This was originally published on my now defunct Atalasoft blog in December 2006.

When showing progress, if you have a list of a known length and processing each item takes about the same time, you can implement it like this pseudocode:

for (int i = 0; i < list.length; ++i) {
    process(list[i]);
    // notifyProgress takes a numerator and denominator to 
    // calculate percent of progress
    notifyProgress(i, list.length);
}

One common problem is not knowing the length beforehand.

A simple solution would be to pick a value for length and then make sure not to go over it.

int lengthGuess = 100;
for (int i=0; list.hasMoreItems(); ++i) {
    process(list.nextItem());
    notifyProgress(min(i, lengthGuess), lengthGuess);
}
notifyProgress(lengthGuess, lengthGuess);

This works ok, if the length is near 100, but if it’s much smaller, it will have to jump at the end, and if it’s much bigger, it will get to 100% way too soon.

To fix this, we might adjust lengthGuess as we learn more:

int lengthGuess = 100;
for (int i=0; list.hasMoreItems(); ++i) {
    process(list.nextItem());
    if (i > 0.8 * lengthGuess) {
        lengthGuess = 2*i;      
    }
    notifyProgress(i, lengthGuess);
}
notifyProgress(lengthGuess, lengthGuess);

In this last example, whenever i is 80% of the way through, we set lengthGuess to 2*i.  This has the effect that the progress goes back and forth between 50% and 80% and then it jumps to the end.  This won’t work. 

What I want is:

1. The progress bar should be monotonically increasing
2. It should get to 100% at the end and not before
3. It should look as smooth as possible, but can jump

An acceptable effect, would be to progress quickly to 50%, then slow down to 75% (50% of the way from 50% to 100%), then slow down again at 87.5% (halfway between 75% and 100%), and so on.  If we keep doing that, we’ll never get to 100% in the loop and can jump to it at the end. This is like Zeno’s Dichotomy paradox (from the Wikipedia).

Suppose Homer wants to catch a stationary bus. Before he can get there, he must get halfway there. Before he can get halfway there, he must get a quarter of the way there. Before traveling a quarter, he must travel one-eighth; before an eighth, one-sixteenth; and so on.

To do that we have to keep a factor to use to adjust the progress we’ve made (playing around with it, I found that using a factor of 1/3 rather than 1/2 was more pleasing).

int lengthGuess = 100;
double begin = 0;
double end = lengthGuess;
double iFactor = 1.0;
double factorAdjust = 1.0/3.0;
for (int i = 0; list.hasMoreItems(); ++i) {
    process(list.nextItem());               
    double progress = begin + (i - begin) * iFactor;
    if (progress > begin + (end-begin) * factorAdjust) {                   
        begin = progress;
        iFactor *= factorAdjust;
    }               
    notifyProgress(progress, lengthGuess);
}
notifyProgress(lengthGuess, lengthGuess);

The choice of lengthGuess is important, I think erring on too small is your best bet.  You don’t want it to be exact, because we’ll slow down when we get 1/3 toward the goal (factorAdjust).  The variables lengthGuess and factorAdjust could be passed in and determined from what information you have about the length of the list.

If you are sending data from the iPhone to the Apple Watch, you might use sendMessage.

func sendMessage(_ message: [String : Any], replyHandler: (([String : Any]) -> Void)?, errorHandler: ((Error) -> Void)? = nil)

If you do this and get the error WCErrorCodePayloadUnsupportedTypes this is because you put an unsupported type in the message dictionary.

The first parameter (message) is a dictionary of String to Any, but the value cannot really be any type. If you read the documentation, it says that message is

A dictionary of property list values that you want to send. You define the contents of the dictionary that your counterpart supports. This parameter must not be nil.

“property list values” means values that can be stored in a Plist. This means you can use simple types like Int, Bool, and String and you can also use arrays and dictionaries as long as they are of those simple types (e.g. an Array of Ints)

I ran into this issue because I tried to use a custom struct in the message dictionary, which is not supported.

Note: I made this post because google is sending people to Programming Tutorials Need to Pick a Type of Learner because it mentions WCErrorCodePayloadUnsupportedTypes incidentally, but isn’t really about that.

One of my projects at Trello was looking into and fixing a reliability problem we were having. Even though it seemed solid in testing, we were getting enough support tickets to know that it must be worse than we thought. We collected data on its success rate and found out that it was successful 99.8% of the time. To move forward with doing more work on it, I had to convince our team and management that 99.8% was bad.

At the time, there was a company-wide push at Atlassian to improve reliability, and there was a line-manager assigned to oversee it across Trello teams, so I spoke to him. I showed him the data, but also told him that there’s an overall feeling on our team that it’s affecting our customers more than it seems.

He suggested that I flip the ratio and instead look at the data as Errors Per Million. When you do that, with the same data, you get 2,000 errors per million attempts. This particular thing happened around 2 million times per day, so that was 4,000 errors per day. That partially explained the issue, because it wouldn’t take much for the support tickets to get out of control. Luckily, not all 4,000 were of the same severity, and many were being retried. Still, that number is too high.

The other thing I found after more analysis, was that the errors were not evenly spread across the user base. They tended to cluster among a smaller cohort, which was experiencing a much worse EPM than then rest of the users.

With that in hand, we greenlit a project to address this by targeting the most severe problems. We found several bugs. Eventually we got the success rate to 99.95%.

It’s not clear that 99.95% is four times better than 99.8%, but the equivalent EPM after the fixes is 500 (as opposed to 2,000 before). What surprised me is how different a reaction a high EPM got compared to equivalent numbers. 2,000 EPM is literally the same as 99.8% success and 0.2% failure rates, but the latter two seem fine. Even if I say we get 2 million attempts per day, it’s hard to intuitively understand what that means.

When I said 2,000 EPM, we instinctively felt that we failed 2,000 users. When I said we get 2 million attempts, everyone knows to double EPM to get 4,000 incidents per day, and we felt even worse. That simple change in reporting made all the difference in our perspective.