I'm trying to understand how a binary addition and logical OR table differs.
does both carry forward 1 or if not which one does carry forward operation and which does not?
The exclusive-or (XOR) operation is like binary addition, except that
there is no carry from one bit position to the next. Thus, each bit
position can be evaluated independently of the rest.
I'll attempt to clarify a few points with a few illustrations.
First, addition. Basically like adding numbers in grade school. But if you have a 1-bit aligned with a 1-bit, you get a 0 with a 1 carry (i.e. 10, essentially analogous to 5 plus 5 in base-10). Otherwise, add them like 'regular' (base-10) numbers. For instance:
₁₁₁
1001
+ 1111
______
11000
Note that in the left-most column two 1's are added to give 10, which with another 1 gives 11 (similar to 5 + 5 + 5).
Now, assuming by "logical OR" you mean something along the lines of bitwise OR (an operation which basically performs the logical OR (inclusive) operation on each pair of corresponding bits), then you have this:
1001
| 1111
______
1111
Only case here you should have a 0 bit is if both bits are 0.
Finally, since you tagged this question xor, which I assume is bitwise as well.
1001
^ 1111
______
0110 = 110₂
In this case, two 1-bits give a 0, and of course two 0-bits give 0.
With a logical OR you get a logical result (Boolean). IOW true OR true is true (anything other than false OR false is true). In some languages (like C) any numeric value other than 0 means true. And some languages use an explicit datatype for true, false (bool, Boolean).
In case of binary OR, you are ORing the bits of two binary values. ie: 1 (which is binary 1) bitwise OR 2 (which is binary 10) is binary 11:
01
10
11
which is 3. Thus binary OR is also an addition when the values do not have shared bits (like flag values).
Related
I need a DFA for a set of all strings beginning with a 1 that, interpreted as the binary representation of an integer, have a remainder of 1 when divided by 3.
For example, the binary number 1010 b is decimal 10. When you divide 10 by 3 you get a remainder of 1, so 1010 is in the language. However, the binary number 1111 b is decimal 15. When you divide 15 by 3 you get a remainder of 0, so 1111 is not in the language.
I've attached my DFA below. Could you please check it?
It looks correct to me.
You could make two simplifications:
q4 represents (mod 0), so you could make it the starting state and get rid of q0 and q5. (Unless you are required to reject strings beginning with a 0? Your question doesn't specify.)
q1 and q3 can be merged. They both represent (mod 1) and have the same transitions.
These two changes would leave you with exactly 3 states, one for each remainder.
I want to set the last three bit of a byte to a specific vaue. What’s the best way to archive this goal?
I can think of at least two solutions… Let’s assume I have the following Byte: 1101 0110 and want to set the three last bits to 011.
Solution 1:
1101 0110
&1111 1000 //and with mask to clear last three bits
|0000 0011 //add the target bits
Solution 2:
1101 0110 >> 3 //shift right to remove last three
0001 1010 << 3 //shift left to clear last three
|0000 0011 //add target bits
Is there a better/shorter/more efficient way?
The best way is to say
b = (b & ~7u) | 3
because 3=0...011 and 7u=0..111 in binary, and the complement of 7u is ~7u=11...1000, so the operation does what you want. It first clears the last three bits (by doing b & ~7u) and then sets the first and the second bits (by doing bitwise-OR with 3).
If the C source code has a >> in it, that does not mean the generated code will have shift instructions.
((x>>3)<<3) | 3 may generate the exact same code as (x & ~7) | 3. Compilers are very sophisticated in their optimization.
Use what is simplest #Martin James.
Recommend #blazs solution, as that is simple to understand and well copes with signed integer issues.
(x & ~7u) | 3
I would recommend you to do your second solution, sure it depends on your hardware architecture but almost always SHIFT operations are faster than ADD.
Given the following C code:
int x = atoi(argv[1]);
int y = (x & -x);
if (x==y)
printf("Wow... they are the same!\n");
What values of x will result in "Wow... they are the same!" getting printed? Why?
So. It generally depends, but I can assume, that your architecture represents numbers with sign in U2 format (everything is false if it's not in U2 format). Let's have an example.
We take 3, which representation will be like:
0011
and -3. which will be:
~ 0011
+ 1
-------
1101
and we make and
1101
& 0011
------
0001
so:
1101 != 0001
that's what is happening underhood. You have to find numbers that fit to this pattern. I do not know what kind of numbers fit it upfront. But basing on this you can predict this.
The question is asking about the binary & operator, and 2's compliment arithmetic.
I would look to how numbers are represented in 2's compliment, and what the binary & symbol does.
Assuming a 2's compliment representation for negative numbers, the only values for which this is true are positive numbers of the form 2^n where n >= 0, and 0.
When you take the 2's compliment of a number, you flip all bits and then add one. So the least significant bit will always match. The next bit won't match unless the prior carried over, and the same for the next bit.
An int is typically 32 bits, however I'll use 5 bits in the following examples for simplicity.
For example, 5 is 00101. Flipping all bits gives us 11010, then adding 1 gives us 11011. Then 00101 & 11011 = 00001. The only bit that matches a set bit is the last one, so 5 doesn't work.
Next we'll try 12, which is 01100. Flipping the bits gives us 10011, then adding 1 gives us 10100. Then 01100 & 10100 = 00100. Because of the carry-over the third bit is set, however the second bit is not, so 12 doesn't work either.
So the most significant bit which is set won't match unless all lower bits carry over when 1 is added. This is true only for numbers with one bit set, i.e. powers of 2.
If we now try 8, which is 01000, flipping the bits gives us 10111 and adding 1 gives us 11000. And 01000 & 11000 = 01000. In this case, the second bit is set, which is the only bit set in the original number. So the condition holds.
Negative numbers cannot satisfy this condition because positive numbers have the most significant bit set to 0, while negative numbers have the most significant bit set to 1. So a bitwise AND of a number and its negative will always have the most significant bit set to 0, meaning this number cannot be negative.
0 is a special case since it is its own negative. 0 & 0 = 0, so it also satisfies this condition.
Another special case is the smallest number you can represent. In the case of a 5-bit number this is -16, which is represented by 10000. Flipping all the bits gives you 01111 and adding 1 gives you 10000, which is the same number. On the surface it seems this number also satisfies the condition, however this is an overflow condition and implementations may not handle this case correctly. See this link for more details.
What does the following condition effectively check in C :
if(a & (1<<b))
I have been wracking my brains but I can't find a pattern.
Any help?
Also I have seen this used a lot in competitive programming, could anyone explain when and why this is used?
It is checking whether the bth bit of a is set.
1<<b will shift over a single set bit b times so that only one bit in the bth position is set.
Then the & will perform a bitwise and. Since we already know the only bit that is set in 1<<b, either it is set in a, in which case we get 1<<b, or it isn't, in which case we get 0.
In mathematical terms, this condition verifies if a's binary representation contains 2b. In terms of bits, this checks if b's bit of a is set to 1 (the number of the least significant bit is zero).
Recall that shifting 1 to the left by b positions produces a mask consisting of all zeros and a single 1 in position b counting from the right. A value of this mask is 2b.
When you perform a bitwise "AND" with such a mask, the result would be non-zero if, and only if, a's binary representation contains 2b.
Lets say for example a = 12 (binary: 1100) and you want to check that the third bit (binaries are read from right to left) is set to 1, to do that you can use & bitwise operator which work as following:
1 & 0 = 0
0 & 1 = 0
0 & 0 = 0
1 & 1 = 1
To check if the third bit in a is set to 1 we can do:
1100
0100 &
------
0100 (4 in decimal) True
if a = 8 (binary: l000) on the other hand:
1000
0100 &
------
0000 (0 in decimal) False
Now to get the 0100 value we can right shift 1 by 2 (1 << 2) wich will append two zeros from the right and we'll get 100, in binaries left trailing zeros doesn't change the value so 100 is the same as 0100.
I know and understand the result.
For example:
<br>
7 (decimal) = 00000111 (binary) <br>
and 7 >> 2 = 00000001 (binary) <br>
00000001 (binary) is same as 7 / 4 = 1 <br>
So 7 >> 2 = 7 / 4 <br>
<br>
But I'd like to know how this logic was created.
Can anyone elaborate on this logic?
(Maybe it just popped up in a genius' head?)
And are there any other similar logics like this ?
It didn't "pop-up" in a genius' head. Right shifting binary numbers would divide a number by 2 and left shifting the numbers would multiply it by 2. This is because 10 is 2 in binary. Multiplying a number by 10(be it binary or decimal or hexadecimal) appends a 0 to the number(which is effectively left shifting). Similarly, dividing by 10(or 2) removes a binary digit from the number(effectively right shifting). This is how the logic really works.
There are plenty of such bit-twiddlery(a word I invented a minute ago) in computer world.
http://graphics.stanford.edu/~seander/bithacks.html Here is for the starters.
This is my favorite book: http://www.amazon.com/Hackers-Delight-Edition-Henry-Warren/dp/0321842685/ref=dp_ob_image_bk on bit-twiddlery.
It is actually defined that way in the C standard.
From section 6.5.7:
The result of E1 >> E2 is E1 right-shifted E2 bit positions. [...]
the value of the result is the integral part of the quotient of E1 / 2E2
On most architectures, x >> 2 is only equal to x / 4 for non-negative numbers. For negative numbers, it usually rounds the opposite direction.
Compilers have always been able to optimize x / 4 into x >> 2. This technique is called "strength reduction", and even the oldest compilers can do this. So there is no benefit to writing x / 4 as x >> 2.
Elaborating on Aniket Inge's answer:
Number: 30710 = 1001100112
How multiply by 10 works in decimal system
10 * (30710)
= 10 * (3*102 + 7*100)
= 3*102+1 + 7*100+1
= 3*103 + 7*101
= 307010
= 30710 << 1
Similarly multiply by 2 in binary,
2 * (1001100112)
= 2 * (1*28 + 1*25 + 1*24 + 1*21 1*20)
= 1*28+1 + 1*25+1 + 1*24+1 + 1*21+1 1*20+1
= 1*29 + 1*26 + 1*25 + 1*22 + 1*21
= 10011001102
= 1001100112 << 1
I think you are confused by the "2" in:
7 >> 2
and are thinking it should divide by 2.
The "2" here means shift the number ("7" in this case) "2" bit positions to the right.
Shifting a number "1"bit position to the right will have the effect of dividing by 2:
8 >> 1 = 4 // In binary: (00001000) >> 1 = (00000100)
and shifting a number "2"bit positions to the right will have the effect of dividing by 4:
8 >> 2 = 2 // In binary: (00001000) >> 2 = (00000010)
Its inherent in the binary number system used in computer.
a similar logic is --- left shifting 'n' times means multiplying by 2^n.
An easy way to see why it works, is to look at the familiar decimal ten-based number system, 050 is fifty, shift it to the right, it becomes 005, five, equivalent to dividing it by 10. The same thing with shifting left, 050 becomes 500, five hundred, equivalent to multiplying it by 10.
All the other numeral systems work the same way.
they do that because shifting is more efficient than actual division. you're just moving all the digits to the right or left, logically multiplying/dividing by 2 per shift
If you're wondering why 7/4 = 1, that's because the rest of the result, (3/4) is truncated off so that it's an interger.
Just my two cents: I did not see any mention to the fact that shifting right does not always produce the same results as dividing by 2. Since right shifting rounds toward negative infinity and integer division rounds to zero, some values (like -1 in two's complement) will just not work as expected when divided.
It's because >> and << operators are shifting the binary data.
Binary value 1000 is the double of binary value 0100
Binary value 0010 is the quarter of binary value 1000
You can call it an idea of a genius mind or just the need of the computer language.
To my belief, a Computer as a device never divides or multiplies numbers, rather it only has a logic of adding or simply shifting the bits from here to there. You can make an algorithm work by telling your computer to multiply, subtract them up, but when the logic reaches for actual processing, your results will be either an outcome of shifting of bits or just adding of bits.
You can simply think that for getting the result of a number being divided by 4, the computer actually right shifts the bits to two places, and gives the result:
7 in 8-bit binary = 00000111
Shift Right 2 places = 00000001 // (Which is for sure equal to Decimal 1)
Further examples:
//-- We can divide 9 by four by Right Shifting 2 places
9 in 8-bit binary = 00001001
Shift right 2 places: 00000010 // (Which is equal to 9/4 or Decimal 2)
A person with deep knowledge of assembly language programming can explain it with more examples. If you want to know the actual sense behind all this, I guess you need to study bit level arithmetic and assembly language of computer.