comp: 39
mask:32
assim:28
Total:99
This is a bad test. It doesnt test for autism, it tests for how much someone tries to hide it. I doubt tryp is autistic but he scored higher than me, and I am autistic
What about people that are openly autistic and doesnt do anything to compensate? Would that mean they get a 0?
If so this test sucks. Its only for self diagnosed autists and not real ones. It only tests how aware you are of your "autism"
My title was misleading.
The test is designed for testing specifically how much you may be hiding traits that can be considered autistic. Therefore, someone with autism could score low if they don't mask a lot or have recently learned not too -- like in your case.
There are other more intensive tests on the website that may pick it up even with someone who does not use masking, although no tests on the website are stated to be for the use of self-diagnosis.
im bored
I can see many scenarios of someone scoring high on this test who is neurotypical because there are many circumstances for people to mask and assimilate.
Schizophrenics, high functioning ASPD & narcs, severely depressed, anyone coming from a different culture or considered an "outsider", and really anyone with severe mental health issues.
Although some questions seemed specifically geared towards autistic traits to me.
im bored
last edit on
5/21/2022 3:13:32 PM
This is my submission, strangely I did not receive a score...
Context
A very good friend wanted to explore an introductory treatment of linear algebra with some hints at theory, I thought it would be fun to work with him as a brush up and I knew I would also learn some new things. We chose Shilov and to our excitement he began with determinants immediately via a definition neither of us were accustomed to. However, something was recognizable about this definition as Shilov began to discuss inversions.....ah....group theory is being hidden under this proverbial presentation.
So.....instead of moving on like my friend, who is now at the end of the chapter, I decided to start from scratch and construct the definition from first principals. I have been neglecting all my other work over the past day to achieve this. Its not all for not, we are sharing notes and solutions and he will not only benefit from the presentation but it will too elucidate a number of ideas that are of some use in a proof to one of the problems at the end of the chapter.
Part 1
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Definition of Determinant (Shilov)
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For a square matrix A where aᵢⱼ i index rows while j index columns (i,j = 1,2,...,n) we define the determinant of A by det||aᵢⱼ|| = D = ∑ (-1)N⁽ᵃ⁽¹⁾ ᵃ⁽²⁾ ᵃ⁽ⁿ⁾⁾aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ
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1.3.1.1 Fundamentals of Permutations
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Permutations ::
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A permutation is a function α:{1,2,...,n} -> {1,2,...,n}
A permutation on some set {1,2,...,n} is an ordering of values from that set, that ordering can be a rearrangement.
A permutation, f : {1, 2, 3, 4, 5} -> {3 5 4 1 2}
Given a permutation is the output of a function α each permutation of n objects corresponds to some function, hence to derive the set of all permutations of n objects is to derive all functions α.
The number of permutations of n objects is always n!.
Ex. If {1, 2, 3} and the permutation is 2 3 1 then α(1) = 2, α(2) = 3 and α(3) = 1. α maps a position to a value from the set, hence α(1) maps the position first position in the order to the value 3 from the set {1,2,3}.
We can obtain additional permutations via the group operation composition
Consider f : {1,2,3,4,5} -> {3,5,4,1,2} and g : {1,2,3,4,5} -> {3,1,5,2,5}
Composition,
f(g) : {1, 2, 3, 4, 5} -> {f(g(1)),f(g(2)),f(g(3)),f(g(4)),f(g(5))} = {f(3),f(1),f(5),f(2),f(4)} = {4,3,2,5,1}
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Symmetric Groups ::
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A symmetric group is the set of all permutations of n objects.
The symmetric group is denoted as Sₙ where n is the number of objects in that group.
Given a set of n objects has n! permutations, the number of elements of Sₙ is n!.
The group operation of a symmetric group is composition.
Ex. S₄ is a set of 4!=24 permutations where its domain is {1,2,3,4}
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Two-Row Notation ::
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The functional form α :{1,2,...n} -> {1,2,...n} is known as One-Row notation.
Two row notation stacks the sets so that the mappings between individual values is more explicit.
Two-Row notation,
α:{1,2,...,n}
{1,2,...,n}
Ex. π : {1,2,3,4} -> {4,2,3,1}
π: {1,2,3,4}
{4,2,3,1}
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Cycle Notation ::
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A cycle is a kind of notation for permutations, it is a way of writing a permutation where the only information we care about are the mappings between values in α.
The rule behind the mappings is that there is a map between any numbers that reference each other in anyway.
Ex. Consider permutation {1,2,3,4,5}
{3,5,4,1,2},
α(1) = 3, 1 -> 3 ;; Now 3 becomes the position of domain,
α(3) = 4, 3 -> 4 ;; Now 4 becomes the position of domain,
α(4) = 1, 4 -> 1
Notice the rule is that any value pointed to becomes the position to find the next mapping, this continues until a value maps to a previously referenced value. This is considered all one large mapping in the following way,
1 -> 3 -> 4 -> 1
This set of mappings is written as a cycle (1 3 4), a cycle with 3 unique values is called a 3-cycle.
Two values have not been referenced yet, 2 and 5.
α(2) = 5 ;; Now 5 becomes the position of domain
α(5) = 2
2 -> 5 -> 2
(2 5) a cycle with 2 unique values is called a 2-cycle or transposition
All together, (1 3 4)(2 5)
We can start at different positions, the order of our cycles may change but the cycles themselves (3-cycle, 2-cycle, etc) always remain the same. Under this notation we only care about the mappings and how they relate to cycles.
Single cycles are called identity cycles.
We can convert cycle notation back to two line notation (or functional) by reversing the process and tracking the maps.
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Products of Cycles ::
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Finding the product of cycles allows us to deal with cycles that have overlap.
We read through our cycles from right to left matching values to our positions via tracking their mappings, the goal being reconstruct the correct permutation.
Ex. Consider the cycles, (3 4 1 5)(2 3 6 1)(3 1)
We note that we are trying to map the set {1,2,3,4,5,6} to the permutation we want to find.
α(1)
1 -> 3 -> 6 ;; α(1) = 6
α(2)
2 -> 3 -> 4 ;; α(2) = 4
α(3)
3 -> 1 -> 2 ;; α(3) = 2
α(4)
4 -> 1 ;; α(4) = 1
α(5)
5 -> 3 ;; α(5) = 3
α(6)
6 -> 1 -> 5 ;; α(6) = 5
α : {1,2,3,4,5,6} -> {6,4,2,1,3,5}
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Permutations as Products of Transpositions ::
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All permutations can be represented as a product of transpositions (2-cycles).
Every cycle can be expressed as a product of transpositions,
(a₁a₂a₃...aₙ₋₁aₙ) = (a₁aₙ)(a₁aₙ₋₁)(a₁aₙ₋₂)...(a₁a₃)(a₁a₂)
Ex. Consider, α : {1,2,3,4,5,6}
{6,4,2,1,3,5}
We convert to cycle notation,
(1 6 5 3 2 4)
Express as product of transpositions,
(1 4)(1 2)(1 3)(1 5)(1 6)
Ex. Consider π : {1,2,3,4,5}
{3,5,4,1,2},
Convert to cycle notation,
(1 3 4)(2 5)
Express as product of transpositions,
(1 4)(1 3)(2 5)
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Inversions ::
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An inversion takes place whenever α(x) > α(y) and x < y.Ex. Given the ordered sequence 2 3 1 we say that α(1) = 2, α(2) = 3, and α(3) = 1.
α(1) < α(2) and 1 < 2, hence by definition no inversion.
α(2) > α(3) and 2 < 3, hence by definition an inversion.
For each α(n) there is number of terms in a ordered sequence less than α(n), we call this value βₙ
Ex. Given an ordered sequence 3 1 2 where α(1) = 3, α(2) = 1, and α(3) = 2 we know α(1) > α(2) while 1 < 2 and
α(1) > α(3) while 1 < 3, hence β₁ = 2.
The number of inversions of a permutation α is defined as N(α(1), α(2), . . ., α(n)) = β₁ + β₂ + . . . + βₙ
Ex.Consider the ordered sequence 3 2 1 where α(1) = 3, α(2) = 2, and α(3) = 1.
α(1) > α(2) and 1 < 2 ; α(1) > α(3) and 1 < 3 ---> β₁ = 2 because two inversions take place in respect to position 1. α(2) > α(3) and 2 < 3 ----> β₂ = 1 because one inversion takes place after position 2.
α(3) has no values that follow it, so β₃ = 0
N(α(1), α(2), α(3)) = β₁ + β₂ + β₃ = 2 + 1 + 0 = 3
A parity is even when N(α(1),α(2),...,α(n)) is even and odd when N(α(1),α(2),...,α(n)) is odd.
Ex.The ordered sequence 3 2 1 has a odd parity because N(α(1), α(2), α(3)) = β₁ + β₂ + β₃ = 2 + 1 + 0 = 3
last edit on
5/26/2022 2:03:08 AM
Part 2
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Transposition and Parity Equivalence ::
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A permutation is even if and only if it can be written as an even number of transpositions and odd if and only if it can be written as an odd number of transpositions.
Hence, the parity of a permutation in terms of inversions is equivalent to product of transpositions.
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Permutations are Even or Odd ::
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All permutations are either even or odd.
An even permutation is one with a positive sign (+) and a negative permutation is one with a negative sign (-).
We denote the sign of a permutation α as sign(α)
sign(α) = +1 if the number of inversions of α is even.
sign(α) = -1 if the number of inversions of α is odd.
so, sign(α) = -1ᴺ⁽ᵃ⁽¹⁾...ᵃ⁽ⁿ⁾⁾
Given the parity of a permutation in terms of inversions is equivalent to the number of transpositions the notion of sign can be re defined as...
sign(α) = -1ᵗ where t is the transpositions in α.
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1.3.1.2 The Determinant
===========================================================
The definition of the Determinant is a specific application of theory of permutations and groups symmetry.
Now that we have all the pieces, lets state them formally and construct the determinant as presented by Shilov.
Let α ∈ Sₙ be a permutation of the symmetric group
Let there be inversions (i,j) of α where i,j ∈ {1,2,...,n}
For each i,j where i < j there is a transposition τᵢⱼ ∈ Sₙ
Let the permutation α be even or odd given τᵢⱼ by sign(α)=-1ᵀ
From a square matrix Aᵢⱼ with aᵢⱼ ∈ Aᵢⱼ where i,j ∈ {1,2,...,n} we can define a product containing just one element from each row and column as aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ where where α(n) corresponds to rows and n corresponds to columns.
Πaₐ₍ₙ₎ₙ
Definition of Determinant
----------------------------------
Given a square matrix Aᵢⱼ where i,j ∈ {1,2,...,n the determinant of Aᵢⱼ is
det||aᵢⱼ|| = ∑ sign(α) Πaₐ₍ₙ₎ₙ
last edit on
5/26/2022 2:03:26 AM
Part 2
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-----------------------------------------------------------
Transposition and Parity Equivalence ::
-----------------------------------------------------------A permutation is even if and only if it can be written as an even number of transpositions and odd if and only if it can be written as an odd number of transpositions.
Hence, the parity of a permutation in terms of inversions is equivalent to product of transpositions.
------------------------------------------------------------
Permutations are Even or Odd ::
------------------------------------------------------------All permutations are either even or odd.
An even permutation is one with a positive sign (+) and a negative permutation is one with a negative sign (-).
We denote the sign of a permutation α as sign(α)
sign(α) = +1 if the number of inversions of α is even.
sign(α) = -1 if the number of inversions of α is odd.so, sign(α) = -1ᴺ⁽ᵃ⁽¹⁾...ᵃ⁽ⁿ⁾⁾
Given the parity of a permutation in terms of inversions is equivalent to the number of transpositions the notion of sign can be re defined as...
sign(α) = -1ᵗ where t is the transpositions in α.
===========================================================
1.3.1.2 The Determinant
===========================================================The definition of the Determinant is a specific application of theory of permutations and groups symmetry.
Now that we have all the pieces, lets state them formally and construct the determinant as presented by Shilov.
Let α ∈ Sₙ be a permutation of the symmetric group
Let there be inversions (i,j) of α where i,j ∈ {1,2,...,n}
For each i,j where i < j there is a transposition τᵢⱼ ∈ Sₙ
Let the permutation α be even or odd given τᵢⱼ by sign(α)=-1ᵀ
From a square matrix Aᵢⱼ with aᵢⱼ ∈ Aᵢⱼ where i,j ∈ {1,2,...,n} we can define a product containing just one element from each row and column as aₐ₍₁₎₁aₐ₍₂₎₂ . . . aₐ₍ₙ₎ₙ where where α(n) corresponds to rows and n corresponds to columns.
Πaₐ₍ₙ₎ₙDefinition of Determinant
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Given a square matrix Aᵢⱼ where i,j ∈ {1,2,...,n the determinant of Aᵢⱼ is det||aᵢⱼ|| = ∑ sign(α) Πaₐ₍ₙ₎ₙ
sup Alice ?