can anybody tell me about even parity checker ?
361852
In computing and telecommunication, a parity bit is a binary digit that indicates whether the number of 1 bits in the preceding data was even or odd. If a single bit is changed in transmission, the message will change parity and the error can be detected at this point. (Note that the bit that changed may have been the parity bit itself!) The most common convention, or parity scheme, is that a parity bit of 1 indicates that there is an odd number of ones in the data, and a parity bit of 0 indicates that there is an even number of ones in the data.
902613
I just plugged this into my calculator.\n\n[1, 0, 2, 0, 2, 0 | 0\n0, 1, -3, 0, -2, 0 | 0\n0, 0, 0, 1, -1, 0 | 0\n0, 0, 0, 0, 0, 1 | 0\n0, 0, 0, 0, 0, 0 | 1 ]\n\nTake it however you want.
1061702
every time i've asked a teacher this, they've told me that its an even number.  The only reason i can come up with is that it is found between  two odd numbers. (-1 and 1)
951662
well the number to the utmost left gets added by 11 and in every row 1 gets added more than the previous row so if u see the difference between any two consecutive row is always 12 !!\nthts wat i thnk
1065380
I won't give you the fomula- figure it out yourself.\nBelow are the possible combinations of 3 bit base 3; you decide on how a forth bit (used as a complement) would affect it.\n\n000\n001\n002\n010\n011\n012\n020\n021\n022\n100\n101\n102\n110\n111\n112\n120\n121\n122\n200\n201\n202\n210\n211\n212\n220\n221\n222
917997
Simply loop through the list of numbers and compute:\n\n'number' modulo '2'\n\nIn C, that would be:\n\nnumber % 2\n\nA modulus of 2 would return a '1' for odd numbers, and '0' for even numbers (since even numbers divide by 2 with no remainder, and odd numbers divide by 2 with a remainder of 1).\n\nExample:\n\nint index;\nint numArray[25];\n\n...(store the 25 numbers in the array)...\n\nfor(index = 0; index < 25; index++) {\n     if(numArray[i] % 2) {\n          printf("%d&#92;n", numArray[i]);\n     }\n}
145253
Take a look at this link. You should find all you need about Fermat Numbers there.
90357
F- even and odds.  There are 3 even sides and 3 odd sides. So the likelihood is the same for both.
864328
binary 11 is 3 decimal
844366
Hmm, I think it's simple addition. The cores are all identical. So if pushed to the limit, it should perform *close* to three times the original, namely 3.2 x 3.
706983
We want to figure out what BER is, just knowing the chance that a symbol (two bits in a row) gets an error.\n\nThe chance that a single bit stays the same is 1-BER.\nAssuming errors are independent, then the chance that two bits in a row are unchanged is (1-BER)^2.  So, set this equal to 0.8 (the chance that an entire symbol is unchanged), and solve for BER.\n\n(I answer this w/o knowing, off-hand, what Amplitude Shift Keying is; this answer just relies on the fact that bit errors will be independent.)
157637
The list may contain more than once the same number.\nthese are all valid lists:\n2 1 3 5 9\n1 1 1\n-5 0 -5 0 -5 1 5\n\nThe algorithm will be O(n) (you have to examine every element at least once and once is enough).\nc code would be something like this\n\nint *find_last_min(int *list, int length)\n{\n  int i, *min = list; /* assuming the list is not empty */\n  for(i = 1; i < length; ++i)\n    if(*min >= list[i])\n      min = list + i;\n  return min;\n}
605811
As Lauren said, it's 1/x.\n\nMemorize this. It will be very useful.
774675
the code for the prg is\n\nmain()\n{\nint no1,no2,min,max,i;\nscanf("%d %d", &no1,&no2);\nif (no1<no2)\n{\nmin=no1;\nmax=no2;\n}\nelse\n{\nmin=no2;\nmax=no1;\n}\n\nfor(i=min;i<=max;i++)\nprintf ("%d&#92;n",i);\n\n}
284585
It depends on the definition that you use for Gray code since there "is no unique definition for Gray code" [Peter Jackel, p. 84]. The most commonly used (at least in a Sobol sequence context) is:\n \nG(n) = n bitxor [n/2],\nwhere [x] is the integer part of x,\nbitxor is the bitwise operation on the binary representation of to integers.\n\nSo you have to express 17 and 8 in binary and perform bitxor.\n\n17 = ___________10001\n8  = ____________01000\nbitxor(17, 8) =____ 11001 = 25\n\n(I did my best to align the 3 numbers above but am not sure how the'll show up in other pcs)\n\nObservation: the bitxor operation can be performed as a "sum without carry on the binary representation".
1098087
Binary code is just 1's and 0's. Reading it is totally arbitrary unless you know what is being coded. The same byte of 8 bits of 1 and zeros could be an ascii character, machine code, or color definition.. ... and on and on. So in actuality your friend is full of poopy.
343578
you search for a common factor, 2s,3s,5s and 10s are the most apparent. Once you look to a number you can immediately decide if it is divisible by 2 ( if the number is even) or by 3 (if the total of the digits is divisble by 3 i.e 168 total = 15 divisible by 3), or by 5 ( the last digit is 0 or 5) or divisble by 10 ( the last digit is 0).. good luck