Simulated Annealing is an optimization algorithm. It is well-suited for combinatoric problems.
Confusion Matrix Ordering
A confusion matrix contains the predictions of a classifier together with their ground truth. The cell \((i, j)\) gives the information how often class \(i\) was predicted to be class \(j\). So the diagonal \(i=j\) contains the number of samples where the classifier got it right.
In this matrix, the order of classes is irrelevant. You can change it. So instead of the order [mouse, cat, dog] you can have [cat, dog, mouse]. But for big matrices it might be interesting to get some more structure. For example, you might want to push the high elements close to the diagonal. This way, you can easier see the errors. An awesome side-effect is that classes which look similar to the classifier are closer together in the list.
Changing the order of the classes can itself be seen as an optimization problem. For example, a matrix can be scored with
I've described the details in Analysis and Optimization of Convolutional Neural Network Architectures, Chapter 5.2.
The simplest way to approach this problem is:
def optimize_brute_force(c, steps=1000):
# initialize
best_permutation = [i for i in range(len(c))]
best_score = calculate_score(best_permutation)
for _ in range(steps):
# Create new candidate
permutation = [i for i in range(len(c))]
random.shuffle(permutation)
# Score candidate
score = calculate_score(permutation)
# Replace if it improved
if score < best_score:
best_score = score
best_permutation = permutation
return best_permutation
Now, this is super inefficient. The reason is that the matrix has sub-structures which you destroy all the time. Instead of creating complete new permutations, you could simply swap two elements of the best permutation so far:
def optimize_greedy_swap(c, steps=1000):
# initialize
best_permutation = [i for i in range(len(c))]
best_score = calculate_score(best_permutation)
for _ in range(steps):
# Create new candidate
i, j = random.sample(best_permutation, 2)
permutation = swap(copy(best_permutation), i, j)
# Score candidate
score = calculate_score(permutation)
# Replace if it improved
if score < best_score:
best_score = score
best_permutation = permutation
return best_permutation
But this is still not good. This problem is of a structure that can lead to
local optima, where a single swap does not improve anything, but only two or
more swaps can. To fix this, one should allow at least equal solutions to change
the currently stored one. So if two solutions \(s_1, s_2\) both have a score of
3141, then the best_solution
should be allowed to switch between the two.
You will quickly figure out that this is not enough. You need to allow the algorithm to make the solution temporarily worse. Still, you want to improve overall. To weight this against each other we have simulated annealing.
To understand the idea, think about cooling metal. There are some constraints on how the atoms can be arranged and what is engergetically preferable / more stable. And while the metal is cooling, the atoms move less. To jump to a more stable setting, the setting has to be closer to the current setting.
In the algorithm, you could say that making the score worse (e.g. from 1234 to 3141), you have to be at the beginning of the optimization process. Some steps later, it might only be possible to make it worse from 1234 to 1300. Even later maybe only to 1250. And in the end it has to be strictly better.
How does that look in code?
def optimize_greedy_swap(c, steps=1000, temperature=1000, cooling_factor=0.995):
# initialize
best_permutation = [i for i in range(len(c))]
best_score = calculate_score(best_permutation)
for _ in range(steps):
# Create new candidate
i, j = random.sample(best_permutation, 2)
permutation = swap(copy(best_permutation), i, j)
# Score candidate
score = calculate_score(permutation)
# Replace if it improved
is_better = score < best_score
prob = exp(-(score - best_score) / temperature)
is_hot_enough = prob > random.random()
if is_better or is_hot_enough:
best_score = score
best_permutation = permutation
temperature *= cooling_factor
return best_permutation
For this specific problem, I noticed that the temperature-trick was way less important than allowing to swap blocks. So once in a while, I don't swap single elements but move a whole range.