In this section we analyze the complexity of \(\mathbf {MColl}\). First, we discuss complexity intuitively in Sect. 5.1 and then give formal arguments and proofs in Sect. 5.2.
5.1 Intuitive Analysis
We intuitively discuss the complexity of our algorithm. In the following, we show that \(\mathbf {MColl}(H,l)\) finds that an l-collision with memory complexity is approximately \(N^{1/3}\) and the expected quantum query complexity is at most approximately \(l! \cdot N^{ (3^{l-1}-1) / (2 \cdot 3^{l-1}) } \).
First, we consider memory complexity. The claim obviously holds for \(l = 1\). In the case \(l \ge 2\), the algorithm uses memory only for storing the list L. The memory size needed for L is \(N^{1/3^{l-1}}\), which is less than or equal to \(N^{1/3}\). Thus, the memory complexity is at most \(N^{1/3}\).
Next, we consider quantum query complexity. We upper bound the expected number of quantum queries by approximately
$$\begin{aligned} Q_l := (N/P_l) \cdot l! \cdot N^{ (3^{l-1}-1) / (2 \cdot 3^{l-1}) }, \end{aligned}$$
where \(P_l\) is the number of the points in Y that have at least l preimages for a fixed H. Regarding \((N/P_l)\) as constant, we obtain the desired bound.
The claim obviously holds for \(l=1\). Assume that the claim holds for \((l-1)\). Since Step 3 makes \(N^{1/{3^{l-1}}}\)-times calls of \(\mathbf {MColl}(H,l-1)\), the number of queries made in operating Step 3 once is approximately
$$\begin{aligned} N^{1/{3^{l-1}}} \cdot Q_{l-1}&= N^{1/{3^{l-1}}} \cdot \left( (N / P_{l-1}) \cdot (l-1)! \cdot N^{(3^{l-2} - 1) / (2 \cdot 3^{l-2})} \right) \nonumber \\&= (N / P_{l-1}) \cdot (l-1)! \cdot N^{(3^{l-1} - 1) / (2 \cdot 3^{l-1})}. \end{aligned}$$
(1)
Note that \(\mathbf {BBHT}(F)\) finds an element x that satisfies \(F(x)=1\) with approximately \(\sqrt{1/p}\) queries to F, where \(p = \Pr _{x \leftarrow X}[F(x)=1]\). Since L contains \(N^{1/{3^{l-1}}}\) elements, here we approximately argue that \(p \approx N^{1/{3^{l-1}}} / |X | \approx N^{(1 - 3^{l-1} ) /3^{l-1} }\) and thus the number of queries to F is approximately \(\sqrt{1/p}\), which is further approximated to \(N^{(3^{l-1} - 1) / ( 2 \cdot 3^{l-1})}\). From the construction of F, the number of queries to H in Step 8 is twice the number of queries to F (see Fig. 3), so the number of queries to H is
$$\begin{aligned} 2 \cdot N^{(3^{l-1} - 1) / ( 2 \cdot 3^{l-1})}. \end{aligned}$$
(2)
Summing up the numbers of queries in Steps 3 and 8 in Eqs. (1) and (2), we obtain the number of queries to H in the case that \(\mathbf {MColl}(H,l)\) does not stop in Steps 5 or 9 as
$$\begin{aligned} \left( (N / P_{l-1}) \cdot (l-1)! + 2 \right) \cdot N^{ (3^{l-1}-1) / (2 \cdot 3^{l-1})} \approx (N / P_{l-1}) \cdot (l-1)! \cdot N^{ (3^{l-1}-1) / (2 \cdot 3^{l-1})}. \end{aligned}$$
(3)
Now let q denote the probability that \(\mathbf {MColl}(H,l)\) outputs without being terminated at Steps 5 or 9. Then the overall quantum query complexity is approximately \((1/q) \cdot \left( (N / P_{l-1}) \cdot (l-1)! \cdot N^{ (3^{l-1}-1) / 2 \cdot 3^{l-1} } \right) \). We assume that q equals the probability that an l-collision is outputted in Step 9, since the probability that Step 5 finds a duplication in L is very small when l is a small constant, and ignoring Step 6 only decreases q and increases the overall complexity. Intuitively, we can assume that q equals the product of two probabilities in Step 9:
-
1.
The probability that the \((l-1)\)-collision \(\{ x^{(i_0)}_1, x^{(i_0)}_2, \dots , x^{(i_0)}_{l-1} \}\) can be extended to an l-collision.
-
2.
The probability that \(\tilde{x} \not \in \{ x^{(i_0)}_1, x^{(i_0)}_2, \dots , x^{(i_0)}_{l-1} \}\) holds, under the condition in which \(\{ x^{(i_0)}_1, x^{(i_0)}_2, \dots , x^{(i_0)}_{l-1} \}\) can be extended to an l-collision.
The probability of 1. is approximately \(P_l / P_{l-1}\), and the probability of 2. is lower bounded by 1 / l. Thus, we have \(q \ge (P_l / P_{l-1}) \cdot (1/l)\). Consequently, we have overall approximated complexity
$$\begin{aligned} (1/q) \cdot \left( (N / P_{l-1}) \cdot (l-1)! \cdot N^{ (3^{l-1}-1) /( 2 \cdot 3^{l-1} )} \right) , \end{aligned}$$
which is at most
$$\begin{aligned} Q_l = (N/P_{l}) \cdot l! \cdot N^{ (3^{l-1}-1) / 2 \cdot 3^{l-1} }. \end{aligned}$$
This validates the claim.
5.2 Precise Analysis
The discussion in the previous section is very informal with many approximations. This section gives the precise bound and proof. The main theorem in this section is as follows:
Theorem 5.1
Let X and Y be finite sets with \(|Y | = N\) and \(|X | \ge l \cdot |Y |\). Let \(H :X \rightarrow Y\) be an arbitrary function. For \(l \ge 1\), \(\mathbf {MColl}(H,l)\) finds an l-collision with expected quantum query complexity at most
$$\begin{aligned} \left( 1 + 18\sqrt{2}e \right) \cdot \left( \frac{2lN^{1/3}}{2lN^{1/3} - 1} \right) ^{l-1} \cdot l \cdot l! \cdot N^{ (3^{l-1}-1) / (2 \cdot 3^{l-1}) } \end{aligned}$$
and memory complexity \(N^{1/3}\).
Remark 5.1
Expected time complexity of \(\mathbf {MColl}(H,l)\) is upper bounded by the product of expected quantum query complexity and \(O(T_H + \lg N)\), where \(T_H\) is the time needed to make a quantum query to H once, using \(O(N^{1/3})\) qubits. See Sect. 6 for details.
Proof
It suffices to show that the claim holds in the case \(|X | = l \cdot |Y |\). The proof for memory complexity is the same as that we described in the previous section. In the following, we consider quantum query complexity.
For \(l \ge 1\), define \(A_l\) as
$$\begin{aligned} A_l := \text {the total number of quantum queries to }H \text { that }\mathbf {MColl}(H,l) \text { makes}, \end{aligned}$$
and for \(l \ge 2\), define \(B_l, C_l\) as
$$\begin{aligned} B_l&:= \text {the number of quantum queries to }H \text { made in Step 3}, \\ C_l&:= \text {the number of quantum queries to }H \text { made in Step 8}. \end{aligned}$$
For \(l \ge 2\), we consider a modification of \(\mathbf {MColl}(H,l)\), denoted by \(\mathbf {MColl'}(H,l)\), which never restarts from Step 3 once it stops in Steps 5 or 9. Let \(D_l\) be the total number of quantum queries to H that \(\mathbf {MColl'}(H,l)\) makes. Let \(\mathsf {success}\) denote the event such that \(\mathbf {MColl'}(H,l)\) outputs an l-collision. Then we have
$$ \mathop {\mathrm {E}}[D_l] = \mathop {\mathrm {E}}[B_l] + \mathop {\mathrm {E}}[C_l] $$
and
$$\begin{aligned} \mathop {\mathrm {E}}[A_l] = \frac{\mathop {\mathrm {E}}[D_l]}{\Pr [\mathsf {success}]} = \frac{\mathop {\mathrm {E}}[B_l] + \mathop {\mathrm {E}}[C_l]}{\Pr [\mathsf {success}]} \end{aligned}$$
(4)
for \(l \ge 2\). In addition, since \(\mathop {\mathrm {E}}[B_l] = N^{1/3^{l-1}} \cdot \mathop {\mathrm {E}}[A_{l-1}]\) for \(l \ge 2\), we have
$$\begin{aligned} \mathop {\mathrm {E}}[A_l] = \frac{N^{1/3^{l-1}} \cdot \mathop {\mathrm {E}}[A_{l-1}] + \mathop {\mathrm {E}}[C_l] }{ \Pr [\mathsf {success}]}. \end{aligned}$$
(5)
We will show two lemmas on bounds for \(\mathop {\mathrm {E}}[C_l]\) and \(\Pr [\mathsf {success}]\) in Sects. 5.3 and 5.4, respectively:
Lemma 5.1
For \(l \ge 2\), \(\mathop {\mathrm {E}}[C_l] \le 18 \cdot \sqrt{\frac{l}{l-1}} \cdot N^{\frac{3^{l-1}-1}{2 \cdot 3^{l-1}}}\) holds.
Lemma 5.2
For \(l \ge 2\), \(\Pr [\mathsf {success}] \ge \frac{l-1}{l} \cdot \frac{1}{l} \cdot \left( 1 - \frac{1}{2l} \cdot N^{\frac{2}{3^{l-1}}-1} \right) \) holds.
Putting them in the inequality (5), we obtain
$$\begin{aligned} \mathop {\mathrm {E}}[A_l] \le \left( N^{1/3^{l-1}} \mathop {\mathrm {E}}[A_{l-1}] + 18 \sqrt{\frac{l}{l-1}} N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}} } \right) \cdot \frac{l^2 f_l}{l-1}, \end{aligned}$$
(6)
where
$$\begin{aligned} f_l = \frac{1}{1 - \frac{1}{2l} \cdot N^{\frac{2}{3^{l-1}}-1}}. \end{aligned}$$
Let \(\{ g_l \}_{1 \le l}\) be a sequence of numbers defined by \(g_1 = 1\) and
$$\begin{aligned} g_l = \left( g_{l-1} + \frac{18\sqrt{l/(l-1)}}{ (l-1) \cdot (l-1)!} \right) f_l \end{aligned}$$
for \(l \ge 2\).
We show the following claims:
Claim
For \(l \ge 1\), \(\mathop {\mathrm {E}}[A_l] \le g_l \cdot l \cdot l! \cdot N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}}}\) holds.
Claim
For \(l \ge 1\), \(g_l \le \left( 1 + 18\sqrt{2}e \right) \cdot \left( \frac{2lN^{1/3}}{2lN^{1/3} - 1} \right) ^{l-1}\) holds.
Combining them, we obtain for \(l \ge 1\),
$$ \mathop {\mathrm {E}}[A_l] \le \left( 1 + 18\sqrt{2}e \right) \cdot \left( \frac{2lN^{1/3}}{2lN^{1/3} - 1} \right) ^{l-1} \cdot l \cdot l! \cdot N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}}} $$
as we wanted. \(\square \)
Proof
(Proof of Claim). We give a proof of this claim by induction on l. Since \(\mathop {\mathrm {E}}[A_1] = 1\), the claim holds for \(l=1\). Now we assume that the claim holds for \((l-1)\). By the induction, we have
$$\begin{aligned} \mathop {\mathrm {E}}[A_l]&\le \left( N^{1/3^{l-1}} \mathop {\mathrm {E}}[A_{l-1}] + 18 \sqrt{\frac{l}{l-1}} N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}} } \right) \cdot \frac{l^2 f_l}{l-1}\\&\le \left( N^{1/3^{l-1}} \left( g_{l-1} \cdot (l-1) \cdot (l-1)! \cdot N^{ \frac{3^{l-2} -1}{2 \cdot 3^{l-2}}} \right) + 18 \sqrt{\frac{l}{l-1}} N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}} } \right) \cdot \frac{l^2 f_l}{l-1} \\&= \left( g_{l-1} + \frac{18\sqrt{l/(l-1)}}{(l-1) \cdot (l-1)!} \right) \cdot f_l \cdot l \cdot l! \cdot N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}} }\\&= g_l \cdot l \cdot l! \cdot N^{ \frac{3^{l-1} -1}{2 \cdot 3^{l-1}}} \end{aligned}$$
and the claim also holds for any \(l \ge 1\). \(\square \)
Proof
(Proof of Claim). Finally, we upper bound \(g_l\). Letting \(h_l = \frac{18\sqrt{l/(l-1)}}{ (l-1) \cdot (l-1)!}\), we have \(g_l = (g_{l-1} + h_l) f_l\). Since \(f_l \ge 1\) holds for \(l \ge 2\), we have
$$\begin{aligned} g_l = \left( g_{l-1} + h_l\right) f_l = \left( \left( g_{l-2} + h_{l-1}\right) f_{l-1} + h_l\right) f_l \le \left( g_{l-2} + h_{l-1} + h_l\right) f_{l-1}f_l. \end{aligned}$$
Continuing calculations, we obtain \( g_l \le \left( 1 + \sum ^l_{i=2} h_i \right) \cdot \prod ^l_{i=2}f_i. \) Thus, we have
$$\begin{aligned} g_l&\le \left( 1 + \sum ^l_{i=2} h_i \right) \cdot \prod ^l_{i=2}f_i \\&= \left( 1 + \sum ^l_{i=2} \frac{18\sqrt{i/(i-1)}}{(i-1) \cdot (i-1)!} \right) \prod ^l_{i=2} \frac{1}{1 - \frac{1}{2l} \cdot N^{\frac{2}{3^{i-1}}-1}}\\&\le \left( 1 + \sum ^l_{i=2} \frac{18 \sqrt{2}}{(i-1)!} \right) \prod ^l_{i=2} \frac{1}{1 - \frac{1}{2l} \cdot N^{-1/3}}\\&\le \left( 1 + 18\sqrt{2} \left( \sum ^l_{i=2} \frac{1}{(i-1)!} \right) \right) \prod ^l_{i=2} \frac{2lN^{1/3}}{2lN^{1/3} - 1}\\&\le \left( 1 {+} 18\sqrt{2} \left( \sum ^\infty _{i=0} \frac{1}{i!} \right) \right) \left( \frac{2lN^{1/3}}{2lN^{1/3} - 1} \right) ^{l-1} {=} \left( 1 + 18\sqrt{2}e \right) \left( \frac{2lN^{1/3}}{2lN^{1/3} - 1} \right) ^{l-1}, \end{aligned}$$
as we wanted. \(\square \)
5.3 Proof of Lemma 5.1
Note that
$$\begin{aligned} \mathop {\mathrm {E}}[C_l] \le \mathop {\mathrm {E}}[C_l \mid \text {Step 8 is operated}] \end{aligned}$$
holds, and we upper bound the conditional expectation \(\mathop {\mathrm {E}}[C_l \mid \text {Step 8 is operated}]\). When the algorithm operates in Step 8, it has already passed Steps 5 and 6. Thus, L has neither duplication nor l-collision. In particular, we can assume that L is a list of completely distinct \((l-1)\) collisions of H, i.e., \(y^{(i_1)} = y^{(i_2)}\) holds if and only if \(i_1 = i_2\). Thus, we have
$$\begin{aligned} \big |F^{-1}(1) \big | = \left|\bigcup _{i = 1}^{N^{1/3^{l-1}}} H^{-1}(y^{(i)}) \right|= \sum _{i=1}^{N^{1/3^{l-1}}} \big |H^{-1}(y^{(i)}) \big | \ge (l-1) \cdot N^{1 / 3^{l-1}} \end{aligned}$$
and
$$\begin{aligned} \frac{\big |F^{-1}(1) \big |}{|X |} \ge \frac{(l-1) \cdot N^{1 / 3^{l-1}}}{l \cdot N}. \end{aligned}$$
Since \(\mathbf {MColl}\) makes two quantum queries to H while making one query to F (See Fig. 3), we have
$$\begin{aligned} \mathop {\mathrm {E}}[C_l \mid \text {Step 8 is operated}] \le 2 \cdot 9 \cdot \sqrt{\frac{l \cdot N}{(l-1) \cdot N^{1 / 3^{l-1}}}} = 18 \cdot \sqrt{\frac{l}{l-1}} \cdot N^{\frac{3^{l-1}-1}{2 \cdot 3^{l-1}}} \end{aligned}$$
by Corollary 2.1 as we wanted.
5.4 Proof of Lemma 5.2
Next, we lower bound \(\Pr [\mathsf {success}]\). Note that
holds.
We need two lemmas. For the proof of Lemma 5.3, we refer readers to Shoup’s textbook [Sho08]. The proof of Lemma 5.4 is given in Appendix A.
Lemma 5.3
([Sho08, Theorem 8.26]). Let [d] be the set of integers \(\{1,2, \dots , d\}\), and \([d]^{\times n}\) be the n-array Cartesian power set of [d] for positive integers d, n. If \(s = (s_1,s_2,\dots ,s_n)\) is chosen uniformly at random from \([d]^{\times n}\), then the probability that \(s_i \ne s_j\) holds for all \(i \ne j\) is lower bounded by \(1 - {n^2}/({2d})\).
Lemma 5.4
Let X and Y be finite sets with \(|Y | = N\) and \(|X | = l N\). Let H be a function from X to Y. Then the number of l-collisions and \((l-1)\)-collisions of H are greater than or equal to N and lN, respectively.
First, we lower bound \(\Pr \left[ c^{(i)} \ne c^{(j)} \text { for }i \ne j \right] \). From the construction of \(\mathbf {MColl}\), we can assume that \(\mathbf {MColl}(H,l-1)\) outputs an \((l-1)\)-collision of H uniformly at random. Thus, we can assume that elements \(c^{(i)} \in L\) are chosen independently and uniformly at random from the set of \((l-1)\)-collisions of H. By Lemma 5.4, the number of \((l-1)\)-collisions of H is at least \(l \cdot N\). Moreover, if n is fixed, \(1 - n^2/2d\) is a monotonically increasing function on d. Therefore, by Lemma 5.3, we have
$$\begin{aligned} \Pr \left[ c^{(i)} \ne c^{(j)} \text { for }i \ne j \right] \ge 1 - \frac{(N^{1/3^{l-1}})^2}{2lN} = 1 - \frac{1}{2l} \cdot N^{\frac{2}{3^{l-1}}-1}. \end{aligned}$$
(7)
Second, we lower bound
. Note that the event \(\mathsf {success}\) occurs if and only if
$$\begin{aligned} y^{(i)} = y^{(j)}\text { for some }i \ne j, \end{aligned}$$
(8)
or
$$\begin{aligned} c^{(i_0)} \text { can be extended to an } l\text {-collision, and } \tilde{x} \not \in \left\{ x^{(i_0)}_1, x^{(i_0)}_2, \dots , x^{(i_0)}_{l-1}\right\} . \end{aligned}$$
(9)
occurs. Recall that \(\tilde{x}\) is the output of Step 8 and \(i_0\) is an index satisfying \(H(\tilde{x})=y^{(i_0)}\). The event (8) corresponds to the event in which \(\mathbf {MColl}\) finds an l-collision in Step 6, and the event (9) corresponds to the event in which \(\mathbf {MColl}\) finds an l-collision in Step 9.
Now, let \({\mathcal {L}}\) be all the possible lists L that satisfy \(c^{(i)} \ne c^{(j)} \text { for } i \ne j\). Let \({\mathcal {L}}_1 \subset {\mathcal {L}}\) denote the set of lists in which there exists l-collisions, i.e., there are two indices \(i \ne j\) such that \(y^{(i)} = y^{(j)}\), and \({\mathcal {L}}_{2} \subset {\mathcal {L}}\) denotes the set of lists in which there is no l-collision, i.e., \(y^{(i)} \ne y^{(j)}\) holds for \(i \ne j\). Then we have \({\mathcal {L}} = {\mathcal {L}}_{1} \coprod {\mathcal {L}}_{2}\). \(\mathbf {MColl}\) finds an l-collision in Step 6 if and only if \(L \in {\mathcal {L}}_{1}\). In the following, we ignore Step 4 and consider that L is not sorted for simplicity.
For a fixed \(L \in {\mathcal {L}}\), let \(A^L\) and \(B^L\) denote the sets of elements in L that can and cannot be extended to l-collisions, respectively. We have that \(L = A^L \coprod B^L\), \(\big |A^L \big |\) equals the number of \(y^{(i)}\) such that \(\big |H^{-1}(y^{(i)}) \big | \ge l\), and \(\big |B^L \big |\) equals the number of \(y^{(i)}\) such that \(\big |H^{-1}(y^{(i)}) \big | = l-1\). Define \(\langle A^L \rangle \), \(\langle B^L \rangle \) by
$$\begin{aligned} \langle A^L \rangle := \left|\bigcup _{c^{(i)} = (..., y^{(i)}) \in A^L } H^{-1}(y^{(i)}) \right|\text { and } \langle B^L \rangle := \left|\bigcup _{c^{(i)} = (..., y^{(i)}) \in B^L } H^{-1}(y^{(i)}) \right|, \end{aligned}$$
which are the numbers of preimages of \(y^{(i)}\)’s in \(A^L\) and \(B^L\), respectively. Note that
$$\begin{aligned} \Pr [\mathsf {success}\mid c^{(i)} \ne c^{(j)} \text { for } i \ne j] = \sum _{L \in {\mathcal {L}}} \Pr [\mathsf {success}\mid L] \Pr [L]. \end{aligned}$$
holds.
If \(L \in {\mathcal {L}}_{2}\), then \(\mathsf {success}\) occurs if and only if the event (9) occurs, that is, \(\tilde{x}\) can be used to construct an l-collision with an \((l-1)\)-collision in L. Note that \(\tilde{x}\) is chosen uniformly at random from the set
$$\begin{aligned} \left( \bigcup _{c^{(i)} = (...,y^{(i)}) \in A^L } H^{-1}(y^{(i)})\right) \bigcup \left( \bigcup _{c^{(i)} = (...,y^{(i)}) \in B^L } H^{-1}(y^{(i)}) \right) , \end{aligned}$$
and the event (9) occurs if and only if
$$\begin{aligned} {\tilde{x} \in \bigcup _{c^{(i)} = (...,y^{(i)}) \in A^L } H^{-1}(y^{(i)})} \wedge {\tilde{x} \ne x^{(i)}_j \text { for all } i \text { and } j} \end{aligned}$$
holds. Now we have
and
which suggests that
In addition, we have \(\big \langle A^L \big \rangle \ge l \cdot |A^L |\) since \(y^{(i)} \ne y^{(j)}\) holds for \(i \ne j\) if \(L \in {\mathcal {L}}_{2}\). Thus, we have \(\big \langle A^L \big \rangle \ge l \cdot \big |A^L \big | \ge (l-1)\cdot \big |A^L \big |\) and \(\big \langle B^L \big \rangle = (l-1)\cdot \big |B^L \big |\). This yields that
$$\begin{aligned} \frac{ \big \langle A^L \big \rangle }{ \big \langle A^L \big \rangle + \big \langle B^L \big \rangle } = \frac{ 1 }{ 1 + \frac{ \big \langle B^L \big \rangle }{ \big \langle A^L \big \rangle } } \ge \frac{ 1 }{ 1 + \frac{ (l-1)\big |B^L \big | }{ (l-1)\big |A^L \big | } } = \frac{\big |A^L \big |}{\big |A^L \big | + \big |B^L \big |}. \end{aligned}$$
Thus, we have
$$\begin{aligned} \Pr [\mathsf {success}\mid L] \ge \frac{ \big |A^L \big | }{ \big |A^L \big | + \big |B^L \big | } \cdot \frac{1}{l} \end{aligned}$$
(10)
for \(L \in {\mathcal {L}}_{2}\). Moreover, since \(\Pr [\mathsf {success}\mid L]=1\) for \(L \in {\mathcal {L}}_{1}\), the inequality (10) also holds for \(L \in {\mathcal {L}}_{1}\). Therefore, we have
Now we use the following lemmas the proofs of which are given in Appendices B and C, respectively.
Lemma 5.5
Let X, Y be finite sets such that \(|X | = l \cdot |Y |\), and H be a function from X to Y. Let A, B denote the sets of \((l-1)\)-collisions of H that can and cannot be extended to l-collisions, respectively. Then we have
$$ \frac{|A |}{|A |+|B |} \ge \frac{l-1}{l}. $$
Lemma 5.6
Let X, Y be finite sets such that \(|X | = l \cdot |Y |\), and H be a function from X to Y. Let A, B denote the sets of \((l-1)\)-collisions of H that can and cannot be extended to l-collisions, respectively. Then we have
$$\begin{aligned} \sum _{L \in {\mathcal {L}}} \frac{ \big |A^L \big | }{ \big |A^L \big | + \big |B^L \big | } \cdot \Pr [L] = \frac{|A |}{|A | + |B |}. \end{aligned}$$
By the above lemmas, we have
Consequently, \(\Pr [\mathsf {success}]\) is lower bounded as \(\Pr [\mathsf {success}] \ge \frac{l-1}{l} \cdot \frac{1}{l} \cdot \left( 1 - \frac{1}{2l} \cdot N^{\frac{2}{3^{l-1}}-1} \right) \), that completes the proof.