# Arduino Adventures: Does 255/256 = 0 or 1?

Max Maxfield tries to wrap his brain around the problems associated with truncating the results from integer division operations.

As you may recall from my previous blogs, I recently purchased a 4x4x4 3D tri-colored LED cube kit powered by an Arduino-compatible controller. Creating different effects for this rascal is helping me reacquaint myself with the C programming language (well, the Arduino version thereof). In turn, this is helping prepare me to program my Arduino-powered robot.

I will be writing a more in-depth column on my cube in the not-so-distant future. Suffice it to say, for the moment, that I started out with some very simple test patterns, and that I've gradually increased the complexity and sophistication of my effects.

Each LED has three color channels: red, green, and blue. Each channel is controlled using pulse-width modulation. The values that can be assigned to each channel range from 0x00 to 0xFF (0 to 255 in decimal). Those two values equate to being fully off and fully on, respectively.

Up until now, my effects have involved driving the various channels hard on or hard off, thus allowing me to assign seven colors (red, green, blue, red + green, red + blue, green + blue, and red + green + blue) to each LED. Now I'm at the stage where I want to experiment with a wider gamut of colors and to transition gradually from one color to another, and therein lies my problem.

For the purpose of this discussion, let's assume that we are interested in a single color channel for a single LED, and that I have two global variables of type "int" (integer): oldColor and newColor. Each will be set initially to either 0 or 255. Let's also assume that I have a function that starts off looking a bit like this.

The idea is that, when this function is called, there are four possible value combinations for oldColor and newColor.

This explains why I'm using integer variables instead of byte (eight-bit) variables. I want to be able to employ negative values. The howMany parameter specifies how many steps we wish to use to transition from the old color value to the new color value. The parameter is also used to calculate the modColor value, which is the amount by which we wish to keep incrementing or decrementing the oldColor (starting) value until it reaches the newColor (ending) value.

Purely for the purposes of these discussions, let's assume that we pass a value of 4 into the howMany parameter when we call out the lightLEDs function. If the old and new colors start off with the same value (both 0 or both 255), then modColor will end up as 0, so nothing will change as we circle around our for loop. Of course, this is what we want in this case.

Now suppose that we call this function when the old color is 0 and the new color is 255. In this case, modColor will be calculated as 255/4 = 63.73. This will be truncated to 63, since modColor was defined as an integer. (We certainly don't want to use floating-point values. They can consume lots of memory, and performing operations on them can involve a lot of compute cycles if our processor doesn't have a floating-point unit.) In this particular case, this is close enough for government work, as they say. This also explains why I used "i <= howMany" as the control in our for loop. If I had used "i < howMany," we would have gone around our for loop four times (i = 0, 1, 2, and 3). Our tmpColor variable would have been 4x63 = 252. By performing the loop an extra cycle while capping the value at 255, we ensure that we leave the LED fully on when we exit the loop.

Now let's consider what happens when the old color is 255 and the new color is 0. Assuming howMany is 4, then modColor will be calculated as -255/4 = -63.73, which will be rounded up (or down, depending on your point of view) to -64. In this case, we really need to perform the loop only four times -- 255-(4x-64) = -1, which will be bottom capped to 0 -- not five times, as the current code would have it. Of course, since we *are* bottom capping the value at 0, the extra cycle won't hurt us, but it's a waste of computing resources, which niggles at me.

The problem is that things get worse the more steps we take. Let's assume the old color is 0, the new color is 255, and we decide to perform the transition using 128 steps. In this case, modColor will be calculated as 255/128 = 1.99, which will be truncated to 1. Even if we perform our for loop 129 times (0 to 128), we'll still end up with a final color value of 129 instead of 255. Or, as an absolute worst case, suppose the old color is 0, the new color is 255, and we perform the transition using 256 steps. Now we end up with modColor as 255/256 = 0.99, which will be truncated to 0. That means we won't change the color at all.

From the above, we know that we don't have to worry about fading the color down from 255 to 0, which involves modColor having negative values, because two's complement values automatically round (truncate) toward negative infinity. It's the positive values we have to worry about. So, how about if I modify my function to look like the following?

By Jove, I think I've solved it. As you can see, if we are fading the LED up (modColor has positive values), we start off multiplying modColor by 2 before dividing it by howMany. Next we extract the least significant bit. If this is a 1, then we know the system will round our value downward, so after we've divided by 2 (to counteract our original multiplication by 2), we add this 1 back in.

Remembering that multiplications and divisions by 2 involve shifting the two's complement value one bit left or right, respectively, this doesn't add much of a computational burden. It also means that we can use "i < howMany" as our loop control.

To see how this works, let's return to our original example (old color = 0, new color = 255, number of steps = 4). When we calculate our modColor, we start by multiplying 255 by 2 to generate 510. Next, we divide this value by 4 to generate 127.5, which will be truncated to 127, or 01111111 in binary. This is the point where we extract the least significant bit -- in this case, 1. Now we divide 127 by 2 to generate 63.5, which will be truncated to 63. Finally, we add the 1, representing the old least significant bit, and get a modColor value of 64. When we proceed around our loop for i = 0, 1, 2, and 3, our corresponding tmpColor values will be 64, 128, 129, and 256. The last one will be capped at 255.

What do you think? Have I missed anything, or have I cracked this problem? Also, can you think of a more elegant way to achieve what I'm trying to do?

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Author

Max The Magnificent 10/28/2013 4:46:34 PM

It's probably not the most efficient option computationally, but perhaps the simplest solution would be to use the Arduino language's map() function.This just goes to show that I really need to spend some time on the Arduino website looking to see what's there. I'm currently ver yhappy with the solution proposed by Javi (Garcia) earlier in these comments, but it's great to know that this (and other) functions are available.

Author

Max The Magnificent 10/28/2013 4:41:18 PM

In the ancient times we used a modified Bresenham algorithm to fade between colours...You make it sound as though you are talking about the last millennium .... Oh, wait a moment... LOL

Author

Max The Magnificent 10/28/2013 4:39:53 PM

Your truncation problem would not be there if you stick with fractions.True enough -- but the problem also goes away if we use the solution proposed by Javi (Garcia) in the early comments to this blog.

Author

Max The Magnificent 10/28/2013 4:38:12 PM

hough the Bresenham algorithm might be appropriate...I know this is not complex -- but I can imagine the problems that woudl ensue trying to explain all of this to an absolute beginner :-)

Author

jpessin 10/28/2013 4:26:42 PM

http://arduino.cc/en/Reference/Map

This gives:

void lightLEDs (int howMany,

int howLong)

{

for (int i = 1; i <= howMany; i++)

{

int tmpColor = map(i, 0, howMany, oldColor, newColor);

// Use current tmpColor to drive LED

delay(howLong);

}

}

Author

Bastian.Schick 10/27/2013 2:50:53 AM

Author

Bastian.Schick 10/27/2013 2:49:14 AM

- The distance between 0 and 255 is (255-0)+1

- In the ancient times we used a modified Bresenham algorithm to fade between colours.

Cheers

Author

another nickname 10/26/2013 8:29:22 PM

Just keep 2 integers for each number numerator and denominator) and rewrite your algorithm using 2 numbers instead of one for all calcluations and do multiplications first and divisions last. It's pretty much the same as Fixed Point arithmetics .

Author

U. Dreher 10/26/2013 7:18:27 PM

The solution is really simple:

result = (dividend + (divisor/2))/divisor or

result = (dividend + (divisor>>1))/divisor

This might lead us to the next level of discussion: what if divisor is odd?

You simply have to decide whether you want to round up (x.5 => x+1) or down (x.5 => x). In one of the cases the above equations will read ((divisor+1)/2) resp. ((divisor+1)>>1).

There's always a last philosophical question or, in other words, the freedom of choice :)

Author

Max The Magnificent 10/26/2013 6:01:09 PM

The standard way to do this efficiently is the Bresenham algorithm...Cool beans -- I will bounce over to peruse and ponder the Wikipedia page on this -- I'm sure I will be doing a lot more of this in the future -- for example, with regard to my current robot project.