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Turing Complete > 指南 > UnsignedRobin 的指南

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Achievements

由 UnsignedRobin 制作

Solution to all achievements

奖励

创建者

UnsignedRobin
在线

类型: 成就, 攻略

Languages: 英语

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更新日期

2022 年 4 月 23 日上午 1:22

2023 年 5 月 19 日下午 12:18

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指南索引

总览

4 NAND = XOR

5 Component Full Adder

Fast Adder

Binary Counter

Symmetric ALU

Condition 10

Turing Complete

Helperbot

Best hat

Black hat

3 Adder Multiply

Tutorial Complete

FastBot

Water world

Binary Racer

留言

4 NAND = XOR

Complete the XOR level using ONLY 4 NAND gates and nothing else

5 Component Full Adder

Complete the full adder level using only 5 blue components

Fast Adder

Complete the byte adder with a delay of 35 or less
(NOTE: This circuit has a delay of 22)

Binary Counter

Complete the Counter level with a 65 nand score or less

NOTE: You need a score of 3/4 in the XOR Gate level for this solution to work (which is better than the depicted solution from the 4NAND=XOR achievement above). This circuit has a gate score of 53

Symmetric ALU

Complete the "Logic Engine" level using only components from the "BYTES" category

Condition 10

Complete the level called "Conditions" using only 10 blue components

Turing Complete

Build a Turing Complete computer

Rewarded upon completing the level TURING COMPLETE

Helperbot

Complete all 4 quests on the ship

All 4 quests can be accessed once The Maze is completed. To access the ship, control the robot manually through the maze and hit ENTER when you reached the door.

1. Push Derpoo into the reactor

Find the gray plush-blob called Derpoo and push it onto the conveyor belt leading to the reactor

⠀

2. Convince Gruliette to go on a date with Dumio

Talk to Dumino and Gruliette

⠀
Talk to Dumino again and find the pink flower in the south-east corner of the map

⠀
Bring the flower back to Gruliette

⠀

3. Poison the boss

Speak to Griffith

⠀
Collect 4 blobs of slime by shooting the enemy slimes (TAB key)

⠀
Put the slime in the barrel behind the chef

4. Find the treasure

Go through the secret door in the south-east corner of the map and follow the path

⠀
The treasure is guarded by Tricksy who is way faster that you. You can not beat it. The trick is to go to the upper narrowing path and wait there (by pressing ENTER), until Tricksy becomes unpatient.

⠀
Once Tricksy is overwhelmed by boredom, it runs around the corner, and you can shoot it. The treasure is now free to be looted.

⠀

Best hat

Obtain the fabled best hat

This achievement can be accessed once The Maze is completed. To access the ship, control the robot manually through the maze and hit ENTER when you reached the door.

Run the solved Maze level, until you obtained 255 coins
Go to the hat man in the south west area of the ship and purchase the best hat

Black hat

Hack the hat seller

This achievement can be accessed once The Maze is completed. To access the ship, control the robot manually through the maze and hit ENTER when you reached the door.

Talk to the hat man (your own coin supply doesn't matter)
Change the number representation from unsigned to signed, so that price numbers greater than 128 become negative
Purchase the Best Hat which costs -1 now (The vendor gives you a coin)

3 Adder Multiply

Complete "The Product of Nibbles" using no more than three 8 bit adders

Tutorial Complete

Complete the "Functions" level

This level can be completed, without solving the actual task. However it is highly adviced to do so.
Approach: In CONDITION Mode, only Bit 1,2,3 has meaning. Now use bit 4 and 5 to decide, if the result leads to a JUMP, a CALL or a RETURN.

JUMP

Set the counter to Arg3

CALL

Push the current counter value + 4 onto the stack
Set the counter to Arg3

RETURN

Prevent Arg3 from accessing the bus
POP last value from stack
Set the counter to popped value

FastBot

Complete the robot race with less than 64 bytes of code

NOTE: The track is a Hilbert Curve with three iterations (if you want to google an actual algorithm)

Instead of coding an algorithm, the path is hard wired without any code

Water world

Beat the level "Water World"

Solution Idea:
1. Read in all values of the landscape and store them in the RAM

2. Go from left to right over the stored landscape values, find the current maximum and calculate the potential water pillar by subtracting the landscape value from the current maximum
Store the water values in the RAM, too (below the landscape values, starting with address 16)
(Screenshot is not the first landscape you see in the level, but is nice to demonstrate the idea)

3. Go from right to left over the stored landscape values, find the current maximum and calculate the potential water pillar by subtracting the landscape value from the current maximum
(Continue with Step 4)

4. If the current water pillar is less that the stored value from 2., then overwrite the value in the RAM (= combine "left->right" and "right->left" water pillars by always using the smaller of the two values)

5. Now the correct water values are stored in the second row of the RAM starting with adress 16.
Add all values together and send the result to the Output

Program: (you have to adjust it to your architecture)

const adr 0 const currVal 1 const max_left 2 const max_right 3 const sum 4 const num_reads 5 const io 7 const ram 0b00001000 set 16 _ num_reads label readValues addi io _ ram addi adr 1 adr subi num_reads 1 num_reads ifi_neq_goto num_reads 0 readValues set 0 _ adr set 16 _ num_reads label readFromLeft addi ram _ currVal if_gt_call currVal max_left setMaxLeft addi adr 16 adr sub max_left currVal ram subi adr 15 adr subi num_reads 1 num_reads ifi_neq_goto num_reads 0 readFromLeft set 15 _ adr set 16 _ num_reads label readFromRight addi ram _ currVal if_gt_call currVal max_right setMaxRight sub max_right currVal currVal addi adr 16 adr if_lt_call currVal ram setActual subi adr 17 adr subi num_reads 1 num_reads ifi_neq_goto num_reads 0 readFromRight set 16 _ adr set 16 _ num_reads label sumValues add sum ram sum addi adr 1 adr subi num_reads 1 num_reads ifi_neq_goto num_reads 0 sumValues addi sum _ io label setMaxLeft addi currVal _ max_left ret _ _ _ label setMaxRight addi currVal _ max_right ret _ _ _ label setActual addi currVal _ ram ret _ _ _

Binary Racer

Beat level 7 in BINARY RACER

(NOTE: In level 7, the sum of the toggled bits are no longer displayed)

Currently not realistically doable as human
Everybody who achieved it so far said:
"I COULD do it if i practiced for 2 hours straight, but i rather wrote a macro which did it for me"

Here is my Java programm that did the trick (you have to adjust the coordinates where to click)
Type the decimal number in the box and hit enter.

import java.awt.AWTException; import java.awt.Robot; import java.awt.event.InputEvent; import java.awt.event.KeyAdapter; import java.awt.event.KeyEvent; import javax.swing.JFrame; import javax.swing.JTextField; public class BinaryRacer extends KeyAdapter { public JFrame window = new JFrame(); public JTextField textField = new JTextField(); public static void main(String[] args) { new BinaryRacer(); } public BinaryRacer(){ textField.addKeyListener(this); window.setDefaultCloseOperation(JFrame.DISPOSE_ON_CLOSE); window.add(textField); window.setSize(300, 100); window.setVisible(true); } @Override public void keyPressed(KeyEvent event) { //on ENTER: calculate and click bits; clear screen if (event.getKeyChar() =='\n') { doClicks(calcBits(textField.getText())); textField.setText(""); } } public boolean[] calcBits(String intString) { boolean[] bits = new boolean[8]; //no error handling :( you have to type numbers int input = Integer.parseInt(intString); for(int j=0;j<bits.length;j++) { if (input % 2 == 1) bits[j] = true; input = input / 2; } return bits; } public void doClicks(boolean [] bits) { //click bits if (bits[7]) click( 570,950); if (bits[6]) click( 670,950); if (bits[5]) click( 770,950); if (bits[4]) click( 870,950); if (bits[3]) click( 970,950); if (bits[2]) click(1070,950); if (bits[1]) click(1170,950); if (bits[0]) click(1270,950); //click submit click( 970,1025); //click back in the window (on second screen) to enter a new number click(window.getX()+50,window.getY()+50); } public void click(int x, int y) { Robot bot; try { bot = new Robot(); bot.mouseMove(x, y); bot.mousePress(InputEvent.BUTTON1_DOWN_MASK); bot.mouseRelease(InputEvent.BUTTON1_DOWN_MASK); } catch (AWTException e) { e.printStackTrace(); } } }

40 条留言

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crumb beaver 8 月 31 日下午 12:55

Thanks for java code.It helps

A Fellow Human 7 月 14 日上午 2:18

I was able to beat Binary Racer after about an hour of practice. If you're trying to do it without using autoclickers/calculators/cheat tables, here are some tricks worth learning:
1. Use the keyboard - 1 for 128, 2 for 64, and so on down to 8 for 1, and space for submit. Takes some time to get used to, but you end up saving a couple seconds and that can be useful on level 7.
2. Think in base 16. By this, I mean that you should, for example, intuit that 213 = 13 x 16 + 5, and use that to input into Binary Racer, rather than working out that 213 = 128 + 64 + 16 + 4 + 1. This does mean that you should be familiar with the binary representations of 0-15 and the 16 times tables up to 16 x 15 (= 240). Knowledge of both does come to you naturally as you grind out Binary Racer, but you should be actively learning your 16 times tables.
3. Make usage of the timerless mode. You can't get the achievement on timerless, but it does help you practice the above two skills.

Kazel 3 月 19 日下午 10:22

@gg.nadrewod
XOR gate can be implemented with a gate score of 3 using one NAND, one OR, and one AND.
MUX can be done with a gate score of 17.
My solution for Binary Counter uses:
1 MUX
3 Byte Splitters
6 AND
7 XOR
1 NOT

gg.nadrewod 1 月 28 日下午 8:04

If the "Binary Counter" achievement requires tight-enough gate logic that the "4 NAND = XOR" setup doesn't work, does anyone have examples of some XOR or MUX setups that can get the achievement?

João Porco da Silva 1 月 5 日下午 5:33

Here is an autohotkey v2 script that solves the binary racer level: using the keypad, you enter the digits in decimal notation and press enter, then, the script converts it to the keys from 1 to 8 and automatically presses the spacebar to send the answer.

https://pastebin.com/raw/EkJZcP9D

plegbot 1 月 3 日上午 11:35

"Currently not realistically doable as human"
Actually, I did it as a human. It is very possible. You just have to recognize binary patterns in your head.
You also have more time in level 7 than in 6, which is helpful.

=^・^= 2024 年 9 月 2 日上午 5:14

uh... the last achievement... I can count but there is so much time pressure and the controlls... I personally would prefer if 1 = 1, 2 = 2, 3 = 4 etc...

ДPoBoSeX 2024 年 5 月 22 日上午 2:13

Binary racer on Python. No need any mouse moves and coords, bits can be input as 1-8 and send by space. In my tool u just press keys and then enter to auto press bits. use numpad / if u mistaken. https://pastebin.com/TqHERMvr

ukey79 2024 年 4 月 14 日上午 3:40

You can use the number keys 1 to 8 to toggle the bits.

arthur111213 2024 年 4 月 6 日下午 3:15

Hey !

I updated the java clicker so that it starts by presenting a screenshot to the user and asking him to click on the buttons.
That way ther's no need to edit the buttons location in the code anymore.

I also added a click on the "start" button so that you just have to hit enter and use the bot

(just click on any button before running the bot so that you can see the "submit" button on the screenshot).

Here is the source code : https://pastebin.com/2v8unKEN

(Sorry for repost. I changed the code and I can't find how to edit comments)

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