24 Agnieszka Patyk we receive the following formula

$$\sum_{s\in S_{k}}\mathbb{P}[s\text{contains at least one of}\{f_{0},\tilde{n}_{1},\ldots,\tilde{n}_{l}\}]=|S_{k}|(1-(1-\frac{L+1}{2^{N}})^{k}).\tag{77}$$
$$\sum_{s\in S_{k}}\mathrm{P}[s\ \mathrm{conta}]$$

Thus, when probing for an answer of f0 + n˜1 + ··· + n˜L, L > 0, we are likely to receive an average of

$$1+\frac{(|S_{1}|-1)(L+1)}{2^{N}}+\sum_{k=2}^{\omega(V)}|S_{k}|(1-(1-\frac{L+1}{2^{N}})^{k})\tag{78}$$

potential answers.

Figure 13 shows test results compared with exact values given by Equation (78)
for noisy answers containing one meaningful blade and three noisy blades. Note that Equation (78) is also valid for right-hand-side questions.

The situation becomes more complex when we are to deal with answers having more than one blade. Although items in S1 are always chosen independently, we cannot say the same about items belonging to Si, 1 < i ≤ ω(V), since the sentence set is chosen by the experimenter. Let us consider the following question

$$(79)$$
$$(b i t e_{a g t}*F i d o+b i t e_{o b j}*P S m i t h)\nmid_{\sharp}b i t e_{o b j}$$
$$I_{\mathrm{O}}.$$
(biteagt ∗Fido+biteob j ∗PSmith) ] bite*ob j* (79)
yielding an answer of four blades name ∗Pat +sex ∗male+age ∗ 66+*bite*+
ob j ∗ biteagt ∗*Fido*. (80)
Clearly, the correct answer (*PSmith*) belongs to S3, but there is one other element of the clean-up memory listed in Table 1 that contains a portion of *PSmith*'s blades —
DogFido

$$({\boldsymbol{81}})$$
$$o+a g e*7+s e x$$
$$0m*p e t,$$

class ∗ animal +type ∗ dog+taste ∗ chickenlike
+name ∗Fido+age ∗ 7+sex ∗male+occupation ∗ pet, (81)
the common blade being sex ∗ *male*. We have two answers that under ideal conditions will surely result in a nonzero inner product: the correct answer in S3 and a potential answer in S7. By calculations analogous to those leading to Equation
(78), the average number of answers giving a nonzero inner product for the above example is

$$2+\frac{4|S_{1}|}{2^{N}}+\sum_{k\in\{3,7\}}(|S_{k}|-1)(1-(1-\frac{4}{2^{N}})^{k})+\sum_{k=2,k\notin\{3,7\}}^{\omega(V)}|S_{k}|(1-(1-\frac{4}{2^{N}})^{k}).\tag{82}$$

The number of meaningful blades in this example is odd, therefore Equation (82) is also valid for right-hand-side questions. Figure 14 shows test results compared with exact values computed by Equation (82).