How to implement bitOR and bitAND operation (on two variable-sized ints, but at least as small as 8 bits) using just basic arithmetics? I don't care about execution speed, the most important is the simplicity and size of code. I've managed to get negation, xor and shifts implemented.
If you already have xor, then I propose the following:
bitAnd(a,b) = ((a+b) - bitXor(a,b)) / 2
where / 2 is (truncated) integer quotient of division by 2 (or bit shift 1 place to the right).
Beware, integers must be sufficiently wide to not overflow!
If we loose highest bit in (a+b) operation, then the operation will not work.
Then you can reconstruct bitOr with one of:
bitOr(a,b) = bitXor(a,b) + bitAnd(a,b)
bitOr(a,b) = (a+b) - bitAnd(a,b)
Related
Figure 10.4 provides an algorithm for converting ASCII strings to binary values. Suppose the decimal number is arbitrarily long. Rather than store a table of 10 values for the thousands-place digit, another table for the 10 ten-thousands-place digit, and so on, design an algorithm to do the conversion without resorting to any tables whatsoever.
I have attached pictures of figure 10.4. I am not looking for an answer to the problem, but rather can someone please explain this problem and perhaps give some direction on how to go about creating the algorithm?
Figure 10.4
Figure 10.4 second image
I am unsure as to what it means by tables and do not know where to start really.
The tables are those global, initialized arrays: one called Lookup10 holding 10, 20, 30, 40, ..., and another called Lookup100 holding 100, 200, 300, 400...
You can ignore the tables: as per the assignment instructions you're supposed to find a different way to accomplish this anyway. Or, you can run that code in simulator or mentally to understand how it works.
The bottom line is that LC-3, while it can do anything (it is turning complete), it can't do much in any one instruction. For arithmetic & logic, it can do add, not, and. That's pretty much it! But that's enough — let's note that modern hardware does everything with only one logic gate, namely NAND, which is a binary operator (so NAND directly available; NOT by providing NAND with the same operand for both inputs; AND by doing NOT after NAND; OR using NOT on both inputs first and then NAND; etc..)
For example, LC-3 cannot multiply or divide or modulus or right shift directly — each of those operations is many instructions and in the general case, some looping construct. Multiplication can be done by repetitive addition, and division/modulus by repetitive subtraction. These are super inefficient for larger operands, and there are much more efficient algorithms that are also substantially more complex, so those greatly increase program complexity beyond that already with the repetitive operation approach.
That subroutine goes backwards through the use input string. It takes a string length count in R1 as parameter supplied by caller (not shown). It looks at the last character in the input and converts it from an ASCII character to a binary number.
(We would commonly do that conversion from ascii character to numeric value using subtraction: moving the character values from the ascii character range of 0x30..0x39 to numeric values in the range 0..9, but they do it with masking, which also works. The subtraction approach integrates better with error detection (checking if not a valid digit character, which is not done here), whereas the masking approach is simpler for LC-3.)
The subroutine then obtains the 2nd last digit (moving backwards through the user's input string), converting that to binary using the mask approach. That yields a number between 0 and 9, which is used as an index into the first table Lookup10. The value obtained from the table at that index position is basically the index × 10. So this table is a × 10 table. The same approach is used for the third (and first or, last-going-backwards) digit, except it uses the 2nd table which is a × 100 table.
The standard approach for string to binary is called atoi (search it) standing for ascii to integer. It moves forwards through the string, and for every new digit, it multiples the existing value, computed so far, by 10 before adding in the next digit's numeric value.
So, if the string is 456, the first it obtains 4, then because there is another digit, 4 × 10 = 40, then + 5 for 45, then × 10 for 450, then + 6 for 456, and so on.
The advantage of this approach is that it can handle any number of digits (up to overflow). The disadvantage, of course, is that it requires multiplication, which is a complication for LC-3.
Multiplication where one operand is the constant 10 is fairly easy even in LC-3's limited capabilities, and can be done with simple addition without looping. Basically:
n × 10 = n + n + n + n + n + n + n + n + n + n
and LC-3 can do those 9 additions in just 9 instructions. Still, we can also observe that:
n × 10 = n × 8 + n × 2
and also that:
n × 10 = (n × 4 + n) × 2 (which is n × 5 × 2)
which can be done in just 4 instructions on LC-3 (and none of these needs looping)!
So, if you want to do this approach, you'll have to figure out how to go forwards through the string instead of backwards as the given table version does, and, how to multiply by 10 (use any one of the above suggestions).
There are other approaches as well if you study atoi. You could keep the backwards approach, but now will have to multiply by 10, by 100, by 1000, a different factor for each successive digit . That might be done by repetitive addition. Or a count of how many times to multiply by 10 — e.g. n × 1000 = n × 10 × 10 × 10.
I have just started doing some binary number exercices to prepare for a class that i will start next month and i got the hang of all the conversion from decimal to binary and viceverca But now with the two letters 'a ' ' b' in this exercise i am not sure how can i apply that knowledge to add the bits with the following exercise
Given two Binary numbers a = (a7a6 ... a0) and b = (b7b6 ... b0).There is a clculator that can add 4-bit binary numbers.How many bits will be used to represent the result of a 4-bit addition? Why?
We would like to use our calculator to calculate a + b. For this we can put as many as eight bits (4 bits of the first and 4 bits of the second number) of our choice in the calculator and continue to use the result bit by bit
How many additions does our calculator have to carry out for the addition of a and b at most? How many bits is the result maximum long?
How many additions does the calculator have to perform at least For the result to be correct for all possible inputs a and b?
The number of bits needed to represent a 4-bit binary addition is 5. This is because there could be a carry-over bit that pushes the result to 5 bits.
For example 1111 + 0010 = 10010.
This can be done the same way as adding decimal numbers. From right to left just add the numbers of the same significance. If the two bits are 1+1, the result is 10 so that place becomes a zero and the 1 carries over to the next pair of bits, just like decimal addition.
With regard to the min/max number of step, these seems more like an algorithm specific question. Look up some different binary addition algorithms, like ripple-carry for instance, and it should give you a better idea of what is meant by the question.
I am using CUDA 4.0 on Geforce GTX 580 (Fermi) . I have numbers as small as 7.721155e-43 . I want to multiply them with each other just once or better say I want to calculate 7.721155e-43 * 7.721155e-43 .
My experience showed me I can't do it just straight forward. Could you please give me suggestion? Do I need to use double precision? How?
The magnitude of the smallest normal IEEE single-precision number is about 1.18e-38, the smallest denormal gets you down to about 1.40e-45. As a consequece an operand of magnitude 7.82e-43 will comprise only about 9 non-zero bits, which in itself may already be a problem, even before you get to the multiplication (whose result will underflow to zero in single precision). So you may also want to look at any up-stream computation that produces these tiny numbers.
If these small numbers are intermediate terms in a mathematical expression, rewriting that expression into a mathematically equivalent one that does not involve tiny intermediates would be one way of addressing the issue. Or you could scale some operands by factors that are powers of two (so as to not incur additional round-off due to the scaling). For example, scale by 2^24 = 16777216.
Lastly, you can switch part of the computation to double precision. To do so, simply introduce temporary variables of type double, perform the computation on them, then convert the final result back to float:
float r, f = 7.721155e-43f;
double d, t;
d = (double)f; // explicit cast is not necessary, since converting to wider type
t = d * d;
[... more intermediate computation, leaving result in 't' ...]
r = (float)t; // since conversion is to narrower type, cast will avoid warnings
In statistics we often have to work with likelihoods that end up being very small numbers and the standard technique is to use logs for everything. Then multiplication on a log scale is just addition. All intermediate numbers are stored as logs. Indeed it can take a bit of getting used to - but the alternative will often fail even when doing relatively modest computations. In R (for my convenience!) which uses doubles and prints 7 significant figures by default btw:
> 7.721155e-43 * 7.721155e-43
[1] 5.961623e-85
> exp(log(7.721155e-43) + log(7.721155e-43))
[1] 5.961623e-85
I know the "<<" is a bit operation. but I do not understand what it exactly functions in TCL, and when should we use it?
can anyone help me on this?
The << operator in Tcl's expressions is an arithmetic bit shift left. It's exceptionally similar to the equivalent in C and many other languages, and would be used in all the same places (it's logically equivalent to a multiply by a suitable power of 2, but it's usually advisable to use a shift when thinking about bits and a multiply when thinking about numbers).
Note that one key difference with many other languages (from Tcl 8.5 onwards) is that it does not “drop bits off the front”; the language implementation automatically uses wider number representations as necessary so that information is never lost. Bits are dropped by using a separate binary mask operation (e.g., & ((1 << $numBits) - 1)).
There are a number of uses for the << shift left operator. Some that come to my mind are :
Bit by bit processing. Shift a number and observe highest order bit etc. It comes in more handy than you might think.
If you add a zero to a number in the decimal number system you effectively multiply it by 10. shifting bits effectively means multiplying by 2. This actually translated into a low level assembly command of bit shifting which has lower compute cycles than multiplication by 2. This is used for efficiency in the gaming industry. Shift if twice (<< 2) to multiply it by 4 and so on.
I am sure there are many others.
The << operation is not much different from C's, for instance. And it's used when you need to shift bits of an integer value to the left. This can be occasionally useful when doing subtle number crunching like implemening a hash function or deserialising something from an input bytestream (but note that [binary scan] covers almost all of what's needed for this sort of thing). For a more general info refer to this Wikipedia article or something like this, this is not really Tcl-related.
The '<<' is a left bit shift. You must apply it to an integer. This arithmetic operator will shift the bits to left.
For example, if you want to shifted the number 1 twice to the left in the Tcl interpreter tclsh, type:
expr { 1 << 2 }
The command will return 4.
Pay special attention to the maximum integer the interpreter hold on your platform.
I was recently asked a question, "how do you multiply without using the multiplication operator, without any sort of looping statements or explicit addition" and realized I wasn't familiar with bitwise operation at all.
There is obviously wikipedia but I need something with more of an explanation geared toward a newbie. There's also this hack guide but I'm not at the level of grasping it yet.
I don't mind if you point out a chapter in a book, as I have access to a good library through Safari Books and other resources.
Knuth, Volume 2 - Seminumerical Algorithms
The crux of this comes down to a "half adder" and a "full adder". A half adder adds two bits of input to produce a single-bit result, and a single-bit carry. A full adder adds three bits of input (two normal inputs plus a carry from a lower bit) to produce a single-bit result and a single-bit carry.
In any case, the result is based on a truth table for addition. For a half adder, that is: 0+0=0, 0+1=1, 1+0=1, 1+1=0+carry.
So, the "normal" part of the result is the XOR of the inputs. The "carry' part of the result is the AND of the inputs. A full adder is pretty much the same, but left as the infamous "exercise for the reader".
Putting those together, you use a half-adder for the least significant bit, and full adders for the other bits to add N bits of input.
Once you can do addition, there are a couple of ways of doing multiplication. The easy (and slow) way to multiply NxM is to add N to itself M times. The faster (but somewhat more difficult to understand) way is to shift and add. For example, Nx5 = Nx4 + Nx1. You can produce NxB, where B = 2L by shifting N left by L bits.