Calculate the characters from a combination
$begingroup$
Given a table with position - 2 character combination pairs like this:
1. aa
2. ab
3. ac
4. ad
...
27. ba
28. bb
29. bc
30. bd
Assuming there are unique combinations with two characters from the alphabet following the same pattern.
How do I calculate what is the first and second character of the combination, if I know the position. That is, how do I find position 28 holds bb?
EDIT:
The answer that also describes how to do the reverse, i.e find the position of a combination based on the pair is going to be the most useful answer.
arithmetic index-notation
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
$endgroup$
add a comment |
$begingroup$
Given a table with position - 2 character combination pairs like this:
1. aa
2. ab
3. ac
4. ad
...
27. ba
28. bb
29. bc
30. bd
Assuming there are unique combinations with two characters from the alphabet following the same pattern.
How do I calculate what is the first and second character of the combination, if I know the position. That is, how do I find position 28 holds bb?
EDIT:
The answer that also describes how to do the reverse, i.e find the position of a combination based on the pair is going to be the most useful answer.
arithmetic index-notation
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
$endgroup$
1
$begingroup$
shouldbabe $27$ if you are starting from $1$ and includingaa?
$endgroup$
– gt6989b
Dec 10 '18 at 14:03
$begingroup$
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
$endgroup$
– Edenia
Dec 10 '18 at 14:05
add a comment |
$begingroup$
Given a table with position - 2 character combination pairs like this:
1. aa
2. ab
3. ac
4. ad
...
27. ba
28. bb
29. bc
30. bd
Assuming there are unique combinations with two characters from the alphabet following the same pattern.
How do I calculate what is the first and second character of the combination, if I know the position. That is, how do I find position 28 holds bb?
EDIT:
The answer that also describes how to do the reverse, i.e find the position of a combination based on the pair is going to be the most useful answer.
arithmetic index-notation
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
$endgroup$
Given a table with position - 2 character combination pairs like this:
1. aa
2. ab
3. ac
4. ad
...
27. ba
28. bb
29. bc
30. bd
Assuming there are unique combinations with two characters from the alphabet following the same pattern.
How do I calculate what is the first and second character of the combination, if I know the position. That is, how do I find position 28 holds bb?
EDIT:
The answer that also describes how to do the reverse, i.e find the position of a combination based on the pair is going to be the most useful answer.
arithmetic index-notation
arithmetic index-notation
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
edited Dec 11 '18 at 0:10
Edenia
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
asked Dec 10 '18 at 14:00
EdeniaEdenia
1085
1085
1
$begingroup$
shouldbabe $27$ if you are starting from $1$ and includingaa?
$endgroup$
– gt6989b
Dec 10 '18 at 14:03
$begingroup$
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
$endgroup$
– Edenia
Dec 10 '18 at 14:05
add a comment |
1
$begingroup$
shouldbabe $27$ if you are starting from $1$ and includingaa?
$endgroup$
– gt6989b
Dec 10 '18 at 14:03
$begingroup$
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
$endgroup$
– Edenia
Dec 10 '18 at 14:05
1
1
$begingroup$
should
ba be $27$ if you are starting from $1$ and including aa?$endgroup$
– gt6989b
Dec 10 '18 at 14:03
$begingroup$
should
ba be $27$ if you are starting from $1$ and including aa?$endgroup$
– gt6989b
Dec 10 '18 at 14:03
$begingroup$
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
$endgroup$
– Edenia
Dec 10 '18 at 14:05
$begingroup$
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
$endgroup$
– Edenia
Dec 10 '18 at 14:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Assuming you have an alphabet A = ['a','b',...'z'] and a permutation list L = ['aa', 'ab', ..., 'az', 'ba', ... , 'zz'], with the first index starting from 1. (BTW the arithmetic would be much cleaner if you start indexing from $0$, like in the C language, for example.)
Then, the permutations come in groups of $26$, so if you are given an index $k$, then
- the second letter is given by the offset of $k$ in the current 26-long block, so it is $A[(k-1)pmod{26} + 1]$ and
- the first letter is given by $Aleft[lfloor(k-1)/26rfloor+1right]$.
If you were to number starting with $0$, the formulae become
- the second letter is given by $A[k pmod{26}]$ and
- the first letter is given by $Aleft[lfloor k/26rfloorright]$.
Another free update you get with C, for example, is that if you declare int k, d = 26; then k/d computes $lfloor k/d rfloor$ automatically, no extra flooring is needed.
UPDATE
The inversion is simple. If first letter has index $F$ and last has index $L$, the offset is given by
$26F + L$ for $0$-based indexing (like C)
$26(F+1) + (L+1) = 26F + L + 27$ for $1$-based indexing
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
$endgroup$
1
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
$begingroup$
For the index $n$, the $(lfloor frac{n-1}{26}rfloor+1)^{th}$ letter of the alphabet is the first character, $(nmod26)^{th}$ letter of the alphabet is the second character.
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
add a comment |
Your Answer
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2 Answers
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2 Answers
2
active
oldest
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oldest
votes
$begingroup$
Assuming you have an alphabet A = ['a','b',...'z'] and a permutation list L = ['aa', 'ab', ..., 'az', 'ba', ... , 'zz'], with the first index starting from 1. (BTW the arithmetic would be much cleaner if you start indexing from $0$, like in the C language, for example.)
Then, the permutations come in groups of $26$, so if you are given an index $k$, then
- the second letter is given by the offset of $k$ in the current 26-long block, so it is $A[(k-1)pmod{26} + 1]$ and
- the first letter is given by $Aleft[lfloor(k-1)/26rfloor+1right]$.
If you were to number starting with $0$, the formulae become
- the second letter is given by $A[k pmod{26}]$ and
- the first letter is given by $Aleft[lfloor k/26rfloorright]$.
Another free update you get with C, for example, is that if you declare int k, d = 26; then k/d computes $lfloor k/d rfloor$ automatically, no extra flooring is needed.
UPDATE
The inversion is simple. If first letter has index $F$ and last has index $L$, the offset is given by
$26F + L$ for $0$-based indexing (like C)
$26(F+1) + (L+1) = 26F + L + 27$ for $1$-based indexing
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
$endgroup$
1
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
$begingroup$
Assuming you have an alphabet A = ['a','b',...'z'] and a permutation list L = ['aa', 'ab', ..., 'az', 'ba', ... , 'zz'], with the first index starting from 1. (BTW the arithmetic would be much cleaner if you start indexing from $0$, like in the C language, for example.)
Then, the permutations come in groups of $26$, so if you are given an index $k$, then
- the second letter is given by the offset of $k$ in the current 26-long block, so it is $A[(k-1)pmod{26} + 1]$ and
- the first letter is given by $Aleft[lfloor(k-1)/26rfloor+1right]$.
If you were to number starting with $0$, the formulae become
- the second letter is given by $A[k pmod{26}]$ and
- the first letter is given by $Aleft[lfloor k/26rfloorright]$.
Another free update you get with C, for example, is that if you declare int k, d = 26; then k/d computes $lfloor k/d rfloor$ automatically, no extra flooring is needed.
UPDATE
The inversion is simple. If first letter has index $F$ and last has index $L$, the offset is given by
$26F + L$ for $0$-based indexing (like C)
$26(F+1) + (L+1) = 26F + L + 27$ for $1$-based indexing
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
$endgroup$
1
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
$begingroup$
Assuming you have an alphabet A = ['a','b',...'z'] and a permutation list L = ['aa', 'ab', ..., 'az', 'ba', ... , 'zz'], with the first index starting from 1. (BTW the arithmetic would be much cleaner if you start indexing from $0$, like in the C language, for example.)
Then, the permutations come in groups of $26$, so if you are given an index $k$, then
- the second letter is given by the offset of $k$ in the current 26-long block, so it is $A[(k-1)pmod{26} + 1]$ and
- the first letter is given by $Aleft[lfloor(k-1)/26rfloor+1right]$.
If you were to number starting with $0$, the formulae become
- the second letter is given by $A[k pmod{26}]$ and
- the first letter is given by $Aleft[lfloor k/26rfloorright]$.
Another free update you get with C, for example, is that if you declare int k, d = 26; then k/d computes $lfloor k/d rfloor$ automatically, no extra flooring is needed.
UPDATE
The inversion is simple. If first letter has index $F$ and last has index $L$, the offset is given by
$26F + L$ for $0$-based indexing (like C)
$26(F+1) + (L+1) = 26F + L + 27$ for $1$-based indexing
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
$endgroup$
Assuming you have an alphabet A = ['a','b',...'z'] and a permutation list L = ['aa', 'ab', ..., 'az', 'ba', ... , 'zz'], with the first index starting from 1. (BTW the arithmetic would be much cleaner if you start indexing from $0$, like in the C language, for example.)
Then, the permutations come in groups of $26$, so if you are given an index $k$, then
- the second letter is given by the offset of $k$ in the current 26-long block, so it is $A[(k-1)pmod{26} + 1]$ and
- the first letter is given by $Aleft[lfloor(k-1)/26rfloor+1right]$.
If you were to number starting with $0$, the formulae become
- the second letter is given by $A[k pmod{26}]$ and
- the first letter is given by $Aleft[lfloor k/26rfloorright]$.
Another free update you get with C, for example, is that if you declare int k, d = 26; then k/d computes $lfloor k/d rfloor$ automatically, no extra flooring is needed.
UPDATE
The inversion is simple. If first letter has index $F$ and last has index $L$, the offset is given by
$26F + L$ for $0$-based indexing (like C)
$26(F+1) + (L+1) = 26F + L + 27$ for $1$-based indexing
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
edited Dec 11 '18 at 4:13
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
answered Dec 10 '18 at 14:16
gt6989bgt6989b
33.9k22455
33.9k22455
1
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
1
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
1
1
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
Haha, I am implementing algorithms in C, bt dubs.
$endgroup$
– Edenia
Dec 10 '18 at 14:17
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
@Edenia see updates
$endgroup$
– gt6989b
Dec 10 '18 at 14:20
$begingroup$
Hmm
A[pos % 26] doesn't return the correct first character. Afirst[(pos / sz) - 1] and Asecond[pos % sz] works though$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
Hmm
A[pos % 26] doesn't return the correct first character. Afirst[(pos / sz) - 1] and Asecond[pos % sz] works though$endgroup$
– Edenia
Dec 10 '18 at 14:32
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
sz being the 26.
$endgroup$
– Edenia
Dec 10 '18 at 14:38
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
$begingroup$
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
$endgroup$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
$begingroup$
For the index $n$, the $(lfloor frac{n-1}{26}rfloor+1)^{th}$ letter of the alphabet is the first character, $(nmod26)^{th}$ letter of the alphabet is the second character.
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
add a comment |
$begingroup$
For the index $n$, the $(lfloor frac{n-1}{26}rfloor+1)^{th}$ letter of the alphabet is the first character, $(nmod26)^{th}$ letter of the alphabet is the second character.
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
add a comment |
$begingroup$
For the index $n$, the $(lfloor frac{n-1}{26}rfloor+1)^{th}$ letter of the alphabet is the first character, $(nmod26)^{th}$ letter of the alphabet is the second character.
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
$endgroup$
For the index $n$, the $(lfloor frac{n-1}{26}rfloor+1)^{th}$ letter of the alphabet is the first character, $(nmod26)^{th}$ letter of the alphabet is the second character.
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
answered Dec 10 '18 at 14:08
Shubham JohriShubham Johri
4,992717
4,992717
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
add a comment |
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
so for the second, index's reminder of the length of the alphabet. Got it. Really useful, thank you. What is this called though? Didn't know what to search for.
$endgroup$
– Edenia
Dec 10 '18 at 14:12
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
$begingroup$
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
$endgroup$
– LSpice
Dec 10 '18 at 14:13
add a comment |
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1
$begingroup$
should
babe $27$ if you are starting from $1$ and includingaa?$endgroup$
– gt6989b
Dec 10 '18 at 14:03
$begingroup$
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
$endgroup$
– Edenia
Dec 10 '18 at 14:05