# Winning at Salt Cartography Using Algorithms and Statistics

I spend way too much of my free time playing Salt - a game where you sail around a procedurally generated ocean, fending off pirates and hunting for treasure. A fairly recent addition to the game was cartography; you can add islands and features upon them to a map to stay oriented in the world. In the world, there exists a cartographer named Hachure, who will evaluate what you've mapped so far and reward you depending on how many islands you've mapped and how accurate your mappings are. So imagine my dismay when I had him evaluate my map for the first time and saw this:

*94%?!*

Unacceptable! I immediately felt a need to find the offending entries and remove them from my map!

… But how to find the culprits?

Fortunately for me, the mapping data is available on disk as a JSON
document ^{1)}. So now this becomes an issue of devising an algorithm to identify a
bad subset of entries, where the only test you have is a black box oracle that
tells you how many entries in the current set are bad. While this *is*
cheating at the game a little bit, I consider it pretty fair, since it's just
saving me time from doing something by hand. ;)

## Time Complexity

After taking a superficial look at this problem, I thought to
myself “Oh, subsets and sets with a test? Sounds NP-complete!”
I suppose I was just jumping at a chance to work on an NP-complete
problem, because not only is this problem solvable in polynomial
time, it's even solvable in `O(n)`

time ^{2)}:

bad entries = {} for item in map entries: set map to contain only item ask oracle how many are bad if more than 0 are bad, then add item to bad entries

Unfortunately, the test for a set of map entries takes a while; it takes a good minute or so for Salt to start and load my save. That means running through the above algorithm would take about 100 minutes. I don't have any qualms about code taking 100 minutes to find me an answer, but since it would be a human operator (ie. me) running these tests by clicking buttons, I want to lower the number of tests as much as possible. I don't even mind if this takes me more than 100 minutes to figure out - it'll be much more fun to stretch my brain!

## Improving Upon the Naïve Algorithm

There's an easy improvement we can make to this algorithm. Instead of testing one entry at a time, we can test them in pairs. After going through 55 tests, if we found a pair with 2 bad entries, we know they're both bad; if we found one with 1 bad entry, we test one of the entries of the pair to see if it or its counterpart is the bad one. So, worst case scenario, 62 tests; a substantial improvement over the original 109!

## Divide and Conquer

It turns out we can do an even better job than 62 tests. If we were just
looking for one, this would be simple; just take half of the entries and try
those. If there's a bad entry, recurse on that half; otherwise, recurse on the
other half. Since we're dividing the problem space in half at each test, this
algorithm is `O(log n)`

, which means this would take about ```
log₂(109) =
6.768184 ≅ 7
```

tests. Unfortunately, we have seven bad entries here, but maybe
this can serve as some inspiration!

Even though we have more than one bad item, we can still apply the divide
and conquer idea. If we find 0 bad items in a half, we can discard it;
we can also discard the other half if we found that the half we're testing
contains *all* the bad items. If we find bad items in both halves, we
can recurse into both halves until we find a subset with a single bad
item, and then apply the algorithm I mentioned in the previous paragraph
to find it.

I'm not even going to try to analyse the time complexity in terms of big-O notation
^{3)}, but it turns out we can
get a pretty reasonable guess for how many tests this should take. Let's
consider the possibilities!

## Is Probability the Answer? Probably.

Let's say we're looking at our first 55 items. What's the probability that those fifty-five items contains 0, 1, 2, 3, 4, 5, 6, or 7 bad items? I'm thankful that I took an introductory statistics course earlier this year, because the ideas presented in that course started to come back to me. It reminded of problems like “what's the probability of 5 out of 20 coin flips coming up heads?”. However, this is a bit different; while a coin flip is not affected by previous flips, this is more like trying to determine how likely it is to draw a certain number of white balls from a collection of white and black ones - once you draw a white ball, the probability of drawing another goes down. Unfortunately, introductory statistics didn't prepare me for this - we only covered binomial (like coin flips), normal, and uniform distributions.

After some poking around on Wikipedia, I discovered what's called the Hypergeometric distribution. It's similar to the binomial distribution in that it involves tests with a binary outcome, but it differs in that it calculates probabilities without replacement. So, exactly what we need!

In R, you can use the `dhyper`

function to calculate probabilities for
a hypergeometric distribution. With that knowledge at hand, we can calculate
the expected number of tests that running through this algorithm should take.

Let's consider the very first test in the program: testing a random subset of 55 entries. We can see the probabilities of the various outcomes of that test like so:

# logic(total=109, nbad=7) > dhyper(0:7, m=7, n=102, k=55) [1] 0.005943616 0.047672750 0.157611948 0.278447776 0.283907536 0.167068665 [7] 0.052537316 0.006810393

We can then recursively apply this logic to determine the expected number of tests we need to take for the subproblem each outcome represents, which would look something like this:

# 0 bad entries in the tested half, 7 in the untested > logic(total=54, nbad=7) # 22.36303 tests # 1 bad entry in the tested half, 6 in the untested > logic(total=55, nbad=1) + logic(total=54, nbad=6) # 26.13599 tests # 2 bad entries in the tested half, 5 in the untested > logic(total=55, nbad=2) + logic(total=54, nbad=5) # 27.7867 tests # 3 bad entries in the tested half, 4 in the untested > logic(total=55, nbad=3) + logic(total=54, nbad=4) # 28.56915 tests # 4 bad entries in the tested half, 3 in the untested > logic(total=55, nbad=4) + logic(total=54, nbad=3) # 28.5926 tests # 5 bad entries in the tested half, 2 in the untested > logic(total=55, nbad=5) + logic(total=54, nbad=2) # 27.85717 tests # 6 bad entries in the tested half, 1 in the untested > logic(total=55, nbad=6) + logic(total=54, nbad=1) # 26.25435 tests # 7 bad entries in the tested half, 0 in the untested > logic(total=55, nbad=7) # 22.53071 tests

We can then multiply these with their probabilities and sum the result (adding 1 for the test we would do at this level) to get the expected number of tests:

> 1 + sum(dhyper(0:7, m=7, n=102, k=55) * c(22.36303, 26.13599, 27.7867, 28.56915, 28.5926, 27.85717, 26.25435, 22.53071)) # 29.01791 tests

Rounding up, we get 30 tests. Not bad! Here is an R script that performs the calcuation for expected tests, and here is a Perl 6 script that implements the algorithm by rearranging the map contents and prompting me for the cartographer's answer. When I actually ran it, it took me 27 tests to finish!

This was a fun little exercise in algorithm design, plus I got to spread my stats wings a bit. I love it when I can use my programming knowledge to improve things like gaming!

^{1)}Sadly, this no longer seems to be the case =(^{2)}An example of an NP-complete version of this would be if some combination of map entries made the map bad, and you wanted to determine the minimum size of such a combination.^{3)}Who am I kidding? ;) Yes I am going to try - I would guess something like`O(b * log(n/b))`

, where b is the number of bad entries.

*Published on 2016-06-16*