How to resolve the algorithm Fractran step by step in the TI-83 BASIC programming language

Published on 12 May 2024 09:40 PM

How to resolve the algorithm Fractran step by step in the TI-83 BASIC programming language

Table of Contents

Problem Statement

FRACTRAN is a Turing-complete esoteric programming language invented by the mathematician John Horton Conway. A FRACTRAN program is an ordered list of positive fractions

P

(

f

1

,

f

2

, … ,

f

m

)

{\displaystyle P=(f_{1},f_{2},\ldots ,f_{m})}

, together with an initial positive integer input

n

{\displaystyle n}

.

The program is run by updating the integer

n

{\displaystyle n}

as follows:

Conway gave a program for primes in FRACTRAN: Starting with

n

2

{\displaystyle n=2}

, this FRACTRAN program will change

n

{\displaystyle n}

to

15

2 × ( 15

/

2 )

{\displaystyle 15=2\times (15/2)}

, then

825

15 × ( 55

/

1 )

{\displaystyle 825=15\times (55/1)}

, generating the following sequence of integers: After 2, this sequence contains the following powers of 2: which are the prime powers of 2.

Write a program that reads a list of fractions in a natural format from the keyboard or from a string, to parse it into a sequence of fractions (i.e. two integers), and runs the FRACTRAN starting from a provided integer, writing the result at each step.
It is also required that the number of steps is limited (by a parameter easy to find).

Use this program to derive the first 20 or so prime numbers.

For more on how to program FRACTRAN as a universal programming language, see:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Fractran step by step in the TI-83 BASIC programming language

Source code in the ti-83 programming language

100->T
2->N
{17,78,19,23,29,77,95,77, 1,11,13,15,15,55}->LA
{91,85,51,38,33,29,23,19,17,13,11,14, 2, 1}->LB
Dim(LA)->U
T->Dim(LC)
For(I,1,T)
	1->J: 1->F
	While J<=U and F=1
		If remainder(N,LB(J))=0
		Then
			Disp N 
			N->LC(I)
			iPart(N/LB(J))*LA(J)->N 
			0->F
		End
		J+1->J
	End
End

  

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