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⊤
⊝
T
2
T
3
T
m
⊛
⊠
.
T
=
Therefore, we can proceed by induction and get a lower bound of
(
2
n
−
1
)
m
.
6.4 Explicit Tensors by Combinatorial Constructions
The currently best lower bound for an explicit tensor is due to [
AF11
]. It improves
on the lower order term of the construction in the last section.
Let
ˇ
=⃒
log
2
n
∞
.For
i
∗{
0
,...,ˇ
}
, we recursively define the followingmatrices:
1.
S
1
,
0
=
(
1
)
.
2. For even
n
=
2
m
>
1,
00
S
m
,
i
0
⊜
⊨
<ˇ
if
i
I
m
S
n
,
i
=
otherwise
.
⊩
0
0
I
m
3. For odd
n
=
2
m
+
1
>
1,
⊜
⊨
⊩
⊤
⊝
000
000
S
m
,
i
00
⊛
⊠
if
i
<ˇ
S
n
,
i
=
⊤
⊝
.
000
I
m
00
0
⊛
⊠
otherwise
I
m
0
Finally, let
T
n
be the tensor consisting of slices
S
n
,
0
,...,
S
n
,ˇ
. The format of
T
n
is
n
.
T
n
is certainly explicit, we can determine the entries by following
the recursive structure.
×
n
×
(ˇ
+
1
)
Theorem 4
R
(
T
n
)
∧
2
n
−
2
h
(
n
)
+
1
where h
(
n
)
is the number of
1
s in the binary
expansion of n.
Note that
h
(
n
)
ↆ
log
n
. We can use the substitution method to prove the theorem.
Certainly
R
2
m
is even, then we can “substitute away” the
two identity matrices, killing
n
products. The remaining tensor is
T
n
/
2
. Therefore,
we get the recurrence
(
T
1
)
=
1 holds. If
n
=
(
T
2
m
)
∧
(
T
m
)
+
.
R
R
2
m
