Int Oddbits ( Void )

The Vedic multiplier is depends on the “UrdhvaTiryagbhyam” sutra (algorithm). These Sutras have been traditionally used for the multiplication of two numbers in the decimal number system. In this work, we apply the same ideas to the binary number system to make the proposed algorithm compatible with the digital hardware. It is a general multiplication formula applicable to all cases of multiplication. It literally means “Vertically and Crosswise”. It is based on a novel concept through which the generation of all partial products can be done with the concurrent addition of these partial products. The algorithm can be generalized for n x n bit number. Since the partial products and their sums are calculated in parallel, the multiplier is independent of the clock frequency of the processor. Due to its regular structure, it can be easily layout in microprocessors and designers can easily circumvent these problems to avoid catastrophic device failures. The processing power of multiplier can easily be increased by increasing the input and output data bus widths since it has a quite a regular structure. Due to its regular structure, it can be easily layout in a silicon chip. The Multiplier based on this sutra has the advantage that as the number of bits increases, gate delay and area increases very slowly as compared to other conventional

So, I went back to the book and looked in chapter two for a while then I went on to chapter three and found on pg. 91 -97 about the if and else statements to create a module. After I played with it for a while and kept getting error messages. I finally realized when I typed if I would type the condition which needed to be int(number) % 2 == 0 : because int is for integer that way the program would know it would be a whole number. The “%” is used in this program because it means remainder or module and I used it for the module part. The double == signs just bean equal I found this on pg. 92. Next on pg. 93 it shows to indent and type your sequence of statements and mine is print(“This si even”). I did this to so when the user types in any even number it will tell they user that it is an even number. Next, when I typed else I hit enter and indented again to type my sequence of statements, which is print(“this is odd”). This is so when the user types in an odd number the program will tell them that it is odd. From our notes I saw how to include the random library which is to type random = random.randint (0, 20000). The first time I entered this I spelled it wrong and immediately received an error message. I went back and changed the spelling and it ran

Count from left to right along the line until you pass 177. Since we have to remember about the 2 address that we subtract I changed the number to 177 to account for them so we make sure we have enough addresses. 128 is not enough so we must go to 256. Now, 256 is the 8th number over so we need to RESERVE 8 bits for the host portion of the address. Since we are solving for host we count those 8 bits from right to left starting at the end of the address with 0’s. In Binary it would look like this:

The mantissa occupies the next 23 bits and is assumed to be preceded by 1 and the radix point.

MGSE3.NBT.5 Multiply a whole number of up to four digits by one-digit whole number, and multiply by two-digit numbers, using strategies based on place value and

Class C: The first and second bits in the first octet become 1s. The third bit must remain a 0. The remaining five bits can be a 1, giving a maximum total of 223. Class

For a 3 bit unsigned number our range is 0 to 7, 0b000 to 0b111. for a 3 bit signed (twos complement) interpretation our range is -4 to +3 (0b100 to 0b011). For now limiting ourselves to 3 bits if you add 7+1, 0b111 + 0b001 = 0b1000 but we only have a 3 bit system so that is 0b000, 7+1 = 8, we cannot represent 8 in our system so that is an overflow, because we happen to be interpreting the bits as unsigned we look at the "unsigned overflow" which is also known as the carry bit or flag. Now if we take those same bits but interpret them as signed, then 0b111 (-1) + 0b001 (+1) = 0b000 (0). Minus one plus one is zero. No overflow, the "signed overflow" is not set...What is the signed overflow?

Suppose that alpha is an int variable and ch is a char variable and the input is:

3 0 , 1 9 8 6 A c tu al ( ‘ 0 0 0 's ) S a le s R e v e n u e - T h i r d P a r t y - D ive r sifie d P r o d u c t G ro u p - T o tal D e l i ve ry C o s t N et S ales R e ve n u e V a ri a b l e C o s t s S tan d ar d C o s t A d ju stm e n ts E f fic ie n c y V a r ia n c e T o tal G ro s s M a rg i n - $ 6 5 ,8 1 6 6 ,0 - 2 ,7 6 3 ,2 7 6 3 9 3 2 0 2 3 9 B u dget ( ‘0 0 0 's ) 6 1 ,0 2 6 1 ,2 - 2 ,6 5 8 ,6 5 1 6 0 6 0 0 0 0 0 D e viatio n ( ‘ 0 0 0 's ) 4 ,8 2 5 4 ,7 2 19 4 ,5 7 - 1 ,1 3 5 24 84 2 0 2 3 9 9 4 1 4

0 0 0 2 4 100 0 0 0 0

Please read the chapter2 of the textbook: "Binary Values and Number Systems" and the the two lectures by Dr. Simona Doboli under the "Module M2: Binary values and number system" on the Blackboard.

13. What is overflow and how can it be detected? How does overflow in unsigned numbers differ from overflow in signed numbers?

# include # include # include # include # include # include # define Black 0 # define White 255 # define Negative 0 # define Positive 1 # define HEIGHT 480 # define WIDTH 640 # define ITERATIONS 200 phtOLOHANDLE hphtolo, hBuffer; UINT32 ImageWidth, ImageHeight, ImageSize; BbTE *pun8ImageBuffer; BbTE *pun8AlignedBuffer; phtOLOSTATUS phtoloStatus; int fb = 0; int f = 0; int a; int b; int pht(640*480); int BWpht(640)(480); int pht(640)(480); int Left(200); int Right(200); int Vertical(200); int ind; float Horz-avg(200); float Horz_avg_r(200); float Valley (30); float Volly_set(30); int Index1(30); int Index11(30); int Index12(30); int ebe_start,ebe_a_end_l,ebe_b_start_l,ebe_b_end_l; int ebs_a_strt_r,ebe_a_end_r,ebe_b_start_r,ebe_b_end_r; int Top; int Index11_left,Index11_Right,Index12_left,Index12_Right; int distance_left, distance_Right; int Top_a, Top_b; int open_dist,distance_new; int distance_old; FILE *stream; FILE *fd; FILE *fd1; FILE *fd2; void phtolo_initialize(); void Acquire(); void Binarization(); int Face_detect(); int Top_detect(int a, int b); void Detect_edge(int a, int b); void Remove_noise(int a1, int Top); void Horz_avg(int *start , int *end, int side); int fuc(int side); int ind(int ind, int side); int Detect_center( int a, int b); void Printebe(int side, int f); main() distance_old==0; open_dist = =0; f==0; int i,j,z; phtolo_initialize(); printf( 'START: '); getchar(); for (z=1:5) Acquire(); Binarization(); a == Face_detect(); Top_b =

A user had opened a word processor, typed the numbers 123456789, and stored the document as a filed called report1. Which of the following determines, in part, what bits the computer

Basic programming in C. In the bubble sort algorithm, there are require to exchange data from