# Bulls and Cows

Bulls and Cows (also known as MOO) is a 2-player game in which one player comes up with a secret and the other has to guess the secret. The secret consists of 4 digits from 0 to 9, where each digit is distinct. Player 2 has to guess these 4 digits, in the right order. At each guess from player 2, player 1 provides feedback as hints. The hints are two numbers: one telling how many digits from the secret player 2 got in the right order (bull), and another telling how many digits they got but in the wrong order (cow).

For example, if player 1 came up with `4271`, and player 2 guessed `1234`, then bull is 1 (digit 2), and cow is 2 (1 and 4 are in the secret, but not in the order guessed by player 2).

The goal of the game is for player 2 to guess the secret with the least amount of guesses.

Bulls sculptures from Cyprus. Dated from sometime between 2000 and 1600 BC. Photo take at the National Archeology Museum in Athens.

In this post we’ll present computational experiments to solve a game of Bulls and Cows optimally.

#### Objective function

We’ll focus on a search strategy to minimize the maximum number of guesses one has to make, for any possible secret. Alternatively, John Francis’s paper [1] proposes heuristics that minimize the expected number of guesses, that is, for the worst case these heuristics might not be optimal, but they perform better on average.

Because the game has turns, our solution is not a single list of guesses, but rather a decision tree where at each node we branch depending on the hint we get back. The metric we are trying to optimize is then the height of such tree.

To break ties between trees of the same height, we consider the smallest tree (the one with least nodes).

#### Brute-force search

Our search algorithm is recursive. At any given recursion level, we are given a set containing all possible numbers that could be secrets. We’ll try all combinations of guesses and hints and see which ones yield the best solution.

When we simulate a given guess and hint, we are restricting the possible numbers that can still be secret, so in the next level of recursion, the set of potential secrets will be smaller.

For example, say that the possible secrets are any 4-digit number with no repeated digits. We then use one of them as a guess, say, `[0, 1, 2, 3]`. One possible hint we could receive is (3, 0). What are the possible secrets that would cause us to receive this hint? `[0, 1, 2, 4]``[0, 1, 2, 5]` and `[7, 1, 2, 3]` are a few of those. If we recurse for this guess and hint, the set of secrets in the next level will be restricted to those that would return a hint (3, 0) for `[0, 1, 2, 3]`.

Continuing our example, is `[0, 1, 2, 3]` the best guess we can make at that point? After recursing for all possible hints, we’ll have a decision tree rooted on `[0, 1, 2, 3]`. The key idea of the search is that we can minimize the height of the final decision tree by minimizing the subtrees at each recursion level (greedy strategy). Thus, we want to find the guess with the shortest subtree.

In pseudo-code code, the search looks like this:

We start with all possible secrets (4-digit number with no repeated digits) and the tree returned should be optimal in minimizing the number of guesses for the worst case.

#### Rust implementation

I initially implemented the search in Python, but it was taking too long, even when artificially restricting the branching. I re-implemented it in Rust and saw some 20x speedups (caveat: I haven’t really tried to optimize the Python version).

The main difference between the pseudo-code and the actual implementation is that we pre-group `possible_secrets` by their hint for guess, which is more efficient than scanning `possible_secrets` for all possible hints:

The function `group_possibilities_by_score()` above makes use of `compute_score` and it also uses a fixed-length array for performance. The set of hints is proportional to the squared size N of the guess, in our case `N=4`.

Turns out that the Rust implementation is still not efficient enough, so we’ll need further optimizations.

#### Optimization – classes of equivalence

What is a good first guess? It doesn’t matter! We don’t have any prior information about the secret and every valid number is equality probable. For example, if we guess `[0, 1, 2, 3]` or `[5, 1, 8, 7]`, the height of the decision tree will be the same. An intuitive way to see why this is the case is that we could relabel the digits such that `[5, 1, 8, 7]` would map to `[0, 1, 2, 3]`.

Francis [1] generalizes this idea for other cases. Say that at a given point we made guesses covering the digits 0, 6, 7, 8 and 9 at least once. Now say we our next guess is `[0, 8, 9, 3]`. In here, 3 is the only digit we haven’t tried yet, but using the re-labeling argument, we can see that `[0, 8, 9, 1]` would yield the same decision tree if we were to swap the labels of 1 and 3. This allow us to skip guesses that belong to the same class, which reduces the branch factor.

We can generate an ID representing a given class. A way to do this is by adding one to each digit we have tried before and  making any digit we haven’t as 0, then converting that number from base 11 to base 10. For example, `[0, 8, 9, 1]` becomes `[1, 9, 10, 0]`. If this was a number in base 11, in base 10 it is 2530 (`(((1*11) + 9)*11 + 10)*11 + 0`). If we do the same with  `[0, 8, 9, 3]`, we’ll get the same number. The code below implements this idea.

In our case, `D = 10` and we store the set of visited digits in a bitset, `visited_bits` (that is, bit `i` is 1 if digit `i` has been visited.

On the search side, we keep a set of classes ids already visited and skip a guess if its class is already in there.

With this optimization the search algorithm runs in under 2 minutes. By inspecting the height of the resulting tree we conclude that the minimum number of guesses necessary for any secret is 7.

The complete code is available on Github.

#### Visualizing

The JSON output by the search algorithm is quite big (the smallest tree with height 7 has almost 7000 nodes). A more interesting way to inspect the data is to create a Bulls and Cows solver. We feed the JSON to this application and ask the user to select the outcome based on the secret they have in mind. We are basically traversing the edges of the decision tree.

I’ve uploaded the application to my personal website and the source code is available on Github.

### Conclusion

I learned about this game pretty recently and was curious to learn of good strategies to solve this problem. From what I’ve been reading, the heuristics that yield good solutions are very complicated for a human to perform. This leads to an question: is there are any solution which is simple to follow but that is reasonably good?

One natural extension for this problem is to use larger numbers of digits (`N` > 4) and have each element be sourced from 0 to `D` – 1. An exhaustive search might be prohibitive, but maybe we can come up with heuristics with constant guarantees. What is a lower bound for the number of guesses for variants with arbitrary N and D?

I struggled to implement this code in Rust, but I found the challenge worthwhile. I learned a bit more about its specifics, especially regarding memory management and data types.

In [1], the author mentions that with optimizations the search took 45 minutes to run on their laptop (2GHz Intel Core 2 Duo). I ran mine on a Intel i7 2.2GHz and was surprised by the running time of 2 minutes. CPUs are not getting exponentially faster these days and my code runs on a single thread.

### References

[1] Strategies for playing MOO or “Bulls and Cows”.
[2] Github – Blog Examples: Bulls and Cows

# Largest sets of subsequences in OCaml

I’ve noticed that there is this set of words in English that look very similar: tough, though, through, thought, thorough, through and trough. Except thought, they have one property in common: they’re all subsequence of thorough. It made me wonder if there are interesting sets of words that are subsequences of other words.

Word cloud made with Jason Davie’s tool (https://www.jasondavies.com/wordcloud/)

This post is an attempt to answer a more general question: given a list of words, what is the largest set of these words such that they’re subsequences of a given word?

A word `A` is a subsequence of a word `B` if `A` can be obtained by removing zero or more characters from B. For example, “ac” is a subsequence of “abc”, so is “bc” and even “abc”, but not “ba” nor “aa”.

A simple algorithm to determine if a word A is a subsequence of another is to start with 2 pointers at the beginning of each word, `pA` and `pB`. We move `pB` forward until `pA` and `pB` point to the same character. In that case we move pA forward. A is a subsequence of B if and only if we reach the end of A before B. We could then iterate over each word `W` and find all the words that are subsequences of W. If the size of the dictionary is `n`, and the size of the largest word is `w`, this algorithm would run in $O(n^2 w)$.

For English words, we can use entries from `/usr/share/dict/words`. In this case, `n` is around 235k (`wc -l /usr/share/dict/words`), so a quadratic algorithm will take a while to run (around 5e10 operations).

Another approach is to generate all subsequences of words for a given word `W` and search the dictionary for the generated word. There are $O(2^w)$ subsequences of a word of length $w$. If we use a hash table, we can then do it in $O(n w 2^w)$. In `/usr/share/dict/words`, the length of the largest word, `w`, is 24.

Running a calculation with the numbers (R script), the number of high-level operations is 4e10, about the same order of magnitude as the quadratic algorithm.

Distribution using ggplot2

A third approach is to use a trie. This data structure allows us to store the words from the dictionary in a space-efficient way and we can search for all subsequences using this structure. The trie will have at most 2e6 characters (sum of characters of all words), less because of shared prefixes. Since any valid subsequence has to be a node in the trie, the cost of search for a given word cannot be more than the size of the trie `t`, so the complexity per word is $O(\min(2^w, t))$. A back of envelope calculation gives us 2e9. But we’re hoping that the size of the trie will be much less than 2e6.

Before implementing the algorithm, let’s define out trie data structure.

### The Trie data structure

A trie is a tree where each node has up to `|E|` children, where `|E|` is the size of the alphabet in consideration. For this problem, we’ll use lower case ascii only, so it’s 26. The node has also a flag telling whether there’s a word ending at this node.

Notice that in this implementation of trie, the character is in the edge of the trie, not in the node. The Map structure from the stlib uses a tree underneath so get and set operations are $O(log |E|)$.

The insertion is the core method of the structure. At a given node we have a string we want to insert. We look at the first character of the word. If a corresponding edge exists, we keep following down that path. If not, we first create a new node.

To decide whether a trie has a given string, we just need to traverse the trie until we either can’t find an edge to follow or after reaching the end node it doesn’t have the `hasEntry` flag set to true:

This and other trie methods are available on github.

### The search algorithm

Given a word `W`, we can search for all its subsequences in a trie with the following recursive algorithm: given a trie and a string we perform two searches: 1) for all the subsequences that contain the first character of current string, in which case we “consume” the first character and follow the corresponding node and 2) for all the subsequences that do not contain the first character of the current string, in which case we “consume” the character but stay at the current node. In pseudo-code:

```Search(t: TrieNode, w: string):
Let c be the first character of w.
Let wrest be w with the first character removed

If t contains a word, it's a subsequence of the
original word. Save it.

// Pick character c
Search(t->child[c], wrest)

// Do not pick character c
Search(t, wrest)
```

The implementation in OCaml is given below:

### Experiments

Our experiment consists in loading the words from `/usr/share/dict/words` into a trie, and then, for each word in the dictionary, look for its subsequences. The full code is available on github.

The code takes 90 seconds to run on my laptop. Not too bad but I’m still exploring ways to improve the performance. One optimization I tried is to, instead of returning an explicit list of strings as mentioned in the search implementation, return them encoded in a trie, since we can save some operations due to shared prefixes. I have that version on github, but unfortunately that takes 240 seconds to run and requires more memory.

Another way is to parallelize the code. The search for subsequences is independent for each word, so it’s an embarrassingly parallel case. I haven’t tried this path yet.

The constructed trie has 8e5 nodes or ~40% of the size of sum of characters.

#### Subsequences of “thorough”

The question that inspired this analysis was finding all the subsequences of thorough. It turns out it has 44 subsequences, but most of them are boring, that is, single letter or small words that look completely unrelated to the original word. The most interesting ones are those that start with `t` and have at least three letters. I selected some of them here:

• tho
• thoo
• thoro
• thorough
• thou
• though
• thro
• throu
• through
• thug
• tog
• tou
• toug
• tough
• trough
• trug
• tug

The word with most subsequences is pseudolamellibranchiate, 1088! The word cloud at the beginning of the post contains the 100 words with the largest number of subsequences. I tried to find interesting words among these, but they’re basically the largest words – large words have exponentially more combination of subsequences, and hence the chance of them existing in the dictionary is greater. I tried to come up with penalization for the score:

1) Divide the number of subsequences by the word’s length. This is not enough, the largest words still show on top.
2) Apply log2 to the number of subsequences and divide by the word’s length. In theory this should account for the exponential number of subsequences of a word. This turns out to be too much of a penalization and the smallest word fare too well in this scenario.

I plotted the distribution of number of subsequences by word lengths. We can see a polynomial curve but with increased variance:

Generated with this ggplot2

In the chart above, we’d see all points with the same x-value in a single vertical line. One neat visualization trick is to add noise (jitter) so we also get a sense of density.

If we use a box plot instead, we can see a quadratic pattern more clearly by looking at the median for each length.

Generated with this ggplot2

Given this result, I tried a final scoring penalization, by dividing the number of subsequences by the square of the length of the word, but it’s still not enough to surface too many interesting words. Among the top 25, streamlined is the most common word, and it has 208 subsequences.

One interesting fact is that the lowest scoring words are those with repeated patterns, for example: kivikivi, Mississippi, kinnikinnick, curucucu and deedeed. This is basically because we only count unique subsequences.

### Conclusion

This was a fun problem to think about and even though it didn’t have very interesting findings, I learned more about OCaml and R. After having to deal with bugs, compilation and execution errors, I like OCaml more than before and I like R less than before.

R has too many ways of doing the same thing and the API is too lenient. That works well for the 80% of the cases which it supports, but finding what went wrong in the other 20% is a pain. OCaml on the other hand is very strict. It doesn’t even let you add an int and a float without an explicit conversion.

I learned an interesting syntax that allows to re-use the qualifier/namespace between several operations when chaining them, for example:

I also used the library `Batteries` for the first time. It has a nice extension for the rather sparse `String` module. It allows us to simply do `open Batteries` but that overrides a lot of the standard modules and that can be very confusing. I was scratching my head for a long time to figure out why the compiler couldn’t find the `union()` function in the `Map` module, even though I seemed to have the right version, until I realized it was being overridden by `Batteries`. From now on, I’ll only use the specific modules, such as `BatString`, so it’s easy to tell which method is coming from which module.

### References

OCaml

• [1] OCaml Tutorials > Map
• [2] Strings – Batteries included
• [3] Using batteries when compiling

R

• [1] R Tutorial – Histogram
• [2] Creating plots in R using ggplot2 – part 10: boxplots
• [3] My R Cheat sheet