HW 2 Solutions

ENGR 21, Fall 2026.

Published

September 17, 2026

Solutions

1 Converting between number systems

1.1 from Decimal to Binary

  1. 100

    TipAnswer

    0b1100100

  2. 1029

    TipAnswer

    0b10000000101

  3. 476

    TipAnswer

    0b111011100

  4. \(1.4 \times 10^{3}\)

    TipAnswer

    0b10101111000

1.2 from Binary to Decimal

  1. 0b1011011

    TipAnswer

    \(91\)

  2. 0b110010101

    TipAnswer

    \(405\)

  3. 0b1000000001

    TipAnswer

    \(513\)

  4. 0b10111

    TipAnswer

    \(23\)

1.3 from Decimal to Hexadecimal

  1. 1000

    TipAnswer

    0x3E8

  2. 176

    TipAnswer

    0xB0

  3. 42678

    TipAnswer

    0xA6B6

  4. 81

    TipAnswer

    0x51

1.4 from Hexadecimal to Decimal

  1. 0x14E3

    TipAnswer

    \(5,347\)

  2. 0xA10B

    TipAnswer

    \(41,227\)

  3. 0x1000

    TipAnswer

    \(4,096\)

  4. 0x1010

    TipAnswer

    \(4,112\)

2 Base systems

2.1 A new base: 7

  1. Write down all numbers from 1 to 16 in base 7.

    TipSolution

    The numbers from one to sixteen in base 7 will be: \(\{1,2,3,4,5,6,\}\) followed by the number \(10\), which equals seven, and then \(11\), which equals eight, and so on. In base 7, the sequence would be \[\{1,2,3,4,5,6,10,11,12,13,14,15,16,20,21,22\}\]

  2. In base 7, what is the meaning of the symbol 66 ? Explain.

    TipSolution

    In base 7, this would need to be interpreted as follows. \[6 \times 7^0 + 6 \times 7^1 = 6_{10} + 42_{10} = 48_{10}\] Therefore, the symbol 66 in base 7 corresponds to the number forty-eight.

  3. What is the largest possible four-‘digit’ number in base 7? Write the number in words, in decimal form, and in base-7 form.

    TipSolution

    The largest possible four-‘digit’ number in base 7 is \(6666_7\). In decimal form, this would be \[6 \times 7^0 + 6 \times 7^1 + 6 \times 7^2 + 6 \times 7^3 = 6_{10} + 42_{10} + 294_{10} + 2058_{10} = 2400_{10}.\] In words, it is the number two thousand four hundred.

2.2 Addition of binary and hexadecimal numbers

In grade school, you learned how to add two multi-digit numbers by hand. In case, you’ve forgotten, here’s a video explaining this to elementary school students.

Your task in this problem is to carry out multi-‘digit’ addition, by hand, to binary and hexadecimal numbers. In the process, you will see the concept of ‘carrying over’ a digit (perhaps you learned the term ‘regrouping’) applies to numbers other than base ten.

As an example, the following sums have been computed for you below.

  • 0b110001 + 0b11011
  • 0xAFC + 0x115

2.3 Hex

  1. 0xE5A + 0x85A

    TipAnswer

    \(5812\) in decimal form and 0x16B4 in hexadecimal form

  2. 0xEE + 0x99

    TipAnswer

    \(391\) in decimal form and 0x187 in hexadecimal form

  3. 0x100 + 0x20

    TipAnswer

    \(288\) in decimal form and 0x120 in hexadecimal form

  4. 0xB0 + 0xB

    TipAnswer

    \(187\) in decimal form and 0xBB in hexadecimal form

2.4 Binary

  1. 0b1011011 + 0b11011001

    TipAnswer

    \(308\) in decimal form and 0b100110100 in hexadecimal form

  2. 0b10011 + 0b100110001

    TipAnswer

    \(324\) in decimal form and 0b101000100 in hexadecimal form

  3. 0b10000000000 + 0b1011111

    TipAnswer

    \(1119\) in decimal form and 0b10001011111 in hexadecimal form

  4. 0b11010001001 + 0b110110110

    TipAnswer

    \(2111\) in decimal form and 0b100000111111 in hexadecimal form

3 Conditionals and loops

3.1 Counter variables

Consider the following program, which counts the number of times pin A7 has been tapped and assigns color1 to pixel number k, where k is the number of times pin A7 has been tapped. The code crashes if pin A7 is touched more than 10 times.

from adafruit_circuitplayground.express import cpx
import time
off = (0,0,0)
color1 = (10,30,10)
color2 = (30,10,10)
cpx.pixels.fill(off)
k = -1
while True:
    time.sleep(0.2)
    if cpx.touch_A7:
        k += 1
        cpx.pixels[k] = color1
        cpx.pixels[k-1] = off

Modify this code so that it has the same behavior for up to 10 taps, but for taps 11 through 20, a different color is assigned to the pth pixel, where p is k-10. Your code will still fail after the 21st tap is detected; that’s fine.

Expected behavior of Circuit Playground Express
TipSolution 
from adafruit_circuitplayground.express import cpx
import time
off = (0,0,0)
color1 = (10,30,10)
color2 = (30,10,10)
cpx.pixels.fill(off)
k = -1
while True:
    time.sleep(0.2)
    if cpx.touch_A7:
        k += 1
        if k < 10:
            cpx.pixels[k] = color1
            cpx.pixels[k-1] = off
        elif k < 20:
            cpx.pixels[k % 10] = color2
            cpx.pixels[(k % 10)-1] = off

3.2 The continue keyword

The following progam was written with the intention of printing all numbers except for 9. However, the program currently fails at doing this. What one change will ensure that printing ‘9’ is skipped? Turn in the new code.

import time
k = 0
while True:
    time.sleep(0.5)
    k += 1
    print(k)
    if k == 9:
        continue
TipSolution 
import time
k = 0
while True:
    time.sleep(0.5)
    k += 1
    if k == 9:
        continue
    print(k)

3.3 The break keyword

In class, we saw that the following code allows us to terminate the inner while loop by touching pin A2.

from adafruit_circuitplayground.express import cpx
import time
k = 0
while True:
    time.sleep(1)
    k += 1
    print(f"Outer iteration {k}")
    if k % 3 == 0: # k ÷ 3 has remainder 0
        m = 0
        while True:
            time.sleep(1)
            m += 1
            print(f"Outer iteration {k}, inner iteration {m} ")
            if cpx.touch_A2:
                print("terminating")
                break

Change the code above so that it meets the following requirements:

  • Touching A2 while in the inner loop should terminate inner loop.
  • Touching A2 while in the outer loop should not do anything.
  • Touching A7 alone while in the inner loop should now also terminate inner loop.
  • Touching A7 while pressing button B while in the outer loop should terminate the outer loop.
TipSolution 
from adafruit_circuitplayground.express import cpx
import time
counter1 = 0
while True:
    time.sleep(1)
    counter1 += 1
    print(f"Outer iteration {counter1}")
    if counter1 % 3 == 0: # k ÷ 3 has remainder 0
        counter2 = 0
        while True:
            time.sleep(1)
            counter2 += 1
            print(f"Outer iteration {counter1}, inner iteration {counter2}")
            if cpx.touch_A2:
                print("terminating inner loop")
                break
            elif cpx.touch_A7 and cpx.button_b:
                print("terminating inner loop")
                break
            elif cpx.touch_A7:
                print("terminating inner loop")
                break
        if cpx.touch_A7 and cpx.button_b:
            print("terminating outer loop")
            break

3.4 The time functions

The function time.monotonic() can be used to measure intervals of time. For example, the following code measures how long after the red LED comes on the user presses button B.

from adafruit_circuitplayground.express import cpx
import time
cpx.red_led = True
start_time = time.monotonic()
while True:
    time.sleep(0.01)
    if cpx.button_b:
        end_time = time.monotonic()
        interval = end_time - start_time
        break
print(f"Interval was {interval} seconds")

Rewrite the code above so that it executes the same functionality without using break or if.

TipSolution 
from adafruit_circuitplayground.express import cpx
import time
cpx.red_led = True
start_time = time.monotonic()
while not cpx.button_b:
    time.sleep(0.01)
end_time = time.monotonic()
interval = end_time - start_time
print(f"Interval was {interval} seconds")

4 The Farmer Was Replaced

For this part of the assignment, you will have to make progress on the game The Farmer Was Replaced. You will be provided a gift card for the game soon.

Start a new game and complete the following tasks in order. Only the marked tasks require a submission; those that are not in bold are not associated with any submission.

Note

It is important that you avoid unlocking any options other than the ones explicitly listed. If you accidentally unlock more features, you should re-start the game.

  1. Harvest (grass) five times

  2. Unlock while

  3. Use the while loop to harvest 200 units of hay.

  4. ☞ Describe, in words, what the drone does using each of these snippets of code.

    • Option 1

      while True:
   do_a_flip()
   while True:
     harvest()
      TipAnswer

      Drone does one flip, and after that continues harvesting for ever.

    • Option 2

      while True:
   harvest()
   while True:
     do_a_flip()
      TipAnswer

      Drone harvests once, then does flips for ever.

    • Option 3

      while True:
   harvest()
   do_a_flip()
      TipAnswer

      Drone harvests and flips in turn until the end of time.

    • Option 4

      while True:
   do_a_flip()
   harvest()
      TipAnswer

      Drone harvests and flips in turn until the end of time, but starts with a flip instead of a harvest.

    • Option 5

      while True:
   do_a_flip()
harvest()
      TipAnswer

      Drone does flips for ever, never harvests.

    • Option 6

      while True:
   harvest()
do_a_flip()
      TipAnswer

      Drone harvests for ever, never does a flip.

  5. Unlock speed (1/5 only)

  6. ☞ Write code that executes a flip if a harvest isn’t available, and harvests if possible.

    TipAnswer

    The following code does the trick.

    if can_harvest():
   harvest()
else:
   do_a_flip()

  7. ☞ Explain why, after unlocking speed, the following code no longer works.

    while True:
   harvest()
    TipAnswer

    Because there is not enough time for the grass to grow before it is harvested.

  8. Unlock bush planting

  9. ☞ Write a program that causes the drone to keep doing flips as it waits for bushes to grow, and harvests them when they’re ready to be harvested. It immediately plants a bush once it is done harvesting.

    TipAnswer

    The following code does the trick.

    while True:
 if can_harvest():
   harvest()
   plant(Entities.Bush)
 else:
   do_a_flip()

  10. Expand the farm. It should now be size 3 x 1.

  11. ☞ Describe in words what the following code does.

    while True:
    if can_harvest():
      harvest()
      plant(Entities.Bush)
    else:
      move(North)
    TipAnswer

    This plants bushes on the entire 3 x 1 farm continuously, harvesting them as they grow.

  12. ☞ We would like to write a program that makes good use of our agricultural resources by planting bushes on two of the plots and grass on the middle plot. It may be helpful to use the following snippet of code for this purpose.

    clear()
plant(Entities.Bush)
move(North)
move(North)
plant(Entities.Bush)
move(North)

    Taking inspiration from the above code if necessary, write a script that, when run, continuously plants bushes on the two outer plots, allows grass to grow on the middle spot, and harvests all three plots one by one for ever. The first line in your code should be clear()

    TipSolution

    The following code does the trick.

    clear()
while(True):
    if can_harvest():
        # Plant bush and harvest
        harvest()
        plant(Entities.Bush)
        move(North)
        # Harvest grass. grows automatically
        harvest()
        move(North)
        # Plant bush and harvest
        harvest()
        plant(Entities.Bush)
        move(North)

  13. Unlock carrots.

  14. Expand the farm one more time, getting to a 3 x 3 size.

  15. ☞ Without using for loops, achieve the functionality shown by the following gif. It is possible to do so using just 9 lines of code. The first line should be clear().

    TipSolution

    The following code does the trick.

    clear()
while True:
  till()
  move(East)
  till()
  move(East)
  till()
  move(East)
  move(North)

  16. ☞ The following code, when executed, produces the functionality shown in the gif below.

    clear()
till()
move(East)
till()
move(East)
till()
move(East)
while True:
  # First row
  if can_harvest():
    harvest()
  plant(Entities.Carrot)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Carrot)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Carrot)
  move(East)
  move(North)
  # Second row: Grass
  if can_harvest():
    harvest()
  move(East)
  if can_harvest():
    harvest()
  move(East)
  if can_harvest():
    harvest()
  move(East)
  move(North)
  # Third row
  if can_harvest():
    harvest()
  plant(Entities.Bush)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Bush)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Bush)
  move(East)
  move(North)

    Modify the above code so that carrots are in the middle row and grass in the bottom row. The bushes should still be in the top row.

    TipSolution 
    clear()
move(North)
till()
move(East)
till()
move(East)
till()
move(East)
move(South)
while True:
  # First row: Grass
  if can_harvest():
    harvest()
  move(East)
  if can_harvest():
    harvest()
  move(East)
  if can_harvest():
    harvest()
  move(East)
  move(North)
  # Second row: Carrots
  if can_harvest():
    harvest()
  plant(Entities.Carrot)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Carrot)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Carrot)
  move(East)
  move(North)
  # Third row
  if can_harvest():
    harvest()
  plant(Entities.Bush)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Bush)
  move(East)
  if can_harvest():
    harvest()
  plant(Entities.Bush)
  move(East)
  move(North)