Edit: Try my own version of Hangman, written with Silverlight and WPF!
In an effort to distract myself from more productive endeavors, I started playing Hangman on The Free Dictionary Web site today. They have an outstanding plugin that you can also add to your Google homepage (which is actually how I discovered it). The interface allows you to simply type the letter into the guess box (if you look below you’ll see that my last guess was the letter “G”), and it updates it as you go:
During game play, the plugin constructs the man you are desperately trying to save. You can only make ten mistakes before the man is hanged:
So, as I was wasting my time today, I started wondering if there’s a strategy to winning Hangman. Assuming that there is, I decided to try and figure it out.
I started with a few basic ideas:
- Certain letters are used more frequently than other letters.
- The frequency of letters are probably different given the total number of letters in the word.
- You can maximize your ability to win Hangman by playing the most common letter first, followed by the next most common letter, and so on.
I realize that there are probably more complicated theories (such as the most common letters changing based on discovered letters), but for this first part I decided that I wanted to keep it somewhat simple. I can complicate it more later on.
(If you aren’t interested in how I explored these ideas than jump to the end of this post to see my findings!)
My first order of business was to find a list of English words. After performing a few searches, I found what I was looking for: Word-List.com. They have word lists in many languages, and the lists are available as zipped text files separated by a newline. This made it very easy to parse.
I unzipped the file and added it to a new Visual Studio 2005 Console Application project. So that the file is always available in the root of the folder, I changed the Copy to Output Directory property to Copy always. Next, I wrote a method that loads all the words into a string array and saves each word to a particular file, based on the length of the word. For example, the words “dog” and “cat” are saved to the file “03.txt” whereas the words “father” and “mother” are saved to the file “06.txt”. Here’s the code:
using (StreamReader input = new StreamReader("words.english.txt"))
{
string contents = input.ReadToEnd().Trim();
string[] wordArray = contents.Split('n');
foreach (string word in wordArray)
{
appendWord(word);
}
}
Using the StreamReader, I loaded the file and split it into a string array based on the character ‘n’ (note the single, rather than double, quotes). This allows me to iterate through an array of words. I pass each word to the appendWord method, which calls the appendFile method. Passed in is a file name that’s based on the length of the word, as well as the word itself.
private static string wordByLength = "{0}.txt";
public static void appendWord(string word)
{
appendFile(String.Format(wordByLength, word.Length.ToString().PadLeft(2, '0')), word);
}
The appendFile method simply creates an instance of a StreamWriter and appends the word to the file (if it exists) or creates a new file (if it doesn’t exist).
public static void appendFile(string file, string text)
{
using (StreamWriter writer = File.AppendText(file))
{
writer.WriteLine(text);
writer.Flush();
}
}
Once I wrote all this and let it run, it took about 20-30 minutes to complete (I don’t know exactly because I went to eat dinner). When I cam back, I had a directory of 23 files filed with sorted and ordered data:
Pretty cool, but not quite there!
Before I moved on, I wanted to know how many words had only two characters, three characters, and so on, all the way to 24. So, I wrote another method that loaded each of the aforementioned files, loaded them into an array, and determined the length of the array.
foreach (string fileByLength in Directory.GetFiles(wordByLengthDirectory))
{
using (StreamReader input = new StreamReader(fileByLength))
{
string contents = input.ReadToEnd().Trim().Replace("rn", "n");
string[] wordArray = contents.Split('n');
int wordLength = wordArray[0].Length;
appendFile(wordCountByLength, wordLength.ToString().PadLeft(2, '0') + ": " + wordArray.Length);
}
}
Nothing really complicated here. I loaded the files into a file array by using the Directory.GetFiles() method, replaced “rn” with “n” since the Environment.NewLine creates carriage return and a newline, split the array, and appended the length of the array (along with the number of characters in the word) into a file. Here were the results:
02: 61
03: 627
04: 2988
05: 7198
06: 14163
07: 20452
08: 27015
09: 29824
10: 29220
11: 25021
12: 19966
13: 14683
14: 9672
15: 5890
16: 3363
17: 1808
18: 838
19: 428
20: 197
21: 81
22: 40
23: 17
24: 5
Kind of interesting that there are more 9-letter words than any other words, eh? It also makes a really pretty looking graph:
While it’s interesting, it doesn’t really help me with my stated goal.
So, with the twenty-three independent files, I decided to find the most common letters, and sort them according to popularity. This, my friends, is the best part. Not only did I decide to use a generic list to contain the details of these files, but I also decided to use predicates and delegates to help me with my task! What fun!?!
First, I needed a collection to start the characters and counts in. For example, when I load the 61 words that contain two characters, I need a collection that keeps track of the frequency each character is used. To do this I defined a class called CharacterCounter, and then created a generic list. First, here’s the class I created.
public class CharacterCounter
{
private char character;
private int count;
public char Character
{
get { return character; }
set { character = value; }
}
public int Count
{
get { return count; }
set { count = value; }
}
}
Very simple class. It has two public properties: Character and Count. When used as a generic list, it allows me to create an object of characters with their associated counts (e.g. frequency used).
In determining the frequency each character is used based on the number of characters in the words, I first iterate through each of the files that contain the sorted words. This allows me to first find the ordered frequency of characters for words with two characters all the way to twenty-four characters. I then load the contents of each file into a string array, so that I can iterate through each of the words. Next, I create a character array out of the word, and iterate through each of the characters. It looks something like this:
foreach (string fileByLength in Directory.GetFiles(wordByLengthDirectory))
{
List<CharacterCounter> characterCounters = new List<CharacterCounter>();
int fileLength = 0;
using (StreamReader input = new StreamReader(fileByLength))
{
string contents = input.ReadToEnd().Trim().Replace("rn", "n");
string[] wordArray = contents.Split('n');
if (fileLength == 0)
{
fileLength = wordArray[0].Length;
}
foreach (string word in wordArray)
{
char[] characterArray = word.ToCharArray();
foreach (char character in characterArray)
{
//...
}
}
}
}
Once I am iterating through each of the characters, I need to start keeping track of the number of times the character appears. Notice that I created a generic list for my CharacterCounter class called characterCounters. Before I increment the character count, I first check to see if the character in question already exists within my list. This is performed using the following code:
CharacterCounter characterCounter = characterCounters.Find(delegate(CharacterCounter d)
{
return d.Character == character;
});
If the character exists, then the characterCounter list returns the CharacterCounter object that has a Character that equals the character I specified. For instance, if the specified character is “a”, and it exists in the CharacterCounter list, then the characterCounter object is returned with an “a” for the Character and the number of times it has been found in the Count property.
If the character does not exist in the characterCounters list, then the characterCounter object is null. If it’s null, I create a new instance of a CharacterCounter, set my values accordingly, and add it to the list. If it’s not null, I increment the count:
if (characterCounter == null)
{
characterCounter = new CharacterCounter();
characterCounter.Character = character;
characterCounter.Count = 1;
characterCounters.Add(characterCounter);
}
else
{
characterCounter.Count += 1;
}
This continues until I’ve gone through every word in the file. At this point, I have a generic CharacterCounter list filled with the number of times a character is found. Since I want to know the most commonly used characters, I need to sort my list so that the most common characters are first.
The following code will sort the characterCounters list by descending count:
characterCounters.Sort(delegate(CharacterCounter cc0, CharacterCounter cc1){ return cc1.Count.CompareTo(cc0.Count);});
Using an anonymous delegate, you can tell the list to sort itself from the largest number to the smallest. Changing the code to the following will sort it from smallest to largest:
characterCounters.Sort(delegate(CharacterCounter cc0, CharacterCounter cc1)
{
return cc0.Count.CompareTo(cc1.Count);
});
All this is done without having to use the IComparable interface! Amazing!
At this point, our list is sorted. All we need to do is create a string based on the ordered characters, and write it to a file:
foreach (CharacterCounter characterCounter in characterCounters)
{
orderedCharacters += characterCounter.Character.ToString();
}
appendFile(orderedFrequencyByLength, fileLength.ToString().PadLeft(2, '0') + ": " + orderedCharacters);
The code simply iterates through the characterCounters list, appends the value of the Character property to a string, and then appends the data to a string.
After all of this work, we can finally show the list.
# of characters in word: list of characters, from most frequent to least frequent
02: aoueyidbrtfslnmhgwzkvjcp
03: aoeuirstdylmghpknbwfczvxj
04: aeoirutslndpkymbhgcwfzvjxq
05: aeriosnltuycdmhpbgkwfvzxjq
06: earinoltsucdmphgbykfwvzxjq
07: eairnoltsucdmpghbykfwvzxjq
08: eaironlstucdmphgybfkwvzxjq
09: eiarontslcudmphygbfkvwzxqj
10: eiaorntslcupdmhygbfvkwzxqj
11: eiaorntslcupmdhygbfvkwzxqj
12: eiaontrslcupmhdygbfvzkxwqj
13: eiaontrslcpumhdygbvfzxkqwj
14: eioantsrlcpuhmdygbvfzxqkwj
15: ieoantsrlcpuhmydgbvfzxqkwj
16: eioatnrslcphumydgbvfzxqkwj
17: ieoatnrsclphumydgbvfzxqkwj
18: oeiatrnsclphmuydgbvfzxqkjw
19: oiaetnrsclhpmyudgbvzfxqkwj
20: oieatrnlchspymugdbvzfxjqk
21: oietanlcrshpymgdubzfvjx
22: oeticrahnslpymdugbxvzjq
23: oiaetlhscnprmdyugbfx
24: oihletapcyrdsnfgmzux
Going from the left to the right, we can tell the frequency of a letter in a word based on the number of characters in the word. For example, in a nine-letter word, the most frequently used characters are: e, i, a, r, o, n, etc.
Now, here’s the real question: does knowing all of this help us to be a better Hangman player?
Well, what don’t you try for yourself. Go to The Free Dictionary and play a game of Hangman. Based on the number of characters in the word, try the list characters above in their given order. From what I’ve seen, it doesn’t work 100% of the time, but it certainly gets it correct more often than not.
In Part II (goodness, a Part II?!?!), I plan to write a test framework that uses the dictionary I downloaded to randomly calculate the accuracy of this method. I’ll use a large set of sample data, and see how often I can guess the word correctly based on the rules above without specifying ten incorrect answers.
In Part III (holy smokes!), I’d like to make the rules more intelligent by changing the most common letters used based on the known letters in the word. For instance, if the word has an S and T, I’d like to know if an E or I is more likely to also be in the word.
Chances are it’ll be awhile before I get to Part II and Part III, but I think I’ll get there eventually.
I truly hope that at least someone else out there has fond some value in this post! Here’s the full source-code, along with all the parsed files (look in the debug directory).
Dictionary.zip (2.15 MB)
Please let me know if you have any thoughts!
