SOLUTION: Please determine the number of bits to encode the part of speech of a sequence of words in English using a tag set of size 40. Assume the tag is a random variable. The following is

Question 1163933: Please determine the number of bits to encode the part of speech of a sequence of words in English using a tag set of size 40. Assume the tag is a random variable. The following is the information that we have:
Among the 40 values of the tag, we know 90% of chances to be one of the four values: ART (article), P (preposition), N(noun), and V(verb). These four values are equally likely to be in a sequence. All other values of the tag has 10% chances which each of them is also equally likely to be in a sequence.
Answer by CPhill(2264) (Show Source): You can put this solution on YOUR website!
```python?code_reference&code_event_index=2
import math
p1 = 0.9 / 4
p2 = 0.1 / 36
entropy = -(4 * p1 * math.log2(p1) + 36 * p2 * math.log2(p2))
print(f"{p1=}")
print(f"{p2=}")
print(f"{entropy=}")

```
```text?code_stdout&code_event_index=2
p1=0.225
p2=0.002777777777777778
entropy=2.7859880937335126
```
To determine the number of bits required to encode the part of speech of a word in a sequence, we calculate the **entropy** ($H$) of the tag set. In information theory, the entropy represents the average minimum number of bits needed to encode a random variable based on its probability distribution.
### 1. Identify the Probabilities
The tag set has a total of 40 values, which are divided into two groups:
* **Group 1 (Frequent Tags):** There are 4 values (ART, P, N, and V). These account for 90% ($0.90$) of the occurrences and are equally likely.
* Probability of each tag ($P_{frequent}$) = $\frac{0.90}{4} = 0.225$
* **Group 2 (Other Tags):** There are $40 - 4 = 36$ remaining values. These account for 10% ($0.10$) of the occurrences and are also equally likely.
* Probability of each tag ($P_{other}$) = $\frac{0.10}{36} \approx 0.002778$ (or $\frac{1}{360}$)
### 2. Calculate the Entropy ($H$)
The formula for entropy is:
$$H(X) = -\sum_{i=1}^{n} p_i \log_2(p_i)$$
Plugging in our values for the two groups:
$$H(X) = - \left[ 4 \times (0.225 \log_2 0.225) + 36 \times \left( \frac{1}{360} \log_2 \frac{1}{360} \right) \right]$$
Breaking it down:
* For the 4 frequent tags: $4 \times 0.225 \times \log_2(0.225) \approx 0.9 \times (-2.152) \approx -1.9368$
* For the 36 other tags: $36 \times \frac{1}{360} \times \log_2(\frac{1}{360}) \approx 0.1 \times (-8.492) \approx -0.8492$
$$H(X) = -(-1.9368 - 0.8492) = 2.786 \text{ bits}$$
### Final Result
The average number of bits required to encode the part of speech for each word in the sequence is approximately **2.786 bits**.
*(Note: If you were using a fixed-length encoding without considering probabilities, you would need $\lceil \log_2(40) \rceil = 6$ bits. However, based on the provided probability distribution, an optimal variable-length encoding like Huffman coding would achieve an average length close to the entropy of 2.786 bits.)*
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