Calculate the characters from a combination












1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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


















1












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    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
















1












1








1





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 0:10







Edenia

















asked Dec 10 '18 at 14:00









EdeniaEdenia

1085




1085








  • 1




    $begingroup$
    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
















  • 1




    $begingroup$
    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










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












2 Answers
2






active

oldest

votes


















1












$begingroup$

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






share|cite|improve this answer











$endgroup$









  • 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










  • $begingroup$
    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










  • $begingroup$
    @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



















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

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






share|cite|improve this answer











$endgroup$









  • 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










  • $begingroup$
    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










  • $begingroup$
    @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












$begingroup$

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






share|cite|improve this answer











$endgroup$









  • 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










  • $begingroup$
    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










  • $begingroup$
    @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








1





$begingroup$

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






share|cite|improve this answer











$endgroup$



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







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 11 '18 at 4:13

























answered Dec 10 '18 at 14:16









gt6989bgt6989b

34.2k22455




34.2k22455








  • 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










  • $begingroup$
    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










  • $begingroup$
    @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




    $begingroup$
    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










  • $begingroup$
    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










  • $begingroup$
    @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











2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















2












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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














2












2








2





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 10 '18 at 14:08









Shubham JohriShubham Johri

5,177717




5,177717












  • $begingroup$
    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












  • $begingroup$
    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


















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Plaza Victoria

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Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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