Two roles of nodes in I2C communication are: master and slave.
Major Difference Between SPI and I2C is: SPI is full Duplex (SPI can be full/half duplex, depending on the hardware, 3 or 4 wire) and I2C is half duplex communication....we cannt send and receive data at a time in I2c..but it possible in SPI ....SPI much faster Than I2C....... Main topic to discuss the difference is 1) speed is difference in case of i2c is 100/400kbps in 7 bit mode max 400kbps (But High speed I2C communication protocols allow speeds upto 3.4Mbps!!) in case of spi speed is upto 1mbps (the higher speed of SPI is due to fact that unlike I2C, SPI interfaces to a slave device using a sperate pin called the slave select + no concept of acknowledgements which means increased Band width..) 2)Connection wise: i2c require less pin then spi (SPI 3wire: 3 IOs, SPI 4 wire: 4 IOs, I2C: 2 IOs) as spi require slave select for individual device... 3)it is better to use i2c in case of if u want to connect more device to connect. (slave addressing advantageous over SPI slave select individual pins for individual slaves) 4)bus arbitration is possible in case of i2c... not in case of spi... (SPI does not require it, since only 1 slave is controlled by the master at any given point) 6)noise sensitivity of i2c is high... there is chance to corrupt the r/w bit... then whole data is corrputed... but in case of spi.. chance is very less as whole word is trasmitted... However, I must say that I2C is more reliable, since the protocol supports slave feedback machanism (ACK) to detect whether was received correctly or corrupt. 8) it is easy to implement the spi...while i2c is little bit complex... (over heads) Formatted Answer @: http://www.bubblews.com/account/65122-kapilddit
The number of even nodes in any list is always half size of the list rounded down to the nearest integer. To round down you simply take the integral portion of the division. E.g., for a list of 5 nodes, nodes 2 and 4 are the even nodes, therefore there are only 2 even nodes. Thus: 5 / 2 = 2.5 = 2. A list of 4 nodes also has 2 even nodes, thus 4 / 2 = 2.0 = 2.
Ne=N2+1Here Ne=no. of leaf nodesN2= no. of nodes of degree 2
yes
lamda /2
ReBoot - 1994 Cross Nodes 4-2 was released on: USA: 10 October 2001
In a binary tree, each level can have a maximum of (2^n) nodes, where (n) is the level number starting from 0. For a binary tree with 3 levels (0, 1, 2), the minimum number of nodes occurs when each level has at least one node. Therefore, the minimum number of nodes is 1 (at level 0) + 1 (at level 1) + 1 (at level 2) = 3 nodes.
If a figure has 2 odd nodes and the rest even nodes then the figure is traceable.
For an s orbital, there are no angular nodes. For a p orbital, there is 1 angular node. For a d orbital, there are 2 angular nodes. The maximum number of angular nodes is given by n-1, where n is the principal quantum number of the orbital.
1014 it is. no of different trees possible with n nodes is (2^n)-n thanx
let suppose total number of nodes/computers = n the formula will be = n(n-1)/2 e.g = 6(6-1)/2 =15 links.
With 3 nodes, there are 5 possible ordered trees. This can be calculated using the formula for the number of ordered trees: n^(n-2), where n is the number of nodes. In this case, 3^(3-2) = 3^1 = 3.