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A source generates six characters, \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E}, \mathrm{F}\), with respective probabilities \(0.05,0.1,0.25,0.3,0.15\), \(0.15\). Calculate the average information per character and the redundancy.

Short Answer

Expert verified
Entropy is approximately 2.389 bits/character, and redundancy is about 0.076.

Step by step solution

01

Determine the formula

The average information, also known as entropy, can be calculated using the formula: \[ H(X) = -\sum_{i=1}^{n} p(x_i) \log_2 p(x_i) \] where \( p(x_i) \) is the probability of each character.
02

Calculate entropy for each character

We calculate the entropy contribution of each character as follows:- For \( \mathrm{A} \): \( 0.05 \log_2 \left(\frac{1}{0.05}\right) = 0.216 \)- For \( \mathrm{B} \): \( 0.1 \log_2 \left(\frac{1}{0.1}\right) = 0.332 \)- For \( \mathrm{C} \): \( 0.25 \log_2 \left(\frac{1}{0.25}\right) = 0.5 \)- For \( \mathrm{D} \): \( 0.3 \log_2 \left(\frac{1}{0.3}\right) = 0.521 \)- For \( \mathrm{E} \): \( 0.15 \log_2 \left(\frac{1}{0.15}\right) = 0.41 \)- For \( \mathrm{F} \): \( 0.15 \log_2 \left(\frac{1}{0.15}\right) = 0.41 \).
03

Sum the contributions to find total entropy

Sum all individual entropy contributions to find the total entropy:\[ H(X) = 0.216 + 0.332 + 0.5 + 0.521 + 0.41 + 0.41 = 2.389 \text{ bits per character} \]
04

Calculate maximum entropy

The maximum entropy for six equally likely characters is calculated as follows:\[ H_{\text{max}} = \log_2(6) \approx 2.585 \text{ bits per character} \]
05

Determine redundancy

Redundancy can be calculated using the formula:\[ R = 1 - \frac{H(X)}{H_{\text{max}}} \]Plugging in the values we have:\[ R = 1 - \frac{2.389}{2.585} \approx 0.076 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Entropy
In the world of information theory, entropy is a measure of uncertainty or randomness associated with a set of possible outcomes. It's often understood as the average amount of information produced by a stochastic source of data. Entropy can help gauge the level of surprise or unpredictability associated with different data events.

When you calculate the entropy of a source, you're finding out how much information, on average, each event (in this case, character) of the source provides. For our six characters, entropy is calculated using the formula: \[ H(X) = -\sum_{i=1}^{n} p(x_i) \log_2 p(x_i) \] where \( p(x_i) \) is the probability of each character. Each character contributes to the overall entropy based on its likelihood of occurrence.
  • High entropy means a lot of uncertainty—like a balanced die roll.
  • Low entropy suggests predictability—like a biased die that often rolls a six.
This measure helps in optimizing the spot an average message occupies and influences how effectively you can compress the data.
Probability
Probability is the backbone concept that drives entropy calculations. It denotes the likelihood of each possible outcome or event. In this context, it tells us how often each character will appear relative to others.

Every character in our exercise has a specific probability attached, ranging from very improbable (like 'A' with 0.05) to more probable (like 'D' with 0.3). Knowing the probability helps us calculate how much useful information a character contributes to the system.
  • A higher probability of a character (e.g., 'D') means it is expected to appear more often.
  • Lower probability (e.g., 'A') means its appearance will carry more information since it is less expected.
Understanding these probabilities allows us to evaluate the unpredictability and the amount of information that the source generates.
Character Encoding
Character encoding refers to the way text is represented in digital form using binary numbers. Every character is assigned a specific number so that computers can process them. In the context of our source generating six characters, encoding plays a role in determining how efficiently they can be stored or transmitted.

Efficiency in character encoding depends on the entropy. A higher entropy might mean employing more bits for detailed encoding, while lower entropy allows some compression.
  • Fixed-length encoding might waste space with lower entropy values.
  • Variable-length encoding, like Huffman encoding, can optimize space by using shorter codes for more frequent characters.
These encoding techniques are essential for reducing storage needs and improving data transmission efficiencies.
Data Redundancy
Data redundancy involves unnecessary repetition of data within a dataset. It's generally seen as wasted space that doesn't add any new information or value. In information theory, redundancy is directly related to the difference between the maximum entropy and the actual entropy of a data set.

The concept of redundancy can be quantified using the formula: \[ R = 1 - \frac{H(X)}{H_{\text{max}}} \] where \( H(X) \) is the entropy of the source and \( H_{\text{max}} \) is the maximum possible entropy if all characters were equally probable. Redundancy quantifies the excess or non-essential elements in the data stream.
  • A higher redundancy means there may be more room for compression.
  • Less redundancy indicates a more efficient and compact dataset.
Understanding redundancy helps refine data storage and transmission strategies, eliminating unnecessary elements that do not enhance the information conveyed.

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Most popular questions from this chapter

A data stream comprises the characters A, B, C, D and \(\mathrm{E}\) with respective probabilities of \(0.12,0.21,0.07\), \(0.31\) and \(0.29\). (a) Which character carries the greatest information content? (b) Which character carries the least information content? (c) Calculate the information associated with the letter D. (d) Calculate the entropy. (e) Calculate the redundancy.

Chips are manufactured by machines \(\mathrm{A}\) and \(\mathrm{B}\). Machine A makes \(65 \%\) of the chips and machine \(\mathrm{B}\) makes the remainder. The probability that a chip is faulty is \(0.03\) when made by machine A and \(0.05\) when made by machine B. A chip is selected at random. Calculate the probability that it is (a) faulty and made by machine A (b) faulty or made by machine \(\mathrm{A}\) (c) faulty or made by machine \(\mathrm{B}\) (d) faulty and made by machine \(\mathrm{B}\) (e) faulty

A fair die is rolled. The events \(E_{1}, \ldots, E_{5}\) are defined as follows: \(E_{1}:\) an even number is obtained \(E_{2}:\) an odd number is obtained \(E_{3}:\) a score of less than 2 is obtained \(E_{4}:\) a 3 is obtained \(E_{5}:\) a score of more than 3 is obtained Find (a) \(P\left(E_{1}\right), P\left(E_{2}\right), P\left(E_{3}\right), P\left(E_{4}\right), P\left(E_{5}\right)\) (b) \(P\left(E_{1} \cap E_{3}\right)\) (c) \(P\left(E_{2} \cap E_{5}\right)\) (d) \(P\left(E_{2} \cap E_{3}\right)\) (e) \(P\left(E_{3} \cap E_{5}\right)\)

A component is manufactured by machines 1 and 2 . Machine 1 manufactures \(72 \%\) of total production of the component. The percentage of components which are acceptable varies, depending upon which machine is used. For machine \(1,97 \%\) of components are acceptable and for machine \(2,92 \%\) are acceptable. \(\mathrm{A}\) component is picked at random. (a) What is the probability it was manufactured by machine 1 ? (b) What is the probability it is not acceptable? (c) What is the probability that it is acceptable and made by machine \(2 ?\) (d) If the component is acceptable what is the probability it was manufactured by machine \(2 ?\) (e) If the component is not acceptable what is the probability it was manufactured by machine \(1 ?\)

A data stream comprises the characters A, B, C, D and \(\mathrm{E}\) with respective probabilities of \(0.23,0.16,0.11\), \(0.37\) and \(0.13\). (a) Calculate the information associated with the character B. (b) Calculate the information associated with the character D. (c) Calculate the entropy, (d) Calculate the redundancy.

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