Here is a snippet of code where I'm using gmp_prob_prime. Even though I'm currently only testing numbers in the 10^6 range this function VERY regularly "fails" my QuickTest and ends up needing to do a manual check of $NumberToTest for primality.
Is gmp_prob_prime not very robust? I didn't expect it to suggest "probable prime" until I was in the 10^9 or even 10^12 range.
Here is the snippet of my code's function that is being called:
function IsPrime($DocRoot, $NumberToTest, $PowOf2)
{
// First a quick test...
// 0 = composite
// 1 = probable prime
// 2 = definite prime
$Reps = 15;
$QuickTest = gmp_prob_prime($NumberToTest,$Reps);
if( $QuickTest == 0 )
{
return 0;
}
if ( $QuickTest == 2 )
{
return 1;
}
// If we get to here then gmp_prob_prime isn't sure whether the $NumberToTest is prime or not.
print "Consider increasing the Reps for gmp_prob_prime.\n";
// Find the sqrt of $NumberToTest;
... code continues ...
I had the same behavior when calling mpz_probab_prime_p directly from C++, but I can't recall if the below information fixed it or not (copied from the manual).
Function: int mpz_probab_prime_p (const mpz_t n, int reps)
Determine whether n is prime. Return 2 if n is definitely prime, return 1 if n is probably prime (without being certain), or return 0 if n is definitely composite.
This function does some trial divisions, then some Miller-Rabin probabilistic primality tests. The argument reps controls how many such tests are done; a higher value will reduce the chances of a composite being returned as “probably prime”. 25 is a reasonable number; a composite number will then be identified as a prime with a probability of less than 2^(-50).
Miller-Rabin and similar tests can be more properly called compositeness tests. Numbers which fail are known to be composite but those which pass might be prime or might be composite. Only a few composites pass, hence those which pass are considered probably prime.
Related
I'm doing a calculation in PHP using bcmath, and need to raise e by a fractional exponent. Unfortunately, bcpow() only accepts integer exponents. The exponent is typically higher precision than a float will allow, so normal arithmetic functions won't cut it.
For example:
$e = exp(1);
$pow = "0.000000000000000000108420217248550443400745280086994171142578125";
$result = bcpow($e, $pow);
Result is "1" with the error, "bc math warning: non-zero scale in exponent".
Is there another function I can use instead of bcpow()?
Your best bet is probably to use the Taylor series expansion. As you noted, PHP's bcpow is limited to raising to integer exponentiation.
So what you can do is roll your own bc factorial function and use the wiki page to implement a Taylor series expansion of the exponential function.
function bcfac($num) {
if ($num==0) return 1;
$result = '1';
for ( ; $num > 0; $num--)
$result = bcmul($result,$num);
return $result;
}
$mysum = '0';
for ($i=0; $i<300; $i++) {
$mysum = bcadd($mysum, bcdiv(bcpow($pow,$i), bcfac($i)) );
}
print $mysum;
Obviously, the $i<300 is an approximation for infinity... You can change it to suit your performance needs.
With $i=20, I got
1.00000000000000000010842021724855044340662275184110560868263421994092888869270293594926619547803962155136242752708629105688492780863293090291376157887898519458498571566021915144483905034693109606778068801680332504212458366799913406541920812216634834265692913062346724688397654924947370526356787052264726969653983148004800229537555582281617497990286595977830803702329470381960270717424849203303593850108090101578510305396615293917807977774686848422213799049363135722460179809890014584148659937665374616
This is comforting since that small of an exponent should yield something really close to 1.0.
Old question, but people might still be interested nonetheless.
So Kevin got the right idea with the Taylor-polynomial, but when you derive your algorithm from it directly, you can get into trouble, mainly your code gets slow for long input-strings when using large cut-off values for $i.
Here is why:
At every step, by which I mean with each new $i, the code calls bcfac($i). Everytime bcfac is called it performs $i-1 calculations. And $i goes all the way up to 299... that's almost 45000 operations! Not your quick'n'easy floating point operations, but slow BC-string-operations - if you set bcscale(100) your bcmul has to handle up to 10000 pairs of chars!
Also bcpow slows down with increasing $i, too. Not as much as bcfac, because it propably uses something akin to the square-and-multiply method, but it still adds something.
Overall the time required grows quadraticly with the number of polynomial terms computed.
So... what to do?
Here's a tip:
Whenever you handle polynomials, especially Taylor-polynomials, use the Horner method.
It converts this: exp(x) = x^0/0! + x^1/1! + x^2/2! + x^3/3! + ...
...into that: exp(x) = ((( ... )*x/3+1 )*x/2+1 )*x/1+1
And suddenly you don't need any powers or factorials at all!
function bc_exp($number) {
$result = 1;
for ($i=299; $i>0; $i--)
$result = bcadd(bcmul(bcdiv($result, $i), $number), 1);
return $result;
}
This needs only 3 bc-operations for each step, no matter what $i is.
With a starting value of $i=299 (to calculate exp with the same precision as kevin's code does) we now only need 897 bc-operations, compared to more than 45000.
Even using 30 as cut-off instead of 300, we now only need 87 bc-operations while the other code still needs 822 for the factorials alone.
Horner's Method saving the day again!
Some other thoughts:
1) Kevin's code would propably crash with input="0", depending on how bcmath handles errors, because the code trys bcpow(0,0) at the first step ($i=0).
2) Larger exponents require longer polynomials and therefore more iterations, e.g. bc_exp(300) will give a wrong answer, even with $i=299, whyle something like bc_exp(3) will work fine and dandy.
Each term adds x^n/n! to the result, so this term has to get small before the polynomial can start to converge. Now compare two consecutive terms:
( x^(n+1)/(n+1)! ) / ( x^n/n! ) = x/n
Each summand is larger than the one before by a factor of x/n (which we used via the Horner method), so in order for x^(n+1)/(n+1)! to get small x/n has to get small as well, which is only the case when n>x.
Inconclusio: As long as the number of iterations is smaller than the input value, the result will diverge. Only when you add steps until your number of iterations gets larger than the input, the algorithm starts to slowly converge.
In order to reach results that can satisfie someone who is willing to use bcmath, your $i needs to be significantly larger then your $number. And that's a huge proplem when you try to calculate stuff like e^346674567801
A solution is to divide the input into its integer part and its fraction part.
Than use bcpow on the integer part and bc_exp on the fraction part, which now converges from the get-go since the fraction part is smaller than 1. In the end multiply the results.
e^x = e^(intpart+fracpart) = e^intpart * e^fracpart = bcpow(e,intpart) * bc_exp(fracpart)
You could even implement it directly into the code above:
function bc_exp2($number) {
$parts = explode (".", $number);
$fracpart = "0.".$parts[1];
$result = 1;
for ($i=299; $i>0; $i--)
$result = bcadd(bcmul(bcdiv($result, $i), $fracpart), 1);
$result = bcmul(bcpow(exp(1), $parts[0]), $result);
return $result;
}
Note that exp(1) gives you a floating-point number which propably won't satisfy your needs as a bcmath user. You might want to use a value for e that is more accurate, in accordance with your bcscale setting.
3) Talking about numbers of iterations: 300 will be overkill in most situations while in some others it might not even be enough. An algorithm that takes your bcscale and $number and calculates the number of required iterations would be nice. Alraedy got some ideas involving log(n!), but nothing concrete yet.
4) To use this method with an arbitrary base you can use a^x = e^(x*ln(a)).
You might want to divide x into its intpart and fracpart before using bc_exp (instead of doing that within bc_exp2) to avoid unneccessary function calls.
function bc_pow2($base,$exponent) {
$parts = explode (".", $exponent);
if ($parts[1] == 0){
$result = bcpow($base,$parts[0]);
else $result = bcmul(bc_exp(bcmul(bc_ln($base), "0.".$parts[1]), bcpow($base,$parts[0]);
return result;
}
Now we only need to program bc_ln. We can use the same strategy as above:
Take the Taylor-polynomial of the natural logarithm function. (since ln(0) isn't defined, take 1 as developement point instead)
Use Horner's method to drasticly improve performance.
Turn the result into a loop of bc-operations.
Also make use of ln(x) = -ln(1/x) when handling x > 1, to guarantee convergence.
usefull functions(don't forget to set bcscale() before using them)
function bc_fact($f){return $f==1?1:bcmul($f,bc_fact(bcsub($f, '1')));}
function bc_exp($x,$L=50){$r=bcadd('1.0',$x);for($i=0;$i<$L;$i++){$r=bcadd($r,bcdiv(bcpow($x,$i+2),bc_fact($i+2)));}return $r;}#e^x
function bc_ln($x,$L=50){$r=0;for($i=0;$i<$L;$i++){$p=1+$i*2;$r = bcadd(bcmul(bcdiv("1.0",$p),bcpow(bcdiv(bcsub($x,"1.0"),bcadd($x,"1.0")),$p)),$r);}return bcmul("2.0", $r);}#2*Sum((1/(2i+1))*(((x-1)/x+1)^(2i+1)))
function bc_pow($x,$p){return bc_exp(bcmul((bc_ln(($x))), $p));}
I've got this spot of code that seems it could be done cleaner with pure math (perhaps a logarigthms?). Can you help me out?
The code finds the first power of 2 greater than a given input. For example, if you give it 500, it returns 9, because 2^9 = 512 > 500. 2^8 = 256, would be too small because it's less than 500.
function getFactor($iMaxElementsPerDir)
{
$aFactors = range(128, 1);
foreach($aFactors as $i => $iFactor)
if($iMaxElementsPerDir > pow(2, $iFactor) - 1)
break;
if($i == 0)
return false;
return $aFactors[$i - 1];
}
The following holds true
getFactor(500) = 9
getFactor(1000) = 10
getFactor(2500) = 12
getFactor(5000) = 13
You can get the same effect by shifting the bits in the input to the right and checking against 0. Something like this.
i = 1
while((input >> i) != 0)
i++
return i
The same as jack but shorter. Log with base 2 is the reverse function of 2^x.
echo ceil(log(500, 2));
If you're looking for a "math only" solution (that is a single expression or formula), you can use log() and then take the ceiling value of its result:
$factors = ceil(log(500) / log(2)); // 9
$factors = ceil(log(5000) / log(2)); // 13
I seem to have not noticed that this function accepts a second argument (since PHP 4.3) with which you can specify the base; though internally the same operation is performed, it does indeed make the code shorter:
$factors = ceil(log(500, 2)); // 9
To factor in some inaccuracies, you may need some tweaking:
$factors = floor(log($nr - 1, 2)) + 1;
There are a few ways to do this.
Zero all but the most significant bit of the number, maybe like this:
while (x & x-1) x &= x-1;
and look the answer up in a table. Use a table of length 67 and mod your power of two by 67.
Binary search for the high bit.
If you're working with a floating-point number, inspect the exponent field. This field contains 1023 plus your answer, except in the case where the number is a perfect power of two. You can detect the perfect power case by checking whether the significand field is exactly zero.
If you aren't working with a floating-point number, convert it to floating-point and look at the exponent like in 3. Check for a power of two by testing (x & x-1) == 0 instead of looking at the significand; this is true exactly when x is a power of two.
Note that log(2^100) is the same double as log(nextafter(2^100, 1.0/0.0)), so any solution based on floating-point natural logarithms will fail.
Here's (nonconformant C++, not PHP) code for 4:
int ceillog2(unsigned long long x) {
if (x < 2) return x-1;
double d = x-1;
int ans = (long long &)d >> 52;
return ans - 1022;
}
This problem is best expressed in code:
$var1 = 286.46; // user input data
$var2 = 3646; // user input data
$var3 = 25000; // minumum amount allowed
$var4 = ($var1 * 100) - $var2; // = 250000
if ($var4 < $var3) { // if 250000 < 250000
print 'This returns!';
}
var_dump($var4) outputs: float(25000) and when cast to int, outputs: int(24999) - and thereby lies the problem.
I don't really know what to do about it though. The issue occurs upon multiplication by 100, and while there are little tricks I can do to get around that (such as *10*10) I'd like to know if there's a 'real' solution to this problem.
Thanks :)
This is a horrible hacky solution and I slightly hate myself for it, but this gives the expected behaviour:
<?php
$var1 = 286.46; // user input data
$var2 = 3646; // user input data
$var3 = 25000; // minumum amount allowed
$var4 = ($var1 * 100) - $var2; // = 250000
if ((string) $var4 < (string) $var3) { // if 250000 < 250000
print 'This returns!';
}
Cast them to strings, and they get converted back to int/float as appropriate for the comparison. I don't like it but it does work.
Really you need BC Math for precise floating point mathematics in PHP.
Its always a good idea to use ceil (or floor based on what you want) when using float number as int
In your case try ceil($var4) before comparison!
That's what floats do sometimes, it is all due to how floats are unable to precisely represent integers from time to time.
Instead of casting it to an int, you can round the number to an integer value and then cast it to an int. (possibly that cast unnecessary, but PHP isn't to clear about how such things happen internally, and even if you know how they happen right now, they may not in the future.
I think you could use bccomp for comparing floating point values but i think it's a function that's not in the PHP Core.
Otherwise i found this function here but i couldn't test it to see if it works
function Comp($Num1,$Num2,$Scale=null) {
// check if they're valid positive numbers, extract the whole numbers and decimals
if(!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num1,$Tmp1)||
!preg_match("/^\+?(\d+)(\.\d+)?$/",$Num2,$Tmp2)) return('0');
// remove leading zeroes from whole numbers
$Num1=ltrim($Tmp1[1],'0');
$Num2=ltrim($Tmp2[1],'0');
// first, we can just check the lengths of the numbers, this can help save processing time
// if $Num1 is longer than $Num2, return 1.. vice versa with the next step.
if(strlen($Num1)>strlen($Num2)) return(1);
else {
if(strlen($Num1)<strlen($Num2)) return(-1);
// if the two numbers are of equal length, we check digit-by-digit
else {
// remove ending zeroes from decimals and remove point
$Dec1=isset($Tmp1[2])?rtrim(substr($Tmp1[2],1),'0'):'';
$Dec2=isset($Tmp2[2])?rtrim(substr($Tmp2[2],1),'0'):'';
// if the user defined $Scale, then make sure we use that only
if($Scale!=null) {
$Dec1=substr($Dec1,0,$Scale);
$Dec2=substr($Dec2,0,$Scale);
}
// calculate the longest length of decimals
$DLen=max(strlen($Dec1),strlen($Dec2));
// append the padded decimals onto the end of the whole numbers
$Num1.=str_pad($Dec1,$DLen,'0');
$Num2.=str_pad($Dec2,$DLen,'0');
// check digit-by-digit, if they have a difference, return 1 or -1 (greater/lower than)
for($i=0;$i<strlen($Num1);$i++) {
if((int)$Num1{$i}>(int)$Num2{$i}) return(1);
else
if((int)$Num1{$i}<(int)$Num2{$i}) return(-1);
}
// if the two numbers have no difference (they're the same).. return 0
return(0);
}
}
}
The problem is that floats just cannot represent some numbers. Since PHP doesn't have a "decimal" (or other fixed-point) type, you can basically only hack your way around these problems.
Assuming the first number in your example $var1 = 286.46 denotes some kind of money, you could just convert that to cents directly after the user entered it (e.g. through stripping the point and reading it as an integer) and thus calculate everything using integer math.
That's not a general solution - and I doubt that one exists (short of using arbitrary precision numbers, which some PHP extensions provide - but I that smells like overkill to me).
I'm trying to design an odds system that goes from 1-100, however it also uses 0-1 for rarer odds.
I was told I should use a floating point format, but I don't know that.
Basically I have..
if (mt_rand(1,1000)/100 == $odds) {} else if (mt_rand(1,100) == $odds) {}
however that only yields the same probability.
I looked up floating point format in the PHP manual, but the answers there couldn't help me.
See Odds to understand how to convert your odds to a probability. If you have odds of 4:1 then there is a 1/5 == 0.2 probability of the event. If your odds are .2:1 then there is a 5/6 (about .833) probability of the event happening. In general, if the odds are m:n against then the probability is n/(m+n).
Now, if you want to simulate whether an event occurs or not, you need to get a random floating point number between 0 and 1 then check if this is less than the probability of the event. You can use something like mt_rand(0,1000)/1000 to get a random number between 0 and 1.
Examples:
$odds1 = 4; // 4:1
$prob1 = 1/($odds1+1); // 1/5
if( mt_rand(0,1000)/1000 <= $prob1 ) {
// event happened
}
$odds2 = .2; // .2:1
$prob2 = 1/($odds2+1); // 5/6
if( mt_rand(0,1000)/1000 <= $prob2 ) {
// event happened
}
Floating point values are inexact. (See Why does `intval(19.9 * 100)` equal `1989`? and search: php floating point inexact.)
You cannot use == for floating point values. A simple 5/10 == 0.5 might already be wrong due to inherent precision loss.
You can either round numbers before comparison, your what I'd advise in your case, pre-convert floats into integers:
# 52 == 100*0.52
if (mt_rand(1,100) == round(100*$odds)) {
Instead of comparing 0.99 with another float, you convert your odds into an integer 99 and compare it with an integer random in the range 1 to 100. If odds already was an integer, not a float, then the *100 multiplication will already cut it out of that first (faux float) comparison.
if($odds < 1){
// floating point math here
if((float)mt_rand(0,100) / 100.0 < $odds){
echo "you're a float winner, harry";
}
}else{
if(mt_rand(0,100) < $odds){
echo "you're an int winner, harry!";
}
}
I have these possible bit flags.
1, 2, 4, 8, 16, 64, 128, 256, 512, 2048, 4096, 16384, 32768, 65536
So each number is like a true/false statement on the server side. So if the first 3 items, and only the first 3 items are marked "true" on the server side, the web service will return a 7. Or if all 14 items above are true, I would still get a single number back from the web service which is is the sum of all those numbers.
What is the best way to handle the number I get back to find out which items are marked as "true"?
Use a bit masking operator. In the C language:
X & 8
is true, if the "8"s bit is set.
You can enumerate the bit masks, and count how many are set.
If it really is the case that the entire word contains bits, and you want to simply
compute how many bits are set, you want in essence a "population count". The absolute
fastest way to get a population count is to execute a native "popcnt" usually
available in your machine's instruction set.
If you don't care about space, you can set up a array countedbits[...] indexed by your value with precomputed bit counts. Then a single memory access computes your bit count.
Often used is just plain "bit twiddling code" that computes bit counts:
(Kernigan's method):
unsigned int v; // count the number of bits set in v
unsigned int c; // c accumulates the total bits set in v
for (c = 0; v; c++)
{
v &= v - 1; // clear the least significant bit set
}
(parallel bit summming, 32 bits)
v = v - ((v >> 1) & 0x55555555); // reuse input as temporary
v = (v & 0x33333333) + ((v >> 2) & 0x33333333); // temp
c = ((v + (v >> 4) & 0xF0F0F0F) * 0x1010101) >> 24; // count
If you haven't seen the bit twiddling hacks before, you're in for a treat.
PHP, being funny, may do funny things with some of this arithmetic.
if (7 & 1) { // if bit 1 is set in returned number (7)
}
Thought the question is old might help someone else. I am putting the numbers in binary as its clearer to understand. The code had not been tested but hope the logic is clear. The code is PHP specific.
define('FLAG_A', 0b10000000000000);
define('FLAG_B', 0b01000000000000);
define('FLAG_C', 0b00100000000000);
define('FLAG_D', 0b00010000000000);
define('FLAG_E', 0b00001000000000);
define('FLAG_F', 0b00000100000000);
define('FLAG_G', 0b00000010000000);
define('FLAG_H', 0b00000001000000);
define('FLAG_I', 0b00000000100000);
define('FLAG_J', 0b00000000010000);
define('FLAG_K', 0b00000000001000);
define('FLAG_L', 0b00000000000100);
define('FLAG_M', 0b00000000000010);
define('FLAG_N', 0b00000000000001);
function isFlagSet($Flag,$Setting,$All=false){
$setFlags = $Flag & $Setting;
if($setFlags and !$All) // at least one of the flags passed is set
return true;
else if($All and ($setFlags == $Flag)) // to check that all flags are set
return true;
else
return false;
}
Usage:
if(isFlagSet(FLAG_A,someSettingsVariable)) // eg: someSettingsVariable = 0b01100000000010
if(isFlagSet(FLAG_A | FLAG_F | FLAG_L,someSettingsVariable)) // to check if atleast one flag is set
if(isFlagSet(FLAG_A | FLAG_J | FLAG_M | FLAG_D,someSettingsVariable, TRUE)) // to check if all flags are set
One way would be to loop through your number, left-shifting it (ie divide by 2) and compare the first bit with 1 using the & operand.
As there is no definite answer with php code, I add this working example:
// returns array of numbers, so for 7 returns array(1,2,4), etc..
function get_bits($decimal) {
$scan = 1;
$result = array();
while ($decimal >= $scan){
if ($decimal & $scan) $result[] = $scan;
$scan<<=1;
}
return $result;
}