I have two simple while loops in my program that I feel ought to be math equations, but I'm struggling to convert them:
float a = someValue;
int b = someOtherValue;
int c = 0;
while (a <= -b / 2) {
c--;
a += b;
}
while (a >= b / 2) {
c++;
a -= b;
}
This code works as-is, but I feel it could be simplified into math equations. The idea here being that this code is taking an offset (someValue) and adjusting a coordinate (c) to minimize the distance from the center of a tile (of size someOtherValue). Any help would be appreciated.
It can be proved that the following is correct:
c = floor((a+b/2)/b)
a = a - c*b
Note that floor means round down, towards negative infinity: not towards 0. (E.g. floor(-3.1)=-4. The floor() library functions will do this; just be sure not to just cast to int, which will usually round towards 0 instead.)
Presumably b is strictly positive, because otherwise neither loop will never terminate: adding b will not make a larger and subtracting b will not make a smaller. With that assumption, we can prove that the above code works. (And paranoidgeek's code is also almost correct, except that it uses a cast to int instead of floor.)
Clever way of proving it:
The code adds or subtracts multiples of b from a until a is in [-b/2,b/2), which you can view as adding or subtracting integers from a/b until a/b is in [-1/2,1/2), i.e. until (a/b+1/2) (call it x) is in [0,1). As you are only changing it by integers, the value of x does not change mod 1, i.e. it goes to its remainder mod 1, which is x-floor(x). So the effective number of subtractions you make (which is c) is floor(x).
Tedious way of proving it:
At the end of the first loop, the value of c is the negative of the number of times the loop runs, i.e.:
0 if: a > -b/2 <=> a+b/2 > 0
-1 if: -b/2 ≥ a > -3b/2 <=> 0 ≥ a+b/2 > -b <=> 0 ≥ x > -1
-2 if: -3b/2 ≥ a > -5b/2 <=> -b ≥ a+b/2 > -2b <=> -1 ≥ x > -2 etc.,
where x = (a+b/2)/b, so c is: 0 if x>0 and "ceiling(x)-1" otherwise. If the first loop ran at all, then it was ≤ -b/2 just before the last time the loop was executed, so it is ≤ -b/2+b now, i.e. ≤ b/2. According as whether it is exactly b/2 or not (i.e., whether x when you started was exactly a non-positive integer or not), the second loop runs exactly 1 time or 0, and c is either ceiling(x) or ceiling(x)-1. So that solves it for the case when the first loop did run.
If the first loop didn't run, then the value of c at the end of the second loop is:
0 if: a < b/2 <=> a-b/2 < 0
1 if: b/2 ≤ a < 3b/2 <=> 0 ≤ a-b/2 < b <=> 0 ≤ y < 1
2 if: 3b/2 ≤ a < 5b/2 <=> b ≤ a-b/2 < 2b <=> 1 ≤ y < 2, etc.,
where y = (a-b/2)/b, so c is: 0 if y<0 and 1+floor(y) otherwise. [And a now is certainly < b/2 and ≥ -b/2.]
So you can write an expression for c as:
x = (a+b/2)/b
y = (a-b/2)/b
c = (x≤0)*(ceiling(x) - 1 + (x is integer))
+(y≥0)*(1 + floor(y))
Of course, next you notice that (ceiling(x)-1+(x is integer)) is same as floor(x+1)-1 which is floor(x), and that y is actually x-1, so (1+floor(y))=floor(x), and as for the conditionals:
when x≤0, it cannot be that (y≥0), so c is just the first term which is floor(x),
when 0 < x < 1, neither of the conditions holds, so c is 0,
when 1 ≤ x, then only 0≤y, so c is just the second term which is floor(x) again.
So c = floor(x) in all cases.
c = (int)((a - (b / 2)) / b + 1);
a -= c * b;
Test case at http://pastebin.com/m1034e639
I think you want something like this:
c = ((int) a + b / 2 * sign(a)) / b
That should match your loops except for certain cases where b is odd because the range from -b/2 to b/2 is smaller than b when b is odd.
Assuming b is positive, abs(c) = floor((abs(a) - b/2) / b). Then, apply sign of a to c.
Related
I am unsure whether or not I better should have posted this question on codereview.stackexchange.com. Anyway, here we go ...
Please consider the following code snippet which is a literal (I have only changed the formatting) excerpt from here and has been printed (in stripped-down form) in the German computer magazine c't, issue 23/2019:
while (lo <= hi) {
std::streamoff pos = std::streamoff((uint64_t(lo) + uint64_t(hi)) / 2);
pos -= pos % std::streamoff(PasswordHashAndCount::size);
pos = std::max<int64_t>(0, pos);
phc.read(mInputFile, pos);
++nReads;
if (hash > phc.hash) {
lo = pos + std::streamoff(PasswordHashAndCount::size);
}
else if (hash < phc.hash) {
hi = pos - std::streamoff(PasswordHashAndCount::size);
}
else {
safe_assign(readCount, nReads);
return phc;
}
}
Why do we need the fourth line pos = std::max<int64_t>(0, pos);?
From the second line, we see that pos is equal or greater to 0, because it is the half of the sum of two numbers which themselves are of type uint64_t.
The third line can't make pos lower than 0. Proof:
For simplicity, replace pos by A and std::streamoff(PasswordHashAndCount::size) by B. Then the third line reads A -= A % B which is equivalent to A = A - (A % B), where A and B are integers, A being equal to or greater than 0, and B being greater than 0 (because ::size is always greater than 0).
First, if A < B, A % B = A. In this case, the third line becomes A = A - A, that is, A = 0.
Secondly, if A == B, A % B becomes A % A which is 0. Therefore, the third line becomes A = A - 0, which is equivalent to a null operation. In other words, A does not change in that case; notably, it remains 0 or greater than that.
Third, if A > B, A - (A % B) is greater than 0. This is because A % B is smaller than B, and thus, A - (A % B) is greater than A - B. The latter in turn is greater than 0, because the condition here was A > B.
Of course, the three cases A > B, A < B and A == B are all cases which can occur. In every case, the third line assigns a new value to A which is 0 or positive.
Coming back to the original variable naming, that means that pos is always 0 or greater than that after execution of the third line.
Given that, I don't understand what the fourth line does. After all, max(0, pos) is always equivalent to pos if pos is 0 or positive.
What am I missing? Is there an error in the reasoning above?
Let's consider what exactly it does:
std::streamoff pos = std::streamoff((uint64_t(lo) + uint64_t(hi)) / 2);
pos = std::max<int64_t>(0, pos);
std::streamoff is some implementation defined signed integer type. Let's consider a case where it is a 64 bit type or smaller: The value of pos will not be changed by the conversion to int64_t because the type is wider, nor when converting back in the assignment because the original value must have been representable.
Let's consider a case where std::streamoff is a 128 bit type or wider: The value comes from (uint64_t(lo) + uint64_t(hi)) / 2 which cannot exceed maximum of int64_t. Thus, the value cannot be changed by the conversion in this case either.
Thus, the use of int64_t has no effect in any case.
The third line can't make pow lower than 0
Is there an error in the reasoning above?
I cannot find any error.
Given that, I don't understand what the fourth line does.
The line has no effect at on the behaviour of the program at all. The program would have equivalent behaviour if the line was written:
;
Besides, on most systems that you find on desktop or server, std::streamoff and int64_t have the same number of bits.
there are 4 unsigned integers(32bits):
a,b,c,d (c > a > d > b)
and a function:
clz(x) (calculate the leading zero numbers of x.eg.clz(2) == 30)
then there are :
n = clz(a^b) // xor
m = clz(c^d)
the question is: can we think m is certainly lesser or equal then n?
thanks in advance!
consider a simplified version: c=0x0111, a=0x0110, d=0x0101, b=0x0001
then c^d=0x0010, a^b=0x0111, so clz(c^d)=2 > clz(a^b)=1
So all you need is a,c,d having the same lz and b having a larger lz to fail the assumption.
I have a mixed integer linear program (MIP or MILP).
In the end I want a boolean variable im my linear program, that has the following properties.
I have two variables:
boolean b.
real x, with x being 0 or larger.
What I want to achieve is:
b == false if x == 0.
b == true if x > 0.
I found a way to depict if x is in specific range (e.g. between 2 and 3) via:
2*b <= x
x <= 3*b
The problem with the above testing formula is, that b will be true if x is in the given range and false if outside that range.
Does anybody know a way to set a boolean variable to false if x == 0 and to true if x is larger than 0?
If U is an upper bound of x then
if x > 0 ==> b == 1
can be made as
x <= U*b
The second part (x == 0 => b == 0) needs to be modified to
x < epsilon ==> b == 0
which can be made as
b <= 1 + x - epsilon
where epsilon is a small number. Other than good practice this is necessary, because solvers do not work in rational arithmetic (although there are some research efforts to make them do so), but with certain precision thresholds, and therefore quantities such as 10e-12 are treated as zero.
I hope this helps!
You could use the signum function http://en.wikipedia.org/wiki/Signum_function take the absolute value and negate it. Since you didn't name a specific programming language I keep it general.
I need to optimize an expression of the form:
(a > b) || (a > c)
I tried several optimized forms one of which is as follows:
(a * 2) > (b + c)
Optimization is not from the compiler's point of view. I would like to reduce the two >s to one.
This is based on the assumption that 1 <= (a, b, c) <= 26
However, this works only for some cases. Is the optimization I am trying to do, really possible? If yes, a start would be really helpful.
The answer is probably: you do not want to optimize that. Moreover, I doubt that there's any way to write this more efficiently. If you say that a, b and c are values between 1 and 26, you shouldn't be using integers (you don't need that precision) if you wanted to be optimal (in size) anyway.
If a > b, the expression a > c will not be executed anyway. So you have at maximum 2 (and at minimum 1) conditional operations, which is really not worth an optimization.
I'm quite doubtful this is even an optimisation in most cases.
a > b || a > c
will evaluate to:
compare a b
jump not greater
compare a c
jump not greater
where
a * 2 > b + c
gives:
shift a left 1 (in temp1)
add b to c (in temp2)
compare temp1 temp2
jump if not greater
As always with performance, it's always much better to base your decision on actual performance measurements (preferably on a selection of processor architectures).
The best I can come up with is this
char a, b, c;
std::cin >> a >> b >> c;
if (((b-a) | (c-a)) & 0x80) {
// a > b || a > c
}
With gcc -O2 this generates only one conditional branch
40072e: 29 c8 sub %ecx,%eax
400730: 29 ca sub %ecx,%edx
400732: 09 d0 or %edx,%eax
400734: a8 80 test $0x80,%al
400736: 74 17 je 40074f <main+0x3f>
This leverages the constraints of the input values, since the values cannot be greater than 26 then subtracting a from b will give you a negative value when a > b, in two's complement you know bit 7 will be set in that case - the same applies to c. I then OR both so that bit 7 indicates whether a > b || a > c, lastly we inspect bit 7 by AND with 0x80 and branch on that.
Update: Out of curiosity I timed 4 different ways of coding this. To generate test data I used a simple linear congruential pseudo-random number generator. I timed it in a loop for 100 million iterations. I assumed for simplicity that if the condition is true we want to add 5 to a counter, do nothing otherwise. I timed it using g++ (GCC) 4.6.3 20120306 (Red Hat 4.6.3-2) on an Intel Xeon X5570 # 2.93GHz using -O2 optimization level.
Here's the code (comment out all but one of the conditional variants):
#include <iostream>
unsigned myrand() {
static unsigned x = 1;
return (x = x * 1664525 + 1013904223);
}
int main() {
size_t count = 0;
for(size_t i=0; i<100000000; ++i ) {
int a = 1 + myrand() % 26;
int b = 1 + myrand() % 26;
int c = 1 + myrand() % 26;
count += 5 & (((b-a) | (c-a)) >> 31); // 0.635 sec
//if (((b-a) | (c-a)) & 0x80) count += 5; // 0.660 sec
//if (a > std::max(b,c)) count += 5; // 0.677 sec
//if ( a > b || a > c) count += 5; // 1.164 sec
}
std::cout << count << std::endl;
return 0;
}
The fastest is a modification on the suggestion in my answer, where we use sign extension to generate a mask that is either 32 1s or 32 0s depending on whether the condition is true of false, and use that to mask the 5 being added so that it either adds 5 or 0. This variation has no branches. The times are in a comment on each line. The slowest was the original expression ( a > b || a > c).
I have the following function:
int mult(int y, int z)
{
if (z == 0)
return 0;
else if (z % 2 == 1)
return mult(2 * y, z / 2) + y;
else
return mult(2 * y, z / 2);
}
What I need to do is prove its correctness by induction. Now the trouble I'm having is that even though I know it works since I ran it I can't follow each individual step.
What is confusing me is that y only shows up as an argument and in no place does it show up in a return except in the recursive part, and yet the function actually returns y as the answer.
How does this happen? I need to be able to follow everything that happens so that I can do the iterations of it for the proof.
Since this is obviously a homework question, I recommend you do what the assinment was likely meant fot you to do. Trace through the code.
1) give a starting value for y and z.
2) either on paper or in a debugger, trace what happens when you call the function.
3) repeat step 2 with your current y/z values until program completion.
#include <iostream>
using namespace std;
int mult(int y, int z)
{
if(z==0) {
cout<<"z is null! - y:"<<y<<" z: "<<z<<endl;
return 0;
}
else if (z%2==1)
{
cout<<"z is odd! - y:"<<y<<" z: "<<z<<endl;
// make z even
return mult(2*y,z/2)+y;
}
else
{
cout<<"z is even! - y:"<<y<<" z: "<<z<<endl;
return mult(2*y,z/2);
}
}
int main() {
cout<<"result: "<<mult(3,13)<<endl;
}
Output:
z is odd! - y:3 z: 13
z is even! - y:6 z: 6
z is odd! - y:12 z: 3
z is odd! - y:24 z: 1
z is null! - y:48 z: 0
result: 39
How it works for 3 and 13:
There's a switch for even and odd numbers (see comment in code).
When z is null, the recursion "starts to return to the initial call". If the number z is odd it adds y to the returned value of the recursive call, if it's even it justs returns the value from the recursive call.
odd: return 0 + 24
odd: return 24 + 12
even: return 36
odd: return 36 + 3
step-by-step analisis
final result: 100
mult(10, 10)
{
makes 100
mult(20, 5)
{
makes 100
mult(40, 2) + 20
{
makes 80
mult(80, 1)
{
makes 80
mult(160, 0) + 80
{
return 0;
}
}
}
}
}
Note: If this is homework, tag it as such.
So, we basically got three recursive cases. To make it all clearer, I'd rewrite the C-code into some functional pseudo-code. Replace mult with an intuitive operator sign and figure out descriptive explanations of low-level expressions like (z%2==1).
You'll come up with something like
a ** b =
| b is 0 -> 0
| b is even -> 2a ** (b/2)
| b is odd -> 2a ** (b/2) + a
Do you get the point now?
One approach would be to translate each line into "English". My translation would be something like this:
if z is zero, return zero
if z is odd, return mult(y*2, z/2) + y
if z is even, return mult(y*2, z/2)
The general pattern is to recursively call mult with the first parameter doubling, and the second parameter halving.
Note that here you're calling mult with z/2, but its arguments are integers, so if your function continues to recurse, the 2nd parameter will halve each time until it gets down to 1, and then finally 1/2 which rounds down to 0 - at which point recursion will stop because z==0.
With those clues, you should be able to understand how this algorithm works.
Demonstrations by induction are based on proving that the result is valid for the first value, and that if the principle is correct for a generic value N, it is provable that it holds for N+1.
To simplify, you can start by proving that it works for z in { 0, 1, 2 } which should be trivial with a manual test. Then to demonstrate the induction step, you start with a generic z=N, and prove that if mult( y, N ) is a valid result, then mult( y, N+1 ) is also a valid result in terms of the previous one. Since there are different branches for even and odd numbers, you will have to prove the induction step for both even and odd N numbers.
ya = ya
a = an even number
b = the next odd number (in other words a + 1)
So, if you want the equation above in terms of only even numbers (an 'a') when given an odd number (a 'b') you can do the following:
yb = y(a+1) = y*a + y
Now confuse everyone by writing 'a' as 2*(z/2).
y*a becomes (2*y)*(z/2)
y*b becomes ((2*y)*(z/2))+y
Since 'z' appears in the formula for both even and odd numbers, we want to think that the code is telling us that (2*y)*(z/2) = (2*y)*(z/2) + y which is obviously MADNESS!
The reason is that we have snuck in the fact that z/2 is an integer and so z can never be odd. The compiler will not let us assign z/2 to an integer when z is odd. If we try to make 'z' odd, the integer we will really be using is (z-1)/2 instead of z/2.
To get around this, we have to test to see if z/2 is odd and pick our formula based on that (eg. either ya or yb in terms of 'a').
In mult(y,z) both 'y' and 'z' are both integers. Using the symbols above mult(2*y,b/2) becomes mult(2*y,a/2) because b/2 will be truncated to a/2 by the compiler.
Since we are always going to get an 'a' as a parameter to 'mult', even when we send a 'b', we have to make sure we are only using formulas that require 'a'. So, instead of yb we use ya+1 as described above.
b/2 = a/2 + 1/2 but 1/2 cannot be represented as part of an int.
Not really an answer, but more of a suggestion.
You may want to reduce the recursion call from 2 to one:
int mult(int y, int z)
{
int result = 0;
if (z == 0)
return result;
result = mult(2 * y, z / 2); // Common between "then" and "else"
if ((z % 2) == 1)
{
result += y;
}
return result;
}
This could be simplified once more by observing the rule "one exit point only":
int mult(int y, int z)
{
int result = 0;
if (z != 0)
{
result = mult(2 * y, z / 2); // Common between "then" and "else"
if ((z % 2) == 1)
{
result += y;
}
}
return result;
}
Although many compilers will perform this simplification automatically, debugging is usually easier when the code is simplified. The debugger will match the code when single-stepping.
Sometimes simplifying will add clarity. Also, adding comments will help you figure out what you are doing as well as the next person who reads the code.