Week 2 · Explore: bits, bytes, ASCII and pixels¶

Introduction to Computer Science (YMT113) · Fall 2026

Nothing here is graded. Run each cell (Shift+Enter), read the output, change something, run it again. Runs in Google Colab with nothing installed.

1 · Counting in binary¶

Python can show any number in binary with bin() and pad it to 8 bits with format.

for n in [0, 1, 2, 5, 73, 128, 255]:
    print(f"{n:>4}  →  {n:08b}")

   0  →  00000000
   1  →  00000001
   2  →  00000010
   5  →  00000101
  73  →  01001001
 128  →  10000000
 255  →  11111111

Read each row from the left: 128 64 32 16 8 4 2 1. Add the place values under the 1s and you get the number back.

bits = "01001001"
value = sum(int(b) * 2 ** (7 - i) for i, b in enumerate(bits))
print(bits, "=", value)

01001001 = 73

2 · Text is numbers: ASCII and Unicode¶

ord() gives a character's code, chr() goes back. ASCII is the first 128 codes; Unicode continues from there.

for c in "Hi!":
    print(f"{c!r:5} {ord(c):4}  {ord(c):08b}")

'H'     72  01001000
'i'    105  01101001
'!'     33  00100001

print(chr(72), chr(105), chr(33))
print("Türkçe:", [ord(c) for c in "Şğü"])
print("😂 is code point", ord("😂"), "and takes", len("😂".encode("utf-8")), "bytes in UTF-8")

H i !
Türkçe: [350, 287, 252]
😂 is code point 128514 and takes 4 bytes in UTF-8

Your name as bytes:

name = "Firat"          # change me
data = name.encode("utf-8")
print(len(name), "characters →", len(data), "bytes →", 8 * len(data), "bits")
print(" ".join(f"{b:08b}" for b in data))

5 characters → 5 bytes → 40 bits
01000110 01101001 01110010 01100001 01110100

3 · The same bits, different meanings¶

Two bytes can be two letters, one 16-bit number, or part of a color. Bits carry no meaning; programs do.

two = "Hi".encode("utf-8")
print("as text   :", two.decode("utf-8"))
print("as number :", int.from_bytes(two, "big"))
print("as color  : R =", two[0], " G =", two[1], " (with B = 0 →", f"#{two[0]:02x}{two[1]:02x}00)")

as text   : Hi
as number : 18537
as color  : R = 72  G = 105  (with B = 0 → #486900)

4 · An image is a grid of numbers¶

Each pixel is three bytes (red, green, blue). Let's draw a 8×8 smiley from a grid of 0/1 and then color it.

import numpy as np
import matplotlib.pyplot as plt

smiley = np.array([
    [0,0,1,1,1,1,0,0],
    [0,1,0,0,0,0,1,0],
    [1,0,1,0,0,1,0,1],
    [1,0,0,0,0,0,0,1],
    [1,0,1,0,0,1,0,1],
    [1,0,0,1,1,0,0,1],
    [0,1,0,0,0,0,1,0],
    [0,0,1,1,1,1,0,0],
])
print(smiley)
plt.figure(figsize=(3, 3)); plt.imshow(smiley, cmap="gray_r"); plt.axis("off"); plt.show()

[[0 0 1 1 1 1 0 0]
 [0 1 0 0 0 0 1 0]
 [1 0 1 0 0 1 0 1]
 [1 0 0 0 0 0 0 1]
 [1 0 1 0 0 1 0 1]
 [1 0 0 1 1 0 0 1]
 [0 1 0 0 0 0 1 0]
 [0 0 1 1 1 1 0 0]]
No description has been provided for this image 
# now in color: every pixel is (R, G, B), 0–255
rgb = np.zeros((8, 8, 3), dtype=np.uint8)
rgb[smiley == 1] = (238, 90, 31)     # orange for the 1s
rgb[smiley == 0] = (246, 246, 242)   # paper for the 0s
print("one pixel:", rgb[0, 2], "→ 3 bytes")
print("whole image:", rgb.size, "bytes")
plt.figure(figsize=(3, 3)); plt.imshow(rgb); plt.axis("off"); plt.show()

one pixel: [238  90  31] → 3 bytes
whole image: 192 bytes
No description has been provided for this image

A 1920×1080 photo is 2 073 600 pixels × 3 bytes ≈ 6.2 MB uncompressed. JPEG makes it ten times smaller by throwing away what your eye cannot see; that is an algorithm, next week's topic.

5 · Why binary search wins¶

Count the steps needed to find one item among n, page by page (linear) and by halving (binary).

import math
print(f"{'n':>15} {'linear':>15} {'binary':>8}")
for n in [1_000, 1_000_000, 1_000_000_000, 8_000_000_000]:
    print(f"{n:>15,} {n:>15,} {math.ceil(math.log2(n)):>8}")

              n          linear   binary
          1,000           1,000       10
      1,000,000       1,000,000       20
  1,000,000,000   1,000,000,000       30
  8,000,000,000   8,000,000,000       33

Eight billion people, 33 steps. That is the wow moment of week 5, seen from here.