time the pile is played on, it cannot be split; additionally, one can only take no more
than \(3\) stones.
A generalization of Nim is the octal game. Players take turns removing several stones from a single heap in each turn, just as in usual Nim; however, the rules that govern when and how a heap can be
taken and/or split are compactly noted as an octal number. For the octal game \(\overline{0.d_1 d_2 d_3 d_4 \ldots}\), the digit \(d_n\) specifies how many heaps are allowed to be left after removing
\(n\) stones from a heap. \(d_n\) is the sum of
• \(1 = 2^0\) if leaving \(0\) heaps is permitted (the heap has exactly \(n\) stones), \(0\) otherwise;
• \(2 = 2^1\) if leaving \(1\) heap is permitted (the heap has more than \(n\) stones), \(0\) otherwise;
• \(4 = 2^2\) if leaving \(2\) heaps is permitted (take \(n\) stones from the heap and split the rest into two non-empty heaps), \(0\) otherwise.
As an example, Dawson's chess has the octal game notation \(0.137\):
• Removing \(1\) stone is only possible when it's the only stone in the pile, so \(d_1 = 1+0+0\).
• Removing \(2\) stones is possible at any time, but it doesn't allow splitting the pile, so \(d_2 = 1+2+0\).
• Removing \(3\) stones is possible at any time, including splitting the pile, so \(d_3 = 1+2+4\).
The regular Nim has octal game notation \(0.3333 \ldots\). Kayles has octal game notation \(0.77\).
Octal games have further variants. The first digit, corresponding to removing \(0\) stones, may be set into \(4\) to indicate that splitting a heap into two (without taking any stones) is a permitted
move. By using a higher base, it's possible to express games that allow splitting a heap into three or more heaps.
Index-\(k\) Nim is a variant of Nim where a player is allowed to take stones from more than one pile in a single turn. This can be combined with subtraction game, limiting the total number of stones
removed or the number of stones removed from a pile.
Wythoff's game is a variant of Index-\(k\) Nim, where the number of stones removed from all piles affected must be the same. Usually, Wythoff's game is played with only two piles. The strategy is
very different; the two-pile game is solved, involving Beatty's sequences using \(\varphi\) and \(\varphi^2\).
Grundy's game is even more different in that stones cannot be taken. The only allowed move is to split a pile into two differently sized piles. This will still end the game since the number of piles
will be limited by the number of stones and it keeps increasing.
Unbalanced Nim has the extra rule that any two heaps may not be equal in size. Heaps of zero objects may or may not be counted as heaps; in the former option, this makes the ending position
different, as \(0, 1, 2, \ldots, n-1,\) where \(n\) is the number of heaps.
Even more quirky variants are greedy Nim, where a player may only take stones from the currently largest pile—any such largest pile if there are many—and building Nim, where the players begin with
placing \(n\) stones into \(p\) piles before playing a game of Nim with the result, so that the starting position is not known until the end of the building phase.
Nimbers and Sprague-Grundy Theorem
The Sprague-Grundy theorem is more fully developed out on this page.
The Sprague-Grundy theorem
Any position of an impartial game is equivalent to a Nim pile of a certain size.
Two players play a game according to the following rules:
• They place three heaps of coins on a table. The first heap has \(10\) coins, the second heap has \(7\) coins, and the third heap has \(9\) coins.
• The second player adds another pile of coins on the table, having at most \(10\) coins.
• The players take turns alternately, starting with the first player. At each move, the player has to remove a positive number of coins from one heap. The player who removes the last coin wins.
It turns out that regardless of the strategy of the first player, the second player always wins with optimal play. How many coins should the second player add in the fourth pile?
Dan and Sam play a game on a \(5\times5\times5\) cube that consists of 125 objects; each one must take at least one object and at most a whole heap, in his turn.
As an explicit example, in the first turn, Dan can take just the bottom front left corner object, or the whole vertical central heap, or the whole bottom front horizontal heap. Thus, he can take any
number of objects of one of the 125 initial heaps (heaps cannot be diagonals).
The winner is the one who takes the last object. If Dan begins, who will win? This means, who has a winning strategy?
This is the fifteenth problem of the set Winning Strategies.
Dan and Sam play a game on a \(4\times9\) grid, in which one takes squares (red) and the other takes circles (blue). Once the game starts, they take turns moving a single piece in their turn. Each
piece can only be moved straight forward or backward, any number of grids (and cannot skip over the opponent's piece). This is the initial position:
A player loses when he is not able to move any of his pieces in his turn. If Dan goes first, who will win? In other words, who has a winning strategy?
This is the twelfth problem of the set Winning Strategies.
Alice and Bob are playing a game called Moving Chips. The game is played on an \(n \times m\) board where some cells have some chips on it. Two players move alternately. Each turn consists of one
person moving a chip to any cell to its left or any cell to its top. For example, the possible movements of chip A on a \(3\times 5\) board are as follows:
The last player who moves all the chips to the top left cell wins.
Consider the configuration below. Alice will move first. Assuming that both players move optimally, who will win the game?
• The chips can be stacked on top of one another.
• The chips can move past any chips.
Do you have a generalized strategy for this problem? Then you might enjoy the Large Data version!
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T Distribution Table - StatCalculators.com (2024)
T Table
Given below is the T Table, otherwise known as the Student’s T-table or T-distribution table. This T table contains both one-tailed T-distribution and two-tailed T-distribution, degrees of freedom up
to 1000, and a confidence level up to 99.9%.
Use this T-Distribution Table to lookup T critical value for confidence level & degrees of freedom for one tail & two-tails.
Related Calculators
Student t-Value Calculator Effect Size (Cohen's d) for a Student t-Test Calculator p-Value Calculator for a Student t-Test T-Statistic and Degrees of Freedom Calculator
What Is a T-Distribution?
Otherwise known as the Student’s T-distribution, the T-distribution is a type of probability distribution, with its bell shape, that is similar to the normal distribution, though it has heavier
tails. T distributions have fatter tails, therefore, a greater chance for extreme values than normal distributions.
What Does a T-Distribution Tell?
A parameter of the T-distribution called degrees of freedom determines tail heaviness. Higher values of the mentioned parameter make the T-distribution resemble a standard normal distribution with a
mean of 0, and a standard deviation of 1. Smaller values of this parameter give heavier tails.
When utilizing the estimated standard deviation, a T-score is calculated as:
T = (m – M)/{d/sqrt(n)}, rather than making the normal distribution with mean 0 and standard deviation 1, the contrast between d and D makes the distribution a T-distribution with (n – 1) degrees of
How To Utilize The T-Table?
Further, we are going to learn how to read the T-Table and map critical values on it using examples, but first, we will require a few things or pre-requisites before we can do that.
The pre-requisites needed to use a T-table are as follows:
The number of tails:
Firstly, you need to know whether the T-test is one-tailed or two-tailed because we will use the respective one-tail or two-tail row to mark the alpha level. The alpha levels are listed at top of the
table [0.50, 0.25, 0.20, 0.15…for the one-tail and 1.00, 0.50, 0.40, 0.30, etc. for the two-tails] and as you can see, they differ based on whether the T-test is one-tailed or two-tailed.
Find out Z Score Table here
Degrees of freedom:
The degrees of freedom [df] show the number of independent values that can differ in an analysis without breaking any constraints. The degrees of freedom will either be explicitly cited in the
problem statement or if it is not explicitly cited, then all you have to do is subtract one from your sample size (n – 1), and the result you get will be your degrees of freedom.
Alpha level:
The significance level, otherwise known as the alpha level (α), is the probability of rejecting the null hypothesis when it is true. The common alpha (α) levels for the T-test are 0.01, 0.05 and 0.10
Once you have all three significance levels, you have to pick the respective column for one-tail or two-tail from the table and map the intersection of the values for the degrees of freedom [df] and
the alpha (α) level.
Example Questions:
Example #1 – Let’s say we want to map a one-tailed t-test for a mean with an alpha level of 0.05. The total number of students involved in this study is 25. To what critical value t should be
Solution – Firstly, we see that there are 25 students involved in this study. We have to subtract 1 from the sample size to get the degrees of freedom [df]. Therefore, df = n – 1 = 25 – 1 = 24.
Example #2 – For a study involving one population and a sample size of 18 (assuming you have a t-distribution), what row of the t-table will you use to find the right-tail – “greater than” –
probability associated with the study results?
A sample size of 18 has n – 1 = 18 – 1 = 17 degrees of freedom when the study involves one population.
Solution – df = 17
Example #3 – For a study involving a paired design with a total of 44 observations, with the results assuming a t-distribution, in order to find the probability affiliated with the study results,
what row of the table will you use?
22 pairs are in a matched-pairs design with 44 total observations. The degrees of freedom [df] is one less than the number of pairs: n – 1 = 22 – 1 = 21.
Solution: df = 21
Example #4 – A t-value of 2.35, from a t-distribution with 14 degrees of freedom, between which two values has an upper-tail – “greater than” – probability on the t-table?
Find the row with 14 degrees of freedom and look for 2.35 utilizing the T-table. However, this exact value doesn’t lie in this row, so look for the values on either side of it: 2.1448 and 2.6245. The
upper-tail probabilities appear in the column headings; the column heading for 2.1448 is 0.025, and the column heading for 2.6245 is 0.010.
Therefore, the upper-tail probability for a T-value of 2.35 must lie between 0.025 and 0.010.
Solution: 0.025 and 0.010.
Check out, Critical Chi-square calculator her
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Corsi di studio e offerta formativa - Università degli Studi di Parma
Learning objectives
Knowledge and understanding:
the theory of vector spaces.
Applying knowledge and understanding:
a) solve systems of linear equations;
b) diagonalize (symmetric) matrices;
c) solve easy problems of analytic geometry;
d) recognize the type of a conic and write its canonical form.
Making judgements:
evaluate the correctness of a simple proof.
Communication and learning skills:
properly express themselves with mathematical language.
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24.3 A* Algorithm | CS61B Textbook
We ended the section on Dijkstra's by discussing a possible way to make Dijkstra's short-circuit and just stop once it hits a given target. Is this good enough?
To answer the above question, we need to sit down and think about how dijkstra's really works. Pictorially, Dijkstra's starts at the source node (imagine the source node being the center of a
circle.) And Dijkstra's algorithm now makes concentric circles around this point, in increasing radii, and 'sweeps' these circles, capturing points.
So... the first node Dijkstra's visits is the city closest to the source, then the city next-closest, then the city next-closest, and so on. This sounds like a good idea. What Dijkstra's is doing is
first visiting all the cities that are 1-unit distance away, then 2 unit-distance away, and so on. In concentric circles.
Now imagine the following: on a map of the US, start somewhere in the center, say, Denver. Now I want you to find me a path to New York using Dijkstra's. You'll end up traversing nodes in 'closest
concentric circle' order.
You'll make a small circle first, just around Denver, visiting all the cities in that circle. Eventually, your circles will get bigger, and you'll make a circle that passes through Las Vegas (and
would have visited, by now, all the other cities that fall within the circle.) Then, your circle will be big enough to engulf Los Angeles and Dallas... but you're nowhere close to New York yet. All
this effort, all these circles, but still... so far from the target. Short-circuiting helps, but only if you actually hit the target node fast.
If only there existed a way to use your prior knowledge: the fact that new-york was eastwards, so you could "hint" your algorithm to prefer nodes that are on the east instead of those that are on the
Introducing: A Star
No, not the sun. It's an algorithm called A*.
Observe the following: Dijkstra's is a "true" (i.e., not an estimate) measure of the distance to a node from the source. So, say, you visit a city in Illinois and your source was Denver, then by that
time, you have a true measure of the distance to Denver. What we're missing is: some janky, rough estimate of the distance from a node to the target node, New York. That would complete the picture.
Because then, if you sum these two things up (the measure from the source to the node + the estimate from the node to the target), you get (an estimate from the source to the target.) Of course, the
better your original estimate from the node to the target, the better your estimate from the source to the target, the better your A* algorithm runs.
So, let's modify our Dijkstra's algorithm slightly. In Dijkstra's, we used bestKnownDistToV as the priority in our algorithm. This time, we'll use bestKnownDistToV+estimateFromVToGoal as our
Here is a demo!
Chicken And Egg
We have a problem. How do we know what the estimate is? I mean, the estimate itself is a distance, and we're using A* to find the distance from some node to some other node.
It seems like we're in an instance of the classic chicken and egg problem. "What came first? The chicken or the egg?" Aside, FYI, one reddit user had an idea about this.
Well, it's called an estimate because it's exactly that. We use A* to get the true shortest path from a source to a target, but the estimate is something we approximate. Coming up with good estimates
is hard sometimes.
But to give you an example in our Denver - New York case. What we might do is just look up the GPS Coordinates of these cities, and calculate the straight line distance between those somehow. Of
course, this wouldn't be correct because there's probably no straight line that one could take from Denver to NYC, but it's a fairly good estimate!
Bad Heuristics
What will happen? Well A* will basically never want to visit this city. (Remember what our priorities are in the priority queue; for this city, the priority will always be ∞∞, even if I visit the
immediate neighbors of this city. The estimated distances from the immediate neighbors of this city to this city were set to ∞∞ after all.)
So... now what? We lose. A* breaks. We get the wrong answer back. Oops.
The takeaway here is that heuristics need to be good. There are two definitions required for goodness.
heuristic(v, target) ≤≤ dist(v, w) + heuristic(w, target)
where dist(v, w) is the weight of the edge from v to w.
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Long multiplication: 2-digits by 2-digits (1) - Multiplication by URBrainy.com
Long multiplication: 2-digits by 2-digits (1)
The standard, or efficient method of multiplying two 2-digit numbers.
5 pages
Long multiplication: 2-digits by 2-digits (1)
The standard, or efficient method of multiplying two 2-digit numbers.
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Cubic Formula Example
Cubic equation Wikipedia
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In algebra, a cubic equation in one variable is an equation of the form
in which a is nonzero.
The solutions of this equation are called roots of the cubic function defined by the left-hand side of the equation. If all of the coefficients a, b, c, and d of the cubic equation are real numbers,
then it has at least one real root (this is true for all …
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Solving Cubic Equations (solutions, examples, videos)
3 hours ago WebIn these lessons, we will consider how to solve cubic equations of the form px 3 + qx 2 + rx + s = 0 where p, q, r and s are constants by using the Factor Theorem and Synthetic
Division. The following diagram shows an …
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Cubic Equations Brilliant Math & Science Wiki
3 hours ago WebA cubic equation is an equation which can be represented in the form ax^3+bx^2+cx+d=0 ax3 +bx2 + cx+d = 0, where a,b,c,d a,b,c,d are complex numbers and a a is non-zero. By …
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Cubic Equations Formula, Examples & Practice Problems
2 hours ago WebNov 21, 2023 · Example 1. Solve the cubic equation and graph the equation using the solutions: 2 x 3 − 9 x 2 + 4 x + 15 = 0. Step 1: Set one side of the equation equal to zero and
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Cubic Formula from Wolfram MathWorld
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Solving The General Cubic Equation Emory University
5 hours ago WebThis gives us the system: m3 = − 27a3b3 n = − (a3 + b3) Now let us hide the detail of the cubing, so we can better see the structure of what is left, by letting A = a3 and B = b3 : m3
= − 27AB n = − (A + B) Now, it is clear …
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THE CUBIC FORMULA Department of Mathematics
5 hours ago WebExample 4: It is clear that x = 1 is a root of the cubic x. 3. + 3 x − 4 = 0 . Use the cubic. formula to obtain a surprising expression for this root. Solution: Comparing our cubic …
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The Cubic Formula Vanderbilt University
1 hours ago WebThe Cubic Formula (Solve Any 3rd Degree Polynomial Equation) But if we apply Cardano's formula to this example, we use a=1, b=0, c=-15, d=-4, and we find that we …
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Cubic Equation Formula: Definition, Derivation, Types, Examples
2 hours ago WebJun 22, 2023 · Cubic Equation Formula: An equation is a mathematical statement with an ‘equal to’ sign between two algebraic expressions with equal values. In algebra, there …
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The Cubic Formula University of Utah
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Polynomials I The Cubic Formula University of California, …
8 hours ago WebPolynomials I - The Cubic Formula Yan Tao Adapted from worksheets by Oleg Gleizer. 1 Cubic Equations by Long Division Definition 1A cubic polynomial (cubic for short) is a …
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4 hours ago WebMar 20, 2024 · Cubic Equation is a mathematical equation in which a polynomial of degree 3 is equated to a constant or another polynomial of maximum degree 2. The standard …
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How To Solve Cubic Equations YouTube
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Volume of a Cube Math Steps, Formula, Examples & Questions
8 hours ago WebFor example, cubic inches (in^3), cubic meters (m^3), or cubic centimeters (cm^3). For example, The volume of this cube is, volume = a^3 . volume = 8^3 . Students should …
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MATH 4552 Cubic equations and Cardano’s formulae Ohio …
6 hours ago Webcubic root of unity.) To obtain (6), change u by multiplying it by a suitable cubic root of unity; then, both (6) and (7) will be satis ed. Formula (5) now gives a solution w= w 1 to …
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Cubic Function Definition, Equation & Examples Study.com
Just Now WebNov 21, 2023 · A cubic function is a polynomial of degree 3, meaning 3 is the highest power of {eq}x {/eq} which appears in the function's formula.The simplest example of such a …
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3 hours ago WebCubic Unit Definition. In geometry, cubic units can be defined as the units used to measure volume. The volume of a unit cube whose length, width, and height are 1 unit each is 1 …
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Normal force in physics: definition with examples and calculation
Imagine that you are sitting quietly on a chair or standing on the floor. Even if you don't think about it, a force is being produced to prevent you from falling or falling through the floor or the
Have you ever wondered why, when you leave an object on a table, it doesn't go through the table and fall to the floor? This is where a "special" force comes into play in our daily life: the normal
In physics, this is one of the most common and essential forces, but it often goes unnoticed. It doesn't have a very flashy name, but without it, our everyday experiences would be very different.
Below I will explain what exactly this force is, how it works and why it is so important for things, including us, to stay in place.
What is normal force?
Normal force is the contact force that a surface exerts on an object when it is resting on it. This force is generated as a reaction to other forces acting on the object, such as its weight due to
gravity, and is responsible for the object not passing through the surface or sinking into it.
In short, this force is the response of the surface that counteracts the forces applied to the object, keeping it in balance.
Graphic explanation
Well, that surface is doing something that we don't notice with the naked eye: it is pushing up against us to keep us balanced. That upward push is what we call the normal force .
It is a force that acts perpendicularly (at a right angle) to the surface on which an object is resting. This force is like the response of the surface to the weight of the object above it. The same
thing happens if we leave a book on a table or if we rest our hand on a wall.
Normal force , like any other force, is measured in Newtons (N).
A Newton is the standard unit of force in the International System of Units (SI), and is defined as the force required to accelerate a mass of 1 kilogram to a rate of 1 meter per second squared.
A simple example
I'll give you an example that you'll probably be familiar with: imagine you're in a park, you're sitting on a bench, and you notice how the bench supports you without you falling to the ground.
What's happening?
The bench is exerting an upward force, which is the normal force , to counteract the downward force that gravity exerts on you, that is, your weight.
So when you are sitting there are two main forces at play:
1. Your weight , which is the force that gravity exerts downwards.
2. The normal force , which is the response of the bench pushing up.
These two forces balance each other out, which is why you don't fall through the floor or go through the bench. So basically, the normal force is responsible for keeping us from sinking into things!
Why is it called "normal"?
The word "normal" in this case has nothing to do with what we use in everyday life (like when we say "that's normal" or "this is weird"). Here, "normal" means perpendicular.
It is a mathematical term used in physics to refer to a specific direction that forms a 90-degree angle with a surface.
So when we say "normal force," we're talking about a force that always points in that direction, outward and perpendicular to the surface.
How exactly does it work?
Now that we understand the basics, let's dig a little deeper into how the normal force works. This force doesn't have a fixed value. It doesn't always push upward with the same intensity. Its value
depends on several factors, and one of the most important is the weight of the object .
If the book weighs 2 kilograms, the table will push upwards with a force equal to the weight of the book. But if you add more books on top, the force exerted by the (normal) table will have to
increase to continue to support the extra weight.
It should be noted that if there are other forces in addition to weight, these also intervene in the normal force. For example, if we push the book downwards with our hand, the normal force will be
the sum of the weight plus the force that we exert downwards.
Calculating normal force with formulas
We distinguish two special cases: when the object is on a horizontal surface and when it is on an inclined plane.
Calculation on a horizontal surface
The basic formula for calculating normal force, in situations where the object is on a horizontal surface with no inclinations or extra forces, is quite simple:
F [n] = m⋅g
• m is the mass of the object in kilograms (kg).
• g is the acceleration of gravity, which on Earth is approximately 9.8 m/s^2.
For example, if you have an object that weighs 10 kg, its weight would be:
Weight=10 kg⋅9.8 m/s ^2 =98N
And the normal force would also be 98 Newtons (N), because the surface has to exert an equal and opposite force to hold the object in balance.
Calculation on an inclined plane
inclined plane. This is a bit more complicated than on a flat surface, because not all of the force of the object's weight is acting directly against the surface, as some of that force is "sliding"
down along the slope.
When you place an object on an inclined surface (such as a ramp or hill), its weight still acts downward due to gravity, but it is broken down into two components:
1. A component parallel to the inclined surface: This is the part of the weight that "pushes" the object down along the slope, which is why objects can slide on a ramp.
2. A component perpendicular to the inclined surface: this is the part of the weight that "presses" directly against the ramp, and this is what generates the normal force.
Let's see how it is calculated in this case.
Weight decomposition on an inclined plane
To calculate the normal force on an inclined plane, the first thing we have to do is break down the force of gravity into these two parts that I mentioned before: one parallel and one perpendicular.
If the angle of inclination of the ramp is θ (the angle between the inclined surface and the flat ground), we can use trigonometry to calculate each of those components.
The total weight of the object is:
Where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s ^2 ).
• Perpendicular component (which is the one that generates the normal force):
This is the part that interests us, because it is the one that "rests" on the ramp and is related to the normal force. The formula to calculate this component is:
Here, cos(θ) is the cosine of the angle of inclination.
• Parallel component (which makes the object tend to slide):
It does not directly affect the normal force, but it is useful to know it to understand the movement of the object on the ramp. It is calculated like this:
Calculation of normal force
Once we have broken down the weight into these two parts, we can calculate the normal force (F [n] ) . The F [n] on an inclined plane will be equal to the perpendicular component of the force of
gravity, since it is this that the surface of the ramp has to counteract.
Therefore, the normal force is:
F [n] = m⋅g⋅cos(θ)
Practical example
Imagine that you have a 5 kg block placed on a ramp inclined at 30 degrees to the ground. We want to calculate the normal force exerted by the ramp on the block.
First, we calculate the weight of the block:
Weight=5 kg⋅9.8 m/s2=49 N
Now we use the formula, with the inclination angle of 30 degrees:
Fnormal=49 N⋅cos(30∘)
The cosine of 30 degrees is approximately 0.866, so:
Fnormal=49 N⋅0.866=42.4 N
Therefore, the normal force that the ramp exerts on the block is 42.4 Newtons .
Why is it smaller than on a flat surface?
Note that the normal force on an inclined ramp is less than if the same block were on a flat surface. If the block were on a horizontal surface (where the angle θ is 0 degrees), the cosine of 0 is 1,
so the normal force would be equal to the entire weight, or 49 N.
In the case of the ramp, since the angle of inclination is 30 degrees, only part of the weight is being "supported" on the ramp, and that is why the normal force is less, 42.4 N instead of the 49 N
that we would have on a flat surface.
The other component of the force is the one that would cause a downward acceleration of the body unless there was a friction force in the opposite direction to compensate it.
This also explains why it is easier for an object to slide down an inclined slope: since the normal force is smaller, there is less friction, and the part of the weight acting down the slope (the
parallel component) helps the object move.
Special situations
So far we have talked about very simple examples, but this force can also behave in interesting ways in more complex situations.
Below are some special examples that will help you better understand this concept:
1. Earrings
Imagine you are walking up a steep hill. Have you noticed that it is easier to slip on a slope? This has to do with the normal force.
When an object is on an inclined surface, the normal force is not as large as when the surface is flat, because the inclination causes part of the gravitational force to act "pushing" down the slope.
On a hill, the normal force doesn't have to offset the entire weight of the object, only part of it. And because it's smaller, there's less resistance and it's easier for you to slip.
2. Pushing against a wall
Now, imagine you are pushing a box against a wall.
In this case, the normal force is not related to the weight of the box, but to the pressure you are exerting. The harder you push the box against the wall, the greater the force the wall will exert
on the box to prevent it from passing through.
If you stop pushing, the normal force disappears.
3. Normal force on an elevator
Have you ever felt like you weigh more or less in a moving elevator?
This also has to do with this type of force. If the elevator goes up at an accelerated rate, the normal force increases (and we feel as if we weigh more), because the floor of the elevator has to
push upwards with greater force to compensate for the acceleration.
If the elevator accelerates downwards, the normal force decreases (and we feel like we weigh less), because the acceleration reduces the need to push upwards as hard.
Relationship with Newton's third law: action and reaction
This law states that for every action, there is a reaction of equal magnitude and in the opposite direction . That is, if one object exerts a force on another, that other object responds with an
equal force but in the opposite direction.
In the case of the normal force, imagine a skater gliding across an ice rink. The skater's weight, which is the action force , pushes downward due to gravity.
In response, the ice exerts an upward force to counteract this weight and prevent it from breaking and the skater from passing through. The magnitude of this force is equal to the skater's weight,
but it acts in the opposite direction, that is, upwards.
In short, normal force is one of those things that is present all the time, but we usually don't realize it exists.
It is the force that prevents us from passing through the ground or objects from sinking into surfaces. It depends on the weight of the object and always acts perpendicular to the surface on which it
is resting. In addition, it is involved in all kinds of everyday situations, from sitting to walking up a slope.
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Figure A to Figure B 5
Google ClassroomGeoGebra Classroom
GeoGebra Classroom
Figure A to Figure B 5
Show that the two triangles are congruent by identifying a series of rigid motion transformations that maps Figure A to Figure B. Use the tools in Geogebra to show that your series of transformations
English / English (United States)
© 2024 GeoGebra®
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Posts about python (old posts, page 7)
It has been way too long without posting a longer item, so... I recicled a script I wrote for a customer, and here it is:
A python script that lists (almost) all email addresses in a qmail system.
Also, a slightly tweaked CSS, thanks to Georg!
Hard Python question
I am trying to do something which is, I think, pretty cool, in python.
However, I like showing things working, and I am having troubles with the last final step on what I am trying to achieve.
Since I know a few better python programmers read this...
Suppose I have this:
What code should be in fun() so that it figures out if it has been called as C.a or as C.b?
I am thinking something like reading the higher step in a backtrace, but I don't know enough python to figure out how to do that.
Custom widgets using PyQt
A short tutorial explaining how to create reusable widgets using PyQt.
In a whim, I checked out kdebindings/dcoppython from KDE's CVS.
I see the README: dcopython is broken
Then I said to myself: maybe I can fix it. And you know what? It seems to be not broken! :-)
At least for simple data types, that is.
dcoppython lets your python program become a DCOP server or client.
A DCOP server is capable of being controlled by KDE's kdcop, and is a very simple way to make your application externally scriptable.
A DCOP client is something that contacts a DCOP server, so that means you can control and script KDE applications (or other DCOP servers) from python scripts.
The neatest thing here is that this stuff doesn't require Qt!
I intend to use it to make some of my apps externally scriptable without PyKDE.
Goats and cars
There's a problem often used to show the unintuitive nature of probability, which has become very well known.
In that problem a contestant in a gameshow has to choose between three doors (A,B,C), on one there is a car, on the other two are goats.
After the contestant chooses, the host opens another door and shows a goat.
Then, the host offers the contestant the chance to switch his closed door for the other closed door.
Should he switch?
The intuitive answer is "it doesn't matter", because there's two doors and one car, so it's a 50-50 chance.
But the real answer is that it does matter, because it's a 33-67 chance!
While it's simple to show this to be the case to a statistically-educated dude, it's somewhat harder for a layman.
In fact, I think most explanations suck.
Here's my shot at it:
If you were offered the chance to switch between your closed door and the other two closed doors, would you take it?
The intuitive answer to that is of course, yes, because it's 67-33 for the car to be on the other two doors.
Now, regardless of where the car is, can the host open one of those two doors and show a goat? Of course, yes.
So, would you feel your odds went down because the host showed one of your two closed doors had a goat behind it? No, because he could always do that, and you know there was (at least) one goat
So, what difference does it make if one door is open or not?
I don't expect this to convince anyone, really, but just in case, I have a python implementation of this problem (goatcar.py :-) if anyone wants it, if empiricism can convince you ;-)
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PIK Interest | How to Model PIK Interest? | How does PIK Interest Accrue
Updated July 11, 2023
What is PIK Interest?
The term “PIK interest” or payment-in-kind interest refers to the option that the borrower is provided to pay the interest on debt instruments or other securities in an alternative way instead of
immediate cash payment.
Bondholders can receive payment in accumulated interest at maturity or the issuance of additional securities or equity. Payment-in-kind interest is very attractive for companies that suffer a cash
crunch or are in the growth phase.
The bondholder does not receive any cash interest payments until the debt instruments are redeemed or reach maturity in the case of payment-in-kind interest. Effectively, payment-in-kind interest
allows the issuers to postpone the cash outflow, and in return, the issuers offer a higher rate of return on these debt instruments.
How to Model PIK Interest?
To model payment-in-kind interest are as follows:
• Firstly, determine the years for payment-in-kind interest for the particular debt instrument. Basically, it signifies the period during which there will be no cash interest payment.
• Next, build the debt schedule initially assuming no payment-in-kind interest. The debt schedule should include the amortization and any other prepayment plans.
• Next, calculate the interest payment for the corresponding period based on the interest rate, opening balance, and outstanding balance.
Outstanding Balance = Opening Balance – Amortization – Prepayment
PIK Interest = Interest Rate * (Outstanding Balance + Opening Balance) / 2
• Next, add the accrued payment-in-kind interest to the outstanding principal at the end of the period, which will increase the ending balance of the debt instrument. So, the ending debt balance is
revised, becoming the opening balance for the next period.
Ending Balance = Outstanding Balance + PIK Interest
• Finally, the non-cash interest expense (PIK interest) must be added to the cash flow statement as the net income includes paying all interest expenses.
Examples of payment-in-kind interest are as follows :
Example #1
Let us take the example of a PIK loan of $15,000 as of January 2020. The loan has a payment-in-kind interest rate of 10% and has to be repaid in equal annual installments over 3 years. Determine the
amount that has to be paid on 31^st December 2022.
• Given the Opening balance of [2020] = $15,000
• Amortization = $15,000 / 3 = $5,000
• Interest rate = 10%
The outstanding balance of [2020] calculates as
Outstanding balance [2020] = Opening balance [2020 ]– Amortization
• Outstanding balance [2020 ]= $15,000 – $5,000
• Outstanding balance [2020 ]= $10,000
payment-in-kind interest accrued for the year 2020 can calculate as,
payment-in-kind interest [2020]= (Opening balance [2020] + Outstanding Balance [2020]) / 2 * Interest Rate
• PIK interest [2020 ]= ($15,000 + $10,000) /2 * 10%
• PIK interest [2020 ]= $1,250
The ending Balance at the end of 2020 can calculate as,
Ending Balance [2020]= Outstanding Balance [2020] + PIK Interest [2020]
• Ending Balance [2020 ]= $10,000 + $1,250
• Ending Balance [2020 ]= $11,250
The outstanding Balance at the end of 2021 can calculate as,
Outstanding Balance [2021] = Opening balance [2021] – Amortization
• Outstanding Balance [2021 ]= $11,250 – $5,000
• Outstanding Balance [2021 ]= $6,250
The payment-in-kind interest accrued for the year 2021 can calculate as,
PIK Interest [2021] = (Opening Balance [2021] + Outstanding Balance [2021]) / 2 * Interest Rate
• PIK Interest [2021 ]= ($11,250 + $6,250) /2 * 10%
• PIK Interest [2021 ]= $875
The ending Balance at the end of 2021 can calculate as,
Ending Balance [2021] = Outstanding balance [2021] + PIK interest [2021]
• Ending Balance [2021 ]= $6,250 + $875
• Ending Balance [2021 ]= $7,125
The outstanding Balance at the end of 2022 can calculate as,
Outstanding Balance [2022] = Opening balance [2022] – Amortization
• Outstanding Balance [2022 ]= $7,125 – $5,000
• Outstanding Balance [2022 ]= $2,125
payment-in-kind interest accrued for the year 2022 can calculate as,
PIK Interest [2022] = (OpenIng Balance [2022] + Outstanding Balance [2022]) / 2 * Interest Rate
• PIK Interest [2022] = ($7,125 + $2,125) /2 * 10%
• PIK Interest [2022] = $462.50
The ending Balance at the end of 2022 can calculate as,
Ending Balance [2022] = Outstanding Balance [2022] + Pik Interest [2022]
• Ending Balance [2022 ]= $2,125 + $462.50
• Ending Balance [2022 ]= $2,587.50
Therefore, the amount that has to be paid on 31^st December 2022 is $2,587.50.
Why is PIK Interest Appealing?
The payment-in-kind interest is very popular among borrowers, especially private equity professionals, for the following two reasons:
• It eases the interest expense burden of a project at its start and hence supports liquidity.
• Lower debt repayment obligations enable the companies to borrow more to support business requirements.
How Does PIK Interest Accrue?
The interest expense incurred during the period is not paid out in cash, rather, it is added back to the outstanding balance, and hence the ending debt balance increases due to accrued interest
expenses. Subsequently, the interest rate is applied to the higher ending balance the following year. Given the compounding nature, a PIK debt results in higher lifetime interest expenses than a
normal debt.
Some of the major advantages of payment-in-kind interest are as follows:
• It is suitable for borrowers facing liquidity issues due to growing business operations. The cash available due to non-payment of interest can be used for various business needs.
• Most PIK loans are unsecured in nature, which means that they don’t have any collateral requirements. It is useful for borrowers.
• The lenders or investors of PIK debt instruments earn a relatively higher return than normal debt instruments.
Some of the major disadvantages of payment-in-kind interest are as follows:
• The lenders don’t receive any cash income until redemption or maturity.
• The lenders are exposed to the risk of huge losses in case of default as these loans are mostly unsecured in nature.
So, it can be seen that despite the high-interest rates, PIK loans enjoy high demand among borrowers in the growth phase. However, the compounding nature of the interest increases the principal
repayment at maturity.
Recommended Articles
This is a guide to PIK Interest. Here we also discuss the introduction and how to model payment-in-kind interest. Along with advantages and disadvantages. You may also have a look at the following
articles to learn more –
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Can ChatGPT help construct non-trivial statistical models? An example with Bayesian "random" splines
I’ve been curious to see how helpful ChatGPT can be for implementing relatively complicated models in R. About two years ago, I described a model for estimating a treatment effect in a
cluster-randomized stepped wedge trial. We used a generalized additive model (GAM) with site-specific splines to account for general time trends, implemented using the mgcv package. I’ve been
interested in exploring a Bayesian version of this model, but hadn’t found the time to try - until I happened to pose this simple question to ChatGPT:
I want to build an R Stan model with cluster specific random splines - is that possible?
And as ChatGPT does, it eagerly let me know that this would not be a problem:
Yes, it’s possible to build a model in R using Stan with cluster-specific random splines. This approach involves using a hierarchical model where each cluster has its own spline, but the splines
share a common structure.
Here is a step-by-step outline to guide you through this process:
1. Data Preparation: Organize your data into clusters and ensure it’s in a suitable format for modeling.
2. Spline Basis Functions: Generate spline basis functions (e.g., B-splines) for your data. This can be done using packages like splines or mgcv in R.
3. Stan Model Code: Write the Stan code for the hierarchical spline model. The model will include cluster-specific random effects for the spline coefficients.
4. Data List for Stan: Create a list of data inputs required by the Stan model, including the spline basis functions and the data itself.
5. Run the Model: Use the rstan package to compile and fit the model.
Maybe I shouldn’t have been, but I was pretty impressed by the code that followed. What follows are the highlights of my interaction with ChatGPT.
Before getting started, here are the libraries needed to run all of the following code:
Data generation
To explore different modeling options, I wanted a simple data generation process to create the simulated data set. ChatGPT suggested using the following non-linear function for \(y_{ik}\), the
outcome for individual \(i\) in cluster \(k\), based on predictor \(x_{ik}\): \[ y_{ik} = \text{sin}(2\pi x_{ik}), \ \ \{ x \in \mathbb{R} \mid 0 \leq x \leq 1 \} \] The code was simple enough:
n <- 1000
k <- 10 # number of clusters
x <- runif(n)
cluster <- sample(1:k, n, replace = TRUE)
y <- sin(2 * pi * x) + rnorm(n, sd = 0.35)
dd <- data.table(y, x, cluster)
dd$cluster <- factor(dd$cluster)
Although the data generation process suggested by ChatGPT was helpful, it had a significant shortcoming. I wanted to model cluster-specific spline curves, but the ChatGPT code generated the same
curve for all clusters. To address this, I used the general formulation and added a cluster-specific effect \(a_k\), which stretches the sin curve differently for each cluster: \[ y_{ik} = \text{sin}
(2\pi a_k x_{ik}), \ \ \{ a \in \mathbb{R} \mid 0.6 \leq a \leq 1.4 \} \]
k <- 10 # number of clusters
defc <- defData(varname = "a", formula = "0.6;1.4", dist = "uniform")
defi <-
defDataAdd(varname = "x", formula = "0;1", dist = "uniform") |>
varname = "y",
formula = "sin(2 * a * ..pi * x)",
variance = 0.35^2
dc <- genData(k, defc, id = "cluster")
dd <- genCluster(dc, "cluster", 100, "id")
dd <- addColumns(defi, dd)
dd[, cluster := factor(cluster)]
Data modeling
The goal is to estimate cluster-specific curves that capture the relationship between \(x\) and \(y\) within each cluster. I am aiming for these curves to reflect the overall trend without
overfitting the data; in other words, we want the estimated function to provide a smooth and interpretable representation of the relationship, balancing flexibility and simplicity.
Although the purpose of my conversation with ChatGPT was to get a Bayesian version of this random spline model, I started off by asking for it to generate a generalized additive model (GAM) to
provide a basis for comparison. This is what it came up with: \[ y_{ik} = \beta_0 + s_k(x_{ik}) + \epsilon_{ik}, \ \ \epsilon \sim N(0, \sigma_y) \]
where \(s_k(x)\) is a smooth spline function of \(x\). The estimated model can be used to provide predictions that can be plotted to describe the relationship between \(x\) and \(y\):
gam <- gamm(
y ~ s(x) + s(x, cluster, bs = "fs", k = 8),
data = dd, method="REML"
dd$g <- predict(gam$gam)
Bayesian spline model
The first Bayesian model that ChatGPT generated can be described using this notation:
• \(N\): number of individuals
• \(K\): number of clusters
• \(M\): number of spline basis functions
• \(y_{ik}\): outcome for individual \(i\) in cluster \(k\), \(i \in 1,\dots,N\), \(k \in 1,\dots ,K\)
• \(\boldsymbol{X} \in \mathbb{R}^{N \times M}\): matrix of spline basis function values
• \(\boldsymbol{\beta_{k}} \in \mathbb{R}^M\): spline coefficients for cluster \(k\) (a vector of length \(M\) for each cluster)
• \(\sigma_y\): standard deviation of the observation noise
• \(\sigma_\beta\): prior standard deviation for the spline coefficients
\[ y_{ik} \sim N\left( \sum_{m=1}^M X_{im} \beta_{km}, \sigma_y \right), \ i \in 1,\dots, N, \ k \in 1, \dots, K\]
\[ \boldsymbol{\beta_{k}} \sim N(0, \sigma_{\beta} \boldsymbol{I_M}), \ \ k \in 1,...,K \\ \sigma_{y} \sim N(0, 1), \ \ \sigma_y \gt 0 \\ \sigma_{\beta} \sim N(0, 1), \ \ \sigma_{\beta} \gt 0 \]
The Stan code provided by ChatGPT aligns with this description. As part of the model, I also requested code to generate outcome predictions for each observation, which is implemented in the generated
quantities block. My goal was to plot the median of those predictions for each individual \(i\) as a comparison to the GAM plot above.”
data {
int<lower=1> N; // number of observations
int<lower=1> K; // number of clusters
int<lower=1> M; // number of basis functions
array[N] int<lower=1, upper=K> cluster; // cluster ids
matrix[N, M] X_spline; // basis function values
vector[N] y; // response variable
parameters {
matrix[K, M] beta; // cluster-specific spline coefficients
real<lower=0> sigma_y; // observation noise
real<lower=0> sigma_beta; // prior standard deviation for beta
model {
sigma_y ~ normal(0, 1);
sigma_beta ~ normal(0, 1);
// Priors for beta
for (k in 1:K) {
beta[k] ~ normal(0, sigma_beta);
// Likelihood
for (n in 1:N) {
y[n] ~ normal(X_spline[n] * beta[cluster[n]]', sigma_y);
generated quantities {
vector[N] y_pred; // Vector of observations.
for (n in 1:N) {
y_pred[n] = normal_rng(X_spline[n] * beta[cluster[n]]', sigma_y);
Spline basis functions
In the likelihood, \(y_i\) is modeled as a function of the vector \(\boldsymbol{X_i}\) rather than the single measurement \(x_i\). While I won’t delve deeply into spline estimation, I want to
conceptually outline how this vector is constructed in the context of cubic splines.
We control the flexibility of the curve by specifying the number of knots. A unique curve is fitted between each pair of knots (as well as at the ends), with constraints ensuring smooth transitions
between these curves. The estimation of these curves is performed using basis functions, specifically B-spline basis functions of \(x\).
The number of basis functions is determined by the number of knots. For instance, the plot below illustrates the \(M=9\) basis functions required for \(K=5\) knots. Each basis function contributes an
element to the vector \(\boldsymbol{X}\) for each value of \(x\). In the case of cubic splines, at most four basis functions can be non-zero between any two knots, as indicated by the intervals on
the x-axis. Consequently, the vector \(\boldsymbol{X}\) consists of the values of each basis function at a given point \(x\), with at most four non-zero entries corresponding to the active basis
functions. (As an example, in the plot below there is a vertical line at a single point \(x\) that passes through four basis functions.)
This example uses \(M = 5\) knots to introduce a slight overfitting of the data, which will allow me to apply another model in the next step that will further smooth the curves. (In a real-world
setting, it may have made more sense to start out with fewer knots.) The bs function (in the splines package) computes the B-spline basis function values for each observed \(x\).
n_knots <- 5
knot_dist <- 1/(n_knots + 1)
probs <- seq(knot_dist, 1 - knot_dist, by = knot_dist)
knots <- quantile(dd$x, probs = probs)
spline_basis <- bs(dd$x, knots = knots, degree = 3, intercept = TRUE)
X_spline <- as.matrix(spline_basis)
Data list for stan
To fit the model, we need to create the data set that Stan will use to estimate the parameters.
stan_data <- list(
N = nrow(dd), # number of observations
K = k, # number of clusters
M = ncol(X_spline), # number of basis functions
cluster = dd$cluster, # vector of cluster ids
X_spline = X_spline, # basis function values
y = dd$y # response variable
Run stan model
ChatGPT provided code to estimate the model using the rstan package. However, I prefer using the cmdstanr package, which I find more stable and generally less finicky. From the plot, you can see that
the estimation was quite good. However, the curves are a bit too wiggly, indicating the data may have been slightly overfit, particularly for clusters 1, 3, and 7.
mod <- cmdstan_model("code/spline.stan")
fit <- mod$sample(
data = stan_data,
chains = 4,
iter_warmup = 500,
iter_sampling = 2000,
parallel_chains = 4,
refresh = 0 # print update every 500 iters
## Running MCMC with 4 parallel chains...
## Chain 2 finished in 5.4 seconds.
## Chain 1 finished in 5.6 seconds.
## Chain 3 finished in 5.6 seconds.
## Chain 4 finished in 5.8 seconds.
## All 4 chains finished successfully.
## Mean chain execution time: 5.6 seconds.
## Total execution time: 5.9 seconds.
draws <- as_draws_df(fit$draws())
ds <- summarize_draws(draws, .fun = median) |> data.table()
dd$np <- ds[substr(variable, 1, 3) == "y_p", 2]
Penalized spline
When I made my initial inquiry to ChatGPT, it provided only a single model and didn’t indicate that there might be alternatives. To elicit another option, I had to specifically ask. To smooth the
estimate provided by the initial model (which admittedly I made too wiggly on purpose), I asked ChatGPT to provide a penalized Bayesian spline model, and it obliged.
The model is just an extension of the spline model, with an added penalization term that is based on the second derivative of the B-spline basis functions. We can strengthen or weaken the
penalization term using a tuning parameter \(\lambda\), that is provided to the model. The Stan model code is unchanged from the original model, except for the added penalization term.
model {
sigma_y ~ normal(0, 1);
sigma_beta ~ normal(0, 1);
// Priors for beta
for (k in 1:K) {
beta[k] ~ normal(0, sigma_beta);
//Penalization <---------------------------------------
for (k in 1:K) {
target += -lambda * sum(square(D2_spline * beta[k]'));
// Likelihood
for (n in 1:N) {
y[n] ~ normal(X_spline[n] * beta[cluster[n]]', sigma_y);
The second derivatives of the B-spline basis functions are estimated using the dbs function in the splines2 package. Like the matrix \(\boldsymbol{X}\), \(\boldsymbol{D_2}\) has dimensions \(N \times
\ M\). Both \(\boldsymbol{D_2}\) and \(\lambda\) are added to the data passed to Stan:
D2 <- dbs(dd$x, knots = knots, degree = 3, derivs = 2, intercept = TRUE)
D2_spline <- as.matrix(D2)
stan_data <- list(
N = nrow(dd),
K = k,
M = ncol(X_spline),
cluster = dd$cluster,
X_spline = X_spline,
D2_spline = D2_spline,
y = dd$y,
lambda = 0.00005
## Running MCMC with 4 parallel chains...
## Chain 2 finished in 16.5 seconds.
## Chain 1 finished in 16.6 seconds.
## Chain 3 finished in 16.7 seconds.
## Chain 4 finished in 16.7 seconds.
## All 4 chains finished successfully.
## Mean chain execution time: 16.6 seconds.
## Total execution time: 16.8 seconds.
The plot directly comparing the penalized Bayesian model with the initial Bayesian model (initial Bayesian model in blue) shows the impact of further smoothing.
A direct comparison between the GAM and penalized Bayesian models (GAM in green) suggests that there might be some differences in the estimation for at least several clusters, particularly those that
change direction twice. The penalized Bayesian model appears to be smoothing more than the GAM:
I was also aware of a third version of the Bayesian spline model that uses a random-walk prior on the \(\beta\text{'s}\) to induce smoothing. Unprompted, ChatGPT did not mention this. But, upon
request it did give me code that I was able to implement successfully. I’ll leave it to you to explore this further on your own—or perhaps ask ChatGPT for assistance.
OpenAI. (2024). ChatGPT (September 30, Version) [Large language model]. https://chat.openai.com/
This work is supported within the National Institutes of Health (NIH) Health Care Systems Research Collaboratory by cooperative agreement UG3/UH3AT009844 from the National Institute on Aging. This
work also received logistical and technical support from the NIH Collaboratory Coordinating Center through cooperative agreement U24AT009676. Support was also provided by the NIH National Center for
Complementary and Integrative Health Administrative Supplement for Complementary Health Practitioner Research Experience through cooperative agreement UH3AT009844 and by the National Center for
Complementary and Integrative Health of the National Institutes of Health under award number UH3AT009844. Work also supported by Memorial Sloan Kettering Cancer Center Support Grant/Core Grant
P30CA008748. The author was the sole writer of this blog post and has no conflicts. The content is solely the responsibility of the author and does not necessarily represent the official views of the
National Institutes of Health.
|
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|
M 2 6 9 4 E S FUJITSU
NO MORE PRODUCED Native| Translation
Form 3.5"/HH Cylinders 1819| | |
Capacity form/unform 973/ 1261 MB Heads 15| | |
Seek time / track 10.0/ 2.5 ms Sector/track | | |
Controller SCSI2 SINGLE-ENDED Precompensation
Cache/Buffer 512 KB FIFO BUFFER Landing Zone
Data transfer rate 5.120 MB/S int Bytes/Sector 256
10.000 MB/S ext SYNC
Recording method RLL 1/7 operating | non-operating
Supply voltage 5/12 V Temperature *C 5 45 | -40 60
Power: sleep W Humidity % 20 80 | 5 95
standby W Altitude km 3.000| 12.000
idle W Shock g 5 | 50
seek 11.5 W Rotation RPM 5400
read/write W Acoustic dBA 43
spin-up W ECC Bit 64
MTBF h 300000
Warranty Month
Lift/Lock/Park YES Certificates
FUJITSU M2691/2692/2693/2694-ES/ESA/ESB/EH/EHA/EHB 41FH5084E/5072E-IM
Single-Ended type SCSI (M269xES)
| 1+-+2 CNH5 CNH10 1+-+2 +--+|XX SCSI
| 3+-+4 5+-+6 | ||XX
| SCSI | ||XX C
| CNH11 +---+ Terminating| ||XX O
| 9+--+1 | | Resistor +--*|XX N CN1
| 10+--+2 | | 1 5 1|XX N
| +---SW1 +++++ |XX E
| 1+-+2 +---CN6 |XX C
| 3+-+4 |XX T
| 1+-CN7 CNH2 |XX O
| 3+-+ |XX R
|1+---+CN5 |XX
|2+---+ |XX
| | 1
| |XX Power
| |XX CN1
| |XX
+---------------------------------------------------------+ 1
Differential Type SCSI (M269xEH)
| 1+-+2 CNH5 1+-+2 CNH10 |XX SCSI
| 3+-+4 5+-+6 |XX
| |XX C
| CNH11 +---+ |XX O
| 9+--+1 | | |XX N CN1
| 10+--+2 | | |XX N
| +---SW1 |XX E
| 1+-+2 CNH2 |XX C
| 3+-+4 |XX T
| 1+-CN7 |XX O
| 3+-+ |XX R
|1+---+CN5 |XX
|2+---+ |XX
| | 1
| +---+|XX Power
| CN6++++|XX CN1
| 5 1 |XX
+---------------------------------------------------------+ 1
FUJITSU M2691/2692/2693/2694-ES/ESA/ESB/EH/EHA/EHB OEM MAN.41FH5084E-
Jumper Setting
x = Jumper set at factory
The user must set the following terminals and SCSI terminating
resistor before installing the IDD in the system.
- Setting terminal: SW1, CNH10, CNH11, CNH2, CNH3
- SCSI terminating resistor (M269xES only)
1. The user must not change the setting of terminals not described in
this section. Do not change setting statuses set at factory ship-
2. Do not change the setting of terminals except SW1 (offline self-
diagnostics)/CNH11(write protect) or do not connect or disconnect
the SCSI terminating resistor module (M269xES only) when power is
turning on.
3. To short the setting terminal, use the short plug attached when
the device is shipped from the factory.
SW1 Switch setting
SW1| x x x x x x x x |
| | | | | | | +------ Motor start mode
| | | | | | +-------- LED display requirement
| | | | | +---------- Synchronous mode transfer request
| | | | +------------ SCSI bus parity
| | | +-------------- Reselection retry count
| | +---------------- UNIT ATTENTION report mode
| +------------------ Offline self-diagnostics
+-------------------- SCSI level
SW1-1 SCSI level setting
| | INQUIRY data | | |
| +----------+-----------+-------------+ | |
| Mode |Byte 2, |Byte 3, | Byte 7 | INQUIRY | SW1-1 |
| |bits 2 to |bits 3 to 0| (Provided | VPD | |
| |0 (ANSI |(Response | function) | informa-| |
| |version) |data format| | tion | |
| SCSI-2| '0,1,0' | '0,0,1,0' |Indicates | Valid | |
x | mode |(SCSI-2) | (SCSI-2) |the function | | OFF |
| | | |of the IDD | | |
| | | |for each bit.| | |
| | '0,0,1,' |'0,0,0,1' | | | |
| SCSI-1|ANSIX3.131|ANSIX3T9.2/|All bits '0' | Invalid | ON |
| CSS |1986 |85-52 | | | |
| mode |(SCSI-1) |(CCS) | | | |
1. Set the display contents of data posted to the initiator from the
IDD with the INQUIRY command according to SW1-2. Select one of the
modes depending on the system software requirements.
2. When the SCSI-1/CCS mode is selected, parameters transferred by
the MODE SENSE command as follows.
a. Page code 3F (all pages equipped in the IDD are transferred)
1 Page 7, page 8 and page A are not transferred
2 Page 1, page 2 and page 4 are transferred with the
specified length of CCS.
b. When page 1, page 2 or page 4 is specified individually, it is
transferred with the specified length of CCS.
c. When page 7, page 8 or page A is specified individuelly, it is
transferred with the specified length of SCSI-2.
d. For the recovery parameter of the VERIFY, the recovery para-
meter in page 1 is used.
3. When SCSI-1/CSS mode is selected and the REQUEST SENSE command is
issued with specifying the transfer byte length to 0, the IDD
transfers the 4-byte sense data.
SW1-2 Offline self-diagnostics
ON Executed (diagnostic mode)
x OFF Stopped (normal operation mode)
Set starting/stopping the IDD offline self-diagnostics. The offline
self-diagnostics tests the IDD controller functions and the basic
read/write operation of the disk drive. In normal operations, this
setting terminal must be OFF.
SW1-3 UNIT ATTENTION report mode setting
x ON For a command other than INQUIRY, REQUEST SENSE, or PRIORITY
RESERVE the IDD responsed with the CHECK CONDITION status.
(SCSI standard)
OFF All received commands are executed normally. (The CHECK
CONDITION status caused by the UNIT ATTENTION condition is
not reported.)
Sets the response method against the received command when the IDD
keeps the UNIT ATTENTION condition. This mode is set for system
requirement, however, it is recommended to use the SCSI standard
setting (setting at factory shipment).
SW1-4 Reselection retry
x ON Retry count of RESELECTION phase = (unlimited)
OFF Retry count of RESELECTION phase = 10
SW1-5 SCSI bus parity setting
x ON SCSI data bus parity check by the IDD Executed
OFF SCSI data bus parity check by the IDD not Executed
SW1-6 Synchronous mode transfer request setting
x ON Synchronous mode transfer enabled
OFF Synchronous mode transfer disabled
Set whethersynchronous mode data transfer request from the TARG is
enabled according to Table. When the synchronous mode data transfer
is enabled, the IDD responds the SYNCHRONOUS DATA TRANSFER REQUEST
message to the INIT that issues the first command after power-on.
1. This setting does not affect asynchronous mode transfer.
2. When synchronous mode transfer request is disabled, the IDD
operates as follows.
- When the SYNCHRONOUS DATA TRANSFER REQUEST message is sent from
the INIT, the IDD reply to that message and the DATA IN and the
DATA OUT phases of the SCSI bus can be executed in synchronous
- The IDD does not send the SYNCHRONOUS DATA TRANSFER REQUEST
message to the INIT.
3. The maximum data transfer rate in synchronous mode is determined
when the SYNCHRONOUS DATA TRANSFER REQUEST message is exchanged
between the IDD and INIT.
The INIT must determine the parameter sent to the IDD when the
message is exchanged in consideration of the signal transfer
characteristics of the system SCSI bus and the data reception
capacity of the INIT.
The IDD can transfer up to 10 MB/s in synchronous mode. However,
since the configuration of the SCSI bus and its transfer
characteristics differ depending on the system, the maximum
transfer rate in which data can be transferred in a stable condi-
tion must be determined for each system.
4. For reference, the IDD data transfer rate in synchronous mode and
the restrictions on the systems configuration of the SCSI bus are
shown below.
Remarks: The following values are rough standards. The values be
evaluated for each system.
| Max. transfer rate | Max. length of | Number of connectable|
| of the IDD (MB/s) | the SCSI cable (m)| SCSI devices |
| 3 to 10 | * | * |
| 2.67 or less | 6(M269xES), | |
| | 25(M269xEH) | 8 |
* The maximum SCSI cable length and the number of connectable SCSI
devices must be determined for each system.
SW1-7 LED display requirement setting
x ON Light when the IDD operates
OFF Light when the IDD is ready
Set the display requirements of the LED on the fron panel or external
SW1-8 Motor start mode setting
x ON The motor is started immediately after power is turned on.
OFF Starting the motor is controlled with the START/STOP command.
This setting only determines the operation mode when power is turned
on. Stopping or restarting the spindle motor can be controlled with
the START/STOP UNIT command for both modes.
CNH2 User Setting Inhibited
|o o|CNH2
|o o|
+-+---- User setting inhibited (OPEN); (Factory test)
CNH10 Power supply to SCSI terminating resistor on IDD (M269xES)
(Single-Ended only)
|x x x|CNH10
|X X X|
| | +-- SCSI terminating resistor
| | power (M269xES only)
| +---- SCSI teminator resistor
| power
+------ User setting inhibited
|SCSI terminating resistor power supply | 5-6 | 3-4 |
|Power is supplied to the terminating resistor | | |
x |from the IDD and TERMPWR pin. Power is supplied |CLOSED|CLOSED|
|to the TERMPWR pin from the IDD. | | |
|The TERMPWR pin is not used. Power is supplied | | |
|to the IDD terminating resistor only from the |OPEN |CLOSED|
|IDD. | | |
|Power is not supplied to the terminating resistor| | |
|from the IDD. Power is supplied to the IDD |CLOSED|OPEN |
|terminating resistor only from the TERMPWR pin. | | |
Note: When the IDD connects to a position other than both ends of the
SCSI cable, do not mount the terminator resistor module on the IDD.
| +------ CNH10 ----+ -+- +5V DC|
+++ | 5 6 4 3 | | |
| TERMPWR +------+ +-+--+-----+----+-+ + Diode |
S *---------+ FUSE +-----+ | | +------+ |
C |26 +------+ +-----+----+ |
S | +------+-----+ |
I *-----------------------------+ Terminator | |
| SIGNAL Mounted | resistor | |
+++ +----+ in socket+------+-----+ |
| +----+IDD + GND |
CNH10 Power supply terminal resistor circuit on SCSI (M269xEH)
(Differential type)
|SCSI terminating resistor power supply | 3-4 |
x |Power is supplied from the IDD |CLOSED|
|Power is supplied from TERMPWR pin only (IDD power|OPEN |
|supply is not used. | |
| + CNH10+ -+- +5V DC|
+++ | 4 3 | | |
| TERMPWR +-+--+-+ + Diode |
S *----------------------+ | +------+ | |
C |25 | +---+ Fuse +------+ |
S |26 | +------+ |
I *----------------------+ |
| TERMPWR |
+++ +---+ |
| +---+ IDD |
CNH11 Setting terminals
| x x x x o |
| X X X X o |
| | | | +-- User setting inhibited
| | | +---- Write Protect
+-+-+------ SCSI ID
CNH11 SCSI ID
| SCSI ID | Jumpers |
| | ID2 | ID1 | ID0 |
|x 0 | OPEN | OPEN | OPEN |
| 1 | OPEN | OPEN | CLOSED|
| 2 | OPEN | CLOSED| OPEN |
| 3 | OPEN | CLOSED| CLOSED|
| 4 | CLOSED| OPEN | OPEN |
| 5 | CLOSED| OPEN | CLOSED|
| 6 | CLOSED| CLOSED| OPEN |
| 7 | CLOSED| CLOSED| CLOSED|
Notes: Set the SCSI ID so that there are no duplicates between SCSI
devices on the same SCSI bus.
The priority of SCSI bus use in ARBITRATION phase is determined by
SCSI ID as follows: 7 > 6 > 5 > 4 > 3 > 2 > 1 > 0
CNH11 Write Protect
7-8 OPEN Write operation is inhibited
x CLOSED Write operation is enabled
Enabling or disabling write protect function is set by pin 7-8 of the
setting terminal CNH11. By setting this write protect function,
writing into disk medium is inhibited.
CN1 DC Power and pin connector assignments
--+ +----CN1-----+| pin 1 +12 VDC
1| | 4 3 2 1 || pin 2 +12 Volts Return (GND)
-+ +------------+| pin 3 + 5 Volts Return (GND)
------------------+ pin 4 + 5 VDC
CN5 External operator panel connector
Pin number Signal
01 LED (V)
02 -LED
CN6 External operator panel connector
When the external operator panel is not connected to CN6, CN6 should
be opened. (default setting: OPEN)
When the external operator panel is connected, pins of CNH11 corres-
ponding to the signals set by the external operator panel should be
Pin number Signal Pin number Signal
01 GND 02 -ID2
03 GND 04 -ID1
05 GND 06 -ID0
CN7 Spindle Sync Connector
Pin number Signal Pin number Signal
01 GND IN 02 SS IN
03 GND OUT 04 SS OUT
FUJITSU M2691/2692/2693/2694-ES/ESA/ESB/EH/EHA/EHB OEM MAN. 41FH5084E
Installation direction
horizontally vertically
+-----------------+ +--+ +--+
| | | +-----+ +-----+ |
| | | | | | | |
+-+-----------------+-+ | | | | | |
+---------------------+ | | | | | |
| | | | | |
x x | | | | | |
+------x------x-------+ | +-----+ +-----+ |
+-+------x--x-------+-+ +--+ +--+
| xx |
| x x |
x x
The permissible orientations of the IDD are shown above, and the
tolerance of the angle is 5* from the horizontal plane.
Mounting frame structure
The disk enclosure (DE) of the IDD serves as a signal ground (SG) and
is insulated from the mounting frame (frame ground: FG). As this
insulation is maintained after the IDD is mounted in the system, the
following precautions must be followed.
Generally, SG an FG are connected at one point in the system
enclosure. Therefore, use following procedure to maintain the
insulation when mounting the IDD:
a) Use the frame with an embossed structure or the like to avoid
contact between the DE base and FG. Mount the IDD with making a
gap of 2.5 mm or more between the IDD and the frame of the system.
b) The inward projection of the screw from the IDD frame wall at the
corner must be 4 mm or less.
Limitatation of side-mounting: When the drive is mounted using side
screw holes, do not use the center hole. (M3 or #6-32 UNC screws)
External magnetic field
The drive should not be installed near the ferromagnetic body like a
speaker to avoid the influence of the external magnetic field.
Ambient temperature
When the IDD is operating, the ambient temperature measured 3 cm from
the disk enclosure (DE) surface and from the PCA surface must satisfy
the specified requirement. For the DE surface temperature at opera-
ting, the contact temperature at the measurement point must satisfy
the specified requirement.
Sequential starting of spindle motors
After the power is turned on to the IDD, a large amount of current
flows in the +12 V DC line when the spindle motor rotation starts.
Therefore, if more than one IDD is used, the spindle motors should
be started sequentially using one of the following procedures to pre-
vent overload of the power supply unit.
a. Issue START/STOP UNIT commands at 20-second intervals to start
the spindle motors.
b. Turn on the +12 V DC power in the power supply unit at 20-second
intervals to start the spindle motors sequentially.
Noise filter
To eliminate AC line noise, a noise filter should be installed at the
AC input terminal on the IDD power supply unit. The specification of
this noise filter is as follows:
Attenuation: 40 dB or more at 10 MHz.
Circuit construction: T-configuration is recommended.
SCSI cable
All SCSI devices on one bus are daisy-chained with an SCSI cable.
A terminating resistor must be mounted in the SCSI device at each end
of the SCSI cable.
Since an SCSI terminating resistor module is mounted in the only IDD
of Single-Ended type SCSI on shipment, it must be removed when the
IDD is not connected at either end of the SCSI cable. Also, a method
for power supply to the terminating resistor must be selected with
the setting terminal on the IDD.
The maximum number of SCSI devices that can be connected to the SCSI
bus is 8, including the host adapter, IDD, and other SCSI equipment.
The connector (socket) for the SCSI cable must be an unshielded 50-
contact socket which has two rows of 25 contacts spaced at 2.54 mm
(0.1 inch) apart. It should also have a key way to prevent insertion
in the wrong direction (bump type connector).
The maximum length of the SCSI cable is as follows. If more than one
SCSI device is connected, the total cable length must not exceed the
6m cable length Single-Ended type SCSI (M269xES)
25m cable length Differential type SCSI (M269xEH)
The use of a 25-pair twisted cable satisfying the following require-
ments is recommended:
Conductor size: 28 American Wire Gauge (AWG) or bigger
Characteristic Impedance: 90 to 132
Each pair of wires in the 25-pair twisted cable must be connected to
pins n and n+1 (where n is an odd number) on the interface connector.
Cables having an identical impedance must be use on the same SCSI bus
to reduce signal reflection and maintain transmission characteristics
When an SCSI device is connected to the SCSI cable except at either
end of the cable, connection to the SCSI connector must be at a
branchpoint of the cable. If a SCSI device is connected to one end of
the SCSI bus, no cable should be connected after the last SCSI device
except when the cable has a terminating resistor.
Power cable
IDDs must be star-connected to the DC power sypply (one to one con-
nection) to reduce the influence of load variations.
DC ground
A DC ground cable may or may not be installed depending on the
system requirements (system installation environment, cabinet
structure, power supply system). This cable is generally connected
to the ground of the power supply unit. It is recommended to connect
with a daisy chain (one by one connection).
Connector for external operator panel
Two types of connectors for the external operator panel are provided
on the IDD. They allow connection of an external LED on the front
panel, and an SCSI ID setting switch.
SCSI terminating resistor (M269xES only)
The SCSI terminating resistor module is installed in the only IDD
of Single-Ended type SCSI (M269xES) when the IDD is shipped from the
factory. The terminating resistor module is mounted in a socket and
must be processed in one of followings.
1. When conecting the IDD to either end of the SCSI cable, do not
demount the terminating resistor module.
2. When connecting the IDD to a position other than both ends of the
SCSI cable, demount terminating resistor module.
When demounting the terminating resistor module, be careful not to
damage the resistor module pins, mount socket and contiguous parts.
When mounting the terminating resistor module, check the mounting
direction and whetherthe module is fixed.
+- Marker shows pin 1 +- Marker shows pin 1
| |
| +-----------------+ |
| |* * * * * * * * *| | +*-*-*-*-*-*-*-*-*+
+- * | +- * Resistor |
|* * * * * * * * *| +*-*-*-*-*-*-*-*-*+
Socket, terminating resistor
module not present
Confirming initial operations
Initial operation in the case of settong so that motor starts at
1. When power is turned on, the LED blinks an instant and the IDD
executes initial self-diagnosis.
2. If an error is detected in the initial self-diagnosis, the LED
on the front panel blinks periodically.
Remark: The spindle motor may or may not start rotating in this stage
3. When the LED display requirements are set to "the IDD is
ready", the LED on the front panel lights 15 seconds after
power is turned on.
4. When the LED display requirements are set to "the IDD operates"
the LED on the front panel remains off (when the initiator
accesses the IDD via the SCSI bus, the LED lights).
2. Initial operation in the case of setting so that motor starts
with START/STOP command.
1. When power is turned on, the LED blinks an instant and the IDD
executes initial self-diagnosis.
2. If an error is detected in the initial self-diagnosis, the LED
on the front panel blinks.
3. The spindle motor does not start rotation until the START/STOP
UNIT command for the start is issued. The INIT needs to issue
the START/STOP UNIT command to start the spindle motor.
4. The disk drive enters the READY status 15 seconds after the
START/STOP UNIT command is issued. At this time, the IDD reads
"system information" from the system space on the disk.
5. When the LED display requirements are set to "the IDD is
ready", the LED on the front panel ligths as in step 4.
6. When the LED display requirements are set to "the IDD operates"
the LED blinks during command execution.
3. Check items at illegal operation
* Check that cables are mounted correctly.
* Check that power and voltages are supplied correctly.
* Check the setting of each jumper setting terminal. The initial
operation depends on the setting of the motor start mode and LED
display requirements.
* If an error is detected in initial self-diagnosis, the LED on
the front panel blinks. In this case, it is recommended to issue
the REQUEST SENSE command from the initiator (host system) to
obtain information (sense data) for error analysis.
When the LED display requirements are set to "the IDD is ready",
the LED is turned off while the drive continues command execution.
However, since the LED is turned off for only one blink, the LED
may seem to be turned on and off or not turned off at all.
When the LED display requirements are set to "the IDD operates",
the LED lights during the IDD is executing a command. However,
in some commands the lighting time is only an instant.
Therefore, it seems that the LED blinks or the LED remains off.
Since the IDD has the automatic readjustment function of positi-
oning (seek) control, it automatically executes the adjustment
operations with seek at specific intervals from power on (first
adjustment: 5 minutes after power on). The seek sound is heard
during the adjustment but this does not indicate a drive error.
Dismounting drives
Take the drive offline.
Disconnect system power before removing the drive. Do not remove
mounting screws holding the cables and drive while power is on.
Do not move the drive until it completely stops (30 seconds after
spindle motor is stopped with START/STOP UNIT command or after
power is turned off).
FUJITSU M2691/2692/2693/2694-ES/ESA/ESB/EH/EHA/EHB OEM MAN.41FH5084E-
Media defects
The number of allowable media defects are as follows.
M2691E : 378 and less
M2692E : 462 and less
M2693E : 546 and less
M2694E : 630 and less
Error recovery
The IDD can try to recover from errors in SCSI bus or the disk drive
using its powerful retry processing. If a recoverable data check
occurs, error-free data can be transferred to the initiator after
being corrected in the data buffer. The initiator software is
released from the complicated error recover processing by these error
recovery functions of the IDD.
It is a 8-byte error detection/correction code for the data field.
It is possible to detect single burst errors with length of up to
44 bits or double burst errors with length of up to 10 bits and to
correct single burst errors with lengths of up to 8 bits.
Defect List
Information of the defect location on the disk is managed by the
defect list. The following are defect lists which the IDD manages.
P list (Primary defect list): This list consists of defect location
information available at the disk drive shipment and is recorded in
a system space. The defects in this list are permanent, so that the
INIT must execute the alternate block allocation using this list
when initializing the disk.
D list (Data defect list): This list consits of defect location
information specified in a FORMAT UNIT command by the INIT at the
initialization of the disk. This information is recorded in the
system space of the disk drive as G list. To execute the alternate
block allocation, the FORMAT UNIT command must be specified.
C list (Certification defect list): This list consist of location
information on defective blocks which are detected by the verifying
operation (certification) of the data block after the initiation
when executing the FORMAT UNIT command. The IDD generates this
information when executing the FORMAT UNIT command, and the alternate
block allocation is made upon the defective block. This information
is recorded in the system space of the disk drive as the G list.
G list (Growth defect list): This list consists of defective logical
data block location information specified in a REASSIGN BLOCKS
command by the INIT, information on defective logical data blocks
assigned alternate blocks by means of IDD automatic alternate block
allocation, information specified as the D list, and information
generated as the C list. They are recorded in the system space on the
disk drive.
The INIT can read out the contents of the P and G lists by the
READ DEFECT DATA command.
Programmable data block length
Data can be accessed with fixed block length unit. Data block length
is programmable, and can be set to the most suitable length, from
180 to 4,160 bytes with 2-byte boundary at formatting time.
High speed data transfer
The data transfer rate on the SCSI bus is 4MB/s maximum in asynchro-
nous mode and 10MB/s in synchronous mode. Such a high data transfer
rate on the SCSI bus can be useful with the large capacity buffer
in the IDD.
The maximum data transfer rate in asynchronous mode may be limited
by the response time of the initiator and the length of SCSI bus
The maximum data transfer in synchronous mode on the single-ended
SCSI bus may be limited by the cable length and transmission charac-
teristics of the SCSI bus and the connected SCSI device number.
512KB programmable multi-segment data buffer
Data is transfered between SCSI bus and disk media through the em-
bedded 512KB data buffer in the IDD. This buffer can be divided into
maximum 32 areas. This feature provides the suitable usage
environment for users.
Since the initiator can control the disconnect/reconnect timing on
the SCSI bus by specifying the condition of stored data in the data
buffer or empty condition of the data buffer, the initiator can per-
form the effective input/output operations with utilizing high data
transfer capability of the SCSI bus regardless of actual data trans-
fer rate of the disk drive.
Read-Ahead cache feature
After executing the READ command, the IDD reads automatically and
stores (prefetches) the subsequent data blocks into the data buffer
(Read-Ahead caching).
The high speed sequential data access can be achieved by transfering
the data from the data buffer without reaccessing the disk in case
the subsequent command requests the prefetched data blocks.
Command queuing feature
The IDD can queue maximum 128 commands, and optimize the issuing
order of queued commands by the reordering function. This feature
realizes the high speed processing.
SCSI bus configuration
Up to eight SCSI devices can be connected to the SCSI bus, with any
combination of the SCSI devices that operate as initiator and that
operate as target.
Each SCSI device on the bus has its own unique address (SCSI-ID: #n).
For input/output operation, a peripheral device attached to the SCSI
bus that operates as target is addressed in unit called as logical
unit. A unique address (LUN: logical unit number) is assigned for
each logical unit.
The initiator selects one SCSI device by specifying that SCSI ID,
then specifies the LUN to select the peripheral device for input/
output operation.
The IDD is constructed so that the whole volume of disk drive is a
single logical unit, the selectable numbers of SCSI ID and LUN are as
SCSI ID: Selectable from 0 to 7
(switch selectable)
LUN: 0 (fixed)
Mean Time Between Failures (MTBF)
MTBF of the IDD during its life time is 300.000 hours.
Operating time (hours) at all field sites
MTBF = -----------------------------------------------------
The number of equipment failures from all field sites
Failure of the equipment means failure that requires repair, adjust-
ments or replacement. Mishandling by the operator, failures due to
bad environmental conditions, power trouble, host system trouble,
cable failures, or other failures not caused by the equipment are not
Mean Time To Repair (MTTR)
MTTR is the average time taken by a well-trined service mechanic to
diagnose and repair a drive malfunction. The drive is designed for a
MTTR of 30 minutes or less.
Service life
Overhaul of the drive is not required for following period which ever
earlier if handled correctly.
- DE surface temperature: 48*C or less......5 years
- DE surface temperature: 48*C and more.....5 years or
20,000 power-on hours
Data security at power failure
Integrity of the data on the disk is guaranteed against all forms of
DC power failure except on blocks where a write operation was being
performed. The above does not applied to formatting disks or assign-
ing alternate blocks.
Failure of the intelligent disk drives is defined as a failure
requiring adjustments, repairs or replacement. Fujitsu is not
responsible for drive failures caused by misuse by the user, poor
environmental conditions, power trouble, host problems, cable fai-
lures, or any failures not caused by the drive itself.
Track format configuration example
|Data block length | 180| 256| 512|1024|2048|4096|
|Zone I | 225| 172| 96| 51| 26| 13|
|Zone II | 213| 163| 91| 48| 25| 12|
|Zone III | 199| 153| 85| 45| 23| 11|
|Zone IV | 187| 143| 80| 42| 22| 11|
|Zone V | 175| 134| 75| 40| 20| 10|
|Zone VI | 162| 124| 69| 37| 19| 9|
|Zone VII | 149| 114| 64| 34| 17| 8|
|Zone VIII | 136| 105| 58| 31| 16| 8|
| |Byte/Sec. |
|Zone I | 56889 |
|Zone II | 53751 |
|Zone III | 50477 |
|Zone IV | 47339 |
|Zone V | 44201 |
|Zone VI | 41064 |
|Zone VII | 37789 |
|Zone VIII | 34652 |
FUJITSU => GDT
Tips & Hints about ICP controllers and other peripherals
Tagged queues and Fujitsu M 26xx harddisks
Fujitsu Harddisks of the M26XX series do not support the SCSI-II
feature tagged queues. You should disable this feature in the
initialize disks menue of the GDTSetup program. An inquiry command
shows, that the harddisk supports tagged queues (this is why you can
enable it in the GDTSetup) but this is not true as stated by Fujitsu
|
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Modeling Evaporating Droplets in Complex Unsteady Flows
Ghosal, Sandip
Apte, Sourabh V.
In many applications, a moving fluid carries a suspension of droplets of a second phase which may change in size due to evaporation or condensation. Examples include liquid fuel drops in engines and
raindrops or ice-crystals in a thunderstorm. If the number of such particles is very large, and, if further, the flow is inhomogeneous, unsteady or turbulent, it may be practically impossible to
explicitly compute all of the fluid and particle degrees of freedom in a numerical simulation of the system. Under such circumstances Lagrangian Particle Tracking (LPT) of a small subset of the
particles is used to reduce the computational effort. The purpose of this paper is to compare the LPT with an alternate method that is based on an approximate solution of the conservation equation of
particle density in phase space by the method of moments (MOM). Closure is achieved by invoking the assumption that the droplet size distribution is locally lognormal. The resulting coupled transport
equations for the local mean and variance of the particle size distribution are then solved in conjunction with the usual equations for the fluid and associated scalar fields. The formalism is
applied to the test case of a uniform distribution of droplets placed in a non homogeneous temperature field and stirred with a decaying Taylor vortex. As a benchmark, we perform a direct numerical
simulation (DNS) of high resolution that keeps track of all the particles together with the fluid flow.
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Re: standard for steganography?
[Date Prev][Date Next][Thread Prev][Thread Next][Date Index][Thread Index]
Re: standard for steganography?
At 0:56 3/1/94 -0500, Sergey Goldgaber wrote:
>On Mon, 28 Feb 1994, Norman Hardy wrote:
>> Has anyone done statistical studies of low bits of pixels or sound samples?
>> I suspect that they are often far from random. A flat 50% distribution in
>> the low bits might standout like a sore thumb. I can imagine the the low
>Yes, pure white noise would be anamalous. I have suggested that one use
>a Mimic function with a "garbage grammar". Implemented correctly, it should
>withstand statistical analysis.
>What is an AD converter? And what are the techniques you speak of that
>mimic those AD converters?
'AD converter' = 'Analog to Digital converter'.
Here are three schemes each with flaws:
Consider an alphabet of 10 bit characters with a probability distribution
such that each bit has an expected value of .6 (instead of the normal .5).
The character 000000000 has a probability of .4^10 = .000105 and
p(1111111111) = .6^10 = .006046. Do a Huffman encoding on this alphabet.
000000000 codes as 13 bits and 1111111111 codes as 7 bits. Take the cipher
stream and execute the Huffman decode(!) operation on the cipher stream.
Out comes a sequence of 10 bit bytes with 60% ones. To retrieve the
original cipher stream execute the normal Huffman coding algorithm and get
the original stream. The flaw here is that Huffman assigns some probability
to each of the 10 bit characters which is 2^-7, 2^-8, ... 2^-13. The
intermediate probabilities are not represented. This would show up without
too much data.
Another scheme is called 'arithmetic coding'. It avoids the above
probability quantization but is tricky to program. I can't find a reference
to it just now but it should appear in any modern book in information
theory. Unlike Huffman it does not code each character into a definite
number of bits but codes a sequence of several characters into a 'real
number'. Adapting this to numbers that real computers can use is tricky.
Again you feed the flat cipher stream into the decoding end of the
algorithm and get biased bits.
The above two schemes are information efficient. With a 60% bias you get
97% efficiency. If you are willing to settle for 80% efficiency you can
merely establish a RNG synchronized at sender and receiver that sends a bit
from the cipher stream with probability .8 and sends a one with probability
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Basic Operators
Circom provides boolean, arithmetic, and bitwise operators. They have the standard semantics but the arithmetic operators applied to numeric values work modulo p.
The precedence and association of the operators are like in Rust (defined here).
Expressions can be built using the next operators, but the conditional operator ?_:_ can only occur at the top level.
Field Elements
A field element is a value in the domain of Z/pZ, where p is the prime number set by default to
p = 21888242871839275222246405745257275088548364400416034343698204186575808495617.
As such, field elements are operated in arithmetic modulo p.
The circom language is parametric to this number, and it can be changed without affecting the rest of the language (using GLOBAL_FIELD_P).
Conditional expressions
Boolean_condition ? true_value : false_value
var z = x>y? x : y;
This conditional expression is not allowed in a nested form, hence can only be used at the top level.
Boolean operators
Next boolean operators are allowed:
Operator Example Explanation
&& a && b Boolean operator AND
|| a || b Boolean operator OR
! ! a Boolean operator NEGATION
Relational operators
The definition of relational operators < , > , <= , >= , == , != depends on the mathematical function val(x) which is defined as follows:
val(z) = z-p if p/2 +1 <= z < p
val(z) = z, otherwise.
According to this function, the definition of the relational operators is as follows:
`x < y` is defined as val(x % p) < val(y % p)
`x > y` is defined as val(x % p) > val(y % p)
`x <= y` is defined as val(x % p) <= val(y % p)
`x >= y` is defined as val(x % p) >= val(y % p)
where <, >, <=, >= are the comparison of integers.
Arithmetic operators
All arithmetic operations work modulo p. We have the next operators:
Operator Example Explanation
+ a + b Arithmetic addition modulo p
- a - b Arithmetic subtraction modulo p
* a * b Arithmetic multiplication modulo p
** a ** b Power modulo p
/ a / b Multiplication by the inverse modulo p
\ a \ b Quotient of the integer division
% a % b Remainder of the integer division
There are operators that combine arithmetic operators with a final assignment.
Operator Example Explanation
+= a += b Arithmetic addition modulo p and assignment
-= a -= b Arithmetic subtraction modulo p and assignment
*= a *= b Arithmetic multiplication modulo p and assignment
**= a ** b Power modulo p and assignment
/= a /= b Multiplication by the inverse modulo p and assignment
\= a \= b Quotient of the integer division and assignment
%= a %= b Remainder of the integer division and assignment
++ a++ Unit increment. Syntactic sugar for a += 1
-- a-- Unit decrement. Syntactic sugar for a -= 1
Bitwise operators
All bitwise operators are performed modulo p.
Operator Example Explanation
& a & b Bitwise AND
| a | b Bitwise OR
~ ~a Complement to the number of bits of the prime number
^ a ^ b Bitwise XOR
>> a >> 4 Right shift operator
<< a << 4 Left shift operator
The shift operations also work modulo p and are defined as follows (assuming p>=7).
For all k with 0=< k <= p/2 (integer division) we have that
• x >> k = x/(2**k)
• x << k = (x*(2**k)~ & ~mask) % p
where b is the number of significant bits of p and mask is 2**b - 1.
For all k with p/2 +1<= k < p we have that
• x >> k = x << (p-k)
• x << k = x >> (p-k)
note that k is also the negative number k-p.
There are operators that combine bitwise operators with a final assignment.
Operator Example Explanation
&= a &= b Bitwise AND and assignment
|= a |= b Bitwise OR and assignment
~= ~=a Complement to the number of bits of the prime number and assignment
^= a ^= b Bitwise XOR and assignment
>>= a >>= 4 Right shift operator and assignment
<<= a <<= 4 Left shift operator and assignment
Examples using operators from the circom library
In the following, there are several examples using combinations of the previous operators.
pragma circom 2.0.0;
template IsZero() {
signal input in;
signal output out;
signal inv;
inv <-- in!=0 ? 1/in : 0;
out <== -in*inv +1;
in*out === 0;
component main {public [in]}= IsZero();
This template checks if the input signal in is 0. In case it is, the value of output signalout is 1. 0, otherwise. Note here that we use the intermediate signal inv to compute the inverse of the
value of in or 0 if it does not exist. If inis 0, then in*inv is 0, and the value of out is 1. Otherwise, in*inv is always 1, then out is 0.
pragma circom 2.0.0;
template Num2Bits(n) {
signal input in;
signal output out[n];
var lc1=0;
var e2=1;
for (var i = 0; i<n; i++) {
out[i] <-- (in >> i) & 1;
out[i] * (out[i] -1 ) === 0;
lc1 += out[i] * e2;
e2 = e2+e2;
lc1 === in;
component main {public [in]}= Num2Bits(3);
This templates returns a n-dimensional array with the value of in in binary. Line 7 uses the right shift >> and operator & to obtain at each iteration the i component of the array. Finally, line 12
adds the constraint lc1 = in to guarantee that the conversion is well done.
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Symbols and Sets of Numbers Equality Symbols Symbols and Sets of Numbers Inequality Symbols. - ppt download
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The Firebird 5.0 Language Reference | Conversion Of Data Types | Implicit Data Type Conversion
Implicit Data Type Conversion
Implicit data conversion is not possible in Dialect 3—the CAST function is almost always required to avoid data type clashes.
In Dialect 1, in many expressions, one type is implicitly cast to another without the need to use the CAST function.For instance, the following statement in Dialect 1 is valid:
SET ADATE = '25.12.2016' + 1
The string literal will be cast to the DATE type implicitly.
In Dialect 3, this statement will raise error 35544569, “Dynamic SQL Error: expression evaluation not supported, Strings cannot be added or subtracted in dialect 3”—a cast will be needed:
SET ADATE = CAST ('25.12.2016' AS DATE) + 1
Or, with a datetime literal:
SET ADATE = DATE '25.12.2016' + 1
In Dialect 1, mixing integer data and numeric strings is usually possible because the parser will try to cast the string implicitly.For example,
will be executed correctly.
In Dialect 3, an expression like this will raise an error, so you will need to write it as a CAST expression:
2 + CAST('1' AS SMALLINT)
The exception to the rule is during string concatenation.
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How to Apply A Function to Every Element In A List In Haskell?
To apply a function to every element in a list in Haskell, you can use various higher-order functions or list comprehensions. Here are a few common approaches:
1. Using map function: The map function takes a function and a list, and applies that function to every element of the list. It returns a new list with the transformed elements. For example: > let
myList = [1, 2, 3, 4, 5] > let increment x = x + 1 > map increment myList [2, 3, 4, 5, 6]
2. Using a list comprehension: List comprehensions provide a way to define a new list using the elements of an existing list. You can apply a function to each element within a comprehension. For
example: > let myList = [1, 2, 3, 4, 5] > let increment x = x + 1 > [increment x | x <- myList] [2, 3, 4, 5, 6]
3. Using recursion: You can also apply a function to every element in a list using recursive functions. Here's an example: applyFunc :: (a -> b) -> [a] -> [b] applyFunc _ [] = [] applyFunc f (x:xs)
= f x : applyFunc f xs -- Usage > let myList = [1, 2, 3, 4, 5] > let increment x = x + 1 > applyFunc increment myList [2, 3, 4, 5, 6]
These methods allow you to apply any desired function to each element of a list without explicitly looping over each element.
What is the eta-reduction technique used in applying a function to every element in a list in Haskell?
The eta-reduction technique in Haskell refers to a reduction where a function is applied to each element in a list using higher-order functions like map or list comprehensions.
In Haskell, the map function takes a function and a list as arguments and applies the function to each element of the list, returning a new list with the transformed values. The eta-reduction
technique can be used to simplify the code when applying a function to every element in a list.
For example, consider the following code:
1 square :: Int -> Int
2 square x = x * x
4 numbers :: [Int]
5 numbers = [1, 2, 3, 4, 5]
7 squaredNumbers :: [Int]
8 squaredNumbers = map square numbers
In the code above, the map function is used to apply the square function to each element in the numbers list, resulting in a new list squaredNumbers that contains the square of each element.
With eta-reduction, the code can be simplified as:
1 square :: Int -> Int
2 square x = x * x
4 numbers :: [Int]
5 numbers = [1, 2, 3, 4, 5]
7 squaredNumbers :: [Int]
8 squaredNumbers = map square numbers
Here, the function application map square numbers can be directly assigned to squaredNumbers without using a separate definition. This is possible because the numbers list is passed as the last
argument to the map function, resulting in eta-reduction.
How to apply a function to every element in a list and produce a new list in Haskell?
In Haskell, you can use the map function to apply a function to every element in a list and produce a new list. Here's an example:
1 multiplyByTwo :: Int -> Int
2 multiplyByTwo x = x * 2
4 main :: IO ()
5 main = do
6 let myList = [1, 2, 3, 4, 5]
7 let multipliedList = map multiplyByTwo myList
8 print multipliedList
In this example, we have a function multiplyByTwo that takes an Int and returns its doubled value. We define a list myList with some integer values.
Using map multiplyByTwo myList, we apply the multiplyByTwo function to every element in myList, resulting in a new list [2, 4, 6, 8, 10]. This new list is assigned to the variable multipliedList,
which we then print using print multipliedList.
The output of this program would be [2, 4, 6, 8, 10], as each element in the original list has been multiplied by 2.
What is the difference between using foldl and map to apply a function to every element in a list in Haskell?
The difference between using foldl and map to apply a function to every element in a list in Haskell lies in how the elements are processed and combined.
map applies a function to every element of a list, producing a new list with the transformed elements. It follows a "mapping" approach where each element is transformed independently. The resulting
list is of the same length as the original list.
On the other hand, foldl (left fold) is a higher-order function that aggregates the elements of a list by repeatedly applying a binary function to the accumulated value (also called the
"accumulator") and the next element. It reduces a list to a single value by iteratively combining the elements. The function takes two arguments: the current accumulated value and the next element,
and returns a new accumulated value. Since foldl aggregates the elements, it returns a single value instead of a list.
Here's an example to illustrate the difference between map and foldl:
1 -- Using map
2 addOne :: [Int] -> [Int]
3 addOne = map (+1)
5 -- Using foldl
6 addOne' :: [Int] -> Int
7 addOne' = foldl (\acc x -> acc + x) 0
9 -- Usage
10 main = do
11 let myList = [1, 2, 3, 4, 5]
12 print (addOne myList) -- Output: [2, 3, 4, 5, 6]
13 print (addOne' myList) -- Output: 15
In the example, addOne applies the (+1) function to every element independently, producing a new list where each element is incremented. On the other hand, addOne' uses foldl to accumulate the sum of
all elements, resulting in a single value 15.
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Two-dimensional limit series in ultraspherical Jacobi polynomials and their approximative properties
For citation:
Guseinov I. G., Gadzhimirzaev R. M. Two-dimensional limit series in ultraspherical Jacobi polynomials and their approximative properties. Izvestiya of Saratov University. Mathematics. Mechanics.
Informatics, 2021, vol. 21, iss. 4, pp. 422-433. DOI: 10.18500/1816-9791-2021-21-4-422-433, EDN: SDAGUK
Two-dimensional limit series in ultraspherical Jacobi polynomials and their approximative properties
Let $C[-1,1]$ be the space of functions continuous on the segment $[-1,1]$, $C[-1,1]^2$ be the space of functions continuous on the square $[-1,1]^2$. We denote by $P_n^\alpha(x)$ the ultraspherical
Jacobi polynomials. Earlier, for function $f$ from the space $C[-1,1]$ limit series were constructed by the system of polynomials $P_n^\alpha(x)$ and the approximative properties of their partial
sums were investigated. In particular, an upper bound for the corresponding Lebesgue function was obtained. Moreover, it was shown that the partial sums of the limit series, in contrast to the
Fourier – Jacobi sums, coincide with the original function at the points $\pm1$. In this paper, for function $f(x, y)$ from the space $C[-1,1]^2$, we construct two-dimensional limit series by the
system of ultraspherical Jacobi polynomials $P_n^\alpha(x)P_m^\beta(y)$ orthogonal on $[-1,1]^2$ with respect to the Jacobi-type weight-function. It is shown that the partial sum of the
two-dimensional limit series coincides with $f(x, y)$ on the set $\{(-1,-1), (-1,1), (1, -1), (1,1)\}$ and is a projection on the subspace of algebraic polynomials $P(x,y)$. Using these properties,
the approximative properties of the partial sums of the two-dimensional limit series are investigated. In particular, the behavior of the corresponding two-dimensional Lebesgue function is studied.
1. Pashkovskiy S. Vychislitel’nye primeneniia mnogochlenov i riadov Chebysheva [Numerical Applications of Polynomials and Tchebychev Series]. Moscow, Nauka, 1983. 384 p. (in Russian).
2. Malvar H. S. Signal Processing with Lapped Transforms. Artech House, 1992. 380 p.
3. Trefethen L. N. Finite Difference and Spectral Methods for Ordinary and Partial Differential Equation. Cornell University, 1996. 299 p.
4. Sharapudinov I. I. Mnogochleny, ortogonal’nye na setkah. Teoriya i prilozheniya [Polynomials Orthogonal on Grids. Theory and Applications]. Makhachkala, Izdatelstvo Dagestan Pedag. Univ., 1997.
255 p. (in Russian).
5. Mukundan R., Ramakrishnan K. R. Moment Functions in Image Analysis. Theory and Applications. Singapore, World Scientific, 1998. 164 p.
6. Dedus F. F., Mahortyh S. A., Ustinin M. N., Dedus A. F. Obobshchennyi spektral’no-analiticheskii metod obrabotki informatsionnykh massivov. Zadachi analiza izobrazhenii i raspoznavaniia obrazov
[Generalized Spectral and Analytic Method of Data Arrays Processing. Problems of Image Analysis and Pattern Recognition]. Moscow, Mashinostroenie, 1999. 356 p. (in Russian).
7. Trefethen L. N. Spectral Methods in Matlab. Philadelphia, SIAM, 2000. 181 p.
8. Sharapudinov I. I. Approximation of discrete functions and Chebyshev polynomials orthogonal on the uniform grid. Mathematical Notes, 2000, vol. 67. iss. 3, pp. 389–397. https://doi.org/10.1007/
9. Sharapudinov I. I. Limit ultraspherical series and their approximation properties. Mathematical Notes, 2013, vol. 94. iss. 2, pp. 281–293. https://doi.org/10.1134/S0001434613070274
10. Szego G. Orthogonal Polynomials. AMS Colloq. Publ., vol. 23, 1939. 440 p. (Russ. ed.: Moscow, Fizmatgiz, 1962. 500 p.).
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Compute minimal number of generators of subring
Compute minimal number of generators of subring
Hi all,
Given a polynomial ring $R = k[x_1,...,x_n]$ over a field $k$ (say $\mathbb{Q}$) and a list of polynomials $p_1,..,p_n$ in $R$ (homogeneous say), is Sage capable of computing the minimal number of
generators of the subring generated by $p_1,...,p_n$ over $k$?
Next, suppose $I$ is a homogeneous ideal of $R$. Can the same approach be used to compute the minimal number of generators of this subring after being projected to $R / I$ ?
Edit: Sorry for the bad formatting (see source), please help me fix it if you can!
Edit2: For example, let $R = \mathbb{Q} [x,y]$ and let $p_1 = x^2, p_2 = x^2 y, p_3 = x^4 y$. This generates a principal ideal, but as a subring it has 2 generators and it is those 2 generators that
I want.
Edit3: In case it is helpful, the case I'm looking at is a ring with an action of a group, and its subring of invariants for which I have generators.
Edit4: Here's a suggested algorithm: sort the generators by degree. Then sequentially add them, and check if they were already in the subring. Does Sage even have suitable subring capabilities?
Welcome to Ask Sage! Thank you for your question!
Hint: a concrete example would help exploring this question.
For the suggested algorithm you might use the SAGBI functionality in Singular:
> LIB "sagbi.lib";
> ring r=0, (x,y), dp;
> ideal A = x2, x2y;
> sagbiReduce(x4y, A);
> sagbiReduce(x5y, A);
> sagbiReduce(x6y, A);
If this is sufficient, then I can add an answer showing how to do it from Sage.
Huh I guess so! I don't know Singular syntax - does it work over a base field like QQ? E.g. does sagbiReduce(x4y+(5/2)x6y, A) return 0? And please also let me know how to deal with projecting it to
$R/I$ !
Singular works over a field of characteristic zero here, and yes sagbiReduce(x4y+(5/2)*x6y, A) returns 0. In SageMath you can do:
from sage.libs.singular.function import singular_function, lib as singular_lib
sagbiReduce = singular_function('sagbiReduce')
R.<x,y> = PolynomialRing(QQ)
A = R.ideal(x^2, x^2*y)
sagbiReduce(x^4*y+(5/2)*x^6*y, A)._sage_()
Unfortunately I don't know how you would deal with $R/I$.
1 Answer
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I understand a minimal generating set to be a set of generators such that no proper subset is a set of generators.
(Is the cardinality of a minimal generating set always the same? For subrings I don't know.)
We suppose a generating set $p_1,\ldots,p_m$ for a subring of $k[x_1,\ldots,x_n]$ is given, and we want to find a subset which is minimal. For this it suffices to solve the subring containment
problem: if we can test whether a given element belongs to some subring, then we can use this to identify redundant generators (and this can be optimized by looking at degrees, as suggested in Edit4
of the question).
In Singular the subring containment problem is solved by inSubring, which can be called from SageMath as follows:
def in_subring(p, A):
R = p.parent()
from sage.libs.singular.function import singular_function, lib as singular_lib
inSubring = singular_function('inSubring')
return inSubring(p, R.ideal(A))[0] == 1
For example:
sage: R.<x,y> = PolynomialRing(QQ)
sage: in_subring(x^4*y - (5/2)*x^6*y, [x^2, x^2*y])
The documentation of inSubring states that it does the same as algebra_containment (using a different algorithm), and the Theory section of the documentation of that procedure describes how it works.
Namely, the trick is to introduce new variables $z_j$ (one for each generator of the subalgebra) and an elimination ordering such that the $x_i$ are all greater than all $z_j$'s; then a polynomial
$f$ belongs to the subalgebra $k[p_1,\ldots,p_m]$ if and only if its normal form $NF(f, J)$ with respect to the ideal $J = \langle z_j - p_j \rangle$ (calculated by the multivariate division
algorithm, using a Groebner basis with respect to the chosen monomial ordering) contains only $z_j$'s.
Repeating the above example using the new method:
sage: R.<x,y,z1,z2> = PolynomialRing(QQ, order='degrevlex(2),degrevlex(2)')
sage: f = x^4*y - (5/2)*x^6*y
sage: J = R.ideal([z1 - x^2, z2 - x^2*y])
sage: f.reduce(J)
-5/2*z1^2*z2 + z1*z2
If I'm not mistaken then this latter method can be adapted to test containment in a subalgebra of $R/I$ as follows:
sage: I = R.ideal([x^6])
sage: (f + x^7).reduce(J + I)
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The minimal generating set $\{x^2,x^2+x\}$ for $k[x]$ shows that the "local minimum" number of generators is not necessarily the global minimum in general, but the answer by Harm Derksen to Minimal
number of generators of a k-algebra over commutative rings states (without a reference) that there is no such issue in the graded/homogeneous case. Moreover he gives the formula $\dim_k(\mathfrak{m}/
\mathfrak{m}^2)$ for the minimal number of generators, where $\mathfrak{m}$ is the ideal consisting of positive degree homogeneous elements in the subalgebra. That seems to check out in the above
example as well.
rburing ( 2021-08-25 18:44:21 +0100 )edit
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Annals of Computer Science and Information Systems
Decision Making in Building Maintenance Using a Graph-Based Knowledge Representation
Wojciech Palacz, Grażyna Ślusarczyk, Andrzej Łachwa, Barbara Strug, Anna Paszyńska, Ewa Grabska
DOI: http://dx.doi.org/10.15439/2017F262
Citation: Communication Papers of the 2017 Federated Conference on Computer Science and Information Systems, M. Ganzha, L. Maciaszek, M. Paprzycki (eds). ACSIS, Vol. 13, pages 17–25 (2017)
Abstract. This paper is an attempt to support effective decision making in building management by assisting maintenance processes. Knowledge about buildings is stored in a graph with many
hierarchies. This representation allows us to express different types of hierarchical dependencies between building parts, like geometrical and functional ones, in one structure. Moreover, such a
structure is useful to extract subgraphs containing information necessary for a given computational task, such as locating a desired place and the shortest path leading to it. As maintanance
processes often require dynamic path target selection, modified indoor navigation methods are proposed. The paper presents the capability of the described knowledge model to cope with complex queries
referring to different types of information. The considered examples show that the proposed approach can be used for various facility maintenance management applications.
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17. M. Minas, “Concepts and realization of a diagram editor generator based on hypergraph transformation,” Science of Computer Programming, vol. 44, no. 2, pp. 157–180, 2002.
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23–38, 2007.
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35. T. H. Cormen, C. E. Leiserson, R. L. Rivest, and C. Stein, Introduction to Algorithms (2nd ed.). MIT Press, 2001.
36. A. Kneidl, A. Borrmann, and D. Hartmann, “Generating sparse navigation graphs for microscopic pedestrian simulation models,” in 18th EG-ICE International Workshop, Twente, Netherlands, 2011.
37. M. Höcker, V. Berkhahn, A. Kneidl, A. Borrmann, and W. Klein, “Graph-based approaches for simulating pedestrian dynamics in building models,” in 8th European Conference on Product & Process
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What is the Heisenberg’s Uncertainty Principle - Info Curiosity
The uncertainty principle is one of the most famous (and probably misunderstood) ideas in physics. It tells us that there is a fuzziness in nature, a fundamental limit to what we can know about the
behavior of Quantum particles and, therefore, the smallest scales of nature. Of the scales, the most we can hope for is to calculate probabilities for where things are and how they will behave.
Unlike Isaac Newton’s clockwork universe, where everything follows clear-cut laws on how to move and prediction is easy if you know starting conditions, the uncertainty principle enshrines a level of
fuzziness into quantum theory.
Werner Heisenberg‘s simple idea tells us why atoms don’t implode, how the sun manages to shine and, strangely, that the vacuum of space is not actually empty.
An early incarnation of the uncertainty principle appeared in a 1927 paper by Heisenberg, a German physicist who working at Niels Bohr’s institute in Copenhagen at the time titled “On the Perceptual
Content of Quantum Theoretical Kinematics and Mechanics”. The more familiar from of the equation came a few years later he had further refined his thoughts in subsequent lectures and papers.
Heisenberg was working through the implications of quantum theory, a strange new way of explaining how atoms behaved that had been developed by physicist, including Niels Bohr, Paul Dirac and Erwin
Schrödinger, over the previous decade. Among its many counter-intuitive ideas, quantum theory proposed that energy was not continuous, but instead came in discrete packets (quanta) and that light
could be described as both a wave and a stream of these Quanta. In fleshing out this radical worldview, Heisenberg discovered a problem in the way the basic physical properties of a particle in a
quantum system could be measured. In one of his regular letters to a colleague, Wolfgang Pauli, he presented the inklings of an idea that has since became a fundamental part of quantum description of
the world.
The uncertainty principle says that we cannot measure the position (x) and the momentum (p) of a particle with absolute precision. The more accurately we know of these values, the less accurately we
know the other. Multiplying together the errors in the measurements of these values (the errors are represented by the triangle symbol in front of each property, the Greek letter “delta”) has to give
a number greater than or equal to half constant called “h-bar”. This is equal to Planck’s constant (usually written as h) divided by 2π. Planck’s constant is an important number in quantum theory, a
way to measure granularity of the world at its smallest scales and it has the value 6.626 x 10^34 joule seconds.
One way to think about the uncertainty principle is an extension of how we see measure things in the world everyday. You can read these words because particles of light, photons, have bounced off or
paper and reached your eyes. Each photon on that path carries with it some information about the surface it has bounced from, at the speed of light. Seeing a subatomic particle, such as an electron,
is not so simple.
The uncertainty principle is at the hearth of many things that we observe but con not explain using classical (non-quantum) physics. Take atoms, for example, where negatively-charged electrons orbit
a positively-charged nucleus. By classic logic, we might expect the two opposite charges to attract each other, leading everything to collapse into, a ball of articles. The uncertainty principle
explains why this doesn’t happen: if an electron got too close to nucleus, then its position in space would be precisely known and, therefore, the error in measuring its position would be minuscule.
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Wide Flange Beam Column
The form on this site allows you to input design information about a wide flange column and calculate which sections are acceptable. Calculations are performed using the equations in the Canadian
Standards Association standard CSA S16-14 Design of Steel Structures. Clauses 11, 13.5 and 13.6 are applied in the calculations.
• The axial capacity calculations are based on the rolled section equations. The x axis is the strong axis of the W section.
• F[y] is the yield strength of the steel section in MPa. F[y] must be between 100 and 1000.
• k[x] and k[y] are the effective length factors for compression about the x and y axes. k[x] and k[y] must be greater than 0.1.
• L[x] and L[y] are the unbraced lengths about the x and y axes. Ly is used as the length of the unbraced compression flange to calculate the lateral torsion buckling moment capacity. L[x] and L[y]
must be greater than 0.1.
• C[f] is the factored applied axial load in kN. C[f] must be greater than 1.
• M[fx] is the factored applied moment about the x axis in kN m. M[fx] must be greater than 1.
• I[reqd] is the required strong axis moment inertia required to meet the deflection limitations for the column. I[reqd] must be greater than 0.1.
• ω[2] is the bending moment gradient factor defined in clause 13.6(a). ω[2] must be between 1.0 and 2.5.
Although no effort has been spared in an attempt to ensure that numerical values are accurate to a degree consistent with current structural design practice, the author and owner of this site do not
assume responsibility for errors or oversights resulting from use of the information contained herein. Anyone making use of this site assumes all liability arising from such uses.
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MAE/MATH joint seminar - Polynomial inclusions: definitions, applications, and open problems
RM 2464, HKUST (2/F, Lift # 25/26)
Predictive modelling in physical science and engineering is mostly based on solving certain partial differential equations where the complexity of solutions is dictated by the geometry of the
domain. Motivated by the broad applications of explicit solutions for spherical and ellipsoidal domains, in particular, the Eshelby's solution in elasticity, we propose a generalization of
ellipsoidal shapes called polynomial inclusions. A polynomial inclusion (or p-inclusion for brevity) of degree k is defined as a smooth, connected, and bounded body whose Newtonian potential is a
polynomial of degree k inside the body. From this viewpoint, ellipsoids are identified as the only p-inclusions of degree two, and many fundamental problems in various physical settings admit
simple closed-form solutions for general p-inclusions as for ellipsoids. Therefore, we anticipate that p-inclusions will be useful for applications including predictive materials models, optimal
designs, and inverse problems. However, the existence of p-inclusions beyond degree two is not obvious, not to mention their direct algebraic parameterizations.
In this work, we explore alternative definitions and properties of p-inclusions in the context of potential theory. Based on the theory of variational inequalities, we show that p-inclusions do exist
for certain polynomials, though a complete characterization remains open. We reformulate the determination of surfaces of p-inclusions as nonlocal geometric flows which are convenient for
numerical simulations and studying geometric properties of p-inclusions. In two dimensions, by the method of conformal mapping we find an explicit algebraic parameterization of p-inclusions. We
also propose a few open problems whose solution will deepen our understanding of relations between domain geometry, Newtonian potentials, and solutions to general partial differential equations. We
conclude by presenting examples of applications of p-inclusions in the context of Eshelby inclusion problems and magnet designs.
講者/ 表演者:
Prof. Liping LIU
Professor, Department of Mathematics & Department of Mechanical Aerospace Engineering, Rutgers University, NJ, USA
Prof. Liping LIU received his bachelor degree in mechanics and engineering science from Beijing University, Beijing in 2000 and PhD in aerospace engineering and mechanics from the University of
Minnesota, Twin Cities in 2006. He joined Rutgers University, New Jersey as an assistant professor in 2012 and currently is a Professor of Mathematics & Mechanical Aerospace Engineering. He received
the Thomas J.R. Hughes Young Investigator Award from the American Society of Mechanical Engineers (ASME) in 2018. His research group at Rutgers University focuses on mechanics and materials, and his
research interests include multiscale-multiphysics analysis and modeling, optimal design of multiphase and multifunctional composites, and theoretical and computational material science.
Department of Mechanical & Aerospace Engineering
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<h2>Spectral methods in GCMs - and some thoughts on CFD.</h2>
There has been a lot of discussion recently on the maths of GCMs. I have summarised some in a series on chaos and the Lorenz equations (
, and
). David Young has commented substiantally on
this thread
, and raised the issue of the use of spectral methods in GCMs. These are used in the dynamical core, which relates pressure and velocity via the Navier-Stokes equations. They are time critical,
because they require resolving sound waves, and so the speed performance here fixes that of the code as a whole. Spectral methods are used because they are fast. But some mystery is made of them,
which I would like to try to dispel. But I'd like to do this in the context of some simplifying observations about CFD.
Discretisation and derivative
Actually this applies to continuum pde (partial diff eqs) in general. To make them amenable to computation, they must be discretised. Fields are represented by discrete values, often at nodal points.
I have worked with many schemes for that - finite difference, finite elements and more recently with meshless methods -
. And there's finite volume, and spectral methods. Each is presented with its own bundle of theory, for Navier-Stokes or whatever. But I have found one simplifying feature. The only thing these
methods need to do is
to provide an operator which gives a discretised first (spatial) derivative from discrete values
. From there on, the linear (or non-) analysis proceeds in the same way, regardless of how discretised. I should add a caveat. If the derivative operator maps on to a different space, as with FEM,
say, then you need the mapping operator for that - called a mass matrix in FEM. This operator usually is quick to apply or even invert, and should cause no numerical difficulty.
By derivative, I generally mean grad, but the divergence is basically the negative transpose of that. There is an important difference in boundary treatment, which I could go into. The derivative
will generally be a large sparse array, which I denote G
. j is the index for space dimension, and a,b for the respective nodes. Again, they may not be space nodes, and a,b may be in different spaces. For finite elements,
= ∫T
dE V
where the T's are the test functions and U's the basis functions (usually the same set). And the mass matrix is
I use the summation convention whereby there is summation over repeated indices.
How to relate these to pde's. Simple. Just replace grad operators by (M
, and div by the transpose, and replace undifferentiated variables by their discretised values. That of course assumes M is invertible, when it may not be square. If not, a Penrose inverse is needed.
Generally the equations will be solved in the range-G space, so the outer multiplication by (M
is omitted.
There's a whole rigmarole of solution methods. But I think they should be reduced to just one - inexact Newton-Raphson. The reason is that this is totally systematic. You just write down the
equations, as discretised above, as in general F(V)=0. Then you substitute a trial solution V
, which isn't right. So you do your best to make it right by linearising:
Ideally, you'd use for B the derivative of F. But that is often very hard to invert, and maybe even to calculate. So B is a compromise that is reasonably close, and so that you can solve for V
. Then you can iterate, although in most CFD you'd have a good starting V
from the previous step, so you would then go on to the next step.
For Navier-Stokes, forming B often involves block LU-factorisation of the N-S equations on v and P. That leaves an awkward Schur complement (
), which is still hopefully symmetric positive definite (depending on how you treat advection). The solution task is to approximate that. In Uzawa itself, you invert it by some iterative process like
conjugate gradient. That just delegates the approximating task to the preconditioner for CG. In augmented Lagrangian, for example, the Schur would be approximated with a multiple of the unit matrix.
And yes, there is upwinding and all that. You can embed that in this framework.
I like inexact N-R conceptually, because it separates the task of deciding what you want (F(V)=0) and of how to make it happen. And if you iterate, you are always checking if what you want has been
satisfied. And there are many ways you can form B. You can diagonalise mass matrices. You can drop terms you think don't matter. All that counts for accuracy is that B is good enough that the
iterations converge. Of course, you may have a preference for speed too.
Spectral methods on the sphere.
As used in GCMs. They have been around for a very long time. I quoted
this GFDL history
"Spectral methods are an alternative to finite difference schemes, the method used by all of the first-generation primitive-equation GCMs. They express the horizontal variation of dynamic model
fields in terms of orthogonal spherical harmonics. The technique simplifies the solution of many of the nonlinear partial differential equations used in general circulation modeling. Its utility had
been explored as early as 1954 (Platzman, 1960; Silberman, 1954).
Heavy calculational demands made spectral methods unsuitable for use in early GCMs. Faster computers, and improvements in algorithms for spectral methods which reduced their calculational intensity,
led to their adoption in GCMs around 1970 (Bourke, 1974; Eliasen et al., 1970; Orszag, 1970; Robert, 1969)."
An often quoted early GFDL paper on implementation, which is also a good intro to GCM maths, is
. An expansive text, which explains in detail the reasons for using spherical harmonics (SH), is
, especially starting about sec 18.10. As Boyd points out, the key issue which makes something like SH necessary is the pole problem. The shrinking of lat-lon elements is not only a nuisance
(unwanted high resolution) but interferes with the Courant condition. Personally, I would first use a
projected cube
to avoid the pole problem, but the spectral decomposition has the virtue of mapping into a much smaller space.
Spherical Harmonics
I've written quite a lot about spherical harmonics, eg
. I use them in TempLS to map anomalies onto a sphere with I think optimal smoothing, and more ambitiously, to
on a sphere for TempLs itself. It's one of the best performing methods for that - comparable to an irregular mesh, and with similar results. So GCM's use them for differentiating.
The spectral transform from lat/lon values to coefficients is done just as you would in Fourier analysis (the SH are orthogonal). You integrate the product of the function with the SH on the grid.
That is a 2D integration which ould be costly, but there is a key saver. The SH is just the product of trig functions over longitude with a Legendre polynomial of cos(lat). So you can do an inner
loop over longitude, using fast FFT. The outer loop over latitude doesn't have this blessing, but that matters much less for overall speed.
The key reason why this is overall faster is that you don't need that many SH functions. The series can be truncated, with loss of only high frequency noise. There is fuss about aliasing etc, but
that has been sorted out. The result is that the mass matrix M
has many fewer rows than columns, so you solve in a lower dimensional space. A downside is loss of sparsity. But sparse methods do not parallelise as well as dense, so it seems the trade-off works.
They don't usually formulate it as I would, with a grad operator G
. I think that is useful, because the derivatives in a lat/lon space carry coefficients to make them correspond to read 3D grad, and I think it is useful to fold them into the G operator. It also
clarifies the trade-off, that G is more compact but with corresponding loss of sparsity.
David Young was concerned that the use of spectral might bring the problems of high order spectral in CFD. I don't think that is right here. It isn't high order in the sense of a
spectral element method
, say (eg hp). It seems to just transform and truncate. So there is no special effort to represent high derivatives, at least not in the form described by
14 comments:
1. Also worth noting that not all GCMs use spectral. Hadley don't, for example.
2. > the key issue which makes something like SH necessary is the pole problem
Which the Hadley folk solve by spectral filtering near the poles.
1. William,
Thanks. It can be quite hard to get details of gridding and filtering. It has puzzled me that quite complicated techniques like spectral transform and filtering were used for the pole problem
before trying other styles of gridding, like cubed sphere, which GFDL now uses.
3. Thanks for mentioning inexact Newton Raphson. We pioneered these methods in the 1980's. It was one of Forester Johnson's big insights. GGNS is currently the only code to reliably implement it for
RANS. That is where new insights will come from.
The problem of course with spectral methods is that you must filter the result to remove oscillations you don't like. First of course you must decide based on some criteria what is noise and what
is signal. Generally for 60 years people have found upwind or stabilized FEM to be better and to be independent of "expert judgment" on a case by case basis. In short the method is the method,
not a mechanism for rationalizing bias.
4. Nick, you say you "would first use a projected cube to avoid the pole problem". Why not use a triangular/icosahedron grid for both weather and climate global modeling? I'm guessing it makes the
math more complicated and thus slows down getting results? But as computers are get faster and faster, maybe that's no longer an issue.
1. Bryan, I think the origin of these methods is deep in the ancient past when FFT's were the fastest solver methods available. There is some good work in the 1970's on fast direct methods using
FFT for PDE's. Paul Swartztrauber at NCAR and Brian Sweet were two of those to look for if you are interested. These methods only work on essentially uniform grids in some coordinate system.
Thus the grids used in GCMs that are essentially uniform in polar coordinates. That's a decision based on 1960's technology and should probably be revisited. I suspect that the problem here
is that a lot of the other "tunable sub grid models" were also optimized for these grids, so there is probably a ripple effect throughout a massive code. In cases like this, starting over is
expensive but allows one to really test the idea that more modern methods have advantages.
Of course there has been huge progress in inexact Newton Krylov methods since the 1970's that are now in fact vastly better than FFT based methods because they are easily parallelized using
domain decomposition and work well for a wide variety of grids and grinding types. Keyes and Knoll have a good survey article on these methods in Journal of Computational Physics about 2001 I
2. Sorry, "Grinding types" should be "gridding types." Darn spell correction.
3. Bryan,
"Why not use a triangular/icosahedron grid"
Yes, you can do even better. But I looked closely at the cubed sphere here; you can tweak it to fairly equal areas. There is a 2007 paper describing GFDL's adoption of cubed sphere. It
completely overcomes the pole singularity issue. Triangles aren't really more difficult, but I suspect would have been a bigger change to their code structure. The good thing about GCM grids
is that you should only have to deal with them once, to create matrices that you then store. That's the point of my claim that discretisation shouldn't matter once you have a first
derivative, and maybe a mass, array. In timestepping, that is what you use.
Regular longitude is useful for the spectral transform FFT. But if you aren't doing that, I'm not sure where FFT would come in.
5. I haven't seen any discussion of the ocean in this series of posts. Coupling of the ocean and the atmosphere is one of the major advantages of GCMs vs other types of modeling. A huge engineering
advance vs 40 years ago. This is a water planet after all.
6. Thanks David and Nick for your insight. I am on old dog still trying to learn new tricks. My first experience in weather modeling was in undergraduate college met lab where we graphically
produced barotropic 500 mb forecasts using hand plotted and analyzed CONUS maps via light tables in about 1974. In graduate school I made a simple baroclinic forecast model using FORTRAN IV in
1977, using punch cards to load and run the program on a mainframe computer that was likely much less powerful than cell phones today. I also had a graduate course in FFT, which I can't honestly
say I mastered very well but managed to pass it somehow. Since that time I have not been involved in weather modeling other than to frequently use the output for weather and air quality
forecasting. So I am much more familiar with the output and its limitations than the detailed workings of the models. Even though I retired last year, I still look at the GFS output every day and
marvel at how well it predicts for a few days compared to what we had in the 1970's.
I suspect it would be worthwhile to start over and build a new weather/climate forecast system from scratch to optimize for modern capabilities like parallel processing. I hope that some well
funded group can do that before too long. I can imagine it might take years to develop and implement. The triangular/icosahedron grid is appealing to me since the grid cells are uniform in area,
although I realize that the earth is not a perfect sphere and there are also issues related to fixing the surface of the model relative to the actual surface of the earth. I remember seeing a
short YouTube video posted in comments at WUWT on determining MSL and I was surprised how complicated that is.
7. Bryan - have you looked at any the information about various models available on NASA websites? Curious as to your impressions.
Obviously, they have models for wings.
1. JCH, thanks for the links. The first link to a NOAA page was helpful for visualizing the cubed sphere grid that Nick has mentioned. I still conceptually prefer a triangular/icosahedron global
grid, but I have never tried to implement one.
The second link regarding turbulence brought back old memories for me. I had a graduate class called "turbulence and diffusion" based on "Turbulent Diffusion in the Environment" by G T Csandy
published in 1974. Some of my fellow students humorously called it "turbulence and confusion". After finishing my graduate classes, I worked at an environmental consulting firm for 9 years
and I was involved in data analysis/QA/QC for atmospheric turbulence measurements at 10 and 60 meters AGL from tall towers and from about 50 to 300 meters AGL from doppler acoustic sounders.
I learned quite a bit from reviewing lots of data over those years and even had a few conference presentations published. In those days I was focused on turbulence as it is involved
dispersion of air pollutants near the ground. But I also seem to recall that it is involved in frictional energy dissipation, which could also play a role in weather and climate.
I have not kept up with turbulence modeling since then, other than in reviewing output from air quality models that include turbulence. I suspect that some aspects of air quality modeling may
be helpful in improving weather and climate modeling, especially regarding atmospheric particles (which are very difficult to model) that will have some small effects on atmospheric energy
budgets and could also be involved in significant effects on snow and ice albedo. The hypothesis that extreme atmospheric dust levels may have contributed to the end of the many glacial
cycles in our current ice age I find very interesting. But I digress.
8. Yes JCH, GFDL has gone with the Woodward Collella finite volume scheme from the 1980's. Somewhat of an advance.
You are out of your depth on the NASA turbulence modeling site. It's all very simple mostly 2D cases. There is far more interesting data and cases in the literature and even some negative
results. I've given on an earlier post here some references that take larger investments of time and are not as easily assessable by outsiders, but that give a fairer and more scientific view.
1. David, my reference to wings was a joke.
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How to Divide Fractions (like ½ ÷ ¼) with Free Online Tutoring in Math
How to Divide Fractions (like ½ ÷ ¼) with Free Online Tutoring in Math
Unlock the mystery of dividing fractions with our latest blog post, the final installment in our comprehensive eleven-part series!
Designed to follow a cyclical learning approach, this article provides a complete lesson on fractions, guiding your child step-by-step through the complexities of dividing fractions like ½ ÷ ¼.
Dividing fractions can be perplexing. Why does ½ ÷ ¼ result in an answer larger than both ½ and ¼? Shouldn’t division make numbers smaller? The key to understanding this lies in illustrating
mathematical models. When children visualize the division process, they can clearly see why ½ ÷ ¼ yields a larger answer. Without this visual aid, many children struggle to grasp the concept, leading
to confusion and frustration in higher levels of math.
In our blog post, your child will:
• Learn to Illustrate Division: Visual models help demystify why dividing fractions often results in larger numbers.
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Articles within this series on Fractions:
Solving problems that deal with fractions is simple when you develop a concrete understanding. I have had incredible results with with the students in my class! The strategies taught within this
article work with children who have ADHD, Dyslexia, and other learning disabilities. Virtually every one of my students who has learned the strategies within this HOW TO DO FRACTIONS article has
passed the standards based assessment for adding, subtracting, multiplying and dividing fractions.
I have scaffold the problems in each lesson.
The first problem in this article is a “Watch Me” problem. The second is a “Work with Me” problem. All the rest are “On Your Own” problems.
*If your child needs a bit more support, they should complete the “On Your Own” problems as a “Work with Me” problem. I have a number of students with gaps in their learning and others with a
variety of learning disabilities. I have had incredible success, by having those students complete 5 to 7 problems within each lesson as a “Work with Me” problem. They play a bit of the video,
then pause it and copy, then watch a bit more, pause it and copy. My students Play – Pause – and Copy until the entire problem is solved. This is like having a personal tutor working through each
and every problem with your child. Every one of my students who has used this strategy has passed the Common Core Proficiency Exam.
How to Divide Fractions
Online Tutoring in Math: Challenge 1
Watch Me
Terrance the Ticklish Mule
Terrance loves his alfalfa, but his owner keeps tickling his nose. His owner loves Terrance, but he can’t help tickling his favorite mule. Each day, Terrance eats 2/3 pounds of alfalfa. His owner
divides the alfalfa into portions that are 2/7 pounds, how many portions will Terrance get to eat?
Hint: How many 2/7’s are in 2/3?
Watch this Free Tutoring for Math Video!
Press PLAY and Watch this Free Tutoring for Math Video below. Then copy these strategies into your notes!
How to Divide Fractions
Online Tutoring in Math: Challenge 2
Work With Me
Singing Camels
Kami & Connie Camel love to sing. They walk around their paddock and sing to all the on lookers. Kami & Connie’s paddock is 8/9 miles long.
If Kami& Connie break the paddock into portions that are 3/4 miles, how many portions will our “Singing Camels” walk and sing?
Hint: How many 3/4’s are in 8/9?
Watch this Free Tutoring for Math Video!
Gather your materials and press PLAY. We’ll solve this problem together, while you watch the math tutorial video below.
Do your children get frustrated when they make a mistake?
We all make mistakes. As a matter of fact, making mistakes is an essential part of the learning process. This is why at the end of each of the following “On Your Own” challenges I encourage
children to fix their mistakes. Finding and fixing your own mistake is the fastest way to learn.
How to Divide Fractions
Online Tutoring in Math: Challenge 3
On Your Own
Honovi the Honey Loving Grizzly Bear
Honovi is always on the lookout for bees. She follows them back to the hive and collects as much honey as possible. Yesterday, Hovoni found a hive. While the bees crawled under her fur and stung
again and again, Honovi stole 4/5 pounds of honey. If Hovoni divides the honey into portions that are each 1/8 pound, how many portions can she make?
Hint: How many 1/8’s are in 4/5?
Watch this Free Tutoring for Math Video!
Once you complete the problem – Hit PLAY on the math tutorial video below. Good Luck!
How to Divide Fractions
Online Tutoring in Math: Challenge 4
On Your Own
Shelby & Sherman Seahorse
Shelby & Sherwood are a couple of hungry seahorses. They love nothing more than a big old mouthful of plankton. The marine biologist that feeds them has a 2/5 bucket of plankton.
A serving of the plankton is 3/8 of the bucket.
How many servings are in the bucket?
Watch this Free Tutoring for Math Video!
Once you complete the problem – Hit PLAY on the math tutorial video below. Good Luck!
How to Divide Fractions
Online Tutoring in Math: Challenge 5
On Your Own
Elyssa the Thirsty Elephant
Elyssa Elephant is thirsty. There is very little water on the African Plains at this time of the year. Luckily for Elyssa and her herd, a local farmer gives a bucket of cool water to each elephant
in Elyssa’s family. Each bucket holds 5/6 gallons of water.
If Elyssa’s trunk holds 2/7 of a gallon, how many trunkfuls of water will Elyssa get to drink?
Watch this Free Tutoring for Math Video!
Once you complete the problem – Hit PLAY on the math tutorial video below. Good Luck!
Want More Tutorials?
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School Of Computer, Data And Mathematical Sciences
Mathematics for Engineers 1Western Sydney University Unit Code: 200237.5
Discipline: MATHEMATICS
Student Contribution Band: 1
Level: 1
Credit Points: 10
Assumed Knowledge
HSC Mathematics achieved at Band 5 or 6. This is the minimum requirement.
Equivalent Units
14505 Engineering Mathematics 1; 200195 Mathematical Methods A; 200196 Mathematical Methods B; 700019 Mathematics for Engineers 1 (WSTC); 700101 Mathematics for Engineers 1 (WSTC Assoc Deg)
Incompatible Units
200031 Mathematics for Business; 200189 Concepts of Mathematics; 300672 Mathematics 1A; 300673 Mathematics 1B
Students enrolled in 3740 Bachelor of Engineering (Honours) or 3689 Bachelor of Engineering must have passed 300743 Mathematics for Engineers Preliminary otherwise permission is required.
About this Unit
This unit is the first of two mathematics units to be completed by all students enrolled in an engineering degree during their first year of study. The content covers a number of topics that underpin
the later-stage engineering mathematics units. The subject matter includes: differential and integral calculus of a single variable, complex numbers, aspects of matrix algebra, vectors, and some
elementary statistics and probability theory. The aim of this unit is to introduce a number of key mathematical concepts needed in the study of Engineering, and to provide a solid foundation for the
follow-on unit Mathematics for Engineers 2.
3621.6 Bachelor of Engineering CONTINUING
3621.7 Bachelor of Engineering CONTINUING
3633.2 Bachelor of Computing CONTINUING
3639.1 Bachelor of Information and Communications Technology CONTINUING
3664.2 Bachelor of Engineering Science CONTINUING
3689.1 Bachelor of Engineering CONTINUING
3689.2 Bachelor of Engineering CONTINUING
3690.4 Bachelor of Engineering Advanced (Honours) CONTINUING
3691.1 Bachelor of Engineering Science CONTINUING
3691.2 Bachelor of Engineering Science CONTINUING
3691.5 Bachelor of Engineering Science CURRENT
3691.6 Bachelor of Engineering Science CURRENT
3728.1 Bachelor of Engineering (Honours)/Bachelor of Business CONTINUING
3728.2 Bachelor of Engineering (Honours)/Bachelor of Business CONTINUING
3728.3 Bachelor of Engineering (Honours)/Bachelor of Business CONTINUING
3728.4 Bachelor of Engineering (Honours)/Bachelor of Business CURRENT
3740.1 Bachelor of Engineering (Honours) CONTINUING
3740.2 Bachelor of Engineering (Honours) CONTINUING
3740.3 Bachelor of Engineering (Honours) CURRENT
3740.4 Bachelor of Engineering (Honours) CURRENT
3771.1 Bachelor of Engineering Advanced (Honours) CURRENT
KP3621CIVI.1 Civil CONTINUING
KP3621ELEC.1 Electrical CONTINUING
KT3026.1 Construction CONTINUING
KT3032.1 Electrical CONTINUING
KT3034.1 Telecommunications CONTINUING
KT3042.1 Mechanical CONTINUING
KT3043.1 Civil CONTINUING
KT3045.1 Robotics and Mechatronics CONTINUING
KT3046.1 Computer CONTINUING
KT3075.1 Civil CONTINUING
KT3077.1 Construction CONTINUING
KT3080.1 Mechanical CONTINUING
KT3081.1 Robotics and Mechatronics CONTINUING
KT3088.1 Electrical CONTINUING
KT3089.1 Environmental CONTINUING
KT3102.1 Electrical CONTINUING
KT3103.1 Telecommunications CONTINUING
KT3104.1 Electrical CONTINUING
KT3105.1 Telecommunications CONTINUING
KT3143.1 Civil CURRENT
KT3144.1 Construction CONTINUING
KT3145.1 Electrical CURRENT
KT3146.1 Mechanical CURRENT
KT3147.1 Robotics and Mechatronics CURRENT
KT3159.1 Civil CURRENT
KT3160.1 Construction CONTINUING
KT3161.1 Electrical CURRENT
KT3162.1 Mechanical CURRENT
KT3163.1 Robotics and Mechatronics CURRENT
KT3166.1 Construction CURRENT
KT3167.1 Civil CURRENT
KT3168.1 Construction CURRENT
KT3169.1 Electrical CURRENT
KT3170.1 Mechanical CURRENT
KT3171.1 Robotics and Mechatronics CURRENT
MT3048.1 Advanced Manufacturing CURRENT
MT3049.1 Materials Engineering CURRENT
MT3050.1 Sustainability Engineering CURRENT
MT3051.1 Civil Engineering CURRENT
MT3052.1 Construction Engineering CURRENT
MT3053.1 Electrical Engineering CURRENT
MT3054.1 Mechanical Engineering CURRENT
MT3055.1 Robotics and Mechatronics Engineering CURRENT
SM3004.1 Formal Systems CONTINUING
SM3005.1 Applied Mathematics CONTINUING
SM3621SOE.1 Soil Engineering CONTINUING
SM3621WATE.1 Water Engineering CONTINUING
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2 -
Woburn Challenge 2018-19 Round 2 - Senior Division
Problem 1: Laser Grid
The IMF (Impossible Mission Force) has dispatched their best agent, Ethan Hunt, to recover a recently stolen microchip. This microchip contains critical Canadian governmental secrets, such as the
Prime Minister's favourite colour, and must be recovered before its captors have time to download its data!
Ethan has tracked the microchip down to an underground base in Saskatchewan. Upon infiltrating it, he's found himself in the middle of a gigantic, square room. When viewed from above, the room can be
represented as a square on a 2D plane, with its bottom-left corner at coordinates (0, 0) and its top-right corner at coordinates (1,000,000, 1,000,000). Ethan has lowered himself down into the room,
and is standing at coordinates (X[E], Y[E]) (1 ≤ X[E], Y[E] ≤ 999,999).
There are N (0 ≤ N ≤ 100,000) vertical lasers extending across the entire room, the i-th of which is a line segment from coordinates (V[i], 0) to (V[i], 1,000,000) (1 ≤ V[i] ≤ 999,999). There are
also M (0 ≤ M ≤ 100,000) horizontal lasers extending across the entire room, the i-th of which is a line segment from coordinates (0, H[i]) to (1,000,000, H[i]) (1 ≤ H[i] ≤ 999,999). All vertical
lasers have distinct V values, all horizontal lasers have distinct H values, and no laser goes directly through Ethan's location (in other words, no V value is equal to X[E], and no H value is equal
to Y[E]).
Ethan was hoping to simply find the stolen microchip, but he's been greeted by a more troubling sight: there are C (1 ≤ C ≤ 100,000) microchips strewn about the room! The i-th microchip is at
coordinates (X[i], Y[i]) (1 ≤ X[i], Y[i] ≤ 999,999). No two microchips are at the same location, no microchip is at Ethan's location, and no laser goes directly through any microchip's location.
One of these C microchips must be the real one, with the rest being decoys, but they all look identical! Unfortunately, Ethan will only have time to go grab at most one of them before getting out of
there. To make matters even worse, Ethan may not pass through any lasers on his way to pick up the microchip of his choice, as they'd trigger an alarm. He'll need to weigh his options and choose his
plan of action carefully!
For each microchip, determine whether or not Ethan would be able to reach its location from (X[E], Y[E]) by following any continuous path on the 2D plane (not necessarily a straight line segment),
without leaving the confines of the room and without passing through any of the N + M lasers.
In test cases worth 5/13 of the points, each integer in the input is no greater than 10.
In test cases worth another 5/13 of the points, N ≤ 2000, M ≤ 2000, and C ≤ 2000.
Input Format
The first line of input consists of two space-separated integers, X[E] and Y[E].
The next line consists of three space-separated integers, N, M, and C.
N lines follow, the i-th of which consists of a single integer, V[i], for i = 1..N.
M lines follow, the i-th of which consists of a single integer, H[i], for i = 1..M.
C lines follow, the i-th of which consists of two space-separated integers, X[i] and Y[i], for i = 1..C.
Output Format
Output C lines with a single character per line, either "Y" if Ethan would be able to reach the i-th microchip, or "N" otherwise, for i = 1..C.
Sample Input
Sample Output
Sample Explanation
The room is illustrated below, with lasers indicated in red, Ethan's location in green, and the microchips in blue. Note that most of the x-coordinates and y-coordinates on the plane (from around 10
to around 999,997) have been collapsed together.
Ethan would only be able to reach the 2nd or 5th microchip.
All Submissions
Best Solutions
Point Value: 7 (partial)
Time Limit: 3.00s
Memory Limit: 32M
Added: Dec 14, 2018
Author: SourSpinach
Languages Allowed:
C++03, PAS, C, HASK, ASM, RUBY, PYTH2, JAVA, PHP, SCM, CAML, PERL, C#, C++11, PYTH3
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Detection of Breast Cancer in Infrared Thermographies Using Stochastic Techniques in a FPGA Platform
Article Information
J. de la Cruz-Alejo^*, Irving Cardiel Alcocer Guillermo, M. B. Arce Vázquez, Ernesto Enciso Contreras
Mechatronic Department, Tecnológico de Estudios Superiores de Ecatepec, Ecatepec, Estado de México
^*Corresponding Author: J. de la Cruz-Alejo, Mechatronic Department, Tecnológico de Estudios Superiores de Ecatepec, Ecatepec, Estado de México
Received: 21 February 2020; Accepted: 20 April 2020; Published: 17 Augst 2020
Citation: J. de la Cruz-Alejo, Irving Cardiel Alcocer Guillermo, M. B. Arce Vázquez, Ernesto Enciso Contreras. Detection of Breast Cancer in Infrared Thermographies Using Stochastic Techniques in a
FPGA Platform. Journal of Bioinformatics and Systems Biology 3 (2020): 045-057.
Share at Facebook
Thermo graphical infrared images to predict the existence of a thermal anomaly according to symmetry in breasts by using stochastic methods and fuzzy logic control is proposed. Statistical results
are established through entropy, kurtosis and media to evaluate symmetry grade between the right and left breast. To predict the grade of breast cancer associated to the tissue and take a decision a
fuzzy controller is designed in base to symmetric assessments distribution. The proposed method is implemented on a FPGA platform optimizing hardware requirements and improves response time. Results
show that the error of the prediction method can be an alternative to detect cancer if the image source are far away from the critical errors, interference to the source or infrared image processed.
Infrared images; FPGA; Stochastic methods; Fuzzy controller breast cancer
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Article Details
1. Introduction
Currently there are several methods to detect cancer, which can detect its appearance and more likely to treat it successfully. Detection methods most commonly used today require complex machines
that do not give a result to the naked eye, but these need to be analyzed by a specialist. These methods are invasive and need machines too expensive. Also, a thermo graphical record of an
individual's body can be created and processed for further analysis to determine the existence of a thermal anomaly. Also, image segmentation involves the use of various techniques to extract and
separate relevant information from an image. [1-7]. Stochastic techniques can be used for detection thermal anomalies, more precisely thermal asymmetry in the chest area to detect cancer. In the
other hand, novel fuzzy control methods have been developed and implemented as a programming platform to control different processes in different areas. Due to their heuristic nature associated with
simplicity and effectiveness for both linear and non-linear systems, fuzzy logic controllers (FLC) has showed their outstanding features in implementations for solar tracking systems, mechatronics
systems, etc. [8-13].
The main aim of this paper is to use stochastic techniques and fuzzy logic controller to detect the thermal asymmetry in the region in a simple and precise way in comparison to the already existing
techniques of breast cancer detection.
The structure of rest of the paper is as follow. Section 2, describe the stochastic processes of the termographical images in order to elaborate a stochastic approach for detecting the breast cancer.
In Section 3, based on the stochastic approach, the grade of breast cancer is determined according to a fuzzy control as alternative to manage the stochastic information. Section 4, shows the
characterization of this propose by simulation and experimental results. The paper concludes with Section 4, where conclusions are drawn.
2. Stochastic techniques for detection thermal asymmetries
Figure 1, represents in a modular way the processes to be performed for the breast cancer detection based on thermographic images. The first block refers to a thermal image obtained by a device
outside the system which is a thermographic camera. Once the thermographic image is obtained, it is processed by a conversion to grayscale taking the luminance of the pixels, a segmentation to
separate the background of the image and the information that correspond to the chest on which the study will be performed. Then, since the body symmetry must be identified, the main image is divided
in two sub-images. These sub-images contain only the tissue corresponding to the left and right chest. Next, stochastic techniques are applied to each breast to determine the asymmetry grade between
these two images without considering the size of the samples. Finally, Linguistic variables are proposed to design the fuzzy control to determine the grade of breast cancer associated to the tissue.
Figure 1: Schematic diagram proposed to detect breast cancer using stochastic techniques.
2.1 Grayscale conversion
To convert the grayscale image, first it is separated into three spectra information RGB. Once the image is separated, it is converted to gray scale and processed to normalize the image values. This
consists in establishing a minimum reference pixel value which will be used to represent different temperatures in the human tissue. Eliminating background noise in the image, it is segmented in
order to isolate both breast tissues and find some key points within the image which are able to provide where begins and ends in the thermal image. Figure 2, shows the key points locations
represented by: Top Right Point (TRP), Left Upper Point (LUP), Right Lower Point (RLP), Left Lower Point (LLP). The first point (TRP), is find through locating the first pixel not belonging to the
bottom of the image, starting with the top left of the image, using:
Where X represents the total width of the image and determines the beginning of the body in the image and Y, represents the total height of the image measured from top to bottom.
In similar way is found the second point (LUP), only that the counting starts from the pixel (1, X), which would be the top row and the final column of the image. Third point (RLP), start with the
first pixel at the left bottom (Y, 1). It starts counting the number of zeros that are in this row from left to right until it reaches X / 2, which is the half of the image, using:
Finally the fourth point (LLP), represent the first pixel corresponding to the left side of the body in the thermal image. It consists in counting the pixels belonging to the bottom of the image or
zero value, starting from the lower left (X, Y), using:
Once the four main points of the image are found, lines are drawn to joins the points TRP with LLP and LUP with RLP, as is shown in Figure 3. The intersection of both points will be called the
geometric center (GC). To obtain this point, is necessary to determine the angles shown in Figure 4.
The angle of the line joining the points LUP and RLP is called α. To determine the value of this angle, the distance between these points on the horizontal axis X is obtained, using:
Then, angle α is calculated by:
Same manner, angle called β, corresponding to the line joining the points TRP and LLP is obtained by:
Third angle φ, is obtained by:
Considering the angles β and φ in addition to the distance from RLP to LLP, which forms the total base of the triangle, the distance from RLP to GC called DC, is found by:
Distance from LLP to GC called LC, is found through the α and φ angles and the distance from RLP to LLP, by:
Once the distance DC and LC have been obtained, the height of the GC point called GCY is obtained by:
Horizontal distance of the GC, called GCX is determined by:
Once these points are located, the image of the breast is separate in two, right and left.
2.2. Stochastic process
Stochastic techniques are applied to the regions of interest to detect thermal asymmetry in base to the pixel values between right and left breasts. Arithmetic mean, is the first function to know
thermal asymmetries which, is known as the mean sample [14, 15, 16], given by:
This value corresponds to the middle of all pixels values in a range from 0 to 255, which correspond to the maximum admissible thermal value. The mean value must be the same for both breasts. In case
of an asymmetry or anomaly, mean value differs. Another function to consider, is the deviation of the pixels from the mean value called variance, given by:
Where S^2 is the variance, Σ(X[i ]- X ^-)^2 is the sum of the variances for each pixel, n number of total pixels.
If the variance moves to the left (being smaller) indicates that the asymmetry is below of the average temperature, in the other case, (being greater) would indicate that the temperature is above the
average. This is relevant to determine a possible cancerous tissue in anyone breast. Another function is the kurtosis which allow to know how the pixels values are concentrated in the standard
central distribution area. Kurtosis coefficient is obtained by:
Where g[2] is the kurtosis coefficient, X[i] is the value of each pixel, X ^- is the mean and n[i ]is the frequency at which the pixel appears.
Finally, entropy is one of the most used function for detection thermal asymmetries. When temperature distribution in both breasts is similar, the entropy value tends to be the same, and is given by:
Where Prob, corresponds to the probability that a certain pixel value will appear in the image, which is represented as
2.3 Fuzzy Control
The fuzzy control structure, is designed to reduce the computational complexity, since the operations performed in each stage like multiplications, sum, divider, power, etc, are not made using
hardware but look-up tables. The look-up tables play a crucial role in all stages of a Mamdani fuzzy control, and thereby, the operations tend to be very few. It is deriving from the principle of
memory that the look-up tables serve to storing the membership values represented in this study by 8 bits which generate 256-levels in a binary data. Figure. 5, shows the block diagram of the Fuzzy
control proposed. It consists of two inputs and one output. To convert the rigid input values to fuzzy values, data from the ADC converter act as a memory addressing while membership values are
stored in look-up tables. For each rigid input value corresponding one membership value. These membership values are the input to the inference stage where they are processed through the Mamdani
max-min implication, having as a result, the conclusion of the fuzzy rules. In the aggregation stage, the conclusion of each rule is combined and summed to obtain a final conclusion. Then,
defuzzification stage converts the membership values to a rigid value. Finally, it is converted to a voltage using a DAC. All stages are implemented in a FPGA platform using description language VHDL
code. The Output of the Fuzzy control will be the Cancer diagnosis.
2.4 Fuzzification
Input variables are defined based on the difference in the Media (DM) and Kurtosis (DK) values between the left and right breast. Each variable is represented by five fuzzy sets which conform to the
linguistic variables. The fuzzy sets for the DM variable are labeled as VL (Very light), S (Small), R (Regular), L (large) and VH (Very large). The fuzzy sets for the DK variable are the same labels.
The output fuzzy variable is called a cancer diagnosis (CD), and also, is represented by five fuzzy sets labeled as H (healthy), C (caution), M (minimal), D (drastic) and VD (very drastic). Each
variable has a universe of discourse in the digital domain in the range from 0 to 1023, which is proposed using 10 bits of resolution. Figure. 6, shows the fuzzy sets proposed in the universe of
discourse for the inputs and output variables. For the membership values, 8 bits of resolution are used, which divides the membership values in a range from 0 to 255 values, which is adequate because
the range is from 0 to 1. So, each fuzzy set contain 256 levels. The base of each fuzzy set is found for those levels whose membership value is zero. This values are especially useful because they
allow to know if the value of the rigid input signal belongs to a given fuzzy set.
2.5 Inference
For the inference and aggregation stages, Mamdani method is used through the max-min operations. For this, an inference matrix is generated to place the output inference values using rules If…Then by
making a correlation between the membership values obtained in the fuzzification stage, using:
Figure 5: Block diagram of the FLC proposed.
Figure 6: Membership functions for a) inputs and b) output variables.
This way, the matrix will have 25 possible output values obtained from the evaluation of the rules shown in Table I. It involves a comparison between two membership values reached in each rule and
selecting the minimum membership value.
The aggregation stage is carried out by the method of maximum membership value obtained in the inference stage whose value is different to 0, given by:
It is the union of the activated rules for each column in the matrix. In this case, five columns. It takes into account those rules that have a nonzero value. For this case, all values contained in a
column are compared and selecting the maximum value and so on. These membership values fall in the universe of discourse of the output fuzzy sets call Diagnosis cancer, which are used for the
defuzzification stage. This stage converts a fuzzy value to a real value.
Table 1: Inference matrix.
3.3 Defuzzification
In order to save processing time and hardware on the FPGA, defuzzification with 256-levels are used to obtain the rigid output value from the aggregation vector, given by:
Where DC[Deff] is the diagnosis cancer, N[,] is the number of fuzzy set activated in the aggregated vector, is the final point of the level where membership value reached in the aggregated vector, is
the initial point of the same level. The schematic design in VHDL for the FLC implemented is shown in Figure 7, which consist of fuzzification, inference (min), aggregation (max) and defuzzification
Figure 7: Schematic design for the FLC implemented.
4.Experimental Results
Measurement results were carried out taking into account a database belonging to the Visual Lab DMR (Database for Mastology Research), which is an online platform that stores mastological images for
the breast cancer by your research group. Registration results of this group were compared with this proposal. To determine if the thermographs correspond to a group of healthy or cancer, a series of
minimum and maximum values for the difference of Mean, Kurtosis and Entropy were taken. Also, and according to the method used by Visual Lab, the following was taking into account:
1. To determine a normal thermal asymmetry, difference between average Media values belonging to the right and left breasts can have a maximum value of 3.
2. Pixels distribution around the average value for each breast must be similar for both breasts. The difference between these values must not be greater than 0.63 to be considered as a normal
3. Difference between the entropy values can be until 0.284, to determine a normal thermal asymmetry.
This manner, for diagnosing cancer, tests were performed on 44 of 72 thermographies. Tests were carried out connecting two probes to the FLC inputs through ADC cables in a digital domain according to
results presented in Table II, which corresponding to the difference between Media and Kurtosis, (only 20 tests values are presented). The internal clock signal to synchronize the FLC process
implemented from advanced VHDL synthesis was establish at 500 KHz. The implementation can be used for five or more linguistic variables, but can be easily expanded repeating sections of code, also
can be used for any type of fuzzy set. This manner, the FLC determine thermal asymmetry between both breasts, diagnosing the grade of cancer. The experimental results were compared in a theoretical
way with the method obtained by the Matlab Fuzzy Logic ToolBox. For example, considering the values of test 2, whose values are DE= 2.3199 and DK= 1.8546 introduced as inputs x and y. The digital
output from the FPGA board was CAUTION whose digital value is 319, corresponding to the caution set, whereas in MatLab Fuzzy Logic ToolBox, the digital value was 333. As can be seen, the FLC
implemented on the FPGA provides very approximate values to those obtained by the Matlab Fuzzy Logic ToolBox. Then, each simulation and experimental result was evaluated with respect to the Visual
Lab researchers result, which was classified as healthy or cancer presented in Table III in the column named Status and in FLC/FPGA column our results. Result differs from the expert group, perhaps
because the values, not represent a specific temperature since the colors are not linked to the temperatures, since these are relative to the maximum and minimum temperature of the object being
photographed. As can be seen, results obtained here (with a relatively large slot of tests), fall into the healthy case where the values for the input signals representing the difference of Media and
Kurtosis. This manner, 23 tests of 44, are similar to the Visual Lab group whose diagnosis were healthy, 2 were cancerous, whereas 8 were Caution, 9 were Medium and 2 were Drastic which differ from
the Visual Lab group whose diagnosis were healthy. One therefore would expect little change in the diagnosis under different pixels’ distribution of the image due to its luminance parameter. Also,
there would likewise be little variation in the image when some interferences are considered. Even though the results obtained cannot be ignored since difference is not greater than 15%, which tells
us that the system proposed has around 85% accuracy. Finally, the mean square error (MSE) is obtained applying (20), to the data values of Table II. The MSE is 1.75%, which is very acceptable
considering that this method uses iterations and 8 bits for sample.
Table 2: Theoretical stochastic values.
Table 3: Comparison between medical (VisualLab) results and those obtained in Matlab and FLC/FPGA.
Test VisualLab Matlab FLC/FPGA Fuzzy Set
2 Healthy 333 319 Caution
4 Healthy 128 127 Healthy
6 Healthy 538 511 Medium
15 Healthy 179 203 Healthy
16 Healthy 538 511 Medium
20 Healthy 128 127 Healthy
24 Healthy 128 127 Healthy
26 Healthy 538 511 Medium
31 Healthy 333 319 Healthy
32 Healthy 333 319 Healthy
36 Healthy 333 319 Healthy
48 Healthy 128 127 Healthy
52 Healthy 128 127 Caution
57 Healthy 128 127 Healthy
62 Healthy 128 127 Healthy
65 Healthy 742 703 Drastic
66 Healthy 128 127 Healthy
70 Healthy 538 511 Medium
91 Healthy 128 127 Healthy
93 Healthy 371 395 Caution
97 Healthy 333 319 Caution
99 Healthy 128 127 Healthy
104 Healthy 333 319 Caution
105 Healthy 128 127 Healthy
106 Healthy 329 311 Caution
110 Healthy 742 703 Drastic
132 Healthy 437 415 Medium
135 Healthy 538 511 Medium
137 Healthy 128 127 Healthy
145 Healthy 128 127 Healthy
147 Healthy 128 127 Healthy
151 Healthy 436 415 Medium
152 Healthy 333 319 Caution
155 Healthy 128 127 Healthy
161 Healthy 128 127 Healthy
163 Healthy 128 127 Healthy
168 Healthy 128 127 Healthy
169 Healthy 210 223 Healthy
174 Healthy 333 319 Caution
180 Cancer 775 799 Drastic
181 Cancer 333 319 Caution
188 Healthy 128 127 Healthy
189 Healthy 538 511 Medium
190 Healthy 538 511 Medium
This paper compares and experimentally confirm breast cancer detection through the modeling, simulation, and measurement using stochastic techniques in a FPGA platform. Stochastic equations were
developed for analysis of infrared images, and a fuzzy logic control was designed and implemented to illustrate the relationship to the stochastic results for detection cancer. The fuzzy controller
provides very approximate values to detect at multiple grades or in a universe of discourse, different grades of cancer. Results are provided in which simulations and measurement show stable
performance in the processes and in the must cases are according to the Visual Lab group diagnosing. The technique has a high percentage of accuracy in terms of cancer detection.
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8. Youssef A, El-Telbany M, Zekry A. The role of artificial intelligence in photo-voltaic systems design and control: a review. Renew Sustain Energy Rev 78 (2017): 72–79
9. Gad HH, Haikal AY, Ali HA. New design of the PV panel control system using FPGA-based MPSoC. Sol Energy 146 (2017): 243–256.
10. Chekired F, Larbes C, Mellit A. Comparative study between two intelligent MPPT-controllers implemented on FPGA: application for photovoltaic systems. Int J Sustain Energy 33 (2014): 483–499.
11. Lughofer E, Sayed-Mouchaweh M. Autonomous data stream clustering implementing split-and-merge concepts: towards a plug-and-play approach. Inf Sci 304 (2015): 54–79.
12. de la Cruz-Alejo J, R. Antonio-Méndez and M Salazar-Pereyra. Fuzzy logic control on FPGA for two axes solar tracking. Neural Computing and Applications 31 (2019): 2469-2483.
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On dynamical reconstruction of an unknown input in the parabolic obstacle problem
A problem of dynamical reconstruction of a control for a parabolic obstacle problem is considered. A solving algorithm for this problem is presented. This algorithm is stable with respect to
informational noise and computational errors. It adaptively takes into account inaccurate measurements of phase trajectories and is regularizing in the sense that the final result becomes better as
the input information becomes more accurate. The algorithm suggested in the paper is based on the theory of positional control. The main element of this algorithm is a procedure of stabilizing some
auxiliary functionals of the Lyapunov type.
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Lesson Plans for Primary, Junior and Intermediate!
See below for math activities for Teachers and Parents!
Some lesson plans were developed from a collaboration between the MKN and London District Catholic School Board.
Primary (K-3) Lesson Plans
WEEK 1 – Probability Game English French
Curriculum Expectation: Describe the probability that an event will occur through investigation with simple games and probability experiments and using mathematical language
WEEK 2 – Measurement Olympics English French
Curriculum Expectation: Estimate and measure distance using standard units (i.e., centimetre, metre) and non-standard units
WEEK 3 – Grocery Shopping Math English French
Curriculum Expectation: Describe the relative locations (e.g., beside, two steps to the right of) and the movements of objects on a map
WEEK 4 – Daily Physical Education Coding English French
Curriculum Expectation: Solve problems involving the addition and subtraction of whole numbers to 18, using a variety of mental strategies
WEEK 5 – Art Attack Math English French
Curriculum Expectation: Identify and describe various polygons (i.e., triangles, quadrilaterals, pentagons, hexagons, heptagons, octagons) and sort and classify them by their geometric properties
(i.e., number of sides or number of vertices), using concrete materials and pictorial representations
WEEK 6 – Race in Space English French
Curriculum Expectation: Describe relationships between quantities by using whole-number addition and subtraction (e.g., “If you ate 7 grapes and I ate 12 grapes, I can say that I ate 5 more grapes
than you did”)
WEEK 7 – Geometry Dance English French
Curriculum Expectation: Describe relationships between quantities by using whole-number addition and subtraction (e.g., “If you ate 7 grapes and I ate 12 grapes, I can say that I ate 5 more grapes
than you did”)
WEEK 8 – Patterning Fish English French
Curriculum Expectation: Demonstrate, through investigation, an understanding that a pattern results from repeating an operation (e.g., addition, subtraction) or making a repeated change to an
attribute (e.g., colour, orientation).
WEEK 9 – Race Against the Clock English French
Curriculum Expectation: Tell and write time to the quarter-hour, using demonstration digital and analogue clocks (e.g., “My clock shows the time recess will start [10:00]).
WEEK 10 – Fraction Bingo English French
Curriculum Expectation: Divide whole objects into parts and identify and describe, through investigation, equal-sized parts of the whole, using fractional names (e.g., halves; fourths or quarters).
WEEK 11 – Roll and Build with Shapes English French
Curriculum Expectation: Compose and describe pictures, designs, and patterns by combining two-dimensional shapes (e.g., “I made a picture of a flower from one hexagon and six equilateral
WEEK 12 – $1 in My Wallet English French
Curriculum Expectation: Estimate, count, and represent (using the ¢ symbol) the value of a collection of coins with a maximum value of one dollar.
Junior (4-6) Lesson Plans
WEEK 1 –My Friend Robot Geometry English French
Learning Outcome: Construct triangles, using a variety of tools (e.g., protractor, compass, dynamic geometry software), given acute or right angles and side measurements
WEEK 2 – Angle Game English French
Learning Outcome: Classify and construct polygons and angles; measure and construct angles up to 180° using a protractor, and classify them as acute, right, obtuse, or straight angle
WEEK 3 – Budget a Dream Vacation English French
Learning Outcome: Add and subtract decimal numbers to hundredths, including money amounts, using concrete materials, estimation, and algorithms
WEEK 4 – Flying Carpet Game English French
Learning Outcome: Explain how a coordinate system represents location, and plot points in the first quadrant of a Cartesian coordinate plane
WEEK 5 – 21 and Out English French
Learning Outcome: Represent, using a common fraction, the probability that an event will occur in simple games and probability experiments
WEEK 6 – Scratch Net Creation English French
Learning Outcome: Construct nets of prisms and pyramids, using a variety of tools
WEEK 7 – One Metre Dash English French
Learning Outcome: Select and justify the most appropriate standard unit (i.e., millimetre, centimetre, decimetre, metre, kilometre) to measure length, height, width, and distance, and to measure the
perimeter of various polygons
WEEK 8 – Create a Parachute English French
Learning Outcome: Collect data by conducting a survey or an experiment…to do with themselves, their environment, issues in their school or community, or content from another subject, and record
observations or measurements
WEEK 9 – Triangle Coding English French
Learning Outcome: Estimate, measure using a variety of tools (e.g., centimetre grid paper, geoboard) and strategies, and record the perimeter and area of polygon
WEEK 10 – Tessellation Art English French
Learning Outcome: Determine, through investigation using a variety of tools… polygons or combinations of polygons that tile a plane, and describe the transformation(s) involved
WEEK 11 – Roll and Build with Shapes English French
Learning Outcome: Solve problems requiring the estimation and calculation of perimeters and areas of rectangles
WEEK 12 – Shooting Hoops English French
Learning Outcome: Select an appropriate type of graph to represent a set of data, graph the data using technology, and justify the choice of graph.
Intermediate (7-8) Lesson Plans
WEEK 1 –My Friend Robot Geometry English French
Learning Outcome: Determine through investigation using a variety of tools … the relationships among area, perimeter, corresponding side lengths and corresponding angles of congruent shapes
WEEK 2 – Dream Trip Planning English French
Learning Outcome: Model real-life relationships involving constant rates where the initial condition starts at 0 (e.g., speed, heart rate, billing rate), through investigation using tables of values
and graphs
WEEK 3 – Flying Carpet Game English French
Learning Outcome: Plot points using all four quadrants of the Cartesian coordinate plane
WEEK 4 – Shopping List English French
Learning Outcome: Solve problems involving the calculation of unit rates
WEEK 5 – Equivalent Volume Coding English French
Learning Outcome: Sketch different polygonal prisms that share the same volume
WEEK 6 – Dilate a Comic Strip English French
Learning Outcome: Identify, perform, and describe dilatations (i.e., enlargements and reductions), through investigation using a variety of tools
WEEK 7 – Recipe Proportions English French
Learning Outcome: Identify and describe real-life situations involving two quantities that are directly proportional
WEEK 8 – Trapezoid Code English French
Learning Outcome: Determine, through investigation using a variety of tools the relationship for calculating the area of a trapezoid, and generalize to develop the formula
WEEK 9 – M&M Probability English French
Learning Outcome: Pose and solve simple probability problems, and solve them by conducting probability experiments and selecting appropriate methods of recording the results (e.g., tally chart, line
plot, bar graph)
WEEK 10 – Popular Data Stats English French
Learning Outcome: Determine, through investigation, the appropriate measure of central tendency (i.e., mean, median, or mode) needed to compare sets of data
WEEK 11 – Weight Estimation English French
Learning Outcome: Solve problems that require conversion between metric units of measure (e.g., millimetres and centimetres, grams and kilograms, millilitres and litres)
WEEK 12 – Geometric Dance English French
Learning Outcome: Sort and classify triangles and quadrilaterals by geometric properties related to symmetry, angles, and sides, through investigation using a variety of tools.
WEEK 13 – Shopping in Foreign Currency English French
Learning Outcome: Identify and compare exchange rates, and convert foreign currencies to Canadian dollars and vice versa.
Intermediate (9) Lesson Plans
WEEK 14 – Shopping for a Phone Plan English French
Learning Outcome: Determine graphically the point of intersection of two linear relations, and interpret the intersection point in the context of an application.
WEEK 15 – Line of Best Fit English French
Learning Outcome: Construct tables of values, scatter plots, and lines or curves of best fit as appropriate using a variety of tools.
WEEK 16 – Interpreting Graphs English French
Learning Outcome: Describe the effects on a linear graph and make the corresponding changes to the linear equation when the conditions of the situation they represent are varied.
WEEK 17 – Architect of a Zoo English French
Learning Outcome: Solve problems involving the areas and perimeters of composite two-dimensional shapes.
WEEK 18 – Brick a House English French
Learning Outcome: Solve problems using the Pythagorean theorem, as required in applications.
WEEK 19 – Changing Linear Graphs English French
Learning Outcome: Describe the effects on a linear graph and make the corresponding changes to the linear equation when the conditions of the situation they represent are varied.
WEEK 20 – Recreation Centre Planning English French
Learning Outcome: Rearrange formulas involving variables in the first degree, with and without substitution (e.g., in analytic geometry, in measurement).
WEEK 21 – Rebound Height Experiment English French
Learning Outcome: Pose problems, identify variables, and formulate hypotheses associated with relationships between two variables.
WEEK 22 – The Next Best Chocolate Bar English French
Learning Outcome: Determine the maximum area of a rectangle with a given perimeter by constructing a variety of rectangles, using a variety of tools (e.g., geoboards, graph paper, tooth- picks, a
pre-made dynamic geometry sketch), and by examining various values of the area as the side lengths change and the perimeter remains constant.
WEEK 23 – Dream House Design Challenge English French
Learning Outcome: Solve problems that require maximizing the area of a rectangle for a fixed perimeter or minimizing the perimeter of a rectangle for a fixed area.
WEEK 24 – Balanced Budget English French
Learning Outcome: Identify different ways to maintain a balanced budget, and use appropriate tools to track all income and spending, for several different scenarios.
WEEK 25 – Large Purchase Budgeting English French
Learning Outcome: Create a financial plan to reach a long-term financial goal, accounting for income, expenses, and tax implications.
Count on MKN filmed videos for a teacher to go along with the Lesson Plans! Click the links below to watch:
Primary (K-3) Lesson Videos
WEEK 1 – Probability Game English
Curriculum Expectation: Describe the probability that an event will occur through investigation with simple games and probability experiments and using mathematical language
WEEK 2 – Measurement Olympics English
Curriculum Expectation: Estimate and measure distance using standard units (i.e., centimetre, metre) and non-standard units
WEEK 3 – Grocery Shopping Math English
Curriculum Expectation: Describe the relative locations (e.g., beside, two steps to the right of) and the movements of objects on a map
WEEK 4 – Daily Physical Education Coding English
Curriculum Expectation: Solve problems involving the addition and subtraction of whole numbers to 18, using a variety of mental strategies
WEEK 6 – Race in Space English
Curriculum Expectation: Describe relationships between quantities by using whole-number addition and subtraction (e.g., “If you ate 7 grapes and I ate 12 grapes, I can say that I ate 5 more grapes
than you did”)
WEEK 10 – Fraction Bingo English
Curriculum Expectation: Divide whole objects into parts and identify and describe, through investigation, equal-sized parts of the whole, using fractional names (e.g., halves; fourths or quarters).
WEEK 11 – Roll and Build With Shapes English
Curriculum Expectation: Compose and describe pictures, designs, and patterns by combining two-dimensional shapes (e.g., “I made a picture of a flower from one hexagon and six equilateral
Junior (4-6) Lesson Videos
WEEK 2 – Angle Game English
Learning Outcome: Classify and construct polygons and angles; measure and construct angles up to 180° using a protractor, and classify them as acute, right, obtuse, or straight angle
WEEK 5 – 21 and Out English
Learning Outcome: Represent, using a common fraction, the probability that an event will occur in simple games and probability experiments
WEEK 7 – One Metre Dash English
Learning Outcome: Select and justify the most appropriate standard unit (i.e., millimetre, centimetre, decimetre, metre, kilometre) to measure length, height, width, and distance, and to measure the
perimeter of various polygons
WEEK 11 – Capture the grid English
Learning Outcome: Solve problems requiring the estimation and calculation of perimeters and areas of rectangles
Intermediate (7-8) Lesson Videos
WEEK 4 – Shopping List English
Learning Outcome: Solve problems involving the calculation of unit rates
Intermediate (9) Lesson Videos
WEEK 16 – Interpreting Graphs English
Learning Outcome: Describe the effects on a linear graph and make the corresponding changes to the linear equation when the conditions of the situation they represent are varied.
WEEK 17 – Architect of a Zoo English
Learning Outcome: Solve problems involving the areas and perimeters of composite two-dimensional shapes.
WEEK 22 – The Next Best Chocolate Bar English
Learning Outcome: Determine the maximum area of a rectangle with a given perimeter by constructing a variety of rectangles, using a variety of tools (e.g., geoboards, graph paper, tooth- picks, a
pre-made dynamic geometry sketch), and by examining various values of the area as the side lengths change and the perimeter remains constant.
For teachers and parents, see below for weekly 15 minute drop-in video sessions designed to help you understand how to implement the lessons and to allow you to ask any questions you may have!
If you are a teacher and would like to have one of our teacher candidates join your virtual classroom to help out with the implementation of any of our lesson plans, please send an email to
bdickso9@uwo.ca for English classrooms and kjohn283@uwo.ca for French classrooms.
Teacher Weekly Video Sessions
Parent Weekly Video Sessions
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Approximation of double Walsh–Fourier series by means of the matrix transform
Approximation of double Walsh–Fourier series by means of the matrix transform
Keywords: Walsh group; Walsh system; Walsh-Fourier series; N\
UDC 517.5
We discuss the rate of approximation of partial sums of the double Walsh–Fourier series in the spaces $L^p(G^2),$ $1\leq p <\infty,$ and $C(G^2)$ by the matrix transform.
L. Baramidze, L.-E. Persson, G. Tephnadze, P. Wall, Sharp $H_{p}-L_{p}$ type inequalities of weighted maximal operators of Vilenkin–Nörlund means and its applications, J. Inequal. and Appl., Article
242 (2016).
I. Blahota, G. Gát, Norm and almost everywhere convergence of matrix transform means of Walsh–Fourier series, Acta Univ. Sapientiae Math., 15, № 2, 244–258 (2023).
I. Blahota, K. Nagy, Approximation by Θ-means of Walsh–Fourier series, Anal. Math., 44, № 1, 57–71 (2018).
I. Blahota, K. Nagy, Approximation by matrix transform of Vilenkin–Fourier series, Publ. Math. Debrecen, 99, № 1-2, 223–242 (2021).
I. Blahota, K. Nagy, Approximation by Marcinkiewicz-type matrix transform of Vilenkin–Fourier series, Mediterr. J. Math., 19, Article 165 (2022).
I. Blahota, K. Nagy, M. Salim, Approximation by Θ-means of Walsh–Fourier series in dyadic Hardy spaces and dyadic homogeneous Banach spaces, Anal. Math., 47, 285–309 (2021).
I. Blahota, K. Nagy, G. Tephnadze, Approximation by Marcinkiewicz Θ-means of double Walsh–Fourier series, Math. Inequal. Appl., 22, № 3, 837–853 (2019).
I. Blahota, G. Tephnadze, A note on maximal operators of Vilenkin–Nörlund means, Acta Math. Acad. Paedagog. Nyházi., 32, № 2, 203–213 (2016).
Á. Chripkó, Weighted approximation via Θ-summations of Fourier–Jacobi series, Studia Sci. Math. Hungar., 47, № 2, 139–154 (2010).
T. Eisner, The Θ-summation on local fields, Ann. Univ. Sci. Budapest. Sect. Comput., 33, 137–160 (2011).
S. Fridli, P. Manchanda, A. H. Siddiqi, Approximation by Walsh–Nörlund means, Acta Sci. Math., 74, 593–608 (2008).
U. Goginava, On the approximation properties of Cesàro means of negative order of Walsh–Fourier series, J. Approx. Theory, 115, 9–20 (2002).
U. Goginava, K. Nagy, Matrix summability of Walsh–Fourier series, Mathematics, MDPI, 10, № 14 (2022).
E. Hewitt, K. Ross, Abstract harmonic analysis, vols. I, II, Springer-Verlag, Heidelberg (1963).
M. A. Jastrebova, On approximation of functions satisfying the Lipschitz condition by arithmetic means of their Walsh–Fourier series, Mat. Sb., 71, 214–226 (1966) (in Russian).
N. Memić, L.-E. Persson, G. Tephnadze, A note on the maximal operators of Vilenkin–Nörlund means with non-increasing coefficients, Stud. Sci. Math. Hungar., 53, № 4, 545–556 (2016).
F. Móricz, B. E. Rhoades, Approximation by weighted means of Walsh–Fourier series, Int. J. Math. Sci., 19, № 1, 1–8 (1996).
F. Móricz, A. Siddiqi, Approximation by Nörlund means of Walsh–Fourier series, J. Approx. Theory, 70, 375–389 (1992).
K. Nagy, Approximation by Nörlund means of quadratical partial sums of double Walsh–Fourier series, Anal. Math., 36, № 4, 299–319 (2010).
K. Nagy, Approximation by Nörlund means of double Walsh–Fourier series for Lipschitz functions, Math. Inequal. Appl., 15, № 2, 301–322 (2012).
F. Schipp, W. R. Wade, Norm convergence and summability of Fourier series with respect to certain product systems, Pure and Appl. Math. Approx. Theory, Marcel Dekker, New York etc. (1992), p.
F. Schipp, W. R. Wade, P. Simon, J. Pál, Walsh series. An introduction to dyadic harmonic analysis, Adam Hilger, Bristol, New York (1990).
V. A. Skvortsov, Certain estimates of approximation of functions by Cesàro means of Walsh–Fourier series, Mat. Zametki, 29, 539–547 (1981) (in Russian).
R. Toledo, On the boundedness of the $L^1$-norm of Walsh–Fejer kernels, J. Math. Anal. and Appl., 457, № 1, 153–178 (2018).
F. Weisz, $Theta$-summability of Fourier series, Acta Math. Hungar., 103, № 1-2, 139–175 (2004).
F. Weisz, $Theta$-summation and Hardy spaces, J. Approx. Theory, 107, 121–142 (2000).
F. Weisz, Several dimensional $Theta$-summability and Hardy spaces, Math. Nachr., 230, 159–180 (2001).
Sh. Yano, On Walsh–Fourier series, Tohoku Math. J., 3, 223–242 (1951).
Sh. Yano, On approximation by Walsh functions}, Proc. Amer. Math. Soc., 2, 962–967 (1951).
A. Zygmund, Trigonometric series, 3rd ed., vols. 1, 2, Cambridge Univ. Press (2015).
How to Cite
BlahotaI. “Approximation of Double Walsh–Fourier Series by Means of the Matrix Transform”. Ukrains’kyi Matematychnyi Zhurnal, Vol. 76, no. 5, June 2024, pp. 664 -79, doi:10.3842/umzh.v76i5.7397.
Research articles
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Change of variables in summations
• Thread starter Lajka
• Start date
In summary, the conversation is about a problem encountered while studying the decimation process in digital signal processing and the lack of information on the definition for changing variables in
summations. The issue is resolved by using an ad-hoc method and it is discussed how this method compares to using a sequence of deltas. The lack of a specific rule for this type of substitution is
also mentioned.
although this may sound trivial, I stumbled upon this problem while studying decimation process in digitial signal processing. I can't find anything on the web about some definition for the change of
variables in sumations (as there is one for integrations), so maybe someone here could help me.
Consider the sum
(the summation range is [-inf, +inf])
If I just do the substitution [itex]m=2n+5[/itex] and get this
it wouldn't be right.
Decimation leads to irreversible changes, aka, I should still have the sum of elements [itex]x[1], x[3], x[5],...[/itex], but, somehow, I now have the sum of
elements of [itex]x[n][/itex] with this simple substitution.
So, I think the right answer would be
I did this ad-hoc, using logic. I was wondering if there is a proper definition for the change of variables in summations, which takes into the account the effects of decimation (which do not exist
in continuous case of course)?
Thanks in advance :)
The problem is that, if n goes from minus to plus infinity, then m = 2n + 5 also goes from minus to plus infinity, but taking only odd values. So you could write
This because the extra term I introduced is 0 for even m and 1 for odd m.
Yeah, that does the same thing as the sequence of deltas, but I kinda like yours more :D
I guess there isn't a rule for this 'substitution' because it's a trivial matter, but I wanted to check it still.
Thank you for your response!
FAQ: Change of variables in summations
1. What is a change of variables in summations?
A change of variables in summations is a mathematical technique used to simplify or evaluate a summation by substituting new variables for the original variables. This can help to make the summation
more manageable and easier to work with.
2. Why is a change of variables useful in summations?
A change of variables can make it easier to evaluate or simplify a summation by allowing us to use known summation formulas or properties. It can also help to identify patterns or relationships
within the summation.
3. How do you perform a change of variables in summations?
To perform a change of variables in summations, you first need to identify the original variables and the new variables you want to use. Then, you can substitute the new variables into the summation
and simplify the expression using algebraic manipulation.
4. What are some common examples of change of variables in summations?
Some common examples of change of variables in summations include using the index shift property, using the geometric series formula, or using trigonometric identities to simplify the summation.
5. How can a change of variables be used to solve real-world problems?
In real-world applications, a change of variables can be used to model and solve problems involving discrete quantities, such as counting the number of possible outcomes or calculating probabilities.
It can also be used to analyze data and identify trends or patterns.
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If you could only learn one major branch of mathematics for the rest of your life, what would it be?
Dylan Torres
If you could only learn one major branch of mathematics for the rest of your life, what would it be?
>inb4 analysis
constructor theory
This is not well-defined.
I was gonna type analysis
Fellow analysis-fag
Too bad analysis is mostly brainlet-tier if we never introduce other branches
Geometry topology and abstract-lie-field-ring-knot algebra have no use whatsoever though
>Geometry topology and abstract-lie-field-ring-knot algebra have no use whatsoever though
[insert a brainlet image macro]
>could give me a countersrgumebt or an example of a use but instead le brainlet wojak :0——
algebra obviously
>>could give me a countersrgumebt or an example of a use but instead le brainlet wojak :0——
Who are you quoting?
Your retardation
cmon how the fuck u gonna say anything else? its only one that applies to shit you see irl
History of Mathematics, especially PM and Godel and subsequent developments. Or "foundations" if I get to change my mind a bit.
Differential geometry. At the end I would have to learn topology, and anlisis so got you.
covering Differential Geometry and Algebraic Geometry,
so requiring Topology, Abstract Algebra & Analysis
My man
Spend three racks of new ring
Group gang group gang group gang
Geometry isn't math, try again
Non-discrete mathematics
Mathematical logic.
Computer Science
Are you memeing? If not, explain yourselves...
Analysis is the non-pajeet answer. Nobody with self esteem would refuse the branch that combines the only parts of topology, measure theory, linear algebra and abstract algebra that actually matter.
Mathematical Physics
>Mathematical Mathematics
So Pure Mathematics?
Homological algebra
Set theory
this is an advanced alien-like joke
Either mechanical engineering or civil engineering
>all these people not choosing number theory
>number theory
Mechanics. Seems useful to any company related to some discipline of engineering.
Complex Analysis
Discrete math is the only interesting math.
>Mathematical logic.
Not a branch of mathematics.
Noah Gutierrez
Blake Martinez
Dylan Davis
Christopher Perez
Samuel Miller
Adam Myers
Joseph Hall
Dylan Smith
Isaac Barnes
Blake Harris
Josiah Sanders
Juan Reyes
Gavin Harris
Matthew Perez
Asher Hill
Lincoln Perez
Gavin Jackson
Julian Ortiz
Adam Roberts
Mason Howard
Brody Lopez
Austin Wilson
Jose Foster
Christian Peterson
Hudson Sanchez
Logan Bennett
Jeremiah Lopez
Cameron Kelly
Gavin Brooks
Oliver Jackson
Thomas Bennett
Gavin Walker
Austin King
Austin Lee
Hudson Perry
Camden Long
Nathan Campbell
Juan Roberts
Grayson Morales
Owen Diaz
Nathaniel Miller
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Weekly Progress Report
Make that Friday.
You know, I love the chickens, but they're not really into the whole exercise scene. I think I need to find some different cartoons.
Anyways...I hope this past week of exercise wan't a painful, soul-destroying experience. Let us know how you're doing in the comments, and don't forget to include your cumulative miles so we can brag
on your progress in the sidebar.
**Ooops...sorry for the delay! I prescheduled, but forgot to set the time.
32 comments:
Slowly chipping away the miles. I'm currently at 59 miles. Tomorrow is the 4-mile race that I've been training for. Wish me luck.
*sigh* Why does blogger have to be difficult with the spacing??
So. 6 miles for me, plus a whole 2 miles from last week, which I never posted, for a grand total of 41 miles. *another sigh*
And Amanda, if you are reading this...I will update the sidebar miles tonight. Promise!
This comment has been removed by the author.
Only 4 measly miles for me this week. The total is now 33. It was in the 100s all week . . .
A bad week. Only up to 27 miles. : )
Well, my pedometer broke while on vacation but I know I have walked a ton! To be conservative, I will say I have walked 5 miles (above and beyond what I would normally do) for a grand total so
far of 54 miles. I will be back home next week and on the treadmill for accurate counts.
hello my lovies!!
I'm leaving my mark with 22 total. Still trying to figure out how to find time to exercise, especially since Elle has gone from sleeping 7-8 hour stretches to THREE hour stretches and this mama
is in a perpetual zombie state. But, glad to be on the board...finally.
Hip Hip Horray!!!!
Another 23 miles for me this week, bringing my total to 118 so far.
Back on track this week with 20 miles for a GRAND TOTAL of 65.
Total of 75 miles...
crappy week. :/
I'm thinking of a word and it starts with L...it ends with...aw heck, it's lazy. I'm even too lazy to finish the game. Just an indication of how this week went. Total of 34 miles for me!
Ugh...a grand total of 27 miles for me. Vacay and weather in the 100's is killing my exercise mojo (but the vacay is great for the soul!)
Yay! I was able to get in 6 miles this week. Way below normal for me, but at this point I'll take anything.
Total 34
I have 47 miles. Have a good week everyone!
DebK 69 miles!!!!!!!!
Let's see...this week 7 miles. So that brings the total to 22. Next week I'll be at the cottage and hope to run or bike every day. Look for a big boost in my numbers in 2 weeks then!
9 for a total of 43.
Got 6 miles in - not quite the goal I had, but defineatly more than if we didn't have this challegne.
Total 30 miles in.
I'm at 99 total - not bad for two weeks of vacation time...
Oh! Jill - just saw your comment - thanks for doing the updates - no idea why the sidebar is looking so funny. Stupid spacing issues...
7 miles this week for a total of 41 miles. It really needs to cool down a bit!
Joy (MO) here: I'm at 38 miles.
Slow week: just 3 miles. Cumulative sits at 37.
Thanks, Jill and Amanda! (BTW I love the chickens ... even if they aren't into exercise.)
I'm at 35 miles now. Hope to get in some cycling this week, if the weather behaves (rain forecast).
12 miles for me this week, for a total of 39 now.
Slow week, but managed to crank out five miles on Thursday. 16 miles for the week, 82 miles altogether!
Melissa 12 more miles!! That does not count all the hills and walking I did at Hershey park, that was my bonus:) 73 miles total!!
One more mile 32 so far...
I'm at 28 miles total. A LONG way to go.
25 miles this week for a total of 52 miles total
Sorry I'm late posting this. I was not on the internet much this past week. My new total is 25. I got to do a lot of walking this week.
School started so it has been difficult to get going with the exercise. And obviously, I've also not been online very much. Sorry for posting this late, but here is my update for last week: 20
That gives me a grand total so far of 59! Woohoo!
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There are so many kinds of numbers we deal with on a regular basis and the C++ Standard Library has a full suite of tools to deal with them. Today we’ll look into random numbers, ratios, mathematical
constants, bit manipulation, complex numbers, and more!
Read the rest of this article »
A couple of years ago I posted a class that generats pseudo-random numbers in a repeatable way. This is useful for a variety of tasks, but a recent comment reminded me that I hadn’t tested its
performance. Today I’ll pit my repeatable random function against the standard Math.random function as well as Skyboy’s repeatable random class. Read on for the results!
Read the rest of this article »
Everybody knows about Math.random(), and for good reason. It’s pretty much the way to get random numbers in AS3, AS2, and JavaScript other than bizarre alternatives like AS3’s BitmapData.noise().
However, it has one critical problem that arises when you want to repeat a certain test or prevent game cheaters from exploiting the randomizer until they get an “easy” setup or desirable outcome.
This problem is the lack of repeatability.
Read the rest of this article »
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A E C1 2013 Q11
4.5 - 2013 Q11
Figure 2 shows a sketch of the curve H with equation
A) Give the coordinates of the point where H crosses the x-axis. (1)
B) Give the equations of the asymptotes to H. (2)
C) Find an equation for the normal to H at the point P(–3, 3). (5)
This normal crosses the x-axis at A and the y-axis at B.
D) Find the length of the line segment AB. Give your answer as a surd. (3)
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Deep holes in vertex operator algebras
Add to your list(s) Download to your calendar using vCal
If you have a question about this talk, please contact nobody.
NCN2 - New connections in number theory and physics
A deep hole of a lattice is a point in the ambient space which has maximal distance to all lattice points. Borcherds showed in 1985 that there is a bijection between the deep holes in the Leech
lattice and the Niemeier lattices with roots. Together with the classification of the deep holes in the Leech lattice by Conway, Parker and Sloane this implies the classification of the Niemeier
lattices with non-trivial root system. In this talk we explain how the notion of a deep hole can be generalised to holomorphic vertex operator algebras of central charge 24 and how they can be used
to classify these vertex operator algebras. This is joint work with Sven Möller.
This talk is part of the Isaac Newton Institute Seminar Series series.
This talk is included in these lists:
Note that ex-directory lists are not shown.
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Cylinder Calculators | List of Cylinder Calculators
List of Cylinder Calculators
Cylinder calculators give you a list of online Cylinder calculators. A tool perform calculations on the concepts and applications for Cylinder calculations.
These calculators will be useful for everyone and save time with the complex procedure involved to obtain the calculation results. You can also download, share as well as print the list of Cylinder
calculators with all the formulas.
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Hey, guys. So now that I know the basics of completely inelastic collisions, I want to show you another type of problem where you have a mass that's added onto a system that's already moving. So
let's go ahead and check this out. The problem we're going to work out down here, we have a sled that's already moving at some speed, and a box gets dropped onto it, so you're adding mass to a system
that's already moving. Basically, the idea is that they're very similar to completely inelastic collisions. Whenever this happens, both objects in the system, right, whatever you had before like the
sled plus the new mass like the box, have to be moving with the same final velocity. So let's go ahead and take a look at our problem, and we'll come back to this in just a second here. So we have
this 70 kilogram sled and a 30 kilogram box. The sled's already moving to the right with 10 meters per second, and then what happens is the box gets dropped onto it. In part a, we want to figure out
the final speed of the system. So let's draw our diagrams for before and after. This is the before. What does the after look like? Well, basically now, this sled is moving to the right, but the box
is on top of it. So the box is on top of it like this, and because they're on top of it and they basically become one system or one object, they're both moving with the same v final. And that's what
we want to figure out here. So let's go ahead and take a look at our energy or, sorry, our momentum conservation equation. So we're going to use \( m_1 v_{1 \text{ initial}} + m_2 v_{2 \text{
initial}} \), and we can use our shortcut for completely inelastic collisions. We know that we can group together the masses because they're both going to have the same \( v_{\text{final}} \). That's
what we want to figure out here. So I'm going to just call this, let's see. I'm going to call the sled 1 and the box 2. Alright? So let's take a look. So we're going to have our sled, that's 70.
That's \( 70 \times 10 + 30 \times \) some speed here equals \( 70 + 30 \times v_{\text{final}} \). So what goes inside of this parenthesis right here? What's the initial speed of the block? Well,
what happens here is that this block gets dropped vertically onto the sled, which means that it has some initial y velocity that I actually don't know. But it turns out it doesn't matter that I don't
know it because the box is only dropping vertically, which means it doesn't contribute any x momentum to the system. Basically, if all the velocity is just vertical, what we can say is that \( v_{2x}
\) is just equal to 0. It doesn't contribute any horizontal momentum to the system. So what we can do is in our momentum conservation equation, we can just cancel this out and there's actually 0
initial speed. So this actually makes our equation even simpler because now we can do is figure out \( v_{\text{final}} \). So now what we can do is this is 700, and when we divide the 100 from the
other side, this becomes our \( v_{\text{final}} \), and this is going to be 7 meters per second. So that's our answer. So it turns out what happens is that you have a sled that's initially moving at
10, and then you're adding mass to the system, and then the final velocity is going to be less. It's going to be 7 meters per second. This should make some sense to you because momentum conservation,
\( p = mv \), says that in order for momentum to be conserved, if your mass is increasing, then in order for you to have the same \( p \) as a system, your \( v \) has to go down. If your mass
increases, your speed has to decrease proportionally so that you keep the same momentum. Alright? So that's what's happening here. Alright. So now let's go ahead and take a look at parts b and c. So
in part b, what we want to do is calculate the change in the momentum of the box. So that's going to be \( \Delta p_2 \). So what's \( \Delta p \)? Remember, it's just going to be \( m_2 (v_{\text
{final}} - v_{\text{initial}}) \). It's \( v_{\text{final}} - v_{\text{initial}} \) here. So what we can do is you can group this together, \( m_2 (v_{\text{final}} - v_{\text{initial}}) \) here.
Okay. So we have the 30 kilogram box, and then what's the \( v_{\text{final}} \)? The \( v_{\text{final}} \) is just the 7 meters per second that we just calculated here. So 7 minus what's the
initial speed of the box? Well, remember, it's just 0. So what ends up happening is you got a change of momentum of 210. Let's do the same exact thing now for part c, except now we want to calculate
the change in momentum of the sled. So this is going to be the same exact equation. I'm just going to skip ahead here. It's going to be \( m_1 (v_{\text{final}} - v_{\text{initial}}) \). So \( m_1 \)
here is going to be 70, not the 30, but now we're going to use the final velocity, which is again 7. What's the initial velocity of the sled? Remember the sled was actually moving already at 10
meters per second, that's the initial. So it turns out what happens is you're going to get negative 210 kilogram meters per second. So if you take a look at these two numbers here, they're the same
but they're opposites and these two things are related. Basically, what we've had here is a situation where we've had momentum conservation, but one object has gained momentum, and that means that
the other one has to lose the same amount. For momentum to be conserved, if one object gains momentum, the other one has to lose the same amount, and the opposite's true. If one object loses
momentum, the other one has to gain the same amount. There's always basically a transfer or an exchange of momentum. The sled has to do some work in accelerating the box to the final speed of 7
meters per second, and so the sled loses some speed, the box gains that speed. Alright? So that's it for this one, guys. Let me know if you have any questions.
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What are some games with fairly simple heuristics to evaluate positions?
I'm teaching a kid programming, and am introducing some basic artificial intelligence concepts at the moment. To begin with we're going to implement a tic-tac-toe game that searches the entire game
tree and as such plays perfectly. Once we finish that I want to apply the same concepts to a game that has too many positions to evaluate every single one so that we need to implement a heuristic to
evaluate intermediate positions.
The best thing I could think of was Dots and Boxes. It has the advantage that I can set the board size arbitrarily large to stop him from searching the entire tree, and I can make a very basic
scoring function be the number of my boxes minus the number of opponent boxes. Unfortunately, this means that for most of the beginning of the game every position will be evaluated equivalently with
a score of 0 because it takes quite a few moves before players actually start making boxes.
Does anyone have any better ideas for games? (Or a better scoring function for dots and boxes)?
Samuel was one of the first to make effective use of heuristic search methods. You can refer the following link:http://incompleteideas.net/book/first/ebook/node109.html
Another game choice could be Reversi aka Othello.
A naive heuristic would be to simply count the number of tiles gained by each valid move and choose the greatest. From there you can factor in a board position and minimizing vulnerable to the
One game you may consider is Connect Four. A simple game with straightforward rules but more complicated than Tic-Tac-Toe.
If you want to know Heuristic Search Method In Artificial Intelligence then visit this Artificial Intelligence Course.
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Pascal's Triangle Calculator - Online Pascals Solver
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Pascal's Triangle
Tool to compute value for Pascal's triangle, an arithmetic list of numbers where each item is either 1 or the sum of the two elements above it.
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Answers to Questions (FAQ)
What is Pascal triangle? (Definition)
Pascal's triangle is a representation in a triangular grid in which each number is the sum of the 2 numbers above it. In more mathematical terms, Pascal's triangle represents the binomial
How to create a pascal triangle?
The principle of the Pascal triangle is based on a pyramidal/triangular construction, write 1 on the first row, and 1 1 on the second row. For the next rows, take two adjacent numbers, add their
values and place this new number directly above. (The missing start and end are equal to 1).
Example: Start of the Pascal Triangle:
Values can be calculated using binomial coefficients, also used in calculation of combinations.
Pascal triangle values can be compared to the Fibonacci sequence where each number is the sum of the two preceding numbers.
Usually, mathematicians call the first row 0, same for the first column.
How to calculate a precise value in the Pascal's triangle?
A value $ V $ of the Pascal triangle at the position (row A, column B, 0-indexed) can be calculated with the binomial coefficients (and thus with factorials) and the formula $$ V = \binom{A}{B} = \
frac{A!}{B!(A-B)!} $$
How to create a Pascal Triangle with Excel?
Write 1 in the cell B1, and =A2+B1 in the cell B2 and copy the contents in as many cells as you wish but do not touch column 1 and row 1. Each row (including zeros) is a new row of the Pascal
Why is the triangle named Pascal?
The triangle is named in honor of Blaise Pascal, who studied it. Although he was not the first to study it, his name is the most used, although it is also called the Khayyam or Tartaglia triangle.
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New Trends in Graph Coloring
Schedule for: 16w5120 - New Trends in Graph Coloring
Beginning on Sunday, October 16 and ending Friday October 21, 2016
All times in Banff, Alberta time, MDT (UTC-6).
Sunday, October 16
16:00 - 17:30 Check-in begins at 16:00 on Sunday and is open 24 hours (Front Desk - Professional Development Centre)
Dinner ↓
17:30 - 19:30 A buffet dinner is served daily between 5:30pm and 7:30pm in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
20:00 - 22:00 Informal gathering (Corbett Hall Lounge (CH 2110))
Monday, October 17
Breakfast ↓
07:00 -
08:45 Breakfast is served daily between 7 and 9am in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
08:45 - Introduction and Welcome by BIRS Station Manager (TCPL 201)
09:00 - Introductions and open problems (TCPL 201)
10:00 - Coffee Break (TCPL Foyer)
Alexander Kostochka: DP-coloring ↓
10:30 - In order to solve a problem in list coloring, Dvorak and Postle introduced the more general notion of DP-coloring (they called it "correspondence coloring"). This new notion seems very
11:00 promising. The goal of the talk is to present the language of DP-coloring and some of its interesting properties. In particular, we will discuss DP coloring of multigraphs and ask some
(TCPL 201)
11:00 - Free discussion and research in groups (TCPL 201)
11:30 - Lunch (Vistas Dining Room)
Guided Tour of The Banff Centre ↓
13:00 -
14:00 Meet in the Corbett Hall Lounge for a guided tour of The Banff Centre campus.
(Corbett Hall Lounge (CH 2110))
Group Photo ↓
14:00 - Meet in foyer of TCPL to participate in the BIRS group photo. The photograph will be taken outdoors, so dress appropriately for the weather. Please don't be late, or you might not be in the
14:20 official group photo!
(TCPL Foyer)
Zdenek Dvorak: Towards exponentially many 3-colorings of triangle-free planar graphs ↓
14:30 - We show that planar triangle-free graphs have exponential number of 3-colorings if and only if the following conjecture holds: There exists $\varepsilon>0$ such that if $G$ is a planar graph
15:00 and $X$ is a set its edges such that $G-X$ is triangle-free, then for some $Y\subseteq X$ of size at least $\varepsilon|X|$, the graph $G-(X\setminus Y)$ is 3-colorable. We also show some
special cases where the latter conjecture holds. Based on joint work with J.-S. Sereni.
(TCPL 201)
15:00 - Coffee Break (TCPL Foyer)
15:30 - Paul Seymour: Tutorial: $\chi$-boundedness part 1 (TCPL 201)
17:00 - Research in groups (TCPL)
Dinner ↓
18:00 -
19:30 A buffet dinner is served daily between 5:30pm and 7:30pm in the Vistas Dining Room, the top floor of the Sally Borden Building.
(Vistas Dining Room)
Tuesday, October 18
- Breakfast (Vistas Dining Room)
David Wood: Non-repetitive colorings ↓
Thue (1906) constructed an arbitrarily long string of three characters containing no substring of even length where the first half of the substring is the same as the second half. We can think
of Thue's theorem as a result about 3-colouring paths, which naturally suggests a generalisation for arbitrary graphs. A nonrepetitive colouring of a graph $G$ is a function that assigns each
vertex of $G$ a colour, such that for every path $P$ of even length in $G$, the sequence of colours on the first half of $P$ is distinct from the sequence of colours on the second half of $P$.
The nonrepetitive chromatic number of $G$ is the minimum number of colours in a nonrepetitive colouring of $G$. It follows from Thue's result that the nonrepetitive chromatic number of a path
(of length at least 4) equals 3. What about nonrepetitive colouring of other graph classes? Alon, Grytczuk, Haluszczak and Riordan (2002) proved that bounded degree graphs have bounded
09:00 nonrepetitive chromatic number; in particular, graphs with maximum degree $\Delta$ are nonrepetitively $O(\Delta^2)$-colourable. The proof is an elegant example of the Lovász Local Lemma. Using
- a recent technique called `entropy compression', Dujmovic, Joret, Kozik and Wood (2015) reduced this bound to $(1+o(1))\Delta^2$. Non-repetitive colourings have been studied for several
09:45 well-structured graph families. For example, Brešar, Grytczuk, Klavžar, Niwczyk and Peterin (2007) proved that every tree is nonrepetitively 4-colourable. Kündgen and Pelsmajer (2008)
generalised this result for graphs of bounded treewidth. Nešetřil, Ossona de Mendez and Wood (2011) studied nonrepetitive colourings of subdivisions, and concluded that graph classes with
bounded nonrepetitive chromatic number have bounded expansion. A challenging open problem, due to Grytczuk (2007), is whether planar graphs have bounded nonrepetitive chromatic number. The best
known upper bound for $n$-vertex planar graphs is $O(\log n)$ due to Dujmovic, Frati, Joret and Wood (2013). This bound was generalised by Dujmovic, Morin and Wood~(2013) for graphs excluding a
fixed topological minor. The proof employs powerful tools such as the Robertson--Seymour graph minor structure theorem. This talk will survey these results, focusing on the tools used, which
should be of wide interest.
(TCPL 201)
Marthe Bonamy: Tight lower bounds for the complexity of multicoloring ↓
In the multicoloring problem, also known as $(a:b)$ or$b$-fold coloring, we are given a graph $G$ and a set of a colors, and the task is to assign a subset of $b$ colors to each vertex of $G$
so that adjacent vertices receive disjoint color subsets. This natural generalization of the classic coloring problem (the $b=1$ case) is equivalent to finding a homomorphism to the Kneser
09:45 graph with parameters $a$ and $b$. It is tightly connected with the fractional chromatic number, and has multiple applications within computer science. We study the complexity of determining
- whether a graph has an $(a:b)$-coloring. Nederlof showed in 2008 a $(b+1)^nn^{O(1)}$-time algorithm for $(a:b)$-coloring. Our main result is that this is essentially optimal: there is no
10:15 algorithm with running time $2^{o(log b)n}$ unless the ETH fails. The crucial ingredient in our hardness reduction is the usage of detecting matrices of Lindström (1965), which is a
combinatorial tool that, to the best of our knowledge, has not yet been used for proving complexity lower bounds. This is joint work with Lukasz Kowalik, Michal Pilipczuk, Arkadiusz Socala and
Marcin Wrochna (University of Warsaw)."
(TCPL 201)
- Coffee Break (TCPL Foyer)
Anton Bernshteyn: The Local Cut Lemma ↓
The entropy compression method is an algorithmic technique that was invented by Moser and Tardos in order to give an effective proof of the Lovász Local Lemma (the LLL for short). It turns out
10:45 that avoiding the LLL and applying the entropy compression method directly can lead to improvements, sometimes significant, in combinatorial bounds. The Local Cut Lemma (the LCL for short) is a
- generalization of the LLL that implies these new combinatorial results. It hides technical parts of the method and thus makes the arguments simpler and shorter. Although the LCL was inspired by
11:15 the entropy compression method, it has a direct probabilistic proof (similar to the classical proof of the LLL); in particular, it not only shows that a certain probability is positive, but
also gives an explicit lower bound for that probability. In this talk, we will discuss the LCL and some of its applications, as well as its connections with the entropy compression.
(TCPL 201)
- Research in groups (TCPL)
- Lunch (Vistas Dining Room)
- Progress reports (TCPL 201)
- Coffee Break (TCPL Foyer)
- Alex Scott: Tutorial: $\chi$-boundedness part 2 (TCPL 201)
- Research in groups (TCPL)
- Dinner (Vistas Dining Room)
Wednesday, October 19
- Breakfast (Vistas Dining Room)
Paul Wollan: Forcing clique immersions with large chromatic number ↓
Hadwiger (infamously) conjectured that there is a tight relationship between the chromatic number of a graph and the maximum size of a clique minor - specifically, every graph with chromatic
09:00 number at least $t$ contains $K_t$ as a minor. Lescure and Meynial and later Abu-Khzam and Langston independently extended Hadwiger's to graph immersions, an alternate model of graph
- containment. Specifically, they conjecture that every graph with chromatic number $t$ contains $K_t$ as an immersion. We present new bounds on the relationship between the chromatic number and
09:30 the maximal size of a clique immersion in a graph. For general graphs, building on the recent work of Dvorak and Yepremyn, we show that every graph with $\chi(G) \ge 3.54t$ contains $K_t$ as an
immersion. A particularly interesting case of Hadwiger's conjecture is when restricted to graphs of small independence number. We show that every graph on $n$ vertices with independence number
$\alpha = 2$ contains a $K_t$ immersion with $t = .4 n$. We conclude with several open questions. This is joint work with Tien-Nam Le and Gregory Gauthier.
(TCPL 201)
Chun-Hung Liu: Characterizations of minimal cycle obstruction sets for balanced and unbalanced partitionable planar graphs ↓
Coloring graphs in such a way that each subgraph induced by a color class has bounded maximum degree receives extensive attention recently. Motivated by Grotzsch's theorem, Steinberg's
09:30 conjecture and other related results, we consider coloring planar graphs with some cycles forbidden. In this talk, we characterize all minimal sets S in which every planar graph G with no cycle
- lengths belonging to S has each of the following properties: (1) V(G) can be partitioned into a stable set and a set that induces a graph of bounded maximum degree, (2) V(G) can be partitioned
10:00 into two sets each induces a graph of bounded maximum degree, and (3) V(G) can be partitioned into two stable sets and one set that induces a graph of bounded maximum degree. This is joint with
Ilkyoo Choi and Sang-il Oum.
(TCPL 201)
- Coffee Break (TCPL Foyer)
Louis Esperet: Coloring pseudo-disks and Jordan curves ↓
10:30 We investigate intersection graphs of pseudo-disks (a pseudo-disk is the homeomorphic image of a disk in the plane) whose interiors are pairwise disjoint (i.e. which may only intersect at their
- boundaries), and in particular we answer a question of Reed and Shepherd (1996) and a question of Hlineny (1998) on their chromatic number. We also consider the chromatic number of intersection
11:00 graphs of pairwise non-crossing Jordan curves in the plane. There is a nice application to the problem of bounding the integrality gap for the maximum (vertex-)packing of cycles in planar
(TCPL 201)
- Research in groups (TCPL)
- Lunch (Vistas Dining Room)
- Free Afternoon (Banff National Park)
- Dinner (Vistas Dining Room)
Thursday, October 20
- Breakfast (Vistas Dining Room)
Robert Samal: 3-Flows with Large Support ↓
09:00 We prove that every 3-edge-connected graph has a 3-flow that takes on a zero value on at most one sixth of the edges. The graph $K_4$ demonstrates that this $\frac{1}{6}$ ratio is best
- possible; there is an infinite family where $\frac 16$ is tight. The proof involves interesting work with connectivity of the graph (relaxing to subdivisions of 2-edge-connected graphs and
09:30 then reducing to cyclically 4-edge-connected ones). Joint work with M. DeVos, J. McDonald, I. Pivotto, and E. Rollova.
(TCPL 201)
Dan Kral: Colorings of plane graphs ↓
09:30 Problems concerning coloring graphs embedded in the plane have always been among the most intensively studied problems in graph theory. In the talk, we will present some recent results, which
- have been obtained with various groups of collaborators, on three classical problems in the area: Steinberg's Conjecture from 1976, the Cyclic Coloring Conjecture of Borodin from 1984, and the
10:00 Cyclic Coloring Conjecture of Plummer and Toft from 1987.
(TCPL 201)
- Coffee Break (TCPL Foyer)
Daniel Cranston: Coloring linegraphs of multigraphs with $\max(\omega, \lceil(5\Delta+3)/6\rceil)$ colors ↓
- We show that linegraphs of multigraphs have chromatic number at most $\max(\omega, \lceil(5\Delta+3)/6\rceil)$. The result is sharp, as shown by line graphs of blow-ups of the 5-cycle. For
11:00 comparison this bound is stronger than Reed's conjecture when $\omega > 2\Delta/3$. The proof uses Tashkinov trees. Based on joint work with Landon Rabern.
(TCPL 201)
- Research in groups (TCPL)
- Lunch (Vistas Dining Room)
- Final progress reports (TCPL 201)
- Coffee Break (TCPL Foyer)
- Luke Postle: Tutorial: potential method (TCPL 201)
- Free discussion and research in groups (TCPL)
- Dinner (Vistas Dining Room)
Friday, October 21
07:00 - Breakfast (Vistas Dining Room)
09:00 - Free discussion and research in groups (TCPL)
10:00 - Coffee Break (TCPL Foyer)
10:30 - Free discussion and research in groups (TCPL)
Checkout by Noon ↓
11:30 - 5-day workshop participants are welcome to use BIRS facilities (BIRS Coffee Lounge, TCPL and Reading Room) until 3 pm on Friday, although participants are still required to checkout of the
12:00 guest rooms by 12 noon.
(Front Desk - Professional Development Centre)
12:00 - Lunch from 11:30 to 13:30 (Vistas Dining Room)
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Evaluating Expressions With Negative Numbers Worksheet
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Evaluating Expressions Date Period Evaluate each using the values given 1 y 2 x use x 1 and y 2 2 a 5 b use a 10 and b 4 3 p2 m use m 1 and p 5 4 y 9 x use x 1 and y 3 5 m p 5 use m
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Negative Number Worksheets
Evaluate Each Expression Worksheet
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Negative Number Worksheet
Evaluating Algebraic Expression Worksheets Math Worksheets 4 Kids
In this set of printable worksheets for 7th grade and 8th grade students evaluate the algebraic expressions containing multi variable The variables may contain whole numbers integers or fractions
Easy Moderate Difficult MCQs based on Equations
Evaluating Algebraic Expressions Super Teacher Worksheets
Basic Level Positive Whole Numbers Evaluate Algebraic Expressions Basic Worksheet 1 FREE Evaluate each algebraic expression Values for the variables are given This level does not include exponents
negative numbers or parentheses 5th and 6th Grades View PDF Evaluate Algebraic Expressions Basic Worksheet 2
In this set of printable worksheets for 7th grade and 8th grade students evaluate the algebraic expressions containing multi variable The variables may contain whole numbers integers or fractions
Easy Moderate Difficult MCQs based on Equations
Basic Level Positive Whole Numbers Evaluate Algebraic Expressions Basic Worksheet 1 FREE Evaluate each algebraic expression Values for the variables are given This level does not include exponents
negative numbers or parentheses 5th and 6th Grades View PDF Evaluate Algebraic Expressions Basic Worksheet 2
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Digital Communication in the Presence of Noise
When we incorporate additive noise into our channel model, so that r(t)=αs[i](t)+n(t), errors can creep in. If the transmitter sent bit 0 using a BPSK signal set (Section 6.14), the integrators'
outputs in the matched filter receiver (Figure 6.14: Optimal receiver structure) would be:
It is the quantities containing the noise terms that cause errors in the receiver's decision-making process. Because they involve noise, the values of these integrals are random quantities drawn from
some probability distribution that vary erratically from bit interval to bit interval. Because the noise has zero average value and has an equal amount of power in all frequency bands, the values of
the integrals will hover about zero. What is important is how much they vary. If the noise is such that its integral term is more negative than αA^2T, then the receiver will make an error, deciding
that the transmitted zero-valued bit was indeed a one. The probability that this situation occurs depends on three factors:
• Signal Set Choice - The difference between the signal-dependent terms in the integrators' outputs (equations (6.45)) defines how large the noise term must be for an incorrect receiver decision to
result. What affects the probability of such errors occurring is the square of this difference in comparison tothe noise term's variability. For our BPSK baseband signal set, the signal-related
value is 4α^2A^4T^2 .
• Variability of the Noise Term - We quantify variability by the average value of its square, which is essentially the noise term's power. This calculation is best performed in the frequency domain
and equals
• A^2T. Thus, the noise term's power is
• Probability Distribution of the Noise Term - The value of the noise terms relative to the signal terms and the probability of their occurrence directly affect the likelihood that a receiver error
will occur. For the white noise we have been considering, the underlying distributions are Gaussian. The probability the receiver makes an error on any bit transmission equals:
Here Q (·) is the integralFigure 6.15 illustrates, Q (·) is a decreasing, very nonlinear function.
The function Q(x) is plotted in semilogarithmic coordinates. Note that it decreases very rapidly for small increases in its arguments. For example, when x increases from 4 to 5, Q(x) decreases by a
factor of 100.
The term A^2T equals the energy expended by the transmitter in sending the bit; we label this term E[b]. We arrive at a concise expression for the probability the matched filter receiver makes a
bit-reception error.
Figure 6.16 shows how the receiver's error rate varies with the signal-to-noise ratio
The probability that the matched-filter receiver makes an error on any bit transmission is plotted against the signal-to-noise ratio of the received signal. The upper curve shows the performance of
the FSK signal set, the lower (and therefore better) one the BPSK signal set.
Exercise 6.17.1
Derive the probability of error expression for the modulated BPSK signal set, and show that its performance identically equals that of the baseband BPSK signal set.
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Ve281 Data Structures and Algorithms Written Assignment One solved
1. Suppose that you and your friend are playing a game of guessing a number. At the
beginning, your friend writes down an integer x in the range [1, N]. Your goal is to
guess that number using the minimum number of guesses. Each time when you guess
a number, your friend will tell you whether the number you guess is equal to, less than,
or greater than the number x he initially sets. Once your guess equals x, the game
Describe a good strategy that you could use to guess the initial number using as few
number of guesses as possible. Suppose that N = 2m − 1. What is the number of
guesses required in the worst case? What is the number of guesses required on average? Give the exact value, not the big-oh, big-omega, or theta notation.
2. Consider the following code
while(n > 1)
if(n % 2 == 1)
n = 3 * n + 1;
n /= 2;
What is the best lower bound f(n) so that the complexity of the above code is in the
set Ω(f(n))?
3. Is the statement n
100 = O(1.001n
) true or not? Prove your answer in a strict way.
4. Prove the following result on the theta notation based on its definition:
Suppose that g1(n) and g2(n) are non-negative functions when n ≥ 0. If f1(n) =
Θ(g1(n)) and f2(n) = Θ(g2(n)), then f1(n)+f2(n) = Θ(h(n)), where h(n) = max{g1(n), g2(n)}.
5. Consider the following code
void func(int n, int a, int b) {
for(i = 1; i <= n; i *= a) for(j = 1; j <= i * b; j++) Statement; } Assume that n ≥ 1, a > 1, and b ≥ 1 are integers. How many times will “Statement”
be called? Write the answer in terms of n, a, and b.
6. A new house is being built down the street. The builders are currently working on
the roof. From observation, there seems to be three different methods for moving the
shingles from the shingle truck to the roof.
Method 1: A builder can carry two packages (known as squares) of shingles on his
shoulder as he climbs up the ladder. He then climbs down and carries two more
squares up the ladder. So on and so forth. Each round trip (up the ladder with
two squares and down the ladder with none) costs $2.
Method 2: A builder rents a lift for $10. The lift can move 20 squares up the roof at
one time at a cost of $1 per round trip.
Method 3: A builder rents a super-lift for $40. Unfortunately, the lift has a slow
leak in its hydraulic system. The lift is able to lift roughly half of the necessary
squares to the roof on the first round trip. However, on the second trip the lift
is only able to lift half of the remaining squares, then half of the remaining, and
so on. To be strict, if the number of the remaining squares is n, the lift lifts b
squares. Even when the super-lift has no hydraulic fluid left in its system, it can
still lift one square of shingles. Each round trip using the super lift costs $2.
Note that in all three methods, it costs $4/square to nail the shingles into the roof.
In the following questions, “cost” refers to total cost, including nailing shingles to the
(a) For each method, write a formula T(n) that expresses the number of round trips
required to move n squares of shingles to the roof.
(b) For each method, write a formula C(n) that expresses the entire cost to install n
squares of shingles on the roof.
(c) Which is cheapest method for a doghouse (8 squares)? How much does it cost?
(d) Which is cheapest method for a shed (128 squares)? How much does it cost?
(e) Which is cheapest method for a house (2048 squares)? How much does it cost?
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Correct spelling for rhomb | Dictionary.net
\ɹˈɒm], \ɹˈɒm], \ɹ_ˈɒ_m]\
Sort: Oldest first
- New Age Dictionary Database
By Oddity Software
- Webster's Revised Unabridged Dictionary
By Noah Webster.
- The Winston Simplified Dictionary
By William Dodge Lewis, Edgar Arthur Singer
- The Concise Standard Dictionary of the English Language
By James Champlin Fernald
- The Clarendon dictionary
By William Hand Browne, Samuel Stehman Haldeman
- The Cabinet Dictionary of the English Language
Word of the day
Colony Stimulating Factor 1
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disulfide-bonded glycoprotein dimer with MW of 70 kDa. It binds to specific high affinity receptor (RECEPTOR, MACROPHAGE COLONY-STIMULATING FACTOR).
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