**Babylonian  Brothers Inheritance Problems Algorithm  and eTCL demo example calculator, numerical analysis**

This page is under development. Comments are welcome, but please load any comments in the comments section at the bottom of the page. Please include your wiki MONIKER  in your comment with the same courtesy that I will give you. Its very hard to reply intelligibly without some background of the correspondent. Thanks,[gold]
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<<TOC>>

[gold] Here is some eTCL starter  code for  Brothers Inheritance Problems Algorithm  in calculator shell. The initial console program was used to check the false shares concept and generate testcases before loading the calculator shell.

The eTCL calculator used Babylonian single false position methods for the brother inheritance problem, but that is probably not the only method the Babylonians could use. The Babylonians did not use algebra notation, so the reader will have to bear some anachronisms in the eTCL pseudocode. Generally, the brothers inherit land or silver pieces in arithmetic progression with the eldest brother receiving the greatest share and the youngest brother the least. In the extant problems, the number (n) of brothers range from 3 to 10 brothers. Each brother received an integral fraction of his next elder in a stair step fashion. The nominal fraction was 1/<integer number of  brothers>.  The share of the younger brothers would be x-x/n or x*(1-1/n) in succession. The normal solution was presented from elder brother to youngest brother and lowest amount, down the staircase. However, there are some brother inheritance solutions that are presented  youngest  to elder brother, up the staircase. In some cases, the beginning or setup of the problem is available, without a worked method of solution. The most consistent error check should be that the sum of the brother shares should equal the original bequest. 

The eTCL calculator computes  a false share for each brother, sums the false shares, and rescales the false shares for the solution. Since the Babylonian problem is set up for integer fractions, the initial false share for elder brother is set to be factored by 1/n. Allowing for multiplication by 1/n or dividing n times for the number (n) of the brothers, the initial false share is set to n**n in the eTCL code (for the eldest brother). From an initial false share of n**n, the share x*(1-1/n) for successive brothers is calculated in an iterative solution. The stored shares are summed and the scale factor used for the false shares is real_estate over the sum of false shares.

Algebraic peg points for the brother shares and solution can be developed. The ratio of brother n-1 over n is   < x*(1-1/n)*(1-1/n)  > /  <x*(1-1/n)> . The share of brother n-1 is  ratio < x*(1-1/n)*(1-1/n)  > /  <x*(1-1/n)>  times the share of brother n.   The average  of brother n-1 and n is << x*(1-1/n)*(1-1/n)  > *  <x*(1-1/n)> > / 2.  The average of two brothers shares reduces to (1-1/n)/2. The average share for all brothers is real_estate / number of brothers, which should be in order of magnitude with the elder shares.

For the initial testcase of 5 brothers, some approximate formulas were developed. The average share for all 5 brothers  was 60/5 or 12. The exact fraction for younger brother5 was <expr (1.-1./5.)*(1.-1./5.)*(1.-1./5.)*(1.-1./5.)> or 0.40960. The share of younger brother5 was  elder brother1 times 0.409 was 7.2802 ( for youngest brother5). The average share between eldest and youngest is  (17.85+7.3)/2 or 12.575. Within an order of magnitude , the average share between eldest and youngest  approximates the average share for all 5 brothers. By observation, the  average share for all 5 brothers  equaled  int (average share between eldest and youngest brother ). The ratios for real_estate/ number_brothers approximated (brother1 + brother5) /2. Rearranging  terms,  brother1 approximated 2*(real_estate/ number_brothers ) - brother5. Rearranging terms,  brother5 approximated 2*(real_estate/ number_brothers ) - brother1.


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**Pseudocode Section**
======
       # using pseudocode 
       # possible problem instances, 3 to 10 brothers
      initialize algorithm_result = 1.
      nominal fraction was 1/<integer number of brothers>
      eldest share is n**n, so factorable n times in Babylonian integer math
      share of the younger brothers would be x-x/n or x*(1-1/n) in succession.
      ??? code [expr xxx*(1.-1./5.)*(1.-1./5.)*(1.-1./5.)*(1.-1./5.)*(1.-1./5.)], ?.32768?
      scale factor = real_estate over the sum of false shares
      check algorithm 
      check_sum = a+b+c+d+e = original real estate
      set answers and printout with resulting values 
      pseudocode: need test cases > small,medium, giant 
      pseudocode: need testcases within range of expected operation.
      pseudocode: are there any cases too small or large to be solved?
      [expr (1.-1./5.)*(1.-1./5.)*(1.-1./5.)*(1.-1./5.)] =0.40960
      elder brother1 * 0.409 = 7.2802 ( youngest brother5)
      average share for all 5 brothers  =  60/5 = 12
      average share between eldest and youngest is  (17.85+7.3)/2 or 12.575
      theorem that  average share for all 5 brothers  =
      int (average share between eldest and youngest )
      within order of magnitude , average share between eldest and youngest
      ~~ average share for all 5 brothers  
      real_estate/ number_brothers ~~~ (brother1 + brother5) /2
      brother1 ~~ 2*(real_estate/ number_brothers ) - brother5
      brother5 ~~ 2*(real_estate/ number_brothers ) - brother1

======
***Testcases Section***  

In planning any software, it is advisable to gather a number of testcases to check the results of the program. The math for the testcases can be checked by pasting statements in the TCL console. Aside from the TCL calculator display, when one presses the report button on the calculator, one will have console show access to the   capacity  functions (subroutines). 

**** Testcase 1 ****
%|table 1|printed in| tcl wiki format|% 
&| quantity| value| comment, if any|& 
&| 1:|testcase_number | |& 
&| 60.0 :|initial real estate   |   |&
&| 5.0 :|number of brothers | |& 
&| 5.0 :|less fraction 1/b (usually number of brothers) | |& 
&| 17.848 :|answers: elder brother   | |&
&| 14.278 :|2nd brother | |&
&| 11.423 :|3ird brother |  |&
&| 9.138 :|4th brother  |  |&
&| 7.310 :|5th brother  |  |&
  
**** Testcase 2 ****
&| 2:|testcase_number | |& 
&| 120.0 :|initial real estate   |   |&
&| 5.0 :|number of brothers | |& 
&| 5.0 :|less fraction 1/b (usually number of brothers) | |& 
&| 35.697 :|answers: elder brother   | |&
&| 28.557 :|2nd brother | |&
&| 22.846 :|3ird brother |  |&
&| 18.277 :|4th brother  |  |&
&| 14.621 :|5th brother  |  |&
 
**** Testcase 3 ****
&| 3:|testcase_number | |& 
&| 3600.0 :|initial real estate   |   |&
&| 5.0 :|number of brothers | |& 
&| 5.0 :|less fraction 1/b (usually number of brothers) | |& 
&| 1070.918 :|answers: elder brother   | |&
&| 856.734 :|2nd brother | |&
&| 685.387 :|3ird brother |  |&
&| 548.310 :|4th brother  |  |&
&| 438.648 :|5th brother  |  |&
 
**** Testcase 4 ****
======
 4 brothers receive 60? at fraction (1-1/7)
 elder brother has 7*3 for false share.
 brother brother4 inherits  294.0 (elder brother)
 brother brother3 inherits  252.0
 brother brother2 inherits  216.0
 brother brother1 inherits  185.142 (youngest brother)
 sum of false shares is 1105.836 
 scale factor for false shares is 0.0332
 elder brother's share is (7**3)*(36.8 / 1105.836) answer = 11.414
 younger brother's share is last*(36.8 / 1105.836) answer = 5.281
 #  hand calculations follow  (good)
 [expr (1.-1./7.)*(1.-1./7.)*(1.-1./7.)]= 0.6297
 youngest brother4 = [expr  0.6297* 11.433]= 7.1993
 1st brother[expr (5.*60+43.)*(2./60.)] = 11.433
 2nd brother=[expr (4.*60+54.)*(2./60.)]= 9.8
 3ird brother=[expr (4.*60+12.)*(2./60.)]=8.4
 4th brother=[expr (3.*60+36.)*(2./60.)]= 7.2
 [expr 11.43+9.8+8.4+7.2] = 36.830
======
**** Testcase 4 ****
======
 10 brothers receive 100? at fraction (1-1/10)?
 brother brother10 inherits  900.0  false shares
 brother brother9 inherits  810.0  false shares
 brother brother8 inherits  729.0  false shares
 brother brother7 inherits  656.10000000000002  false shares
 brother brother6 inherits  590.49000000000001  false shares
 brother brother5 inherits  531.44100000000003  false shares
 brother brother4 inherits  478.29690000000005  false shares
 brother brother3 inherits  430.46721000000002  false shares
 brother brother2 inherits  387.42048900000003  false shares
 brother brother1 inherits  348.67844010000005  false shares
 sum of false shares is 5861.8940391000006 
 scale factor for false shares is 0.017059332586529214
 elder brother's share is (10**3)*(100. / 5861.894) 17.059332586529216
 younger brother's share is last*(100. / 5861.894) 5.9482214754181051
 #  hand calculations follow  (good)
 average = 120/10 = 12
 [expr (1.-1./10.)*(1.-1./10.)*(1.-1./10.)*(1.-1./10.)*(1.-1./10.)*(1.-1./10.)*(1.-1./10.)*(1.-1./10.)*(1.-1./10.)]= 
 0.38742 = factor for youngest brother.
 8th brother cited as 8 silver pieces
 8th brother calculated as 430.46*0.01705
 or 7.3 silver pieces as computer solution stands above.
 elder brother1 approximates [expr 2*(100./10.)-3.] or 17, order of magnitude
 This problem has factor 1+1/n omitted or obscured, 
 Another problem cited factor of eleventh part or algebraic 1+1/11.
 Maybe this factor should be 11 instead of 10.
 Note that the 1+1/n factor may be set at a prime number, 
 even when there are 4 brothers, the factor is seventh or 1+1/7. 
======
**** Testcase 5 ****
======
 4 brothers receive 72? at fraction (1-1/6)?
 This problem has the inheritance omitted or obscured. 
 However, almost certain number is 1 mina (60 bucks) and some 1/6 fraction multiple,
 like 72 silver pieces. Solution is not repeat not given.
 brother brother4 inherits  213.333  false shares
 brother brother3 inherits  177.777  false shares
 brother brother2 inherits  148.148  false shares
 brother brother1 inherits  123.456  false shares
 sum of false shares is 662.716
 scale factor for false shares is 0.108
 elder brother's share is (10**3)*(72. / 662.716)
 27.812 silver pieces for elder brother
 younger brother's share is last*(72. / 662.716)
 13.412 silver pieces for youngest brother
 #  hand calculations follow  (good)
 average = 72/4 =  18 silver pieces
 [expr (1.-1./6.)*(1.-1./6.)*(1.-1./6.)] or fraction 0.5787037 for youngest brother
 elder brother1 approximates [expr 2*(72./4.)-3.] or 33 silver pieces, order of magnitude
 younger brother4 approximates  [expr  33*.5787] or 19.097 silver pieces, order of magnitude
======
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***Screenshots Section***
****figure 1.****

[Babylonian Brothers Inheritance Problems Algorithm and eTCL demo example screenshot png]

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***References:***
   * The Algebra of Inheritance: A Rehabilitation of Al-Khuwarizmi,
   * Solomon Gandz Source: Osiris, Vol. 5 (1938), pp. 319-391 
   * A Note to Kazuo Muroi,  Inheritance Problems in
   * Susa Mathematical Text No. 26, Jens Hoyrupt
   * Mathematical Cuneiform Texts,  Neugebauer and Sachs
   * Extraction of Cube Roots in Babylonian Mathematics, Kazuo Muroi, Centaurus Volume 31, issue 3, 1988
   * Babylonian Mathematical Texts II-III Author(s): A. Sachs Source: Journal of Cuneiform Studies, Vol. 6, No. 4   
   * (1952), pp. 151-156 Published by: The American Schools of Oriental Research 
   * Computing the Cube Root,  Ken Turkowski, Apple Computer Technical Report #KT-32 10 February 1998
   * Approximating Square Roots and Cube Roots , Ali Ibrahim Hussenom,  2014/11/04
   * Aryabhata’s Root Extraction Methods, Abhishek Parakh , Louisiana State University,  Aug 31st 2006  
   * Another Method for Extracting Cube Roots, Brian J. Shelburne, 
   * Dept of Math and Computer, Science Wittenberg University 
   * Jeanette C. Fincke* and Mathieu Ossendrijver* BM 46550 – a Late Babylonian  Mathematical Tablet with 
   * Computations of Reciprocal Numbers,Zeitschrift für Assyriologie 2016; 106(2): 185–197
   * Interpretation of reverse algorithms in several mesopotamian texts, Christine Proust
   * A Geometric Algorithm with Solutions to Quadratic Equations
   * in a Sumerian Juridical Document from Ur III Umma
   * Joran Friberg,  Chalmers University of Technology, Gothenburg, Sweden
   * google search engine <Trapezoid area bisection> 
   * Wikipedia search engine <Trapezoid area > 
   * mathworld.wolfram.com, Trapezoid and right trapezoid
   * Mathematical Treasure: Old Babylonian Area Calculation, uses ancient method
   * Frank J. Swetz , Pennsylvania State University
   * Wikipedia, see temple of Edfu, area method used as late as 200 BC in Egypt.
   * [Oneliner's Pie in the Sky] 
   * [One Liners] 
   * [Category Algorithm]
   * [Babylonian Number Series and eTCL demo example calculator]
   * [Brahmagupta Area of Cyclic Quadrilateral and eTCL demo example calculator]
   * [Gauss Approximate Number of Primes and eTCL demo example calculator]
   * Land surveying in ancient Mesopotamia, M. A. R. Cooper
   * [Sumerian Approximate Area Quadrilateral and eTCL Slot Calculator Demo Example , numerical analysis] 
   * Thomas G. Edwards, Using the Ancient Method of False Position to Find Solutions
   * Joy B. Easton, rule of double false position
   * Vera Sanford, rule of false position
   * www.britannica.com, topic, mathematics trapezoid
   * [Sumerian Equivalency Values, Ratios, and the Law of Proportions with Demo Example Calculator]
   * [Babylonian Sexagesimal Notation for Math on Clay Tablets in Console Example] 
   * Babylonians Tracked Jupiter With Advanced Tools: Trapezoids, Michael Greshko, news.nationalgeographic.com
   * Geometry in Babylonian Astronomy, Cluster of Excellence Topology, Humboldt University of Berlin
   * Mathieu Ossendrijver: „Ancient Babylonian astronomers calculated Jupiter’s position
   * from the area under a time-velocity graph“, in: Science, January 29, 2016.
   * Late Babylonian Field Plans in the British Museum, books.google.com/books 
   * Karen Rhea Nemet-Nejat
   * Late Babylonian Surface Mensuration Author(s): Marvin A. Powell Source: jstor
   * translation:  trapezoid in two babylonian astronomical cuneiform
   * texts for jupiter (act 813 & act 817) from the seleucid era , 310 BC -75 AD 
   * Otto Neugebauer, Astronomical Cuneiform Texts, 3 Vols. 
   * Lund Humphreys, London, 1955:405,430-31. 
   * DeSegnac, MS 3908 A RE-CONSTRUCTION,  D.A.R. DeSegnac
   * A draft for an essay 
   * DeSegnac, MENTAL COMPUTING OF THREE ARCHAIC
   * MESOPOTAMIAN PUZZLES W 20044, 35, W 20044, 20 & W 20214, essay draft
   * DeSegnac, HARMONY OF NUMBERS I and II,  D.A.R. DeSegnac, A draft for an essay
----
**Appendix Code**

***appendix TCL programs and scripts ***

======
        # pretty print from autoindent and ased editor
        # Babylonian Brothers Inheritance Problems Algorithm calculator
        # written on Windows XP on eTCL
        # working under TCL version 8.5.6 and 1.0.1
        # gold on TCL WIKI, 25jan2017
        package require Tk
        package require math::numtheory
        namespace path {::tcl::mathop ::tcl::mathfunc math::numtheory }
        set tcl_precision 17
        frame .frame -relief flat -bg aquamarine4
        pack .frame -side top -fill y -anchor center
        set names {{} { initial estate  :} }
        lappend names { number of brothers (b) :}
        lappend names { next brother etc, his elder share less fraction (usually 1/b ) : }
        lappend names { answers: elder brother : }
        lappend names { 2nd brother :}
        lappend names { 3ird brother : }
        lappend names { 4th brother : }
        lappend names { 5th brother   :} 
        foreach i {1 2 3 4 5 6 7 8} {
            label .frame.label$i -text [lindex $names $i] -anchor e
            entry .frame.entry$i -width 35 -textvariable side$i
            grid .frame.label$i .frame.entry$i -sticky ew -pady 2 -padx 1 }
        proc about {} {
            set msg "Calculator for Babylonian Brothers Inheritance Problems Algorithm 
            from TCL WIKI,
            written on eTCL "
            tk_messageBox -title "About" -message $msg } 
       proc calculate {     } {
            global answer2
            global side1 side2 side3 side4 side5
            global side6 side7 side8 
            global testcase_number 
            incr testcase_number 
            set side1 [* $side1 1. ]
            set side2 [* $side2 1. ]
            set side3 [* $side3 1. ]
            set side4 [* $side4 1. ]
            set side5 [* $side5 1. ]
            set side6 [* $side6 1. ]
            set side7 [* $side7 1. ]
            set side8 [* $side8 1. ] 
            set real_estate $side1 
            set number_of_brothers $side2    
            set fraction2 $side3      
            # initialize placeholder answer
            set result 1.
            set side4 0.
            set side5 0.
            set side6 0.
            set side7 0.
            set side8 0.
        set brothers $number_of_brothers 
        set decrement [/ 1. $brothers ] 
        set inheritance [** $number_of_brothers $number_of_brothers]
        set keeper [int [- $number_of_brothers 1] ]
        set i [int  $number_of_brothers ]
        set total_false_position 0.
        while {$i>0} {
        set inheritance [- $inheritance [* $inheritance $decrement]]
        if { $inheritance < 0 } { break } 
        set side$keeper $inheritance
        set total_false_position [+ $total_false_position $inheritance  ]
        incr keeper 1
        incr i -1
        }
            set scale_factor [/ $real_estate $total_false_position ]
            set side4 [* $scale_factor $side4 ]
            set side5 [* $scale_factor $side5 ]
            set side6 [* $scale_factor $side6 ]
            set side7 [* $scale_factor $side7 ]
            set side8 [* $scale_factor $side8 ]
        }
        proc fillup {aa bb cc dd ee ff gg hh} {
            .frame.entry1 insert 0 "$aa"
            .frame.entry2 insert 0 "$bb"
            .frame.entry3 insert 0 "$cc"
            .frame.entry4 insert 0 "$dd"
            .frame.entry5 insert 0 "$ee"
            .frame.entry6 insert 0 "$ff" 
            .frame.entry7 insert 0 "$gg"
            .frame.entry8 insert 0 "$hh" 
             }
        proc clearx {} {
            foreach i {1 2 3 4 5 6 7 8 } {
                .frame.entry$i delete 0 end } }
        proc reportx {} {
            global side1 side2 side3 side4 side5
            global side6 side7 side8
            global testcase_number 
            console show;
            puts "%|table $testcase_number|printed in| tcl wiki format|% "
            puts "&| quantity| value| comment, if any|& "
            puts "&| $testcase_number:|testcase_number | |& "
            puts "&| $side1 :|initial real estate   |   |&"
            puts "&| $side2 :|number of brothers | |& "  
            puts "&| $side3 :|less fraction 1/b (usually number of brothers) | |& "
            puts "&| $side4 :|answers: elder brother   | |&"
            puts "&| $side5 :|2nd brother | |&"
            puts "&| $side6 :|3ird brother |  |&"
            puts "&| $side7 :|4th brother  |  |&"
            puts "&| $side8 :|5th brother  |  |&" 
            }
        frame .buttons -bg aquamarine4
        ::ttk::button .calculator -text "Solve" -command { calculate   }
        ::ttk::button .test2 -text "Testcase1" -command {clearx;fillup 60.   5.   5.0  17.  14.  11. 9. 7.0}
        ::ttk::button .test3 -text "Testcase2" -command {clearx;fillup 120.  5.0  5.0  35.  28.  22. 18. 14.0 }
        ::ttk::button .test4 -text "Testcase3" -command {clearx;fillup 3600. 5.0  5.0  1070. 856.  685. 548. 438.0 }
        ::ttk::button .clearallx -text clear -command {clearx }
        ::ttk::button .about -text about -command {about}
        ::ttk::button .cons -text report -command { reportx }
        ::ttk::button .exit -text exit -command {exit}
        pack .calculator  -in .buttons -side top -padx 10 -pady 5
        pack  .clearallx .cons .about .exit .test4 .test3 .test2   -side bottom -in .buttons
        grid .frame .buttons -sticky ns -pady {0 10}
               . configure -background aquamarine4 -highlightcolor brown -relief raised -border 30
        wm title . "Babylonian Brothers Inheritance Problems Algorithm"   
======
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*** Pushbutton Operation***

For the push buttons, the recommended procedure is push testcase and fill frame, change first three entries etc, push solve, and then push report. Report allows copy and paste from console. 

For  testcases in a computer session, the eTCL calculator increments a new testcase number internally, eg. TC(1), TC(2) , TC(3) , TC(N). The testcase number is internal to the calculator and will not be printed until the report button is pushed for the current result numbers. The current result numbers will be cleared on the next solve button.   The command { calculate; reportx  } or { calculate ; reportx; clearx  } can be added or changed to report automatically. Another wrinkle would be to print out the current text,  delimiters,  and numbers in a TCL wiki style table as

======
  puts " %| testcase $testcase_number | value| units |comment |%"
  puts " &| volume| $volume| cubic meters |based on length $side1 and width $side2   |&"  
======
----
** Console program for testing concept **
======
        # pretty print from autoindent and ased editor
        # initial console program,  Babylonian Brothers false shares
        # written on Windows XP on eTCL
        # working under TCL version 8.5.6 and eTCL 1.0.1
        # gold on TCL WIKI, 10feb2017
        package require Tk
        package require math::numtheory
        namespace path {::tcl::mathop ::tcl::mathfunc math::numtheory }
        set tcl_precision 17
        console show
        set brothers 5
        set i 5   
        set decrement [/ 1. $brothers ] 
        set inheritance  [expr 5**5]
        set false_sum 0
        set real_estate 60.
        while {$i>0} {
        set inheritance [- $inheritance [* $inheritance $decrement]]
        if { $inheritance < 0 } { break } 
        puts "brother brother$i inherits  $inheritance"
        set side$i $inheritance
        set false_sum [+ $false_sum $inheritance]
        incr i -1
        }
        puts " sum of false shares is $false_sum "
        puts " scale factor for false shares is [expr $real_estate / $false_sum ]"
        puts " elder brother's share is (5**5)*($real_estate / $false_sum) [expr (5**5)*($real_estate / $false_sum) ]"
        puts " younger brother's share is last*($real_estate / $false_sum) [expr $inheritance *($real_estate / $false_sum) ]"
======
output for false shares
======
brother brother5 inherits  2500.0
brother brother4 inherits  2000.0
brother brother3 inherits  1600.0
brother brother2 inherits  1280.0
brother brother1 inherits  1024.0
 sum of false shares is 8404.0 
 scale factor for false shares is 0.00713
 elder brother's share is (5**5)*(60. / 8404.0) 22.310
 younger brother's share is last*(60. / 8404.0) 7.3108
======
[gold] This page is copyrighted under the TCL/TK license terms, [http://tcl.tk/software/tcltk/license.html%|%this license]. 

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