How to express an integer as the sum of powers-of-two? - php

Question

I have this number:
0101
and I know it is the binary sum of
0100
0001
How can I get those values from the first one?
EDIT
I have create a snippet of code based on the logic by JoeCortopassi. The snippet its this:
protected function _getSitesPublished($value) {
static $bits = null;
if (!$bits) {
$bits = array ();
$bit = 0;
$x = 0;
while ($bit < 4294967295) {
$bit = pow (2, $x);
$bits [$bit] = $bit;
++ $x;
}
}
$sites = array ();
foreach ($bits as $bit) {
if (($value & $bit) == $bit) {
$sites [] = $bit;
}
}
return $sites;
}
It only create the bits the first time the method its called. I have to make the comprobation
if (($value & $bit) == $bit)
since $value & $bit will return an int (may be other than 0, as in 6 & 3) and because of that I can't use only if ($value & $bit)
Thanks to all for your help.
EDIT 2
Oops! I had a little bug... forgot to increase the $x XD

Answer 1

$values = bindec('0101');
$bar = 1; // 0001
$fizz = 2; // 0010
$foo = 4; // 0100
$baz = 8; // 1000
if ( $values & $bar )
{
//returns true
}
if ( $values & $fizz )
{
//returns false
}
if ( $values & $foo )
{
//returns true
}
if ( $values & $baz )
{
//returns false
}
EDIT:
Is this more of what you're looking for? Not able to run it right now to test, but it should convey the message:
function bitCheck($original, $num, $return)
{
if ( $num == 0 )
{
return $return;
}
if ($original & $num)
{
$return[] = $num;
}
return bitCheck($original, $num-1,$return);
}

Answer 2

Using JoeCortopassi's code:
$value= bindec($binary);
$sums=array();
$counter=1;
while($counter<=$value){
if($counter & value)
$sums[]=$counter;
$counter*=2;
}
print_r($sums);

Answer 3

A Mathematica solution:
k[number_] :=
ReplacePart[ConstantArray[0, Length#number], # -> 1] & /#
(Position[number, 1] // Flatten)
Gives a list of the single bit components of the binary input:
(* k[{1, 0, 1, 1, 0}] -> {{1, 0, 0, 0, 0}, {0, 0, 1, 0, 0}, {0, 0, 0, 1, 0}} *)
A Mathematica solution which takes a list of binary digits and returns a list of digit values.
Clear[s];
Options[s] := {Base -> 2, Totalled -> False};
s[number_, OptionsPattern[]] :=
With[{digitVals =
Reverse#Flatten#
NestList[# OptionValue#Base &, {1}, Length#number - 1] number},
If[OptionValue#Totalled, Total#digitVals, digitVals]]
(* s[{1, 0, 1, 0, 1}] -> {16, 0, 4, 0, 1} *)

Answer 4

The following works in java, which you could use similar logic in php or mathematica:
public class IntAsPowerOfTwo {
public static void main(String[] args) {
printIntAsSumPowerOf2(256);
printIntAsSumPowerOf2(15);
printIntAsSumPowerOf2(-1023);
}
/**
* Prints an integer as sum of powers of 2.
*
* #param valueToEvaluate
*/
public static void printIntAsSumPowerOf2(int valueToEvaluate) {
if (valueToEvaluate < 0) {
System.out.println("Integer to evaluate must be non negative.");
}
int runningValue = valueToEvaluate;
int currPower = 0;
// Increase until larger than current value.
while (Math.pow(2, currPower) < runningValue) {
currPower++;
}
// Output sum of power of 2s.
boolean firstOutput = true;
while (currPower >= 0) {
if (runningValue >= Math.pow(2, currPower)) {
if (firstOutput) {
System.out.print(valueToEvaluate + " = 2^" + currPower);
firstOutput = false;
} else {
System.out.print(" + 2^" + currPower);
}
runningValue = runningValue - (int) Math.pow(2, currPower);
}
currPower--;
}
System.out.print("\n");
}
}