Help for combinatorial question?

  • Context: Undergrad 
  • Thread starter Thread starter erogol
  • Start date Start date
Click For Summary

Discussion Overview

The discussion revolves around a combinatorial problem involving "codewords" formed from the alphabet A={0,1,2,3}. Participants are exploring the criteria for legitimacy based on the number of zeros in the codewords, specifically focusing on how many n-letter codewords contain an even number of zeros.

Discussion Character

  • Homework-related, Mathematical reasoning, Debate/contested

Main Points Raised

  • One participant defines a legitimate codeword as one that contains an even number of zeros, providing examples of legitimate and illegitimate codewords.
  • Another participant expresses uncertainty about how to approach the problem but states a proposed answer: 2^(2n-1) + 2^n - 1.
  • A suggestion is made to consider a summation over all possible even numbers of zeros to find the total number of legitimate codewords.
  • Some participants question the willingness of others to engage with the problem-solving process.

Areas of Agreement / Disagreement

There is no consensus on the solution to the problem, and participants express varying levels of engagement and willingness to attempt solving it.

Contextual Notes

Some assumptions about the problem-solving approach and the definitions of legitimacy may be implicit, and the mathematical steps to derive the proposed answer remain unresolved.

erogol
Messages
14
Reaction score
0
a "codeword" from the alphabet A={0,1,2,3) is said to be legitimate if it contains even number of zeros. Thus for instance the codeword 31020 is legitimated and 0002 is not. How many n - letter codewords are legitimated ?
 
Physics news on Phys.org
erogol said:
a "codeword" from the alphabet A={0,1,2,3) is said to be legitimate if it contains even number of zeros. Thus for instance the codeword 31020 is legitimated and 0002 is not. How many n - letter codewords are legitimated ?

Hi erogol! :wink:

Show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
 
i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1
 
erogol said:
i have no sense to solve it i just know answer is 2^(2n-1) + 2^n -1

ok, start by making a sum over all possible (even) numbers of 0s …

the total number of legitimated words is ∑ what ? :smile:
 
So you are not even going to try?
 
HallsofIvy said:
So you are not even going to try?

Are you talking to me? :biggrin:
 

Similar threads

MHB What Is an Encoding Matrix in Coding Theory?
  • · Replies 6 ·
Replies
6
Views
2K
MHB A simple (?) combinatorial problem
  • · Replies 2 ·
Replies
2
Views
2K
Where Does 10^{n-1} Come From in the Recurrence Relation Example?
  • · Replies 6 ·
Replies
6
Views
4K
MHB Solve GRE Combinations Question: 24 x 9 x 10 Options
  • · Replies 5 ·
Replies
5
Views
2K
MDS/cyclic code, binary string question
  • · Replies 5 ·
Replies
5
Views
3K
Undergrad Making Sense of Notation Confusion in Statistical Digital Signal Processing
  • · Replies 6 ·
Replies
6
Views
2K
Undergrad Counting Quadruples of Integers: A Combinatorial Problem
  • · Replies 3 ·
Replies
3
Views
2K
MHB Graph Grammer Function For Two Exotic Graphs Question please.
  • · Replies 4 ·
Replies
4
Views
2K
Undergrad Help me understand skewness in QQ-plots please
  • · Replies 1 ·
Replies
1
Views
2K
Graduate Proof that if the alphabet set is at most countable, then strings cnt
  • · Replies 4 ·
Replies
4
Views
3K