By construction, fuzzers create inputs that may be hard to read. This causes issues during *debugging*, when a human has to analyze the exact cause of the failure. In this chapter, we present techniques that *automatically reduce and simplify failure-inducing inputs to a minimum* in order to ease debugging.

**Prerequisites**

- The simple "delta debugging" technique for reduction has no specific prerequisites.
- As reduction is typically used together with fuzzing, reading the chapter on basic fuzzing is a good idea.
- The later grammar-based techniques require knowledge on derivation trees and parsing.

To use the code provided in this chapter, write

```
>>> from fuzzingbook.Reducer import <identifier>
```

and then make use of the following features.

A *reducer* takes a failure-inducing input and reduces it to the minimum that still reproduces the failure. This chapter provides `Reducer`

classes that implement such reducers.

Here is a simple example: An arithmetic expression causes an error in the Python interpreter:

```
>>> !python -c 'x = 1 + 2 * 3 / 0'
Traceback (most recent call last):
File "<string>", line 1, in <module>
ZeroDivisionError: division by zero
```

Can we reduce this input to a minimum? To use a `Reducer`

, one first has to build a `Runner`

whose outcome is `FAIL`

if the precise error occurs. We therefore build a `ZeroDivisionRunner`

whose `run()`

method will specifically return a `FAIL`

outcome if a `ZeroDivisionError`

occurs.

```
>>> from Fuzzer import ProgramRunner
>>> class ZeroDivisionRunner(ProgramRunner):
>>> """Make outcome 'FAIL' if ZeroDivisionError occurs"""
>>> def run(self, inp=""):
>>> result, outcome = super().run(inp)
>>> if result.stderr.find('ZeroDivisionError') >= 0:
>>> outcome = 'FAIL'
>>> return result, outcome
```

If we feed this expression into a `ZeroDivisionRunner`

, it will produce an outcome of `FAIL`

as designed.

```
>>> python_input = "x = 1 + 2 * 3 / 0"
>>> python_runner = ZeroDivisionRunner("python")
>>> result, outcome = python_runner.run(python_input)
>>> outcome
'FAIL'
```

Delta Debugging is a simple and robust reduction algorithm. We can tie a `DeltaDebuggingReducer`

to this runner, and have it determine the substring that causes the `python`

program to fail:

```
>>> dd = DeltaDebuggingReducer(python_runner)
>>> dd.reduce(python_input)
'3/0'
```

The input is reduced to the maximum: We get the essence of the division by zero.

At this point, we have seen a number of test generation techniques that all in some form produce inputs in order to trigger failures. If they are successful – that is, the program actually fails – we must find out why the failure occurred and how to fix it.

`MysteryRunner`

class with a `run()`

method that – given its code – can occasionally fail. But under which circumstances does this actually happen? We have deliberately obscured the exact condition in order to make this non-obvious.

In [4]:

```
class MysteryRunner(Runner):
def run(self, inp):
x = inp.find(chr(0o17 + 0o31))
y = inp.find(chr(0o27 + 0o22))
if x >= 0 and y >= 0 and x < y:
return (inp, Runner.FAIL)
else:
return (inp, Runner.PASS)
```

Let us fuzz the function until we find a failing input.

In [5]:

```
mystery = MysteryRunner()
random_fuzzer = RandomFuzzer()
while True:
inp = random_fuzzer.fuzz()
result, outcome = mystery.run(inp)
if outcome == mystery.FAIL:
break
```

In [6]:

```
failing_input = result
failing_input
```

Out[6]:

' 7:,>((/$$-/->.;.=;(.%!:50#7*8=$&&=$9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#'

Something in this input causes `MysteryRunner`

to fail. But what is it?

One important step in the debugging process is *reduction* – that is, to identify those circumstances of a failure that are relevant for the failure to occur, and to *omit* (if possible) those parts that are not. As Kernighan and Pike \cite{Kernighan1999} put it:

For every circumstance of the problem, check whether it is relevant for the problem to occur. If it is not, remove it from the problem report or the test case in question.

Specifically for inputs, they suggest a *divide and conquer* process:

Proceed by binary search. Throw away half the input and see if the output is still wrong; if not, go back to the previous state and discard the other half of the input.

This is something we can easily try out. For instance, we can see whether the error still occurs if we only feed in the first half:

In [7]:

```
half_length = len(failing_input) // 2 # // is integer division
first_half = failing_input[:half_length]
mystery.run(first_half)
```

Out[7]:

(" 7:,>((/$$-/->.;.=;(.%!:50#7*8=$&&=$9!%6(4=&69':", 'PASS')

Nope – the first half alone does not suffice. Maybe the second half?

In [8]:

```
second_half = failing_input[half_length:]
mystery.run(second_half)
```

Out[8]:

('\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#', 'PASS')

*smaller chunks* – say, one character after another. If our test is deterministic and easily repeated, it is clear that this process eventually will yield a reduced input. But still, it is a rather inefficient process, especially for long inputs. What we need is a *strategy* that effectively minimizes a failure-inducing input – a strategy that can be automated.

*delta debugging* \cite{Zeller2002}. Delta Debugging implements the "binary search" strategy, as listed above, but with a twist: If neither half fails (also as above), it keeps on cutting away smaller and smaller chunks from the input, until it eliminates individual characters. Thus, after cutting away the first half, we cut away
the first quarter, the second quarter, and so on.

Let us illustrate this on our example, and see what happens if we cut away the first quarter.

In [9]:

```
quarter_length = len(failing_input) // 4
input_without_first_quarter = failing_input[quarter_length:]
mystery.run(input_without_first_quarter)
```

Out[9]:

('50#7*8=$&&=$9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#', 'FAIL')

Ah! This has failed, and reduced our failing input by 25%. Let's remove another quarter.

In [10]:

```
input_without_first_and_second_quarter = failing_input[quarter_length * 2:]
mystery.run(input_without_first_and_second_quarter)
```

Out[10]:

('\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#', 'PASS')

This is not too surprising, as we had that one before:

In [11]:

```
second_half
```

Out[11]:

'\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#'

In [12]:

```
input_without_first_and_second_quarter
```

Out[12]:

'\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#'

How about removing the third quarter, then?

In [13]:

```
input_without_first_and_third_quarter = failing_input[quarter_length:
quarter_length * 2] + failing_input[quarter_length * 3:]
mystery.run(input_without_first_and_third_quarter)
```

Out[13]:

("50#7*8=$&&=$9!%6(4=&69':<7+1<2!4$>92+$1<(3%&5''>#", 'PASS')

Ok. Let us remove the fourth quarter.

In [14]:

```
input_without_first_and_fourth_quarter = failing_input[quarter_length:quarter_length * 3]
mystery.run(input_without_first_and_fourth_quarter)
```

Out[14]:

('50#7*8=$&&=$9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+', 'FAIL')

Yes! This has succeeded. Our input is now 50% smaller.

However, this is comething we happily let a computer do for us. We first introduce a `Reducer`

class as an abstract superclass for all kinds of reducers. The `test()`

methods runs a single test (with logging, if so wanted); the `reduce()`

method will eventually reduce an input to the minimum.

In [15]:

```
class Reducer(object):
def __init__(self, runner, log_test=False):
"""Attach reducer to the given `runner`"""
self.runner = runner
self.log_test = log_test
self.reset()
def reset(self):
self.tests = 0
def test(self, inp):
result, outcome = self.runner.run(inp)
self.tests += 1
if self.log_test:
print("Test #%d" % self.tests, repr(inp), repr(len(inp)), outcome)
return outcome
def reduce(self, inp):
self.reset()
# Default: Don't reduce
return inp
```

The `CachingReducer`

variant saves test results, such that we don't have to run the same tests again and again:

In [16]:

```
class CachingReducer(Reducer):
def reset(self):
super().reset()
self.cache = {}
def test(self, inp):
if inp in self.cache:
return self.cache[inp]
outcome = super().test(inp)
self.cache[inp] = outcome
return outcome
```

Here comes the *Delta Debugging* reducer. Delta Debugging implements the strategy sketched above: It first removes larger chunks of size $\frac{1}{2}$; if this does not fail, then we proceed to chunks of size $\frac{1}{4}$, then $\frac{1}{8}$ and so on.

`Runner`

framework. The variable `n`

(initially 2) indicates the granularity – in each step, chunks of size $\frac{1}{n}$ are cut away. If none of the test fails (`some_complement_is_failing`

is False), then `n`

is doubled – until it reaches the length of the input.

In [17]:

```
class DeltaDebuggingReducer(CachingReducer):
def reduce(self, inp):
self.reset()
assert self.test(inp) != Runner.PASS
n = 2 # Initial granularity
while len(inp) >= 2:
start = 0
subset_length = len(inp) / n
some_complement_is_failing = False
while start < len(inp):
complement = inp[:int(start)] + \
inp[int(start + subset_length):]
if self.test(complement) == Runner.FAIL:
inp = complement
n = max(n - 1, 2)
some_complement_is_failing = True
break
start += subset_length
if not some_complement_is_failing:
if n == len(inp):
break
n = min(n * 2, len(inp))
return inp
```

To see how the `DeltaDebuggingReducer`

works, let us run it on our failing input. With each step, we see how the remaining input gets smaller and smaller, until only three characters remain:

In [18]:

```
dd_reducer = DeltaDebuggingReducer(mystery, log_test=True)
dd_reducer.reduce(failing_input)
```

Test #1 ' 7:,>((/$$-/->.;.=;(.%!:50#7*8=$&&=$9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#' 97 FAIL Test #2 '\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#' 49 PASS Test #3 " 7:,>((/$$-/->.;.=;(.%!:50#7*8=$&&=$9!%6(4=&69':" 48 PASS Test #4 '50#7*8=$&&=$9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+<7+1<2!4$>92+$1<(3%&5\'\'>#' 73 FAIL Test #5 "50#7*8=$&&=$9!%6(4=&69':<7+1<2!4$>92+$1<(3%&5''>#" 49 PASS Test #6 '50#7*8=$&&=$9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+' 48 FAIL Test #7 '\'<3+0-3.24#7=!&60)2/+";+' 24 PASS Test #8 "50#7*8=$&&=$9!%6(4=&69':" 24 PASS Test #9 '9!%6(4=&69\':\'<3+0-3.24#7=!&60)2/+";+' 36 FAIL Test #10 '9!%6(4=&69\':=!&60)2/+";+' 24 FAIL Test #11 '=!&60)2/+";+' 12 PASS Test #12 "9!%6(4=&69':" 12 PASS Test #13 '=&69\':=!&60)2/+";+' 18 PASS Test #14 '9!%6(4=!&60)2/+";+' 18 FAIL Test #15 '9!%6(42/+";+' 12 PASS Test #16 '9!%6(4=!&60)' 12 FAIL Test #17 '=!&60)' 6 PASS Test #18 '9!%6(4' 6 PASS Test #19 '6(4=!&60)' 9 FAIL Test #20 '6(460)' 6 FAIL Test #21 '60)' 3 PASS Test #22 '6(4' 3 PASS Test #23 '(460)' 5 FAIL Test #24 '460)' 4 PASS Test #25 '(0)' 3 FAIL Test #26 '0)' 2 PASS Test #27 '(' 1 PASS Test #28 '()' 2 FAIL Test #29 ')' 1 PASS

Out[18]:

'()'

Now we know why `MysteryRunner`

fails – it suffices that the input contains two matching parentheses with a number between them. Delta Debugging determines this is 29 steps. Its result is *1-minimal*, meaning that every character contained is required to produce the error; removing any (as seen in tests `#27`

and `#29`

, above) no longer makes the test fail. This property is guaranteed by the delta debugging algorithm, which in its last stage always tries to delete characters one by one.

A reduced test case such as the one above has many advantages:

A reduced test case

**reduces the**. The test case is shorter and focused, and thus does not burden the programmer with irrelevant details. A reduced input typically leads to shorter executions and smaller program states, both of which reduce the search space as it comes to understanding the bug. In our case, we have eliminated lots of irrelevant input – only the two characters the reduced input contains are relevant.*cognitive load*of the programmerA reduced test case

**is easier to communicate**. All one needs here is the summary:`MysteryRunner fails on "()"`

, which is much better than`MysteryRunner fails on a 4100-character input (attached)`

.A reduced test case helps in

**identifying duplicates**. If similar bugs have been reported already, and all of them have been reduced to the same cause (namely that the input contains matching parentheses), then it becomes obvious that all these bugs are different symptoms of the same underlying cause – and would all be resolved at once with one code fix.

How effective is delta debugging? In the best case (when the left half or the right half fails), the number of tests is logarithmic proportional to the length $n$ of an input (i.e., $O(\log_2 n)$); this is the same complexity as binary search. In the worst case, though, delta debugging can require a number of tests proportional to $n^2$ (i.e., $O(n^2)$) – this happens in the case when we are down to character granularity, and we have to repeatedly tried to delete all characters, only to find that deleting the last character results in a failure \cite{Zeller2002}. (This is a pretty pathological situation, though.)

If the input language is syntactically complex, delta debugging may take several attempts at reduction, and may not be able to reduce inputs at all. In the second half of this chapter, we thus introduce an algorithm named *Grammar-Based Reduction* (or GRABR for short) that makes use of *grammars* to reduce syntactically complex inputs.

Despite its general robustness, there are situations in which delta debugging might be inefficient or outright fail. As an example, consider some *expression input* such as `1 + (2 * 3)`

. Delta debugging requires a number of tests to simplify the failure-inducing input, but it eventually returns a minimal input

In [19]:

```
expr_input = "1 + (2 * 3)"
dd_reducer = DeltaDebuggingReducer(mystery, log_test=True)
dd_reducer.reduce(expr_input)
```

Test #1 '1 + (2 * 3)' 11 FAIL Test #2 '2 * 3)' 6 PASS Test #3 '1 + (' 5 PASS Test #4 '+ (2 * 3)' 9 FAIL Test #5 '+ ( 3)' 6 FAIL Test #6 ' 3)' 3 PASS Test #7 '+ (' 3 PASS Test #8 ' ( 3)' 5 FAIL Test #9 '( 3)' 4 FAIL Test #10 '3)' 2 PASS Test #11 '( ' 2 PASS Test #12 '(3)' 3 FAIL Test #13 '()' 2 FAIL Test #14 ')' 1 PASS Test #15 '(' 1 PASS

Out[19]:

'()'

Looking at the tests, above, though, only few of them actually represent syntactically valid arithmetic expressions. In a practical setting, we may want to test a program which actually *parses* such expressions, and which would *reject* all invalid inputs. We define a class `EvalMysteryRunner`

which first *parses* the given input (according to the rules of our expression grammar), and *only* if it fits would it be passed to our original `MysteryRunner`

. This simulates a setting in which we test an expression interpreter, and in which only valid inputs can trigger the bug.

In [22]:

```
class EvalMysteryRunner(MysteryRunner):
def __init__(self):
self.parser = EarleyParser(EXPR_GRAMMAR)
def run(self, inp):
try:
tree, *_ = self.parser.parse(inp)
except SyntaxError as exc:
return (inp, Runner.UNRESOLVED)
return super().run(inp)
```

In [23]:

```
eval_mystery = EvalMysteryRunner()
```

In [24]:

```
dd_reducer = DeltaDebuggingReducer(eval_mystery, log_test=True)
dd_reducer.reduce(expr_input)
```

Test #1 '1 + (2 * 3)' 11 FAIL Test #2 '2 * 3)' 6 UNRESOLVED Test #3 '1 + (' 5 UNRESOLVED Test #4 '+ (2 * 3)' 9 UNRESOLVED Test #5 '1 2 * 3)' 8 UNRESOLVED Test #6 '1 + ( 3)' 8 UNRESOLVED Test #7 '1 + (2 *' 8 UNRESOLVED Test #8 ' + (2 * 3)' 10 UNRESOLVED Test #9 '1+ (2 * 3)' 10 UNRESOLVED Test #10 '1 (2 * 3)' 9 UNRESOLVED Test #11 '1 + 2 * 3)' 10 UNRESOLVED Test #12 '1 + ( * 3)' 10 UNRESOLVED Test #13 '1 + (2 3)' 9 UNRESOLVED Test #14 '1 + (2 *3)' 10 UNRESOLVED Test #15 '1 + (2 * ' 9 UNRESOLVED Test #16 '1 (2 * 3)' 10 UNRESOLVED Test #17 '1 +(2 * 3)' 10 UNRESOLVED Test #18 '1 + (2* 3)' 10 UNRESOLVED Test #19 '1 + (2 3)' 10 UNRESOLVED Test #20 '1 + (2 * )' 10 UNRESOLVED Test #21 '1 + (2 * 3' 10 UNRESOLVED

Out[24]:

'1 + (2 * 3)'

This behavior is possible if the program under test has several constraints regarding input validity. Delta debugging is not aware of these constraints (nor of the input structure in general), so it might violate these constraints again and again.

To reduce inputs with high syntactical complexity, we use another approach: Rather than reducing the input string, we reduce the *tree* representing its structure. The general idea is to start with a *derivation tree* coming from parsing the input, and then *substitute subtrees by smaller subtrees of the same type*. These alternate subtrees can either come

- From the tree itself, or
- By applying an alternate grammar expansion using elements from the tree.

Let us show these two strategies using an example. We start with a derivation tree from an arithmetic expression:

In [26]:

```
derivation_tree, *_ = EarleyParser(EXPR_GRAMMAR).parse(expr_input)
display_tree(derivation_tree)
```

Out[26]:

`<expr>`

symbol up in the tree with some `<expr>`

subtree down in the tree. For instance, we could replace the uppermost `<expr>`

with its right `<expr>`

subtree, yielding the string `(2 + 3)`

:

In [28]:

```
new_derivation_tree = copy.deepcopy(derivation_tree)
# We really should have some query language
sub_expr_tree = new_derivation_tree[1][0][1][2]
display_tree(sub_expr_tree)
```

Out[28]:

In [29]:

```
new_derivation_tree[1][0] = sub_expr_tree
display_tree(new_derivation_tree)
```

Out[29]:

In [30]:

```
all_terminals(new_derivation_tree)
```

Out[30]:

'(2 * 3)'

Replacing one subtree by another only works as long as individual elements such as `<expr>`

occur multiple times in our tree. In the reduced `new_derivation_tree`

, above, we could replace further `<expr>`

trees only once more.

*alternative expansions*. That is, for a symbol, we check whether there is an alternative expansion with a smaller number of children. Then, we replace the symbol with the alternative expansion, filling in needed symbols from the tree.

As an example, consider the `new_derivation_tree`

above. The applied expansion for `<term>`

has been

```
<term> ::= <term> * <factor>
```

Lat us replace this with the alternative expansion:

`<term> ::= <factor>`

In [31]:

```
term_tree = new_derivation_tree[1][0][1][0][1][0][1][1][1][0]
display_tree(term_tree)
```

Out[31]:

In [32]:

```
shorter_term_tree = term_tree[1][2]
display_tree(shorter_term_tree)
```

Out[32]:

In [33]:

```
new_derivation_tree[1][0][1][0][1][0][1][1][1][0] = shorter_term_tree
display_tree(new_derivation_tree)
```

Out[33]:

In [34]:

```
all_terminals(new_derivation_tree)
```

Out[34]:

'(3)'

We introduce the `GrammarReducer`

class, which is again a `Reducer`

. Note that we derive from `CachingReducer`

, as the strategy will produce several duplicates.

In [35]:

```
class GrammarReducer(CachingReducer):
def __init__(self, runner, parser, log_test=False, log_reduce=False):
super().__init__(runner, log_test=log_test)
self.parser = parser
self.grammar = parser.grammar()
self.start_symbol = parser.start_symbol()
self.log_reduce = log_reduce
self.try_all_combinations = False
```

We define a number of helper functions, which we will need for our strategy. `tree_list_to_string()`

does what the name suggest, creating a string from a list of derivation trees:

In [36]:

```
def tree_list_to_string(q):
return "[" + ", ".join([all_terminals(tree) for tree in q]) + "]"
```

In [37]:

```
tree_list_to_string([derivation_tree, derivation_tree])
```

Out[37]:

'[1 + (2 * 3), 1 + (2 * 3)]'

`possible_combinations()`

takes a list of lists $[[x_1, x_2], [y_1, y_2], \dots]$ and creates a list of combinations $[[x_1, y_1], [x_1, y_2], [x_2, y_1], [x_2, y_2], \dots]$.

In [38]:

```
def possible_combinations(list_of_lists):
if len(list_of_lists) == 0:
return []
ret = []
for e in list_of_lists[0]:
if len(list_of_lists) == 1:
ret.append([e])
else:
for c in possible_combinations(list_of_lists[1:]):
new_combo = [e] + c
ret.append(new_combo)
return ret
```

In [39]:

```
possible_combinations([[1, 2], ['a', 'b']])
```

Out[39]:

[[1, 'a'], [1, 'b'], [2, 'a'], [2, 'b']]

The functions `number_of_nodes()`

and `max_height()`

return the number of nodes and the maximum height of the given tree, respectively.

In [40]:

```
def number_of_nodes(tree):
(symbol, children) = tree
return 1 + sum([number_of_nodes(c) for c in children])
```

In [41]:

```
number_of_nodes(derivation_tree)
```

Out[41]:

25

In [42]:

```
def max_height(tree):
(symbol, children) = tree
if len(children) == 0:
return 1
return 1 + max([max_height(c) for c in children])
```

In [43]:

```
max_height(derivation_tree)
```

Out[43]:

12

Let us now implement our two simplification strategies – replacing subtrees and alternate expansions.

The method `subtrees_with_symbol()`

returns all subtrees in the given tree whose root is equal to the given symbol. If `ignore_root`

is set (default), then the root node of `tree`

is not compared against. (The `depth`

parameter will be discussed below.)

In [44]:

```
class GrammarReducer(GrammarReducer):
def subtrees_with_symbol(self, tree, symbol, depth=-1, ignore_root=True):
# Find all subtrees in TREE whose root is SYMBOL.
# If IGNORE_ROOT is true, ignore the root note of TREE.
ret = []
(child_symbol, children) = tree
if depth <= 0 and not ignore_root and child_symbol == symbol:
ret.append(tree)
# Search across all children
if depth != 0 and children is not None:
for c in children:
ret += self.subtrees_with_symbol(c,
symbol,
depth=depth - 1,
ignore_root=False)
return ret
```

Here's an example: These are all subtrees with `<term>`

in our derivation tree `derivation_tree`

.

In [45]:

```
grammar_reducer = GrammarReducer(
mystery,
EarleyParser(EXPR_GRAMMAR),
log_reduce=True)
```

In [46]:

```
all_terminals(derivation_tree)
```

Out[46]:

'1 + (2 * 3)'

In [47]:

```
[all_terminals(t) for t in grammar_reducer.subtrees_with_symbol(
derivation_tree, "<term>")]
```

Out[47]:

['1', '(2 * 3)', '2 * 3', '3']

`<term>`

subtrees to simplify the tree, these are the subtrees we could replace them with.

`subtrees_with_symbols()`

, above). We then pick the first possible combination (or *all* combinations, if the attribute `try_all_combinations`

is set).

In [48]:

```
class GrammarReducer(GrammarReducer):
def alternate_reductions(self, tree, symbol, depth=-1):
reductions = []
expansions = self.grammar.get(symbol, [])
expansions.sort(
key=lambda expansion: len(
expansion_to_children(expansion)))
for expansion in expansions:
expansion_children = expansion_to_children(expansion)
match = True
new_children_reductions = []
for (alt_symbol, _) in expansion_children:
child_reductions = self.subtrees_with_symbol(
tree, alt_symbol, depth=depth)
if len(child_reductions) == 0:
match = False # Child not found; cannot apply rule
break
new_children_reductions.append(child_reductions)
if not match:
continue # Try next alternative
# Use the first suitable combination
for new_children in possible_combinations(new_children_reductions):
new_tree = (symbol, new_children)
if number_of_nodes(new_tree) < number_of_nodes(tree):
reductions.append(new_tree)
if not self.try_all_combinations:
break
# Sort by number of nodes
reductions.sort(key=number_of_nodes)
return reductions
```

In [49]:

```
grammar_reducer = GrammarReducer(
mystery,
EarleyParser(EXPR_GRAMMAR),
log_reduce=True)
```

In [50]:

```
all_terminals(derivation_tree)
```

Out[50]:

'1 + (2 * 3)'

Here's *all* combinations for `<term>`

:

In [51]:

```
grammar_reducer.try_all_combinations = True
print([all_terminals(t)
for t in grammar_reducer.alternate_reductions(derivation_tree, "<term>")])
```

The default, though, is simply to return the first of these:

In [52]:

```
grammar_reducer.try_all_combinations = False
[all_terminals(t) for t in grammar_reducer.alternate_reductions(
derivation_tree, "<term>")]
```

Out[52]:

['1', '1 * 1']

`subtrees_with_symbol()`

); then we go for alternate expansions (using `alternate_expansions()`

).

In [53]:

```
class GrammarReducer(GrammarReducer):
def symbol_reductions(self, tree, symbol, depth=-1):
"""Find all expansion alternatives for the given symbol"""
reductions = (self.subtrees_with_symbol(tree, symbol, depth=depth)
+ self.alternate_reductions(tree, symbol, depth=depth))
# Filter duplicates
unique_reductions = []
for r in reductions:
if r not in unique_reductions:
unique_reductions.append(r)
return unique_reductions
```

In [54]:

```
grammar_reducer = GrammarReducer(
mystery,
EarleyParser(EXPR_GRAMMAR),
log_reduce=True)
```

In [55]:

```
all_terminals(derivation_tree)
```

Out[55]:

'1 + (2 * 3)'

`<expr>`

nodes. Note how we first return subtrees (`1 + (2 * 3)`

, `(2 * 3)`

, `2 * 3`

) before going for alternate expansions of `<expr>`

(`1`

).

In [56]:

```
reductions = grammar_reducer.symbol_reductions(derivation_tree, "<expr>")
tree_list_to_string([r for r in reductions])
```

Out[56]:

'[1 + (2 * 3), (2 * 3), 2 * 3, 1]'

`<term>`

nodes. Again, we first have subtrees of the derivation tree, followed by the alternate expansion `1 * 1`

.

In [57]:

```
reductions = grammar_reducer.symbol_reductions(derivation_tree, "<term>")
tree_list_to_string([r for r in reductions])
```

Out[57]:

'[1, (2 * 3), 2 * 3, 3, 1 * 1]'

We are now able to return a number of alternatives for each symbol in the tree. This is what we apply in the core function of our reduction strategy, `reduce_subtree()`

. Starting with `subtree`

, for every child, we find possible reductions. For every reduction, we replace the child with the reduction and test the resulting (full) tree. If it fails, our reduction was successful; otherwise, we put the child back into place and try out the next reduction. Eventually, we apply `reduce_subtree()`

on all children, reducing these as well.

In [58]:

```
class GrammarReducer(GrammarReducer):
def reduce_subtree(self, tree, subtree, depth=-1):
symbol, children = subtree
if len(children) == 0:
return False
if self.log_reduce:
print("Reducing", all_terminals(subtree), "with depth", depth)
reduced = False
while True:
reduced_child = False
for i, child in enumerate(children):
(child_symbol, _) = child
for reduction in self.symbol_reductions(
child, child_symbol, depth):
if number_of_nodes(reduction) >= number_of_nodes(child):
continue
# Try this reduction
if self.log_reduce:
print(
"Replacing",
all_terminals(
children[i]),
"by",
all_terminals(reduction))
children[i] = reduction
if self.test(all_terminals(tree)) == Runner.FAIL:
# Success
if self.log_reduce:
print("New tree:", all_terminals(tree))
reduced = reduced_child = True
break
else:
# Didn't work out - restore
children[i] = child
if not reduced_child:
if self.log_reduce:
print("Tried all alternatives for", all_terminals(subtree))
break
# Run recursively
for c in children:
if self.reduce_subtree(tree, c, depth):
reduced = True
return reduced
```

All we now need is a few drivers. The method `reduce_tree()`

is the main entry point into `reduce_subtree()`

:

In [59]:

```
class GrammarReducer(GrammarReducer):
def reduce_tree(self, tree):
return self.reduce_subtree(tree, tree)
```

The custom method `parse()`

turns a given input into a derivation tree:

In [60]:

```
class GrammarReducer(GrammarReducer):
def parse(self, inp):
tree, *_ = self.parser.parse(inp)
if self.log_reduce:
print(all_terminals(tree))
return tree
```

The method `reduce()`

is the one single entry point, parsing the input and then reducing it.

In [61]:

```
class GrammarReducer(GrammarReducer):
def reduce(self, inp):
tree = self.parse(inp)
self.reduce_tree(tree)
return all_terminals(tree)
```

`expr_input`

and the `mystery()`

function. How quickly can we reduce it?

In [62]:

```
expr_input
```

Out[62]:

'1 + (2 * 3)'

In [63]:

```
grammar_reducer = GrammarReducer(
eval_mystery,
EarleyParser(EXPR_GRAMMAR),
log_test=True)
grammar_reducer.reduce(expr_input)
```

Test #1 '(2 * 3)' 7 FAIL Test #2 '2 * 3' 5 PASS Test #3 '3' 1 PASS Test #4 '2' 1 PASS Test #5 '(3)' 3 FAIL

Out[63]:

'(3)'

Success! In only five steps, our `GrammarReducer`

reduces the input to the minimum that causes the failure. Note how all tests are syntactically valid by construction, avoiding the `UNRESOLVED`

outcomes that cause delta debugging to stall.

`#2`

, where the input (tree) is reduced to `2 * 3`

, our `GrammarReducer`

first tries to replace the tree with `2`

and `3`

, which are the alternate `<term>`

subtrees. This may work, of course; but if there are many possible subtrees, our strategy will spend quite some time trying one after the other.

Delta debugging, as introduced above, follows the idea of trying to cut inputs approximately in half, and thus quickly proceeds towards a minimal input. By replacing a tree with much smaller subtrees, we *could* possibly reduce a tree significantly, but may need several attempts to do so. A better strategy is to only consider *large* subtrees first – both for the subtree replacement as well as for alternate expansions. To find such *large* subtrees, we limit the *depth* by which we search for possible replacements in the subtree – first, by looking at the direct descendants, later at lower descendants.

`depth`

parameter used in `subtrees_with_symbol()`

and passed through the invoking functions. If set, *only* symbols at the given depth are returned. Here's an example, starting again with our derivation tree `derivation_tree`

:

In [64]:

```
grammar_reducer = GrammarReducer(
mystery,
EarleyParser(EXPR_GRAMMAR),
log_reduce=True)
```

In [65]:

```
all_terminals(derivation_tree)
```

Out[65]:

'1 + (2 * 3)'

In [66]:

```
display_tree(derivation_tree)
```

Out[66]:

At a depth of 1, there is no `<term>`

symbol:

In [67]:

```
[all_terminals(t) for t in grammar_reducer.subtrees_with_symbol(
derivation_tree, "<term>", depth=1)]
```

Out[67]:

[]

At a depth of 2, we have the `<term>`

subtree on the left hand side:

In [68]:

```
[all_terminals(t) for t in grammar_reducer.subtrees_with_symbol(
derivation_tree, "<term>", depth=2)]
```

Out[68]:

['1']

At a depth of 3, we have the `<term>`

subtree on the right hand side:

In [69]:

```
[all_terminals(t) for t in grammar_reducer.subtrees_with_symbol(
derivation_tree, "<term>", depth=3)]
```

Out[69]:

['(2 * 3)']

The idea is now to start with a depth of 0, subsequently increasing it as we proceed:

In [70]:

```
class GrammarReducer(GrammarReducer):
def reduce_tree(self, tree):
depth = 0
while depth < max_height(tree):
reduced = self.reduce_subtree(tree, tree, depth)
if reduced:
depth = 0 # Start with new tree
else:
depth += 1 # Extend search for subtrees
return tree
```

In [71]:

```
grammar_reducer = GrammarReducer(
mystery,
EarleyParser(EXPR_GRAMMAR),
log_test=True)
grammar_reducer.reduce(expr_input)
```

Test #1 '(2 * 3)' 7 FAIL Test #2 '(3)' 3 FAIL Test #3 '3' 1 PASS

Out[71]:

'(3)'

We see that a depth-oriented strategy needs even fewer steps in our setting.

We close by demonstrating the difference between text-based delta debugging and our grammar-based reduction. We build a very long expression:

In [73]:

```
long_expr_input = GrammarFuzzer(EXPR_GRAMMAR, min_nonterminals=100).fuzz()
long_expr_input
```

Out[73]:

'++---((-2 / 3 / 3 - -+1 / 5 - 2) * ++6 / +8 * 4 / 9 / 2 * 8 + ++(5) * 3 / 8 * 0 + 3 * 3 + 4 / 0 / 6 + 9) * ++++(+--9 * -3 * 7 / 4 + --(4) / 3 - 0 / 3 + 5 + 0) * (1 * 6 - 1 / 9 * 5 - 9 / 0 + 7) * ++(8 - 1) * +1 * 7 * 0 + ((1 + 4) / 4 * 8 * 9 * 4 + 4 / (4) * 1 - (4) * 8 * 5 + 1 + 4) / (+(2 - 1 - 9) * 5 + 3 + 6 - 2) * +3 * (3 - 7 + 8) / 4 - -(9 * 4 - 1 * 0 + 5) / (5 / 9 * 5 + 2) * 7 + ((7 - 5 + 3) / 1 * 8 - 8 - 9) * --+1 * 4 / 4 - 4 / 7 * 4 - 3 / 6 * 1 - 2 - 7 - 8'

With grammars, we need only a handful of tests to find the failure-inducing input:

In [75]:

```
grammar_reducer = GrammarReducer(eval_mystery, EarleyParser(EXPR_GRAMMAR))
with Timer() as grammar_time:
print(grammar_reducer.reduce(long_expr_input))
```

(9)

In [76]:

```
grammar_reducer.tests
```

Out[76]:

10

In [77]:

```
grammar_time.elapsed_time()
```

Out[77]:

1.603279961971566

In [78]:

```
dd_reducer = DeltaDebuggingReducer(eval_mystery)
with Timer() as dd_time:
print(dd_reducer.reduce(long_expr_input))
```

((2 - 1 - 2) * 8 + (5) - (4)) / ((2) * 3) * (9) / 3 / 1 - 8

In [79]:

```
dd_reducer.tests
```

Out[79]:

900

In [80]:

```
dd_time.elapsed_time()
```

Out[80]:

45.202309913001955

A *reducer* takes a failure-inducing input and reduces it to the minimum that still reproduces the failure. This chapter provides `Reducer`

classes that implement such reducers.

Here is a simple example: An arithmetic expression causes an error in the Python interpreter:

In [81]:

```
!python -c 'x = 1 + 2 * 3 / 0'
```

`Reducer`

, one first has to build a `Runner`

whose outcome is `FAIL`

if the precise error occurs. We therefore build a `ZeroDivisionRunner`

whose `run()`

method will specifically return a `FAIL`

outcome if a `ZeroDivisionError`

occurs.

In [83]:

```
class ZeroDivisionRunner(ProgramRunner):
"""Make outcome 'FAIL' if ZeroDivisionError occurs"""
def run(self, inp=""):
result, outcome = super().run(inp)
if result.stderr.find('ZeroDivisionError') >= 0:
outcome = 'FAIL'
return result, outcome
```

`ZeroDivisionRunner`

, it will produce an outcome of `FAIL`

as designed.

In [84]:

```
python_input = "x = 1 + 2 * 3 / 0"
python_runner = ZeroDivisionRunner("python")
result, outcome = python_runner.run(python_input)
outcome
```

Out[84]:

'FAIL'

`DeltaDebuggingReducer`

to this runner, and have it determine the substring that causes the `python`

program to fail:

In [85]:

```
dd = DeltaDebuggingReducer(python_runner)
dd.reduce(python_input)
```

Out[85]:

'3/0'

The input is reduced to the maximum: We get the essence of the division by zero.

- Reducing failure-inducing inputs to a minimum is helpful for testing and debugging.
*Delta debugging*is a simple and robust algorithm to easily reduce test cases.- For syntactically complex inputs,
*grammar-based reduction*is much faster and yields better results.

Our next chapter focuses on Web GUI Fuzzing, another domain where generating and reducing test cases is central.

The delta debugging algorithm discussed here stems from \cite{Zeller2002}; actually, this is the exact Python implementation as used by Zeller in 2002. The idea of systematically reducing inputs has been discovered a number of times, although not as automatic and generic as delta debugging. \cite{Slutz1998}, for instance, discusses systematic reduction of SQL statements for SQL databases; the general process as manual work is well described by \cite{Kernighan1999}.

The deficits of delta debugging as it comes to syntactically complex inputs were first discussed in *compiler testing*, and *reducing tree inputs* rather than string inputs was quickly discovered as an alternative. *Hierarchical Delta Debugging* (*HDD*) \cite{Misherghi2006} applies delta debugging on subtrees of a parse tree, systematically reducing a parse tree to a minimum. *Generalized Tree Reduction* \cite{Herfert2017} generalizes this idea to apply arbitrary *patterns* such as replacing a term by a compatible term in a subtree, as `subtrees_with_symbol()`

does.

None of the existing approaches, though, makes use of grammars to apply expansion alternatives (as `alternate_reductions()`

does); nor do they strive towards replacing larger subtrees first (as `reduce_tree()`

does). The present chapter thus makes an original contribution.

While `GrammarReducer`

is a generic approach that can be parameterized with an arbitrary grammar, *language-specific* approaches can do a much better job for the language at hand. *C-Reduce* \cite{Regehr2012} is a reducer specifically targeting the reduction of programming languages. Besides reductions in the style of delta debugging or tree transformations, C-Reduce comes with more than 30 source-to-source transformations that replace aggregates by scalars, remove function parameters at a definition and all call sites, change functions to return `void`

and deleting all `return`

statements, and many more. While specifically instantiated for the C language (and used for testing C compilers), these principles extend to arbitrary programming languages following an ALGOL-like syntax. When testing a compiler, C-Reduce is the tool to go for.

This blog post by David McIver contains lots of insights on how to apply reduction in practice, in particular multiple runs with different abstraction levels.

How to best reduce inputs is still an underdeveloped field of research, with lots of opportunities.

When fuzzing with a population, it can be useful to occasionally *reduce* the length of each element, such that future descendants are shorter, too, which typically speeds up their testing.

Consider the `MutationFuzzer`

class from the chapter on mutation-based fuzzing.
Extend it such that whenever a new input is added to the population, it is first reduced using delta debugging.

Grammar-based input reduction, as sketched above, might be a good algorithm, but is by no means the only alternative. One interesting question is whether "reduction" should only be limited to elements already present, or whether one would be allowed to also create *new* elements. These would not be present in the original input, yet still allow to produce a much smaller input that would still reproduce the original failure.

As an example, consider the following grammar:

```
<number> ::= <float> | <integer> | <not-a-number>
<float> ::= <digits>.<digits>
<integer> ::= <digits>
<not-a-number> ::= NaN
<digits> ::= [0-9]+
```

Assume the input `100.99`

fails. We might be able to reduce it to a minimum of, say, `1.9`

. However, we cannot reduce it to an `<integer>`

or to `<not-a-number>`

, as these symbols do not occur in the original input. By allowing to *create* alternatives for these symbols, we could also tests inputs such as `1`

or `NaN`

and further generalize the class of inputs for which the program fails.

Create a class `GenerativeGrammarReducer`

as subclass of `GrammarReducer`

; extend the method `reduce_subtree()`

accordingly.

Create a *benchmark* for the grammars already defined earlier, consisting of:

- A set of
*inputs*, produced from these very grammars using`GrammarFuzzer`

and derivatives; - A set of
*tests*which check for the occurrence of individual symbols as well as pairs and triples of these symbols:- Tests should be
*unresolved*if the input is not syntactically valid; - Tests should
*fail*if the symbols (or pairs or triples thereof) occur; - Tests should
*pass*in all other cases.

- Tests should be

Compare delta debugging and grammar-based debugging on the benchmark. Implement HDD \cite{Misherghi2006} and *Generalized Tree Reduction* \cite{Herfert2017} and add them to your comparison. Which approach performs best, and under which circumstances?