How to implement LoRaWAN class A, B and C in OMNeT++ using FLoRa or other?
I want to run an example of the FLoRa framework.
Related
Suppose three devices A, B and C in a CDMA network with the following 8-bit orthogonal codes:
A = 10101010
B = 11001100
C = 10010110
The transmission power of B is twice as compared to A and C. Perform all the steps to send and
detect 00 for Sender A, 10 for Sender B and 0X for Sender C (where X means that the sender
doesn’t transmit in this interval).
I got stuck by a simple box puzzle in a mobile game called Genshin impact. Here is the scenario:
There are n boxes in front of the player. Each box has one side marked in blue, and the player has to turn one box at a time until all the blue sides facing the player. (There are some relationships between boxes, for example, if you turn box A, the boxes C and D will turn as well). The box can only be turned horizontally so we only need to consider 4 sides.
I normally apply brutal force to solve this kind of a puzzle but now I am curious about the hidden logic.
Here is a concrete example:
n = 5 (We have 5 boxes in front of us) named A, B, C, D, E
if you turn A, C will move as well A->C
if you turn B, A, C will move B->(A,C)
if you turn C, A, E will move C->(A,E)
if you turn D, C, E will move D-> (C,E)
if you turn E, C will move E-> C
All the turns are synced with the same pace, which means turn B one unit, A, C will turn one unit as well. To simply the scenario, in this game, the player turns one unit at a time and always turns clockwise.
The initial state is:
A = 3
B = 0
C = 2
D = 1
E = 2
What are these numbers represent? when the blue mark is in front of me, I assign that position 0, the blue mark to the left side, I assign that position 1, and so on. So the number 2 means the blue mark is in the back. Therefore, we can use position number mod 4 to determine the blue mark's location.
Any ideas? Any coding simulation or mathematic interpretation will be greatly appreciated.
Consider two computers A and B. A has uses G and B uses G' where G' != G and of different degrees. Computer A wants to send a data D and it uses CRC for that cause.
The claim says that there couldn't be a scenario where A sends a CRC-message corresponding to data D and computer B will accept it as a valid message. Why is that?
We know that computer A sends D*2^r XOR R (r is the degree of G) and computer B divides by G'. So in other words, why can't G', accidentally, divide D*2^r XOR R?
Obviously it has something to do with the fact that deg(G) != deg(G') but I didn't figure it out.
Thanks!
I am fairly new to the Z80 and machine code, so please don't assume I know anything.
Basically, what I want to know is this: If you load register H with a value (I'll call it y), will HL then be 0xy0? e.g. if H was loaded with 0xAF would HL be 0xAF00?
And would the same be true for loading L with y?
Thanks in advance.
The H and L 8-bit registers can be treated independently. Loading a value in H, will not affect the value in L, and vice versa. The two registers H and L can also be treated as a 16-bit register pair. The following source FIRST STEPS IN MACHINE CODE describes this.
two single register transfers, e.g.
LD H, B
LD L, C
to copy BC into
HL.
and
You can, if you wish, load a register pair directly with a single
instruction, rather than use two instructions. From last time, you
will recall that the H and L, B and C, and D and E registers can be
paired such that they effectively can hold any number between 0 and
65535 (00 to FFFF hex). C, E, and L form the low byte of the pair,
while B, D, and H are the high bytes.
I want to find a formula with it I can find the number of the communication that can occur at the same time (different combination) in f port router for network on chip.
In network on chip's router there are 5 port and each port has input and output.
5 is the maximum number of communication that can happen at the same time. Here an example:
input 1 -> output 2
input 2 -> output 3
input 3 -> output 4
input 4 -> output 5
input 5 -> output 1
I want to calculate how many combinaison I can have.
I could only figure out that the question is based on mathematics and has nothing to do with the networking and probability tags!
As you mentioned
In each port the input cannot communicate with other inputs and cannot communicate with the output in the same ports.
So, assuming one port communicating via input,
the port can communicate with only 4 output of other ports and not with the 4 input of the other ports and also not with the output of the same port.
So,net communication in this case while considering 5 inputs = 5*4 = 20.
Similarly,for the output case, the overall communications = 5*4 = 20.
Therefore,net communication done= 20 + 20 = 40.
If duplication while communicating is not considered(taken into account), then the total communication for input & output for all ports= 2 * 5 Combination 2 = 2 * 10 = 20.