Last time we looked at how we solve a problem generally by considering the riddle of the fox, chicken and grain. Today we'll attempt to implement this in Python.

As discussed last time, there are two main parts to solving a problem:

- Translating the problem from natural language into the model domain
- Finding a solution by manipulating the model

We're going to cheat a bit and skirt around the first of these by translating our problem into the model domain by hand. That means we can focus on the second part of the problem.

But what is the model? We'll start with something very simple and see how far it can stretch. We can solve our riddle using just sets and actions that modify them.

## Initial state

We start by creating a set for the near bank of the river and one for the far bank. Initially we'll populate the near bank and leave the far bank empty:

```
start_state = {
'near bank': {
'you',
'fox',
'chicken',
'sack of grain'},
'far bank': set()
}
```

## Win state

The objective is to cross the river with all intact so we can define the win state by reversing the banks:

```
win_state = {
'far bank': {
'you',
'fox',
'chicken',
'sack of grain'},
'near bank': set()
}
```

## Actions and consequences

To get from the start state to the win state, we need to execute a sequence of actions. Each action encapsulates the changes to the sets from making a particular move and the conditions that must be true for the move to be possible.

For example, let's consider the simplest possible move our riddle allows - crossing the river alone. If we are on the near bank, we cease to be on the near bank and instead exist on the far bank. Similarly, if we are on the far bank we cease to be on the far bank and instead exist on the near bank. In each case, we have a condition and two set operations. We can represent this as follows:

```
cross_alone' = [{
'if': 'near bank',
'contains': ['you'],
'then': [{
'modify': 'near bank',
'remove': ['you']
}, {
'modify': 'far bank',
'add': ['you']
}]
}, {
'if': 'far bank',
'contains': ['you'],
'then': [{
'modify': 'far bank',
'remove': ['you']
}, {
'modify': 'near bank',
'add': ['you']
}]
}]
```

The same approach can be used to define the moves with the fox, chicken or sack of grain. We just need to check for the presence of both 'you' and the item (e.g. fox) and the remove both from one bank and add both to another.

Consequences are essentially just actions that are executed automatically in response to any action. These capture things that happen as a result of our actions such as the chicken getting eaten if it's left with an unsupervised fox.

## Problem spec

We're now in a position to lay out the entire problem spec:

```
problem_spec = {
'start_state': {
'near bank': {
'you',
'fox',
'chicken',
'sack of grain'},
'far bank': set()
},
'win_state': {
'far bank': {
'you',
'fox',
'chicken',
'sack of grain'},
'near bank': set()
},
'actions': {
'cross with fox': [{
'if': 'near bank',
'contains': ['you', 'fox'],
'then': [{
'modify': 'near bank',
'remove': ['you', 'fox']
}, {
'modify': 'far bank',
'add': ['you', 'fox']
}]
}, {
'if': 'far bank',
'contains': ['you', 'fox'],
'then': [{
'modify': 'far bank',
'remove': ['you', 'fox']
}, {
'modify': 'near bank',
'add': ['you', 'fox']
}]
}],
'cross with chicken': [{
'if': 'near bank',
'contains': ['you', 'chicken'],
'then': [{
'modify': 'near bank',
'remove': ['you', 'chicken']
}, {
'modify': 'far bank',
'add': ['you', 'chicken']
}]
}, {
'if': 'far bank',
'contains': ['you', 'chicken'],
'then': [{
'modify': 'far bank',
'remove': ['you', 'chicken']
}, {
'modify': 'near bank',
'add': ['you', 'chicken']
}]
}],
'cross with sack of grain': [{
'if': 'near bank',
'contains': ['you', 'sack of grain'],
'then': [{
'modify': 'near bank',
'remove': ['you', 'sack of grain']
}, {
'modify': 'far bank',
'add': ['you', 'sack of grain']
}]
}, {
'if': 'far bank',
'contains': ['you', 'sack of grain'],
'then': [{
'modify': 'far bank',
'remove': ['you', 'sack of grain']
}, {
'modify': 'near bank',
'add': ['you', 'sack of grain']
}]
}],
'cross alone': [{
'if': 'near bank',
'contains': ['you'],
'then': [{
'modify': 'near bank',
'remove': ['you']
}, {
'modify': 'far bank',
'add': ['you']
}]
}, {
'if': 'far bank',
'contains': ['you'],
'then': [{
'modify': 'far bank',
'remove': ['you']
}, {
'modify': 'near bank',
'add': ['you']
}]
}]
},
'consequences': {
'fox eats chicken': [{
'if': 'near bank',
'contains': ['fox', 'chicken'],
'not_contains': ['you'],
'then': [{
'modify': 'near bank',
'remove': ['chicken']
}]
}, {
'if': 'far bank',
'contains': ['fox', 'chicken'],
'not_contains': ['you'],
'then': [{
'modify': 'far bank',
'remove': ['chicken']
}]
}],
'chicken eats grain': [{
'if': 'near bank',
'contains': ['chicken', 'sack of grain'],
'not_contains': ['you'],
'then': [{
'modify': 'near bank',
'remove': ['sack of grain']
}]
}, {
'if': 'far bank',
'contains': ['chicken', 'sack of grain'],
'not_contains': ['you'],
'then': [{
'modify': 'far bank',
'remove': ['sack of grain']
}]
}]
}
}
```

Not quite as concise as the original text description, is it? Let's remind ourselves:

You have a fox, a chicken and a sack of grain. You must cross a river with only one of them at a time. If you leave the fox with the chicken he will eat it; if you leave the chicken with the grain he will eat it. How can you get all three across safely?

Our formal description suffers from a lot of repetition because we have no notion of abstraction and we're also capturing some additional information that was implicit. We also have field names and more whitespace and structural elements to contend with. There's certainly a lot of scope to condense it while retaining a formal syntax.

## The algorithm

Now we have a formal specification of the start state, end state and all the possible moves in-between, there just remains the task of finding an optimal sequence of moves from start to finish.

We could do a simple breadth first search of the tree of possible moves but this doesn't scale very well. An interesting alternative that I already used in a previous post is the monte-carlo tree search algorithm. Essentially, the idea is to select a random move based on previous success. At each node in the tree we record the number of plays through that node and the number of times it led to a win. The ratio of wins:plays is used to weight the probability of selecting that path again in the future.

Here I have adapted the `explored_move`

class to work with set state:

```
class explored_move:
initial_wins = 1
initial_plays = 2
def __init__(self, num_actions, prev_move=None):
self.num_actions = num_actions
self.prev_move = prev_move
self.next_moves = {}
self.wins = explored_move.initial_wins
self.plays = explored_move.initial_plays
# play this move
def play(self):
self.plays += 1
# mark move as illegal (don't try again)
def illegal(self):
self.wins = 0
# this is a win - propagate up to root
def win(self):
self.wins += 1
if self.prev_move is not None:
self.prev_move.win()
# get number of wins for next move i
def num_wins(self, i):
return self.next_moves[i].wins if i in self.next_moves else explored_move.initial_wins
# get number of plays for next move i
def num_plays(self, i):
return self.next_moves[i].plays if i in self.next_moves else explored_move.initial_plays
# create a move following the this one
def play_next(self, i):
if i not in self.next_moves:
self.next_moves[i] = explored_move(self.num_actions, self)
next_move = self.next_moves[i]
next_move.play()
return next_move
# choose move weighted by previous wins
def choose_next_move(self):
# calculate a common denominator
d = 1
for m in range(self.num_actions):
if self.num_wins(m) > 0:
d *= self.num_plays(m)
# accumulate win ratio * common denominator
max_i = 0
for m in range(self.num_actions):
max_i += self.num_wins(m) * (d // self.num_plays(m))
# if all moves have zero wins then return -1 (no valid move)
if max_i == 0:
return -1
# select a random point in the weighted move probability space
j = random.randrange(0, max_i)
# accumulate the move intervals to determine the selected move m for index i
i = 0
for m in range(self.num_actions):
i += self.num_wins(m) * (d // self.num_plays(m))
if i > j:
return m
# should never get here unless the above maths went wrong!
raise Exception('Unreachable: ' + str(i) + ' > ' + str(max_i))
# return the best of the explored next moves
def best_move(self):
# calculate a common denominator
d = 1
for i in range(self.num_actions):
if self.num_wins(i) > 0:
d *= self.num_plays(i)
# start with ratio zero and no valid move
best_ratio = 0
best_m = (-1, None)
# for each move if the ratio is better store it as the new best along with the best move
for item in self.next_moves.items():
m, node = item
ratio = node.wins * (d // node.plays)
if ratio > best_ratio:
best_ratio = ratio
best_m = (m, node)
# return the best move
return best_m
```

There are no significant changes here from the previous version - really just renaming rules to actions so I won't elaborate further. Suffice it to say that this class represents a move that has been explored by the monte-carlo tree search and contains some methods to update the counts as well as to find the move with the highest probability of a winning outcome.

At the core of the algorithm is a relatively simple function to apply an action. This updates the state based on the changes in the action definition provided that the action's conditions are met:

```
def apply_action(state, action):
for a in action:
# check for match
match = True
if 'contains' in a:
for c in a['contains']:
if c not in state[a['if']]:
match = False
break
if 'not_contains' in a:
for n in a['not_contains']:
if n in state[a['if']]:
match = False
break
if not match:
continue
# apply modifications
for m in a['then']:
if 'remove' in m:
for x in m['remove']:
state[m['modify']].remove(x)
elif 'add' in m:
for x in m['add']:
state[m['modify']].add(x)
return True
return False
```

We use this in our modified `try_play_move`

function that instead of making string substitutions, now updates the set state based on a selected random move:

```
def try_play_move(actions, consequences, cur_move, state):
# try to select a move
i = cur_move.choose_next_move()
# if no valid move return Nones
if i == -1:
return None, None
# make the move
next_move = cur_move.play_next(i)
next_state = copy_state(state)
if not apply_action(next_state, actions[i]):
next_move.illegal()
return cur_move, state
# apply consequences
for c in consequences:
apply_action(next_state, c)
return next_move, next_state
```

This depends on a little helper to duplicate the set state:

```
def copy_state(state):
new_state = {}
for name, contents in state.items():
new_state[name] = set(contents)
return new_state
```

Finally, we modify the play function to use set state and to mark a move as winning when the current state equals the win state:

```
def play(spec, paths, max_depth=1000):
actions = [a for _,a in spec['actions'].items()]
consequences = [c for _,c in spec['consequences'].items()]
start_state = spec['start_state']
win_state = spec['win_state']
# start with root move
root = explored_move(len(actions))
# explore random paths
for p in range(paths):
# reset state for new path
state = start_state
cur_move = root
# make depth number of moves
for d in range(max_depth):
# play a random move
next_move, next_state = try_play_move(actions, consequences, cur_move, state)
# if no legal move then stop this path
if next_move is None:
break
# illegal move
if next_move == cur_move:
continue
# check if matches any win state
if equal_state(next_state, win_state):
next_move.win()
break
# move on to next move
cur_move = next_move
state = next_state
# replay move sequence with most wins
route = []
i, node = root.best_move()
while node is not None and node.wins > 1:
route.append(i)
i, node = node.best_move()
# return the route
return route
```

When all the paths have been probed, this function generates the optimal route based on the recorded win ratios. We invoke it as follows:

```
route = play(problem_spec, 500)
print(route)
```

And this runs 500 paths, producing the route:

`[1, 3, 2, 1, 0, 3, 1]`

Which in english means:

- Cross with chicken
- Cross alone
- Cross with sack of grain
- Cross with chicken
- Cross with fox
- Cross alone
- Cross with chicken

Hurrah! It has found a solution!

## What next?

We have solved a problem in a general way but what kind of problems does this set model lend itself to? What can't be done this way? Is there a better choice of model and can we adapt our monte-carlo tree search to work in that domain? We'll start thinking about some of these questions next time.

## Source

You can download the complete source for this example here: https://gist.github.com/jeremyorme/aaaf03535546bdbed2193c72ea3c03ca