Calculate the characters from a combination
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
Edenia
1085
add a comment |
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
Edenia
1085
1
shouldbabe $27$ if you are starting from $1$ and includingaa?
– gt6989b
Dec 10 '18 at 14:03
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
– Edenia
Dec 10 '18 at 14:05
add a comment |
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
Edenia
1085
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
Edenia
1085
asked Dec 10 '18 at 14:00
Edenia
1085
edited Dec 11 '18 at 0:10
asked Dec 10 '18 at 14:00
Edenia
1085
asked Dec 10 '18 at 14:00
Edenia
1085
asked Dec 10 '18 at 14:00
Edenia
1085
1085
1
shouldbabe $27$ if you are starting from $1$ and includingaa?
– gt6989b
Dec 10 '18 at 14:03
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
– Edenia
Dec 10 '18 at 14:05
add a comment |
1
shouldbabe $27$ if you are starting from $1$ and includingaa?
– gt6989b
Dec 10 '18 at 14:03
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
– Edenia
Dec 10 '18 at 14:05
1
1
should
ba be $27$ if you are starting from $1$ and including aa?– gt6989b
Dec 10 '18 at 14:03
should
ba be $27$ if you are starting from $1$ and including aa?– gt6989b
Dec 10 '18 at 14:03
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
– Edenia
Dec 10 '18 at 14:05
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
– Edenia
Dec 10 '18 at 14:05
add a comment |
2 Answers
2
active
oldest
votes
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
gt6989b
33k22452
1
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
– Edenia
Dec 10 '18 at 14:32
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
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 Johri
3,943716
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.
– Edenia
Dec 10 '18 at 14:12
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
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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
gt6989b
33k22452
1
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
– Edenia
Dec 10 '18 at 14:32
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
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
gt6989b
33k22452
1
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
– Edenia
Dec 10 '18 at 14:32
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
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
gt6989b
33k22452
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
gt6989b
33k22452
edited Dec 11 '18 at 4:13
answered Dec 10 '18 at 14:16
gt6989b
33k22452
answered Dec 10 '18 at 14:16
gt6989b
33k22452
answered Dec 10 '18 at 14:16
gt6989b
33k22452
33k22452
1
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
– Edenia
Dec 10 '18 at 14:32
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
1
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
HmmA[pos % 26]doesn't return the correct first character.Afirst[(pos / sz) - 1]andAsecond[pos % sz]works though
– Edenia
Dec 10 '18 at 14:32
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
1
1
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
Haha, I am implementing algorithms in C, bt dubs.
– Edenia
Dec 10 '18 at 14:17
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
@Edenia see updates
– gt6989b
Dec 10 '18 at 14:20
Hmm
A[pos % 26] doesn't return the correct first character. Afirst[(pos / sz) - 1] and Asecond[pos % sz] works though– Edenia
Dec 10 '18 at 14:32
Hmm
A[pos % 26] doesn't return the correct first character. Afirst[(pos / sz) - 1] and Asecond[pos % sz] works though– Edenia
Dec 10 '18 at 14:32
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
sz being the 26.
– Edenia
Dec 10 '18 at 14:38
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
@gt6989b I think you wrote the reverse: the first letter is given by $A[lfloor k/26rfloor]$
– Shubham Johri
Dec 10 '18 at 14:46
|
show 7 more comments
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 Johri
3,943716
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.
– Edenia
Dec 10 '18 at 14:12
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
add a comment |
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 Johri
3,943716
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.
– Edenia
Dec 10 '18 at 14:12
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
add a comment |
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 Johri
3,943716
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 Johri
3,943716
answered Dec 10 '18 at 14:08
Shubham Johri
3,943716
answered Dec 10 '18 at 14:08
Shubham Johri
3,943716
answered Dec 10 '18 at 14:08
Shubham Johri
3,943716
3,943716
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.
– Edenia
Dec 10 '18 at 14:12
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
add a comment |
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.
– Edenia
Dec 10 '18 at 14:12
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
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.
– Edenia
Dec 10 '18 at 14:12
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.
– Edenia
Dec 10 '18 at 14:12
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
It's even nicer with 0-based numbering (and doesn't require the unusual interpretation of $26 bmod 26$ as 26).
– LSpice
Dec 10 '18 at 14:13
add a comment |
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should
babe $27$ if you are starting from $1$ and includingaa?– gt6989b
Dec 10 '18 at 14:03
@gt6989b Oh, I missed one character when counting. (Missed to write it). Sorry.
– Edenia
Dec 10 '18 at 14:05