Using a matroid to model emotional temperament.
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An idiot crank devised a model of emotional temperament using a bitmask, $XYZ$. $X$ represent the bit that means "able to handle extreme negative emotions." $Y$ represents the bit that means "able to handle neutral emotions" (e.g. boredom, interest, contentment). $Z$ represents the bit that means "able to handle extreme positive emotions." So in this model a human could have the following eight emotional temperaments.
$000$: can't handle any emotions.
$001$: can only handle extreme happiness (manic temperament).
$010$: can only handle neutral emotions.
$011$: can handle both neutral emotions and extreme happiness, but not negative emotions.
$100$: can handle only extreme negative emotions (depressive temperament).
$101$: can handle either extreme negative emotions or extreme positive emotions, but not neutral ones (bipolar temperament).
$110$: can handle both extreme negative emotions and neutral emotions, but not extreme happiness.
$111$: can handle the entire emotional spectrum.
For the transition ${000} to {100, 010, 001}$ this model behaves like a matroid because you only need to learn one more temperament to advance to the next level. For the transition ${100, 010, 001} to {110, 101, 011
}$ this model still behaves like a matroid because you only need to learn one more temperament to advance to the next level. But for the transition ${100, 010, 001} to {111}$ this model does not behave like a matroid, because, for example, a person who is $011$ temperament would need to learn three temperaments ($100$, $110$, $101$) to advance to the next level. Is there a way to fix this model so that it is turned into a matroid?
matroids
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
$endgroup$
add a comment |
$begingroup$
An idiot crank devised a model of emotional temperament using a bitmask, $XYZ$. $X$ represent the bit that means "able to handle extreme negative emotions." $Y$ represents the bit that means "able to handle neutral emotions" (e.g. boredom, interest, contentment). $Z$ represents the bit that means "able to handle extreme positive emotions." So in this model a human could have the following eight emotional temperaments.
$000$: can't handle any emotions.
$001$: can only handle extreme happiness (manic temperament).
$010$: can only handle neutral emotions.
$011$: can handle both neutral emotions and extreme happiness, but not negative emotions.
$100$: can handle only extreme negative emotions (depressive temperament).
$101$: can handle either extreme negative emotions or extreme positive emotions, but not neutral ones (bipolar temperament).
$110$: can handle both extreme negative emotions and neutral emotions, but not extreme happiness.
$111$: can handle the entire emotional spectrum.
For the transition ${000} to {100, 010, 001}$ this model behaves like a matroid because you only need to learn one more temperament to advance to the next level. For the transition ${100, 010, 001} to {110, 101, 011
}$ this model still behaves like a matroid because you only need to learn one more temperament to advance to the next level. But for the transition ${100, 010, 001} to {111}$ this model does not behave like a matroid, because, for example, a person who is $011$ temperament would need to learn three temperaments ($100$, $110$, $101$) to advance to the next level. Is there a way to fix this model so that it is turned into a matroid?
matroids
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
$endgroup$
2
$begingroup$
What do you mean by "behaves like a matroid"? Also, who is the crank? Can we get a reference?
$endgroup$
– Aaron Dall
Dec 9 '18 at 21:20
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@AaronDall if you really insist on a reference: 72archetypes.com/wiki/table_formats You should know that he reverses it. XYZ in my post corresponds to {-3, -2} {-1, 0, 1} {2, 3} while his goes {3, 2} {1, 0, -1} {-2, -3}. Mine corresponds to the number line. That's why it is the way it is: so the isomorphism to the number line is preserved.
$endgroup$
– Tomislav Ostojich
Dec 9 '18 at 23:36
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@AaronDall my intuitive understanding of a matroid is that it's a structure on which "greedy" algorithms work. You can add bit by bit and get the optimal situation.
$endgroup$
– Tomislav Ostojich
Dec 10 '18 at 0:47
$begingroup$
Let's put intuition aside and get concrete. A matroid is a pair $(E, I)$ where $E$ is a finite set (the ground set) and $I$ is a collection of subsets of $E$ (the independent sets of the matroid) such that for any weight function $w$ on $E$ the greedy algorithm selects a cardinality maximal element of $I$ that is also $w$-maximal. What is the ground set you're considering? Is it $E = {1,2,3}$? What are the independent sets you are considering? Is it all subsets of $E$?
$endgroup$
– Aaron Dall
Dec 11 '18 at 8:52
$begingroup$
@AaronDall Yes and yes.
$endgroup$
– Tomislav Ostojich
Dec 11 '18 at 18:25
add a comment |
$begingroup$
An idiot crank devised a model of emotional temperament using a bitmask, $XYZ$. $X$ represent the bit that means "able to handle extreme negative emotions." $Y$ represents the bit that means "able to handle neutral emotions" (e.g. boredom, interest, contentment). $Z$ represents the bit that means "able to handle extreme positive emotions." So in this model a human could have the following eight emotional temperaments.
$000$: can't handle any emotions.
$001$: can only handle extreme happiness (manic temperament).
$010$: can only handle neutral emotions.
$011$: can handle both neutral emotions and extreme happiness, but not negative emotions.
$100$: can handle only extreme negative emotions (depressive temperament).
$101$: can handle either extreme negative emotions or extreme positive emotions, but not neutral ones (bipolar temperament).
$110$: can handle both extreme negative emotions and neutral emotions, but not extreme happiness.
$111$: can handle the entire emotional spectrum.
For the transition ${000} to {100, 010, 001}$ this model behaves like a matroid because you only need to learn one more temperament to advance to the next level. For the transition ${100, 010, 001} to {110, 101, 011
}$ this model still behaves like a matroid because you only need to learn one more temperament to advance to the next level. But for the transition ${100, 010, 001} to {111}$ this model does not behave like a matroid, because, for example, a person who is $011$ temperament would need to learn three temperaments ($100$, $110$, $101$) to advance to the next level. Is there a way to fix this model so that it is turned into a matroid?
matroids
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
$endgroup$
An idiot crank devised a model of emotional temperament using a bitmask, $XYZ$. $X$ represent the bit that means "able to handle extreme negative emotions." $Y$ represents the bit that means "able to handle neutral emotions" (e.g. boredom, interest, contentment). $Z$ represents the bit that means "able to handle extreme positive emotions." So in this model a human could have the following eight emotional temperaments.
$000$: can't handle any emotions.
$001$: can only handle extreme happiness (manic temperament).
$010$: can only handle neutral emotions.
$011$: can handle both neutral emotions and extreme happiness, but not negative emotions.
$100$: can handle only extreme negative emotions (depressive temperament).
$101$: can handle either extreme negative emotions or extreme positive emotions, but not neutral ones (bipolar temperament).
$110$: can handle both extreme negative emotions and neutral emotions, but not extreme happiness.
$111$: can handle the entire emotional spectrum.
For the transition ${000} to {100, 010, 001}$ this model behaves like a matroid because you only need to learn one more temperament to advance to the next level. For the transition ${100, 010, 001} to {110, 101, 011
}$ this model still behaves like a matroid because you only need to learn one more temperament to advance to the next level. But for the transition ${100, 010, 001} to {111}$ this model does not behave like a matroid, because, for example, a person who is $011$ temperament would need to learn three temperaments ($100$, $110$, $101$) to advance to the next level. Is there a way to fix this model so that it is turned into a matroid?
matroids
matroids
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
614414
614414
2
$begingroup$
What do you mean by "behaves like a matroid"? Also, who is the crank? Can we get a reference?
$endgroup$
– Aaron Dall
Dec 9 '18 at 21:20
$begingroup$
@AaronDall if you really insist on a reference: 72archetypes.com/wiki/table_formats You should know that he reverses it. XYZ in my post corresponds to {-3, -2} {-1, 0, 1} {2, 3} while his goes {3, 2} {1, 0, -1} {-2, -3}. Mine corresponds to the number line. That's why it is the way it is: so the isomorphism to the number line is preserved.
$endgroup$
– Tomislav Ostojich
Dec 9 '18 at 23:36
$begingroup$
@AaronDall my intuitive understanding of a matroid is that it's a structure on which "greedy" algorithms work. You can add bit by bit and get the optimal situation.
$endgroup$
– Tomislav Ostojich
Dec 10 '18 at 0:47
$begingroup$
Let's put intuition aside and get concrete. A matroid is a pair $(E, I)$ where $E$ is a finite set (the ground set) and $I$ is a collection of subsets of $E$ (the independent sets of the matroid) such that for any weight function $w$ on $E$ the greedy algorithm selects a cardinality maximal element of $I$ that is also $w$-maximal. What is the ground set you're considering? Is it $E = {1,2,3}$? What are the independent sets you are considering? Is it all subsets of $E$?
$endgroup$
– Aaron Dall
Dec 11 '18 at 8:52
$begingroup$
@AaronDall Yes and yes.
$endgroup$
– Tomislav Ostojich
Dec 11 '18 at 18:25
add a comment |
2
$begingroup$
What do you mean by "behaves like a matroid"? Also, who is the crank? Can we get a reference?
$endgroup$
– Aaron Dall
Dec 9 '18 at 21:20
$begingroup$
@AaronDall if you really insist on a reference: 72archetypes.com/wiki/table_formats You should know that he reverses it. XYZ in my post corresponds to {-3, -2} {-1, 0, 1} {2, 3} while his goes {3, 2} {1, 0, -1} {-2, -3}. Mine corresponds to the number line. That's why it is the way it is: so the isomorphism to the number line is preserved.
$endgroup$
– Tomislav Ostojich
Dec 9 '18 at 23:36
$begingroup$
@AaronDall my intuitive understanding of a matroid is that it's a structure on which "greedy" algorithms work. You can add bit by bit and get the optimal situation.
$endgroup$
– Tomislav Ostojich
Dec 10 '18 at 0:47
$begingroup$
Let's put intuition aside and get concrete. A matroid is a pair $(E, I)$ where $E$ is a finite set (the ground set) and $I$ is a collection of subsets of $E$ (the independent sets of the matroid) such that for any weight function $w$ on $E$ the greedy algorithm selects a cardinality maximal element of $I$ that is also $w$-maximal. What is the ground set you're considering? Is it $E = {1,2,3}$? What are the independent sets you are considering? Is it all subsets of $E$?
$endgroup$
– Aaron Dall
Dec 11 '18 at 8:52
$begingroup$
@AaronDall Yes and yes.
$endgroup$
– Tomislav Ostojich
Dec 11 '18 at 18:25
2
2
$begingroup$
What do you mean by "behaves like a matroid"? Also, who is the crank? Can we get a reference?
$endgroup$
– Aaron Dall
Dec 9 '18 at 21:20
$begingroup$
What do you mean by "behaves like a matroid"? Also, who is the crank? Can we get a reference?
$endgroup$
– Aaron Dall
Dec 9 '18 at 21:20
$begingroup$
@AaronDall if you really insist on a reference: 72archetypes.com/wiki/table_formats You should know that he reverses it. XYZ in my post corresponds to {-3, -2} {-1, 0, 1} {2, 3} while his goes {3, 2} {1, 0, -1} {-2, -3}. Mine corresponds to the number line. That's why it is the way it is: so the isomorphism to the number line is preserved.
$endgroup$
– Tomislav Ostojich
Dec 9 '18 at 23:36
$begingroup$
@AaronDall if you really insist on a reference: 72archetypes.com/wiki/table_formats You should know that he reverses it. XYZ in my post corresponds to {-3, -2} {-1, 0, 1} {2, 3} while his goes {3, 2} {1, 0, -1} {-2, -3}. Mine corresponds to the number line. That's why it is the way it is: so the isomorphism to the number line is preserved.
$endgroup$
– Tomislav Ostojich
Dec 9 '18 at 23:36
$begingroup$
@AaronDall my intuitive understanding of a matroid is that it's a structure on which "greedy" algorithms work. You can add bit by bit and get the optimal situation.
$endgroup$
– Tomislav Ostojich
Dec 10 '18 at 0:47
$begingroup$
@AaronDall my intuitive understanding of a matroid is that it's a structure on which "greedy" algorithms work. You can add bit by bit and get the optimal situation.
$endgroup$
– Tomislav Ostojich
Dec 10 '18 at 0:47
$begingroup$
Let's put intuition aside and get concrete. A matroid is a pair $(E, I)$ where $E$ is a finite set (the ground set) and $I$ is a collection of subsets of $E$ (the independent sets of the matroid) such that for any weight function $w$ on $E$ the greedy algorithm selects a cardinality maximal element of $I$ that is also $w$-maximal. What is the ground set you're considering? Is it $E = {1,2,3}$? What are the independent sets you are considering? Is it all subsets of $E$?
$endgroup$
– Aaron Dall
Dec 11 '18 at 8:52
$begingroup$
Let's put intuition aside and get concrete. A matroid is a pair $(E, I)$ where $E$ is a finite set (the ground set) and $I$ is a collection of subsets of $E$ (the independent sets of the matroid) such that for any weight function $w$ on $E$ the greedy algorithm selects a cardinality maximal element of $I$ that is also $w$-maximal. What is the ground set you're considering? Is it $E = {1,2,3}$? What are the independent sets you are considering? Is it all subsets of $E$?
$endgroup$
– Aaron Dall
Dec 11 '18 at 8:52
$begingroup$
@AaronDall Yes and yes.
$endgroup$
– Tomislav Ostojich
Dec 11 '18 at 18:25
$begingroup$
@AaronDall Yes and yes.
$endgroup$
– Tomislav Ostojich
Dec 11 '18 at 18:25
add a comment |
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$begingroup$
What do you mean by "behaves like a matroid"? Also, who is the crank? Can we get a reference?
$endgroup$
– Aaron Dall
Dec 9 '18 at 21:20
$begingroup$
@AaronDall if you really insist on a reference: 72archetypes.com/wiki/table_formats You should know that he reverses it. XYZ in my post corresponds to {-3, -2} {-1, 0, 1} {2, 3} while his goes {3, 2} {1, 0, -1} {-2, -3}. Mine corresponds to the number line. That's why it is the way it is: so the isomorphism to the number line is preserved.
$endgroup$
– Tomislav Ostojich
Dec 9 '18 at 23:36
$begingroup$
@AaronDall my intuitive understanding of a matroid is that it's a structure on which "greedy" algorithms work. You can add bit by bit and get the optimal situation.
$endgroup$
– Tomislav Ostojich
Dec 10 '18 at 0:47
$begingroup$
Let's put intuition aside and get concrete. A matroid is a pair $(E, I)$ where $E$ is a finite set (the ground set) and $I$ is a collection of subsets of $E$ (the independent sets of the matroid) such that for any weight function $w$ on $E$ the greedy algorithm selects a cardinality maximal element of $I$ that is also $w$-maximal. What is the ground set you're considering? Is it $E = {1,2,3}$? What are the independent sets you are considering? Is it all subsets of $E$?
$endgroup$
– Aaron Dall
Dec 11 '18 at 8:52
$begingroup$
@AaronDall Yes and yes.
$endgroup$
– Tomislav Ostojich
Dec 11 '18 at 18:25