Home > Python > Solving Scramble With Friends – a tale of three data structures

## Solving Scramble with Friends – A tale of three data structures

This post aims to illustrate how to solve Scramble With Friends/Boggle and the techniques and data structures necessary. In particular, we will see how to prune an enormous search space, how to use recursion to simplify the task, and the relative performance of three different data structures: an unsorted list, a sorted list with binary search, and a trie.

# Problem statement

Given an N x N board, (4×4 in case of Boggle and Scramble with Friends), find all of the paths through the board which create a valid word. These paths must use each tile at most once and be contiguous; tile N must be adjacent to tile N – 1 in any of 8 directions (North, Northeast, East, Southeast, South, Southwest, West, Northwest).

In this example, ANT is a valid solution in the bottom left corner, but NURSE is not because the Ns are not located next to the U.

# Most naive approach

First, imagine that we have a method that, given a path through the board returns all valid words that starts with that path, called `DoSolve`. For instance, `DoSolve(['Qu'])` would return all the valid words (and their locations) that start with Qu. `DoSolve(['N', 'E'])` would return all the valid paths that start with N followed by an E. With such a method, it is trivial to solve the problem. In pseudocode:

``````Solve(board):
solutions = empty list
for each location on board:
append DoSolve(location) to solutions
return solutions
``````

The tricky part is, how do we make DoSolve work? Remember that it must

1. return only valid words
2. return only words that are formed from contiguous tiles
3. must not repeat any tiles.

The easiest way to solve this problem is through recursion. As you probably remember, recursion is simply when a method calls itself. In order to avoid an infinite loop (and eventual stack overflow), there must be some sort of stopping criteria. As it was taught to me, you have to be sure that the problem becomes smaller at each step, until it becomes so small as to be trivially solved. Here’s that basic algorithm:

``````DoSolve(board, tiles_used)
solutions = empty list

current word = empty string
for letter in tiles_used:
append letter to current word

# 1 - ensure that the word is valid
if the current word is a valid word:
append current word to solutions

most recent tile = last tile in tiles_used

# 2 - ensure that the next tile we use is adjacent to the last one
for each tile adjacent to most recent tile:

# 3 - ensure that we do not repeat any tiles
if tile is on the board and tile hasn't been used previously:
new_path = append tile to copy of tiles_used list
solutions_starting_from_tile = DoSolve(board, new_path)
append solutions_starting_from_tile to solutions

return solutions
``````

This will work, but it suffers from an incredible flaw. Do you see it?

The problem is that this method will waste an inordinate amount of time exhaustively searching the board for solutions even when the current path is completely useless. It will continue to search for words starting with QuR, QuRE, etc., etc., even though no words in the English language start with these letters. The algorithm is still correct, but it can be optimized very simply.

# Pruning the search space

Those with some algorithms background might recognize the code above as a modified form of a depth first search. As described previously, a depth first search will exhaustively explore every possible path through the board. We can vastly improve the efficiency of this algorithm by quitting the search early when we know it won’t be fruitful. If we know all of the valid words, we can quit when we know that no word starts with the current string. Thus in the previous example, if the current word built up so far were “QuR”, the algorithm could determine that no words start with QuR and thus fail fast and early. This optimization will not affect the correctness of the algorithm because none of the potential paths we eliminate could possibly lead to a valid word; by constraint #1 this means that none of those paths would have ended up in the list of solutions in the first place.

How much does this save? On random 3×3 boards, The fastest implementation I have is sped up by a factor of 75. Solving 4×4 boards without pruning is infeasible.

# Basic Python implementation

Assume as a black box we have two methods `IsWord(string)` and `HasPrefix(string)`. The pseudocode above can be expressed with Python (forgive the slightly modified parameter list; I found it easier to write it this way):

```def DoSolve(self, board, previous_locations, location, word_prefix):
"""Returns iterable of FoundWord objects.

Args:
previous_locations: a list of already visited locations
location: the current Location from where to start searching
word_prefix: the current word built up so far, as a string
"""
solutions = []

new_word = word_prefix + board[location.row][location.col]
previous_locations.append(location)

if PREFIX_PRUNE and not self.HasPrefix(new_word):
if DEBUG:
print 'No words found starting with "%s"' %(new_word)
return solutions

# This is a valid, complete words.
if self.IsWord(new_word):
new_solution = FoundWord(new_word, previous_locations)
if DEBUG:
print 'Found new solution: %s' %(str(new_solution))
solutions.append(new_solution)

# Recursively search all adjacent tiles
for new_loc in location.Adjacent():
if board.IsValidLocation(new_loc) and new_loc not in previous_locations:
# make a copy of the previous locations list so our current list
# is not affected by this recursive call.
defensive_copy = list(previous_locations)
solutions.extend(self.DoSolve(board, defensive_copy, new_loc, new_word))
else:
if DEBUG:
print 'Ignoring %s as it is invalid or already used.' %(str(new_loc))

return solutions

```

# Data structures

The data structure and algorithms used to implement the `IsWord` and `HasPrefix` methods are incredibly important.

I will examine three implementations and discuss the performance characteristics of each:

• Unsorted list
• Sorted list with binary search
• Trie

# Unsorted list

An unsorted list is a terrible data structure to use for this task. Why? Because it leads to running time proportional to the number of elements in the word list.

```class UnsortedListBoardSolver(BoardSolver):
def __init__(self, valid_words):
self.valid_words = valid_words

def HasPrefix(self, prefix):
for word in self.valid_words:
if word.startswith(prefix):
return True
return False

def IsWord(self, word):
return word in self.valid_words
```

While this is easy to understand and reason about, it is extremely slow, especially for a large dictionary (I used approximately 200k words for testing).

``````Took 207.697 seconds to solve 1 boards; avg 207.697 with <__main__.UnsortedListBoardSolver object at 0x135f170>
``````

(This time is with the pruning turned on).

# Sorted list

Since we know the valid words ahead of time, we can take advantage of this fact and sort the list. With a sorted list, we can perform a binary search and cut our running time from O(n) to O(log n).

Writing a binary search from scratch is very error prone; in his classic work Programming Pearls, Jon Bently claims that fewer than 10% of programmers can implement it correctly. (See blog post for more).

Fortunately, there’s no reason whatsoever to write our own binary search algorithm. Python’s standard library already has an implementation in its bisect module. Following the example given in the module documentation, we get the following implementation:

```class SortedListBoardSolver(BoardSolver):
def __init__(self, valid_words):
self.valid_words = sorted(valid_words)

# http://docs.python.org/library/bisect.html#searching-sorted-lists
def index(self, a, x):
'Locate the leftmost value exactly equal to x'
i = bisect.bisect_left(a, x)
if i != len(a) and a[i] == x:
return i
raise ValueError

def find_ge(self, a, x):
'Find leftmost item greater than or equal to x'
i = bisect.bisect_left(a, x)
if i != len(a):
return a[i]
raise ValueError

def HasPrefix(self, prefix):
try:
word = self.find_ge(self.valid_words, prefix)
return word.startswith(prefix)
except ValueError:
return False

def IsWord(self, word):
try:
self.index(self.valid_words, word)
except ValueError:
return False
return True
```

Running the same gauntlet of tests, we see that this data structure performs far better than the naive, unsorted list approach.

``````Took 0.094 seconds to solve 1 boards; avg 0.094 with <__main__.SortedListBoardSolver object at 0x135f1d0>
Took 0.361 seconds to solve 10 boards; avg 0.036 with <__main__.SortedListBoardSolver object at 0x135f1d0>
Took 2.622 seconds to solve 100 boards; avg 0.026 with <__main__.SortedListBoardSolver object at 0x135f1d0>
Took 25.065 seconds to solve 1000 boards; avg 0.025 with <__main__.SortedListBoardSolver object at 0x135f1d0>
``````

# Trie

The final data structure I want to illustrate is that of the trie. The Wikipedia article has a lot of great information about it.

From Wikipedia:

In computer science, a trie, or prefix tree, is an ordered tree data structure that is used to store an associative array where the keys are usually strings. Unlike a binary search tree, no node in the tree stores the key associated with that node; instead, its position in the tree defines the key with which it is associated. All the descendants of a node have a common prefix of the string associated with that node, and the root is associated with the empty string. Values are normally not associated with every node, only with leaves and some inner nodes that correspond to keys of interest.

Snip:

Looking up a key of length m takes worst case O(m) time. A BST performs O(log(n)) comparisons of keys, where n is the number of elements in the tree, because lookups depend on the depth of the tree, which is logarithmic in the number of keys if the tree is balanced. Hence in the worst case, a BST takes O(m log n) time. Moreover, in the worst case log(n) will approach m. Also, the simple operations tries use during lookup, such as array indexing using a character, are fast on real machines

Again, I don’t want to reinvent what’s already been done before, so I will be using Jeethu Rao’s implementation of a trie in Python rather than rolling my own.

Here is a demonstration of its API via the interactive prompt:

```>>> import trie
>>> t = trie.Trie()
>>> t.find_full_match('h')
>>> t.find_full_match('hell')
1
>>> t.find_full_match('hello')
2
>>> t.find_prefix_matches('hello')
[2]
>>> t.find_prefix_matches('hell')
[2, 1]
```

Unfortunately, there’s a bug in his code:

```>>> t = trie.Trie()
>>> 'fail' in t
False
>>> 'hello' in t
True
# Should be false; 'h' starts a string but
# it is not contained in data structure
>>> 'h' in t
True
```

His `__contains__` method is as follows:

```def __contains__( self, s ) :
if self.find_full_match(s,_SENTINEL) is _SENTINEL :
return False
return True
```

The `find_full_match` is where the problem lies.

```def find_full_match( self, key, fallback=None ) :
"'
Returns the value associated with the key if found else, returns fallback
"'
r = self._find_prefix_match( key )
if not r[1] and r[0]:
return r[0][0]
return fallback
```

`_find_prefix_match` returns a tuple of node that the search terminated on, remainder of the string left to be matched. For instance,

```>>> t._find_prefix_match('f')
[[None, {'h': [None, {'e': [None, {'l': [None, {'l': [None, {'o': [0, {}]}]}]}]}]}], 'f']
>>> t.root
[None, {'h': [None, {'e': [None, {'l': [None, {'l': [None, {'o': [0, {}]}]}]}]}]}]
```

This makes sense, ‘f’ doesn’t start any words in the trie containing just the word ‘hello’, so the root is returned with the ‘f’ string that doesn’t match. `find_full_match` correctly handles this case, since r[1] = ‘f’, not r[1] = False, and the fallback is returned. That fallback is used by contains to signify that the given string is not in the trie.

The problem is when the string in question starts a valid string but is itself not contained in the trie. As we saw previously, ‘h’ is considered to be in the trie.

```>>> r = t._find_prefix_match('h')
>>> r
[[None, {'e': [None, {'l': [None, {'l': [None, {'o': [0, {}]}]}]}]}], "]
>>> r[0]
[None, {'e': [None, {'l': [None, {'l': [None, {'o': [0, {}]}]}]}]}]
>>> r[1]
"
>>> bool(not r[1] and r[0])
True
>>> r[0][0]
# None
```

The issue is that his code does not check that there is a value stored in the given node. Since no such value has been stored, the code returns None, which is not equal to the SENTINEL_ value that his `__contains__` method expects. We can either change findfullmatch to handle this case correctly, or change the `__contains__` method to handle the None result as a failure. Let’s modify the `find_full_match` method to obey it’s implied contract (and be easier to understand):

```def find_full_match( self, key, fallback=None ) :
"'
Returns the value associated with the key if found else, returns fallback
"'
curr_node, remainder = self._find_prefix_match(key)
stored_value = curr_node[0]
has_stored_value = stored_value is not None
if not remainder and has_stored_value:
return stored_value
return fallback
```

Let’s make sure this works:

```>>> reload(trie)
<module 'trie' from 'trie.py'>
>>> t = trie.Trie()
>>> 'f' in t
False
>>> 'hell' in t
False
>>> 'hello' in t
True
```

OK, with that minor patch, here’s a first pass implementation of the solution using the trie:

```class TrieBoardSolver(BoardSolver):
def __init__(self, valid_words):
self.trie = trie.Trie()
for index, word in enumerate(valid_words):
# 0 evaluates to False which screws up trie lookups; ensure value is 'truthy'.

def HasPrefix(self, prefix):
return len(self.trie.find_prefix_matches(prefix)) > 0

def IsWord(self, word):
return word in self.trie
```

Unfortunately, this is slow. How slow?

``````Took 2.626 seconds to solve 1 boards; avg 2.626 with <__main__.TrieBoardSolver object at 0x135f070>
Took 22.681 seconds to solve 10 boards; avg 2.268 with <__main__.TrieBoardSolver object at 0x135f070>
``````

While this isn’t as bad as the unsorted list, it’s still orders of magnitudes slower than the binary search implementation.

Why is this slow? Well, it’s doing a whole lot of unnecessary work. For instance, if we want to determine if ‘h’ is a valid prefix, this implementation will first construct the list of all words that start with h, only to have all that work thrown away when we see that the list is not empty.

A much more efficient approach is to cheat a little and use the previously discussed method `_find_prefix_match` which returns the node in the tree that the search stopped at and how much of the search string was unmatched.

By using this method directly, we can avoid creating the lists of words which we then throw away. We modify the HasPrefix method to the following:

```def HasPrefix(self, prefix):
curr_node, remainder = self.trie._find_prefix_match(prefix)
return not remainder
```

With this optmization, the trie performance becomes competitive with the binary search:

``````Took 0.027 seconds to solve 1 boards; avg 0.027 with <__main__.TrieBoardSolver object at 0x135f230>
Took 0.019 seconds to solve 1 boards; avg 0.019 with <__main__.SortedListBoardSolver object at 0x135f330>
Took 0.199 seconds to solve 10 boards; avg 0.020 with <__main__.TrieBoardSolver object at 0x135f230>
Took 0.198 seconds to solve 10 boards; avg 0.020 with <__main__.SortedListBoardSolver object at 0x135f330>
Took 2.531 seconds to solve 100 boards; avg 0.025 with <__main__.TrieBoardSolver object at 0x135f230>
Took 2.453 seconds to solve 100 boards; avg 0.025 with <__main__.SortedListBoardSolver object at 0x135f330>
``````

It is still slower, but not nearly as bad as before.

# Solving the board in the screenshot

With all this machinery in place, we can run the original board in the screenshot through the algorithms:

```words = ReadDictionary(sys.argv[1])
print 'Read %d words' %(len(words))

trie_solver = TrieBoardSolver(words)
sorted_list_solver = SortedListBoardSolver(words)
b = Board(4, 4)
b.board = [
['l', 'qu', 'r', 'e'],
['s', 'l', 'u', 's'],
['a', 't', 'i', 'c'],
['n', 'r', 'e', 'n']
]
print 'Solving with binary search'
sorted_solutions = sorted_list_solver.Solve(b)
print 'Solving with trie'
trie_solutions = trie_solver.Solve(b)

# Results should be exactly the same
assert sorted_solutions == trie_solutions

for solution in sorted(sorted_solutions):
print solution
words = sorted(set([s.word for s in sorted_solutions]))
print words
```

The results appear after the conclusion.

# Conclusion

I have presented both pseudocode and Python implementations of an algorithm for solving the classic Boggle/Scramble With Friends game. In the process, we saw such concepts as recursion, depth first search, optimizing code through pruning unfruitful search branches, and the importance of using the right data structure. This post does not aim to be exhaustive; I hope I have piqued your interest for learning more about tries and other lesser known data structures.

For results, the patched trie implementation, and the driver program, see below. To run the driver program, pass it a path to a list of valid words, one per line. e.g.

``````python2.6 solver.py /usr/share/dict/words
``````

https://gist.github.com/2833942#file_trie.py

For mobile users, the link can be found at https://gist.github.com/2833942

Categories: Python
1. May 31, 2012 at 10:32 am

Nice writeup. I went through almost the exact same series of approaches last year when I was writing a boggle solver. The only exception being that I implemented the final trie-based version in C. I wrote about mine here: http://thraxil.org/users/anders/posts/2011/04/17/A-Boggling-Return-to-C/

2. June 1, 2012 at 9:15 am

Nice post. I remember Chown gave us this for CS210 back in the day. My solution wasn’t very good, if i remember correctly.

3. June 1, 2012 at 9:21 am

Yup, I remember doing it in CS210 as well, but I certainly didn’t use a trie or even a sorted list back then. I don’t know if I even used prefix pruning.. but it’s nice to be able to go back to old assignments that took hours and hours and be able to bang them out much more quickly

4. June 24, 2012 at 11:23 pm

Very nice!

I actually implemented (and deployed) my own solution to this (scramble with friends cheat), also developed in Python, earlier this year. You’re on the right path with a trie structure – however, Python can be a memory hog in production. There are some graph algorithms which address this issue – the scrabble guys solved this issue in the late 80’s.

Now, the second issue you run into when building a scramble with friends solver for production use is the UI challenge. With about two minutes per round, you can’t have a user flipping back and forth between screens – and you can’t just show the word, you need to show a path through the grid. I wrote a blog post covering this topic – this was actually the hard part of the project – designing a physically fast UI experience… link to post: adventures in ui – time limits in a scramble with friends cheat

BTW – performance is less of an issue for this project in the production environment than you might think. Assuming you have done a decent job thinking through your basic architecture, you will spend more cycle time / resources handling basic IO / UI functions than calculating solutions. The solver is almost always faster than the underlying web server.

5. June 24, 2012 at 11:33 pm

One additional thought / question, since you guys seem interested in the topic.

I noticed some other word game solvers on the web were based on PHP and – I assume – MySQL. Anyone know how you could write a reasonably fast word game solver which runs inside a database using SQL?

I went straight to Python and never really explored this route… most of the obvious SQL solutions seemed to involve a lot of full table scans or convoluted indexing..

One more link you might find interesting – a counter-point to my prior comment about algorithm speed being secondary in a production implementation of a word game solver. It does matter when you’re trying to crank through every word in the dictionary….

Results of testing # of tries to guess each hangman word in the dictionary – with cheating allowed (solves a conditional probability problem with each guess, using perfect knowledge of all remaining eligible words in the dictionary). Required us to map reduce this puppy across eight processors on our development box…

link to my blog article: hard words for hanging with friends

• June 26, 2012 at 10:51 pm

Thanks for the info – glad to see I wasn’t alone in my solution. Nice writeup

6. November 4, 2012 at 8:37 am

Very interesting. Has anyone used this in monte carlo simulations to come up with a probability sorted word list as a study aid for frequent players? That seems like something useful that is not actually cheating. Does SWF use a 16x six-sided die algorithm to pick the letters it puts on the board (like boggle)?

• November 5, 2012 at 9:32 pm

Well, the problem is that I don’t actually have the list of words that SWF accepts. If you’ll notice, I’m using the dict that’s builtin to UNIX systems; I’m fairly sure that’s not the same as what Scramble with Friends uses.

I have no idea what algorithm they use, but some of the arrangements of letters (constantly seeing ING adjacent e.g.) lead me to believe there’s something going on that’s not entirely random.

• November 8, 2012 at 7:35 am

Yeah, after playing a bit more, it’s definitely not completely random. It seems like it consistently generates easier boards than the Boggle app I have. Maybe I’ll try to collect some data. Thanks!

• November 22, 2012 at 10:52 am

Sooo, I went through my SWF history and copied down a bunch of the boards that had been generated by SWF (about 40 of them). I ran those through your solver with the SOWPODS dictionary, which doesn’t match the SWF dictionary, but provides some comparison points. I’m working in windows, otherwise, I probably would have used the UNIX built in dictionary.

Using this dictionary, the average number of words per board is about 396. I then generated 100 boards from each of the SWF boards by just shifting the position of the letters in the boards around. In every case, the number of solutions in the original SWF board exceeds the average of the 100 random boards using the same letter set. In most cases (about 85% or 34 of 40) the number of SWF solutions exceeds the reordered board solutions by at least one standard deviation.

So to be concise, there are way more possible word solutions in the boards SWF generates than a random ordering of the same letters sets on the board would create. 🙂 Thanks for putting this algorithm up. I had fun using it. If I poke around at the numbers any more, I’ll keep you updated.

1. June 26, 2012 at 10:23 pm