(15,11) Hamming Code, too work out how to calculate check digits

  • Thread starter Thread starter jackscholar
  • Start date Start date
  • Tags Tags
    Code Work
AI Thread Summary
In a (15,11) Hamming code, there are four check digits and eleven information digits, totaling fifteen bits. The equations provided, n=4+c and n=2^c - 1, help determine the number of check digits (c) needed for error detection and correction. The relationship 4+c=2^c-5 leads to the conclusion that c must satisfy this equation to maintain the integrity of the code. The confusion arises from understanding how to manipulate these equations to derive the number of check digits. Ultimately, the (15,11) Hamming code is structured to ensure reliable data transmission through its calculated check digits.
jackscholar
Messages
73
Reaction score
0

Homework Statement


I need to figure out how to find out the amount of check digits in a (15,11) hamming code. I know the amount of check digits is four and the amount of information digits is 7 but I don't know how to figure out how to get the answer, I only have the answer.

Homework Equations


I was given the equation n=4+c
n=2^c - 1
4+c=2^c-5
c=2^c-5

The Attempt at a Solution


I do not understand how the equation above is possible, so i could not figure out how the result c=2^c-5 was achieved
 
Last edited:
Physics news on Phys.org
Normally a (15,11) hamming code would mean 15 total bits, 11 of which are data.
 

Thread 'Family of lines that are at a distance of 5 from the origin'

I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
View full post »

Thread 'Making the denominator of a certain fraction real'

Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
View full post »

Similar threads

Converting number base of decimals
Replies
11
Views
2K
Application of N choose K, unsure of what is happening
Replies
6
Views
2K
How many different teams can be formed ...?
Replies
7
Views
2K
Product of Missing Digits in a Number
Replies
2
Views
2K
Algorithm analysis problem turned into a math problem
Replies
6
Views
2K
Advanced Arithmetic: Find Min Digits for Fraction Decimal
Replies
7
Views
3K
Finding Angle C in an ABC Triangle
Replies
19
Views
3K
B How to pick some numbers out of 13 integers, by a 4 digits code
Replies
4
Views
2K
How do I put this series into a formula?
Replies
3
Views
2K
Equation of a circle from given conditions
Replies
11
Views
2K

Hot Threads

Recent Insights

Back
Top