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yep, that's the fix. i figured it out as well
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i like the idea of level count. I was inpired by it. then i belived i took a different approach to solve the problem.
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i ah... i think i did it in a wrong way. Since i know that the character can jump, so i just ask one charactor jump and sweep the map for required gems. and it worked.but if the character can't jump at all, then i haven't figured out how to solve it yet.=====update======for the sake of seeking the true, i just did my thought through version which uses two experts, no jumping. and i'm super lazy, so i created several functions for the computer to determind what to do. i only lower or raise the level like several time until task complete. in a word, i let it automate.
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i've done 2 versions since the game asked me to exhaust all resource acquired previously. so here are my two solutions:==============cheating code==============let xp=Expert()
let ch=Character()
world.place(xp, facing:.north, atColumn: 0, row: 4)
world.place(ch, facing:.north, atColumn: 2, row: 0)
var countGem=0
func aCycle(){
for i in 1...6{
if countGem < totalGems {
if ch.isOnGem{
ch.collectGem()
countGem += 1
}
ch.jump()
}
}
}
func halfMission(){
aCycle()
ch.turnRight()
ch.jump()
ch.turnRight()
aCycle()
ch.turnLeft()
ch.jump()
ch.turnLeft()
aCycle()
}
halfMission()
ch.turnLeft()
ch.turnLeft()
halfMission()===============thought through code========let totalGems = randomNumberOfGems
let xp1 = Expert()
let xp2 = Expert()
var levelCount=0 //count where the levels are, to determind to move up or down
var gemCount=0
world.place(xp1, facing:.north, atColumn:0, row:4)
world.place(xp2, facing:.north, atColumn:3, row:0)
// place expert 1 and 2 on the map
func collectCount (){
if xp2.isOnGem{
xp2.collectGem()
gemCount += 1
}
}
//create a function that determins whether isOnGem, then collect and countGem
func solveSpot(){
if xp2.isOnGem{
collectCount()
}
if !xp2.isBlockedLeft{
xp2.turnLeft()
xp2.moveForward()
collectCount()
xp2.turnLeft()
xp2.turnLeft()
xp2.moveForward()
xp2.turnLeft()
}
if !xp2.isBlockedRight{
xp2.turnRight()
xp2.moveForward()
collectCount()
xp2.turnRight()
xp2.turnRight()
xp2.moveForward()
xp2.turnRight()
}
else if xp2.isBlockedLeft && xp2.isBlockedRight {
}
if !xp2.isBlocked{xp2.moveForward()}
else if xp2.isBlocked {xp2.turnLeft()
xp2.turnLeft() //turn around at the end of the road
}
}
//this function solve each row, if it's not block left or right, go and check gems, yes gem? get it and return, no gem? just return.
func solveColumn(){
for i in 1...7 {
if gemCount < totalGems {
solveSpot()
}
}
}
//solve the whole column automatically
func upCycle(){
xp1.turnLockUp()
levelCount += 1
solveColumn()
}
//move the lock up and solve a column automatically
func downCycle(){
xp1.turnLockDown()
levelCount -= 1
solveColumn()
}
//move the lock down and solve a column automatically
while gemCount < totalGems {
xp1.turnLockUp()
if levelCount == 1{
for i in 1...3 {
upCycle()
}
}
if levelCount == 3{
for i in 1...3 {
downCycle()
}
}
}
//b4 required gems are collected, move up level then sweep the whole 3 column for accessible avilable gems. then at the end, if still need more gems, move down and repeat the cycle.
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hi, regarding this, i had another problem.when calling the method "moveForward()" i have to define which instance is using it, so it would be" expertOne.moveForward()"but if i want to create a new function, to tell the instance to move forward a certain number of steps, i would write:let expertOne = Expert()func moveSteps(steps : Int){for i in 1...steps {expertOne.moveForward()}}however if i have two or more instance of Expert(), i don't want to make different functions for each instance, so is there a way to code like this:func moveSteps(who: [some-data-type], steps: Int){for i in 1...steps {who.moveForward()}}so that everytime i can just callmoveSteps (who: expertTwo; steps: 4)then expertTwo will execute "expertTwo.moveForward()" 4 times,my code was definitely unacceptable, it's just for discussion sake. Any idea if i can realize this? like letting String becoming an instance name? Much appriciated.
