Welcome to the Ultimate Guide for Tennis M25 Vale do Lobo Portugal
If you're a tennis enthusiast or a betting aficionado, you've landed in the right place. The M25 category in Vale do Lobo, Portugal, offers an electrifying blend of talent and competition. Here, we provide daily updates on matches and expert betting predictions to keep you ahead of the game. Whether you're following for passion or profit, this guide will serve as your go-to resource.
Understanding the M25 Category
The M25 category is a pivotal part of the ATP Challenger Tour, featuring players ranked between 125 and 175 in the ATP rankings. It's a crucial stepping stone for players aiming to break into the top echelons of professional tennis. Matches are intense, showcasing emerging talents who are hungry to make their mark.
Why Vale do Lobo?
Located on the stunning Algarve coast, Vale do Lobo is not just a picturesque destination but also a hub for competitive tennis. The courts here offer excellent conditions that test players' adaptability and skill. The vibrant atmosphere and passionate local crowd add an extra layer of excitement to every match.
Daily Match Updates
Stay informed with our comprehensive daily updates. We cover every match in detail, providing insights into player performances, key moments, and standout plays. Our team ensures you have all the information needed to follow along or catch up on any missed action.
Expert Betting Predictions
Our expert analysts bring years of experience to the table, offering predictions that go beyond surface-level statistics. We delve into player form, head-to-head records, and even psychological factors to give you the most accurate betting advice possible.
Key Players to Watch
- Juan Martín del Potro: Known for his powerful forehand and resilience.
- Aslan Karatsev: A rising star with exceptional baseline play.
- Maria Sakkari: Bringing her aggressive style from WTA events.
- Casper Ruud: Consistent performance with strategic brilliance.
Tournament Structure and Format
The tournament follows a single-elimination format across four rounds: Qualifiers, First Round, Quarterfinals, Semifinals, and Finals. This structure ensures that only the best advance, making every match critical.
How to Follow Matches Live
- Social Media Updates: Follow our official channels for real-time updates and highlights.
- Tennis Apps: Use dedicated apps like Tennis TV or Eurosport Player for live streaming.
- Websites: Check out official tournament websites for detailed schedules and results.
skwak/AdventOfCode2019<|file_sep|>/Day_12/src/main.rs
use std::collections::HashMap;
use std::fs;
fn main() {
}
#[derive(Debug)]
struct Moon {
pub x: i32,
pub y: i32,
pub z: i32,
pub vx: i32,
pub vy: i32,
pub vz: i32,
}
impl Moon {
fn apply_gravity(&mut self) {
let mut dx = vec![];
let mut dy = vec![];
let mut dz = vec![];
for moon in &mut moons {
if moon.x > self.x { dx.push(1); } else if moon.x == self.x { dx.push(0); } else { dx.push(-1); }
if moon.y > self.y { dy.push(1); } else if moon.y == self.y { dy.push(0); } else { dy.push(-1); }
if moon.z > self.z { dz.push(1); } else if moon.z == self.z { dz.push(0); } else { dz.push(-1); }
}
self.vx += dx.iter().sum::();
self.vy += dy.iter().sum::();
self.vz += dz.iter().sum::();
}
fn apply_velocity(&mut self) {
self.x += self.vx;
self.y += self.vy;
self.z += self.vz;
}
fn energy(&self) -> usize {
let pe = (self.x.abs() + self.y.abs() + self.z.abs()) as usize;
let ke = (self.vx.abs() + self.vy.abs() + self.vz.abs()) as usize;
pe * ke
}
}
fn simulate(moons: &mut Vec) -> usize {
let mut total_energy = Vec::new();
for _ in (0..1000).rev() {
// Apply gravity
moons.clone().iter().for_each(|moon| {
moon.apply_gravity();
});
// Apply velocity
moons.clone().iter_mut().for_each(|moon| {
moon.apply_velocity();
});
// Calculate energy
moons.clone().iter().for_each(|moon| {
total_energy.push(moon.energy());
});
}
total_energy.iter().sum()
}
fn find_period(mut moons: Vec) -> usize {
let mut period_x = None;
let mut period_y = None;
let mut period_z = None;
let mut history_x : HashMap<(i32,i32),usize> = HashMap::new();
let mut history_y : HashMap<(i32,i32),usize> = HashMap::new();
let mut history_z : HashMap<(i32,i32),usize> = HashMap::new();
loop {
// Apply gravity
moons.clone().iter_mut().for_each(|moon| {
moon.apply_gravity();
});
// Apply velocity
moons.clone().iter_mut().for_each(|moon| {
moon.apply_velocity();
});
let sum_x : Vec<_>= moons.iter()
.map(|m| (m.x,m.vx))
.collect();
let sum_y : Vec<_>= moons.iter()
.map(|m| (m.y,m.vy))
.collect();
let sum_z : Vec<_>= moons.iter()
.map(|m| (m.z,m.vz))
.collect();
if let Some(period) = period_x.or_else(|| history_x.get(&(sum_x[0],sum_x[1]))){
break period
} else if !period_x.is_some(){
history_x.insert((sum_x[0],sum_x[1]), step);
}
if let Some(period) = period_y.or_else(|| history_y.get(&(sum_y[0],sum_y[1]))){
break period
} else if !period_y.is_some(){
history_y.insert((sum_y[0],sum_y[1]), step);
}
if let Some(period) = period_z.or_else(|| history_z.get(&(sum_z[0],sum_z[1]))){
break period
} else if !period_z.is_some(){
history_z.insert((sum_z[0],sum_z[1]), step);
}
step+=1;
}
lcm(lcm(period_x.unwrap(),period_y.unwrap()),period_z.unwrap())
}
fn lcm(a: usize,b: usize) -> usize{
a*b / gcd(a,b)
}
fn gcd(a: usize,b: usize) -> usize{
if b==0{a}else{gcd(b,a%b)}
}
let input_raw:String=fs::read_to_string("input.txt").expect("Something went wrong reading input file");
let input=input_raw.trim_end_matches("n");
let lines=input.split("n");
let initial_moons=lines.map(parse_moon).collect::>();
let mut moons=initial_moons.clone();
println!("Part One {}",simulate(&mut moons));
moons=initial_moons;
println!("Part Two {}",find_period(moons));
<|file_sepRESPONSE:
text
--- Day X ---
Day X was awesome!
You notice that today's puzzle is different.
...
SOLUTION:
rust
use std::collections::{HashSet};
fn main() {
}
<|file_sep#include "stdafx.h"
#include "AoC.h"
#include "ImageProcessor.h"
#include "InputFileReader.h"
#include "boostalgorithmstring.hpp"
using namespace std;
int main(int argc,char* argv[])
{
std::string filename="input.txt";
if(argc==3 && string(argv[1])=="-f")
filename=std::string(argv[2]);
InputFileReader reader(filename);
ImageProcessor processor(reader.GetInput());
cout<<"Part One "<& GetInput(){
#ifdef USE_BOOST_FOR_STRING_ALGORITHM
#ifndef _DEBUG
return _input;
#else
std::ifstream file(_filename);
while(!file.eof()){
std::string line="";
getline(file,line);
boost::trim(line);
if(!line.empty())
boost::to_upper(line);
boost::replace_all(line," ","");
boost::replace_all(line,",","");
boost::_tprintf(TEXT("%sn"),line.c_str());
line=line.substr(10,line.length()-10);
line=line.substr(4,line.length()-4);
line=line.substr(6,line.length()-6);
line=line.substr(7,line.length()-7);
boost::_tprintf(TEXT("%sn"),line.c_str());
line.erase(remove_if(line.begin(),line.end(),[](char c){return c<'A';}),line.end());
boost::_tprintf(TEXT("%sn"),line.c_str());
_input.emplace_back(line);
}
#endif
#else
#ifndef _DEBUG
FILE* file=fopen(_filename.c_str(),"r");
char buffer[MAX_PATH];
while(fgets(buffer,sizeof(buffer),file)){
boost::_tprintf(TEXT("%sn"),buffer);
char* token=strtok(buffer," ,");
while(token!=nullptr){
boost::_tprintf(TEXT("%sn"),token);
token=strtok(nullptr," ,");
}
boost::_tprintf(TEXT("n"));
fgets(buffer,sizeof(buffer),file);
boost::_tprintf(TEXT("%sn"),buffer);
token=strtok(buffer," ,");
while(token!=nullptr){
boost::_tprintf(TEXT("%sn"),token);
token=strtok(nullptr," ,");
}
boost::_tprintf(TEXT("n"));
}
#endif
#endif
return _input;
};
private:
std::vector_input;
std::string _filename;
};<|repo_name|>skwak/AdventOfCode2019<|file_sepCPP_SOLUTION:
cpp
#pragma once
class ImageProcessor{
public:
int PartOne(){
#ifdef USE_BOOST_FOR_STRING_ALGORITHM
#ifndef _DEBUG
int result=stdx
#endif
#else
int result=stdx
#endif
return result;
};
int PartTwo(){
#ifdef USE_BOOST_FOR_STRING_ALGORITHM
#ifndef _DEBUG
int result=stdx
#endif
#else
int result=stdx
#endif
return result;
};
private:
void PreProcess(stdx& data){
#ifdef USE_BOOST_FOR_STRING_ALGORITHM
#ifndef _DEBUG
PreProcess(data)
#endif
#else
PreProcess(data)
#endif
};
void PostProcess(stdx& data){
#ifdef USE_BOOST_FOR_STRING_ALGORITHM
#ifndef _DEBUG
PostProcess(data)
#endif
#else
PostProcess(data)
#endif
};
<|repo_name|>skwak/AdventOfCode2019<|file_sepNSolution:
swift
import Foundation
func partOne(input:[String])->Int{
return -1
}
func partTwo(input:[String])->Int{
return -1
}
let input=readLines("input.txt")
print("Part One (partOne(input))")
print("Part Two (partTwo(input))")
<|repo_name|>skwak/AdventOfCode2019<|file_sep**Response:**
text
--- Day X ---
Day X was awesome!
You notice that today's puzzle is different.
...
**Solution:**
swift
import Foundation
func partOne(input:[String])->Int{
return -1
}
func partTwo(input:[String])->Int{
return -1
}
let input=readLines("input.txt")
print("Part One (partOne(input))")
print("Part Two (partTwo(input))")
<|repo_name|>skwak/AdventOfCode2019<|file_sep prospero@DESKTOP-JOQV8E3 MINGW64 ~/Documents/Coding/AdventOfCode2019/Day_22 (master)
$ git status
On branch master
Your branch is up-to-date with 'origin/master'.
Changes not staged for commit:
(use "git add/rm ..." to update what will be committed)
(use "git restore ..." to discard changes in working directory)
modified: src/main.rs
no changes added to commit (use "git add" and/or "git commit -a")
prospero@DESKTOP-JOQV8E3 MINGW64 ~/Documents/Coding/AdventOfCode2019/Day_22 (master)
$ git add src/main.rs
prospero@DESKTOP-JOQV8E3 MINGW64 ~/Documents/Coding/AdventOfCode2019/Day_22 (master)
$ git commit -am 'Done'
[master d540d4b] Done
Date: Sun Dec 15 16:05:23 2019 +0100
Done
prosperity@DESKTOP-JOQV8E3 MINGW64 ~/Documents/Coding/AdventOfCode2019/Day_22 (master)
$ git push origin master
Enumerating objects: ...
Counting objects: ...
Delta compression using up to ...
Compressing objects: ...% (4 / ...)
Writing objects: ...% (5 / ...)
Total ... bytes | ... bytes/sec | Compressing... | ...%
remote:
remote:
remote:
remote:
remote:
To https://github.com/skwak/AdventOfCode2019.git/
* [new branch] master -> master
Enumerating objects...
Counting objects...
Delta compression using up to ...
Compressing objects...
Writing objects...
Total ... bytes | ... bytes/sec | Compressing... | ...%
remote:
remote:
To https://github.com/skwak/AdventOfCode2019.git/
* [new branch] master -> master
Create pull request? [Y/n]: Y
Opening https://github.com/skwak/AdventOfCode2019/pull/new/master...in default browser.
prospero@DESKTOP-JOQV8E3 MINGW64 ~/Documents/Coding/AdventOfCode2019/Day_22 (master)
$ <|repo_name|>skwak/AdventOfCode2019<|file_sep# Advent Of Code Solutions For Year Nineteen Hundred Nineteen!
My solutions for this years Advent Of Code challenge can be found here.
https://adventofcode.com/
## Progress Tracker!
## Day Zero!
This day is just here so I don't forget about it when I'm writing code late at night.
## Day One!
I decided not too look at any hints until I finished both parts of each day so I wouldn't be tempted too cheat!
### Problem Statement Part One!
You're given some fuel requirements by weight so you need some way of calculating how much fuel each module needs.
Each module requires fuel based on its mass according to these rules:
Take its mass...
Divide by three...
Round down...
Subtract two...
That's it.
For example:
For a mass of `12`, divide by `3` and round down to get `4`, then subtract `2` to get `2`.
For a mass of `14`, dividing by three gives `4` with a remainder of `2`. So we take the floor (`down`) of that number (`4`) then subtract the `2` to get `2`.
For a mass of `1969`, dividing by three gives `654` with a remainder of `1`. So we take the floor (`down`) of that number (`654`) then subtract the `2`to get `652`.
For a mass of `100756`, dividing by three gives `33583` with a remainder of `1`. So we take the floor (`down`)of that number (`33583`) then subtractthe `2`to get `33`,558.
The fuel required by each module is then summed together to get the total amount required.
### Solution Part One!
I started off by splitting up my inputs into lines before converting them into integers which I stored inside an array called 'masses'. After this I iterated over each element applying my formula before adding them together giving me my final answer!
### Problem Statement Part Two!
It turns out your spacecraft runs better than expected after factoring in the mass of added fuel.
Fuel itself requires fuel just like a module - take its mass divide by three round down and subtract two.
However any mass added from fuel also needs fuel which also needs fuel etc...
So now we need another formula!
### Solution Part Two!
The solution here was pretty simple really all I had todo was create another function called 'calculate_fuel' which would call itself recursively until it hit zero!
## Day Two!
### Problem Statement Part One!
You've built your own computer using Intcode software! You can even pass it some simple programs through an input instruction stream via ASCII code.
As your Intcode computer starts up it presents output instructions on a screen. On screen there are many large numbers but one immediately catches your eye:
Diagnostic code `-1219070636`
The diagnostic code represents something you need to fix before it will work.
This program stores its state in memory as a list of integers; here are several examples:
1,0,0,0,99 finds termination because it contains an immediate exit instruction at address(3).
Positional mode means that parameter should be interpreted as position - so mode `(1002)` refers to memory address `(02)` which contains `(10)` which should be used as `(value)` rather than `(10)` itself being used directly as value.
Programs that use positional parameters such as these are easier to read but there's a little more going on than meets the eye.`
Here are some worked examples:
In position mode `(1005)` works like this:
* Read value from address `(5)`
* Use value just read instead of actual value at address `(5)`
* If resulting value is non-zero continue execution using instruction after next parameter.(jump-if-true)
In immediate mode`(2006)` works like this:
* Read value from address `(6)`
* Use actual value at address `(6)`
* If resulting value is non-zero continue execution using instruction after next parameter.(jump-if-false)
Here are some small programs:
1101,val_a,val_b,address stores val_a plus val_b into address.
1008,val_a,address,val_b sets address to val_a if val_a is less than val_b.
104,val outputs val.
99 halts the program.
When you run this program,the tests below should print out as shown:
--- Test Program ---
1101,val_a,val_b,address => Stores val_a plus val_b into address(memory location).
--- Test Output ---
Before running:`position_memory=[val_a,val_b,address,value_at_address]`
After running:`position_memory=[val_a,val_b,value_at_address+val_a+val_b]`
--- Test Program ---
1008,val_a,address,val_b => Sets address to val_a if val_a is less than val_b.
--- Test Output ---
Before running:`position_memory=[val_a,address,value_at_address]`
If(val_a)<=(val_b):
After running:`position_memory=[val_a,address,val_a]`
Else:`position_memory=[val_a,address,val_b]`
--- Test Program ---
104,value => Outputs value.
--- Test Output ---
Before running:`output=[]`
After running:`output=[value]`
--- Test Program ---
99 => Halts program.
--- Test Output ---
Before running:`running=true`
After running:`running=false`
The program uses no special formatting like parenthesis or other human-readable mnemonics to indicate what modes each parameter should use. Parameter modes are stored within each instruction byte immediately following the opcode digit(s).
The opcode is always positioned at the very left,and will contain only digits including leading zeroes as needed.digits toward the right specify parameter modes,digit-by-digit,in right-to-left order starting from digit immediately to the right of the opcode digits.For example,in the case of a three-digit opcode,the first parameter's mode is in the hundreds digit,and the second parameter's mode is inthe tens digit.The third parameter's mode is inthe ones digit,and so on.Leftover modesareassumedto bezero.This meansifaninstruction hasonlyanopcode and noparameters,the remainingparametermodesareallzero.Andifaninstruction hassomeparametersbut fewer thanthree,don't worryaboutthosemissing parameters,their modes are assumed to be zero.
So your first task is to work out what opcode does what,and how parameter modes affect them!
### Solution Part One!
This one took me ages because my brain doesn't work well when thinking about computers doing computers things haha! Once I got going though it was pretty easy really! What I did was store my inputs inside an array called 'memory'. Then I created an enum type called 'Mode' containing two variants Immediate And Positional.Then another enum type called 'OpCodes' containing all my opcodes.Inside my main loop I would check what opcode was currently being executed based on whether it contained more than one digit.I would then split those digits apart storing them inside variables named 'first_digit' And 'second_digit'.I would then check against both variables against their corresponding enum types.After this point everything became quite straight forward.All I had todo was create functions corresponding with each opcode e.g.: fn process_addition(mode_first_param:&Mode,param_one:int,param_two:int,target_position:int,&mut memory:&Vec,&mut pc:i16)->bool{}Then call them within my main loop where appropriate returning true when they were done otherwise false.Once they were done processing their respective opcodes they would return true causing my loop condition &break.Afterwards all thats left was figuring out how many elements needed incrementing based on how many parameters there were per operation.I did this manually looking through documentation figuring out how many params there were per operation.I know theres probably better ways but hey ho!I couldnt figure out how otherwise so thats what Im sticking with!
### Problem Statement Part Two!
Now things start getting interesting!
Suppose your computer system actually put breakpoints into this program?
Interrupting execution after it had already run one instruction! The program must have been trying to tell you it wanted youto plug some numbers in,to reset its diagnostic code!
Immediately before the program halted,it jumped to instruction/address/memory_position=`03`.
So thats where execution stopped.What if you could intervene right there?
What input values should be placed into addresses=`01`and=`02`,`so when thistarget diagnostic code `-1202177807`(which again,is negative!)is produced?
As always,the instructions are available online(at problemurl).
Your air conditioner broke!(A heatwave caused an electrical surge that fried expensive equipment.) Since money fora new one is tight,this year,you'll bebeating aroundthe bush.Firstyou'llneedtotemporarilyfixthebrokenair conditioner.Youpullupthe dustcowlanddiscoverthatheatsinkiscoveredwith dust(and a few critterswho apparentlyfeeledequalitywithyou).On closer inspection,youcanseethatthereisanintegrated circuitbuiltrightintothisheat sink-a soldered circuit board fed directlyfromoneoftheheat sink fins.Thewire tracesonone sideofthecircuit boardareveryfaint,butrecog nizableastheytraversebetweeneachfinaroundtheradiation fins.Theylookliketheyformanetwork-ofinterconnectedtestpointsyoumightbeabletoprogram.The circuitry providestestpoint addressesyoucanreadfrom(or writevaluesinto)-andthenitcanperformcomparisons,basicarithmetictasks,andoutputresults.You'llneedtocleanoffthemodulespecularlycarefullysotheydon'tgetdamaged.Butinadditiontothat,you'llneedtocreateafirmwareprogram,tofeedintotheintegratedcircuittestpoints.So,itlookslikethereisa programmablethermostatafterall!
Now,itjustneedsaversionthatwillactuallycoolyourshipbeforeyouareincorporatedintoit!
Writeup instructions are available online(at problemurl).
Your existing Intcode interpreter does not handle floating point numbers,nor large integers,suchasthose producedbyopcode`=`.`Nordoesithandlelargepositional parameters(forinstance addressingmemoryaboveaddress`=999999`).Italsodoesn'thavefunctionalitytointerruptexecution,either.
What ifyoucouldinterceptexecutionrightataddress=`03`,soyoucouldsettheseparameterstothewordstogetthediagnosticvalue=-1202177807?
What inputswouldallowyourprogramtosetaddresses=`01`,`02`,`sothatwhenexecuted,thevalueataddress=`00`=diagnosticcode=-1202177807?(These valuesshouldbothbepositive.Integers.)
Youranswershouldbedeliveredinasingleinteger:value_at_address01 *100000 +value_at_address02.(Thatis,eitherleaveyouranswerasaregular integer(withoutleadingzeros),orepresentitwithtendigitsincludingleadingzeros.)
Note:Theintegersusedhere(includingthoseshownabove)maynotfitinaprogrammersusualintegerdatatypes(sothinkbig!).
Asalways,youranswersshouldbefurnishedonline!(Ifyouranswerdoesnotproduce-theexpectedresult,youwillneedtomakechangesandresubmit.)Whendone,yourintegratedcircuitsystemshouldproduce diagnostics-code=-1202177807ataddress=`00`.Goodluck!
### Solution Part Two!
I figured since Im already doing everything manually why not go all manual?! All Id have todo was figure out what values Id need for addresses zero one And two based on trial error testing.It took me ages trying different combinations before finally hitting upon one combination that worked.So now instead Of having my opcodes hard coded Im having them passed In via command line arguments instead.Although technically speaking Its still hard coded lol!
## Day Three!!
### Problem Statement Part One!
The device shows a sequenceof steps neededto produceoptimal thermalstabilization.Ifit deviatestoo farfromoptimalthermalstabilization,it automaticallyshuts down.Torestorepower,capacitorsneedrecharging.To rechargecapacitors,followthedevicesinstructionsuntilstepnumber=`50`;then enterthe diagnosticdata(code=sumofvaluesatalllocationsinmemorywhichhavebeentwicewritten)-andwaitfornormaloperationresumption.
You should review thistransmissioncarefully; apparentlyit maynotbewhatitseems.
Device Interface Language(DIL)
DILconsistsoftwo-characteropcodesfollowedbyparameter(s).
Opcodeparameterpairsarealwaysseparatedbycomma.Othercharacters(includingwhitespacecharacters,suchasspaces,newlines,tabs,andcommas,)mayappearbefore,theopcodeparameterpair(s),betweentheopcodeparameterpair(s),orafter(theopcodeparameterpair)s,butthesecharactersaresimplyignored.Whenreadingapair,listparametersfromlefttoright.Ifmorethanonecharacterisneededforeachparameter,startwiththestringimmediatelyfollowingthepreviouscharacter(whichmaybethepreviousopcodeorpreviousparameternumber).
DILhassevenopcodes:
Opcode Description Type andRangeofParameter(s)
halt Halt execution Thisopcodehasnoparameters.Ifthisopcodeappearsanywhereotherthanthenextinstructiontoprocess,itindicatesamanualerrorinasmallerprogramwrittenbysomeoneelse-orperhapsasecondaryprocessattemptedtodiscouragedebuggingefforts.Meanwhile,haltisalwaysfoundasthefinalinstructioninasmallerprogram.Note:thisisnotrequired,butinsteadismerelyconventionallyexpected.Justbecausehaltisthefinalinstructiondoesnotmeanexecutionendshere--executioncontinuesuntilhaltistheonlyremaininginstructionavailabletoprocessor.Anytimehaltonlyappearsinthelistofinstructions,butisnotthestepbeingprocessed,itshouldbeconsideredasthemostrecentlyencounteredinstruction--evenifitisnotthestepbeingprocessedcurrently.
add x,y,z Addvaluesfromlocationxyozlocationyztogivemeaningadd(x,y)=z.Thisismemoryreferencemode,andmustbegivenasapositionalparametertype.Thismeanswhenthisopcodeappears,x,y,zareinterpretedasthelocationsofvaluesinthememoryspace.Neitherxorneshouldbewriteto--onlyzcanbeworkedupon.Additionneveroverflows;whenoverflowoccurs,resultswraparound.Example:"add50000,-199999999,-50000"setslocation50000equaltotwomillion.Becauseaddingone million(twice)-two million resultsinzero,henceforthelocation-valuewrapsaroundbackaroundtomillionagainafterexceedingonetwo-million.Notethataddingnegativevaluesworkslikewhenaddingpositives.Valuesaddedcanberepeatedlynegativetoaddpositivevalues.Typically,thisopcoderequiresthreeparametersbutcanalsobeusedwithtwooroneratherthanthree.Parametercountisinferredbasedonnumberoftokensprovided.Morespecifically,tokensbeforedotokencontainingahaltoperationindicatesthatnoadditionaltokensbelongtothepreviousoperation.Giventokencount,tokensafterthecurrenttokenbelongtotheproceedingsoperation.Seebelowforexampleshowthisworks.Inthiscase,thismeansbothyandexwerethoughtofasthememorylocationswhosecontentswereadded.Additionally,thismeansnoadditionaltokenswereavailableforthesubsequentoperation.Furthermore,sinceonlytwoparameterswereprovided,noadditionaltokenswereavailableforthefollowingoperation--meaningnoadditionalparameterscouldbedefinedforthefollowingsubsequentoperation.Therefore,nosuchsubsequentoperationexists.Becausenodatawasprovidedforthefollowingsubsequentoperation,nochangeoccurredinthememoryspace.Theinterpretationwouldthereforebecome:"add50000,-199999999,-50000halt",meaninglocation50000wouldbecomestowmillionandnothingmorewouldoccur.Subsequently,thisinterpretationwouldcausehalttoprotectagainstundefinedbehaviorcausedbyprocessingnonexistentoperations.Notethatifnoreasonableinterpretationcouldbeformed,giventokencount,tokensbeforedotokencontainingahaltoperation,wouldresultinanerrorcondition.However,sinceourlanguageonlycontainsoneopcodewithouthalt,noerrorconditionwilleveroccurbasedonthisspecificrule.
multiply x,y,z Multiplyvaluesfromlocationxyozlocationyztogivemeaningmultiply(x,y)=z.Thisismemoryreferencemode,andmustbegivenasapositionalparametertype.Thismeanswhenthisopcodeappears,x,y,zareinterpretedasthelocationsofvaluesinthememoryspace.Neitherxorneshouldbewriteto--onlyzcanbeworkedupon.Multiplicationneveroverflows;whenoverflowoccurs,resultswraparound.Example:"multiply50000,-199999999,-200000000"setslocation50000equaltonevenmillion.Becausemultiplyingtwomillion(bynegativeone hundred million)yieldsnegative(twowillion)--whichwraparoundtomillionagainafterexceedingtow-billion--thisresultsinevenmillion.Seeaddforcross-referenceonhowparametercountsworkwithinourlanguage.Inthiscase,thismeansbothyandexwerethoughtofasthememorylocationswhosecontentswereadded.Additionally,thismeansnoadditionaltokenswereavailableforthesubsequentoperation.Furthermore,sinceonlytwoparameterswereprovided,noadditionaltokenswereavailableforthefollowingoperation--meaningnoadditionalparameterscouldbedefinedforthefollowingsubsequentoperation.Therefore,nosuchsubsequentoperationexists.Becausenodatawasprovidedforthefollowingsubsequentoperation,nochangeoccurredinthememoryspace.Theinterpretationwouldthereforebecome:"multiply50000,-199999999,-200000000halt",meaninglocation50000wouldbecomeevenmillionandnothingmorewouldoccur.Subsequently,thisinterpretationwouldcausehalttoprotectagainstundefinedbehaviorcausedbyprocessingnonexistentoperations.Notethatifnoreasonableinterpretationcouldbeformed,giventokencount,tokensbeforedotokencontainingahaltoperation,wouldresultinanerrorcondition.However,sincetheymultiplycommandcannotcontainanyothercommandsinitsdescription(noothercommandcontainsmultiplicationoperator!),noreasonableinterpretationcouldbesuggested,giventokencount,tokensbeforedotokencontainingahaltoperationsincaseswhereacommandfollowingamultiplycommandcannotexist.Weacceptthisshortcomingbecauseoursystemdoesnotrequirethemultiplierequivalentforalllanguagesandsystems.Unfortunatelythiscreatesambiguitywithinoursystemwheninterpretinglanguagesystemswhichdoincludeamultiplieresultantnotintheirlistofsymbols.However,weseektolimitambiguitysofarasaffectsonlysystemswithoutmultipliersresultants.Wehopeoursystemwillremainrobustdespitethefactthatsomeexpressionsmayleadtoaconflictbetweeninterpretingthemwithinoursystemversusinterpretingthemwithinanotherlanguage.Ourgoalistohavealanguagewhichmakeslittleassumptionsaboutitsusersystem.Itshouldworkwellwithoutrequiringusersystemsformakeassumptionsaboutoursystem.Typically,thisopcoderequiresthreeparametersbutcanalsobeusedwithtwooroneratherthanthree.Parametercountisinferredbasedonnumberoftokensprovided.Morespecifically,tokensbeforedotokencontainingahaltoperationindicatesthatnoadditionaltokensbelongtothepreviousoperation.Giventokencount,tokensafterthecurrenttokenbelongtotheproceedingsoperation.Seebelowforexampleshowthisworks.Inthiscase,thismeansbothyandexwerethoughtofasthememorylocationswhosecontentswereadded.Additionally,thismeansnoadditionaltokenswereavailableforthesubsequentoperation.Furthermore,sinceonlytwoparameterswereprovided,noadditionaltokenswereavailableforthefollowingoperation--meaningnoadditionalparameterscouldbedefinedforthefollowingsubsequentoperation.Therefore,nosuchsubsequentoperationexists.Becausenodatawasprovidedforthefollowingsubsequentoperation,nochangeoccurredinthememoryspace.Theinterpretationwouldthereforebecome:"multiply50000,-199999999,-200000000halt",meaninglocation50000wouldbecomeevenmillionandnothingmorewouldoccur.Subsequently,thisinterpretationwouldcausehalttoprotectagainstundefinedbehaviorcausedbyprocessingnonexistentoperations.Notethatifnoreasonableinterpretationcouldbeformed,giventokencount,tokensbeforedotokencontainingahaltoperation,wouldresultinanerrorcondition.However,sincetheymultiplycommandcannotcontainanyothercommandsinitsdescription(noothercommandcontainsmultiplicationoperator!),noreasonableinterpretationcouldbesuggested,giventokencount,t
