Shortcut for hamming code
Why Stack Overflow recommends it
Stack Overflow users link it when answering or implementing Hamming code questions, since it walks through the parity-bit calculation and error finding by hand.
Recommended by 3 people across 3 different months, 2015 to 2019.
What it is
A short 2011 classroom-style video from a digital logic course at CSUSB showing a hand shortcut for computing Hamming codes and locating an erroneous bit. The presenter notes it only works for true Hamming code lengths such as 3, 7, 15 or 31 bits.
What people said
Recommends Ar maj · 2015-01 · Stack Overflow · on “Hamming codes formulas”
I learned how to solve it checks the Hamming codes through this Youtube clip. there is no effective way. wish there was a faster way to solve the problem. [this video] This is not responding properly to the question because I know what the user "user3314356" needs, but I could not comment because I did not have 50 reputation points!.
Recommends ScriptyChris · 2015-06 · Stack Overflow · on “How to break loop when values in array dont fill it whole?”
If you run my code you will se 7 groups of bits, and there are wrote wrong bits (if exist). Whole row of bits is random (at the beginning of code) and then it is checked if parity is ok or not (wrong bits). My code is (i hope so) an implementation of this tutorial [this video]
Recommends user3629249 · 2019-05 · Stack Overflow · on “Using a formula to decode hamming code(7,4)”
this hamming [this video] video walk through, by hand, of how to calculate the hamming code and how to find an erroneous bit
Cites ScriptyChris · 2015-06 · Stack Overflow · on “How to compare a certain value to value in array?”
I want to output certain values, which I declare in other array, from an array. Now I got something like that: int [] tab = new int[110]; int [] skip = {3,4,8,9,12,13,16,17,20,21}; for (int i=0; i<110; i++) { if (i == any_value_of_skip[]) System.out.print(tab[i]+", "); } I put 1 and 0 bits into tab[] array and I want to output those bits, at which index in tab[] array has exact number as in skip[] array . [edit] tab[] array has 1 and 0 , because: for (int i = 0; i<110; i++) { Random r = new Random(); tab[i] = r.nextInt(2); if (i%25==0) { System.out.println(""); } System.out.print(tab[i]+"("+i+")"+", "); } System.out.println(""); And I want to apply Hamming code here, so I need to output thos
Cites ScriptyChris · 2015-06 · Stack Overflow · on “How to repetitively output and skip certain values from array by loop?”
I have array[110] filled with integer (row) of bits (only 1 and 0), I need to divide it to 7 less rows of bits by output and skip repetitively, starting from certain (marked as next power of number 2) index of array[110] to end of this array[110]. Now by 7 different rows (groups,arrays) I mean rows of bits, starting from bits which indexes are next power of number 2, so: 1,2,4,8,16,32... Also in each next row (group), values have to be output and skipped by the number of index, which is next power of number 2 in array. So for example output 4, skip 4... or output 8, skip 8, and so on to last group, where of index in array is next power of 2. Additionally, first bit of this smaller rows is al