#
Hangman
*
*

# Hangman

Simple script to solve hangman.

Does the following:

- Selects a word a random from the dictionary
- Given the length of the word, finds the most common letter occurances across all words of that length
- Simple algorithm
- Guesses most common letters until solution is found

- Improved algorithm
- Filters for most common letters at every iteration for dictionary words that match pattern of the guessed letters
- Continues until solution is found

Note: Syntax is nearly Ruby compatable, however tuple types need to be changed to Hash and type decleration needs to be removed from the array.

## Usage

```
$ crystal hangman.cr
```

## Output

### Simple Algorithm

```
$ crystal hangman.cr --error-trace
Selecting a word from the dictionary at random...
The word is --------- (9 of 9 digits remaining)
Solving using simple algorithm...
Trying with e...
The word is ------e-- (8 of 9 digits remaining)
Trying with i...
The word is ------e-- (8 of 9 digits remaining)
Trying with a...
The word is --a---e-- (7 of 9 digits remaining)
...
Trying with y...
The word is crac-less (1 of 9 digits remaining)
Trying with g...
The word is crac-less (1 of 9 digits remaining)
Trying with b...
The word is crac-less (1 of 9 digits remaining)
Trying with f...
The word is crac-less (1 of 9 digits remaining)
Trying with v...
The word is crac-less (1 of 9 digits remaining)
Trying with k...
The word is crackless (0 of 9 digits remaining)
Solved after 22 iterations with simple algorithm
```

### Improved Algorithm

```
Solving using improved algorithm...
Trying with e...
The word is ------e-- (8 of 9 digits remaining)
Trying with s...
The word is ------ess (6 of 9 digits remaining)
Trying with n...
The word is ------ess (6 of 9 digits remaining)
Trying with l...
The word is -----less (5 of 9 digits remaining)
Trying with r...
The word is -r---less (4 of 9 digits remaining)
Trying with t...
The word is -r---less (4 of 9 digits remaining)
Trying with a...
The word is -ra--less (3 of 9 digits remaining)
Trying with c...
The word is crac-less (1 of 9 digits remaining)
Trying with k...
The word is crackless (0 of 9 digits remaining)
Solved after 10 iterations with improved algorithm
```

## Notes

- Build on
`Crystal 0.23.1 (2017-09-08) LLVM 4.0.1`

- Requires list of words to be under
`/usr/local/share/dict`

as is default on macOS

Owner

github statistic

- 1
- 0
- 0
- 0
- over 3 years ago
- October 7, 2017

License

Links

Synced at

Sat, 16 Jan 2021 17:46:23 GMT