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

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










  • $begingroup$
    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

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


















  • $begingroup$
    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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Berounka

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Berounka.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Caractéristiques Longueur 139  km Bassin 8 861  km 2 Bassin collecteur Elbe Débit moyen ? Cours Source Confluence de la Mže et de la Radbuza · Localisation Pilsen · Altitude 305  m · Coordonnées 49° 45′ 13″ N, 13° 23′ 24″ E Confluence Vltava · Localisation Prague · Altitude 187  m · Coordonnées 49° 59′ 43″ N, 14° 24′ 04″ E Géographie Pays traversés   République tchèque modifier   La Berounka est une rivière de la République tchèque. Elle porte le nom de Mže depuis sa source en Allemagne près de la frontière jusqu'à sa confluence avec la Radbuza à Plzeň. Elle continue ensuite sous le nom de Berounka jusqu'à sa confluence avec la Vltava, près de Prague. La rivière est une destination prisée pour les amateurs de canoë, qui appréc...
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Sphinx de Gizeh

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Pour les articles homonymes, voir Sphinx. Sphinx de Gizeh.mw-parser-output .entete.egypte-antique{background-image:url("//upload.wikimedia.org/wikipedia/commons/c/cc/Picto_infobox_ancient_Egypt.png")} Site d'Égypte antique Le visage du sphinx de Gizeh. Noms en arabe أبو الهول , Abou al-Hôl Localisation Région Basse-Égypte Coordonnées 29° 58′ 31″ nord, 31° 08′ 16″ est Géolocalisation sur la carte : Égypte Sphinx de Gizeh modifier   Le sphinx de Gizeh est la statue thérianthrope qui se dresse devant les grandes pyramides du plateau de Gizeh, en Basse-Égypte. Sculpture monumentale monolithique la plus grande du monde [ 1 ] avec 73,5 mètres de longueur, 14 mètres de largeur et 20,22 mètres de hauteur, elle représente un sphinx couchant. Réalisée vers -2500, elle serait attribuée à Djédefrê, l'un des pharaons de la IV e  dynastie, représentant son père le pharaon Khéops. Sommaire 1 Étymologie 2 Desc...
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Different font size/position of beamer's navigation symbols template's content depending on regular/plain...

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3 Let's suppose I'd like to divert the navigation symbols template (e.g. in order replace them by the frame numbers, see this bug report ;) I would be facing troubles regarding the font size/position, depending on regular/plain frames, as shown by the following MCE. How could I get rid of these troubles? documentclass{beamer} usetheme{PaloAlto} setbeamertemplate{navigation symbols}{insertframenumber/inserttotalframenumber} begin{document} begin{frame} frametitle{A regular frame} end{frame} begin{frame}[plain] frametitle{A plain frame} end{frame} end{document} beamer header-footer share | improve this question asked Dec 10 '18 a...
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