How to solve a large system of linear algebra?












3












$begingroup$


A friend has given me the following puzzle to solve, however, I lack the linear algebra knowledge to calculate the solution, and my attempts to brute force the solution have been foiled by a large number of combinations.



The Problem:



Every letter in the alphabet is assigned a whole number from $1-26$. No two letters have the same number. Below is a list of $44$ words and the value of their letters added up. For example, if $O=11$, $H=23$, $I=2$, OHIO would equal $11+23+2+11 = 47$ (these values are not necessarily correct).



enter image description here



find the value of ALBUQUERQUE (added in the same manner).
Thanks for any solutions or ideas.










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  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    With a computer you could run Gaussian elimination, though it's not obvious to me that the solution over vectors of 26 real numbers is actually unique (because if any solution exists, then at least 18 of the equations are redundant, so what's to stop more than 18 of them from being redundant?)
    $endgroup$
    – Ian
    3 hours ago








  • 3




    $begingroup$
    Didn't noticed Albuquerque was so close to Jamestown.
    $endgroup$
    – zwim
    2 hours ago


















3












$begingroup$


A friend has given me the following puzzle to solve, however, I lack the linear algebra knowledge to calculate the solution, and my attempts to brute force the solution have been foiled by a large number of combinations.



The Problem:



Every letter in the alphabet is assigned a whole number from $1-26$. No two letters have the same number. Below is a list of $44$ words and the value of their letters added up. For example, if $O=11$, $H=23$, $I=2$, OHIO would equal $11+23+2+11 = 47$ (these values are not necessarily correct).



enter image description here



find the value of ALBUQUERQUE (added in the same manner).
Thanks for any solutions or ideas.










share|cite|improve this question









New contributor




Klaus234 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    With a computer you could run Gaussian elimination, though it's not obvious to me that the solution over vectors of 26 real numbers is actually unique (because if any solution exists, then at least 18 of the equations are redundant, so what's to stop more than 18 of them from being redundant?)
    $endgroup$
    – Ian
    3 hours ago








  • 3




    $begingroup$
    Didn't noticed Albuquerque was so close to Jamestown.
    $endgroup$
    – zwim
    2 hours ago
















3












3








3


2



$begingroup$


A friend has given me the following puzzle to solve, however, I lack the linear algebra knowledge to calculate the solution, and my attempts to brute force the solution have been foiled by a large number of combinations.



The Problem:



Every letter in the alphabet is assigned a whole number from $1-26$. No two letters have the same number. Below is a list of $44$ words and the value of their letters added up. For example, if $O=11$, $H=23$, $I=2$, OHIO would equal $11+23+2+11 = 47$ (these values are not necessarily correct).



enter image description here



find the value of ALBUQUERQUE (added in the same manner).
Thanks for any solutions or ideas.










share|cite|improve this question









New contributor




Klaus234 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




A friend has given me the following puzzle to solve, however, I lack the linear algebra knowledge to calculate the solution, and my attempts to brute force the solution have been foiled by a large number of combinations.



The Problem:



Every letter in the alphabet is assigned a whole number from $1-26$. No two letters have the same number. Below is a list of $44$ words and the value of their letters added up. For example, if $O=11$, $H=23$, $I=2$, OHIO would equal $11+23+2+11 = 47$ (these values are not necessarily correct).



enter image description here



find the value of ALBUQUERQUE (added in the same manner).
Thanks for any solutions or ideas.







linear-algebra systems-of-equations puzzle






share|cite|improve this question









New contributor




Klaus234 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|cite|improve this question









New contributor




Klaus234 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|cite|improve this question




share|cite|improve this question








edited 3 hours ago









dantopa

6,57942244




6,57942244






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asked 3 hours ago









Klaus234Klaus234

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Klaus234 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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Check out our Code of Conduct.












  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    With a computer you could run Gaussian elimination, though it's not obvious to me that the solution over vectors of 26 real numbers is actually unique (because if any solution exists, then at least 18 of the equations are redundant, so what's to stop more than 18 of them from being redundant?)
    $endgroup$
    – Ian
    3 hours ago








  • 3




    $begingroup$
    Didn't noticed Albuquerque was so close to Jamestown.
    $endgroup$
    – zwim
    2 hours ago




















  • $begingroup$
    Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
    $endgroup$
    – dantopa
    3 hours ago










  • $begingroup$
    With a computer you could run Gaussian elimination, though it's not obvious to me that the solution over vectors of 26 real numbers is actually unique (because if any solution exists, then at least 18 of the equations are redundant, so what's to stop more than 18 of them from being redundant?)
    $endgroup$
    – Ian
    3 hours ago








  • 3




    $begingroup$
    Didn't noticed Albuquerque was so close to Jamestown.
    $endgroup$
    – zwim
    2 hours ago


















$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
3 hours ago




$begingroup$
Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. For equations, please use MathJax.
$endgroup$
– dantopa
3 hours ago












$begingroup$
With a computer you could run Gaussian elimination, though it's not obvious to me that the solution over vectors of 26 real numbers is actually unique (because if any solution exists, then at least 18 of the equations are redundant, so what's to stop more than 18 of them from being redundant?)
$endgroup$
– Ian
3 hours ago






$begingroup$
With a computer you could run Gaussian elimination, though it's not obvious to me that the solution over vectors of 26 real numbers is actually unique (because if any solution exists, then at least 18 of the equations are redundant, so what's to stop more than 18 of them from being redundant?)
$endgroup$
– Ian
3 hours ago






3




3




$begingroup$
Didn't noticed Albuquerque was so close to Jamestown.
$endgroup$
– zwim
2 hours ago






$begingroup$
Didn't noticed Albuquerque was so close to Jamestown.
$endgroup$
– zwim
2 hours ago












3 Answers
3






active

oldest

votes


















3












$begingroup$

To solve it by hand you need to look for words that have similar sets of letters. Using OREGON and RENO you know $G+O=28$. It's too bad they didn't give you ARKANSAS. RENO and NOME give $M=R+3$. Can you find $D+A-O=17?$ MONTGOMERY and MONTEREY are interesting. It is supposed to be a certain amount of work.






share|cite|improve this answer











$endgroup$





















    1












    $begingroup$

    Linear System



    Craft a linear system. ALASKA represents $3times A + L + K + S = 73$ ...



    Definitions:



    $n=44$, number of names



    $m=26$, number of letters in alphabet



    Linear System



    Create the linear system $mathbf{A}x = b$ where $mathbf{A}inmathbb{R}^{44times 26}$



    System matrix



    Rules for building $mathbf{A}$:



    Each row corresponds to a name. E.g. ALASKA is in row $1$, WICHITA in row $44$.



    Each column corresponds to a letter. E.g. A is $1$, Z is $26$.



    Example: ALASKA has 3 A, 1 K, 1 L, 1 S. The non-zero entries of the first row are:
    $$mathbf{A}_{1,1} = 3, quad mathbf{A}_{1,11} = 1, quad mathbf{A}_{1,12} = 1, quad mathbf{A}_{1,19} = 1$$



    Array plots



    The system matrix and data vector are plotted below.



    Ab



    Solution via Gaussian Elimination



    Thanks to astute reader @FredH, the system can be solved exactly by elementary means.



    begin{array}{cc}
    text{A} & 10 \
    text{B} & 20 \
    text{C} & 3 \
    text{D} & 12 \
    text{E} & 6 \
    text{F} & 15 \
    text{G} & 23 \
    text{H} & 13 \
    text{I} & 24 \
    text{J} & 9 \
    text{K} & 16 \
    text{L} & 19 \
    text{M} & 21 \
    text{N} & 4 \
    text{O} & 5 \
    text{P} & 22 \
    text{Q} & 0 \
    text{R} & 18 \
    text{S} & 8 \
    text{T} & 14 \
    text{U} & 7 \
    text{V} & 26 \
    text{W} & 25 \
    text{X} & 11 \
    text{Y} & 17 \
    text{Z} & 2 \
    end{array}



    Raw data



    To spare others from tedious typing, here is the data in cut and paste form.



    b = (73,56,134,64,73,64,78,88,81,64,129,85,56,102,68,77,91,56,105,77,81,83,91,49,109,111,36,61,72,157,47,58,93,65,61,44,70,122,33,69,106,91,99,113)



    (ALASKA, HOUSTON, MONTGOMERY, SALEM, ARIZONA, IDAHO, NANTUCKET, SAN ANTONILO, ATLANTA, IOWA, NASHVILLE, SAVANNAH, BOSTON, JAMESTOWN, NEVADA, SEATTLE, BUFFALO, KANSAS, NEW ORLEANS, TAMPA, CHICAGO, KENTUCKY, NEW YORK, TEXAS, COLUMBIA, LOUISIANA, NOME, TOLEDO, DENVER, LOUISVILLE, OHIO, TULSA, DETROIT, MAINE, OREGON, UTAH, EL PASO, MICHIGAN, RENO, VENICE, HAWAII, MONTEREY, SACRAMENTO, WICHITA)



    Albuquerque



    This is a trick question because we have no Q values in the data table (matrix rank = 25). So we can not make any statement about the value of Q.



    The row vector for ALBUQUERQUE is
    $${1,1,0,0,2,0,0,0,0,0,0,1,0,0,0,0,2,1,0,0,3,0,0,0,0,0}$$. The dot product of this vector with the solution vector $=100$.



    Final answer: ALBUQUERQUE $= 100 + 2Q$ where $Q$ is unknown.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
      $endgroup$
      – FredH
      2 hours ago






    • 2




      $begingroup$
      The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
      $endgroup$
      – FredH
      2 hours ago



















    0












    $begingroup$

    This question has already been answered sufficiently by dantopa with the addition of the comment by FredH. However, I'll just put some Java/MATLAB code here for the sake of completeness, and so you can see how to solve a problem like this using a computer.



    Java:



    import java.util.*;

    public static void main(String args) {

    String names = {"alaska", "arizona", "atlanta", "boston", "buffalo", "chicago", "columbia", "denver", "detroit", "elpaso", "hawaii", "houston", "idaho", "iowa",
    "jamestown", "kansas", "kentucky", "louisiana", "louisville", "maine", "michigan", "monterey", "montgomery", "nantucket", "nashville", "nevada", "neworleans", "newyork", "nome",
    "ohio", "oregon", "reno", "sacramento", "salem", "sanantonio", "savannah", "seattle", "tampa", "texas", "toledo", "tulsa",
    "utah", "venice", "wichita"};



    int x = new int[44][26];


    int count = 0;
    int value = 0;

    for(String y :names) {
    for(int i = 0; i < y.length(); i++) {
    value = (int)y.charAt(i) - 97; // I looked up an ASCII table because chars are stored as integers and subtracted 97 to that a would be at the first index.
    x[count][value]++;
    }
    count++;
    }

    System.out.print("["); // This is just printing out in a convenient form so I we can copy it into MATLAB easily
    for(int i = 0; i < 44; i++) {
    for(int j = 0; j < 26; j++) {
    System.out.print(x[i][j]);
    if(j < 25) System.out.print(",");
    }
    System.out.println(";");
    }

    System.out.println("]");
    }}


    Ok copy the output MATLAB (you could do it in some Java library or done this first part in MATLAB however I don't like doing normal programming on MATLAB or doing math in Java)



    In MATLAB:



    a = \paste the output from java here:
    b = [73, 73, 81, 56, 91, 81, 109, 72, 93, 70, 106, 56, 64, 64, 102, 56, 83, 111, 157, 65, 122, 91, 134, 78, 129, 68, 105, 91, 36, 47, 61, 33, 99, 64, 88, 85, 77, 77, 49, 61, 58, 44, 69, 113];
    b = b';
    x = ab



    The output will be all the solutions in alphabetical order, however $Q$ will be $0$. Since the problem called for numbers between $1$ and $26$, just replace it with the number which is not included already. It is $1$, so you can deduce that $Q = 1$ and use it to calculate the value of Albuquerque.






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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      3












      $begingroup$

      To solve it by hand you need to look for words that have similar sets of letters. Using OREGON and RENO you know $G+O=28$. It's too bad they didn't give you ARKANSAS. RENO and NOME give $M=R+3$. Can you find $D+A-O=17?$ MONTGOMERY and MONTEREY are interesting. It is supposed to be a certain amount of work.






      share|cite|improve this answer











      $endgroup$


















        3












        $begingroup$

        To solve it by hand you need to look for words that have similar sets of letters. Using OREGON and RENO you know $G+O=28$. It's too bad they didn't give you ARKANSAS. RENO and NOME give $M=R+3$. Can you find $D+A-O=17?$ MONTGOMERY and MONTEREY are interesting. It is supposed to be a certain amount of work.






        share|cite|improve this answer











        $endgroup$
















          3












          3








          3





          $begingroup$

          To solve it by hand you need to look for words that have similar sets of letters. Using OREGON and RENO you know $G+O=28$. It's too bad they didn't give you ARKANSAS. RENO and NOME give $M=R+3$. Can you find $D+A-O=17?$ MONTGOMERY and MONTEREY are interesting. It is supposed to be a certain amount of work.






          share|cite|improve this answer











          $endgroup$



          To solve it by hand you need to look for words that have similar sets of letters. Using OREGON and RENO you know $G+O=28$. It's too bad they didn't give you ARKANSAS. RENO and NOME give $M=R+3$. Can you find $D+A-O=17?$ MONTGOMERY and MONTEREY are interesting. It is supposed to be a certain amount of work.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          Ross MillikanRoss Millikan

          298k23198371




          298k23198371























              1












              $begingroup$

              Linear System



              Craft a linear system. ALASKA represents $3times A + L + K + S = 73$ ...



              Definitions:



              $n=44$, number of names



              $m=26$, number of letters in alphabet



              Linear System



              Create the linear system $mathbf{A}x = b$ where $mathbf{A}inmathbb{R}^{44times 26}$



              System matrix



              Rules for building $mathbf{A}$:



              Each row corresponds to a name. E.g. ALASKA is in row $1$, WICHITA in row $44$.



              Each column corresponds to a letter. E.g. A is $1$, Z is $26$.



              Example: ALASKA has 3 A, 1 K, 1 L, 1 S. The non-zero entries of the first row are:
              $$mathbf{A}_{1,1} = 3, quad mathbf{A}_{1,11} = 1, quad mathbf{A}_{1,12} = 1, quad mathbf{A}_{1,19} = 1$$



              Array plots



              The system matrix and data vector are plotted below.



              Ab



              Solution via Gaussian Elimination



              Thanks to astute reader @FredH, the system can be solved exactly by elementary means.



              begin{array}{cc}
              text{A} & 10 \
              text{B} & 20 \
              text{C} & 3 \
              text{D} & 12 \
              text{E} & 6 \
              text{F} & 15 \
              text{G} & 23 \
              text{H} & 13 \
              text{I} & 24 \
              text{J} & 9 \
              text{K} & 16 \
              text{L} & 19 \
              text{M} & 21 \
              text{N} & 4 \
              text{O} & 5 \
              text{P} & 22 \
              text{Q} & 0 \
              text{R} & 18 \
              text{S} & 8 \
              text{T} & 14 \
              text{U} & 7 \
              text{V} & 26 \
              text{W} & 25 \
              text{X} & 11 \
              text{Y} & 17 \
              text{Z} & 2 \
              end{array}



              Raw data



              To spare others from tedious typing, here is the data in cut and paste form.



              b = (73,56,134,64,73,64,78,88,81,64,129,85,56,102,68,77,91,56,105,77,81,83,91,49,109,111,36,61,72,157,47,58,93,65,61,44,70,122,33,69,106,91,99,113)



              (ALASKA, HOUSTON, MONTGOMERY, SALEM, ARIZONA, IDAHO, NANTUCKET, SAN ANTONILO, ATLANTA, IOWA, NASHVILLE, SAVANNAH, BOSTON, JAMESTOWN, NEVADA, SEATTLE, BUFFALO, KANSAS, NEW ORLEANS, TAMPA, CHICAGO, KENTUCKY, NEW YORK, TEXAS, COLUMBIA, LOUISIANA, NOME, TOLEDO, DENVER, LOUISVILLE, OHIO, TULSA, DETROIT, MAINE, OREGON, UTAH, EL PASO, MICHIGAN, RENO, VENICE, HAWAII, MONTEREY, SACRAMENTO, WICHITA)



              Albuquerque



              This is a trick question because we have no Q values in the data table (matrix rank = 25). So we can not make any statement about the value of Q.



              The row vector for ALBUQUERQUE is
              $${1,1,0,0,2,0,0,0,0,0,0,1,0,0,0,0,2,1,0,0,3,0,0,0,0,0}$$. The dot product of this vector with the solution vector $=100$.



              Final answer: ALBUQUERQUE $= 100 + 2Q$ where $Q$ is unknown.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
                $endgroup$
                – FredH
                2 hours ago






              • 2




                $begingroup$
                The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
                $endgroup$
                – FredH
                2 hours ago
















              1












              $begingroup$

              Linear System



              Craft a linear system. ALASKA represents $3times A + L + K + S = 73$ ...



              Definitions:



              $n=44$, number of names



              $m=26$, number of letters in alphabet



              Linear System



              Create the linear system $mathbf{A}x = b$ where $mathbf{A}inmathbb{R}^{44times 26}$



              System matrix



              Rules for building $mathbf{A}$:



              Each row corresponds to a name. E.g. ALASKA is in row $1$, WICHITA in row $44$.



              Each column corresponds to a letter. E.g. A is $1$, Z is $26$.



              Example: ALASKA has 3 A, 1 K, 1 L, 1 S. The non-zero entries of the first row are:
              $$mathbf{A}_{1,1} = 3, quad mathbf{A}_{1,11} = 1, quad mathbf{A}_{1,12} = 1, quad mathbf{A}_{1,19} = 1$$



              Array plots



              The system matrix and data vector are plotted below.



              Ab



              Solution via Gaussian Elimination



              Thanks to astute reader @FredH, the system can be solved exactly by elementary means.



              begin{array}{cc}
              text{A} & 10 \
              text{B} & 20 \
              text{C} & 3 \
              text{D} & 12 \
              text{E} & 6 \
              text{F} & 15 \
              text{G} & 23 \
              text{H} & 13 \
              text{I} & 24 \
              text{J} & 9 \
              text{K} & 16 \
              text{L} & 19 \
              text{M} & 21 \
              text{N} & 4 \
              text{O} & 5 \
              text{P} & 22 \
              text{Q} & 0 \
              text{R} & 18 \
              text{S} & 8 \
              text{T} & 14 \
              text{U} & 7 \
              text{V} & 26 \
              text{W} & 25 \
              text{X} & 11 \
              text{Y} & 17 \
              text{Z} & 2 \
              end{array}



              Raw data



              To spare others from tedious typing, here is the data in cut and paste form.



              b = (73,56,134,64,73,64,78,88,81,64,129,85,56,102,68,77,91,56,105,77,81,83,91,49,109,111,36,61,72,157,47,58,93,65,61,44,70,122,33,69,106,91,99,113)



              (ALASKA, HOUSTON, MONTGOMERY, SALEM, ARIZONA, IDAHO, NANTUCKET, SAN ANTONILO, ATLANTA, IOWA, NASHVILLE, SAVANNAH, BOSTON, JAMESTOWN, NEVADA, SEATTLE, BUFFALO, KANSAS, NEW ORLEANS, TAMPA, CHICAGO, KENTUCKY, NEW YORK, TEXAS, COLUMBIA, LOUISIANA, NOME, TOLEDO, DENVER, LOUISVILLE, OHIO, TULSA, DETROIT, MAINE, OREGON, UTAH, EL PASO, MICHIGAN, RENO, VENICE, HAWAII, MONTEREY, SACRAMENTO, WICHITA)



              Albuquerque



              This is a trick question because we have no Q values in the data table (matrix rank = 25). So we can not make any statement about the value of Q.



              The row vector for ALBUQUERQUE is
              $${1,1,0,0,2,0,0,0,0,0,0,1,0,0,0,0,2,1,0,0,3,0,0,0,0,0}$$. The dot product of this vector with the solution vector $=100$.



              Final answer: ALBUQUERQUE $= 100 + 2Q$ where $Q$ is unknown.






              share|cite|improve this answer











              $endgroup$









              • 1




                $begingroup$
                Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
                $endgroup$
                – FredH
                2 hours ago






              • 2




                $begingroup$
                The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
                $endgroup$
                – FredH
                2 hours ago














              1












              1








              1





              $begingroup$

              Linear System



              Craft a linear system. ALASKA represents $3times A + L + K + S = 73$ ...



              Definitions:



              $n=44$, number of names



              $m=26$, number of letters in alphabet



              Linear System



              Create the linear system $mathbf{A}x = b$ where $mathbf{A}inmathbb{R}^{44times 26}$



              System matrix



              Rules for building $mathbf{A}$:



              Each row corresponds to a name. E.g. ALASKA is in row $1$, WICHITA in row $44$.



              Each column corresponds to a letter. E.g. A is $1$, Z is $26$.



              Example: ALASKA has 3 A, 1 K, 1 L, 1 S. The non-zero entries of the first row are:
              $$mathbf{A}_{1,1} = 3, quad mathbf{A}_{1,11} = 1, quad mathbf{A}_{1,12} = 1, quad mathbf{A}_{1,19} = 1$$



              Array plots



              The system matrix and data vector are plotted below.



              Ab



              Solution via Gaussian Elimination



              Thanks to astute reader @FredH, the system can be solved exactly by elementary means.



              begin{array}{cc}
              text{A} & 10 \
              text{B} & 20 \
              text{C} & 3 \
              text{D} & 12 \
              text{E} & 6 \
              text{F} & 15 \
              text{G} & 23 \
              text{H} & 13 \
              text{I} & 24 \
              text{J} & 9 \
              text{K} & 16 \
              text{L} & 19 \
              text{M} & 21 \
              text{N} & 4 \
              text{O} & 5 \
              text{P} & 22 \
              text{Q} & 0 \
              text{R} & 18 \
              text{S} & 8 \
              text{T} & 14 \
              text{U} & 7 \
              text{V} & 26 \
              text{W} & 25 \
              text{X} & 11 \
              text{Y} & 17 \
              text{Z} & 2 \
              end{array}



              Raw data



              To spare others from tedious typing, here is the data in cut and paste form.



              b = (73,56,134,64,73,64,78,88,81,64,129,85,56,102,68,77,91,56,105,77,81,83,91,49,109,111,36,61,72,157,47,58,93,65,61,44,70,122,33,69,106,91,99,113)



              (ALASKA, HOUSTON, MONTGOMERY, SALEM, ARIZONA, IDAHO, NANTUCKET, SAN ANTONILO, ATLANTA, IOWA, NASHVILLE, SAVANNAH, BOSTON, JAMESTOWN, NEVADA, SEATTLE, BUFFALO, KANSAS, NEW ORLEANS, TAMPA, CHICAGO, KENTUCKY, NEW YORK, TEXAS, COLUMBIA, LOUISIANA, NOME, TOLEDO, DENVER, LOUISVILLE, OHIO, TULSA, DETROIT, MAINE, OREGON, UTAH, EL PASO, MICHIGAN, RENO, VENICE, HAWAII, MONTEREY, SACRAMENTO, WICHITA)



              Albuquerque



              This is a trick question because we have no Q values in the data table (matrix rank = 25). So we can not make any statement about the value of Q.



              The row vector for ALBUQUERQUE is
              $${1,1,0,0,2,0,0,0,0,0,0,1,0,0,0,0,2,1,0,0,3,0,0,0,0,0}$$. The dot product of this vector with the solution vector $=100$.



              Final answer: ALBUQUERQUE $= 100 + 2Q$ where $Q$ is unknown.






              share|cite|improve this answer











              $endgroup$



              Linear System



              Craft a linear system. ALASKA represents $3times A + L + K + S = 73$ ...



              Definitions:



              $n=44$, number of names



              $m=26$, number of letters in alphabet



              Linear System



              Create the linear system $mathbf{A}x = b$ where $mathbf{A}inmathbb{R}^{44times 26}$



              System matrix



              Rules for building $mathbf{A}$:



              Each row corresponds to a name. E.g. ALASKA is in row $1$, WICHITA in row $44$.



              Each column corresponds to a letter. E.g. A is $1$, Z is $26$.



              Example: ALASKA has 3 A, 1 K, 1 L, 1 S. The non-zero entries of the first row are:
              $$mathbf{A}_{1,1} = 3, quad mathbf{A}_{1,11} = 1, quad mathbf{A}_{1,12} = 1, quad mathbf{A}_{1,19} = 1$$



              Array plots



              The system matrix and data vector are plotted below.



              Ab



              Solution via Gaussian Elimination



              Thanks to astute reader @FredH, the system can be solved exactly by elementary means.



              begin{array}{cc}
              text{A} & 10 \
              text{B} & 20 \
              text{C} & 3 \
              text{D} & 12 \
              text{E} & 6 \
              text{F} & 15 \
              text{G} & 23 \
              text{H} & 13 \
              text{I} & 24 \
              text{J} & 9 \
              text{K} & 16 \
              text{L} & 19 \
              text{M} & 21 \
              text{N} & 4 \
              text{O} & 5 \
              text{P} & 22 \
              text{Q} & 0 \
              text{R} & 18 \
              text{S} & 8 \
              text{T} & 14 \
              text{U} & 7 \
              text{V} & 26 \
              text{W} & 25 \
              text{X} & 11 \
              text{Y} & 17 \
              text{Z} & 2 \
              end{array}



              Raw data



              To spare others from tedious typing, here is the data in cut and paste form.



              b = (73,56,134,64,73,64,78,88,81,64,129,85,56,102,68,77,91,56,105,77,81,83,91,49,109,111,36,61,72,157,47,58,93,65,61,44,70,122,33,69,106,91,99,113)



              (ALASKA, HOUSTON, MONTGOMERY, SALEM, ARIZONA, IDAHO, NANTUCKET, SAN ANTONILO, ATLANTA, IOWA, NASHVILLE, SAVANNAH, BOSTON, JAMESTOWN, NEVADA, SEATTLE, BUFFALO, KANSAS, NEW ORLEANS, TAMPA, CHICAGO, KENTUCKY, NEW YORK, TEXAS, COLUMBIA, LOUISIANA, NOME, TOLEDO, DENVER, LOUISVILLE, OHIO, TULSA, DETROIT, MAINE, OREGON, UTAH, EL PASO, MICHIGAN, RENO, VENICE, HAWAII, MONTEREY, SACRAMENTO, WICHITA)



              Albuquerque



              This is a trick question because we have no Q values in the data table (matrix rank = 25). So we can not make any statement about the value of Q.



              The row vector for ALBUQUERQUE is
              $${1,1,0,0,2,0,0,0,0,0,0,1,0,0,0,0,2,1,0,0,3,0,0,0,0,0}$$. The dot product of this vector with the solution vector $=100$.



              Final answer: ALBUQUERQUE $= 100 + 2Q$ where $Q$ is unknown.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited 2 hours ago

























              answered 2 hours ago









              dantopadantopa

              6,57942244




              6,57942244








              • 1




                $begingroup$
                Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
                $endgroup$
                – FredH
                2 hours ago






              • 2




                $begingroup$
                The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
                $endgroup$
                – FredH
                2 hours ago














              • 1




                $begingroup$
                Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
                $endgroup$
                – FredH
                2 hours ago






              • 2




                $begingroup$
                The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
                $endgroup$
                – FredH
                2 hours ago








              1




              1




              $begingroup$
              Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
              $endgroup$
              – FredH
              2 hours ago




              $begingroup$
              Your data contains a typo: "SAN ANTONILO" instead of "SAN ANTONIO". Perhaps that will work better?
              $endgroup$
              – FredH
              2 hours ago




              2




              2




              $begingroup$
              The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
              $endgroup$
              – FredH
              2 hours ago




              $begingroup$
              The problem says the values are distinct whole numbers from $1$ to $26$, so $Q = 1$ and ALBUQUERQUE $= 102$.
              $endgroup$
              – FredH
              2 hours ago











              0












              $begingroup$

              This question has already been answered sufficiently by dantopa with the addition of the comment by FredH. However, I'll just put some Java/MATLAB code here for the sake of completeness, and so you can see how to solve a problem like this using a computer.



              Java:



              import java.util.*;

              public static void main(String args) {

              String names = {"alaska", "arizona", "atlanta", "boston", "buffalo", "chicago", "columbia", "denver", "detroit", "elpaso", "hawaii", "houston", "idaho", "iowa",
              "jamestown", "kansas", "kentucky", "louisiana", "louisville", "maine", "michigan", "monterey", "montgomery", "nantucket", "nashville", "nevada", "neworleans", "newyork", "nome",
              "ohio", "oregon", "reno", "sacramento", "salem", "sanantonio", "savannah", "seattle", "tampa", "texas", "toledo", "tulsa",
              "utah", "venice", "wichita"};



              int x = new int[44][26];


              int count = 0;
              int value = 0;

              for(String y :names) {
              for(int i = 0; i < y.length(); i++) {
              value = (int)y.charAt(i) - 97; // I looked up an ASCII table because chars are stored as integers and subtracted 97 to that a would be at the first index.
              x[count][value]++;
              }
              count++;
              }

              System.out.print("["); // This is just printing out in a convenient form so I we can copy it into MATLAB easily
              for(int i = 0; i < 44; i++) {
              for(int j = 0; j < 26; j++) {
              System.out.print(x[i][j]);
              if(j < 25) System.out.print(",");
              }
              System.out.println(";");
              }

              System.out.println("]");
              }}


              Ok copy the output MATLAB (you could do it in some Java library or done this first part in MATLAB however I don't like doing normal programming on MATLAB or doing math in Java)



              In MATLAB:



              a = \paste the output from java here:
              b = [73, 73, 81, 56, 91, 81, 109, 72, 93, 70, 106, 56, 64, 64, 102, 56, 83, 111, 157, 65, 122, 91, 134, 78, 129, 68, 105, 91, 36, 47, 61, 33, 99, 64, 88, 85, 77, 77, 49, 61, 58, 44, 69, 113];
              b = b';
              x = ab



              The output will be all the solutions in alphabetical order, however $Q$ will be $0$. Since the problem called for numbers between $1$ and $26$, just replace it with the number which is not included already. It is $1$, so you can deduce that $Q = 1$ and use it to calculate the value of Albuquerque.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                This question has already been answered sufficiently by dantopa with the addition of the comment by FredH. However, I'll just put some Java/MATLAB code here for the sake of completeness, and so you can see how to solve a problem like this using a computer.



                Java:



                import java.util.*;

                public static void main(String args) {

                String names = {"alaska", "arizona", "atlanta", "boston", "buffalo", "chicago", "columbia", "denver", "detroit", "elpaso", "hawaii", "houston", "idaho", "iowa",
                "jamestown", "kansas", "kentucky", "louisiana", "louisville", "maine", "michigan", "monterey", "montgomery", "nantucket", "nashville", "nevada", "neworleans", "newyork", "nome",
                "ohio", "oregon", "reno", "sacramento", "salem", "sanantonio", "savannah", "seattle", "tampa", "texas", "toledo", "tulsa",
                "utah", "venice", "wichita"};



                int x = new int[44][26];


                int count = 0;
                int value = 0;

                for(String y :names) {
                for(int i = 0; i < y.length(); i++) {
                value = (int)y.charAt(i) - 97; // I looked up an ASCII table because chars are stored as integers and subtracted 97 to that a would be at the first index.
                x[count][value]++;
                }
                count++;
                }

                System.out.print("["); // This is just printing out in a convenient form so I we can copy it into MATLAB easily
                for(int i = 0; i < 44; i++) {
                for(int j = 0; j < 26; j++) {
                System.out.print(x[i][j]);
                if(j < 25) System.out.print(",");
                }
                System.out.println(";");
                }

                System.out.println("]");
                }}


                Ok copy the output MATLAB (you could do it in some Java library or done this first part in MATLAB however I don't like doing normal programming on MATLAB or doing math in Java)



                In MATLAB:



                a = \paste the output from java here:
                b = [73, 73, 81, 56, 91, 81, 109, 72, 93, 70, 106, 56, 64, 64, 102, 56, 83, 111, 157, 65, 122, 91, 134, 78, 129, 68, 105, 91, 36, 47, 61, 33, 99, 64, 88, 85, 77, 77, 49, 61, 58, 44, 69, 113];
                b = b';
                x = ab



                The output will be all the solutions in alphabetical order, however $Q$ will be $0$. Since the problem called for numbers between $1$ and $26$, just replace it with the number which is not included already. It is $1$, so you can deduce that $Q = 1$ and use it to calculate the value of Albuquerque.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  This question has already been answered sufficiently by dantopa with the addition of the comment by FredH. However, I'll just put some Java/MATLAB code here for the sake of completeness, and so you can see how to solve a problem like this using a computer.



                  Java:



                  import java.util.*;

                  public static void main(String args) {

                  String names = {"alaska", "arizona", "atlanta", "boston", "buffalo", "chicago", "columbia", "denver", "detroit", "elpaso", "hawaii", "houston", "idaho", "iowa",
                  "jamestown", "kansas", "kentucky", "louisiana", "louisville", "maine", "michigan", "monterey", "montgomery", "nantucket", "nashville", "nevada", "neworleans", "newyork", "nome",
                  "ohio", "oregon", "reno", "sacramento", "salem", "sanantonio", "savannah", "seattle", "tampa", "texas", "toledo", "tulsa",
                  "utah", "venice", "wichita"};



                  int x = new int[44][26];


                  int count = 0;
                  int value = 0;

                  for(String y :names) {
                  for(int i = 0; i < y.length(); i++) {
                  value = (int)y.charAt(i) - 97; // I looked up an ASCII table because chars are stored as integers and subtracted 97 to that a would be at the first index.
                  x[count][value]++;
                  }
                  count++;
                  }

                  System.out.print("["); // This is just printing out in a convenient form so I we can copy it into MATLAB easily
                  for(int i = 0; i < 44; i++) {
                  for(int j = 0; j < 26; j++) {
                  System.out.print(x[i][j]);
                  if(j < 25) System.out.print(",");
                  }
                  System.out.println(";");
                  }

                  System.out.println("]");
                  }}


                  Ok copy the output MATLAB (you could do it in some Java library or done this first part in MATLAB however I don't like doing normal programming on MATLAB or doing math in Java)



                  In MATLAB:



                  a = \paste the output from java here:
                  b = [73, 73, 81, 56, 91, 81, 109, 72, 93, 70, 106, 56, 64, 64, 102, 56, 83, 111, 157, 65, 122, 91, 134, 78, 129, 68, 105, 91, 36, 47, 61, 33, 99, 64, 88, 85, 77, 77, 49, 61, 58, 44, 69, 113];
                  b = b';
                  x = ab



                  The output will be all the solutions in alphabetical order, however $Q$ will be $0$. Since the problem called for numbers between $1$ and $26$, just replace it with the number which is not included already. It is $1$, so you can deduce that $Q = 1$ and use it to calculate the value of Albuquerque.






                  share|cite|improve this answer









                  $endgroup$



                  This question has already been answered sufficiently by dantopa with the addition of the comment by FredH. However, I'll just put some Java/MATLAB code here for the sake of completeness, and so you can see how to solve a problem like this using a computer.



                  Java:



                  import java.util.*;

                  public static void main(String args) {

                  String names = {"alaska", "arizona", "atlanta", "boston", "buffalo", "chicago", "columbia", "denver", "detroit", "elpaso", "hawaii", "houston", "idaho", "iowa",
                  "jamestown", "kansas", "kentucky", "louisiana", "louisville", "maine", "michigan", "monterey", "montgomery", "nantucket", "nashville", "nevada", "neworleans", "newyork", "nome",
                  "ohio", "oregon", "reno", "sacramento", "salem", "sanantonio", "savannah", "seattle", "tampa", "texas", "toledo", "tulsa",
                  "utah", "venice", "wichita"};



                  int x = new int[44][26];


                  int count = 0;
                  int value = 0;

                  for(String y :names) {
                  for(int i = 0; i < y.length(); i++) {
                  value = (int)y.charAt(i) - 97; // I looked up an ASCII table because chars are stored as integers and subtracted 97 to that a would be at the first index.
                  x[count][value]++;
                  }
                  count++;
                  }

                  System.out.print("["); // This is just printing out in a convenient form so I we can copy it into MATLAB easily
                  for(int i = 0; i < 44; i++) {
                  for(int j = 0; j < 26; j++) {
                  System.out.print(x[i][j]);
                  if(j < 25) System.out.print(",");
                  }
                  System.out.println(";");
                  }

                  System.out.println("]");
                  }}


                  Ok copy the output MATLAB (you could do it in some Java library or done this first part in MATLAB however I don't like doing normal programming on MATLAB or doing math in Java)



                  In MATLAB:



                  a = \paste the output from java here:
                  b = [73, 73, 81, 56, 91, 81, 109, 72, 93, 70, 106, 56, 64, 64, 102, 56, 83, 111, 157, 65, 122, 91, 134, 78, 129, 68, 105, 91, 36, 47, 61, 33, 99, 64, 88, 85, 77, 77, 49, 61, 58, 44, 69, 113];
                  b = b';
                  x = ab



                  The output will be all the solutions in alphabetical order, however $Q$ will be $0$. Since the problem called for numbers between $1$ and $26$, just replace it with the number which is not included already. It is $1$, so you can deduce that $Q = 1$ and use it to calculate the value of Albuquerque.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 1 hour ago









                  Jack PfaffingerJack Pfaffinger

                  364112




                  364112






















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