What is the partition of the Set S induced by R in this example?

up vote
0
down vote

favorite

A company makes charm bracelets with 4 charms: A, B, C, and D. There are 24 different orders they can enter the machine:

ABCD ABDC ACBD ACDB ADBC ADCB

BCAD BCDA BDAC BDCA CABD CADB

CDAB CDBA DABC DACB DBAC DBCA

BACD BADC CBAD CBDA DCAB DCBA

Once the charms are threaded on a bracelet many of the arrangements look the same. If the relation R on the set S of line arrangements with the rule (X,Y) ∈ R if arrangements X and Y look the same when threaded on a bracelet, what is the partition of the set S induced by R?

Would this be right?

The partition of the set S induced by R is
{{ABCD, BCDA, CDAB, DABC}, {ABCD, BDCA, DCAB, CABD}, {ACBD, CDBA, DBAC, BACD},{ADBC, DBCA, BCAD, CADB}, {ADCB, DCBA, CBAD, BADC}}

asked Nov 19 at 18:16

happysaint

add a comment |

up vote
0
down vote

favorite

A company makes charm bracelets with 4 charms: A, B, C, and D. There are 24 different orders they can enter the machine:

ABCD ABDC ACBD ACDB ADBC ADCB

BCAD BCDA BDAC BDCA CABD CADB

CDAB CDBA DABC DACB DBAC DBCA

BACD BADC CBAD CBDA DCAB DCBA

Would this be right?

The partition of the set S induced by R is
{{ABCD, BCDA, CDAB, DABC}, {ABCD, BDCA, DCAB, CABD}, {ACBD, CDBA, DBAC, BACD},{ADBC, DBCA, BCAD, CADB}, {ADCB, DCBA, CBAD, BADC}}

asked Nov 19 at 18:16

happysaint

add a comment |

up vote
0
down vote

favorite

A company makes charm bracelets with 4 charms: A, B, C, and D. There are 24 different orders they can enter the machine:

ABCD ABDC ACBD ACDB ADBC ADCB

BCAD BCDA BDAC BDCA CABD CADB

CDAB CDBA DABC DACB DBAC DBCA

BACD BADC CBAD CBDA DCAB DCBA

Would this be right?

The partition of the set S induced by R is
{{ABCD, BCDA, CDAB, DABC}, {ABCD, BDCA, DCAB, CABD}, {ACBD, CDBA, DBAC, BACD},{ADBC, DBCA, BCAD, CADB}, {ADCB, DCBA, CBAD, BADC}}

asked Nov 19 at 18:16

happysaint

A company makes charm bracelets with 4 charms: A, B, C, and D. There are 24 different orders they can enter the machine:

ABCD ABDC ACBD ACDB ADBC ADCB

BCAD BCDA BDAC BDCA CABD CADB

CDAB CDBA DABC DACB DBAC DBCA

BACD BADC CBAD CBDA DCAB DCBA

Would this be right?

The partition of the set S induced by R is
{{ABCD, BCDA, CDAB, DABC}, {ABCD, BDCA, DCAB, CABD}, {ACBD, CDBA, DBAC, BACD},{ADBC, DBCA, BCAD, CADB}, {ADCB, DCBA, CBAD, BADC}}

discrete-mathematics

asked Nov 19 at 18:16

happysaint

asked Nov 19 at 18:16

happysaint

asked Nov 19 at 18:16

happysaint

asked Nov 19 at 18:16

happysaint

asked Nov 19 at 18:16

happysaint

add a comment |

1 Answer
1

active

oldest

votes

up vote
0
down vote

Almost. Note that you can flip around a bracelet.

... or would a bracelet and its flipped counterpart not 'look the same'? Then again, if I rotate a bracelet, one could argue that that does not look the same either ... Hmmm ... maybe you shpould look for some clarification on what 'look the same' exactly means ...

(BTW: you're missing $4$ orders... Also, first entry of second set should be ABDC, and first entry of third set ACDB...)

edited Nov 19 at 18:33

answered Nov 19 at 18:26

Bram28

58.7k44185

"look the same" means the same is an equivalence relation and the charms of one are in the reverse order of the charms of the other, wouldn't there only be 5 subsets with 20 members?
– happysaint
Nov 19 at 18:51

@happysaint OK, so then you should get 3 groups of 8 ... (e.g. your first group of 4 is the flipped versions of the last group of 4)
– Bram28
Nov 19 at 18:55

Could it be 6 groups of 4? So: {{ABCD, DABC, CDAB, BCDA}, {ABDC, CABD, DCAB, BDCA}, for 2 groups and {ACBD, DACB, BDAC, CBDA}, {ACDB, BACD, DBAC, CDBA}, {ADBC, CADB, BCAD, DBCA}, {ADCB, BADC, CBAD, DCBA}}
– happysaint
Nov 19 at 19:50

If flipping a bracelet around means it still 'looks the same', then ABCD is the same as BCDA,CDAB, DABC, DCBA, CBAD, BADC, and ADCB. So, three group of 8. if flipping around means it no longer looks the same, you get 6 groups of 4.
– Bram28
Nov 19 at 20:24

I guessed what "look the same means" in my comment above and am confused now what it actually means. Wouldn't shifting letters to the next place, as with 6 groups of 4 'look the same'? For example from the first group of 4, ABCD is related to DABC, DABC is related to CDAB and CDAB is related to BCDA, they are reflexive, symmetric, and transitive as an equivalence relation and would all 'look similar' since it's shifting letters to the next place to get similar looking arrangements?
– happysaint
Nov 19 at 20:32

|
show 1 more comment

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',
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
});

}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3005300%2fwhat-is-the-partition-of-the-set-s-induced-by-r-in-this-example%23new-answer', 'question_page');
}
);

Post as a guest

Name

Required, but never shown

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

up vote
0
down vote

Almost. Note that you can flip around a bracelet.

(BTW: you're missing $4$ orders... Also, first entry of second set should be ABDC, and first entry of third set ACDB...)

edited Nov 19 at 18:33

answered Nov 19 at 18:26

Bram28

58.7k44185

"look the same" means the same is an equivalence relation and the charms of one are in the reverse order of the charms of the other, wouldn't there only be 5 subsets with 20 members?
– happysaint
Nov 19 at 18:51

@happysaint OK, so then you should get 3 groups of 8 ... (e.g. your first group of 4 is the flipped versions of the last group of 4)
– Bram28
Nov 19 at 18:55

Could it be 6 groups of 4? So: {{ABCD, DABC, CDAB, BCDA}, {ABDC, CABD, DCAB, BDCA}, for 2 groups and {ACBD, DACB, BDAC, CBDA}, {ACDB, BACD, DBAC, CDBA}, {ADBC, CADB, BCAD, DBCA}, {ADCB, BADC, CBAD, DCBA}}
– happysaint
Nov 19 at 19:50

If flipping a bracelet around means it still 'looks the same', then ABCD is the same as BCDA,CDAB, DABC, DCBA, CBAD, BADC, and ADCB. So, three group of 8. if flipping around means it no longer looks the same, you get 6 groups of 4.
– Bram28
Nov 19 at 20:24

I guessed what "look the same means" in my comment above and am confused now what it actually means. Wouldn't shifting letters to the next place, as with 6 groups of 4 'look the same'? For example from the first group of 4, ABCD is related to DABC, DABC is related to CDAB and CDAB is related to BCDA, they are reflexive, symmetric, and transitive as an equivalence relation and would all 'look similar' since it's shifting letters to the next place to get similar looking arrangements?
– happysaint
Nov 19 at 20:32

|
show 1 more comment

up vote
0
down vote

Almost. Note that you can flip around a bracelet.

(BTW: you're missing $4$ orders... Also, first entry of second set should be ABDC, and first entry of third set ACDB...)

edited Nov 19 at 18:33

answered Nov 19 at 18:26

Bram28

58.7k44185

"look the same" means the same is an equivalence relation and the charms of one are in the reverse order of the charms of the other, wouldn't there only be 5 subsets with 20 members?
– happysaint
Nov 19 at 18:51

@happysaint OK, so then you should get 3 groups of 8 ... (e.g. your first group of 4 is the flipped versions of the last group of 4)
– Bram28
Nov 19 at 18:55

Could it be 6 groups of 4? So: {{ABCD, DABC, CDAB, BCDA}, {ABDC, CABD, DCAB, BDCA}, for 2 groups and {ACBD, DACB, BDAC, CBDA}, {ACDB, BACD, DBAC, CDBA}, {ADBC, CADB, BCAD, DBCA}, {ADCB, BADC, CBAD, DCBA}}
– happysaint
Nov 19 at 19:50

If flipping a bracelet around means it still 'looks the same', then ABCD is the same as BCDA,CDAB, DABC, DCBA, CBAD, BADC, and ADCB. So, three group of 8. if flipping around means it no longer looks the same, you get 6 groups of 4.
– Bram28
Nov 19 at 20:24

I guessed what "look the same means" in my comment above and am confused now what it actually means. Wouldn't shifting letters to the next place, as with 6 groups of 4 'look the same'? For example from the first group of 4, ABCD is related to DABC, DABC is related to CDAB and CDAB is related to BCDA, they are reflexive, symmetric, and transitive as an equivalence relation and would all 'look similar' since it's shifting letters to the next place to get similar looking arrangements?
– happysaint
Nov 19 at 20:32

|
show 1 more comment

up vote
0
down vote

Almost. Note that you can flip around a bracelet.

(BTW: you're missing $4$ orders... Also, first entry of second set should be ABDC, and first entry of third set ACDB...)

edited Nov 19 at 18:33

answered Nov 19 at 18:26

Bram28

58.7k44185

Almost. Note that you can flip around a bracelet.

(BTW: you're missing $4$ orders... Also, first entry of second set should be ABDC, and first entry of third set ACDB...)

edited Nov 19 at 18:33

answered Nov 19 at 18:26

Bram28

58.7k44185

edited Nov 19 at 18:33

answered Nov 19 at 18:26

Bram28

58.7k44185

answered Nov 19 at 18:26

Bram28

58.7k44185

answered Nov 19 at 18:26

Bram28

58.7k44185

"look the same" means the same is an equivalence relation and the charms of one are in the reverse order of the charms of the other, wouldn't there only be 5 subsets with 20 members?
– happysaint
Nov 19 at 18:51

@happysaint OK, so then you should get 3 groups of 8 ... (e.g. your first group of 4 is the flipped versions of the last group of 4)
– Bram28
Nov 19 at 18:55

Could it be 6 groups of 4? So: {{ABCD, DABC, CDAB, BCDA}, {ABDC, CABD, DCAB, BDCA}, for 2 groups and {ACBD, DACB, BDAC, CBDA}, {ACDB, BACD, DBAC, CDBA}, {ADBC, CADB, BCAD, DBCA}, {ADCB, BADC, CBAD, DCBA}}
– happysaint
Nov 19 at 19:50

If flipping a bracelet around means it still 'looks the same', then ABCD is the same as BCDA,CDAB, DABC, DCBA, CBAD, BADC, and ADCB. So, three group of 8. if flipping around means it no longer looks the same, you get 6 groups of 4.
– Bram28
Nov 19 at 20:24

I guessed what "look the same means" in my comment above and am confused now what it actually means. Wouldn't shifting letters to the next place, as with 6 groups of 4 'look the same'? For example from the first group of 4, ABCD is related to DABC, DABC is related to CDAB and CDAB is related to BCDA, they are reflexive, symmetric, and transitive as an equivalence relation and would all 'look similar' since it's shifting letters to the next place to get similar looking arrangements?
– happysaint
Nov 19 at 20:32

|
show 1 more comment

"look the same" means the same is an equivalence relation and the charms of one are in the reverse order of the charms of the other, wouldn't there only be 5 subsets with 20 members?
– happysaint
Nov 19 at 18:51

@happysaint OK, so then you should get 3 groups of 8 ... (e.g. your first group of 4 is the flipped versions of the last group of 4)
– Bram28
Nov 19 at 18:55

Could it be 6 groups of 4? So: {{ABCD, DABC, CDAB, BCDA}, {ABDC, CABD, DCAB, BDCA}, for 2 groups and {ACBD, DACB, BDAC, CBDA}, {ACDB, BACD, DBAC, CDBA}, {ADBC, CADB, BCAD, DBCA}, {ADCB, BADC, CBAD, DCBA}}
– happysaint
Nov 19 at 19:50

If flipping a bracelet around means it still 'looks the same', then ABCD is the same as BCDA,CDAB, DABC, DCBA, CBAD, BADC, and ADCB. So, three group of 8. if flipping around means it no longer looks the same, you get 6 groups of 4.
– Bram28
Nov 19 at 20:24

I guessed what "look the same means" in my comment above and am confused now what it actually means. Wouldn't shifting letters to the next place, as with 6 groups of 4 'look the same'? For example from the first group of 4, ABCD is related to DABC, DABC is related to CDAB and CDAB is related to BCDA, they are reflexive, symmetric, and transitive as an equivalence relation and would all 'look similar' since it's shifting letters to the next place to get similar looking arrangements?
– happysaint
Nov 19 at 20:32

"look the same" means the same is an equivalence relation and the charms of one are in the reverse order of the charms of the other, wouldn't there only be 5 subsets with 20 members?
– happysaint
Nov 19 at 18:51

@happysaint OK, so then you should get 3 groups of 8 ... (e.g. your first group of 4 is the flipped versions of the last group of 4)
– Bram28
Nov 19 at 18:55

Could it be 6 groups of 4? So: {{ABCD, DABC, CDAB, BCDA}, {ABDC, CABD, DCAB, BDCA}, for 2 groups and {ACBD, DACB, BDAC, CBDA}, {ACDB, BACD, DBAC, CDBA}, {ADBC, CADB, BCAD, DBCA}, {ADCB, BADC, CBAD, DCBA}}
– happysaint
Nov 19 at 19:50

If flipping a bracelet around means it still 'looks the same', then ABCD is the same as BCDA,CDAB, DABC, DCBA, CBAD, BADC, and ADCB. So, three group of 8. if flipping around means it no longer looks the same, you get 6 groups of 4.
– Bram28
Nov 19 at 20:24

I guessed what "look the same means" in my comment above and am confused now what it actually means. Wouldn't shifting letters to the next place, as with 6 groups of 4 'look the same'? For example from the first group of 4, ABCD is related to DABC, DABC is related to CDAB and CDAB is related to BCDA, they are reflexive, symmetric, and transitive as an equivalence relation and would all 'look similar' since it's shifting letters to the next place to get similar looking arrangements?
– happysaint
Nov 19 at 20:32

|
show 1 more comment

draft saved

draft discarded

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.

Some of your past answers have not been well-received, and you're in danger of being blocked from answering.

Please pay close attention to the following guidance:

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.

To learn more, see our tips on writing great answers.

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Post as a guest

Name

Required, but never shown

Name

Required, but never shown

Name

Required, but never shown

This page is only for reference, If you need detailed information, please check here

搜尋此網誌

Krdytkyu