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How many ambiguous combinations?
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$begingroup$
I am not a mathematician so please forgive my ignorance.
I am examining a body of data in which animals are identified by marking their toes. These animals have four feet, with differing number of markable toes per foot depending on morphology.
Each mark is ascribed an integer value for its position (i.e., the first toe counts X, the second toe counts Y, the third toe counts Z, and so on) such that the sum of the values of all marked toes on an animal is the identity number of the individual.
Various numbering schemes are used. In the case I am currently looking at the values given to the four sets (one set per foot) of marks are: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000.
I believe I can calculate the number of possible combinations of marks $binomAB$ where A is the number of toes per foot that could be marked, and B is the number of toes per foot that are marked.
So, for example, marking up to two (0, 1 or 2) of five toes per foot produces $(binom50 + binom51 + binom52)^4 -1 = 65,534$ possible combinations of marks (with the $-1$ eliminating the completely unmarked combination).
However, the numbering scheme shown above (and other schemes) can produce ambiguous identity numbers where more than one combination of marks can produce the same identity number. I.e., (trivially) the marks 4+7 and marks 1+10 both sum to 11.
I have two questions:
How can I calculate the number of ambiguous combinations of marks for a given $binomAB$ and numbering scheme?
Without resorting to exhaustive iteration, is it possible to determine whether an arbitrary identity number (sum of marks) is ambiguous or unique?
Any guidance much appreciated!
p.s. I tagged this question with "combinatronics"; please advise if other tags would be more appropriate.
combinatorics
asked yesterday
TodeTode
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 6 more comments
$begingroup$
I am not a mathematician so please forgive my ignorance.
I am examining a body of data in which animals are identified by marking their toes. These animals have four feet, with differing number of markable toes per foot depending on morphology.
Each mark is ascribed an integer value for its position (i.e., the first toe counts X, the second toe counts Y, the third toe counts Z, and so on) such that the sum of the values of all marked toes on an animal is the identity number of the individual.
Various numbering schemes are used. In the case I am currently looking at the values given to the four sets (one set per foot) of marks are: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000.
I believe I can calculate the number of possible combinations of marks $binomAB$ where A is the number of toes per foot that could be marked, and B is the number of toes per foot that are marked.
So, for example, marking up to two (0, 1 or 2) of five toes per foot produces $(binom50 + binom51 + binom52)^4 -1 = 65,534$ possible combinations of marks (with the $-1$ eliminating the completely unmarked combination).
However, the numbering scheme shown above (and other schemes) can produce ambiguous identity numbers where more than one combination of marks can produce the same identity number. I.e., (trivially) the marks 4+7 and marks 1+10 both sum to 11.
I have two questions:
How can I calculate the number of ambiguous combinations of marks for a given $binomAB$ and numbering scheme?
Without resorting to exhaustive iteration, is it possible to determine whether an arbitrary identity number (sum of marks) is ambiguous or unique?
Any guidance much appreciated!
p.s. I tagged this question with "combinatronics"; please advise if other tags would be more appropriate.
combinatorics
asked yesterday
TodeTode
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Why not just do an exhaustive iteration? There is only about a million possibilities afterall.
$endgroup$
– Jens
yesterday
$begingroup$
Thanks @Jens. Yes, I can do that, but as I mentioned I have described just one case. I ultimately need to consider various $binomAB$ and various numbering schemes. So I could make these inputs into a software algorithm, but a straight forward mathematical solution would likely be a more elegant.
$endgroup$
– Tode
yesterday
$begingroup$
Given the screw... uh...."unconventional" nature of these ID schemes, I seriously doubt any elegant mathematical solution exists.
$endgroup$
– Jens
yesterday
$begingroup$
I agree the numbering scheme is "unconventional". So what if we arrange the existing four physical sets: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000 into five logical sets: 1, 2, 4, 7, 10, 20, 40, 70, 100, 200, 400, 700, 1000, 2000, 4000, 7000, and 10000, 20000, 40000, 70000? With this arrangement ambiguity exists for any combination of marks within a set whose value is $>=$ an element of the next set.
$endgroup$
– Tode
yesterday
$begingroup$
Following my own logic, there are no combinations of $binom40$ or $binom41$ that satisfy the condition for ambiguity. There is one ambiguous combination of $binom42$ which is $4+7 >= 10$, and there are three ambiguous combinations of $binom43$ which are $1+2+7 >= 10$, $1+4+7 >= 10$, and $2+4+7 >= 10$. I can subtract these from the numbers of unambiguous combinations to derive the total number of ambiguous combinations.
$endgroup$
– Tode
yesterday
|
show 6 more comments
$begingroup$
I am not a mathematician so please forgive my ignorance.
I am examining a body of data in which animals are identified by marking their toes. These animals have four feet, with differing number of markable toes per foot depending on morphology.
Each mark is ascribed an integer value for its position (i.e., the first toe counts X, the second toe counts Y, the third toe counts Z, and so on) such that the sum of the values of all marked toes on an animal is the identity number of the individual.
Various numbering schemes are used. In the case I am currently looking at the values given to the four sets (one set per foot) of marks are: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000.
I believe I can calculate the number of possible combinations of marks $binomAB$ where A is the number of toes per foot that could be marked, and B is the number of toes per foot that are marked.
So, for example, marking up to two (0, 1 or 2) of five toes per foot produces $(binom50 + binom51 + binom52)^4 -1 = 65,534$ possible combinations of marks (with the $-1$ eliminating the completely unmarked combination).
However, the numbering scheme shown above (and other schemes) can produce ambiguous identity numbers where more than one combination of marks can produce the same identity number. I.e., (trivially) the marks 4+7 and marks 1+10 both sum to 11.
I have two questions:
How can I calculate the number of ambiguous combinations of marks for a given $binomAB$ and numbering scheme?
Without resorting to exhaustive iteration, is it possible to determine whether an arbitrary identity number (sum of marks) is ambiguous or unique?
Any guidance much appreciated!
p.s. I tagged this question with "combinatronics"; please advise if other tags would be more appropriate.
combinatorics
asked yesterday
TodeTode
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am not a mathematician so please forgive my ignorance.
I am examining a body of data in which animals are identified by marking their toes. These animals have four feet, with differing number of markable toes per foot depending on morphology.
Each mark is ascribed an integer value for its position (i.e., the first toe counts X, the second toe counts Y, the third toe counts Z, and so on) such that the sum of the values of all marked toes on an animal is the identity number of the individual.
Various numbering schemes are used. In the case I am currently looking at the values given to the four sets (one set per foot) of marks are: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000.
I believe I can calculate the number of possible combinations of marks $binomAB$ where A is the number of toes per foot that could be marked, and B is the number of toes per foot that are marked.
So, for example, marking up to two (0, 1 or 2) of five toes per foot produces $(binom50 + binom51 + binom52)^4 -1 = 65,534$ possible combinations of marks (with the $-1$ eliminating the completely unmarked combination).
However, the numbering scheme shown above (and other schemes) can produce ambiguous identity numbers where more than one combination of marks can produce the same identity number. I.e., (trivially) the marks 4+7 and marks 1+10 both sum to 11.
I have two questions:
How can I calculate the number of ambiguous combinations of marks for a given $binomAB$ and numbering scheme?
Without resorting to exhaustive iteration, is it possible to determine whether an arbitrary identity number (sum of marks) is ambiguous or unique?
Any guidance much appreciated!
p.s. I tagged this question with "combinatronics"; please advise if other tags would be more appropriate.
combinatorics
combinatorics
asked yesterday
TodeTode
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
TodeTode
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
TodeTode
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
TodeTode
1
asked yesterday
TodeTode
1
1
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Tode is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Why not just do an exhaustive iteration? There is only about a million possibilities afterall.
$endgroup$
– Jens
yesterday
$begingroup$
Thanks @Jens. Yes, I can do that, but as I mentioned I have described just one case. I ultimately need to consider various $binomAB$ and various numbering schemes. So I could make these inputs into a software algorithm, but a straight forward mathematical solution would likely be a more elegant.
$endgroup$
– Tode
yesterday
$begingroup$
Given the screw... uh...."unconventional" nature of these ID schemes, I seriously doubt any elegant mathematical solution exists.
$endgroup$
– Jens
yesterday
$begingroup$
I agree the numbering scheme is "unconventional". So what if we arrange the existing four physical sets: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000 into five logical sets: 1, 2, 4, 7, 10, 20, 40, 70, 100, 200, 400, 700, 1000, 2000, 4000, 7000, and 10000, 20000, 40000, 70000? With this arrangement ambiguity exists for any combination of marks within a set whose value is $>=$ an element of the next set.
$endgroup$
– Tode
yesterday
$begingroup$
Following my own logic, there are no combinations of $binom40$ or $binom41$ that satisfy the condition for ambiguity. There is one ambiguous combination of $binom42$ which is $4+7 >= 10$, and there are three ambiguous combinations of $binom43$ which are $1+2+7 >= 10$, $1+4+7 >= 10$, and $2+4+7 >= 10$. I can subtract these from the numbers of unambiguous combinations to derive the total number of ambiguous combinations.
$endgroup$
– Tode
yesterday
|
show 6 more comments
$begingroup$
Why not just do an exhaustive iteration? There is only about a million possibilities afterall.
$endgroup$
– Jens
yesterday
$begingroup$
Thanks @Jens. Yes, I can do that, but as I mentioned I have described just one case. I ultimately need to consider various $binomAB$ and various numbering schemes. So I could make these inputs into a software algorithm, but a straight forward mathematical solution would likely be a more elegant.
$endgroup$
– Tode
yesterday
$begingroup$
Given the screw... uh...."unconventional" nature of these ID schemes, I seriously doubt any elegant mathematical solution exists.
$endgroup$
– Jens
yesterday
$begingroup$
I agree the numbering scheme is "unconventional". So what if we arrange the existing four physical sets: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000 into five logical sets: 1, 2, 4, 7, 10, 20, 40, 70, 100, 200, 400, 700, 1000, 2000, 4000, 7000, and 10000, 20000, 40000, 70000? With this arrangement ambiguity exists for any combination of marks within a set whose value is $>=$ an element of the next set.
$endgroup$
– Tode
yesterday
$begingroup$
Following my own logic, there are no combinations of $binom40$ or $binom41$ that satisfy the condition for ambiguity. There is one ambiguous combination of $binom42$ which is $4+7 >= 10$, and there are three ambiguous combinations of $binom43$ which are $1+2+7 >= 10$, $1+4+7 >= 10$, and $2+4+7 >= 10$. I can subtract these from the numbers of unambiguous combinations to derive the total number of ambiguous combinations.
$endgroup$
– Tode
yesterday
$begingroup$
Why not just do an exhaustive iteration? There is only about a million possibilities afterall.
$endgroup$
– Jens
yesterday
$begingroup$
Why not just do an exhaustive iteration? There is only about a million possibilities afterall.
$endgroup$
– Jens
yesterday
$begingroup$
Thanks @Jens. Yes, I can do that, but as I mentioned I have described just one case. I ultimately need to consider various $binomAB$ and various numbering schemes. So I could make these inputs into a software algorithm, but a straight forward mathematical solution would likely be a more elegant.
$endgroup$
– Tode
yesterday
$begingroup$
Thanks @Jens. Yes, I can do that, but as I mentioned I have described just one case. I ultimately need to consider various $binomAB$ and various numbering schemes. So I could make these inputs into a software algorithm, but a straight forward mathematical solution would likely be a more elegant.
$endgroup$
– Tode
yesterday
$begingroup$
Given the screw... uh...."unconventional" nature of these ID schemes, I seriously doubt any elegant mathematical solution exists.
$endgroup$
– Jens
yesterday
$begingroup$
Given the screw... uh...."unconventional" nature of these ID schemes, I seriously doubt any elegant mathematical solution exists.
$endgroup$
– Jens
yesterday
$begingroup$
I agree the numbering scheme is "unconventional". So what if we arrange the existing four physical sets: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000 into five logical sets: 1, 2, 4, 7, 10, 20, 40, 70, 100, 200, 400, 700, 1000, 2000, 4000, 7000, and 10000, 20000, 40000, 70000? With this arrangement ambiguity exists for any combination of marks within a set whose value is $>=$ an element of the next set.
$endgroup$
– Tode
yesterday
$begingroup$
I agree the numbering scheme is "unconventional". So what if we arrange the existing four physical sets: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000 into five logical sets: 1, 2, 4, 7, 10, 20, 40, 70, 100, 200, 400, 700, 1000, 2000, 4000, 7000, and 10000, 20000, 40000, 70000? With this arrangement ambiguity exists for any combination of marks within a set whose value is $>=$ an element of the next set.
$endgroup$
– Tode
yesterday
$begingroup$
Following my own logic, there are no combinations of $binom40$ or $binom41$ that satisfy the condition for ambiguity. There is one ambiguous combination of $binom42$ which is $4+7 >= 10$, and there are three ambiguous combinations of $binom43$ which are $1+2+7 >= 10$, $1+4+7 >= 10$, and $2+4+7 >= 10$. I can subtract these from the numbers of unambiguous combinations to derive the total number of ambiguous combinations.
$endgroup$
– Tode
yesterday
$begingroup$
Following my own logic, there are no combinations of $binom40$ or $binom41$ that satisfy the condition for ambiguity. There is one ambiguous combination of $binom42$ which is $4+7 >= 10$, and there are three ambiguous combinations of $binom43$ which are $1+2+7 >= 10$, $1+4+7 >= 10$, and $2+4+7 >= 10$. I can subtract these from the numbers of unambiguous combinations to derive the total number of ambiguous combinations.
$endgroup$
– Tode
yesterday
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Per your request, here is the program I made. It was done in Visual Basic for Excel. Below is a screenshot of Excel after the macro executed:
For the program to work, it is necessary that the "toe values" in row 3 are entered into Excel beforehand, exactly as shown. The X's below the toe values show which toes were chosen for each sum. To count the number of sums created or the number of ambiguous sums, insert a filter at row 3. The screenshot can be enlarged by clicking on it.
Below are screenshots showing the program:
The program takes about 20 seconds to run.
answered 14 mins ago
JensJens
3,90021031
$endgroup$
add a comment |
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$begingroup$
Per your request, here is the program I made. It was done in Visual Basic for Excel. Below is a screenshot of Excel after the macro executed:
For the program to work, it is necessary that the "toe values" in row 3 are entered into Excel beforehand, exactly as shown. The X's below the toe values show which toes were chosen for each sum. To count the number of sums created or the number of ambiguous sums, insert a filter at row 3. The screenshot can be enlarged by clicking on it.
Below are screenshots showing the program:
The program takes about 20 seconds to run.
answered 14 mins ago
JensJens
3,90021031
$endgroup$
add a comment |
$begingroup$
Per your request, here is the program I made. It was done in Visual Basic for Excel. Below is a screenshot of Excel after the macro executed:
For the program to work, it is necessary that the "toe values" in row 3 are entered into Excel beforehand, exactly as shown. The X's below the toe values show which toes were chosen for each sum. To count the number of sums created or the number of ambiguous sums, insert a filter at row 3. The screenshot can be enlarged by clicking on it.
Below are screenshots showing the program:
The program takes about 20 seconds to run.
answered 14 mins ago
JensJens
3,90021031
$endgroup$
add a comment |
$begingroup$
Per your request, here is the program I made. It was done in Visual Basic for Excel. Below is a screenshot of Excel after the macro executed:
For the program to work, it is necessary that the "toe values" in row 3 are entered into Excel beforehand, exactly as shown. The X's below the toe values show which toes were chosen for each sum. To count the number of sums created or the number of ambiguous sums, insert a filter at row 3. The screenshot can be enlarged by clicking on it.
Below are screenshots showing the program:
The program takes about 20 seconds to run.
answered 14 mins ago
JensJens
3,90021031
$endgroup$
Per your request, here is the program I made. It was done in Visual Basic for Excel. Below is a screenshot of Excel after the macro executed:
For the program to work, it is necessary that the "toe values" in row 3 are entered into Excel beforehand, exactly as shown. The X's below the toe values show which toes were chosen for each sum. To count the number of sums created or the number of ambiguous sums, insert a filter at row 3. The screenshot can be enlarged by clicking on it.
Below are screenshots showing the program:
The program takes about 20 seconds to run.
answered 14 mins ago
JensJens
3,90021031
answered 14 mins ago
JensJens
3,90021031
answered 14 mins ago
JensJens
3,90021031
answered 14 mins ago
JensJens
3,90021031
3,90021031
add a comment |
add a comment |
Tode is a new contributor. Be nice, and check out our Code of Conduct.
Tode is a new contributor. Be nice, and check out our Code of Conduct.
Tode is a new contributor. Be nice, and check out our Code of Conduct.
Tode is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Why not just do an exhaustive iteration? There is only about a million possibilities afterall.
$endgroup$
– Jens
yesterday
$begingroup$
Thanks @Jens. Yes, I can do that, but as I mentioned I have described just one case. I ultimately need to consider various $binomAB$ and various numbering schemes. So I could make these inputs into a software algorithm, but a straight forward mathematical solution would likely be a more elegant.
$endgroup$
– Tode
yesterday
$begingroup$
Given the screw... uh...."unconventional" nature of these ID schemes, I seriously doubt any elegant mathematical solution exists.
$endgroup$
– Jens
yesterday
$begingroup$
I agree the numbering scheme is "unconventional". So what if we arrange the existing four physical sets: 70000, 1, 2, 4, 7, 40000, 10, 20, 40, 70, 20000, 100, 200, 400, 700 and 10000, 1000, 2000, 4000, 7000 into five logical sets: 1, 2, 4, 7, 10, 20, 40, 70, 100, 200, 400, 700, 1000, 2000, 4000, 7000, and 10000, 20000, 40000, 70000? With this arrangement ambiguity exists for any combination of marks within a set whose value is $>=$ an element of the next set.
$endgroup$
– Tode
yesterday
$begingroup$
Following my own logic, there are no combinations of $binom40$ or $binom41$ that satisfy the condition for ambiguity. There is one ambiguous combination of $binom42$ which is $4+7 >= 10$, and there are three ambiguous combinations of $binom43$ which are $1+2+7 >= 10$, $1+4+7 >= 10$, and $2+4+7 >= 10$. I can subtract these from the numbers of unambiguous combinations to derive the total number of ambiguous combinations.
$endgroup$
– Tode
yesterday