I am trying to decipher some assembly code that involves multiple left rotations on an 8-bit binary number.
For reference, the code is:
lab: rol dl,1
rol dl,1
dec ecx
jnz lab
The dec and jnz isn't an issue, but is there to show that the 2 rols are executed several times.
What I am trying to do is figure out a mathematical equivalent of this code, such as a formula. I'm certainly not looking for a complete formula to tell me the whole code, but I would like to know if there is a formula that gives the equivalent (in denary) of a single left rotation.
I've tried figuring this out with a couple of different numbers, but cannot see a link between the two results. For example: if the start number is 115 it comes out as 220, but if the start number is 99 it comes out as 216.
Given your sample results, I assume we are treating the 8-bit quantity as unsigned.
The 7 low-order bits are shifted left, multiplying that part of the number by 2; and the high-order bit is swapped around to the beginning.
Thus, (x % 128) * 2 + (x / 128), using the usual integer div/mod operators.
Shifting a byte containing number X by one bit (position) left is equal to multiplying the number X by 2:
x << 1 <==> x = x * 2
Related
Interview question: you're given a file of roughly one billion unique numbers, each of which is a 32-bit quantity. Find a number not in the file.
When I was approaching this question, I tried a few examples with 3-bit and 4-bit numbers. For the examples I tried, I found that when I XOR'd the set of numbers, I got a correct answer:
a = [0,1,2] # missing 3
b = [1,2,3] # missing 0
c = [0,1,2,3,4,5,6] # missing 7
d = [0,1,2,3,5,6,7] # missing 4
functools.reduce((lambda x, y: x^y), a) # returns 3
functools.reduce((lambda x, y: x^y), b) # returns 0
functools.reduce((lambda x, y: x^y), c) # returns 7
functools.reduce((lambda x, y: x^y), d) # returns 4
However, when I coded this up and submitted it, it failed the test cases.
My question is: in an interview setting, how can I confirm or rule out with certainty that an approach like this is not a viable solution?
In all your examples, the array is missing exactly one number. That's why XOR worked. Try not to test with the same property.
For the problem itself, you can construct a number by taking the minority of each bit.
EDIT
Why XOR worked on your examples:
When you take the XOR for all the numbers from 0 to 2^n - 1 the result is 0 (there are exactly 2^(n-1) '1' in each bit). So if you take out one number and take XOR of all the rest, the result is the number you took out because taking XOR of that number with the result of all the rest needs to be 0.
Assuming a 64-bit system with more than 4gb free memory, I would read the numbers into an array of 32-bit integers. Then I would loop through the numbers up to 32 times.
Similarly to an inverse ”Mastermind” game, I would construct a missing number bit-by-bit. In every loop, I count all numbers which match the bits, I have chosen so far and a subsequent 0 or 1. Then I add the bit which occurs less frequently. Once the count reaches zero, I have a missing number.
Example:
The numbers in decimal/binary are
1 = 01
2 = 10
3 = 11
There is one number with most-significant-bit 0 and two numbers with 1. Therefore, I take 0 as most significant bit.
In the next round, I have to match 00 and 01. This immediately leads to 00 as missing number.
Another approach would be to use a random number generator. Chances are 50% that you find a non-existing number as first guess.
Proof by counterexample: 3^4^5^6=4.
I am grappling with this problem Codeforces 276D. Initially I used a brute force approach which obviously failed for large inputs(It started when inputs were 10000000000 20000000000). In the tutorials Fcdkbear(turtor for the contest) talks about a dp solution where a state is d[p][fl1][fr1][fl2][fr2].
Further in tutorial
We need to know, which bits we can place into binary representation of number а in p-th position. We can place 0 if the following condition is true: p-th bit of L is equal to 0, or p-th bit of L is equal to 1 and variable fl1 shows that current value of a is strictly greater then L. Similarly, we can place 1 if the following condition is true: p-th bit of R is equal to 1, or p-th bit of R is equal to 0 and variable fr1 shows that current value of a is strictly less then R. Similarly, we can obtain, which bits we can place into binary representation of number b in p-th position.
This is going over my head as when ith bit of L is 0 then how come we can place a zero in a's ith bit. If L and R both are in same bucket(2^i'th boundary like 16 and 24) we will eventually place a 0 at 4th whereas we can place a 1 if a = 20 because i-th bit of R is 0 and a > R. I am wondering what is the use of checking if a > L or not.
In essence I do not get the logic of
What states are
How do we recur
I know that might be an overkill but could someone explain it in descriptive manner as editorial is too short to explain anything.
I have already looked in here but suggested solution is different from one given in editorial. Also I know this can be solved with binary search but I am concerned with DP solution only
If I got the problem right: Start to compare the bits of l and r from left (MSB) to right(LSB). As long as these bits are equal there is no freedom of choice, the same bits must appear in a and b. the first bit differing must be 1 in r and 0 in l. they must appear also in a (0) and b(1). from here you can maximise the XOR result. simply use zeros for b an ones for a. that gives a+1==b and the xor result is a+b which is always 2^n-1.
I'm not following the logic as written above but the basic idea is to look bit by bit.
If L and R have different values in the same bit position then we have already found candidates that would maximize the xor'd value of that position (0 xor 1 = 1 xor 0 = 1). The other case to consider is whether the span of R-L is greater than the position value of that bit. If so then there must be two different values of A and B falling between L and R where that bit position has opposite values (as well as being able to generate any combinations of values in the lower bits.)
I am reading about middle square method in hashing at followiong location.
http://www.brpreiss.com/books/opus4/html/page212.html
Here author mentioned that keys which have a large number of leading zeroes will collide.
A similar line of reasoning applies for keys which have a large number of trailing zeroes.
Request to give an example and explanation what does author mean in above two statements.
as the method says, the key x are being divided by 2^(w-k). at the bit level this actually means that the bits of x gets shifted by to the right by (w-k).
now take a scenario where x = 7, (w-k) = 3. so x = 00000111 and shifting those bits to the right by 3 means x = 00000000.
now, u will notice that for all the values of x from 7 (00000111) to 0 (00000000), the bit shift would result in 00000000 (shifting bits to right always appends 0 to the left side). thus values that have large number of leading 0's (7 to 0) will collide to the same hash.
On Page 140 of Programming Pearls, 2nd Edition, Jon proposed an implementation of sets with bit vectors.
We'll turn now to two final structures that exploit the fact that our sets represent integers. Bit vectors are an old friend from Column 1. Here are their private data and functions:
enum { BITSPERWORD = 32, SHIFT = 5, MASK = 0x1F };
int n, hi, *x;
void set(int i) { x[i>>SHIFT] |= (1<<(i & MASK)); }
void clr(int i) { x[i>>SHIFT] &= ~(1<<(i & MASK)); }
int test(int i) { return x[i>>SHIFT] &= (1<<(i & MASK)); }
As I gathered, the central idea of a bit vector to represent an integer set, as described in Column 1, is that the i-th bit is turned on if and only if the integer i is in the set.
But I am really at a loss at the algorithms involved in the above three functions. And the book doesn't give an explanation.
I can only get that i & MASK is to get the lower 5 bits of i, while i>>SHIFT is to move i 5 bits toward the right.
Anybody would elaborate more on these algorithms? Bit operations always seem a myth to me, :(
Bit Fields and You
I'll use a simple example to explain the basics. Say you have an unsigned integer with four bits:
[0][0][0][0] = 0
You can represent any number here from 0 to 15 by converting it to base 2. Say we have the right end be the smallest:
[0][1][0][1] = 5
So the first bit adds 1 to the total, the second adds 2, the third adds 4, and the fourth adds 8. For example, here's 8:
[1][0][0][0] = 8
So What?
Say you want to represent a binary state in an application-- if some option is enabled, if you should draw some element, and so on. You probably don't want to use an entire integer for each one of these- it'd be using a 32 bit integer to store one bit of information. Or, to continue our example in four bits:
[0][0][0][1] = 1 = ON
[0][0][0][0] = 0 = OFF //what a huge waste of space!
(Of course, the problem is more pronounced in real life since 32-bit integers look like this:
[0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0][0] = 0
The answer to this is to use a bit field. We have a collection of properties (usually related ones) which we will flip on and off using bit operations. So, say, you might have 4 different lights on a piece of hardware that you want to be on or off.
3 2 1 0
[0][0][0][0] = 0
(Why do we start with light 0? I'll explain this in a second.)
Note that this is an integer, and is stored as an integer, but is used to represent multiple states for multiple objects. Crazy! Say we turn lights 2 and 1 on:
3 2 1 0
[0][1][1][0] = 6
The important thing you should note here: There's probably no obvious reason why lights 2 and 1 being on should equal six, and it may not be obvious how we would do anything with this scheme of information storage. It doesn't look more obvious if you add more bits:
3 2 1 0
[1][1][1][0] = 0xE \\what?
Why do we care about this? Do we have exactly one state for each number between 0 and 15?How are we going to manage this without some insane series of switch statements? Ugh...
The Light at the End
So if you've worked with binary arithmetic a bit before, you might realize that the relationship between the numbers on the left and the numbers on the right is, of course, base 2. That is:
1*(23) + 1*(22) + 1*(21) +0 *(20) = 0xE
So each light is present in the exponent of each term of the equation. If the light is on, there is a 1 next to its term- if the light is off, there is a zero. Take the time to convince yourself that there is exactly one integer between 0 and 15 that corresponds to each state in this numbering scheme.
Bit operators
Now that we have this done, let's take a second to see what bitshifting does to integers in this setup.
[0][0][0][1] = 1
When you shift bits to the left or the right in an integer, it literally moves the bits left and right. (Note: I 100% disavow this explanation for negative numbers! There be dragons!)
1<<2 = 4
[0][1][0][0] = 4
4>>1 = 2
[0][0][1][0] = 2
You will encounter similar behavior when shifting numbers represented with more than one bit. Also, it shouldn't be hard to convince yourself that x>>0 or x<<0 is just x. Doesn't shift anywhere.
This probably explains the naming scheme of the Shift operators to anyone who wasn't familiar with them.
Bitwise operations
This representation of numbers in binary can also be used to shed some light on the operations of bitwise operators on integers. Each bit in the first number is xor-ed, and-ed, or or-ed with its fellow number. Take a second to venture to wikipedia and familiarize yourself with the function of these Boolean operators - I'll explain how they function on numbers but I don't want to rehash the general idea in great detail.
...
Welcome back! Let's start by examining the effect of the OR (|) operator on two integers, stored in four bit.
OR OPERATOR ON:
[1][0][0][1] = 0x9
[1][1][0][0] = 0xC
________________
[1][1][0][1] = 0xD
Tough! This is a close analogue to the truth table for the boolean OR operator. Notice that each column ignores the adjacent columns and simply fills in the result column with the result of the first bit and the second bit OR'd together. Note also that the value of anything or'd with 1 is 1 in that particular column. Anything or'd with zero remains the same.
The table for AND (&) is interesting, though somewhat inverted:
AND OPERATOR ON:
[1][0][0][1] = 0x9
[1][1][0][0] = 0xC
________________
[1][0][0][0] = 0x8
In this case we do the same thing- we perform the AND operation with each bit in a column and put the result in that bit. No column cares about any other column.
Important lesson about this, which I invite you to verify by using the diagram above: anything AND-ed with zero is zero. Also, equally important- nothing happens to numbers that are AND-ed with one. They stay the same.
The final table, XOR, has behavior which I hope you all find predictable by now.
XOR OPERATOR ON:
[1][0][0][1] = 0x9
[1][1][0][0] = 0xC
________________
[0][1][0][1] = 0x5
Each bit is being XOR'd with its column, yadda yadda, and so on. But look closely at the first row and the second row. Which bits changed? (Half of them.) Which bits stayed the same? (No points for answering this one.)
The bit in the first row is being changed in the result if (and only if) the bit in the second row is 1!
The one lightbulb example!
So now we have an interesting set of tools we can use to flip individual bits. Let's go back to the lightbulb example and focus only on the first lightbulb.
0
[?] \\We don't know if it's one or zero while coding
We know that we have an operation that can always make this bit equal to one- the OR 1 operator.
0|1 = 1
1|1 = 1
So, ignoring the rest of the bulbs, we could do this
4_bit_lightbulb_integer |= 1;
and know for sure that we did nothing but set the first lightbulb to ON.
3 2 1 0
[0][0][0][?] = 0 or 1? \\4_bit_lightbulb_integer
[0][0][0][1] = 1
________________
[0][0][0][1] = 0x1
Similarly, we can AND the number with zero. Well- not quite zero- we don't want to affect the state of the other bits, so we will fill them in with ones.
I'll use the unary (one-argument) operator for bit negation. The ~ (NOT) bitwise operator flips all of the bits in its argument. ~(0X1):
[0][0][0][1] = 0x1
________________
[1][1][1][0] = 0xE
We will use this in conjunction with the AND bit below.
Let's do 4_bit_lightbulb_integer & 0xE
3 2 1 0
[0][1][0][?] = 4 or 5? \\4_bit_lightbulb_integer
[1][1][1][0] = 0xE
________________
[0][1][0][0] = 0x4
We're seeing a lot of integers on the right-hand-side which don't have any immediate relevance. You should get used to this if you deal with bit fields a lot. Look at the left-hand side. The bit on the right is always zero and the other bits are unchanged. We can turn off light 0 and ignore everything else!
Finally, you can use the XOR bit to flip the first bit selectively!
3 2 1 0
[0][1][0][?] = 4 or 5? \\4_bit_lightbulb_integer
[0][0][0][1] = 0x1
________________
[0][1][0][*] = 4 or 5?
We don't actually know what the value of * is now- just that flipped from whatever ? was.
Combining Bit Shifting and Bitwise operations
The interesting fact about these two operations is when taken together they allow you to manipulate selective bits.
[0][0][0][1] = 1 = 1<<0
[0][0][1][0] = 2 = 1<<1
[0][1][0][0] = 4 = 1<<2
[1][0][0][0] = 8 = 1<<3
Hmm. Interesting. I'll mention the negation operator here (~) as it's used in a similar way to produce the needed bit values for ANDing stuff in bit fields.
[1][1][1][0] = 0xE = ~(1<<0)
[1][1][0][1] = 0xD = ~(1<<1)
[1][0][1][1] = 0xB = ~(1<<2)
[0][1][1][1] = 0X7 = ~(1<<3)
Are you seeing an interesting relationship between the shift value and the corresponding lightbulb position of the shifted bit?
The canonical bitshift operators
As alluded to above, we have an interesting, generic method for turning on and off specific lights with the bit-shifters above.
To turn on a bulb, we generate the 1 in the right position using bit shifting, and then OR it with the current lightbulb positions. Say we want to turn on light 3, and ignore everything else. We need to get a bit shifting operation that ORs
3 2 1 0
[?][?][?][?] \\all we know about these values at compile time is where they are!
and 0x8
[1][0][0][0] = 0x8
Which is easy, thanks to bitshifting! We'll pick the number of the light and switch the value over:
1<<3 = 0x8
and then:
4_bit_lightbulb_integer |= 0x8;
3 2 1 0
[1][?][?][?] \\the ? marks have not changed!
And we can guarantee that the bit for the 3rd lightbulb is set to 1 and that nothing else has changed.
Clearing a bit works similarly- we'll use the negated bits table above to, say, clear light 2.
~(1<<2) = 0xB = [1][0][1][1]
4_bit_lightbulb_integer & 0xB:
3 2 1 0
[?][?][?][?]
[1][0][1][1]
____________
[?][0][?][?]
The XOR method of flipping bits is the same idea as the OR one.
So the canonical methods of bit switching are this:
Turn on the light i:
4_bit_lightbulb_integer|=(1<<i)
Turn off light i:
4_bit_lightbulb_integer&=~(1<<i)
Flip light i:
4_bit_lightbulb_integer^=(1<<i)
Wait, how do I read these?
In order to check a bit we can simply zero out all of the bits except for the one we care about. We'll then check to see if the resulting value is greater than zero- since this is the only value that could possibly be nonzero, it will make the entire integer nonzero if and only if it is nonzero. For example, to check bit 2:
1<<2:
[0][1][0][0]
4_bit_lightbulb_integer:
[?][?][?][?]
1<<2 & 4_bit_lightbulb_integer:
[0][?][0][0]
Remember from the previous examples that the value of ? didn't change. Remember also that anything AND 0 is 0. So, we can say for sure that if this value is greater than zero, the switch at position 2 is true and the lightbulb is zero. Similarly, if the value is off, the value of the entire thing will be zero.
(You can alternately shift the entire value of 4_bit_lightbulb_integer over by i bits and AND it with 1. I don't remember off the top of my head if one is faster than the other but I doubt it.)
So the canonical checking function:
Check if bit i is on:
if (4_bit_lightbulb_integer & 1<<i) {
\\do whatever
}
The specifics
Now that we have a complete set of tools for bitwise operations, we can look at the specific example here. This is basically the same idea- except a much more concise and powerful way of executing it. Let's look at this function:
void set(int i) { x[i>>SHIFT] |= (1<<(i & MASK)); }
From the canonical implementation I'm going to make a guess that this is trying to set some bits to 1! Let's take an integer and look at what's going on here if i feed the value 0x32 (50 in decimal) into i:
x[0x32>>5] |= (1<<(0x32 & 0x1f))
Well, that's a mess.. let's dissect this operation on the right. For convenience, pretend there are 24 more irrelevant zeros, since these are both 32 bit integers.
...[0][0][0][1][1][1][1][1] = 0x1F
...[0][0][1][1][0][0][1][0] = 0x32
________________________
...[0][0][0][1][0][0][1][0] = 0x12
It looks like everything is being cut off at the boundary on top where 1s turn into zeros. This technique is called Bit Masking. Interestingly, the boundary here restricts the resulting values to be between 0 and 31... Which is exactly the number of bit positions we have for a 32 bit integer!
x[0x32>>5] |= (1<<(0x12))
Let's look at the other half.
...[0][0][1][1][0][0][1][0] = 0x32
Shift five bits to the right:
...[0][0][0][0][0][0][0][1] = 0x01
Note that this transformation exactly destroyed all information from the first part of the function- we have 32-5 = 27 remaining bits which could be nonzero. This indicates which of 227 integers in the array of integers are selected. So the simplified equation is now:
x[1] |= (1<<0x12)
This just looks like the canonical bit-setting operation! We've just chosen
So the idea is to use the first 27 bits to pick an integer to shift and the last five bits indicate which bit of the 32 in that integer to shift.
The key to understanding what's going on is to recognize that BITSPERWORD = 2SHIFT. Thus, x[i>>SHIFT] finds which 32-bit element of the array x has the bit corresponding to i. (By shifting i 5 bits to the right, you're simply dividing by 32.) Once you have located the correct element of x, the lower 5 bits of i can then be used to find which particular bit of x[i>>SHIFT] corresponds to i. That's what i & MASK does; by shifting 1 by that number of bits, you move the bit corresponding to 1 to the exact position within x[i>>SHIFT] that corresponds to the ith bit in x.
Here's a bit more of an explanation:
Imagine that we want capacity for N bits in our bit vector. Since each int holds 32 bits, we will need (N + 31) / 32 int values for our storage (that is, N/32 rounded up). Within each int value, we will adopt the convention that bits are ordered from least significant to most significant. We will also adopt the convention that the first 32 bits of our vector are in x[0], the next 32 bits are in x[1], and so forth. Here's the memory layout we are using (showing the bit index in our bit vector corresponding to each bit of memory):
+----+----+-------+----+----+----+
x[0]: | 31 | 30 | . . . | 02 | 01 | 00 |
+----+----+-------+----+----+----+
x[1]: | 63 | 62 | . . . | 34 | 33 | 32 |
+----+----+-------+----+----+----+
etc.
Our first step is to allocate the necessary storage capacity:
x = new int[(N + BITSPERWORD - 1) >> SHIFT]
(We could make provision for dynamically expanding this storage, but that would just add complexity to the explanation.)
Now suppose we want to access bit i (either to set it, clear it, or just to know its current value). We need to first figure out which element of x to use. Since there are 32 bits per int value, this is easy:
subscript for x = i / 32
Making use of the enum constants, the x element we want is:
x[i >> SHIFT]
(Think of this as a 32-bit-wide window into our N-bit vector.) Now we have to find the specific bit corresponding to i. Looking at the memory layout, it's not hard to figure out that the first (rightmost) bit in the window corresponds to bit index 32 * (i >> SHIFT). (The window starts afteri >> SHIFT slots in x, and each slot has 32 bits.) Since that's the first bit in the window (position 0), then the bit we're interested in is is at position
i - (32 * (i >> SHIFT))
in the windows. With a little experimenting, you can convince yourself that this expression is always equal to i % 32 (actually, that's one definition of the mod operator) which, in turn, is always equal to i & MASK. Since this last expression is the fastest way to calculate what we want, that's what we'll use.
From here, the rest is pretty simple. We start with a single bit in the least-significant position of the window (that is, the constant 1), and move it to the left by i & MASK bits to get it to the position in the window corresponding to bit i in the bit vector. This is where the expression
1 << (i & MASK)
comes from. With the bit now moved to where we want it, we can use this as a mask to set, clear, or query the value of the bit at that position in x[i>>SHIFT] and we know that we're actually setting, clearing, or querying the value of bit i in our bit vector.
If you store your bits in an array of n words you can imagine them to be layed out as a matrix with n rows and 32 columns (BITSPERWORD):
3 0
1 0
0 xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx
1 xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx
2 xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx
....
n xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx xxxxxxxxxx
To get the k-th bit you divide k by 32. The (integer) result will give you the row (word) the bit is in, the reminder will give you which bit is within the word.
Dividing by 2^p can be done simply by shifting p postions to the right. The reminder can be obtained by getting the p rightmost bits (i.e the bitwise AND with (2^p - 1)).
In C terms:
#define div32(k) ((k) >> 5)
#define mod32(k) ((k) & 31)
#define word_the_bit_is_in(k) div32(k)
#define bit_within_word(k) mod32(k)
Hope it helps.
I need to do a linear interpolation over time between two values on an 8 bit PIC microcontroller (Specifically 16F627A but that shouldn't matter) using PIC assembly language. Although I'm looking for an algorithm here as much as actual code.
I need to take an 8 bit starting value, an 8 bit ending value and a position between the two (Currently represented as an 8 bit number 0-255 where 0 means the output should be the starting value and 255 means it should be the final value but that can change if there is a better way to represent this) and calculate the interpolated value.
Now PIC doesn't have a divide instruction so I could code up a general purpose divide routine and effectivly calculate (B-A)/(x/255)+A at each step but I feel there is probably a much better way to do this on a microcontroller than the way I'd do it on a PC in c++
Has anyone got any suggestions for implementing this efficiently on this hardware?
The value you are looking for is (A*(255-x)+B*x)/255. It requires only 8x8 multiplication, and a final division by 255, which can be approximated by simply taking the high byte of the sum.
Choosing x in range 0..128, no approximation is needed: take the high byte of (A*(128-x)+B*x)<<1.
Assuming you interpolate a sequence of values where the previous endpoint is the new start point:
(B-A)/(x/255)+A
sounds like a bad idea. If you use base 255 as a fixedpoint representation, you get the same interpolant twice. You get B when x=255 and B as the new A when x=0.
Use 256 as the fixedpoint system. Divides become shifts, but you need 16-bit arithmetic and 8x8 multiplication with a 16-bit result. The previous issue can be fixed by simply ignoring any bits in the higher-bytes as x mod 256 becomes 0. This suggestion uses 16-bit multiplication, but can't overflow. and you don't interpolate over the same x twice.
interp = (a*(256 - x) + b*x) >> 8
256 - x becomes just a subtract-with-borrow, as you get 0 - x.
The PIC lacks these operations in its instruction set:
Right and left shift. (both logical and arithmetic)
Any form of multiplication.
You can get right-shifting by using rotate-right instead, followed by masking out the extra bits on the left with bitwise-and. A straight-forward way to do 8x8 multiplication with 16-bit result:
void mul16(
unsigned char* hi, /* in: operand1, out: the most significant byte */
unsigned char* lo /* in: operand2, out: the least significant byte */
)
{
unsigned char a,b;
/* loop over the smallest value */
a = (*hi <= *lo) ? *hi : *lo;
b = (*hi <= *lo) ? *lo : *hi;
*hi = *lo = 0;
while(a){
*lo+=b;
if(*lo < b) /* unsigned overflow. Use the carry flag instead.*/
*hi++;
--a;
}
}
The techniques described by Eric Bainville and Mads Elvheim will work fine; each one uses two multiplies per interpolation.
Scott Dattalo and Tony Kubek have put together a super-optimized PIC-specific interpolation technique called "twist" that is slightly faster than two multiplies per interpolation.
Is using this difficult-to-understand technique worth running a little faster?
You could do it using 8.8 fixed-point arithmetic. Then a number from range 0..255 would be interpreted as 0.0 ... 0.996 and you would be able to multiply and normalize it.
Tell me if you need any more details or if it's enough for you to start.
You could characterize this instead as:
(B-A)*(256/(x+1))+A
using a value range of x=0..255, precompute the values of 256/(x+1) as a fixed-point number in a table, and then code a general purpose multiply, adjust for the position of the binary point. This might not be small spacewise; I'd expect you to need a 256 entry table of 16 bit values and the multiply code. (If you don't need speed, this would suggest your divison method is fine.). But it only takes one multiply and an add.
My guess is that you don't need every possible value of X. If there are only a few values of X, you can compute them offline, do a case-select on the specific value of X and then implement the multiply in terms of a fixed sequence of shifts and adds for the specific value of X. That's likely to be pretty efficient in code and very fast for a PIC.
Interpolation
Given two values X & Y , its basically:
(X+Y)/2
or
X/2 + Y/2 (to prevent the odd-case that A+B might overflow the size of the register)
Hence try the following:
(Pseudo-code)
Initially A=MAX, B=MIN
Loop {
Right-Shift A by 1-bit.
Right-Shift B by 1-bit.
C = ADD the two results.
Check MSB of 8-bit interpolation value
if MSB=0, then B=C
if MSB=1, then A=C
Left-Shift 8-bit interpolation value
}Repeat until 8-bit interpolation value becomes zero.
The actual code is just as easy. Only i do not remember the registers and instructions off-hand.