[tbas] Chinese Year of the ...

The author of tbas, Antonio Maschio, loves DATA and READ statements, and after putting together a Chinese Zodiac submission for RosettaCode, I can see why. And then there's TAB() which is also very helpful.

DATA "甲","乙","丙","丁","戊","己","庚","辛","壬","癸"
 DECLARE celestial$(10)
 MAT READ celestial$
 
 DATA "子","丑","寅","卯","辰","巳","午","未","申","酉","戌","亥"
 DECLARE terrestrial$(12)
 MAT READ terrestrial$
 
 DATA "Rat","Ox","Tiger","Rabbit","Dragon","Snake","Horse","Goat","Monkey","Rooster","Dog","Pig"
 DECLARE animals$(12)
 MAT READ animals$
 
        DATA "Wood","Fire","Earth","Metal","Water"
 DECLARE elements$(5)
 MAT READ elements$
 
 DATA "yang","yin"
 DECLARE aspects$(2)
 MAT READ aspects$
 
 DATA "jiă","yĭ","bĭng","dīng","wù","jĭ","gēng","xīn","rén","gŭi"
 DATA "zĭ","chŏu","yín","măo","chén","sì","wŭ","wèi","shēn","yŏu","xū","hài"
 DECLARE celestialpinyin$(UBOUND(celestial$(),1))
 DECLARE terrestrialpinyin$(UBOUND(terrestrial$(),1))
 MAT READ celestialpinyin$
 MAT READ terrestrialpinyin$
 
 DATA 1935,1938,1931,1961,1963,1991,1993,1996,2001
 DECLARE years(9)
 MAT READ years
 
 DECLARE _base = 4  
 DECLARE _year 
 DECLARE cycleyear 
 DECLARE stemnumber 
 DECLARE stemhan$    
 DECLARE stempinyin$ 
 DECLARE elementnumber 
 DECLARE element$       
 DECLARE branchnumber 
 DECLARE branchhan$    
 DECLARE branchpinyin$ 
 DECLARE animal$       
 DECLARE aspectnumber 
 DECLARE aspect$       
 DECLARE index 
 
 DECLARE i 
 DECLARE top = UBOUND(years(),1)
 FOR i = 1 TO top
  _year = years(i)
  cycleyear = _year - _base
  stemnumber = MOD(cycleyear, 10) 
  stemhan$    = celestial$(stemnumber + 1)
  stempinyin$ = celestialpinyin$(stemnumber + 1)
  elementnumber = div(stemnumber, 2) + 1
  element$       = elements$(elementnumber)
  branchnumber = MOD(cycleyear, 12)  
  branchhan$    = terrestrial$(branchnumber + 1)
  branchpinyin$ = terrestrialpinyin$(branchnumber + 1)
  animal$       = animals$(branchnumber + 1)
  aspectnumber = MOD(cycleyear, 2)
  aspect$       = aspects$(aspectnumber + 1)
  index = MOD(cycleyear, 60) + 1  
  PRINT _year; 
  PRINT TAB(5);stemhan$+branchhan$;
  PRINT TAB(12);stempinyin$;"-";branchpinyin$;
  PRINT TAB(25);element$;" ";animal$;" ("+aspect$+")";
  PRINT TAB(50);"year";index;"of the cycle"  
 NEXT

Running the program gives

$ tbas chinZod.bas
 1935 乙亥 yĭ-hài     Wood Pig (yin)           year 12 of the cycle
 1938 戊寅 wù-yín     Earth Tiger (yang)       year 15 of the cycle
 1931 辛未 xīn-wèi    Metal Goat (yin)         year 8 of the cycle
 1961 辛丑 xīn-chŏu   Metal Ox (yin)           year 38 of the cycle
 1963 癸卯 gŭi-măo    Water Rabbit (yin)       year 40 of the cycle
 1991 辛未 xīn-wèi    Metal Goat (yin)         year 8 of the cycle
 1993 癸酉 gŭi-yŏu    Water Rooster (yin)      year 10 of the cycle
 1996 丙子 bĭng-zĭ    Fire Rat (yang)          year 13 of the cycle
 2001 辛巳 xīn-sì     Metal Snake (yin)        year 18 of the cycle

© Copyright Bruce M. Axtens, 2018

[tbas] tbas basic interpreter

In my ample (NOT!) spare time, I'm helping out with a programming language project called tbas. I saw it advertised on comp.lang.basic.misc in September 2017 (yep, usenet is still going strong.)

The author is one Antonio Maschio from Italy.

You can get source, documentation and a CygWin binary from the main site. There are some examples on RosettaCode.

To give you an idea of the flavour of the language, the following is a simple CAESAR cipher implementation.

    LET MSG$ = ""
    LET OFFS = 0
    LINE INPUT "Message"; MSG$
    INPUT "Offset"; OFFS
    DIM TARG(LEN(MSG$))
    CHANGE MSG$ TO TARG
    FOR I = 1 TO UDIM(TARG)
        TARG(I) = TARG(I) + OFFS
    NEXT
    CHANGE TARG TO MSG$
    PRINT MSG$

An example of running the above

    >tbas CAESAR.BAS

    Message ? My dog's got fleas

    Offset ? 1

    Nz!eph(t!hpu!gmfbt

© Copyright Bruce M. Axtens, 2018

[COBOL] Sum multiples of 3 and 5

Last night I posted the following code on RosettaCode as another solution to the Sum multiples of 3 and 5 challenge. (Mine is the third one down after all the more-mathematical solutions.)

My solution is as brute-force as it gets, using only adds and comparisons.

       IDENTIFICATION DIVISION.
       PROGRAM-ID. SUM35.

       DATA DIVISION.
       WORKING-STORAGE SECTION.
       01  THREE-COUNTER   USAGE BINARY-CHAR value 1.
           88 IS-THREE VALUE 3.
       01  FIVE-COUNTER    USAGE BINARY-CHAR value 1.
           88 IS-FIVE VALUE 5.
       01  SUMMER          USAGE BINARY-DOUBLE value zero. 
       01  I               USAGE BINARY-LONG.
       01  N               USAGE BINARY-LONG.

       PROCEDURE DIVISION.
       10-MAIN-PROCEDURE.
           MOVE 1000000000 TO N.
           MOVE 1 TO I.
           PERFORM 20-INNER-LOOP WITH TEST AFTER UNTIL I >= N.
           DISPLAY SUMMER.
           STOP RUN.
       20-INNER-LOOP.
           IF IS-THREE OR IS-FIVE 
               ADD I TO SUMMER END-ADD
               IF IS-THREE
                   MOVE 1 TO THREE-COUNTER
               ELSE
                   ADD 1 TO THREE-COUNTER
               END-IF
               IF IS-FIVE
                   MOVE 1 TO FIVE-COUNTER
               ELSE    
                   ADD 1 TO FIVE-COUNTER
               END-IF
           ELSE
               ADD 1 TO FIVE-COUNTER END-ADD
               ADD 1 TO THREE-COUNTER END-ADD
           END-IF.
           ADD 1 TO I.
           EXIT.
       END PROGRAM SUM35.

The above code compiles on GnuCOBOL 2.0 and MicroFocus Visual COBOL 2.3. In the latter environment, I was able to get a run time of 7.3 seconds for the 1,000,000,000 iterations (AMD A6-5200 APU running at 2.00 GHz.)

I ended up in COBOL via the usual circuitous route through other programming languages after figuring out a solution using mLite. The next postings will demonstrate the mLite and perhaps others.

© Copyright Bruce M. Axtens., 2015

[Euphoria] Decimal to Binary

Having solved the Binary Digits task in mLite and FBSL, I thought I'd have a go in OpenEuphoria. Kudos to David Craig of Rapid Deployment Software for the original implementation and to the subsequent team of heroes for a very powerful Windows/Linux/Mac programming (interpreted and/or compiled) language.

The code below, expressed in v4.1.0 form, follows the FBSL fairly closely.

include std/math.e 
include std/convert.e

function Bin(integer n, sequence s = "")
  if n > 0 then
   return Bin(floor(n/2),(mod(n,2) + #30) & s)
  end if
  if length(s) = 0 then
   return to_integer("0")
  end if
  return to_integer(s)
end function

printf(1, "%d\n", Bin(0))
printf(1, "%d\n", Bin(5))
printf(1, "%d\n", Bin(50))
printf(1, "%d\n", Bin(9000))

There's already an Euphoria solution to the task. Some enterprising soul might want to check which of the two solutions is the faster.

© Copyright Bruce M. Axtens, 2015

[FBSL] Decimal to Binary

Having solved the Binary Digits task in mLite, I thought I'd have a go in FBSL. Kudos to the FBSL team for a very useful and powerful Windows scripting language. Even ABAP users can leverage it!

The code below, expressed in v3.5 RC form, follows the mLite fairly closely.

#AppType Console
function Bin(byval n as integer, byval s as string = "") as string
 if n > 0 then return Bin(n \ 2, (n mod 2) & s)
 if s = "" then return "0"
 return s
end function

print Bin(5)
print Bin(50)
print Bin(9000)

pause

It was during the testing of this code that it dawned on me that the mLite wasn't handling the binary 0; appropriately. Thus those who notice such things will discover a subtle change in the mLite posting.

© Copyright Bruce M. Axtens, 2015

[mLite] Convert to Binary Notation

A short diversion: the Standard ML dialect called mLite is where I go for the intellectual challenge of thinking in a functional programming way. Kudos to Nils M. Holm for developing it. mLite is one of the languages I use to solve RosettaCode tasks, in this case, Binary Digits.

This turned out to be a fairly simple task and only took me a few minutes to solve. Granted, I reused some code from the logarithm calculator of a few postings ago, but that was just for the conversion of a list of numbers to their printable form.

fun binary
   (0, []) = "0"
 | (0, x) = implode ` map (fn x = if int x then chr (x + 48) else x) x
 | (n, b) = binary (n div 2, n mod 2 :: b)
 | n = binary (n, [])
; 

The gist of that is

• pass in a number and recurse with the number and an empty list.
  
• With a number that isn't zero recurse with the number divved by 2 and the list prepended with the number modded by 2.
  
• Keep going until the number is zero. At that point, convert the answer's digits to string representation and implode them into a single string.
  
• If the number is zero and there are no digits, return "0"

Might get back to some VB6 next. Or javascript. © Copyright Bruce M. Axtens, 2015

[mLite] Arbitrary Precision Integers and Logarithms

Yeah yeah, there's something 'bout you baby I like

-- Status Quo, 1981.

It's a bit worrisome when I quote musicians from last century, but I do like mLite. Granted, I don't dream about it, but it's fun to figure out how to do things in a functional way.

In this case I was wanting to solve the RosettaCode challenge Arbitrary-precision integers (included) which has one calculate 5^4^3^2, find out how many digits there are and then display the top 20 digits and bottom 20 digits. Bignums weren't a challenge, as they are supported natively in mLite. No, the biggest challenge was figuring out how many digits. 5^4^3^2 is an extremely large number, with 183231 digits and all of my first attempts did not complete even after running for a day.

mLite's creator, Nils M. Holm, suggested I use logarithms. However, logarithms aren't built in to mLite yet, so I had to cook up my own logarithm library. So how do you calculate logarithms by hand? I dug around a bit and found an article on Quora entitled How can we calculate the logarithms by hand without using any calculator?. Harald Overbeek's description formed the basis for the mLite code below.

fun ntol (0, x) = if len x < 1 then [0] else x
       | (n, x) = ntol (n div 10, (n mod 10) :: x)
       | n      = ntol (n, [])
;

ntol converts a number to a list.

fun powers_of_10 9 = 1000000000
               | 8 = 100000000
               | 7 = 10000000
               | 6 = 1000000
               | 5 = 100000
               | 4 = 10000
               | 3 = 1000
               | 2 = 100
               | 1 = 10
               | 0 = 1
;

powers_of_10 is a precomputation of all the powers I knew I'd encounter during the calculation. This alone sped up the code a lot.

fun size (c, 0) = c
       | (c, n > 9999999999) = size (c + 10, trunc (n / 10000000000))
       | (c, n)              = size (c +  1, trunc (n / 10))
       | n                   = size (     0, trunc (n / 10))
;

size works out the number of digits by keeping track of how many calculations it takes to divide the number by 10 until the remainder is zero.

fun makeVisible L = map (fn x = if int x then chr (x + 48) else x) L

makeVisible turns an array of digits into a string, with handling for non-numeric elements

fun log10 (n, 0, x) = ston  implode  makeVisible ` rev x
        | (n, decimals, x) =
            let val n' = n^10;
              val size_n' = size n'
            in 
              log10 (n' / powers_of_10 size_n', decimals - 1, size_n' :: x)
   end
        | (n, decimals) =
            let
              val size_n = size n
            in
              log10 (n / 10^size_n, decimals, #"." :: rev (ntol size_n) @ [])
            end
;

Then log10 ties it all together. The second parameter specifies the number of digits precision. Below I've specified 6.

Being somewhat mathematically challenged, I still had to figure out how to use the library I had made to calculate the digits. Thankfully, there's a mathematics community on StackExchange. They helped me out a lot. Now I could work out the number of digits in the number by rounding up the result of log10(5) * 4^9 (seeing as 3^2 is 9).

Thus I ended up with

val fourThreeTwo = 4^3^2;
val fiveFourThreeTwo = 5^fourThreeTwo;

val digitCount = trunc (log10(5,6) * fourThreeTwo + 0.5);
print "Count  = "; println digitCount;

val end20 = fiveFourThreeTwo mod (10^20);
print "End 20 = "; println end20;

val top20 = fiveFourThreeTwo div (10^(digitCount - 20)); 
print "Top 20 = "; println top20;

Which, after 1 hour and 9 minutes on an AMD A6 cpu, gave

>mlite -f 5p4p3p2.m
 Count = 183231
 End 20 = 92256259918212890625
 Top 20 = 62060698786608744707 

A nice change from my day job which centres around JavaScript, Peloton, HTML, PHP and the odd bit of T-SQL. Enjoy!

© Copyright Bruce M. Axtens, 2015

[mLite] Learning an ML dialect

I've tried to get my head into functional languages for some time. This time I'm having a go at mLite, which the author, Nils M Holm, describes as "a lightweight (and slightly odd) inhabitant of the ML universe ... Much like ML, but with dynamic typing, guards, and a Haskell-style apply operator."

I've created a presence on RosettaCode for mLite and solved the Ackermann function challenge.

© Bruce M. Axtens, 2014.

[Batch File] Rediscovering CMD.EXE batch scripts

Insanity must be setting in; I’ve been trying to solve RosettaCode tasks with Windows CMD.EXE batch scripts. For example, their A+B task:

A+B - in programming contests, classic problem, which is given so contestants can gain familiarity with online judging system being used.

A+B is one of few problems on contests, which traditionally lacks fabula.

Problem statement Given 2 integer numbers, A and B. One needs to find their sum.

Input data

Two integer numbers are written in the input stream, separated by space.

Output data

The required output is one integer: the sum of A and B.

Example:

Input Output

2 2 4

3 2 5

These are what I posted as solutions:

Prompts version

::aplusb.cmd
@echo off
setlocal
set /p a="A: "
set /p b="B: "
set /a c=a+b
echo %c%
endlocal

All on the commandline version

::aplusb.cmd
@echo off
setlocal
set a=%1
set b=%2
set /a c=a+b
echo %c%
endlocal

Formula on the command line version

::aplusb.cmd
@echo off
setlocal
set /a c=%~1
echo %c%
endlocal

Example of 'Formula on the command line version'

>aplusb 123+456
579
>aplusb "1+999"
1000

Parse the input stream version (thanks to Tom Lavedas on alt.msdos.batch.nt)

::aplusb.cmd
@echo off
setlocal
set /p a="Input stream: "
call :add %a%
echo %res%
endlocal
goto :eof

:add
set /a res=res+%1
shift
if "%1" neq "" goto :add

Example of 'parse the input stream version'

>aplusb
Input stream: 1234 5678
6912
>aplusb
Input stream: 123 234 345 456 567 678 789 890
4082

Batch files are a bit arcane, it’s true, but the extensions added post Windows 2000 make it a lot more entertaining.

[VBScript] Small contributions to RosettaCode

I recently discovered RosettaCode. The initial blurb from the site is as follows:

Rosetta Code is a programming chrestomathy site. The idea is to present solutions to the same task in as many different languages as possible, to demonstrate how languages are similar and different, and to aid a person with a grounding in one approach to a problem in learning another. Rosetta Code currently has 370 tasks, and covers 195 languages, though we do not (and cannot) have solutions to every task in every language.

A variety of tasks are listed, and visitors to this site are invited to solve the tasks in the language of their choice. The tasks cover everything from the mundane Empty Program to the classic Towers of Hanoi, the practical User Input, the mathematically-inclined Lucas-Lehmer test, and the involved yet entertaining RCRPG.

I have made some small contributions to its collection of VBScript programs, as below:

99 Bottles of Beer Array concatenation Assertions Delete a file Execute a Markov algorithm Fibonacci sequence Flatten a list Function definition Generic swap Palindrome detection Pangram checker Program termination Sorting algorithms/Gnome sort Tokenize a string

Feel free to edit what I’ve done and add your own in whatever language takes your fancy.

© Copyright Bruce M. Axtens, 2010.