Question

in natural languages, there are big differences in what letters are the most common across words. in the english language, the letter \e\ is the most common, occurring 12.7% of the time, and the letter \z\ is the least common, occurring only 0.05% of the time.

that’s why the huffman coding algorithm can help to compress natural language documents. huffman coding reduces the amount of bits required to represent data by choosing new bit codes for each source symbol, and choosing short codes for the most frequent source symbols.

here’s an english word encoded with huffman coding:

0101100111110101111000110

these are the codes for the letters in the word, a subset of the table of codes for all the letters in the english alphabet:

letter bit code
a 1110
d 11111
e 100
r 0101
w 00110

decode the word according to the bit codes in the table.

what is the original english word represented by that encoding?

choose 1 answer:

a) redraw
b) reward
c) drawer
d) warder
e) warred

Explanation:

Step1: Analyze the bit string

The encoded bit string is \( 0101100111110101111000110 \). We need to match substrings of this string to the bit codes in the table. The bit codes are: \( a: 1110 \), \( d: 11111 \), \( e: 100 \), \( r: 0101 \), \( w: 00110 \) (wait, let's re - check the table. Wait, the table has: \( a: 1110 \), \( d: 11111 \), \( e: 100 \), \( r: 0101 \), \( w: 00110 \)? Wait, no, looking at the table again (as per the image):

Wait the table is:

letter | bit code

a | 1110

d | 11111

e | 100

r | 0101

w | 00110? Wait no, maybe I misread. Wait the encoded string is \( 0101100111110101111000110 \). Let's split it step by step.

First, look for the shortest bit codes first (since Huffman coding uses shorter codes for more frequent symbols). The code for \( e \) is \( 100 \) (3 bits), \( r \) is \( 0101 \) (4 bits), \( a \) is \( 1110 \) (4 bits), \( d \) is \( 11111 \) (5 bits), \( w \) is \( 00110 \) (5 bits)? Wait no, let's re - examine the encoded string: \( 0101\ 100\ 11111\ 0101\ 1110\ 00110 \). Wait let's check:

- \( 0101 \): matches \( r \) (bit code \( 0101 \))
- Then \( 100 \): matches \( e \) (bit code \( 100 \))
- Then \( 11111 \): matches \( d \) (bit code \( 11111 \))
- Then \( 0101 \): matches \( r \)
- Then \( 1110 \): matches \( a \)
- Then \( 00110 \): matches \( w \)

Putting them together: \( r - e - d - r - a - w \)? No, that's not right. Wait maybe I split the string wrong. Let's write the encoded string as a sequence: \( 0\ 1\ 0\ 1\ 1\ 0\ 0\ 1\ 1\ 1\ 1\ 1\ 0\ 1\ 0\ 1\ 1\ 1\ 1\ 0\ 0\ 0\ 1\ 1\ 0 \)

Wait another approach: Let's list the bit codes with their lengths:

- \( e \): 3 bits (\( 100 \))
- \( r \): 4 bits (\( 0101 \))
- \( a \): 4 bits (\( 1110 \))
- \( d \): 5 bits (\( 11111 \))
- \( w \): 5 bits (\( 00110 \))

Now, let's parse the encoded string \( 0101100111110101111000110 \) from left to right:

1. First 4 bits: \( 0101 \) → \( r \)
2. Next 3 bits: \( 100 \) → \( e \)
3. Next 5 bits: \( 11111 \) → \( d \)
4. Next 4 bits: \( 0101 \) → \( r \)
5. Next 4 bits: \( 1110 \) → \( a \)
6. Next 5 bits: \( 00110 \) → \( w \)

Wait that gives \( r - e - d - r - a - w \) → "redraw"? Wait no, "redraw" is r - e - d - r - a - w? Wait "redraw" is spelled r - e - d - r - a - w? Let's check the option A: redraw.

Wait let's verify the bit codes again. Let's check the encoded string length. The encoded string is \( 0101100111110101111000110 \). Let's count the bits: \( 0(1),1(2),0(3),1(4),1(5),0(6),0(7),1(8),1(9),1(10),1(11),1(12),0(13),1(14),0(15),1(16),1(17),1(18),1(19),0(20),0(21),0(22),1(23),1(24),0(25) \). Wait maybe my initial split was wrong.

Wait another way: Let's check the option "reward". Let's see the letters in "reward": r, e, w, a, r, d.

Check their bit codes:

- r: \( 0101 \)
- e: \( 100 \)
- w: \( 00110 \)
- a: \( 1110 \)
- r: \( 0101 \)
- d: \( 11111 \)

Now, let's concatenate the bit codes: \( 0101\ 100\ 00110\ 1110\ 0101\ 11111 \)

Let's count the bits: \( 0101 \) (4) + \( 100 \) (3) + \( 00110 \) (5) + \( 1110 \) (4) + \( 0101 \) (4) + \( 11111 \) (5) = \( 4 + 3+5 + 4+4 + 5=25 \) bits. The encoded string is 25 bits (let's count: \( 0101100111110101111000110 \) has 25 characters). Let's check the concatenated code for "reward":

\( 0101\ 100\ 00110\ 1110\ 0101\ 11111 \) → let's write it out: \( 0101100001101110010111111 \)? No, that's not matching. Wait maybe "drawer": d, r, a, w, e, r.

Bit codes:

- d: \( 11111 \)
- r: \( 0101 \)
- a: \( 1110 \)
- w: \( 00110 \)
- e: \( 100 \)
- r: \( 0101 \)

Concatenated: \( 11111\ 0101\ 1110\ 00110\ 100\ 0101 \)

Count bits: \( 5 + 4+4 + 5+3 + 4=25 \) bits. Let's write this out: \( 1111101011110001101000101 \)? No, not matching.

Wait let's go back to the encoded string: \( 0101100111110101111000110 \)

Let's split it as follows:

- \( 0101 \) (r)
- \( 100 \) (e)
- \( 11111 \) (d)
- \( 0101 \) (r)
- \( 1110 \) (a)
- \( 00110 \) (w)

Concatenated: \( 0101100111110101111000110 \), which matches the encoded string. So the letters are r, e, d, r, a, w → "redraw", which is option A.

Answer:

A. redraw