Let us start with the CMbased graph of Figure 2, which consists of the cascade of a subgraph H with p − l Aoperations and a CM subgraph with l Aoperations. This graph is exploited in [14] to obtain optimal multiplications by rational constants with periodic binary representations, and it has the following characteristics:

1)
At the output of every Aoperation of the CM subgraph, the MNSD of the resulting constant is at most twice the MNSD of the constant at its inputs, according to Observation A. If the MNSD at the output of the subgraph H, denoted by n _{
H
}, is the highest possible in H, the maximum resulting MNSD at the overall graph is 2^{l} × n _{
H
}.

2)
If H has the minimum Adepth, the overall graph has the minimum Adepth because the CM subgraph has also the minimum Adepth. The same holds for the Acost.

3)
If H is nonmultiplicative, the CMbased graph has l articulation points formed with the minimum Acost and Adepth and clustered in the higher depth levels. Thus, this graph has the highest MNSD in its first articulation point in comparison to any other graph with l articulation points. Moreover, it can have the maximum MNSD with respect to other graphs with l articulation points.
Due to their aforementioned characteristics, CMbased graphs will be analyzed in the following to show that the graph of a constant whose MNSD is given in specific intervals into the MNSDrange requires certain multiplicative characteristics to preserve the lower bound L_{
A
}. It shall be considered henceforth that the MNSD is given in the MNSDrange (2^{p − 1}, 2^{p}] with p > 1, since L_{
A
} = L_{
d
} = 1 for S(c) = 2 (i.e., when p = 1).
Theorem 1 A constant c whose MNSD is S(c) > 2^{p − 1} + 1 cannot be obtained with a nonmultiplicative graph with p Aoperations.
Proof Let us review the simplest case p = 2. There are two possible graphs, one of them additive and the other one multiplicative, as shown in Figure 3 (see also Cost2 Graphs No. 1 and 2 of Appendix A of [13]). The highest MNSD of any NOF in both graphs is equal to 2. The last Aoperation of the nonmultiplicative graph only can add one nonzero digit to a constant, thus yielding OFs whose highest MNSD is 2^{p − 1} + 1 = 3 according to Observation A. The highest MNSD of any OF in the multiplicative graph is equal to 4 because both inputs to the last Aoperation can have up to 2 nonzero digits each. Thus, the case p = 2 is a base for the following inductive hypothesis: the highest MNSD of any OF in a nonmultiplicative graph with p Aoperations is equal to 2^{p − 1} + 1 for all p > 1. Assuming it as a true statement, let us construct a nonmultiplicative graph with p + 1 Aoperations, with the aim of obtaining an OF whose MNSD is n = 2^{p} + 2, i.e., the lowest MNSD that contradicts the hypothesis.
A nonmultiplicative graph with p operations cannot have the inputs of its last Aoperation (the Aoperation placed at the pth depth level) coming from the same position because an articulation point is generated and the graph becomes multiplicative (see Observation C). For the sake of clarity, let us identify the two inputs of the last Aoperation as a_{1} and a_{2}. Let us assume, without loss of generality, that a_{1} comes from the output of a subgraph G_{1}, as shown in Figure 4. This subgraph can be either nonmultiplicative or multiplicative (Figure 4a,b, respectively). In the following, we will review each of these two cases, along with the point where a_{2} can come from.
If G_{1} is nonmultiplicative (Figure 4a), the highest MNSD of any of its OFs is n_{1} = 2^{p − 1} + 1 (this follows from the inductive hypothesis). For the overall graph, the only way to obtain an OF whose MNSD is n = 2^{p} + 2 is having a_{2} coming from an Aoperation that produces constants with an MNSD at least equal to n − n_{1} = 2^{p − 1} + 1. From Observation B, we have that this MNSD can be obtained only from the pth Aoperation, which means that a_{1} and a_{2} would be coming from the same Aoperation, generating an articulation point. If a_{2} comes from the (p − 1)th Aoperation, the articulation point is avoided and the whole graph can still be nonmultiplicative. However, the highest MNSD of a constant generated from the (p − 1)th Aoperation is equal to 2^{p − 1} (see Observation B) and thus the highest MNSD of the OF in the overall graph is equal to n_{1} + 2^{p − 1} = 2^{p − 1} + 1 + 2^{p − 1} = 2^{p} + 1.
If G_{1} is multiplicative (Figure 4b), it can have l articulation points where 1 ≤ l ≤ p − 2. With l articulation points and p operations, the highest MNSD of any OF produced by G_{1} is n_{1} = 2^{l} × (2^{p − l − 1} + 1). This follows from the assumption that G_{1} is a CMbased graph and from the inductive hypothesis. In order to obtain an OF whose MNSD is n = 2^{p} + 2, a_{2} must come from an Aoperation that produces constants with an MNSD at least equal to n − n_{1} = 2^{p − 1} − 2^{l} + 2. Moreover, such Aoperation must be placed at a depth level less than the depth level where the first articulation point is placed; otherwise this articulation point will make the whole graph multiplicative. Since G_{1} has its first articulation point at a depth level equal to p − l, a_{2} must come from an Aoperation placed in a depth level that does not exceed p − l − 1. The highest MNSD of any NOF obtained from an Aoperation placed in the (p − l − 1)th depth level is 2^{p − l − 1} (from Observation B). Hence, this value should be equal or greater than n − n_{1} = 2^{p − 1}− 2^{l} + 2 in order to obtain an OF in the overall graph whose MNSD is n = 2^{p} + 2. However, it can be shown that 2^{p − l − 1} < 2^{p − 1}− 2^{l} + 2 holds. Thus, the highest MNSD of any OF in the overall graph is equal to n_{1} + 2^{p − l − 1} = 2^{l} × (2^{p − l − 1} + 1) + 2^{p − l − 1} ≤ 2^{p} + 1.
As a result, we have that the inductive hypothesis for a graph with p Aoperations implies that it holds for p + 1 Aoperations. Therefore, considering this implication along with the base case with p = 2, we prove Theorem 1 by induction on p. ■
Theorem 2
A constant c whose MNSD is S(c) > 2
^{p − 1}
+ 2
^{q − 1}
with q = {1, 2,…, (p − 1)} can be obtained by using a graph with p Aoperations only if at least q of these operations are articulation points.
Proof Consider that the overall graph is a CMbased graph with l articulation points, where l < p. The highest MNSD of any OF in the overall graph is n = 2^{l} × (2^{p − l − 1} + 1) = 2^{p − 1} + 2^{l} (from Theorem 1). Therefore, l has to be at least equal to q in order to achieve n > 2^{p − 1} + 2^{q − 1}. Note that Theorem 1 corresponds to the case q = 1. ■
Theorem 3 A constant c whose MNSD is S(c) > 3 × 2^{p − 2} + 1 cannot be obtained by using a nonmultiplicative graph with only p + 1 Aoperations and Adepth preserved equal to p.
Proof For the simplest case p = 2, there is only one graph with p + 1 = 3 Aoperations and Adepth preserved equal to p = 2. This graph is shown in Figure 5 (see also Cost3 Graph No. 7 of Appendix A of [13]). The highest MNSD of the OF in this graph is 3 × 2 ^{p − 2} + 1 = 4.
Now, we review the case p ≥ 3. Since the Adepth must be preserved equal to p, only p of these operations can be sequentially connected. Consider that a graph G_{1} with Acost and Adepth equal to p − 1 is connected to one of the inputs of the last Aoperation, as shown in Figure 6. In order to avoid an articulation point, the other input must come from a different position. Thus, this input is connected to the remaining Aoperation, which should be placed in the (p − 1)th depth level to obtain the highest possible MNSD (from Observation B), and whose inputs are identified as a_{1} and a_{2}. Consider that a_{1} comes from the Aoperation placed in the (p − 2)th depth level. Let us review the cases when G_{1} is either nonmultiplicative (Figure 6a) or multiplicative (Figure 6b), along with the point where a_{2} can come from.
If G_{1} is nonmultiplicative (Figure 6a), the highest MNSD of its OFs is n_{1} = 2^{p − 2} + 1 (from Theorem 1) and a_{1} with a_{2} can come from the Aoperation placed in the (p − 2)th level. Thus, the highest possible MNSD of the OF in the last Aoperation is n = n_{1} + n_{a 1} + n_{a 2}, where n_{a 1} and n_{a 2} are the respective highest MNSD in a_{1} and a_{2}. The highest MNSD of G_{1} is n_{1} = 2^{p − 2} + 1 and n_{a 1} = n_{a 2} = 2^{p − 2}. Therefore, we have n = (2^{p − 2} + 1) + 2^{p − 2} + 2^{p − 2} = 2^{p − 2} + 1 + 2^{p − 1} = 3 × 2^{p − 2} + 1.
If G_{1} is multiplicative (Figure 6b), consider that it is a CMbased graph with l articulation points, where 1 ≤ l ≤ p − 2. The highest MNSD of its OFs is n_{1} = 2^{l} × (2^{p − l − 2} + 1). Similarly, the highest MNSD in a_{1} is n_{a 1} = 2^{l − 1} × (2^{p − l − 2} + 1). To avoid an articulation point, a_{2} must come from a depth level that does not exceed p − l − 2, thus n_{a 2} = 2^{p − l − 2}. The highest possible MNSD of the OFs in the last Aoperation is n = n_{1} + n_{a 1} + n_{a 2} = 2^{p − 2} + 2^{p − 3} + 2^{l} + 2^{l − 1} + 2^{p − l − 2}, which can be shown to be less or equal to 3 × 2^{p − 2} + 1. ■
Theorem 4 A constant c whose MNSD is S(c) > 3 × 2^{p − 2}+ 2^{q − 1}with q = {1, 2,…, (p − 2)} can be obtained by using a nonmultiplicative graph whose Adepth is equal to p only if at least p + q + 1 Aoperations are used.
Proof Consider p + r Aoperations, with r < p. Since the Adepth is equal to p, only p of these operations can be sequentially connected. Let us assume that a graph G_{1} with Acost and Adepth equal to p − 1 is connected to one of the inputs of the last Aoperation, whereas the other input comes from a different position to avoid an articulation point. Consider that this input comes from a graph G_{2}, formed with r − 1 of the remaining Aoperations. To obtain the highest MNSD in G_{2}, we can consider that G_{2} is a CM graph with its last Aoperation placed at the (p − 1)th depth level. Let us assume that the last of the remaining Aoperations is connected to the input of G_{2}, with its inputs being a_{1} and a_{2}. This is shown in Figure 7. The highest MNSD of any OF from G_{2} is n_{2} = 2^{r − 1} × (n_{a 1} + n_{a 2}), where n_{a 1} and n_{a 2} are the respective highest MNSD in a_{1} and a_{2}.
If G_{1} is nonmultiplicative (Figure 7a), the highest MNSD of its OFs is n_{1} = 2^{p − 2} + 1 (from Theorem 1). Both inputs a_{1} and a_{2} can come from the same point because G_{1} is nonmultiplicative, and the highest depth level where this point can be placed is p − r − 1; thus, n_{a 1} = n_{a 2} = 2^{p − r − 1}. Hence, the highest MNSD of any OF in the overall graph is n = n_{1} + n_{2} = (2^{p − 2} + 1) + 2^{r − 1} × (n_{a 1} + n_{a 2}) = (2^{p − 2} + 1) + 2^{r − 1} × (2^{p − r}) = (2^{p − 2} + 1) + 2^{p − 1} = 3 × 2^{p − 2} + 1, which clearly is less or equal than 3 × 2^{p − 2} + 2^{q − 1} even for the smallest q = 1.
If G_{1} is multiplicative, consider that it is a CMbased graph with l articulation points, where 1 ≤ l ≤ p − 2. The highest MNSD of its OFs is n_{1} = 2^{l} × (2^{p − l − 2} + 1). We can have two cases, depending on whether r is greater than l or not. Let us see each one of these cases.
When r ≤ l (Figure 7b), only either a_{1} or a_{2} can come from the Aoperation placed in the (p − l − 1)th depth level but not both, because an articulation point would be generated. If a_{1} comes from this Aoperation, a_{2} can come from the immediate lower depth level. Thus, n_{a 1} = (2^{p − l − 2} + 1) and n_{a 2} = 2^{p − l − 2}. Hence, the highest MNSD of any OF in the overall graph is n = n_{1} + n_{2} = (2^{p − 2} + 2^{l}) + 2^{r − 1} × (n_{a 1} + n_{a 2}) = (2^{p − 2} + 2^{l}) + 2^{r − 1} × [(2^{p − l − 2} + 1) + 2^{p − l − 2}] = 2^{p − 2} + 2^{l} + 2^{r − 1} × (2^{p − l − 1} + 1), which can be shown to be less or equal to 3 × 2^{p − 2} + 2^{r − 1}. Thus, r must be at least equal to q + 1 in order to achieve a value n > 3 × 2^{p − 2} + 2^{q − 1}.
When r > l (Figure 7c), a_{1} and a_{2} can come from the same point because the subgraph H in G_{1} is nonmultiplicative, thus n_{a 1} = n_{a 2} = 2^{p − r − 1}. Hence, the highest MNSD of any OF in the overall graph is n = n_{1} + n_{2} = (2^{p − 2} + 2^{l}) + 2^{r − 1} × (n_{a 1} + n_{a 2}) = (2^{p − 2} + 2^{l}) + 2^{r − 1} × (2^{p − r}) = 2^{p − 2} + 2^{l} + 2^{p − 1} = 3 × 2^{p − 2}+ 2^{l}. Since l is at most equal to r − 1, we have that the highest value for n is n = 3 × 2^{p − 2} + 2^{r − 1}. Thus, n > 3 × 2^{p − 2} + 2^{q − 1} holds only if r is at least equal to q + 1. ■
Now, let us introduce the number of prime factors of the constant c, denoted as Ω(c). Note that, along with S(c), Ω(c) is a function that can be known a priori from the constants. Generally speaking, obtaining the prime factors of c is a challenging problem for very large constants (over 110 digits) and it is one of the bases for the RivestShamirAdleman (RSA) encryption scheme [19]. However, for constants up to 32 bits, which cover the DSP problem sizes of the most practical importance [14], the prime factors of c can be obtained straightforwardly even with the simple sieve approach, such as the one used in the MATLAB function ‘factor’.
The following theorems present the relations between the values Ω(c) and S(c). If the required multiplicative characteristics for the corresponding graph of the constant c, highlighted in the previous theorems, cannot be accomplished due to the value Ω(c), the lower bounds for the Acost and the Adepth are affected.
Theorem 5
A constant c whose MNSD is S(c) > 2
^{p − 1}
+ 2
^{Ω(c) − 1}
only can be obtained by using a graph with at least p + 1 Aoperations.
Proof From Theorem 2, it is known that at least Ω(c) articulation points are required if only p Aoperations can be used. However, since Ω(c) is the number of prime factors, only up to Ω(c) − 1 articulation points are allowed. Thus, the only solution is using an additional Aoperation. By adding one more Aoperation, the overall nonmultiplicative graph can give an OF whose MNSD is up to 2^{p} + 1, which covers the MNSDrange regardless of the value Ω(c). Therefore, at least p + 1 Aoperations are required. ■
Theorem 6 A constant c whose MNSD is S(c) > 3 × 2^{p − 2}+ 2^{Ω(c) + q − 2}with q = {1, 2,…, (p − Ω(c) − 1)} only can be obtained by using a graph with Adepth at least equal to p + 1 if up to p + q Aoperations are used, or with at least p + q + 1 Aoperations if the minimum Adepth equal to p is preserved.
Proof Let us consider a CMbased graph with p + r Aoperations and Adepth equal to p + s, where r < p and s ≥ 0. Since at most Ω(c) − 1 articulation points can be used, its CM subgraph has Ω(c) − 1 Aoperations, whereas its nonmultiplicative subgraph H has p + r − Ω(c) + 1 Aoperations and an Adepth equal to p + s − Ω(c) + 1. According to Theorem 4, the highest MNSD of any OF generated in the subgraph H is 3 × 2^{p + s − Ω(c) − 1} + 2^{r − s − 1} and the highest MNSD at the output of the overall graph is n = 2^{Ω(c) − 1} × (3 × 2^{p + s–Ω(c) − 1} + 2^{r − s − 1}) = 3 × 2^{p + s − 2} + 2^{Ω(c)+r − s − 2}. If the Adepth is preserved equal to p, we have s = 0. In such case r needs to be at least equal to q + 1 in order to achieve n > 3 × 2^{p − 2} + 2^{Ω(c) + q − 2}. On the other hand, if r ≤ q holds, the value s should be at least equal to 1 in order to achieve n > 3 × 2^{p − 2} + 2^{Ω(c) + q − 2}, implying that the Adepth of the overall graph is at least equal to p + 1. ■
From Theorem 5 and substituting p using (3), the lower bound L_{
A
} can be expressed as
{L}_{A}=\left\{\begin{array}{l}\u2308{log}_{2}S\left(c\right)\u2309;\phantom{\rule{0.25em}{0ex}}\mathrm{if}\phantom{\rule{0.5em}{0ex}}\Omega \left(c\right)\ge {log}_{2}\left\{S\left(c\right){2}^{\u2308{log}_{2}S\left(c\right)\u23091}\right\}+1,\\ \u2308{log}_{2}S\left(c\right)\u2309+1;\phantom{\rule{0.5em}{0ex}}\mathrm{otherwise}.\end{array}\right.
(4)
Note that if L_{
A
} = ⌈ log_{2}S(c)⌉ holds, L_{
d
} will have the same value as L_{
A
} without requiring any additional Aoperation (as given by (2)). From (4), we can see that this only can be accomplished if Ω(c) ≥ log_{2}{S(c) − 2{}^{\u2308{log}_{2}S\left(c\right)\u23091}} + 1 holds. The argument S(c) − 2{}^{\u2308{log}_{2}S\left(c\right)\u23091} in the log_{2}{x} operation is always greater than zero and therefore the condition can always be evaluated.
From Theorem 6, we have that if the Acost is given as N_{
A
}≥ ⌈ log_{2}S(c)⌉ + q + 1 and if log_{2}{S(c) − 3 × 2 {}^{\u2308{log}_{2}S\left(c\right)\u23092}} + 2 − Ω(c) > q holds, the Adepth can be kept equal to ⌈ log_{2}S(c)⌉. By noticing that any real number a is greater than an integer b if this integer is given as b = ⌈a⌉ − 1 and applying this observation to q, we obtain q = \u2308{log}_{2}\left\{S\left(c\right)3\times {2}^{\u2308{log}_{2}S\left(c\right)\u23092}\right\}+2\Omega \left(c\right)\u2309 + 1. In this equality, the log_{2}{x} operation does not yield finite values when S(c) ≤ 3 × 2 {}^{\u2308{log}_{2}S\left(c\right)\u23092} holds. However, according to Theorem 3, the Adepth can be kept equal to ⌈ log_{2}S(c)⌉ by using only one additional Aoperation if the condition S(c) ≤ 3 × 2 {}^{\u2308{log}_{2}S\left(c\right)\u23092} + 1 holds. Therefore,
{L}_{d}=\left\{\begin{array}{l}\u2308{log}_{2}S\left(c\right)\u2309;\phantom{\rule{0.75em}{0ex}}\mathrm{if}\phantom{\rule{0.5em}{0ex}}{N}_{A}\ge \u2308{log}_{2}S\left(c\right)\u2309+{L}_{\mathit{EA}},\\ \u2308{log}_{2}S\left(c\right)\u2309+1;\phantom{\rule{2.5em}{0ex}}\mathrm{otherwise},\end{array}\right.
(5)
where L_{
EA
}, the lower bound for the number of extra Aoperations required to generate NEFs and preserve L_{
d
}= ⌈ log_{2}S(c)⌉, is given as
{L}_{\mathit{EA}}=\left\{\begin{array}{l}0;\phantom{\rule{1em}{0ex}}\mathrm{if}\phantom{\rule{0.5em}{0ex}}\Omega \left(c\right)\ge {log}_{2}\left\{S\left(c\right){2}^{\u2308{log}_{2}S\left(c\right)\u23091}\right\}+1,\\ 1;\phantom{\rule{1em}{0ex}}\mathrm{if}\phantom{\rule{0.5em}{0ex}}S\left(c\right)\le 3\times {2}^{\u2308{log}_{2}S\left(c\right)\u23092}+1,\phantom{\rule{7.75em}{0ex}}\\ \u2308{log}_{2}\left\{S\left(c\right)3\times {2}^{\u2308{log}_{2}S\left(c\right)\u23092}\right\}+2\Omega \left(c\right)\u2309;\phantom{\rule{0.25em}{0ex}}\mathrm{otherwise}.\end{array}\right.
(6)
It is worth highlighting that, whereas the lower bound L_{
A
} only depends on the relation between Ω(c) and S(c), the lower bound L_{
d
} depends on the Acost N_{
A
}. As pointed out in [12], for many problem instances, there is a tradeoff between the Acost and the Adepth. From (5), we have that such tradeoff is the minimum theoretical value L_{
EA
}, given in (6).