text:
## Stream: new members

### Topic: zero divisors and nilpotents

#### Aniruddh Agarwal (May 17 2020 at 17:53):

Which definition would be considered better:

def is_nilpotent [comm_ring α] (a : α) := ∃ n : ℕ, a ^ n = 0
def is_nilpotent' [comm_ring α] {n : ℕ} (a : α) := a ^ n = 0

Top one

#### Kevin Buzzard (May 17 2020 at 17:54):

The bottom one would be hard to use because in most cases Lean would not be able to guess n.

#### Patrick Stevens (May 17 2020 at 17:54):

The bottom one would force the user to know the witness to nilpotence, right?

^^

#### Aniruddh Agarwal (May 17 2020 at 19:10):

I've almost proved that nonzero nilpotents are zero divisors, I just need to prove

theorem nonzero_nilpotent_has_nonzero_power [comm_ring α] :
∀ a : α, a ≠ 0 → is_nilpotent a → ∃ m : ℕ, a^m ≠ 0 ∧ ∀ n : ℕ, n > m → a^n = 0
:= sorry

to get it, but I'm stumped. I want to find m by taking the minimum of the set (type?) { n : ℕ | a^n = 0 }, but I'm not sure how to set that up. Here's where I'm at so far:

import algebra.associated
import algebra.group_power

variables {α : Type*}

/- An element a of a ring α is a zero divisor if there exists a b ∈ α
such that b ≠ 0 and ab = 0. -/
def is_zero_divisor [comm_ring α] (a : α) := ∃ b : α, b ≠ 0 ∧ a * b = 0

/- An element a of a ring α is nilpotent if there exists a n ∈ ℕ such
that n ≠ 0 and a^n = 0 -/
def is_nilpotent [comm_ring α] (a : α) := ∃ n : ℕ, n ≠ 0 ∧ a ^ n = 0

/- If a is a nonzero nilpotent, then we can find a nonzero natural m
such that a^m ≠ 0 and a^n = 0 for all n > m. -/
theorem nonzero_nilpotent_has_nonzero_power [comm_ring α] :
∀ a : α, a ≠ 0 → is_nilpotent a → ∃ m : ℕ, a^m ≠ 0 ∧ ∀ n : ℕ, n > m → a^n = 0
:=
begin
unfold is_nilpotent,
intros a hanz hanlp,
have set := { n : ℕ | a^n = 0 },
sorry,
end

I apparently can't just write min set to get the minimum because lean can't find an instance of a decidable_linear_order for N. Any advice?

#### Bhavik Mehta (May 17 2020 at 19:13):

Try going classical, it might be the decidability that causes issues

#### Aniruddh Agarwal (May 17 2020 at 19:15):

Sorry, can you be more specific?

#### Bhavik Mehta (May 17 2020 at 19:15):

In this specific case though, nat.find is what you probably should use

#### Bhavik Mehta (May 17 2020 at 19:18):

begin
classical,
unfold is_nilpotent,
intros a hanz hanlp,
let property : ℕ → Prop := λ n, a^n = 0,
have there_is_some : ∃ n, property n,
sorry,
let m := nat.find there_is_some,
have : a^m = 0 := nat.find_spec there_is_some,

end

#### Bhavik Mehta (May 17 2020 at 19:19):

Here's a sketch of how you might use nat.find

#### Bhavik Mehta (May 17 2020 at 19:19):

(in your case, the property maybe should be a^n ≠ 0, I haven't checked)

#### Mario Carneiro (May 17 2020 at 19:20):

I guess it could just be the existential in is_nilpotent: ∃ n : ℕ, n ≠ 0 ∧ a ^ n = 0

#### Bhavik Mehta (May 17 2020 at 19:20):

That could work too

#### Bhavik Mehta (May 17 2020 at 19:21):

@Aniruddh Agarwal, you might also like to note that I used let instead of have sometimes - have doesn't keep track of the actual definition, just its type (so it's the right sort of thing for properties, since we don't care about what the actual proof was) but let does keep track of how it was defined

#### Reid Barton (May 17 2020 at 19:29):

n ≠ 0 is unnecessary; is there a reason to include it?

#### Mario Carneiro (May 17 2020 at 19:39):

In this case you probably want to find the least m such that a^(m+1)=0

#### Aniruddh Agarwal (May 17 2020 at 20:53):

@Bhavik Mehta I've slightly modified your sketch to get to

begin
classical,
unfold is_nilpotent,
intros a hanz hanlp,
have there_is_some : ∃ n, a^(n+1) = 0,
sorry,
have m := nat.find_min' there_is_some,
end

which is almost what I want, except that m's type

m : (λ (n : ℕ), a ^ (n + 1) = 0) ?m_1 → nat.find there_is_some ≤ ?m_1

contains a dummy variable ?m_1

#### Bhavik Mehta (May 17 2020 at 20:55):

The dummy variable represents a natural you haven't chosen yet

#### Bhavik Mehta (May 17 2020 at 20:55):

Once the dummy variable is picked (I'll call it t), m would have type a^(t + 1) = 0 → nat.find there_is_some ≤ t

#### Bhavik Mehta (May 17 2020 at 20:58):

Have a look at the functions around nat.find to see what you're given to work with:

protected def find : ℕ := nat.find_x.1

protected theorem find_spec : p nat.find := nat.find_x.2.left

protected theorem find_min : ∀ {m : ℕ}, m < nat.find → ¬p m := nat.find_x.2.right

protected theorem find_min' {m : ℕ} (h : p m) : nat.find ≤ m :=
le_of_not_gt (λ l, find_min l h)

#### Aniruddh Agarwal (May 17 2020 at 22:11):

I'm still confused and unable to get the minimum of that set out using those functions around nat.find :/

#### Kevin Buzzard (May 17 2020 at 22:15):

Look at the type of nat.find. That tells you what it does. What is your exact problem? Post some working code

#### Aniruddh Agarwal (May 17 2020 at 22:46):

There is where I'm currently stuck:

import algebra.associated
import algebra.group_power

variables {α : Type*}

/- An element a of a ring α is nilpotent if there exists a n ∈ ℕ such
that a^n = 0. -/
def is_nilpotent [comm_ring α] (a : α) := ∃ n : ℕ, n ≠ 0 ∧ a ^ n = 0

/- If a is a nonzero nilpotent, then we can find a nonzero natural m
such that a^m ≠ 0 and a^n = 0 for all n > m. -/
theorem nonzero_nilpotent_has_nonzero_power [comm_ring α] :
∀ a : α, a ≠ 0 → is_nilpotent a → ∃ m : ℕ, a^m ≠ 0 ∧ ∀ n : ℕ, n > m → a^n = 0
:=
begin
classical,
unfold is_nilpotent,
intros a hanz hanlp,
have there_is_some : ∃ n, a^(n+1) = 0,
cases hanlp with n hn,
cases n with,
constructor,
exact hn.right,
have m := nat.find there_is_some,
have hm := nat.find_min' there_is_some,
constructor,
split,
-- I want to show that m satisfies a^(m+1)=0
sorry,
sorry,
sorry,
sorry,
end

#### Kevin Buzzard (May 17 2020 at 22:47):

Can we please change the definition of nilpotent by dropping the n != 0 condition?

#### Kevin Buzzard (May 17 2020 at 22:47):

It just makes things harder

#### Kevin Buzzard (May 17 2020 at 22:49):

oh but then your proof will need refactoring

#### Mario Carneiro (May 17 2020 at 22:50):

You forgot to use nat.find_spec

#### Reid Barton (May 17 2020 at 22:51):

Do you need this statement? How about there exists m such that a^m = 0 and a^n \ne 0 for n < m?

#### Kevin Buzzard (May 17 2020 at 22:52):

Yes what you're doing does look like an unnatural thing to prove

#### Kevin Buzzard (May 17 2020 at 22:53):

let m := nat.find there_is_some, not have

#### Kevin Buzzard (May 17 2020 at 22:57):

Reid's suggestion is better because it doesn't involve this unnatural hypothesis about a being non-zero.

#### Kevin Buzzard (May 17 2020 at 22:59):

Why not fix the definition of nilpotent and then prove what Reid suggests?

#### Kevin Buzzard (May 17 2020 at 23:02):

The trick with these things is to find the right statement which doesn't have special cases such as "doesn't work if a=0".

#### Mario Carneiro (May 17 2020 at 23:05):

How about \exists m, \forall n, a^n = 0 <-> m <= n

#### Mario Carneiro (May 17 2020 at 23:06):

everything is better with iff

#### Kevin Buzzard (May 17 2020 at 23:07):

import algebra.associated
import algebra.group_power

variables {α : Type*}

open_locale classical

/- An element a of a ring α is nilpotent if there exists a n ∈ ℕ such
that a^n = 0. -/
def is_nilpotent [comm_ring α] (a : α) := ∃ n : ℕ, a ^ n = 0

/- If a is a nonzero nilpotent, then we can find a nonzero natural m
such that a^m ≠ 0 and a^n = 0 for all n > m. -/
theorem nonzero_nilpotent_has_nonzero_power [comm_ring α] :
∀ a : α, is_nilpotent a → ∃ m : ℕ, a^m = 0 ∧ ∀ n : ℕ, n < m → a^n ≠ 0
:=
begin
rintros a hna,
use nat.find hna,
split,
{ exact nat.find_spec hna},
{ intro n,
exact nat.find_min hna}
end

#### Kevin Buzzard (May 17 2020 at 23:07):

Reid's version comes out really nicely.

#### Kevin Buzzard (May 17 2020 at 23:08):

Mario's version will also come out nicely

#### Reid Barton (May 17 2020 at 23:20):

-  such that a^m ≠ 0 and a^n = 0 for all n > m. -/
+  such that a^m = 0 and a^n ≠ 0 for all n < m. -/

#### Kevin Buzzard (May 18 2020 at 09:49):

@Aniruddh Agarwal here is a way of thinking about all this. You want to find a way of expressing, in Lean, the following idea: if a is nilpotent, and you look at the function from $\mathbb{N}$ to $\{T,F\}$ which sends $n$ to the truth-value of the question "Is $a^n=0$?", then this function looks like "F,F,F,...,F,...,F,T,T,T,...,T,T,...". This is an idea, not a formalised statement. There are lots of different ways to formalise it. You chose one way, which treated the case a=0 as a special case (where the function looks like "T,T,T,T,..." or "F,T,T,T,..." depending on whether R is the zero ring or not) but one could just as well prove the lemma that said that a=0 iff a^n=0 for all n>=1, which would be a much nicer way of dealing with a=0 rather than trying to formalise something general and then leaving out the a=0 case. Leaving in a=0 for now, we want to formalise the idea that there will be 0 or more F's, but a finite number, followed by a bunch of T's. Because 0 F's is possible (if R=0 and a=0) we should let the "key" natural number be the first T, rather than the last F (which might not exist in general). If you are in a situation where you want the index of the last F assuming a isn't zero, you'll be able to get it by subtracting 1 from the first T. But the more natural lemma is the one which applies for all a, rather than just non-zero a.

In short, the lemmas which one wants to formalise are the ones with the most natural statements and the fewest edge cases.

#### Jalex Stark (May 18 2020 at 13:23):

of course the things kevin says about picking natural statements is also true when doing "paper maths", but human mathematicians can elaborate away the "trivial" edge-cases

#### Aniruddh Agarwal (May 18 2020 at 18:42):

:sigh: I finally finished the proof that nonzero nilpotents are zero divisors: https://github.com/anrddh/commutative-algebra-playground/blob/master/src/zero_divisors_and_nilpotents.lean#L45

#### Aniruddh Agarwal (May 18 2020 at 18:45):

Thanks all for your substantial help

#### Kevin Buzzard (May 18 2020 at 18:54):

theorem nz_nilpotent_is_zerodivisor [comm_ring α] :
∀ a : α, is_nilpotent a → a ≠ 0 → is_zero_divisor a
:=
begin
rintros a ha ha',
rcases exists_min_pow_zero_of_nilpotent a ha with ⟨m, hm1, hm2⟩,
cases m with m,
{ -- a^0=0 can't happen
exfalso,
apply ha',
apply eq_zero_of_zero_eq_one,
rw ←hm1,
refl
},
{ use a^m,
split,
{ apply hm2,
},
rw ←hm1,
rw pow_succ
}
end

#### Kevin Buzzard (May 18 2020 at 18:55):

You shouldn't really use simp in the middle of proofs, they will break later on as simp gets smarter.

#### Kevin Buzzard (May 18 2020 at 19:02):

/- All nonzero nilpotents are zerodivisors. -/
theorem nz_nilpotent_is_zerodivisor [comm_ring α] (a : α) (ha : is_nilpotent a) (ha' : a ≠ 0) : is_zero_divisor a
:=
let ⟨m, hm1, hm2⟩ := exists_min_pow_zero_of_nilpotent a ha in
begin
cases m with m,
{ exact false.elim (ha' (eq_zero_of_zero_eq_one _ hm1.symm _))},-- a^0=0 can't happen
{ exact ⟨a^m, hm2 _ (lt_add_one m), hm1⟩}
end

#### Kevin Buzzard (May 18 2020 at 19:06):

I can't figure out how to get the cases into term mode, but if I could this could be a completely term mode proof. The term proof is the same as the tactic proof, each tactic can be turned into a term.

#### Patrick Stevens (May 18 2020 at 19:06):

Note that it's perfectly fine to use simp if you phrase things so that simp is doing something specific. For example, have something : foo, by simp is safe: even if simp gets smarter, it's not going to start doing awful things to your proof state, because it's been restricted to this one little fragment of proof.

#### Patrick Stevens (May 18 2020 at 19:08):

(Although for efficiency, mathlib uses squeeze_simp to discover a more minimal subset of things that simp would use, and then restricts the call to simp using only so that it only uses those things)

#### Patrick Stevens (May 18 2020 at 19:16):

Kevin Buzzard said:

I can't figure out how to get the cases into term mode, but if I could this could be a completely term mode proof. The term proof is the same as the tactic proof, each tactic can be turned into a term.

I'm not sure it's possible, although my confidence here is weak - exists_min_pow_zero_of_nilpotent has output a Prop existential, but to extract anything from this in term mode you'd need a Type, wouldn't you?

#### Kevin Buzzard (May 18 2020 at 19:26):

Patrick Stevens said:

I'm not sure it's possible, although my confidence here is weak - exists_min_pow_zero_of_nilpotent has output a Prop existential, but to extract anything from this in term mode you'd need a Type, wouldn't you?

I don't know, this is all a bit CS for me.

#### Alex J. Best (May 18 2020 at 19:33):

/- All nonzero nilpotents are zerodivisors. -/
theorem nz_nilpotent_is_zerodivisor' [comm_ring α] (a : α) (ha : is_nilpotent a) (ha' : a ≠ 0) : is_zero_divisor a
:= match exists_min_pow_zero_of_nilpotent a ha with
| ⟨0, hm1, hm2⟩ := false.elim (ha' (eq_zero_of_zero_eq_one _ hm1.symm _))
| ⟨(m+1), hm1, hm2⟩ := ⟨a^m, hm2 _ (lt_add_one m), hm1⟩
end

terms are love, terms are life

#### Mario Carneiro (May 18 2020 at 20:07):

Patrick Stevens said:

I'm not sure it's possible, although my confidence here is weak - exists_min_pow_zero_of_nilpotent has output a Prop existential, but to extract anything from this in term mode you'd need a Type, wouldn't you?

exists_min_pow_zero_of_nilpotent can be proven in Type, since it has an explicit witness, namely nat.find of some stuff

Last updated: May 10 2021 at 18:22 UTC


text:
# Determine the RSA private key from public key

I was told that you can determine the private key of an RSA encryption with the public key.

I thought the point of a public key was to avoid letting people find the private key, but anyway, they said I could find it by simply factoring the public key modulus? Were they joshing me or can it be done? If so, how would I do that (need to acquire $$p,q$$ somehow?)?

So with with example Public Key $$( e, n ) = ( 13, 119 )$$ or any other valid key you want to use...

• Note that solving a Sudoku is also considered a hard problem, given large enough dimensions of the Sudoku. Hard problems are not impossible problems; they are just hard to solve for large enough puzzles. – Maarten Bodewes Oct 21 '19 at 14:26

format: plain


text:
s3_decomp.i

## Changes to input file:

We are going to change four things: The initial conditions, the solution time, we are going to add mesh adaptivity, and we will add more post-processors.

### Initial Conditions

We know that the initial conditions should be 46.774 mol% chromium with minor variations. We will input this using random initial conditions.

[ICs]
[./concentrationIC]   # 46.774 mol% Cr with variations
type = RandomIC
min = 0.44774
max = 0.48774
seed = 210
variable = c
[../]
[]

Note that the seed is the random number generator seed. It should not effect how the simulation behaves.

### Solution Time

Before we were just running the simulation long enough to begin to see the behavior. Now we want to run it for the full seven days. We change this in the executioner block.

    end_time = 604800   # 7 days

### Mesh Adaptivity

In order to accurately calculate the phase interfaces, we need a relatively fine mesh. However, in the bulk of the phases we can significantly speed up the solution by using a coarse mesh. Luckily, MOOSE comes equipped to let us refine the mesh where we need extra accuracy and coarsen it where we don't. We change the mesh block to look like this:

[Mesh]
type = GeneratedMesh
dim = 2
distribution = DEFAULT
elem_type = QUAD4
nx = 25
ny = 25
nz = 0
xmin = 0
xmax = 25
ymin = 0
ymax = 25
zmin = 0
zmax = 0
uniform_refine = 2
[]

The basic mesh is 25 × 25 elements. This defines the coarsest the mesh can be. The uniform_refine option refines the mesh for the first time step to be a 100 × 100 element mesh. If we do not turn on mesh adaptivity, the mesh would remain 100 × 100 throughout the simulation and it would be the exact same mesh we have used already.

To add mesh adaptivity, we go back to the executioner block, add the adaptivity sub-block, and tell it we want the maximum refinement level to be two factors finer than the coarsest mesh allowed.

[Executioner]
type = Transient
solve_type = NEWTON
l_max_its = 30
l_tol = 1e-6
nl_max_its = 50
nl_abs_tol = 1e-9
end_time = 604800   # 7 days
petsc_options_iname = '-pc_type -ksp_grmres_restart -sub_ksp_type
-sub_pc_type -pc_asm_overlap'
petsc_options_value = 'asm      31                  preonly
ilu          1'
[./TimeStepper]
type = IterationAdaptiveDT
dt = 10
cutback_factor = 0.8
growth_factor = 1.5
optimal_iterations = 7
[../]
[./Adaptivity]
coarsen_fraction = 0.1
refine_fraction = 0.7
max_h_level = 2
[../]
[]

### Post-processors

With the mesh adaptivity turned on, we are now interested in how many elements the surface has. We can calculate this by turning the NumNodes post-processor on.

We are also interested in how the size of the time step changes throughout the simulation, so we add a TimeStepSize post-processor. We can also look at the number of iterations at each time step.

[Postprocessors]
[./step_size]             # Size of the time step
type = TimestepSize
[../]
[./iterations]            # Number of iterations needed to converge timestep
type = NumNonlinearIterations
[../]
[./nodes]                 # Number of nodes in mesh
type = NumNodes
[../]
[./evaluations]           # Cumulative residual calculations for simulation
type = NumResidualEvaluations
[../]
[./active_time]           # Time computer spent on simulation
type = RunTime
time_type = active
[../]
[]

## Simulation Results:

Decomposition Results - Mesh shown

The image to the right is of the results of this simulation. The mesh is shown to see how adaptivity effects the mesh. The mesh remains fine where there is a large concentration gradient, but coarsens in large areas with little change.

We can see that the alloy did decompose into the two phases and that the phases are forming circles. This means our model is likely good. Depending on your random variations in the initial condition, you may also have circles or you could have a stripe through the surface.

Next we are ready to input the mobility as a function and to see and how much of the surface is devoted to each phase.

## Continue:

step 4: Make the Mobility a Function


text:
Back to School '17: Sour Candy

View as PDF

Points: 15 (partial)
Time limit: 1.4s
Memory limit: 256M

Authors:
Problem type
Allowed languages
Ada, Assembly, Awk, Brain****, C, C#, C++, COBOL, CommonLisp, D, Dart, F#, Forth, Fortran, Go, Groovy, Haskell, Intercal, Java, JS, Kotlin, Lisp, Lua, Nim, ObjC, OCaml, Octave, Pascal, Perl, PHP, Pike, Prolog, Python, Racket, Ruby, Rust, Scala, Scheme, Sed, Swift, TCL, Text, Turing, VB, Zig

One summer day, April bought pieces of sour candy to eat. Each piece of candy has a unique sourness value which indicates how sour it is.

Initially, the candy pieces were lined up in a straight line. However, April wants to change the order of the candies by taking any piece of candy, and placing it either at the front or back of the line. Given the order April wants to place the candies in, find the minimum number of moves needed to place them in that arrangement.

Input Specification

The first line contains an integer , which is the amount of sour candy April has.

The next line contains a sequence of space separated integers , where indicates the sourness of the candy.

The final line contains space separated integers, representing the new order April wants to place her candy in.

Output Specification

On the first line, output a single integer , representing the minimum number of moves it takes for April to arrange her sour candy.

On the next lines, output the instructions followed to arrange the sour candy. Output F i to move the element (1-indexed) to the front and B i to move the element to the back.

If there are multiple possible ordering of operations you may output any of them. It is guaranteed that it is always possible to rearrange the candy.

Sample Input 1

5
1 2 3 4 5
5 2 3 4 1

Sample Output 1

2
F 5
B 2

Explanation for Sample Output 1

April can move the candy with sourness value to the front of the lineup. She can then move the candy with sourness value to the back of the line. Note that the candy with sourness value was shifted to index before being moved.

The only other solution is

2
B 1
F 4

Sample Input 2

4
4 3 5 2
5 3 4 2

Sample Output 2

2
F 2
F 3

Explanation for Sample Output 2

April can move the candy with sourness value to the front, then move the candy with sourness value to the front.


text:
English
Europe
Middle East
Asia

VT Markets APP

Trade CFDs on FX, Gold and more

Daily market analysis

Amazon.com Inc. and Apple Inc. jumped as greater revenues relieved profit growth concerns. Bull market?￼

August 1, 2022

Lead by big tech, US stock advanced on Friday and notched the best month since 2020. Amazon.com Inc. and Apple Inc. soared as higher revenues from the pair’s iconic powerhouses relieved fears of profit slowdown on profit growth. Despite some concerning indications from economic indicators, the earnings season as a whole has outperformed expectations, with over 75% of S&P500 corporations reporting results that exceeded analyst expectations. Investors cannot resist but predict that Corporate America will be able to weather the storm of inflation, massive rate hikes, and slowing economic growth.

However, two key US price indexes reported higher than forecast, especially the personal consumption expenditures index, which Federal Reserve highly focused on, climbing at the fastest pace since 2005. Overall, the rebound of stocks this month has not yet convinced investors that a new bull market has begun.

Due to a stronger-than-expected earnings season, the S&P500 and the Dow Jones Industrial Average rose on Friday. The S&P500 increased 1.42% daily, while the Dow Jones Industrial Average rose 1% daily. Energy and Consumers Discretion have the best performance of all sectors, increasing by 4.51% and 4.27%, respectively. Meanwhile, the Nasdaq 100 increased by 1.8% on Friday, while the MSCI World index increased by 1.2%.

Main Pairs Movement

On Friday, the US dollar fell to a three-week low below 105.6 as speculation swirled over a possible recession in the wake of conflicting economic data. The DXY index maintained its bearish trend and dropped to a three-week low during the first half of the trading day, before rebounding to a daily high over 106.6 during the middle of the European trading session. Inflation in the United States remained inexorable in June, according to the US price index, and an increase in interest rates is not inconceivable.

The GBPUSD was steady on Friday, losing 0.07%. The cables witnessed some new transactions and reached a daily high level above 1.224 during the Asia trading session amid a weak US dollar across the board. However, the cables fell to a daily low level below 1.207 following the release of a strong US Personal Consumption Expenditures report ahead of the New York trading session. During the middle of the Asia session, EURUSD reached a daily high above 1.025 before falling to a low below 1.015. The pair increased by 0.23% on the day.

Gold climbed 0.58% daily and is now firmly established at over $1760 as US yields erase prior gains and inflation fears. In the meantime, WTI crude oil gained bullish momentum throughout the late Asian session and surpassed$99 per barrel on Friday.

Technical Analysis

XAUUSD(4-Hour Chart)

Gold gathers bullish move following a dip earlier in the session as the US yields turn south. From the technical perspective, gold continues to build its upside momentum beyond the 1746 support zone and gains positive traction for the third consecutive day on Friday at the time of writing. The four-hour outlook is bullish as gold remains way above the ascending trend line. Despite the RSI indicator having reached beyond the overbought territory, the positive MACD seems to continue to support the bulls. As long as gold sustains its momentum above the trend line, its positive outlook remains unchanged. To the upside, the breakout of the immediate hurdle of 1766 would give gold another rally.

Resistance: 1766, 1786, 1811

Support: 1746, 1721, 1680

USDJPY (4-Hour Chart)

USDJPY plunges badly to near 133.00 as the US dollar confronts selling pressure following the fading aggressive Fed tightening expectations. From the technical perspective, it is now seen as in correction to the four-hour outlook up trend. Since USDJPY has officially broken the bullish channel, the bias becomes a downside. Strong support is expected to see at 132.50 for the rebound. A breakout of the support level of 132.50 would accelerate the downside, heading further south. For now, for the bull to reclaim its position, USDJPY needs to at least rise above 135.94.

Resistance: 134.13, 135.13, 135.94

Support: 132.50

EURUSD (4-Hour Chart)

EURUSD edges higher near 1.0200 as the US dollar loses traction. From the technical aspect, the euro dollar starts a fresh recovery against the dollar after hitting the midline of the Bollinger band and the 20 Simple Moving Average. EURUSD looks to oscillate in the range of 1.0109- 1.0284 on the four-hour chart. The outlook of EURUSD is currently neutral as the RSI indicator hovers around the midline and the MACD remains neutral. On the upside, the resistance of 1.0284 would be the first major hurdle to break to begin the bullish move of the euro dollar. On the contrary, if EURUSD falls below 1.0109, the pair might decline towards the 0.9952 level.

Resistance: 1.0284, 1.0362, 1.0459

Support: 1.0109, 0.9952

Economic Data


text:
# UnicodeDecodeError: ASCII Codec Can’t Decode¶

Unicode is currently non fully supported by VNL and ATK (2016). We are working to have complete unicode support in the next releases. If you get an error message similar to:

Traceback (most recent call last):
File “.\zipdir\NL\GUI\Main\Main.py”, line 114, in main
File “.\zipdir\NL\GUI\MainWindow\MainStack.py”, line 21, in init
File “.\zipdir\NL\GUI\MainWindow\MainStack.py”, line 114, in readSettings
UnicodeEncodeError: ‘ascii’ codec can’t encode character u’\x86’ in position 14: ordinal not in range(128)

you will have to ensure that your username, the project folder, and the installation folder do not contain any non-ascii character.

You can also try the following solution: #. Open the vnl (vnl.exe) executable present in the bin folder in your ATK installation folder. #. Change the export LANG=C line with export LANG=C.UTF-8.


text:
# printString -- lowlevel function to print a string, net, or symbol

## Description

printString(o,s) -- send the string s to the output file o.

This function is intended for internal use only.

The argument s may also be a sequence or list, in which case its elements are printed. If an integer is encountered, then it specifies a number of spaces to be printed. If a symbol or indeterminate is encountered, its name is printed. If null is encountered, nothing is printed.

 i1 : printString(stdio, (a,10,"b",20,c)) a b c o1 = stdio o1 : File

## For the programmer

The object printString is .


text:
Headbutt trees are special trees found in the Johto and Kanto regions that a Pokémon can use Headbutt on to encounter wild Pokémon. They are only found in the games that feature Johto: Pokémon Gold, Silver, and Crystal and Pokémon HeartGold and SoulSilver.

## Generation II

 This section is incomplete.Please feel free to edit this section to add missing information and complete it. Reason: Confirm that formulas/algorithms apply to both Gold/Silver and Crystal (and not only to one)

In Generation II, most outdoor areas in Johto contain a number of special trees () that Pokémon can use Headbutt on. These trees are distinct from other trees that either take up two vertical paces or that have rounded tops. When a Pokémon uses Headbutt on a tree, a wild Pokémon may drop out of the tree. Depending on the time of day, the Pokémon that falls out may be asleep. Some Pokémon can only be caught by this method (e.g., Pineco, Heracross).

All headbuttable trees contain wild Pokémon, but different trees have different chances of creating an encounter after Headbutt is used. The headbuttable trees in an area also generate their encounters from different sets of Pokémon depending on whether they have moderate encounter chances or low encounter chances. For example, on Route 44, trees with moderate encounter chances may only contain Spearow or Aipom, while trees with a high encounter chance may also contain Heracross.

### Encounter mechanics

The encounter rate and encounter table of each tree depends on the tree's index and the player's Trainer ID number.

The tree's index is an integer from 0 to 9, which depends on its X and Y coordinates on the map—that is, its distance from the westernmost and northernmost edges, respectively. Specifically, the tree's index is calculated using the following formula.

${\displaystyle TreeIndex=\left\lfloor {\frac {X\cdot Y+X+Y}{5}}\right\rfloor {\bmod {1}}0}$

The encounter rate and tree type depends on the last digit of the player's Trainer ID.

• If a tree's index is equal to that ID digit, the tree is a "high-encounter tree" and its encounter rate is 80%.
• If the tree's index is one of the next four indices after that ID digit (wrapping back around to 0 after 9), the tree is a "moderate-encounter tree" and its encounter rate is 50%.
• Otherwise, the tree is a "moderate-encounter tree" and its encounter rate is 10%.

#### Encounter rate by index and Trainer ID

The following is a table depicting the encounter rate of the tree, based on the tree index and the last digit of the player's Trainer ID. Tree indexes are displayed in rows, while Trainer ID digits are displayed in columns.

An 80% encounter rate indicates the tree is a "high-encounter tree". Otherwise it is a "moderate-encounter tree".

0 1 2 3 4 5 6 7 8 9
0 80% 10% 10% 10% 10% 10% 50% 50% 50% 50%
1 50% 80% 10% 10% 10% 10% 10% 50% 50% 50%
2 50% 50% 80% 10% 10% 10% 10% 10% 50% 50%
3 50% 50% 50% 80% 10% 10% 10% 10% 10% 50%
4 50% 50% 50% 50% 80% 10% 10% 10% 10% 10%
5 10% 50% 50% 50% 50% 80% 10% 10% 10% 10%
6 10% 10% 50% 50% 50% 50% 80% 10% 10% 10%
7 10% 10% 10% 50% 50% 50% 50% 80% 10% 10%
8 10% 10% 10% 10% 50% 50% 50% 50% 80% 10%
9 10% 10% 10% 10% 10% 50% 50% 50% 50% 80%

#### Analysis

Since X and Y are interchangeable in the tree index formula, it is possible to "fix" one dimension to consider traveling along the other. Substituting "Z" for the fixed axis and "n" for the axis that will be traversed, the formula becomes:

{\displaystyle {\begin{aligned}TreeIndex&=\left\lfloor {\frac {Z\cdot n+Z+n}{5}}\right\rfloor {\bmod {1}}0\\&=\left\lfloor {\frac {(Z+1)\cdot n+Z}{5}}\right\rfloor {\bmod {1}}0\end{aligned}}}

This result shows that, if a single row or column of trees is traversed, moving to an adjacent tree increases the tree's index by ${\textstyle {\tfrac {Z+1}{5}}}$ (modulo 10), where Z is the distance of that row or column from its origin edge (north or west). This means that the closer a row or column is to the edge, the slower the indices of those trees change as the row or column is traversed.

### Pokémon

#### Pokémon Gold and Silver

Wild Pokémon encounters from headbuttable trees in Pokémon Gold and Silver fall into two simple groups.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 65%
G S C 10 15%
G S C 10 5%
G S C 10 65%
G S C 10 15%
G S C 10 5%
G S C 10 15%
High-encounter trees
G S C 10 50%
G S C 10 5%
G S C 10 50%
G S C 10 5%
G S C 10 15%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 80%
G S C 10 20%
High-encounter trees
G S C 10 50%
G S C 10 20%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

#### Pokémon Crystal

In Pokémon Crystal, wild Pokémon encounters from headbuttable trees fall into six different groups.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 65%
G S C 10 15%
G S C 10 20%
High-encounter trees
G S C 10 50%
G S C 10 20%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 5%
G S C 10 5%
G S C 10 50%
G S C 10 10%
G S C 10 30%
High-encounter trees
G S C 10 15%
G S C 10 5%
G S C 10 15%
G S C 10 5%
G S C 10 60%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 15%
G S C 10 20%
G S C 10 65%
High-encounter trees
G S C 10 20%
G S C 10 50%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 15%
G S C 10 20%
G S C 10 65%
High-encounter trees
G S C 10 20%
G S C 10 50%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 20%
G S C 10 50%
G S C 10 15%
G S C 10 15%
High-encounter trees
G S C 10 20%
G S C 10 50%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Moderate-encounter trees
G S C 10 80%
G S C 10 20%
High-encounter trees
G S C 10 50%
G S C 10 20%
G S C 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

## Generation IV

In Pokémon HeartGold and SoulSilver, Pokémon are now pre-defined for each eligible location. Due to this, the levels of Pokémon vary between locations, and the player can now find Pokémon in Kanto beyond Routes 26 and 27. The player can Headbutt trees while Surfing, and in fact certain encounter trees are otherwise unreachable.

Each location with trees has two groups of six encounter slots - a tree can be assigned to one of them, or none at all. The trees that potentially can contain encounters are pre-defined and will always be the same. The trees that are not part of the game's Headbutt array will never contain encounters - those that are, may contain an encounter depending on the player's Trainer ID and secret ID number combination. Unlike in Generation II, a Headbutt tree that contains encounters will always yield one; Repel cannot prevent Pokémon from appearing when Headbutting a tree.

Additionally, Pewter City, Cherrygrove City, Routes 25, 38, and 47, and National Park have a third group of encounter slots assigned to certain trees, with the ones on Route 38 and in Cherrygrove City and National Park being blocked off by a rocky wall and thus not available until much later.

The amount of trees with Pokémon can range from as few as 2, to as many as 44 or 45 in Ilex Forest.

### Encounter mechanics

Which of the trees with possible encounters are assigned to have them is influenced by the following formula.

${\displaystyle X=((ID_{secret}\ll 16)|ID_{normal})\mod 10}$

The X is used to determine encounter slots for a given defined tree; where TreeID is tree's index number in the location's Headbutt array (for example, this equals to 0 for the first defined tree).

• One potential encounter tree (unused): The tree will have group A encounters if X is even, and group B if X is odd.
• Two (Route 16 half by Celadon City), three (Routes 14, 16, 21, 39 and 45, as well as Mt. Silver's mountainside) or four (Routes 3, 5 and 33) potential encounter trees: ${\displaystyle Encounter_{ID}=Tree_{ID}+(x*n)}$, where n is amount of potential encounter trees (that aren't part of Group C) in the location

In the following table, rows refer to n while columns refer to EncounterID. - signifies lack of encounters:

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39
2 A B A B A B A B A B A N/A
3 A B - A - B A - B - A - A B - B A A B - A - B A - B - A N/A
4 - A B - A B - A B - A B - A B A B - A B A B - A B A B - A B A B

These can be simplified into cases:

${\displaystyle {\begin{cases}{\text{ For }}n=2{\begin{cases}A,&{\text{if }}Encounter_{ID}\mod 4=\{0,3\}\\B,&{\text{if }}Encounter_{ID}\mod 4=\{1,2\}\end{cases}}\\{\text{ For }}n=3{\begin{cases}A,&{\text{if }}Encounter_{ID}\mod 18=\{0,3,7,11,13,17\}\\B,&{\text{if }}Encounter_{ID}\mod 18=\{1,5,6,9,14,16\}\end{cases}}\\{\text{ For }}n=4{\begin{cases}A,&{\text{if }}Encounter_{ID}\mod 5=\{1,2\},&{\text{for }}Encounter_{ID}<20\\B,&{\text{if }}Encounter_{ID}\mod 5=\{3,4\},&{\text{for }}Encounter_{ID}<20\\A,&{\text{if }}Encounter_{ID}\mod 5=\{1,3\},&{\text{for }}Encounter_{ID}\geq 20\\B,&{\text{if }}Encounter_{ID}\mod 5=\{2,4\},&{\text{for }}Encounter_{ID}\geq 20\end{cases}}\end{cases}}}$

• More than four trees: ${\displaystyle Encounter_{ID}=(Tree_{ID}\mod 5)+(x*5)}$
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
- A B - A B - A B - A B -
25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49
- A B A B - A B A B - A B A B - A B A B -

Due to the way the trees are assigned encounters, certain IDs have one less tree with encounters. For example, locations with 4 potential encounter trees have one tree with no encounters unless the ID combination results in X of 4 and 9. Similarly, in Ilex Forest, if an ID combination results in 0 or 5, there will be 12 trees without encounters instead of 11.

Group C trees are defined separately from normal trees, and as such are not influenced by the Trainer ID numbers.

#### Example

For Trainer ID of 19410 and secret ID of 21243:

• 1941010 = 4BD216
• 2124310 = 52FB16

{\displaystyle {\begin{aligned}X&=((52FB\ll 16)|4BD2)\mod A\\X&=(52FB0000|4BD2)\mod A\\X&=52FB4BD2\mod A\\X&=8\end{aligned}}}

Effectively, X is a modulo operation by 10 of a 32-bit number where secret ID is responsible for its 16 upper bits and regular ID is responsible for its 16 lower bits.

With X of 8, EncounterID for first tree in the location is 16 for areas with 2 potential encounter trees, 24 for areas with 3, 32 for areas with 4, and 40 for areas with more than 4. In the first case, the tree has encounter group A; in all other cases it has encounter group B.

### Pokémon

Pokémon Games Location Levels Rate
Group A
New Bark Town, Routes 29 and 30: Lv. 2-3
Ecruteak City and Route 37: Lv. 12-14
Safari Zone Gate: Lv. 14-16
Routes 26 and 27: Lv. 28-29
Route 31: Lv. 3-4
Route 34: Lv. 9-10
Route 35: Lv. 10-11
Route 36: Lv. 4-5
Route 39: Lv. 14-15
Route 48: Lv. 15-16
HG SS Varies 20%
HG SS Varies 50%
HG SS Varies 30%
Group B
New Bark Town, Routes 29 and 30: Lv. 4-5
Ecruteak City and Route 37: Lv. 15-17
Safari Zone Gate: Lv. 17-19
Routes 26 and 27: Lv. 31-32
Route 31: Lv. 5-6
Route 34: Lv. 11-12
Route 35: Lv. 13-14
Route 36: Lv. 6-7
Route 39: Lv. 16-17
Route 48: Lv. 17-18
HG SS Varies 20%
HG SS Varies 50%
HG SS Varies 30%
HG SS Varies 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 2-3 20%
HG SS 2-3 50%
HG SS 2-3 30%
Group B
HG SS 4-5 20%
HG SS 4-5 50%
HG SS 4-5 30%
HG SS 4-5 30%
Group C
The four trees on the hill by the sea. Accessible by surfing west into the water, then using Rock Climb.
HG SS 18-20 65%
HG SS 18-20 30%
HG SS 21-25 5%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 10 20%
HG SS 2-3 50%
HG SS 10 30%
Group B
HG SS 10 20%
HG SS 4-5 50%
HG SS 10 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Azalea Town and Route 33: Lv. 3-5
Route 42: Lv. 13-14
Route 44: 21-22
Route 46: Lv. 2-3
HG SS Varies 50%
HG SS Varies 20%
HG SS Varies 30%
Group B
Azalea Town and Route 33: Lv. 6-8
Route 42: Lv. 15-16
Route 44: 23-24
Route 46: Lv. 4-5
HG SS Varies 80%
HG SS Varies 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Pallet Town and Route 1: Lv. 2-3
Viridian City and Routes 2 and 22: Lv. 3-4
HG SS Varies 50%
HG SS Varies 30%
HG SS Varies 20%
Group B
Pallet Town and Route 1: Lv. 5-6
Viridian City and Routes 2 and 22: Lv. 5-7
HG SS Varies 50%
HG SS Varies 30%
HG SS Varies 30%
HG SS Varies 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 3-5 50%
HG SS 3-5 30%
HG SS 3-5 20%
Group B
HG SS 7-9 80%
HG SS 7-9 20%
Group C
Two uppermost trees by the fence, reachable from Route 2
HG SS 5-9 65%
HG SS 5-9 30%
HG SS 8-12 5%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Cerulean City, Routes 5 and 6: Lv. 11-13
Route 8: Lv. 16-17
HG SS Varies 50%
HG SS Varies 30%
HG SS Varies 20%
Group B
Cerulean City, Routes 5 and 6: Lv. 14-16
Route 8: Lv. 18-19
HG SS Varies 80%
HG SS Varies 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Vermilion City: Lv. 11-13
Celadon City and Route 7: 15-17
Route 11: Lv. 13-14
Route 16: Lv. 26-27
HG SS Varies 50%
HG SS Varies 30%
HG SS Varies 20%
Group B
Vermilion City: Lv. 14-16
Celadon City and Route 7: 18-20
Route 11: Lv. 15-16
Route 16: Lv. 29-30
HG SS Varies 80%
HG SS Varies 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Fuchsia City and Route 15: Lv. 21-23
Routes 12 and 13: Lv. 22-23
Routes 14 and 18: Lv. 26-27
Route 21: Lv. 24-26
HG SS Varies 50%
HG SS Varies 30%
HG SS Varies 30%
HG SS Varies 20%
Group B
Fuchsia City: Lv. 26-34
Routes 12 and 13: Lv. 24-25
Routes 14 and 18: Lv. 29-30
Route 15: Lv. 24-26
Route 21: Lv. 27-30
HG SS Varies 15%
HG SS Varies 65%
HG SS Varies 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 6-7 50%
HG SS 6-7 30%
HG SS 6-7 20%
Group B
HG SS 9-10 80%
HG SS 9-10 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 10-11 50%
HG SS 10-11 30%
HG SS 10-11 20%
Group B
HG SS 13-14 80%
HG SS 13-14 20%
Group C
The four trees at the northwest corner of Route 25.
HG SS 10-14 30%
HG SS 15-18 5%
HG SS 10-14 65%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Route 28 and Mt. Silver (exterior): Lv. 39-40
Mt. Silver (mountainside): Lv. 42-44
HG SS Varies 20%
HG SS Varies 30%
HG SS Varies 50%
Group B
Route 28 and Mt. Silver (exterior): Lv. 41-43
Mt. Silver (mountainside): Lv. 45-48
HG SS Varies 20%
HG SS Varies 30%
HG SS Varies 50%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 4-5 20%
HG SS 4-5 50%
HG SS 4-5 30%
Group B
HG SS 6-7 20%
HG SS 6-7 80%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 13-14 20%
HG SS 13-14 50%
HG SS 13-14 30%
Group B
HG SS 15-16 20%
HG SS 15-16 50%
HG SS 15-16 30%
HG SS 15-16 30%
Group C
The easternmost tree at top of the hill accessible via Route 39 by Rock Climb.
HG SS 18-20 65%
HG SS 18-20 30%
HG SS 21-25 5%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
Route 43: Lv. 15-16
Lake of Rage: Lv. 14-16
HG SS Varies 20%
HG SS Varies 50%
HG SS Varies 30%
Group B
Route 43: Lv. 17-18
Lake of Rage: Lv. 17-19
HG SS Varies 15%
HG SS Varies 20%
HG SS Varies 65%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 23-24 50%
HG SS 23-24 20%
HG SS 23-24 15%
HG SS 23-25 15%
Group B
HG SS 26-27 80%
HG SS 26-27 20%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 15-16 20%
HG SS 15-16 50%
HG SS 15-16 30%
Group B
HG SS 17-18 20%
HG SS 17-18 50%
HG SS 17-18 30%
HG SS 17-18 30%
Group C
Leftmost and rightmost trees at top of the east waterfall and the two upmost trees on top of the hill on the western side
HG SS 25-28 50%
HG SS 25-28 30%
HG SS 25-28 50%
HG SS 25-28 30%
HG SS 25-28 15%
HG SS 27-30 5%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 3-5 30%
HG SS 3-5 10%
HG SS 3-5 30%
HG SS 3-5 10%
HG SS 3-5 60%
Group B
HG SS 6-8 10%
HG SS 6-8 10%
HG SS 6-8 50%
HG SS 6-8 10%
HG SS 6-8 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 10-12 20%
HG SS 10-12 50%
HG SS 10-12 30%
Group B
HG SS 13-15 20%
HG SS 13-15 50%
HG SS 13-15 30%
HG SS 13-15 30%
Group C
Five trees at top of the hill in the north-east section. Requires Rock Climb.
HG SS 18-20 65%
HG SS 18-20 30%
HG SS 21-25 5%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.

Pokémon Games Location Levels Rate
Group A
HG SS 3-4 50%
HG SS 3-4 10%
HG SS 3-4 10%
HG SS 3-4 30%
HG SS 3-4 5%
HG SS 3-4 5%
HG SS 3-4 30%
Group B
HG SS 5-7 50%
HG SS 5-7 10%
HG SS 5-7 10%
HG SS 5-7 30%
HG SS 5-7 5%
HG SS 5-7 5%
HG SS 5-7 30%
A colored background means that the Pokémon can be found in this location in the specified game. A white background with a colored letter means that the Pokémon cannot be found here.


text:
# mathlibdocumentation

analysis.normed_space.int

# The integers as normed ring #

This file contains basic facts about the integers as normed ring.

Recall that ∥n∥ denotes the norm of n as real number. This norm is always nonnegative, so we can bundle the norm together with this fact, to obtain a term of type nnreal (the nonnegative real numbers). The resulting nonnegative real number is denoted by ∥n∥₊.

@[simp]
theorem int.nnnorm_coe_nat (n : ) :
@[simp]
theorem int.norm_coe_nat (n : ) :


text:
# Gina Ripley, president of Dearing Company, is considering the purchase of a computer-aided manufacturing system. The annual net cash benefits and savings associated with the system are described as follows: The system will cost $9,000,000 and last 10 years. The company’s cost of capital is 12 percent. Required: 1. Calculate the payback period for the system. Assume that the company has a policy of only accepting projects with a payback of five years or less. Would the system be acquired? 2. Calculate the NPV and IRR for the project. Should the system be purchased—even if it does not meet the payback criterion? 3. The project manager reviewed the projected cash flows and pointed out that two items had been missed. First, the system would have a salvage value, net of any tax effects, of$1,000,000 at the end of 10 years. Second, the increased quality and delivery performance would allow the company to increase its market share by 20 percent. This would produce an additional annual net benefit of $300,000. Recalculate the payback period, NPV, and IRR given this new information. (For the IRR computation, initially ignore salvage value.) Does the decision change? Suppose that the salvage value is only half what is projected. Does this make a difference in the outcome? Does salvage value have any real bearing on the company’s decision? BuyFindarrow_forward ### Cornerstones of Cost Management (C... 4th Edition Don R. Hansen + 1 other Publisher: Cengage Learning ISBN: 9781305970663 #### Solutions Chapter Section BuyFindarrow_forward ### Cornerstones of Cost Management (C... 4th Edition Don R. Hansen + 1 other Publisher: Cengage Learning ISBN: 9781305970663 Chapter 19, Problem 15E Textbook Problem 45 views ## Gina Ripley, president of Dearing Company, is considering the purchase of a computer-aided manufacturing system. The annual net cash benefits and savings associated with the system are described as follows:The system will cost$9,000,000 and last 10 years. The company’s cost of capital is 12 percent.Required: 1. Calculate the payback period for the system. Assume that the company has a policy of only accepting projects with a payback of five years or less. Would the system be acquired? 2. Calculate the NPV and IRR for the project. Should the system be purchased—even if it does not meet the payback criterion? 3. The project manager reviewed the projected cash flows and pointed out that two items had been missed. First, the system would have a salvage value, net of any tax effects, of $1,000,000 at the end of 10 years. Second, the increased quality and delivery performance would allow the company to increase its market share by 20 percent. This would produce an additional annual net benefit of$300,000. Recalculate the payback period, NPV, and IRR given this new information. (For the IRR computation, initially ignore salvage value.) Does the decision change? Suppose that the salvage value is only half what is projected. Does this make a difference in the outcome? Does salvage value have any real bearing on the company’s decision?

1.

To determine

Calculate the payback period for the given system and state whether system would be acquired or not.

### Explanation of Solution

Payback period: Payback period is the expected time period which is required to recover the cost of investment. It is one of the capital investment method used by the management to evaluate the proposal of long-term investment (fixed assets) of the business. But payback method has high risk than other method, because it does not follow the time value of money concept in valuing the cash inflows.

Calculate the payback period for the given system and state whether system would be acquired or not as follows:

When the estimated annual net cash is equal (even cash flow), the cash payback period is calculated as below:

PaybackPeriod=Amount to be investedEstimated annual net cash inflow =$9,000,000$1,500,000 (1)=6.00years

Working note (1):

Calculate the annual cash flow of the investment

2.

To determine

Calculate the net present value (NPV) and internal rate of return (IRR) for the given project, and indicate whether the given project is better for the investment or not.

3.

To determine

Recalculate the payback period, net present value and internal rate of return of the project based on the given changes.

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text:
## January 10, 2006

### Killing Horizons

You know what a black hole is, right?  Not so fast, these cul de sacs in space-time may not exist at all.

We’ve all read about the enormous black hole at the center of our galaxy, the Milky Way.  However, black holes by their nature can’t be seen since light doesn’t escape from them.  What is observed is matter falling towards what appears to be almost a black hole.  These observations would be the same whether or not actual black holes exist.

Son Jarah, a string theory post-doc at the Free University of Brussels, is more concerned over his latest paper than he’s been over the dozens that he’s written before.  The other papers have properly academic titles like “K-Theory and S-Duality: Starting Over from Square 3”, “IIB Soliton Spectra with All Fluxes Activated”, and “Nonabelian Superconductors: Vortices and Confinement in ${\cal N}=2$ SQCD”.  Despite intense curiosity, I have no idea what any of them are about.

His latest paper is simply “Killing Horizons”.  Even though I don’t understand the math, I do know what this paper is about.

It’s both a radical claim and one that is not, strictly speaking, in his field of string theory: two reasons why a young man who will someday be looking for tenure has cause to be concerned.

Jarah’s claim is that event horizons introduce anomalies into physics which both contradict the usual ways that we formulate physical principals and are unhelpful in proceeding towards a Grand Unified Field Theory (GUT).  If there are no event horizons, then there are essentially no black holes.  He puts his claim a little more cautiously: “…while the elimination of horizons is a possibility which may be a consequence of a hypothesized principle, we have presented no evidence that either this possibility or this principle is realized.”  He goes on to suggest observational data which may soon be available which would tend to validate (or contradict) his hypothesis.

Black holes have been getting grey for a while.  Hawking long ago suggested that they leak “Hawking radiation”.  Lately Hawking has agreed with others that information trapped inside a black hole can leak out.  The black holes of string theory aren’t really black

An implication of Jarah’s argument is that physics researchers have overlooked the anomalies introduced by the concepts of event horizons and black holes; their energy is better spent in speculating on a universe which contains neither.

If this argument is correct, for example, the strange concept of a period of “inflationary expansion” in the very early life of our universe is unnecessary and unhelpful.  According to Jarah, there is a simpler quantum explanation for the fact that the entire observed universe seems to have the same level of background radiation despite the observation that some parts of the universe would seem to be across the event horizon from other parts.

Jarah agonized over how strong a claim to make in the abstract for his paper.  “I rarely read papers with radical claims, myself,” he said.  On the other hand, the abstract should be interesting or no one will read the paper.  And it should describe what’s inside.

Not very long ago, the referees of a scientific journal would have decided whether or not this paper should be published.  Such gatekeepers have been notoriously conservative about papers which upset the conventional wisdom.  On the other hand, they’ve probably saved many hot-headed young scientists from career-damaging mistakes.  For better or worse (I think better), Jarah and his contemporaries publish their papers simply by posting them on a dedicated website.

That doesn’t mean you and I can self-publish our speculations there, however; it’s Jarah’s undergraduate degree from Caltech and doctorate from Berkeley which give him write privileges to this repository.  Jarah’s paper will only get whatever attention it gets because of his credentials, other papers he’s published, talks he’s given, and conferences he’s participated in.

Science like business advances in fits and starts.  A new theory fits all known facts.  Then new observations, usually first thought to be observational errors, fall outside the penumbra of the current theory.  The current theory is patched and patched like an old computer program to fit the new facts.  Finally someone takes the bold step of replacing the patched-up old theory with a new, cleaner one that explains the then known facts.  And the cycle starts again.

Just as most new businesses fail, most new ideas are wrong.  It is only with hindsight that it is clear what the breakthroughs are.  I’m proud of Jarah for this intellectual startup.  If I could, I would short black holes; they’re history.


text:
# Gauss's Lemma on Primitive Rational Polynomials

## Theorem

Let $\Q$ be the field of rational numbers.

Let $\Q \sqbrk X$ be the ring of polynomials over $\Q$ in one indeterminate $X$.

Let $\map f X, \map g X \in \Q \sqbrk X$ be primitive polynomials.

Then their product $f g$ is also a primitive polynomial.

## Proof 1

Let $f$ and $g$ be as follows:

 $\displaystyle f$ $=$ $\displaystyle \sum_{k \mathop \in \Z} a_k X^k$ $\displaystyle g$ $=$ $\displaystyle \sum_{k \mathop \in \Z} b_k X^k$

From the definition of primitive polynomial, the coefficients of $f$ and $g$ are all integers.

From the definition of polynomial product:

$\displaystyle f g = \sum_{k \mathop \in \Z} c_k \mathbf X^k$

where:

$\displaystyle c_k = \sum_{\substack {p \mathop + q \mathop = k \\ p, q \mathop \in \Z}} a_p b_q$

it is clear that the coefficients of $f g$ are also all integers.

Aiming for a contradiction, suppose $f g$ is not primitive.

Then its coefficients must have a GCD greater than $1$.

Therefore there exists some prime $p$ which divides all the coefficients of $fg$.

Now $p$ can not divides all the coefficients of either $f$ or $g$, because they are primitive polynomials.

So:

Let $i$ be the smallest integer such that $p$ does not divide $a_i$
Let $j$ be the smallest integer such that $p$ does not divide $b_j$.

Consider the coefficient $c_{i+j}$ of $f g$:

$\displaystyle c_{i + j} = \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k} = a_0 b_{i + j} + a_1 b_{i + j - 1} + \cdots + a_i b_j + \cdots + a_{i + j - 1} b_1 + a_{i + j} b_0$

From the assumption, it follows that $p \divides c_{i + j}$, and so:

$\displaystyle p \divides \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$

where $\divides$ denotes divisibility.

Also, from the choice of $i$ and $j$, we have:

$p \divides a_m$ whenever $m < i$
$p \divides b_n$ whenever $n < j$.

Now all the terms of $\displaystyle \sum_{k \mathop = 0}^{i + j} a_k b_{i + j - k}$, except for $a_i b_j$, contain a factor from either $\set {a_0, a_1, \ldots, a_{i - 1} }$ or $\set {b_0, b_1, \ldots, b_{j - 1} }$.

It follows that we have:

$p \divides \paren {\displaystyle \sum_{k \mathop = 0}^{i - 1} a_k b_{i + j - k} + \sum_{k \mathop = i + 1}^{i + j} a_k b_{i + j - k} }$

But then, as also $p \divides c_{i + j}$, it follows that $p \divides a_i b_j$ as well.

From Euclid's Lemma for Prime Divisors, $p$ has to divide one or the other.

This contradicts the definition of $i$ and $j$.

So $i$ and $j$ cannot both exist.

It follows that $p$ divides at least one of $f$ and $g$, one of which is therefore not primitive.

From this contradiction, we conclude that $f g$ must be primitive.

Hence the result.

$\blacksquare$

## Proof 2

By hypothesis $f$ and $g$ have content $1 \in \Z$.

Therefore:

$f, g \in \Z \sqbrk X$

Aiming for a contradiction, suppose that $f g$ is not primitive, say:

$\cont {f g} = d \ne 1$

By the Fundamental Theorem of Arithmetic we can choose a prime $p$ dividing $d$.

Let $\pi : \Z \to \Z / p \Z$ be the canonical epimorphism to the ring of integers modulo $p$.

From Ring of Integers Modulo Prime is Field, $\Z / p \Z$ is a field.

Let $\Pi : \Z \sqbrk X \to \paren {\Z / p \Z} \sqbrk X$ be the induced homomorphism of the polynomial rings.

By construction, $p$ divides each coefficient of $f g$, so:

$\map \Pi {f g} = \map \Pi f \, \map \Pi g = 0$

From Polynomial Forms over Field form Integral Domain, $\paren {\Z / p \Z} \sqbrk X$ is an integral domain.

Thus:

$\map \Pi f = 0$ or $\map \Pi g = 0$

After possibly exchanging $f$ and $g$, we may assume that $\map \Pi f = 0$.

$f = a_0 + a_1 X + \cdots + a_n X^n$

we must have:

$\map \pi {a_i} = 0$

for $i = 0, \ldots, n$.

That is:

$a_i \in p \Z$

for $i = 0, \ldots, n$.

But this says precisely that $p$ divides each $a_i$, $i = 0, \ldots, n$.

Therefore $p$ divides the content of $f$, a contradiction.

Hence our assumption that $f g$ is not primitive was invalid.

The result follows by Proof by Contradiction.

$\blacksquare$

## Source of Name

This entry was named for Carl Friedrich Gauss.


text:
## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$$0=x^2-3$$
To write a quadratic equation with the given values as solutions, we find the corresponding factors. We then FOIL out the factors to find the original quadratic equation: $$\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)=0 \\ x^2-\left(\sqrt{3}\right)^2 =0 \\ 0=x^2-3$$


text:
## Astronomy C

Name
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### Re: Astronomy C

OrigamiPlanet wrote:This problem set from this particular PA finals is really confusing me, problems 80-83. Could someone help me understand what formula or formulas I need to be using for this?

https://scioly.org/wiki/images/7/73/Pen ... stions.pdf

https://scioly.org/wiki/images/1/12/Ans ... states.pdf

Thanks!
It's kinda weird. I can't figure out how they got the answer to number 80, I tried alotta different weird things, nothing worked. But when I calculated escape velo of the B0 star, it is greater then the original speed, making me think the answer should be never? They probably did something along the line of accelerating the gas (by a factor of 278) due to the gravitational acceleration of the neutron star, I don't know what.
81 assuming you have the answer to 80, set 2.48E16=1/2mv^2, solve for v, and divide by c to get 74%
82, also assuming you have the answer to 80, find rest energy with e=mc^2 and divide 2.48E16 by 9E16 to get 28%
83 I did that your given the energy is 8.5E29 and find mass with e=mc^2 and divide 8.5E29/9E16 to get 9.4E12, but this disagrees with the key that says 3.4E13.

Either the key on this test is wrong on some parts, or I'm being very stupid right now.
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### Re: Astronomy C

Name wrote: It's kinda weird. I can't figure out how they got the answer to number 80, I tried alotta different weird things, nothing worked. But when I calculated escape velo of the B0 star...They probably did something along the line of accelerating the gas (by a factor of 278) due to the gravitational acceleration of the neutron star, I don't know what...Either the key on this test is wrong on some parts, or I'm being very stupid right now.
Upfront, I find the key is right.

Yeah, this is overcomplicating, but good ideas when you're unsure. Off the top of my head it might not be trivial to show how the gas is accelerated, which I guess one might figure as follows: (work by gravity) = F * dist = mg * dist. Like I said, might require harder math to do that in practice since that value of g there would change with distance. It would be nice to do with plain kinetic energy (1/2 m v^2)...if you knew the velocity of the gas on impact...that would be hard because accelerations and kinematics of swirling things will be hard (though, technically a valid approach). Even then, you'd still have to account for gravitational potential energy. In a way, you intuited this with the gravitational acceleration pulling stuff off the B0 star even though gas may not be able to exceed the escape velocity of the star by itself. Maybe there's an equation for these ideas, but there's other ways to do this.

This might be a bit tricky to those that don't have more solid physics backgrounds. Still doable, of course! You definitely don't want to deal with acceleration, which often makes problems involving energy harder (this is sort of the point of defining energy). I'll hint for now you have to think about conservation of energy, KE, and GPE, but even then you might mess up what you're using for mass and radius (conventions are a pain but useful!).
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### Re: Astronomy C

Name wrote:
OrigamiPlanet wrote:This problem set from this particular PA finals is really confusing me, problems 80-83. Could someone help me understand what formula or formulas I need to be using for this?

https://scioly.org/wiki/images/7/73/Pen ... stions.pdf

https://scioly.org/wiki/images/1/12/Ans ... states.pdf

Thanks!
It's kinda weird. I can't figure out how they got the answer to number 80, I tried alotta different weird things, nothing worked. But when I calculated escape velo of the B0 star, it is greater then the original speed, making me think the answer should be never? They probably did something along the line of accelerating the gas (by a factor of 278) due to the gravitational acceleration of the neutron star, I don't know what.
81 assuming you have the answer to 80, set 2.48E16=1/2mv^2, solve for v, and divide by c to get 74%
82, also assuming you have the answer to 80, find rest energy with e=mc^2 and divide 2.48E16 by 9E16 to get 28%
83 I did that your given the energy is 8.5E29 and find mass with e=mc^2 and divide 8.5E29/9E16 to get 9.4E12, but this disagrees with the key that says 3.4E13.

Either the key on this test is wrong on some parts, or I'm being very stupid right now.
Their answer is correct by my math.

By Conservation of Energy, we know that K_1 + U_1 = K_2 + U_2 where K and U are the kinetic energy and gravitational potential energy respectively. Thus,

$\frac{1}{2}mv_1^2 - \frac{GMm}{r_1} = K_2 - \frac{GMm}{r_2}$

note that v_1 = 800,000 m/s, r_1 = 0.227 AU = 4e10 m, m=1kg, M=1.88Msol = 3.7e30kg, so K2 = 2.49e16 J.

A faster way of doing this is recognizing that since we're near the bottom of a potential well, the initial energies are essentially zero, since U_2 has such great magnitude, so K_2=U_2. This yield the same answer.
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### Re: Astronomy C

I'm being thrown into this event with 5 days until states (counting today). We have bad notes (but we have notes at least...) How do you suggest I prepare for this event?

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### Re: Astronomy C

diplodocus wrote:I'm being thrown into this event with 5 days until states (counting today). We have bad notes (but we have notes at least...) How do you suggest I prepare for this event?
Hi, Diplodocus!

Math is what really makes the difference for doing well in astronomy. Try and look over the math notes that you have (if you have them). If you don't have them, a good general reference sheet may be found on the wiki here. Especially look at the math that pertains to section C) of the topics on the rule sheet.

Also, make sure you have enough of the of the DSOs (preferably IR, UV, visible) that you can identify them–you can't answer questions if you don't know what it is.
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### Re: Astronomy C

TheMysteriousMapMan wrote:
diplodocus wrote:I'm being thrown into this event with 5 days until states (counting today). We have bad notes (but we have notes at least...) How do you suggest I prepare for this event?
Hi, Diplodocus!

Math is what really makes the difference for doing well in astronomy. Try and look over the math notes that you have (if you have them). If you don't have them, a good general reference sheet may be found on the wiki here. Especially look at the math that pertains to section C) of the topics on the rule sheet.

Also, make sure you have enough of the of the DSOs (preferably IR, UV, visible) that you can identify them–you can't answer questions if you don't know what it is.
I disagree with this, actually. You can learn some of the very basic math early (i.e. parallax, distance modulus, basically what's here) but your focus should be on DSOs and other general information. Most likely, the bulk of points will be on these, and they won't be as time-intensive as the more involved math will be. Only when you have a good conceptual grasp on Astronomy does it become worth it to dive deep into the math in an effort to go from "good" to "great".

If you're competing in PA, I recommend looking at past PA state astronomy tests, as the test will likely follow a similar format: https://scioly.org/wiki/index.php/2019_ ... #Astronomy.
East Stroudsburg South Class of 2012, Alumnus of JT Lambert, Drexel University Class of 2017

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### Re: Astronomy C

EastStroudsburg13 wrote:
TheMysteriousMapMan wrote:
diplodocus wrote:I'm being thrown into this event with 5 days until states (counting today). We have bad notes (but we have notes at least...) How do you suggest I prepare for this event?
Hi, Diplodocus!

Math is what really makes the difference for doing well in astronomy. Try and look over the math notes that you have (if you have them). If you don't have them, a good general reference sheet may be found on the wiki here. Especially look at the math that pertains to section C) of the topics on the rule sheet.

Also, make sure you have enough of the of the DSOs (preferably IR, UV, visible) that you can identify them–you can't answer questions if you don't know what it is.
I disagree with this, actually. You can learn some of the very basic math early (i.e. parallax, distance modulus, basically what's here) but your focus should be on DSOs and other general information. Most likely, the bulk of points will be on these, and they won't be as time-intensive as the more involved math will be. Only when you have a good conceptual grasp on Astronomy does it become worth it to dive deep into the math in an effort to go from "good" to "great".

If you're competing in PA, I recommend looking at past PA state astronomy tests, as the test will likely follow a similar format: https://scioly.org/wiki/index.php/2019_ ... #Astronomy.
I guess it just might be my experiences then, East. The tests I have taken have been pretty math heavy, with maybe 1-5 questions about DSOs.

Edit: Grammar.
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### Re: Astronomy C

TheMysteriousMapMan wrote:
EastStroudsburg13 wrote:
TheMysteriousMapMan wrote:
Hi, Diplodocus!

Math is what really makes the difference for doing well in astronomy. Try and look over the math notes that you have (if you have them). If you don't have them, a good general reference sheet may be found on the wiki here. Especially look at the math that pertains to section C) of the topics on the rule sheet.

Also, make sure you have enough of the of the DSOs (preferably IR, UV, visible) that you can identify them–you can't answer questions if you don't know what it is.
I disagree with this, actually. You can learn some of the very basic math early (i.e. parallax, distance modulus, basically what's here) but your focus should be on DSOs and other general information. Most likely, the bulk of points will be on these, and they won't be as time-intensive as the more involved math will be. Only when you have a good conceptual grasp on Astronomy does it become worth it to dive deep into the math in an effort to go from "good" to "great".

If you're competing in PA, I recommend looking at past PA state astronomy tests, as the test will likely follow a similar format: https://scioly.org/wiki/index.php/2019_ ... #Astronomy.
I guess it just might be my experiences then, East. The tests I have taken have been pretty math heavy, with maybe 1-5 questions about DSOs.

Edit: Grammar.
That seems likely, because that strikes me as a rather unusual Astronomy question distribution. I do not see a test on the Test Exchange that shares that distribution, but then again, I haven't done a thorough check so perhaps one is in there (and perhaps syo can provide input due to his greater experience with Michigan).
East Stroudsburg South Class of 2012, Alumnus of JT Lambert, Drexel University Class of 2017

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### Re: Astronomy C

EastStroudsburg13 wrote:
TheMysteriousMapMan wrote:
EastStroudsburg13 wrote: I disagree with this, actually. You can learn some of the very basic math early (i.e. parallax, distance modulus, basically what's here) but your focus should be on DSOs and other general information. Most likely, the bulk of points will be on these, and they won't be as time-intensive as the more involved math will be. Only when you have a good conceptual grasp on Astronomy does it become worth it to dive deep into the math in an effort to go from "good" to "great".

If you're competing in PA, I recommend looking at past PA state astronomy tests, as the test will likely follow a similar format: https://scioly.org/wiki/index.php/2019_ ... #Astronomy.
I guess it just might be my experiences then, East. The tests I have taken have been pretty math heavy, with maybe 1-5 questions about DSOs.

Edit: Grammar.
That seems likely, because that strikes me as a rather unusual Astronomy question distribution. I do not see a test on the Test Exchange that shares that distribution, but then again, I haven't done a thorough check so perhaps one is in there (and perhaps syo can provide input due to his greater experience with Michigan).
It also might be that I was not clear–math is what, in my opinion/experience, separates the medalers from the rest–because it requires a little more intuition and knowledge than familiarity with basic DSO facts. Thinking about it again, DSOs might be the best place to start with only five days. While some tests have been mostly DSO/conceptual that I have taken, the distribution of scores on those invite tests (my coach wrote them) has been such that the schools who had prepped with math knowledge were clumped significantly higher than the others.
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### Re: Astronomy C

Note: Most you's in this most are just meant colloquially, like to everyone.

In my experience, you can actually do really well on tests either way. Keep in mind I haven't done any strict correlation. But the top medalers I see usually have a solid *overall* grasp of the event. Sometimes if there's more points on the math section people who are better at math win. Sometimes if there are more points on the concepts, people who are more prepared with their DSOs/concepts win.

As for particular states, I'm honestly unsure where MI astro is at right now for the writers. I recall years ago MI had a prof. writing that liked paragraph explanations and some nuclear stuff. In that way, I could see why you (to MapMan) might say math is more important. I've found some different writers (some profs and teachers) that prefer those kinds of tests. That said, even then you'd need to know your stellar evolution basics and vocab, which is not as easy to pull up on the spot as some equations (at least in my opinion). If I had to guess the *majority* of tests ARE mostly made up of conceptual/DSO questions...and people usually mess up a ton of them at that (maybe you'd be surprised). I would say maybe a few more state tests than average involve paragraphs for whatever reason (e.g. maybe a prof writes and thinks that's the way to go). This is mostly guessing keep in mind.

So you really do have to both. It then comes down to: are you better at math or concepts / organizing notes? Are you aiming for a medal or not? How is your partner? What state are you / do people have insights about that state.

I think East was just saying that this person probably wants to get moving, medal or not. So getting a math equation sheet and a few practice examples to work off of for a test goes pretty quickly. On the other hand, the massive amount of info and DSO/concept points will definitely require more time (as MapMan said in the last post).
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text:
# CMC - Problem 2

Problem 2. (5 points) Let

$N=\binom{2013}{0}+\binom{2013}{3}+\binom{2013}{6}+\dots+\binom{2013}{2013}.$

If $$3N=a^{2013}+b^{2013}+c^{2013}$$ where $$a+b+c=3$$, find the value of $$abc.$$

Note by Cody Johnson
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2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

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I did in this way:

$$(1 + x)^{2013} = {2013 \choose 0} + {2013 \choose 1}x + {2013 \choose 2}x^2+ \dots {2013 \choose 2012}x^{2012} + {2013 \choose 2013}x^{2013}$$

Hence , in $$(1 + x)^{2013} + (1 + \omega x)^{2013} + (1+ \omega^2 x)^{2013},$$ coeff. of $$x^{3k + 1}$$ and $$x^{3k + 2}$$ becomes 0 , and coeff. of $$x^{3k}$$ triples

Now, Replace x = 1,

$$(1 + 1)^{2013} + (1 + \omega )^{2013} + (1+ \omega^2 )^{2013}$$

= $$3\bigg({2013 \choose 0} + {2013 \choose 3} + {2013 \choose 6} + \dots + {2013 \choose 2013}\bigg)$$

= $$3N$$

Hence, $$3N = (1 + 1)^{2013} + (1 + \omega)^{2013} + ( 1+ \omega^2)^{2013}$$

Hence , $$a = 2, b = 1+\omega, c = 1 + \omega^2$$, $$a + b+ c = 3 + (1 + \omega + \omega^2) = 3$$, as required.

$$abc = 2(1 + \omega)(1 + \omega^2) = \fbox{2}$$

- 5 years, 1 month ago

Exactly how I did it. Good work, you get the 5 points!

Edit: Pi Han Goh is actually the winner.

- 5 years, 1 month ago

The same process is followed be me!

- 5 years, 1 month ago

Where did the omega come from?

- 5 years, 1 month ago

$$\omega$$ is usually used to denote $$e^{\pi i /3} = \text{cis} \dfrac{\pi}{3}$$ (although he should have specified).

- 5 years, 1 month ago

Michael, do you mean $$\omega=e^{2\pi i/3}$$?

- 5 years, 1 month ago

Thanks. Now it makes sense.

- 5 years, 1 month ago

The way your question is currently phrased, there are (likely to be) numerous answers. Think of a (complex) plane cutting a sucked in ball. We know that there are 6 points of intersection, and this plane is unlikely to be tangential at each of these points.

Staff - 5 years, 1 month ago

- 5 years, 1 month ago

Tell me this - if an integer answer is required by the problem, doesn't that force the user to find the integer answer only? A line in the solution would be like, "And, since we're guaranteed an integer answer, then $$\boxed{2}$$ is the only number that satisfies the problem condition."

Note that I've computed that $$2$$ is the only integer answer for this problem.

- 5 years, 1 month ago

The problem should stand along by itself, without requiring the reader to know that "only integer answers are allowed". If this is a necessary assumption, it should be clearly stated in the question that you are only looking for integer answers. The question, as stated, could have several answers in a neighborhood of 2. It could be interpreted as finding the range of possible values of $$abc$$.

Note that Jatin's solution only shows that $$abc=2$$ is a possible answer, you should explain why no other integer answers are possible.

Staff - 5 years, 1 month ago

So would it be better to say, "find the only possible integer value of $$abc$$?" Or maybe I should've asked for some completely different thing, like to find $$N\pmod{2011}$$?

- 5 years, 1 month ago

I think it would have been better to ask for a closed form expression for $$N$$, and $$\frac{2^{2013} -2 } {3}$$ would have been a good answer. There isn't a strong reason for you to stick to "integer answers from 0 to 999". We do that simply because it is the simplest to explain, without having to deal with various other explanations / considerations.

By asking for "the only possible integer value of $$abc$$", you then require a proof that it is the only possible integer value, which need not necessarily be the case. Yes, there is a natural choice of $$a, b, c$$ in your question, but this need not be the only solution. It is possible for $$abc=0$$, by setting $$c=0$$ and $$a^{2013} + (3-a)^{2013} = 2^{2013} -2$$, and showing that a solution must exist by applying the intermediate value theorem on $$[0, 1.5]$$ (real interval). However, this strays away from the original intention of your question.

Note: I do not know if $$abc=1$$ is possible. It most likely is, but I can't think of an immediate argument for it.

Staff - 5 years, 1 month ago

Yes, but that's not a good problem-solving tactic neither is it good problem-writing practice to do that.

- 5 years, 1 month ago

- 4 years, 10 months ago

$$3N = 2^{2013} -2$$, one choice for $$a,b,c$$ are $$2, -x, -x^2$$ where $$x^2+x+1=0$$. So $$abc=2$$ in this case. But ..

- 5 years, 1 month ago

$$3N\neq2^{2013}-2$$

- 5 years, 1 month ago

Actually, $$3N = 2^{2013} - 2$$, I have proved that in my deleted comment. I've removed my comment because I couldn't prove that $$abc=2$$ only like what Jatin did.

- 5 years, 1 month ago

You are right, $$3N = 2^{2013} - 2$$

Actually, $$3N = 2^{2013} + (1 + \omega)^{2013} + (1 + \omega^2)^{2013}$$

= $$2^{2013} + 2Re(1 + \omega)^{2013} = 2^{2013} + 2 Re(e^{i\frac{\pi}{3} \times 2013})$$

= $$2^{2013} - 2$$

How did you prove it?

- 5 years, 1 month ago

I first consider $$N_0 = {0 \choose 0}, N_1 = {3 \choose 0} + { 3 \choose 3 }, N_2 = {6 \choose 0 } + {6 \choose 3 } + { 6 \choose 6} , \ldots , N_{671} = {2013 \choose 0 } + {2013 \choose 3 } + \ldots {2013 \choose 2013}$$

And I find that $$3N_0, 3N_1, 3N_2, 3N_3$$ are very close to the powers of $$8$$, so I made the conjecture $$3 N_j = 8^j + 2(-1)^j$$ and I proved it by induction with the help of Pascal's identities, it was a little tedious.

Great job by the way!

- 5 years, 1 month ago

Hi Jatin!

How do you get $$(1+\omega)^{2013}+(1+\omega^2)^{2013}=2Re(1+\omega)^{2013}$$?

I did it this way, since $$1+\omega+\omega^2=0$$, hence we have $$1+\omega=-\omega^2$$ and $$1+\omega^2=-\omega$$.

Therefore, $$(1+\omega)^{2013}+(1+\omega^2)^{2013}=-2$$.

Many thanks!

- 5 years, 1 month ago

Hi, $$(1+\omega)^{2013}$$ is conjugate of $$(1 + \omega^2)^{2013}$$,

and $$z + \bar{z} = 2Re(z)$$, well known identity in complex nos.

- 5 years, 1 month ago

Thanks Jatin! :)

- 5 years, 1 month ago

Whoops, my bad. I guess it does.

- 5 years, 1 month ago

1

- 5 years, 1 month ago

abc = 1 ?

- 5 years, 1 month ago


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# Find the median for each list of numbers.$100,114,125,135,150,172$

## 130

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### Video Transcript

this question gives us a list of numbers and asks us to find the median. The data that it gives is 100 114 125 135 150 and 172. Now, we know when we have a data set of in, um, points that the median will be at the point and plus 1/2. So here we have six data points to an equal six. Which means our data, our median will be at 0.6 plus 1/2. Just 7/2, which is 3.5. Well, how do we have a data point at 3.5? How does that make sense? Well, when you have something like this, that your median is right in the middle of your data, Uh, with no point attached to it, this will happen when you have an even number of data points. What you do is you take point the point right in front of it. So three in the point right after it for and you average them so well, uh, add them together 1 25 plus 1 35 and divide by two. That will give us the median equal to 130. And that's your final answer.


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# A forecast system uses the model { 1 , 1 , V t , W t } as a base model for forecasting sales of a

1. A forecast system uses the model { 1 , 1 , V t , W t } as a base model for

forecasting sales of a group of products. Intervention information is available to incorporate into the model analysis. Initially, ( µ 0 | D 0 ) N[400 , 20]. Produce forecast distributions p ( Y t | D 0 ,I ) for each of t =

1 , 2 , 3 and 4 given the following intervention information I available

at t = 0 :

1. V 1 = 100 and ω 1 N[100 , W 1 ] with W 1 = 400 . This non-zero

mean evolution error describes the likely impact on sales level

of the receipt of a new export licence in the ﬁrst time period. Verify that ( Y 1 | D 0 ,I ) N[500 , 520] .

1. ν 2 N[500 , V 2 ] with V 2 = 200 describing the anticipated eﬀect

of an additional spot order at t = 2 , and W 2 = 5 . Verify that ( Y 2 | D 0 ,I ) N[1000 , 625] .

1. The expected eﬀect of a price increase at t = 3 is modelled via a transfer response function, with

Y t = µ t + E t + ν t , ( t = 3 , 4 ,... ) ,

where V t = 100, W t = 5 and

E t = 0 . 8 E t 1 , ( t = 4 , 5 ,... ) ,

with

( E 3 | D 0 ,I ) N[ 40 , 200] ,

independently of ( µ 0 | D 0 ).

Verify that ( Y 3 | D 0 ,I ) N[460 , 730] .

1. At t = 4, there is a prior probability of 0.4 that a further export licence will be granted. If so, then ω 4 N[200 , 600], otherwise ω 4 N[0 , 5] as usual. Show that p ( Y 4 | D 0 ,I ) is a mixture of

two normals,

( Y 4 | D 0 ,I )

( N[668 , 1258] , with probability 0 . 4;

N[468 , 663] , with probability 0 . 6 .

Plot the corresponding forecast density over the range 400 Y 4 750 . Calculate the forecast mean E[ Y 4 | D 0 ,I ] and vari- ance V[ Y 4 | D 0 ,I ]. From (c), calculate the joint distribution of ( Y 3 , E 3 | D 0 ,I ) and deduce the posterior ( E 3 | Y 3 , D 0 ,I ). Verify that ( E 3 | Y 3 = 387 , D 0 ,I ) N[ 60 , 145] .

1. Calculate the joint posterior p ( µ 4 , E 4 | Y 4 , D 0 ,I ) as a function of

Y 4 , and deduce the marginal posterior for µ 4 .

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## File: wise3arch.tex

package info (click to toggle)
wise 2.4.1-21
 123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263646566676869707172737475767778798081828384858687888990919293949596979899100101102103104105106107108109110111112113114115116117118119120121122123124125126127128129130131132133134135136137138139140141142143144145146147148149150151152153154155156157158159160161162163164165166167168169 \documentstyle{article} \begin{document} \newcommand{\programtext}[1]{{\tt #1}} \title{Wise3 Open Architecture} \author{Ewan Birney, Guy Slater Sanger Centre\\ Wellcome Trust Genome Campus\\ Hinxton, Cambridge CB10 1SA,\\ England.\\ Email: birney@sanger.ac.uk,gslater@hgmp.mrc.ac.uk} \maketitle \newpage \tableofcontents \newpage \section{Introduction} The aim of this paper is to lay out some architecture goals for the next generation of the Wise package, Wise3. In addition we would like to lay out some of the changes to the software of Wise. The architecture is designed to be open and provide additional code to work seemlessly with the Wise package. There are two main groups of people this open architecture is aimed at. FIXME: the document should be more focused at general wise3 issues. \begin{itemize} \item Large sites with databases which are not kept in simple fasta databases. \item Hardware manufacturers, including specialised hardware, who would like to improve on the speed or sensitivity aspects of the database search. \end{itemize} The aim is also to encourage people to work on a consistent framework for using genewise and genewise type algorithms sensibly by encouraging people to conform to standards whenever it is sensible. The main goal is to prevent the annoying habit of hardware manufactures being asked to implement genewise'' without a clear definition of what that means, and in addition, being forced not merely to implement genewise, but also the entire supporting framework, such as alignments and post processing. This is a huge additional strain on the hardware manufactures which does not help anyone. For consumers of hardware or database systems, this provides a single document which you can point to to indicate what you want from the system. It should clear up a considerable amount of confusion for providing compliant systems. \subsection{Committment to open source, freely available code} The Wise2 package has been licensed under the GNU General Public License since its inception. In addition, parts of the package has even less restrictive Licenses. I have a strong committment to keep Wise a freely available, open source package. The aim of the open architecture is to allow Wise to be compiled with additional extensions provided by third party sources: As the users will be compiling in the additional extensions, this will not require the 3rd parties to License their code under GPL. They will be free to license their code under any license they see fit, including keep their source code closed and charging for it. \subsection{Potential conflict of interests} I have been a consultant to Compugen (a company which builds specialised hardware) and am currently a consultant to Paracel (again, a specialised hardware company) and I have been a consultant to a number of bioinformatics and pharmaceutical companies worldwide. I do \emph{not believe} that my involvement with these companies prejuidices the open source nature of the Wise package, nor has this architecture been written to favour one particular company over another. I'd like to point out both that my committment to the scientific endeavour involved in the Wise package is far greater than any committment to any company, and that if I wanted to make money out of the Wise package in a serious manner I would have gone private \emph{myself} some time ago. I am sympathetic to people who are concerned about any potential conflict of interests that I have: I would welcome people to email their concerns and we can discuss it. I am very happy to allow the involvement of independent researchers who can verify the open nature Wise - all suggestions welcome. \subsection{A collaborative approach} This document is the first stab at a definition of an open architecture. I would very much welcome feedback from both manufactures, corporate users and academic users as to what would help them make better use of Wise in a large scale software environment. Please feel free to make your own suggestions and corrections. \section{Overview of the architecture} The architecture will define 3 main interfaces \begin{itemize} \item A C based interface of opening, iterating over and closing a database of sequences. \item A CORBA based interface of opening, iterating over and closing a database of sequences: this reuses the BioSource idl from bioperl for greater code reuse between packages. \item A C based interface of running a database search of a particular algorithm type, of a single query structure against a database of sequences as give by the above, C interface \end{itemize} In addition there needs to be additional rules for propagating command line arguments into the intialisation of the database and search routines, and also conventions for how to find and link against libaries containing this information, regardless of where they came from. Although at first the expected mode of action will be that the Wise source code will be compiled and then linked to the additional functionalities provided as C libraries, eventually dynamic loading routines might be considered. It was also tempting to focus on a CORBA only based interface of the code: however, even with free, lightweight effective ORBs such as ORBit, the technology is both an overkill for what is effectively a series of very simple interfaces and ties the portability of the code to the portability of the ORBs. In addition, people may be concerned that performance issues would come into play, even though ORBit will do direct function calls for in process object requests. A CORBA defintion has been provided for the database layer, but will always sit behind the pure C interface (ORBit will be our ORB of choice to do the mapping). The C interfaces will be designed with the following features in mind \begin{itemize} \item All the definitions will focus on passing opaque pointers to functions \item When functions return status values, they will return an integer, with 0 being success and non zero indicating an error of some sort \end{itemize} \section{Database Access, C definition} \begin{verbatim} typedef struct WOpenSeq WOpenSeq; typedef struct WOpenSeqStream WOpenSeqStream; typedef struct WOpenSeqDB WOpenSeqDB; typedef int WOpenStatus; char * WOpenSeq_seq(WOpenSeq * seq); char * WOpenSeq_subseq(WOpenSeq * seq,long start,long end); long WOpenSeq_length(WOpenSeq * seq); char * WOpenSeq_desc(WOpenSeq * seq); char * WOpenSeq_identifier(WOpenSeq * seq); WOpenStatus WOpenSeqStream_open(WOpenSeqStream * stream); \end{verbatim} \section{Error Libraries} The idea is that we ditch the error handling libraries of Wise2 (the /base stuff) and switch over to using glib routines \end{document}