Given an input file with four billion integers, provide an algorithm to generate an integer which is not contained in the file. Another interesting fact is that if we sum all the red numbers in the 2nd column, it is none other than the original number, 238(=128+64+32+8+4+2). So, how many 1's in the 2nd column? Yes, 6 times operations for addition. 1st column is for x, the 2nd column indicates whether odd or even, and the 3rd column is the representation of the binary bit of the 2nd column: x Notice that we're adding y to val only when x is odd. Whenever x is odd, the value of y is added to val until x equals to zero, and then it returns 3094. Here is an example that shows output of 3 integers with field width 5 and left aligned.Ĭout << left << setw(5) << x << left << setw(5) << y << left << setw(5) << val << endl Īctually, the multiplication is known as Russian Peasant Multiplication. It first saves the integer to bitset, and the compare (xor) the bit pattern starting from both ends.įor(int i = 0 i (m) (m) (m) (m) (m) << endl Ĭout << "palindrome: " << isPalindrome(m) << endl * if the dereferenced pointer is 1, the machine is little-endianĪs we see from the output, LSB (0x39) was written first at the lower address, which indicate my machine is little endian.įollowing example tells if the bit pattern of an interger is a palindrome or not. Hint: adding 1 should turn off all the bits (because -1 + 1 = 0) and short needs 2 bytes. What's the representation of bit pattern for short -1? Invert all of the bits to the left of that one Starting from the right, find the first '1' Here is a simple way of getting two's complement: action So, the two's complement satisfies basic arithmetic, but one's complement (The resulting number by changing just the sign bit) does not. In other words, the bit pattern for -(x+1) can be described as the complement of the bits in x(aka ~x). To get negative representation of any integer, take bitwise complement and then add one! ~x + 1 = -x The resulting number is the two's complement of the number. Invert a bit pattern of a number, and then add 1. Then, how we make the sum of the two be zero?īy adding 1 after inverting all the bits in 00000111 (7): So, we need to change it to the system like the picture below: We have two issues: (two 0s and (+1)+(-1) is not zero). Let's look at the diagram for 4-bit number system: However, when we do an addition of the two, it does not become 0: Then, -7 will have the following bit pattern: How about -7? If we use the Most Significant Bit (MSB) as a sign bit, and let the value of 1 represent (-) sign. The number 7 is expressed by the following bit pattern: Let's limit our discussion to 8 bits (1 byte). In computer, every bit is mapped representing something. The code below shows how to set or clear a bit of an integer. #define CLEARBIT(a, pos) (a &= ~(1 >8) & 0xff // left most (most significant) byteĪssuming 16 bit, 2-byte short integer, two's complement: Note that a bitwise right-shift will be the equivalent of integer division by 2:ġst arg: int, 2nd arg: bit position to clear */ ![]() ![]() >): Moving all the bits of a number a specified number of places to the right. ![]() X^y = (x|y) & !(x&y ) : either x or y but not both I made this article quite a while ago to make it easier to work with binary strings, you may find it of some use.Bit n of ~x is the opposite of bit n of xīit n of x&y is 1 if bit n of x and bit n of y is 1.īit n of x|y is 1 if bit n of x or bit n of y is 1.īit n of x^y is 1 if bit n of x or bit n of y is 1 but not if both are 1. If you want to work with binary values directly, unfortunately there is no binary literal in C#. By ANDing that with the original integer, we can check if the bit is setĬonsole.WriteLine( " Base Value = ", otherValue & ( byte)(Bits.Bit0 | Bits.Bit1)) So, shifting 1 left by N will give us a binary pattern where only the Nth bit is set. Those numbers are in binary format by the way. So 1 << 5 would result in 100000, 1 << 2 would result in 100, etc. It will perform a left binary shift on the left operand. To get this, we just use the << operator. Now we just have to get a binary pattern to use against the AND. So, we can apply this further, and say that if q AND n is equal to n, then bit n was set. Now, unfortunately, you don't get a true or false value. You use the AND against a bit pattern to see if that bit is set. ![]() For example, 39 AND 4 would be worked out like so:Īs you can see, you end up with a binary sequence, which is represented by 4 in decimal (base-10). To get a bit, you have to AND it, which gets all the bits which are set in both of the numbers given to it. For example, 39 would be represented by 100111. Basically, you use the formula:īitNSet = (originalInteger & (1 << N) = 1 << N)Įffectively, every integer is represented by a binary sequence. It's a matter of using ANDs, and other boolean stuff (if you want to read multiple bits).
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