# CollatzASM

by Felix: Y12 Age ~16

## Introduction

The Collatz Conjecture is that you can start with any positive integer and end up with 1 by (1) halving each even number in the chain and (2) tripling each odd number then adding 1. (See the program comments). Felix had the good idea of writing in-line assembler code to determine the Collatz chain length of an integer supplied by the user. He also had the ability and perseverance to implement it! The program and the linked Wikipedia page both give the result that the number 27 has a chain length of 111. The program should output the correct chain length as long as no register overflows, and Felix assures us that it will work for all inputs up to 100000.

## The Program

```program CollatzASM;
{

use this file except in compliance with the License, as described at
}
{\$ASMMODE INTEL}
uses
SysUtils;
var
StartInt : Integer;
Mem1, Mem2, Mem3, Mem4 : Integer;
//Mem1 is N
//Mem2 is for determining if N is odd/even
//Mem3 is for halving N
//Mem4 is a running total for chain length
begin
//The following code finds the length of a given number's Collatz chain
//A Collatz chain is generated by the sequence;
//          If N is odd    N = (3*N) +1
//          If N is even   N = 1/2 * N
//This sequence continues until N = 1
Writeln('Enter an integer to find the length of its Collatz chain.');
asm
Mov EAX, StartInt
Mov Mem1, EAX              //Storing the given number into Mem1
@Start:
MOV Mem3, 1
MOV ECX, Mem1
MOV Mem2, ECX              //Mem2:=Mem1 for determing odd/even
CMP Mem1, 1                //Checks if N=1
JE @End
ADD Mem4, 1                //As N<>1, there will be one stage of sequence
@Mod2:
AND Mem2, 1                //If the rightmost bit is 1 (odd), Mem2 will become 1 else 0
CMP Mem2, 1
JE @Odd                    //This loop decreases Mem2 by 2 each time-
JL @Even                   //until is either 1 or 0, if 0 Mem1 is even, if 1 then Mem1 is odd
@Odd:
Mov ECX, Mem1
ADD ECX, Mem1              //You are sent here if Mem1 (N) is odd-
ADD ECX, Mem1              //we put Mem1 into ECX, and add Mem1 twice to it-
ADD ECX, 1                 //tripling it, we then add one and make Mem1 equal this
MOV Mem1, ECX
JMP @Start
@Even:
SHR  Mem1, 1               //Divide by 2 by shifting one place to the right
JMP @Start
@End:
end;
writeln(Mem4);               //Finally we output the number of cycles it took for Mem1=1