How do you implement XOR using +-*/? - language-agnostic

How can the XOR operation (on two 32 bit ints) be implemented using only basic arithmetic operations? Do you have to do it bitwise after dividing by each power of 2 in turn, or is there a shortcut? I don't care about execution speed so much as about the simplest, shortest code.
Edit:
This is not homework, but a riddle posed on a hacker.org. The point is to implement XOR on a stack-based virtual machine with very limited operations (similar to the brainfuck language and yes - no shift or mod). Using that VM is the difficult part, though of course made easier by an algorithm that is short and simple.
While FryGuy's solution is clever, I'll have to go with my original ideal (similar to litb's solution) because comparisons are difficult to use as well in that environment.

I would do it the simple way:
uint xor(uint a, uint b):
uint ret = 0;
uint fact = 0x80000000;
while (fact > 0)
{
if ((a >= fact || b >= fact) && (a < fact || b < fact))
ret += fact;
if (a >= fact)
a -= fact;
if (b >= fact)
b -= fact;
fact /= 2;
}
return ret;
There might be an easier way, but I don't know of one.

I don't know whether this defeats the point of your question, but you can implement XOR with AND, OR, and NOT, like this:
uint xor(uint a, uint b) {
return (a | b) & ~(a & b);
}
In english, that's "a or b, but not a and b", which maps precisely to the definition of XOR.
Of course, I'm not sticking strictly to your stipulation of using only the arithmetic operators, but at least this a simple, easy-to-understand reimplementation.

I'm sorry i only know the straight forward one in head:
uint32_t mod_op(uint32_t a, uint32_t b) {
uint32_t int_div = a / b;
return a - (b * int_div);
}
uint32_t xor_op(uint32_t a, uint32_t b) {
uint32_t n = 1u;
uint32_t result = 0u;
while(a != 0 || b != 0) {
// or just: result += n * mod_op(a - b, 2);
if(mod_op(a, 2) != mod_op(b, 2)) {
result += n;
}
a /= 2;
b /= 2;
n *= 2;
}
return result;
}
The alternative in comments can be used instead of the if to avoid branching. But then again, the solution isn't exactly fast either and it makes it look stranger :)

It's easier if you have the AND because
A OR B = A + B - (A AND B)
A XOR B = A + B - 2(A AND B)
int customxor(int a, int b)
{
return a + b - 2*(a & b);
}

Related

Equal numbers in arrays are not equal in Octave [duplicate]

I'm currently writing some code where I have something along the lines of:
double a = SomeCalculation1();
double b = SomeCalculation2();
if (a < b)
DoSomething2();
else if (a > b)
DoSomething3();
And then in other places I may need to do equality:
double a = SomeCalculation3();
double b = SomeCalculation4();
if (a == 0.0)
DoSomethingUseful(1 / a);
if (b == 0.0)
return 0; // or something else here
In short, I have lots of floating point math going on and I need to do various comparisons for conditions. I can't convert it to integer math because such a thing is meaningless in this context.
I've read before that floating point comparisons can be unreliable, since you can have things like this going on:
double a = 1.0 / 3.0;
double b = a + a + a;
if ((3 * a) != b)
Console.WriteLine("Oh no!");
In short, I'd like to know: How can I reliably compare floating point numbers (less than, greater than, equality)?
The number range I am using is roughly from 10E-14 to 10E6, so I do need to work with small numbers as well as large.
I've tagged this as language agnostic because I'm interested in how I can accomplish this no matter what language I'm using.
TL;DR
Use the following function instead of the currently accepted solution to avoid some undesirable results in certain limit cases, while being potentially more efficient.
Know the expected imprecision you have on your numbers and feed them accordingly in the comparison function.
bool nearly_equal(
float a, float b,
float epsilon = 128 * FLT_EPSILON, float abs_th = FLT_MIN)
// those defaults are arbitrary and could be removed
{
assert(std::numeric_limits<float>::epsilon() <= epsilon);
assert(epsilon < 1.f);
if (a == b) return true;
auto diff = std::abs(a-b);
auto norm = std::min((std::abs(a) + std::abs(b)), std::numeric_limits<float>::max());
// or even faster: std::min(std::abs(a + b), std::numeric_limits<float>::max());
// keeping this commented out until I update figures below
return diff < std::max(abs_th, epsilon * norm);
}
Graphics, please?
When comparing floating point numbers, there are two "modes".
The first one is the relative mode, where the difference between x and y is considered relatively to their amplitude |x| + |y|. When plot in 2D, it gives the following profile, where green means equality of x and y. (I took an epsilon of 0.5 for illustration purposes).
The relative mode is what is used for "normal" or "large enough" floating points values. (More on that later).
The second one is an absolute mode, when we simply compare their difference to a fixed number. It gives the following profile (again with an epsilon of 0.5 and a abs_th of 1 for illustration).
This absolute mode of comparison is what is used for "tiny" floating point values.
Now the question is, how do we stitch together those two response patterns.
In Michael Borgwardt's answer, the switch is based on the value of diff, which should be below abs_th (Float.MIN_NORMAL in his answer). This switch zone is shown as hatched in the graph below.
Because abs_th * epsilon is smaller that abs_th, the green patches do not stick together, which in turn gives the solution a bad property: we can find triplets of numbers such that x < y_1 < y_2 and yet x == y2 but x != y1.
Take this striking example:
x = 4.9303807e-32
y1 = 4.930381e-32
y2 = 4.9309825e-32
We have x < y1 < y2, and in fact y2 - x is more than 2000 times larger than y1 - x. And yet with the current solution,
nearlyEqual(x, y1, 1e-4) == False
nearlyEqual(x, y2, 1e-4) == True
By contrast, in the solution proposed above, the switch zone is based on the value of |x| + |y|, which is represented by the hatched square below. It ensures that both zones connects gracefully.
Also, the code above does not have branching, which could be more efficient. Consider that operations such as max and abs, which a priori needs branching, often have dedicated assembly instructions. For this reason, I think this approach is superior to another solution that would be to fix Michael's nearlyEqual by changing the switch from diff < abs_th to diff < eps * abs_th, which would then produce essentially the same response pattern.
Where to switch between relative and absolute comparison?
The switch between those modes is made around abs_th, which is taken as FLT_MIN in the accepted answer. This choice means that the representation of float32 is what limits the precision of our floating point numbers.
This does not always make sense. For example, if the numbers you compare are the results of a subtraction, perhaps something in the range of FLT_EPSILON makes more sense. If they are squared roots of subtracted numbers, the numerical imprecision could be even higher.
It is rather obvious when you consider comparing a floating point with 0. Here, any relative comparison will fail, because |x - 0| / (|x| + 0) = 1. So the comparison needs to switch to absolute mode when x is on the order of the imprecision of your computation -- and rarely is it as low as FLT_MIN.
This is the reason for the introduction of the abs_th parameter above.
Also, by not multiplying abs_th with epsilon, the interpretation of this parameter is simple and correspond to the level of numerical precision that we expect on those numbers.
Mathematical rumbling
(kept here mostly for my own pleasure)
More generally I assume that a well-behaved floating point comparison operator =~ should have some basic properties.
The following are rather obvious:
self-equality: a =~ a
symmetry: a =~ b implies b =~ a
invariance by opposition: a =~ b implies -a =~ -b
(We don't have a =~ b and b =~ c implies a =~ c, =~ is not an equivalence relationship).
I would add the following properties that are more specific to floating point comparisons
if a < b < c, then a =~ c implies a =~ b (closer values should also be equal)
if a, b, m >= 0 then a =~ b implies a + m =~ b + m (larger values with the same difference should also be equal)
if 0 <= λ < 1 then a =~ b implies λa =~ λb (perhaps less obvious to argument for).
Those properties already give strong constrains on possible near-equality functions. The function proposed above verifies them. Perhaps one or several otherwise obvious properties are missing.
When one think of =~ as a family of equality relationship =~[Ɛ,t] parameterized by Ɛ and abs_th, one could also add
if Ɛ1 < Ɛ2 then a =~[Ɛ1,t] b implies a =~[Ɛ2,t] b (equality for a given tolerance implies equality at a higher tolerance)
if t1 < t2 then a =~[Ɛ,t1] b implies a =~[Ɛ,t2] b (equality for a given imprecision implies equality at a higher imprecision)
The proposed solution also verifies these.
Comparing for greater/smaller is not really a problem unless you're working right at the edge of the float/double precision limit.
For a "fuzzy equals" comparison, this (Java code, should be easy to adapt) is what I came up with for The Floating-Point Guide after a lot of work and taking into account lots of criticism:
public static boolean nearlyEqual(float a, float b, float epsilon) {
final float absA = Math.abs(a);
final float absB = Math.abs(b);
final float diff = Math.abs(a - b);
if (a == b) { // shortcut, handles infinities
return true;
} else if (a == 0 || b == 0 || diff < Float.MIN_NORMAL) {
// a or b is zero or both are extremely close to it
// relative error is less meaningful here
return diff < (epsilon * Float.MIN_NORMAL);
} else { // use relative error
return diff / (absA + absB) < epsilon;
}
}
It comes with a test suite. You should immediately dismiss any solution that doesn't, because it is virtually guaranteed to fail in some edge cases like having one value 0, two very small values opposite of zero, or infinities.
An alternative (see link above for more details) is to convert the floats' bit patterns to integer and accept everything within a fixed integer distance.
In any case, there probably isn't any solution that is perfect for all applications. Ideally, you'd develop/adapt your own with a test suite covering your actual use cases.
I had the problem of Comparing floating point numbers A < B and A > B
Here is what seems to work:
if(A - B < Epsilon) && (fabs(A-B) > Epsilon)
{
printf("A is less than B");
}
if (A - B > Epsilon) && (fabs(A-B) > Epsilon)
{
printf("A is greater than B");
}
The fabs--absolute value-- takes care of if they are essentially equal.
We have to choose a tolerance level to compare float numbers. For example,
final float TOLERANCE = 0.00001;
if (Math.abs(f1 - f2) < TOLERANCE)
Console.WriteLine("Oh yes!");
One note. Your example is rather funny.
double a = 1.0 / 3.0;
double b = a + a + a;
if (a != b)
Console.WriteLine("Oh no!");
Some maths here
a = 1/3
b = 1/3 + 1/3 + 1/3 = 1.
1/3 != 1
Oh, yes..
Do you mean
if (b != 1)
Console.WriteLine("Oh no!")
Idea I had for floating point comparison in swift
infix operator ~= {}
func ~= (a: Float, b: Float) -> Bool {
return fabsf(a - b) < Float(FLT_EPSILON)
}
func ~= (a: CGFloat, b: CGFloat) -> Bool {
return fabs(a - b) < CGFloat(FLT_EPSILON)
}
func ~= (a: Double, b: Double) -> Bool {
return fabs(a - b) < Double(FLT_EPSILON)
}
Adaptation to PHP from Michael Borgwardt & bosonix's answer:
class Comparison
{
const MIN_NORMAL = 1.17549435E-38; //from Java Specs
// from http://floating-point-gui.de/errors/comparison/
public function nearlyEqual($a, $b, $epsilon = 0.000001)
{
$absA = abs($a);
$absB = abs($b);
$diff = abs($a - $b);
if ($a == $b) {
return true;
} else {
if ($a == 0 || $b == 0 || $diff < self::MIN_NORMAL) {
return $diff < ($epsilon * self::MIN_NORMAL);
} else {
return $diff / ($absA + $absB) < $epsilon;
}
}
}
}
You should ask yourself why you are comparing the numbers. If you know the purpose of the comparison then you should also know the required accuracy of your numbers. That is different in each situation and each application context. But in pretty much all practical cases there is a required absolute accuracy. It is only very seldom that a relative accuracy is applicable.
To give an example: if your goal is to draw a graph on the screen, then you likely want floating point values to compare equal if they map to the same pixel on the screen. If the size of your screen is 1000 pixels, and your numbers are in the 1e6 range, then you likely will want 100 to compare equal to 200.
Given the required absolute accuracy, then the algorithm becomes:
public static ComparisonResult compare(float a, float b, float accuracy)
{
if (isnan(a) || isnan(b)) // if NaN needs to be supported
return UNORDERED;
if (a == b) // short-cut and takes care of infinities
return EQUAL;
if (abs(a-b) < accuracy) // comparison wrt. the accuracy
return EQUAL;
if (a < b) // larger / smaller
return SMALLER;
else
return LARGER;
}
The standard advice is to use some small "epsilon" value (chosen depending on your application, probably), and consider floats that are within epsilon of each other to be equal. e.g. something like
#define EPSILON 0.00000001
if ((a - b) < EPSILON && (b - a) < EPSILON) {
printf("a and b are about equal\n");
}
A more complete answer is complicated, because floating point error is extremely subtle and confusing to reason about. If you really care about equality in any precise sense, you're probably seeking a solution that doesn't involve floating point.
I tried writing an equality function with the above comments in mind. Here's what I came up with:
Edit: Change from Math.Max(a, b) to Math.Max(Math.Abs(a), Math.Abs(b))
static bool fpEqual(double a, double b)
{
double diff = Math.Abs(a - b);
double epsilon = Math.Max(Math.Abs(a), Math.Abs(b)) * Double.Epsilon;
return (diff < epsilon);
}
Thoughts? I still need to work out a greater than, and a less than as well.
I came up with a simple approach to adjusting the size of epsilon to the size of the numbers being compared. So, instead of using:
iif(abs(a - b) < 1e-6, "equal", "not")
if a and b can be large, I changed that to:
iif(abs(a - b) < (10 ^ -abs(7 - log(a))), "equal", "not")
I suppose that doesn't satisfy all the theoretical issues discussed in the other answers, but it has the advantage of being one line of code, so it can be used in an Excel formula or an Access query without needing a VBA function.
I did a search to see if others have used this method and I didn't find anything. I tested it in my application and it seems to be working well. So it seems to be a method that is adequate for contexts that don't require the complexity of the other answers. But I wonder if it has a problem I haven't thought of since no one else seems to be using it.
If there's a reason the test with the log is not valid for simple comparisons of numbers of various sizes, please say why in a comment.
You need to take into account that the truncation error is a relative one. Two numbers are about equal if their difference is about as large as their ulp (Unit in the last place).
However, if you do floating point calculations, your error potential goes up with every operation (esp. careful with subtractions!), so your error tolerance needs to increase accordingly.
The best way to compare doubles for equality/inequality is by taking the absolute value of their difference and comparing it to a small enough (depending on your context) value.
double eps = 0.000000001; //for instance
double a = someCalc1();
double b = someCalc2();
double diff = Math.abs(a - b);
if (diff < eps) {
//equal
}

Free-Pascal Implementation of the Sieve of Eratosthenes

My teacher gave me an assignment like this:
Using the number n given, find the largest prime number p with p<=n and n<=10^9.
I tried doing this by using the following function:
Const amax=1000000000
Var i,j,n:longint;
a:array [1..amax] of boolean;
Function lp(n:longint):longint;
Var max:longint;
Begin
For i:=1 to n do a[i]:=true;
For i:=2 to round(sqrt(n)) do
If (a[i]=true) then
For j:=1 to n div i do
If (i*i+(j-1)*i<=n) then
a[i*i+(j-1)*i]:=false;
max:=0;
i:=n;
While max=0 do
Begin
If a[i]=true then max:=i;
i:=i-1;
End;
lp:=max;
End;
This code worked flawlessly for numbers such as 1 million, but when i tried n=10^9, the program took a long time to print the output. So here's my question: Are there any ways to improve my code for lower delay? Or maybe a different code?
The most important aspect here is that the greatest prime that is not greater than n must be fairly close to n. A quick look at The Gaps Between Primes (at The Prime Pages - always worth a look for everything to do with primes) shows that for 32-bit numbers the gaps between primes cannot be greater than 335. This means that the greatest prime not greater than n must be in the range [n - 335, n]. In other words, at most 336 candidates need to be checked - for example via trial division - and this is bound to be lots faster than sieving a billion numbers.
Trial division is a reasonable choice for tasks of this kind, because the range to be scanned is so small. In my answer to Prime sieve implementation (using trial division) in C++ I analysed a couple of ways for speeding it up.
The Sieve of Eratosthenes is also a good choice, it just needs to be modified to sieve only the range of interest instead of all numbers from 1 to n. This is called a 'windowed sieve' because it sieves only a window. Since the window will most likely not contain all the primes up to the square root of n (i.e. all the primes that could be potential least prime factors of composites in the range to be scanned) it is best to sieve the factor primes via a separate, simple Sieve of Eratosthenes.
First I'm showing a simple rendition of normal (non-windowed) sieve, as a baseline for comparing the windowed code to. I'm using C# in order to show the algorithm more clearly than would be possible with Pascal.
List<uint> small_primes_up_to (uint n)
{
if (n == uint.MaxValue)
throw new ArgumentOutOfRangeException("n", "n must be less than UINT32_MAX");
var eliminated = new bool[n + 1]; // +1 because indexed by numbers
eliminated[0] = true;
eliminated[1] = true;
for (uint i = 2, sqrt_n = (uint)Math.Sqrt(n); i <= sqrt_n; ++i)
if (!eliminated[i])
for (uint j = i * i; j <= n; j += i)
eliminated[j] = true;
return remaining_unmarked_numbers(eliminated, 0);
}
The fuction has 'small' in its name because it is not really suited for sieving big ranges; I use similar code (with a few bells and whistles) only for sieving the small factor primes needed by more advanced sieves.
The code for extracting the sieved primes is equally simple:
List<uint> remaining_unmarked_numbers (bool[] eliminated, uint sieve_base)
{
var result = new List<uint>();
for (uint i = 0, e = (uint)eliminated.Length; i < e; ++i)
if (!eliminated[i])
result.Add(sieve_base + i);
return result;
}
Now, the windowed version. One difference is that the potential least factor primes need to be sieved separately (by the function just shown) as explained earlier. Another difference is that the starting point of the crossing-off sequence for a given prime may lie outside the range to be sieved. If the starting point lies before the start of the window then a bit of modulo magic is necessary to find the first 'hop' that lands in the window. From then on everything proceeds as usual.
List<uint> primes_between (uint m, uint n)
{
m = Math.Max(m, 2);
if (m > n)
return new List<uint>(); // empty range -> no primes
// index overflow in the inner loop unless `(sieve_bits - 1) + stride <= UINT32_MAX`
if (n - m > uint.MaxValue - 65521) // highest prime not greater than sqrt(UINT32_MAX)
throw new ArgumentOutOfRangeException("n", "(n - m) must be <= UINT32_MAX - 65521");
uint sieve_bits = n - m + 1;
var eliminated = new bool[sieve_bits];
foreach (uint prime in small_primes_up_to((uint)Math.Sqrt(n)))
{
uint start = prime * prime, stride = prime;
if (start >= m)
start -= m;
else
start = (stride - 1) - (m - start - 1) % stride;
for (uint j = start; j < sieve_bits; j += stride)
eliminated[j] = true;
}
return remaining_unmarked_numbers(eliminated, m);
}
The two '-1' terms in the modulo calculation may seem strange, but they bias the logic down by 1 to eliminate the inconvenient case stride - foo % stride == stride that would need to be mapped to 0.
With this, the greatest prime not exceeding n could be computed like this:
uint greatest_prime_not_exceeding (uint n)
{
return primes_between(n - Math.Min(n, 335), n).Last();
}
This takes less than a millisecond all told, including the sieving of the factor primes and so on, even though the code contains no optimisations whatsoever. A good overview of applicable optimisations can be found in my answer to prime number summing still slow after using sieve; with the techniques shown there the whole range up to 10^9 can be sieved in about half a second.

Algorithm for subtraction of binary numbers

I am doing a project in C++ and I'm having hard time with substraction of binary numbers. It's easy doing it on paper but in code it's a bit hard for me.
Can someone please give me the algorithm for the subtraction of two binary numbers? It's supposed to be done WITHOUT conversion to the decimal system.
Thanks!
int subtractBinaries(int x, int y) {
while (y != 0) {
int borrow = (~x) & y;
x = x ^ y;
y = borrow << 1;
}
return x;
}

How should I do floating point comparison?

I'm currently writing some code where I have something along the lines of:
double a = SomeCalculation1();
double b = SomeCalculation2();
if (a < b)
DoSomething2();
else if (a > b)
DoSomething3();
And then in other places I may need to do equality:
double a = SomeCalculation3();
double b = SomeCalculation4();
if (a == 0.0)
DoSomethingUseful(1 / a);
if (b == 0.0)
return 0; // or something else here
In short, I have lots of floating point math going on and I need to do various comparisons for conditions. I can't convert it to integer math because such a thing is meaningless in this context.
I've read before that floating point comparisons can be unreliable, since you can have things like this going on:
double a = 1.0 / 3.0;
double b = a + a + a;
if ((3 * a) != b)
Console.WriteLine("Oh no!");
In short, I'd like to know: How can I reliably compare floating point numbers (less than, greater than, equality)?
The number range I am using is roughly from 10E-14 to 10E6, so I do need to work with small numbers as well as large.
I've tagged this as language agnostic because I'm interested in how I can accomplish this no matter what language I'm using.
TL;DR
Use the following function instead of the currently accepted solution to avoid some undesirable results in certain limit cases, while being potentially more efficient.
Know the expected imprecision you have on your numbers and feed them accordingly in the comparison function.
bool nearly_equal(
float a, float b,
float epsilon = 128 * FLT_EPSILON, float abs_th = FLT_MIN)
// those defaults are arbitrary and could be removed
{
assert(std::numeric_limits<float>::epsilon() <= epsilon);
assert(epsilon < 1.f);
if (a == b) return true;
auto diff = std::abs(a-b);
auto norm = std::min((std::abs(a) + std::abs(b)), std::numeric_limits<float>::max());
// or even faster: std::min(std::abs(a + b), std::numeric_limits<float>::max());
// keeping this commented out until I update figures below
return diff < std::max(abs_th, epsilon * norm);
}
Graphics, please?
When comparing floating point numbers, there are two "modes".
The first one is the relative mode, where the difference between x and y is considered relatively to their amplitude |x| + |y|. When plot in 2D, it gives the following profile, where green means equality of x and y. (I took an epsilon of 0.5 for illustration purposes).
The relative mode is what is used for "normal" or "large enough" floating points values. (More on that later).
The second one is an absolute mode, when we simply compare their difference to a fixed number. It gives the following profile (again with an epsilon of 0.5 and a abs_th of 1 for illustration).
This absolute mode of comparison is what is used for "tiny" floating point values.
Now the question is, how do we stitch together those two response patterns.
In Michael Borgwardt's answer, the switch is based on the value of diff, which should be below abs_th (Float.MIN_NORMAL in his answer). This switch zone is shown as hatched in the graph below.
Because abs_th * epsilon is smaller that abs_th, the green patches do not stick together, which in turn gives the solution a bad property: we can find triplets of numbers such that x < y_1 < y_2 and yet x == y2 but x != y1.
Take this striking example:
x = 4.9303807e-32
y1 = 4.930381e-32
y2 = 4.9309825e-32
We have x < y1 < y2, and in fact y2 - x is more than 2000 times larger than y1 - x. And yet with the current solution,
nearlyEqual(x, y1, 1e-4) == False
nearlyEqual(x, y2, 1e-4) == True
By contrast, in the solution proposed above, the switch zone is based on the value of |x| + |y|, which is represented by the hatched square below. It ensures that both zones connects gracefully.
Also, the code above does not have branching, which could be more efficient. Consider that operations such as max and abs, which a priori needs branching, often have dedicated assembly instructions. For this reason, I think this approach is superior to another solution that would be to fix Michael's nearlyEqual by changing the switch from diff < abs_th to diff < eps * abs_th, which would then produce essentially the same response pattern.
Where to switch between relative and absolute comparison?
The switch between those modes is made around abs_th, which is taken as FLT_MIN in the accepted answer. This choice means that the representation of float32 is what limits the precision of our floating point numbers.
This does not always make sense. For example, if the numbers you compare are the results of a subtraction, perhaps something in the range of FLT_EPSILON makes more sense. If they are squared roots of subtracted numbers, the numerical imprecision could be even higher.
It is rather obvious when you consider comparing a floating point with 0. Here, any relative comparison will fail, because |x - 0| / (|x| + 0) = 1. So the comparison needs to switch to absolute mode when x is on the order of the imprecision of your computation -- and rarely is it as low as FLT_MIN.
This is the reason for the introduction of the abs_th parameter above.
Also, by not multiplying abs_th with epsilon, the interpretation of this parameter is simple and correspond to the level of numerical precision that we expect on those numbers.
Mathematical rumbling
(kept here mostly for my own pleasure)
More generally I assume that a well-behaved floating point comparison operator =~ should have some basic properties.
The following are rather obvious:
self-equality: a =~ a
symmetry: a =~ b implies b =~ a
invariance by opposition: a =~ b implies -a =~ -b
(We don't have a =~ b and b =~ c implies a =~ c, =~ is not an equivalence relationship).
I would add the following properties that are more specific to floating point comparisons
if a < b < c, then a =~ c implies a =~ b (closer values should also be equal)
if a, b, m >= 0 then a =~ b implies a + m =~ b + m (larger values with the same difference should also be equal)
if 0 <= λ < 1 then a =~ b implies λa =~ λb (perhaps less obvious to argument for).
Those properties already give strong constrains on possible near-equality functions. The function proposed above verifies them. Perhaps one or several otherwise obvious properties are missing.
When one think of =~ as a family of equality relationship =~[Ɛ,t] parameterized by Ɛ and abs_th, one could also add
if Ɛ1 < Ɛ2 then a =~[Ɛ1,t] b implies a =~[Ɛ2,t] b (equality for a given tolerance implies equality at a higher tolerance)
if t1 < t2 then a =~[Ɛ,t1] b implies a =~[Ɛ,t2] b (equality for a given imprecision implies equality at a higher imprecision)
The proposed solution also verifies these.
Comparing for greater/smaller is not really a problem unless you're working right at the edge of the float/double precision limit.
For a "fuzzy equals" comparison, this (Java code, should be easy to adapt) is what I came up with for The Floating-Point Guide after a lot of work and taking into account lots of criticism:
public static boolean nearlyEqual(float a, float b, float epsilon) {
final float absA = Math.abs(a);
final float absB = Math.abs(b);
final float diff = Math.abs(a - b);
if (a == b) { // shortcut, handles infinities
return true;
} else if (a == 0 || b == 0 || diff < Float.MIN_NORMAL) {
// a or b is zero or both are extremely close to it
// relative error is less meaningful here
return diff < (epsilon * Float.MIN_NORMAL);
} else { // use relative error
return diff / (absA + absB) < epsilon;
}
}
It comes with a test suite. You should immediately dismiss any solution that doesn't, because it is virtually guaranteed to fail in some edge cases like having one value 0, two very small values opposite of zero, or infinities.
An alternative (see link above for more details) is to convert the floats' bit patterns to integer and accept everything within a fixed integer distance.
In any case, there probably isn't any solution that is perfect for all applications. Ideally, you'd develop/adapt your own with a test suite covering your actual use cases.
I had the problem of Comparing floating point numbers A < B and A > B
Here is what seems to work:
if(A - B < Epsilon) && (fabs(A-B) > Epsilon)
{
printf("A is less than B");
}
if (A - B > Epsilon) && (fabs(A-B) > Epsilon)
{
printf("A is greater than B");
}
The fabs--absolute value-- takes care of if they are essentially equal.
We have to choose a tolerance level to compare float numbers. For example,
final float TOLERANCE = 0.00001;
if (Math.abs(f1 - f2) < TOLERANCE)
Console.WriteLine("Oh yes!");
One note. Your example is rather funny.
double a = 1.0 / 3.0;
double b = a + a + a;
if (a != b)
Console.WriteLine("Oh no!");
Some maths here
a = 1/3
b = 1/3 + 1/3 + 1/3 = 1.
1/3 != 1
Oh, yes..
Do you mean
if (b != 1)
Console.WriteLine("Oh no!")
Idea I had for floating point comparison in swift
infix operator ~= {}
func ~= (a: Float, b: Float) -> Bool {
return fabsf(a - b) < Float(FLT_EPSILON)
}
func ~= (a: CGFloat, b: CGFloat) -> Bool {
return fabs(a - b) < CGFloat(FLT_EPSILON)
}
func ~= (a: Double, b: Double) -> Bool {
return fabs(a - b) < Double(FLT_EPSILON)
}
Adaptation to PHP from Michael Borgwardt & bosonix's answer:
class Comparison
{
const MIN_NORMAL = 1.17549435E-38; //from Java Specs
// from http://floating-point-gui.de/errors/comparison/
public function nearlyEqual($a, $b, $epsilon = 0.000001)
{
$absA = abs($a);
$absB = abs($b);
$diff = abs($a - $b);
if ($a == $b) {
return true;
} else {
if ($a == 0 || $b == 0 || $diff < self::MIN_NORMAL) {
return $diff < ($epsilon * self::MIN_NORMAL);
} else {
return $diff / ($absA + $absB) < $epsilon;
}
}
}
}
You should ask yourself why you are comparing the numbers. If you know the purpose of the comparison then you should also know the required accuracy of your numbers. That is different in each situation and each application context. But in pretty much all practical cases there is a required absolute accuracy. It is only very seldom that a relative accuracy is applicable.
To give an example: if your goal is to draw a graph on the screen, then you likely want floating point values to compare equal if they map to the same pixel on the screen. If the size of your screen is 1000 pixels, and your numbers are in the 1e6 range, then you likely will want 100 to compare equal to 200.
Given the required absolute accuracy, then the algorithm becomes:
public static ComparisonResult compare(float a, float b, float accuracy)
{
if (isnan(a) || isnan(b)) // if NaN needs to be supported
return UNORDERED;
if (a == b) // short-cut and takes care of infinities
return EQUAL;
if (abs(a-b) < accuracy) // comparison wrt. the accuracy
return EQUAL;
if (a < b) // larger / smaller
return SMALLER;
else
return LARGER;
}
The standard advice is to use some small "epsilon" value (chosen depending on your application, probably), and consider floats that are within epsilon of each other to be equal. e.g. something like
#define EPSILON 0.00000001
if ((a - b) < EPSILON && (b - a) < EPSILON) {
printf("a and b are about equal\n");
}
A more complete answer is complicated, because floating point error is extremely subtle and confusing to reason about. If you really care about equality in any precise sense, you're probably seeking a solution that doesn't involve floating point.
I tried writing an equality function with the above comments in mind. Here's what I came up with:
Edit: Change from Math.Max(a, b) to Math.Max(Math.Abs(a), Math.Abs(b))
static bool fpEqual(double a, double b)
{
double diff = Math.Abs(a - b);
double epsilon = Math.Max(Math.Abs(a), Math.Abs(b)) * Double.Epsilon;
return (diff < epsilon);
}
Thoughts? I still need to work out a greater than, and a less than as well.
I came up with a simple approach to adjusting the size of epsilon to the size of the numbers being compared. So, instead of using:
iif(abs(a - b) < 1e-6, "equal", "not")
if a and b can be large, I changed that to:
iif(abs(a - b) < (10 ^ -abs(7 - log(a))), "equal", "not")
I suppose that doesn't satisfy all the theoretical issues discussed in the other answers, but it has the advantage of being one line of code, so it can be used in an Excel formula or an Access query without needing a VBA function.
I did a search to see if others have used this method and I didn't find anything. I tested it in my application and it seems to be working well. So it seems to be a method that is adequate for contexts that don't require the complexity of the other answers. But I wonder if it has a problem I haven't thought of since no one else seems to be using it.
If there's a reason the test with the log is not valid for simple comparisons of numbers of various sizes, please say why in a comment.
You need to take into account that the truncation error is a relative one. Two numbers are about equal if their difference is about as large as their ulp (Unit in the last place).
However, if you do floating point calculations, your error potential goes up with every operation (esp. careful with subtractions!), so your error tolerance needs to increase accordingly.
The best way to compare doubles for equality/inequality is by taking the absolute value of their difference and comparing it to a small enough (depending on your context) value.
double eps = 0.000000001; //for instance
double a = someCalc1();
double b = someCalc2();
double diff = Math.abs(a - b);
if (diff < eps) {
//equal
}

How to store a symmetric matrix?

Which is the best way to store a symmetric matrix in memory?
It would be good to save half of the space without compromising speed and complexity of the structure too much. This is a language-agnostic question but if you need to make some assumptions just assume it's a good old plain programming language like C or C++..
It seems a thing that has a sense just if there is a way to keep things simple or just when the matrix itself is really big, am I right?
Just for the sake of formality I mean that this assertion is always true for the data I want to store
matrix[x][y] == matrix[y][x]
Here is a good method to store a symmetric matrix, it requires only N(N+1)/2 memory:
int fromMatrixToVector(int i, int j, int N)
{
if (i <= j)
return i * N - (i - 1) * i / 2 + j - i;
else
return j * N - (j - 1) * j / 2 + i - j;
}
For some triangular matrix
0 1 2 3
4 5 6
7 8
9
1D representation (stored in std::vector, for example) looks like as follows:
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
And call fromMatrixToVector(1, 2, 4) returns 5, so the matrix data is vector[5] -> 5.
For more information see http://www.codeguru.com/cpp/cpp/algorithms/general/article.php/c11211/TIP-Half-Size-Triangular-Matrix.htm
I find that many high performance packages just store the whole matrix, but then only read the upper triangle or lower triangle. They might then use the additional space for storing temporary data during the computation.
However if storage is really an issue then just store the n(n+1)/2 elements making the upper triangle in a one-dimensional array. If that makes access complicated for you, just define a set of helper functions.
In C to access a matrix matA you could define a macro:
#define A(i,j, dim) ((i <= j)?matA[i*dim + j]:matA[j*dim + i])
then you can access your array nearly normally.
Well I would try a triangular matrix, like this:
int[][] sym = new int[rows][];
for( int i = 0; i < cols; ++i ) {
sym=new int[i+1];
}
But then you wil have to face the problem when someone wants to access the "other side". Eg he wants to access [0][10] but in your case this val is stored in[10][0] (assuming 10x10).
The probably "best" way is the lazy one - dont do anything until the user requests. So you could load the specific row if the user types somethin like print(matrix[4]).
If you want to use a one dimensional array the code would look something like this:
int[] new matrix[(rows * (rows + 1 )) >> 1];
int z;
matrix[ ( ( z = ( x < y ? y : x ) ) * ( z + 1 ) >> 1 ) + ( y < x ? y : x ) ] = yourValue;
You can get rid of the multiplications if you create an additional look-up table:
int[] new matrix[(rows * (rows + 1 )) >> 1];
int[] lookup[rows];
for ( int i= 0; i < rows; i++)
{
lookup[i] = (i * (i+1)) >> 1;
}
matrix[ lookup[ x < y ? y : x ] + ( x < y ? x : y ) ] = yourValue;
If you're using something that supports operator overloading (e.g. C++), it's pretty easy to handle this transparently. Just create a matrix class that checks the two subscripts, and if the second is greater than the first, swap them:
template <class T>
class sym_matrix {
std::vector<std::vector<T> > data;
public:
T operator()(int x, int y) {
if (y>x)
return data[y][x];
else
return data[x][y];
}
};
For the moment I've skipped over everything else, and just covered the subscripting. In reality, to handle use as both an lvalue and an rvalue correctly, you'll typically want to return a proxy instead of a T directly. You'll want a ctor that creates data as a triangle (i.e., for an NxN matrix, the first row will have N elements, the second N-1, and so on -- or, equivalantly 1, 2, ...N). You might also consider creating data as a single vector -- you have to compute the correct offset into it, but that's not terribly difficult, and it will use a bit less memory, run a bit faster, etc. I'd use the simple code for the first version, and optimize later if necessary.
You could use a staggered array (or whatever they're called) if your language supports it, and when x < y, switch the position of x and y. So...
Pseudocode (somewhat Python style, but not really) for an n x n matrix:
matrix[n][]
for i from 0 to n-1:
matrix[i] = some_value_type[i + 1]
[next, assign values to the elements of the half-matrix]
And then when referring to values....
if x < y:
return matrix[y][x]
else:
return matrix[x][y]