the size of the tree may be as important as the overall number of multiplication and addition operations. On subplot 3 of Figure 2.7 the comparison of the total number of points kept in the tree by SD and SDsdp algorithms is shown. As expected the SDsdp algorithm keeps significantly less points in the tree than the SD algorithm.
Finally on subplot 4 of Figure 2.7, the comparison of the bit error rate (BER) perfor- mance of the exact ML detector (SDsdp algorithm) and the approximate MMSE nulling and cancelling with optimal ordering heuristic is shown. Over the range of SNRs considered here, the ML detector outperforms the MMSE detector significantly, thereby justifying our efforts in constructing more efficient ML algorithms.
Remark: Recall that the lower bound introduced in this section is valid only if the origi- nal problem is binary, i.e.,D={−12,12}k−1. A generalization to caseD={−32,−12, 12,32}k−1 can be found in [106]. It is not difficult to generalize it to anyD={−L−21,−L−22, . . . ,L−23,L−21}k−1 by noting that any k −1-dimensional vector whose elements are numbers from {−L+ 1,−L+ 2, . . . , L−2, L−1}can be represented as a linear transformation of a (k−1)(L−1)- dimensional vector from D = {−12,12}(k−1)(L−1). (The interested reader can find more on this in [69]). However, this significantly increases the dimension of the SDP problem in (2.14), which may cause our algorithm to be inefficient. Motivated by this, in the following section we consider a different framework, based onH∞ estimation theory, which will (as we will see in Section 2.8) produce as a special case a general lower bound applicable for any D.
4 6 8 10 106
108
SNR [db]
flop count
Flop count
SDsdp SD SDsdp−sdp
0 20 40 60
100 102 104 106
level
number of points per level
Distribution of points, SNR=4 db
SDsdp SD
4 6 8 10
102 104 106 108
SNR [db]
total number of points
Total number of points SD SDsdp
4 6 8 10
10−3 10−2 10−1 100
SNR [db]
bit error rate
BER performance ML null−can MMSE
Figure 2.7: Computational complexity of the SD and SDsdp algorithms, m = 50, D = {−12,12}50
To simplify the notation, we rewrite (2.11) as
min
a∈D⊂Zk−1kb−Lak2, (2.19)
where we introduceda=s1:k−1,b=z1:k−1, and L=R1:k−1,1:k−1.
Consider an estimation problem where a and b−La are unknown vectors, b is the observation, and the quantities we want to estimate are a and b. In the H∞ framework, the goal is to construct estimators ˆa =f1(b) and ˆb =f2(b), such that for some given γ, some β ≥0, and some diagonal matrixD >0, we have
β||a−aˆ||2+||b−bˆ||2
a∗Da+kb−Lak2 ≤γ2 (2.20)
forall a andb (see, e.g., [48]).
Obtaining a desired lower bound from (2.20) is now straightforward. Note that for all
a andb we can write
kb−Lak2 ≥γ−2
β||a−ˆa||2+||b−bˆ||2
−a∗Da, (2.21)
and, in particular,
mina∈Dkb−Lak2 ≥min
a∈D γ−2β||a−aˆ||2−a∗Da
+γ−2||b−bˆ||2. (2.22)
Note that the minimization on the right-hand side (RHS) of (2.22) is straightforward since it can be done componentwise (which is why we chose D > 0 diagonal). Thus, for any H∞ estimators ˆa =f1(b) and ˆb =f2(b), (2.22) provides a readily computable lower bound. The issue, of course, is how to obtain the best ˆaand ˆb (andD and γ). To this end, let us assume that the estimators are linear, i.e., ˆa=K1b and ˆb=K2b for some matrices K1 and K2 (see Figure 2.8).
- - ?
-
-
-
-
a L
b−La K1
K2 ˆ a
bˆ
Figure 2.8: An H∞ estimation analogy used in deriving a lower bound on integer least- squares problem.
Introducing c=
D1/2a b−La
and noting that
T =
D−1/2 0 LD−1/2 I
−
K1 K2
LD−1/2 I
=
√β(I−K1L)D−1/2 −√ βK1 (I−K2)LD−1/2 I−K2
maps c to
√β(a−ˆa) b−bˆ
, from (2.21) we see that for all cit must hold that
c∗T∗Tc≤γ2c∗Ic
(see [48]). SinceT is square, this implies either of the equivalent inequalities
T T∗ ≤γ2I or T∗T ≤γ2I. (2.23)
The tighter the bound in (2.23), the tighter the bound in (2.22). In other words, the closer γ−1T is to a unitary matrix, the tighter (2.22) becomes. Hence we attempt to choose K1
and K2 to make γ−2T T∗ as close to identity as possible.
To this end, post multiply T with the unitary matrix
Φ =
∇−1 D−1/2L∗∆−∗
−LD−1/2∇−1 ∆−∗
.
∇and ∆ are found via the factorizations
D−1/2L∗LD−1/2+I =∇∗∇ and LD−1L∗+I = ∆∆∗, (2.24)
to obtain
TΦ =
A B
0 C
(2.25)
where
A=p
βD−1/2∇−1, B=p
β(D−1L∗∆−∗−K1∆), and C= (I−K2)∆. (2.26)
ThusT T∗ ≤γ2I implies
AA∗+BB∗ BC∗ CB∗ CC∗
≤γ2I. (2.27)
Note that we have many degrees of freedom when choosing K1 and K2, and wish to make judicious choices. So, to simplify things, let us choose K2 such that CC∗ = γ12I for some 0≤γ1≤γ. (Clearly, this can always be done, since from (2.24) we have that ∆ is invertible, and the simple choice K2 = I −γ1∆−1 will do the job.) To make half the eigenvalues of γ−2T T∗ unity, we set the Schur complement of the (2,2) entry of (2.27) to zero, i.e.,
AA∗+BB∗−γ2I−BC∗(CC∗−γ2I)−1CB∗= 0. (2.28)
Using CC∗=C∗C=γ12I, it easily follows that
BB∗= (1− γ12
γ2)(γ2I−AA∗). (2.29)
Using the definitions of Aand B from (2.26), we obtain pβK1 =p
βD−1L∗(LD−1L∗+I)−1−B∆−1. (2.30)
From the (1,1) entry of (2.27) it follows that
γ2I−(AA∗+BB∗)≥0,
which is the only constraint on γ. Combining this constraint with the definition of A from (2.26), the definition of ∇ from (2.24), and the expression forBB∗ from (2.29), we obtain that
γ2 ≥ β
λmin(D+L∗L).
We summarize the results of this section in the following theorem:
Theorem 2.1. Consider the integer least-squares problem (2.19). Then for any γ2 ≥
β
λmin(D+L∗L),0≤γ1 ≤γ, and any matricesD ≥0,B, and∆satisfying∆∆∗=I+LD−1L∗
and BB∗= (1−γγ122)(γ2I−β(D+L∗L)−1),
mina∈Dkb−Lak2 ≥min
a∈Dγ−2||p βa−p
βD−1L∗(LD−1L∗+I)−1b+B∆−1b||2−a∗Da+γ12
γ2||∆−1b||2.
Proof. Follows from the previous discussion, noting that
||b−bˆ||2 =||(I −K2)b||2 =||C∆−1b||2 =γ21||∆−1b||2
and
AA∗ =β(D+L∗L)−1.
The next corollary directly follows from Theorem 2.1.
Corollary 2.1. Consider the setting of the Theorem 1 and let β= 1. Then
mina∈Dkb−Lak2 ≥ min
a∈Dγ−2||a−D−1L∗(LD−1L∗ +I)−1b+Bφ||2−a∗Da+ γ12 γ2||φ||2,
(2.31)
where B is the unique symmetric square root of(1−γγ122)(γ2I−(D+L∗L)−1), and φisany vector of the squared length b∗(I+LD−1L∗)−1b.
It should be noted that we have several degrees of freedom in choosing the parameters (γ1, γ, D, φ), and we can exploit that to tighten the bound in (2.31) as much as possible.
Optimizing simultaneously over all these parameters appears to be rather difficult. However, we can simplify the problem and letγ1 →γ. This has two benefits: it maximizes the third term in (2.31) and it sets B= 0 so that we need not worry about the vector φ. Finally, to
maximize the first term, we need to take γ as its smallest possible value, i.e., we set
γ2 = 1
λmin(D+L∗L).
This leads to the following result:
Corollary 2.2. Consider the setting of the Theorem 2.1 and letβ = 1. Then
mina∈Dkb−Lak2 ≥λmin(L∗L+D)||a−(L∗L+D)−1L∗b||2−a∗Da+b∗(I−L((L∗L+D)−1)L∗)b (2.32)
Remark: We would like to note that the bound given in the previous Corollary could have been also obtained in a faster way. Below we show a possible derivation that an anonymous reviewer has provided to us.
Let D be a diagonal matrix such that D≥0. Then we have
||b−La||2 =a∗L∗La−2b∗La+b∗b=a∗(L∗L+D)a−2b∗La+b∗b−a∗Da
= (a−(L∗L+D)−1L∗b)∗(L∗L+D)(a−(L∗L+D)−1L∗b)−b∗L((L∗L+D)−1)L∗b+b∗b−a∗Da
≥λmin(L∗L+D)||a−(L∗L+D)−1L∗b||2−a∗Da+b∗(I−L((L∗L+D)−1)L∗)b
It is not difficult to see that this is precisely the same bound as the bound given in Corollary 2.2. The interested reader can find more on this type of bound in [96] and [79].
In the following sections we show how various choices of the free parameters in the general lower bound from Theorem 2.1 yield several interesting special cases of lower bounds. In particular, in Section 2.5 we show that the lower bound obtained by solving a related convex optimization problem, where the search space is relaxed from integers to a sphere, can be deduced as a special case of the lower bound from Theorem 2.1. Then, in Section 2.6, we show that the lower bound obtained by solving another convex optimization problem, where
the search space is now relaxed from integers to a polytope, can also be deduced as a special case of the lower bound from Theorem 2.1. Finally, in Section 2.8, we use (2.32) to deduce the lower bound based on the minimum eigenvalue ofL∗L.