Why is this weird space appearing after my minus sign

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I'm using MathJax-Node to turn TeX maths equations into both MathML and SVG on my server. Mostly this works fine. But when I render Tupper's self-referential formula, a weird space appears after the second minus sign in the exponent: Rendered formula.



The space also appears on mathURL (which is where the above image comes from). When I examine the MathML generated, I can see that MathJax has explicitly inserted a mspace element.



The TeX I'm using is as follows:



frac12<leftlfloormodleft(leftlfloorfrac y17rightrfloor2^-17leftlfloor xrightrfloor-modleft(leftlfloor yrightrfloor,17right),2right)rightrfloor


Any idea what's going on here?






spacing math-operators







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  • 3




    Although this is not exactly a TeX question, but have you looked at the first mod? It is not about the minus sign, but rather the mod. The mod has, by design, a preceding quad space and an appending half quad space.
    – Ruixi Zhang
    1 hour ago















up vote
1
down vote

favorite












I'm using MathJax-Node to turn TeX maths equations into both MathML and SVG on my server. Mostly this works fine. But when I render Tupper's self-referential formula, a weird space appears after the second minus sign in the exponent: Rendered formula.



The space also appears on mathURL (which is where the above image comes from). When I examine the MathML generated, I can see that MathJax has explicitly inserted a mspace element.



The TeX I'm using is as follows:



frac12<leftlfloormodleft(leftlfloorfrac y17rightrfloor2^-17leftlfloor xrightrfloor-modleft(leftlfloor yrightrfloor,17right),2right)rightrfloor


Any idea what's going on here?






spacing math-operators







share|improve this question









New contributor




Sora2455 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.















  • 3




    Although this is not exactly a TeX question, but have you looked at the first mod? It is not about the minus sign, but rather the mod. The mod has, by design, a preceding quad space and an appending half quad space.
    – Ruixi Zhang
    1 hour ago













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I'm using MathJax-Node to turn TeX maths equations into both MathML and SVG on my server. Mostly this works fine. But when I render Tupper's self-referential formula, a weird space appears after the second minus sign in the exponent: Rendered formula.



The space also appears on mathURL (which is where the above image comes from). When I examine the MathML generated, I can see that MathJax has explicitly inserted a mspace element.



The TeX I'm using is as follows:



frac12<leftlfloormodleft(leftlfloorfrac y17rightrfloor2^-17leftlfloor xrightrfloor-modleft(leftlfloor yrightrfloor,17right),2right)rightrfloor


Any idea what's going on here?






spacing math-operators







share|improve this question









New contributor




Sora2455 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I'm using MathJax-Node to turn TeX maths equations into both MathML and SVG on my server. Mostly this works fine. But when I render Tupper's self-referential formula, a weird space appears after the second minus sign in the exponent: Rendered formula.



The space also appears on mathURL (which is where the above image comes from). When I examine the MathML generated, I can see that MathJax has explicitly inserted a mspace element.



The TeX I'm using is as follows:



frac12<leftlfloormodleft(leftlfloorfrac y17rightrfloor2^-17leftlfloor xrightrfloor-modleft(leftlfloor yrightrfloor,17right),2right)rightrfloor


Any idea what's going on here?






spacing math-operators




spacing math-operators






share|improve this question









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Sora2455 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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share|improve this question









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share|improve this question








edited 1 hour ago









egreg

683k8418193065




683k8418193065






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asked 1 hour ago









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Sora2455 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.







  • 3




    Although this is not exactly a TeX question, but have you looked at the first mod? It is not about the minus sign, but rather the mod. The mod has, by design, a preceding quad space and an appending half quad space.
    – Ruixi Zhang
    1 hour ago













  • 3




    Although this is not exactly a TeX question, but have you looked at the first mod? It is not about the minus sign, but rather the mod. The mod has, by design, a preceding quad space and an appending half quad space.
    – Ruixi Zhang
    1 hour ago








3




3




Although this is not exactly a TeX question, but have you looked at the first mod? It is not about the minus sign, but rather the mod. The mod has, by design, a preceding quad space and an appending half quad space.
– Ruixi Zhang
1 hour ago





Although this is not exactly a TeX question, but have you looked at the first mod? It is not about the minus sign, but rather the mod. The mod has, by design, a preceding quad space and an appending half quad space.
– Ruixi Zhang
1 hour ago











2 Answers
2






active

oldest

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up vote
3
down vote













You're using “mod” as a math operator like “exp” or “sin”. The command mod is for notation such as



5 equiv 2 mod3


and the inserted space is obviously wanted here.



Use operatornamemod instead.



documentclassarticle
usepackageamsmath

begindocument

[
frac12<
 leftlfloor
 operatornamemodleft(
 leftlfloorfracy17rightrfloor
 2^-17lfloor xrfloor-operatornamemod(lfloor yrfloor,17),2
 right)
 rightrfloor
]

enddocument


I've removed useless (and wrong) left and right pairs.



enter image description here



It works equally well in MathJax (tested on Math.SE):



enter image description here






share|improve this answer





























    up vote
    3
    down vote













    Do you mean this:



    enter image description here



    documentclassarticle

    usepackagexintexpr

    % Tupper

    % q = floor(y/17)
    % r = mod(y, 17) (modulo for floor)

    % Tupper formula boils down to say:

    % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

    % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

    % xintiiexpr notations

    % floored division is //
    % associated modulo is /: (or 'mod')
    % divmod(,) is both // and /: (as in Python)

    begindocument

    % 0 <= x < 106
    % k <= y < k + 17

    xintdefiivar TupperK := 
    960939379918958884971672962127852754715004339660129306651505519271702802395266
    424689642842174350718121267153782770623355993237280874144307891325963941337723
    487857735749823926629715517173716995165232890538221612403238855866184013235585
    136048828693337902491454229288667081096184496091705183454067827731551705405381
    627380967602565625016981482083418783163849115590225610003652351370343874461848
    378737238198224849863465033159410054974700593138339226497249461751545728366702
    369745461014655997933798537483143786841806593422227898388722980000748404719;

    setlengthunitlength1pt

    defparseTupperXY#1#2% #1 = x, #2=y, 
     xintdefiivar Yq, Yr = divmod(#2, 17);%
     edefresultTupperXYxinttheiiexpr odd(Yq // 2**(17 * #1 + Yr))relax%
    %

    % This is a bit slow, be patient (big powers of 2 are computed...)
    % Much faster would be to convert TupperK to binary...
    beginpicture(106,17)
     xintFor* #1 in xintSeq0105do % x
     xintFor* #2 in xintSeq016do % y - k
     parseTupperXY#1#2+TupperK
     if1resultTupperXY
     put(numexpr105-#1, numexpr16-#2)rule1pt1pt
     fi
     
     
    endpicture

    enddocument


    I hoped this



    documentclassarticle

    usepackagexintexpr

    % Tupper

    % q = floor(y/17)
    % r = mod(y, 17) (modulo for floor)

    % Tupper formula boils down to say:

    % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

    % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

    % xintiiexpr notations

    % floored division is //
    % associated modulo is /: (or 'mod')
    % divmod(,) is both // and /: (as in Python)

    begindocument

    % 0 <= x < 106
    % k <= y < k + 17

    xintdefiivar TupperK := 
    960939379918958884971672962127852754715004339660129306651505519271702802395266
    424689642842174350718121267153782770623355993237280874144307891325963941337723
    487857735749823926629715517173716995165232890538221612403238855866184013235585
    136048828693337902491454229288667081096184496091705183454067827731551705405381
    627380967602565625016981482083418783163849115590225610003652351370343874461848
    378737238198224849863465033159410054974700593138339226497249461751545728366702
    369745461014655997933798537483143786841806593422227898388722980000748404719;

    setlengthunitlength1pt

    xintdefiivar twoToThe17 := 2**17;
    xintdefiivar twoToThe17timesX := 1;

    beginpicture(106,17)
     xintFor* #1 in xintSeq0105do % x
     xintFor* #2 in 012345678910111213141516do % y - k
     xintdefiivar Yq, Yr = divmod(#2 + TupperK, 17);%
     xintifbooliiexpr odd(Yq // (twoToThe17timesX * 2**Yr ))
     put(numexpr105-#1, numexpr16-#2)rule1pt1pt
     
     
     xintdefiivar twoToThe17timesX := twoToThe17 * twoToThe17timesX;%
     
    endpicture

    enddocument


    would be faster, but it is only a bit faster. Hence the bottleneck is in the big divisions, less so in the powers of two.



    It would be faster to cheat a bit and compute binary representation of k via xintDecToBin from xintbinhex for example.






    share|improve this answer






















    • loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
      – jfbu
      31 mins ago










    • provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
      – jfbu
      14 mins ago











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    You're using “mod” as a math operator like “exp” or “sin”. The command mod is for notation such as



    5 equiv 2 mod3


    and the inserted space is obviously wanted here.



    Use operatornamemod instead.



    documentclassarticle
    usepackageamsmath

    begindocument

    [
    frac12<
     leftlfloor
     operatornamemodleft(
     leftlfloorfracy17rightrfloor
     2^-17lfloor xrfloor-operatornamemod(lfloor yrfloor,17),2
     right)
     rightrfloor
    ]

    enddocument


    I've removed useless (and wrong) left and right pairs.



    enter image description here



    It works equally well in MathJax (tested on Math.SE):



    enter image description here






    share|improve this answer


























      up vote
      3
      down vote













      You're using “mod” as a math operator like “exp” or “sin”. The command mod is for notation such as



      5 equiv 2 mod3


      and the inserted space is obviously wanted here.



      Use operatornamemod instead.



      documentclassarticle
      usepackageamsmath

      begindocument

      [
      frac12<
       leftlfloor
       operatornamemodleft(
       leftlfloorfracy17rightrfloor
       2^-17lfloor xrfloor-operatornamemod(lfloor yrfloor,17),2
       right)
       rightrfloor
      ]

      enddocument


      I've removed useless (and wrong) left and right pairs.



      enter image description here



      It works equally well in MathJax (tested on Math.SE):



      enter image description here






      share|improve this answer
























        up vote
        3
        down vote










        up vote
        3
        down vote









        You're using “mod” as a math operator like “exp” or “sin”. The command mod is for notation such as



        5 equiv 2 mod3


        and the inserted space is obviously wanted here.



        Use operatornamemod instead.



        documentclassarticle
        usepackageamsmath

        begindocument

        [
        frac12<
         leftlfloor
         operatornamemodleft(
         leftlfloorfracy17rightrfloor
         2^-17lfloor xrfloor-operatornamemod(lfloor yrfloor,17),2
         right)
         rightrfloor
        ]

        enddocument


        I've removed useless (and wrong) left and right pairs.



        enter image description here



        It works equally well in MathJax (tested on Math.SE):



        enter image description here






        share|improve this answer














        You're using “mod” as a math operator like “exp” or “sin”. The command mod is for notation such as



        5 equiv 2 mod3


        and the inserted space is obviously wanted here.



        Use operatornamemod instead.



        documentclassarticle
        usepackageamsmath

        begindocument

        [
        frac12<
         leftlfloor
         operatornamemodleft(
         leftlfloorfracy17rightrfloor
         2^-17lfloor xrfloor-operatornamemod(lfloor yrfloor,17),2
         right)
         rightrfloor
        ]

        enddocument


        I've removed useless (and wrong) left and right pairs.



        enter image description here



        It works equally well in MathJax (tested on Math.SE):



        enter image description here







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        egreg

        683k8418193065




        683k8418193065




















            up vote
            3
            down vote













            Do you mean this:



            enter image description here



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            defparseTupperXY#1#2% #1 = x, #2=y, 
             xintdefiivar Yq, Yr = divmod(#2, 17);%
             edefresultTupperXYxinttheiiexpr odd(Yq // 2**(17 * #1 + Yr))relax%
            %

            % This is a bit slow, be patient (big powers of 2 are computed...)
            % Much faster would be to convert TupperK to binary...
            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in xintSeq016do % y - k
             parseTupperXY#1#2+TupperK
             if1resultTupperXY
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             fi
             
             
            endpicture

            enddocument


            I hoped this



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            xintdefiivar twoToThe17 := 2**17;
            xintdefiivar twoToThe17timesX := 1;

            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in 012345678910111213141516do % y - k
             xintdefiivar Yq, Yr = divmod(#2 + TupperK, 17);%
             xintifbooliiexpr odd(Yq // (twoToThe17timesX * 2**Yr ))
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             
             
             xintdefiivar twoToThe17timesX := twoToThe17 * twoToThe17timesX;%
             
            endpicture

            enddocument


            would be faster, but it is only a bit faster. Hence the bottleneck is in the big divisions, less so in the powers of two.



            It would be faster to cheat a bit and compute binary representation of k via xintDecToBin from xintbinhex for example.






            share|improve this answer






















            • loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
              – jfbu
              31 mins ago










            • provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
              – jfbu
              14 mins ago















            up vote
            3
            down vote













            Do you mean this:



            enter image description here



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            defparseTupperXY#1#2% #1 = x, #2=y, 
             xintdefiivar Yq, Yr = divmod(#2, 17);%
             edefresultTupperXYxinttheiiexpr odd(Yq // 2**(17 * #1 + Yr))relax%
            %

            % This is a bit slow, be patient (big powers of 2 are computed...)
            % Much faster would be to convert TupperK to binary...
            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in xintSeq016do % y - k
             parseTupperXY#1#2+TupperK
             if1resultTupperXY
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             fi
             
             
            endpicture

            enddocument


            I hoped this



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            xintdefiivar twoToThe17 := 2**17;
            xintdefiivar twoToThe17timesX := 1;

            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in 012345678910111213141516do % y - k
             xintdefiivar Yq, Yr = divmod(#2 + TupperK, 17);%
             xintifbooliiexpr odd(Yq // (twoToThe17timesX * 2**Yr ))
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             
             
             xintdefiivar twoToThe17timesX := twoToThe17 * twoToThe17timesX;%
             
            endpicture

            enddocument


            would be faster, but it is only a bit faster. Hence the bottleneck is in the big divisions, less so in the powers of two.



            It would be faster to cheat a bit and compute binary representation of k via xintDecToBin from xintbinhex for example.






            share|improve this answer






















            • loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
              – jfbu
              31 mins ago










            • provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
              – jfbu
              14 mins ago













            up vote
            3
            down vote










            up vote
            3
            down vote









            Do you mean this:



            enter image description here



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            defparseTupperXY#1#2% #1 = x, #2=y, 
             xintdefiivar Yq, Yr = divmod(#2, 17);%
             edefresultTupperXYxinttheiiexpr odd(Yq // 2**(17 * #1 + Yr))relax%
            %

            % This is a bit slow, be patient (big powers of 2 are computed...)
            % Much faster would be to convert TupperK to binary...
            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in xintSeq016do % y - k
             parseTupperXY#1#2+TupperK
             if1resultTupperXY
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             fi
             
             
            endpicture

            enddocument


            I hoped this



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            xintdefiivar twoToThe17 := 2**17;
            xintdefiivar twoToThe17timesX := 1;

            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in 012345678910111213141516do % y - k
             xintdefiivar Yq, Yr = divmod(#2 + TupperK, 17);%
             xintifbooliiexpr odd(Yq // (twoToThe17timesX * 2**Yr ))
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             
             
             xintdefiivar twoToThe17timesX := twoToThe17 * twoToThe17timesX;%
             
            endpicture

            enddocument


            would be faster, but it is only a bit faster. Hence the bottleneck is in the big divisions, less so in the powers of two.



            It would be faster to cheat a bit and compute binary representation of k via xintDecToBin from xintbinhex for example.






            share|improve this answer














            Do you mean this:



            enter image description here



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            defparseTupperXY#1#2% #1 = x, #2=y, 
             xintdefiivar Yq, Yr = divmod(#2, 17);%
             edefresultTupperXYxinttheiiexpr odd(Yq // 2**(17 * #1 + Yr))relax%
            %

            % This is a bit slow, be patient (big powers of 2 are computed...)
            % Much faster would be to convert TupperK to binary...
            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in xintSeq016do % y - k
             parseTupperXY#1#2+TupperK
             if1resultTupperXY
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             fi
             
             
            endpicture

            enddocument


            I hoped this



            documentclassarticle

            usepackagexintexpr

            % Tupper

            % q = floor(y/17)
            % r = mod(y, 17) (modulo for floor)

            % Tupper formula boils down to say:

            % bit (x, y) is ON <===> floor(q, 2**(17x+r)) is ODD

            % https://fr.wikipedia.org/wiki/Formule_autor%C3%A9f%C3%A9rente_de_Tupper

            % xintiiexpr notations

            % floored division is //
            % associated modulo is /: (or 'mod')
            % divmod(,) is both // and /: (as in Python)

            begindocument

            % 0 <= x < 106
            % k <= y < k + 17

            xintdefiivar TupperK := 
            960939379918958884971672962127852754715004339660129306651505519271702802395266
            424689642842174350718121267153782770623355993237280874144307891325963941337723
            487857735749823926629715517173716995165232890538221612403238855866184013235585
            136048828693337902491454229288667081096184496091705183454067827731551705405381
            627380967602565625016981482083418783163849115590225610003652351370343874461848
            378737238198224849863465033159410054974700593138339226497249461751545728366702
            369745461014655997933798537483143786841806593422227898388722980000748404719;

            setlengthunitlength1pt

            xintdefiivar twoToThe17 := 2**17;
            xintdefiivar twoToThe17timesX := 1;

            beginpicture(106,17)
             xintFor* #1 in xintSeq0105do % x
             xintFor* #2 in 012345678910111213141516do % y - k
             xintdefiivar Yq, Yr = divmod(#2 + TupperK, 17);%
             xintifbooliiexpr odd(Yq // (twoToThe17timesX * 2**Yr ))
             put(numexpr105-#1, numexpr16-#2)rule1pt1pt
             
             
             xintdefiivar twoToThe17timesX := twoToThe17 * twoToThe17timesX;%
             
            endpicture

            enddocument


            would be faster, but it is only a bit faster. Hence the bottleneck is in the big divisions, less so in the powers of two.



            It would be faster to cheat a bit and compute binary representation of k via xintDecToBin from xintbinhex for example.







            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited 6 mins ago

























            answered 35 mins ago









            jfbu

            42k63135




            42k63135











            • loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
              – jfbu
              31 mins ago










            • provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
              – jfbu
              14 mins ago

















            • loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
              – jfbu
              31 mins ago










            • provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
              – jfbu
              14 mins ago
















            loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
            – jfbu
            31 mins ago




            loop is very inefficient because for each #1=x, on computes again and again same power of 2 seventeen times, I will upload improved code.
            – jfbu
            31 mins ago












            provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
            – jfbu
            14 mins ago





            provides a good test of xint big integers capabilities... but too slow for incorporating into test suite...
            – jfbu
            14 mins ago











            Sora2455 is a new contributor. Be nice, and check out our Code of Conduct.









             

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