recurrsive Modulus Usage

ikenD

Hi there,

I am looking for help with the following :

I have cameras.
I have dvr's - 3 types - 4 channel, 8 channel, and 16 channel.
Someone on the web page enters the amount of cameras -
I need to work out how many dvr's (dependent on channels) they need

My code so far is pretty poor

if($qty % 4 >= 0){
   $dvr = 4;
}
if ($qty > 4 && $qty < 9)
{ 
   $dvr = 8;
}
if ($qty > 8 && $qty < 17)
{ 
   $dvr = 16;
}
if ($qty > 16)
{ 
   $mod = $qty % 4;
   echo "mod = ". $mod."<br><br>";
}

echo "You need a ".$dvr. " port dvr";

Example - if $qty=19;
It should read "1 16 port dvr and 1 4 port dvr"

Example - if $qty=24;
It should read "1 16 port dvr and 1 8 port dvr"

Example - if $qty=30;
It should read "2 * 16 port dvr"

I tried mucking around with the mod function - and probably need some sort of recursion - but am a bit lost now.

Any help greatly appreciated

kindest Regards

ikenD

laserlight

ikenD wrote:
I have cameras.
I have dvr's - 3 types - 4 channel, 8 channel, and 16 channel.
Someone on the web page enters the amount of cameras -
I need to work out how many dvr's (dependent on channels) they need

Sorry, I lack knowledge of your problem domain, so I have no idea what is a dvr and how, given the number of cameras, one can work out the number of dvr's required.

However, looking at your sample output, my guess is that the sum of the channels of the dvrs is supposed to be no less than the quantity, and you are trying to minimise the number of dvrs.

If so, looking at the channel distribution among the dvr types, you can use a greedy algorithm: while the quantity is greater than 16, subtract 16 from the quantity, and increase the count of 16 channel dvrs. You can do away with a loop by using integer division. Then repeat for the reduced quantity with 8, and finally with 4. If there is any remainder, add 1 to the number of 4 channel dvrs required.

ikenD

Hi Laserlight -

Many thanks for this - the logic is probably just what I needed

I will give it a try later (going for some food now)

kindest regards

ikenD

laserlight

Unfortunately, my suggested solution is flawed: $qty = 30 is a counterexample.

We could modify the algorithm: you still perform integer division of the quantity by 16 to get the 16 channel dvr count, but when you perform integer division of the balance by 8 to get the 8 channel dvr count, you check if the result is greater than 1. If it is, then you increase the 16 channel dvr count by 1, reset the 8 channel dvr count to 0 and stop. Otherwise, you perform integer division of the new balance by 4. Again, you check if the result if greater than 1. If it is, then you increase the 8 channel dvr count by 1, reset the 4 channel dvr count to 0 and stop. Otherwise, you check if there is any remaining quantity, and if there is, you increase the 4 channel dvr count by 1.

halojoy

As the scale is binary
16,8,4
One way would be to convert the total as a binary number
and then figure out by the BITS SET in the binary word
what the dvrs needed.

BINARY 8 bits number:
11111111
where bits are representing:
128,64,32,16,8,4,2,1
32= two 16 dvr
if either of 1 or 2 is set, we need another 4 dvr

<?php

$total = 14;

$binword = decbin( $total );

echo $binword; // gives 1110 ( 8+4+2+0 )

?>

ikenD

Hi Laserlight and Halojoy,

I got this :

$qty=160;

// $qty your variable
$dvrs = array(16=>0,8=>0,4=>0);

while ($qty > 0)
{

if ($qty > 8)
{
$dvr[16]++;
$qty-=16;
}
else if ($qty > 4)
{
$dvr[8]++;
$qty-=8;
}
else
{
$dvr[4]++;
$qty-=4;
}
}

This works -

However, the bit from halojoy is really nice indeed - seems somehow correct - and I will perhaps use that (after testing it)

Many thanks to both of you - laserlight for the logic and halojoy for the funky mechanism

pax et lux

AikenD

halojoy

I have been doing mindwork
on my binar - bitwise idea while I been surfing.

Now I have got down to test my idea in code.
But I am sorry 😕
Spending 4 lines to get 3 values must be regarded as sort of a failure
😃

<?php

$qty = 57; // Amount entered.

// Evaluation
$n   = ($qty & 3) ? $qty+4 : $qty; // add 4 if needed
$four  = ($n & 4) ? 1 : 0;   // test the 4-bit
$eight = ($n & 8) ? 1 : 0;   // test the 8-bit
$sixt  = intval($n/16);     // express higher in 16-units

// Display
echo 'quantity: '.$qty.'<br><br>';
echo 'four: '.$four.'<br>';
echo 'eight: '.$eight.'<br>';
echo 'sixteen: '.$sixt.'<br>';

?>

ikenD

halojoy !!

Nice bit of code - very impressed - compared to my looping (which actually has bugs in it of course)

Can't wait to write it into the page -

Many thanks indeed -

Pax et Lux

ikenD

halojoy

Yes I am happy with it.
Both you ikenD and I had the feeling this was the right way.
Looks like we were right.
Now, any coding in PHP can be done in many ways. And they will work.
But in this case I feel I am close to the ideal.
Greetings 🙂

laserlight

halojoy wrote:
But in this case I feel I am close to the ideal.

I think so too, since the types of dvr is presumably fixed, hence making good use of bitwise operations is a neat trick, not an inflexible hack.

That said, what impressed me most was this line:

$n   = ($qty & 3) ? $qty+4 : $qty; // add 4 if needed

Effectively, this addresses the problem that I detected in my first suggestion, and can be applied to simplify my second suggestion. Strictly speaking, you do not need to add 4: you only need to add enough to make the result a multiple of 4; but it is simpler to just add 4.