Using a matroid to model emotional temperament.
$begingroup$
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
668617
$endgroup$
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
668617
$endgroup$
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
$begingroup$
@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 |
$begingroup$
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
668617
$endgroup$
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
668617
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
668617
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
668617
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
668617
asked Dec 7 '18 at 3:45
Tomislav OstojichTomislav Ostojich
668617
668617
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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029437%2fusing-a-matroid-to-model-emotional-temperament%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3029437%2fusing-a-matroid-to-model-emotional-temperament%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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