International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020 Available
Online www.ijtsrd.com e-ISSN 2456 - 6470
Basic Maximal Total Strong Dominating Functions
Dr. M. Kavitha1,∗ ,
1
Department of Mathematics, KPR Institute of Engineering and Technology, Coimbatore - 641407, India.
Let G = (V, E) be a simple graph. A subset D of V (G) is called a total strong dominating set of G, if for every
u ∈ V (G), there exists a v ∈ D such that u and v are adjacent and deg(v) ≥ deg(u). The minimum cardinality
of a total strong dominating set of G is called total strong domination number of G and is denoted by γts (G).
Corresponding to total strong dominating set of G, total strong dominating function can be defined. The minimum
weight of a total strong dominating function is called the fractional total strong domination number of G and is
t (G). A study of total strong dominating functions is carried out in this paper.
denoted by γsf
Keywords: Total Strong Dominating Function, Maximal Total Strong Dominating Function, Basic Maximal Total
Strong Dominating Function
Introduction: Corresponding to total strong dominating sets in a graph, total strong dominating functions may be
defined. The minimality of a total strong dominating function can be characterised. Convex combination of minimal
total strong dominating function is defined and studied. A total strong dominating function is basic, if it cannot
be expressed as a covex combination of minimal total strong dominating functions. Basic total strong dominating
functions are characterised.
Definition 0.1:
Let G = (V, E) be a simple graph without strong isolates. A function
f : V (G) → [0, 1] is called a Total Strong Dominating Function (TSDF), if f (Ns (u)) =
X
f (v) ≥ 1, ∀ u ∈
v∈Ns (u)
V (G), where
Ns (u) = {x ∈ N (u) : deg (x) ≥ deg (u)}.
Definition 0.2:
A TSDF is called a Minimal Total Strong Dominating Function (MTSDF), if whenever g : V (G) → [0, 1] and
g < f , g is not a TSDF.
Definition 0.3: Let G be a graph without isolated vertex. A TSDF (MTSDF) is called a basic TSDF (basic
MTSDF) denoted by BTSDF (BMTSDF) if it cannot be expressed as a proper convex combination of two distinct
TSDFs (MTSDFs ).
Remark 0.4: A BTSDF need not be a BMTSDF.
71
*Corresponding author: M. Kavitha, E-mail: kavinandhu7204@gmail.com
International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020
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Lemma 0.5: Let f and g be two distinct MTSDFs of a graph G with Bfs = Bgs and Pf = Pg . Let δ(v) = f (v)−g(v),
for every v ∈ V . Then
(i) If f (v) = 0 or f (v) = 1, then δ(v) = 0.
X
(ii)
δ(v) = 0, ∀ v ∈ Bfs .
u∈Ns (v)
(iii)
X
δ(v) = 0, ∀ v ∈ Bgs .
u∈Ns (v)
Proof:
(i) We have, f (v) = 1 if and only if g(v) = 1 and
f (v) = 0 if and only if g(v) = 0.
Therefore, (i) follows.
(ii) Let v ∈ Bfs .
Then v ∈ Bgs .
X
X
δ(v)
=
(f (u) − g(u))
v∈Ns (v)
u∈Ns (v)
X
=
X
f (u) −
u∈Ns (v)
g(u)
u∈Ns (v)
= 1 − 1 (since v ∈ Bfs and v ∈ Bgs )
= 0.
Therefore, (ii) follows.
(iii) is similar to (ii).
Lemma 0.6: Let f and g be convex linear combination of MTSDFs g1 , g2 ,...,gn such that f is minimal. Then
n
n
\
[
Bfs = Bgs =
Bgsi , Pf = Pg =
Pgi and g is minimal.
i=1
i=1
Proof:
Let v ∈ Pf .
Then f (v) > 0.
n
n
X
X
Let f =
λi gi , 0 < λi < 1,
λi = 1.
i=1
i=1
Suppose gi (v) = 0, ∀ i. Then f (v) = 0, a contradiction.
n
[
Therefore, gi (v) > 0, for at least one i. Therefore, v ∈
Pgi .
Therefore, Pf ⊆
n
[
Pgi . Suppose v ∈
i=1
i=1
n
[
Pgi .
i=1
Then g(vi ) > 0, for some i. Therefore, f (v) > 0.
n
[
Therefore, v ∈ Pf . Therefore,
Pgi ⊆ Pf .
Therefore, Pf =
n
[
i=1
Pgi . Similarly, Pg =
i=1
Therefore, Pf = Pg =
n
[
n
[
Pgi .
i=1
Pgi . Let v ∈ Bfs .
i=1
Then f (Ns (v)) = 1. Therefore,
n
X
λi gi (Ns (v)) = 1.
i=1
Suppose v ∈
/ Bgsi , for some i, 1 ≤ i ≤ n. Then gi (Ns (v)) > 1.
Since gi (Ns (v)) ≥ 1, for all j , 1 ≤ j ≤ n,
72
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n
X
λi gi (Ns (v)) >
i=1
λi .
i=1
= 1, a contradiction.
Therefore, v ∈
Bgsi ,
Therefore, Bfs ⊆
Let v ∈
n
X
∀ i. Therefore, v ∈
n
\
Bgsi .
i=1
Bgsi .
i=1
n
\
Bgsi . Then gi (Ns (v)) = 1, for all i, 1 ≤ i ≤ n.
i=1
Therefore,
n
\
n
X
λi gi (Ns (v)) =
i=1
n
X
λi = 1.
i=1
Therefore, f (Ns (v)) = 1. Therefore, v ∈ Bfs .
n
n
\
\
Therefore,
Bgsi ⊆ Bfs . Therefore, Bfs =
Bgsi .
i=1
Similarly,
Bgs
=
n
\
Bgsi .
Hence
Bfs
=
i=1
Bgs
i=1
n
\
=
Bgsi .
i=1
Since f is minimal, Bfs weakly dominates Pf .
n
n
\
[
s
That is,
Bgi weakly dominates
Pgi .
i=1
i=1
That is, Bgs weakly dominates Pg .
Hence g is a MTSDF.
Theorem 0.7: Let f be a MTSDF. Then f is a BMTSDF if and only if there does not exist an MTSDF g such
that Bfs = Bgs and Pf = Pg .
Proof:
Suppose f is a BMTSDF.
Suppose there exists a MTSDF g such that Bfs = Bgs and Pf = Pg .
Let S = {a ∈ < : ha = (1 + a)g − af } be a TSDF and Bhs a = Bfs and Pha = Pf .
Then S is a bounded open interval.
Let S = (k1 , k2 ).
Then k1 < −1 < 0 < k2 .
Also hk1 and hk2 are MTSDFs .
hk1 = (1 + k1 )g − k1 f .
(1 + k1 )g hk1
−
.
Therefore, f =
k1
k1
1 + k1
−1
and λ2 =
.
Let λ1 =
k1
k1
Therefore, λ1 and λ2 are positive and λ1 + λ2 = 1.
Therefore, f is a convex combination of g and hk1 .
Therefore, f is not a BMTSDF, a contradiction.
Therefore, there does not exist a MTSDF g such that Bfs = Bgs and Pf = Pg .
Conversely, suppose f is not a BMTSDF.
n
X
Then there exists MTSDFs g1 , g2 ,...,gn such that f =
λi gi ,
where 0 < λi < 1 and
n
X
i=1
λi = 1.
i=1
73
International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020
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Let g =
n
X
µi gi , 0 < µi < 1 and
i=1
Then by lemma 0.6, Bfs = Bgs =
n
X
µi = 1.
i=1
n
\
n
[
i=1
i=1
Bgsi and Pf = Pg =
Pgi and since f is a MTSDF, g is a MTSDF.
Thus there exists a MTSDF g such that Bfs = Bgs and Pf = Pg .
Hence the theorem.
Theorem 0.8: Let f be a MTSDF of a graph G = (V, E) with Bfs = {v1 , v2 , ..., vm } and Pf0 = {u ∈ V : 0 <
f (u) < 1} = {u1 , u2 , ..., un }.
Let A 
= [aij ] be a m × n matrix defined by
 1, if v weakly dominates u
i
j
aij =
 0, otherwise.
n
X
Consider the system of linear equations given by
aij xj = 0, 1 ≤ i ≤ m.
j=1
Then f is a BMTSDF if and only if the above system does not have a non-trivial solution.
Proof:
Suppose f is not a BMTSDF.
Then there exists a MTSDF g such that Bfs = Bgs and Pf = Pg .
Let xj = f (uj ) − g(uj ), 1 ≤ j ≤ n.
Suppose xj = 0, ∀ j , 1 ≤ j ≤ n.
If f (v) = 0, then v ∈
/ Pf = Pg .
Therefore, g(v) = 0.
Therefore, f (v) − g(v) = 0, ∀ v ∈
/ Pf .
That is, f (v) = g(v), ∀ v ∈
/ Pf .
If f (v) = 1, then by Theorem ??, g(v) = 1 and hence f (v) − g(v) = 0.
Therefore, f = g , a contradiction.
Therefore, there exists, some j , 1 ≤ j ≤ n such that xj 6= 0.
Let f (uj1 ) − g(uj1 ) 6= 0.
n
n
X
X
aij xj =
aij (f (uj ) − g(uj ))
j=1
j=1X
=
(f (u) − g(u))
u∈Ns (vi )
(since aij = 1 because vi weakly dominates u)
= 0, by lemma0.5.
Since xj 6= 0, the left hand side has a non-trivial solution.
Conversely, let {x1 , x2 , ..., xn } be a non-trivial solution for the system of
linear equations.
Define g: V (G) → [0, 1] as follows:
 f (v),
if v ∈
/ Pf0
g(v) =
f (v) + xj , if v = uj , 1 ≤ j ≤ n, where M is to be suitably chosen.
M
Since {x1 , x2 , ..., xn } is a non-trivial solution, g 6= f .
xj
Since 0 < f (uj ) < 1, choose Mj > 0 such that 0 < (f (uj ) + M
) < 1, for each j, 1 ≤ j ≤ n.
j
0
Let M = max{M1 , M2 , ..., Mn }.
Choose M to be equal to M’.
74
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X
For any v ∈ V , g(Ns (v)) =
g(u)
u∈Ns (v)
=
X
g(u) +
u∈Ns (v)∩P
=
X
0
f
g(u)
0
f
u∈Ns (v)−P
X
xi
(f (vi ) + 0 ) +
M
u∈Ns (v)−Pf0
ui ∈Ns (v)∩Pf0
X
f (u)
1 X
xi
M0
i
u∈Ns (u)
X
= f (Ns (v)) + M1 0
xi
=
If v ∈ Bfs , then
X
xi =
n
X
i
X
f (u) +
i
aij xi = 0.
j=1
(since v weakly dominates Pf and hence aij = 1).
Therefore, g(Ns (v)) = f (Ns (v)) = 1.
Suppose v ∈
/ Bfs .
Then f (Ns (v)) > 1.
Choose M 00 > 0 such that g(Ns (v)) > 1, ∀ v ∈
/ Bfs .
Let M = max{M 0 , M 00 }.
For this choice of M , we have
X
0 ≤ g(v) ≤ 1 and
g(v) ≥ 1, ∀ v ∈ V
u∈Ns (v)
Therefore, g is a TSDF.
From what we have seen above,
Bfs = Bgs and Pf = Pg .
Since f is a MTSDF, Bfs weakly dominates Pf .
Therefore, Bgs weakly dominates Pg . Therefore, g is a MTSDF.
Hence f is not a BMTSDF.
Corollary 0.9: Let G = (V, E) be a graph without isolated vertices. Let S be a minimal total strong dominating
set of G. Then χs is a BMTSDF.
Proof:
Clearly, χs is a MTSDF.
Let f = χs .
Pf0 = φ.
Therefore from the above theorem, χs is a BMTSDF.
Example 0.10:
Consider Hajo’s Graph H3 :
vs 1
sv3
v2 s
v4 s
s
v5
Define f1 and f2 as follows:
f1 (v1 ) = f1 (v4 ) = f1 (v2 ) = f1 (v6 ) = 0.
f1 (v3 ) = f1 (v5 ) = 1.
75
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0 vs 1
f1 :
v2 0s
0
v4 s
f2 (v1 ) = f2 (v4 ) = f2 (v6 ) = 0.
1
f2 (v2 ) = f2 (v3 ) = f2 (v5 ) = .
2
sv13
s
sv6
0
1 v5
0vs 1
1
1
2v
v22s
f2 :
0
v4 s
s3
s
v5
1
2
s0v6
f1 is a TSDF.
Bfs1 = {v1 , v3 , v4 , v5 }.
Pf1 = {v3 , v5 }.
Bfs1 weakly dominates Pf1 .
Therefore, f1 is a MTSDF.
Therefore, f2 is a TSDF.
Bfs2 = V .
Bfs2 weakly dominates Pf2 .
f2 is a MTSDF.
Pf01 = φ.
Let A = [aij ]m×n be a m × n matrix defined by



1, if vi ∈ Bfs weakly dominates uj in Pf0


aij =
.



 0, otherwise.
Consider the system of linear equations given by
a11 x1 + a12 x2 + . . . + a1n xn = 0
.
.
.
am1 x1 + am2 x2 + . . . + a1n xn = 0.
Since Pf01 = φ, the system of equations does not occur and hence does not have a non-trivial solution.
Therefore, f1 is a BMTSDF.
f2 is a TSDF.
Bfs2 = V, Pf2 = {v2 , v3 , v4 }.
Pf02 = {v2 , v3 , v4 }.
76
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A = [aij ]6×3

1

0

0
=
1


0
0
1
1
0
0
0
1

0

0

0

1


0
1
6×3


x1 + x2 = 0




x =0
2
imply x1 = 0 , x2 = 0


x
=
0
1




x2 = 0
Therefore, f2 is a BMTSDF.
Example 0.11:
0
1
u
P5 :
1
u
v1
v2
1
0
u
u
u
v3
v4
v5
f1 is a TSDF.
Bfs1 = {v1 , v2 , v4 , v5 }.
Pf1 = {v2 , v3 , v4 }.
Bfs1 weakly dominates Pf1 .
v2 v3 v4
A = [aij ]4×3 =
v1
v2
v4
v5
1
0
0
0
0
1
1
0
0
0
0
x1
x2
x3
1
x1 = 0, x2 = 0, x3 = 0.
Therefore, f1 is a BMTSDF.
Example 0.12:
0
t
P9 :
Consider
v1
1
1
0
0
1
1
1
0
v2
v3
v4
v5
v6
v7
v8
v9
t
t
t
t
t
t
Define f1 on V (P6 ) as follows:
f1 (v1 ) = f1 (v4 ) = f1 (v5 ) = f1 (v9 ) = 0.
f1 (v2 ) = f1 (v3 ) = f1 (v6 ) = f1 (v7 ) = f1 (v8 ) = 1.
f1 is a TSDF.
Bfs1 = {v1 , v2 , v3 , v4 , v5 , v6 , v8 , v9 }.
Pf1 = {v2 , v3 , v6 , v7 , v8 }.
Bfs1 weakly dominates Pf1 .
Therefore, f1 is a MTSDF.
77
t
t
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A =
v2 v3 v6 v7
v8
v1
1
0
0
0
0
v2
v3
0
0
0
0
1
1
0
0
0
0
v4
0
1
0
0
0
v5
0
0
1
0
0
v6
0
0
0
1 0
v8
0
0
0
1
0
v9
0
0
0
0
1
xi = 0, ∀ i, 1 ≤ i ≤ 5.
Therefore, f1 is a BMTSDF.
Example 0.13:
0
t
P9 :
Consider
v1
1
1
0
0
v2
v3
v4
v5
t
t
t
t
x1
x2
x3
x4
x5
1
1
1
0
v6
v7
v8
v9
t
0
0
0
0
0
0
0
0
x1
x2
x1
x2
x3
x4
x4
x5
t
t
t
Define f1 on V (P9 ) as follows:
f1 (v1 ) = f1 (v4 ) = f1 (v5 ) = f1 (v9 ) = 0.
f1 (v2 ) = f1 (v3 ) = f1 (v6 ) = f1 (v7 ) = f1 (v8 ) = 1.
f1 is a TSDF.
Bfs1 = {v1 , v2 , v3 , v4 , v5 , v6 , v7 , v8 , v9 }.
Pf1 = {v2 , v3 , v6 , v7 , v8 }.
Bfs1 weakly dominates Pf1 .
Therefore, f1 is a MTSDF.
 
x1
 
x2   


 
1 0 0 0 0
0
 

   x1   
0 1 0 0 0
  0

 x
x   
1 0 0 0 0  1   2  0

 x2  x3   
0 1 0 0 0     0

     
A=
 x3  = x4  =  
0 0 1 0 0 
   

     0
0 0 0 1 0 x4  x5  0


   

 x5
x6   
  0
0 0 0 1 0
x7 
 
0 0 0 0 1
0
x 
 8
x9
xi = 0, ∀ i, 1 ≤ i ≤ 5.
Therefore, f1 is a BMTSDF.
Example 0.14:
1
1
0
0
t
t
t
t
Let P11 :
v1 v2 v3 v4
0
1
1
0
1
1
0
v5
v6
v7
v8
v9
v10
v11
t
t
Define f1 on V (P11 ) as follows:
f1 (v1 ) = f1 (v4 ) = f1 (v5 ) = f1 (v8 ) = f1 (v11 ) = 0.
78
t
t
t
t
t
International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020
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f1 (v2 ) = f1 (v3 ) = f1 (v6 ) = f1 (v7 ) = f1 (v9 ) = f1 (v10 ) = 1.
f1 is a TSDF.
Bfs1 = {v1 , v2 , v3 , v4 , v5 , v6 , v7 , v9 , v10 , v11 }.
Pf1 = {v2 , v3 , v6 , v7 , v9 , v10 }.
Bfs1 weakly dominates Pf1 .
Therefore, f1 is a MTSDF.
A =
v2 v23 v6 v7
v8
v1
1
0
0
0
0
v2
v3
0
0
0
0
1
1
0
0
0
0
v4
0
1
0
0
0
v5
0
0
1
0
0
v6
0
0
0
1 0
v8
0
0
0
1
0
v9
0
0
0
0
1
1
1
1
0
0
1
1
1
1
0
v2
v3
v4
v5
v6
v7
v8
v9
v10
v11
xi = 0, ∀ i, 1 ≤ i ≤ 6.
Therefore, f1 is a BMTSDF.
Again consider
0
t
P11 :
v1
t
t
t
t
x1
x2
x3
x4
x5
t
t
t
Define f2 on V (P11 ) as follows:
f2 (v1 ) = f2 (v5 ) = f2 (v6 ) = f2 (v11 ) = 0.
f2 (v2 ) = f2 (v3 ) = f2 (v4 ) = f2 (v7 ) = f2 (v8 ) = f2 (v9 ) = f2 (v10 ) = 1.
f2 is a TSDF.
Bfs2 = {v1 , v2 , v4 , v5 , v6 , v7 , v10 , v11 }.
Pf2 = {v2 , v3 , v4 , v7 , v8 , v9 , v10 }.
Bfs2 weakly dominates Pf2 .
79
0
0
0
0
0
0
0
0
x1
x2
x1
x2
x3
x4
x4
x5
t
t
t
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v2
v3
v4
v7 v8 v9 v10
v1
1
0
0
0
0 0
0
v2
0
1
0
0
0 0
0
v4
v5
0
0
1
0
0
0
1
0
0 0
0 0 0
v6
0
0
0
1
0
v7
0
0
0
v10
0
0
0
0
0
1 0 0
0 1 0
v11
0
0
0
0
0 0 1
B =
x1
x2
x3
x4
x5
x6
x7
0
0 0
0
0
0
0
0
0
0
xi = 0, ∀ i, 1 ≤ i ≤ 7.
Therefore, f2 is a BMTSDF.
Example 0.15:
Let G = C2n+1 .
Let V (G) = {v1 , v2 , ..., v2n+1 }.
Case (i)
Let n ≡ 0 (mod 2).
Let n = 2k .
Then 2n + 1
= 4k + 1.
 1, if i ≡ 1, 2 (mod 4)
Let f (vi ) =
 0, if i ≡ 3, 0 (mod 4).
Then f is a TSDF.
Bfs = {v2 , v3 , v4 , v5 , ..., v4k+1 }
Pf = {v1 , v5 , ..., v4k+1 , v2 , v6 , ..., v4k−2 }
Clearly, Bfs weakly dominates Pf .
Therefore, f is a MTSDF.
v1 v2 v5
A =
v2
1 0
0 0
v3
0 1
0 0
v4
0 0
0 0
.
1 0
0 1
v5
.
v6 . . .
...
...
...
...
v4k−1v4k−2
v4k+1
0
0
0
x1
0
0
0
0
x2
0
0
0
0
x3
0
0
0
0
x4
.
0
.
.
.
.
.
.
.
.
.
v4k+1
1
x2k+1
0
0
0 0
...
0
0
0
80
International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020
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Therefore, xi = 0, ∀ i, 1 ≤ i ≤ 2k + 1.
Therefore, f is a BTMSDF.
Case (ii)
Let n ≡ 1 (mod 2).
Let n = 2k + 1.
Then 2n + 1
= 4k + 3.
 1, if i ≡ 1, 2 (mod 4)
Let f (vi ) =
 0, if i ≡ 3, 0 (mod 4).
Then f is a TSDF.
Bfs = {v1 , v2 , v5 , v6 , ..., v4k+2 }.
Pf = {v1 , v2 , v5 , v6 , ..., v4k+1 , v4k+2 }.
Clearly, Bfs weakly dominates Pf .
Therefore, f is a MTSDF.
v2
v1 v2 v5 v6 . . .
0 1 0 0 ...
1 0 0 0 ...
v5
0
0
v6
0.
0
v1
A =
v4k+1 v4k+2
0
0
x1
0
x2
0
0
x3
0
0
x4
.
0
.
.
.
.
x2k+1
0
0
0
0
1
...
0
1
0 ...
0
.
.
.
.
.
v4k+2
0
0
1
0 0 ...
Therefore, xi = 0, ∀ i, 1 ≤ i ≤ 2k + 2.
Therefore, f is a BTMSDF.
Case (iii)
Let G = C2n .
Let V (G) = 
{v1 , v2 , ..., v2n }.
 1, if i ≡ 1, 2 (mod 4)
Let f (vi ) =
 0, otherwise.
Clearly,
f is a TSDF.
 V,
if n ≡ 2 (mod 4)
Bfs =
 V − {v1 , v2n }, if n ≡ 1 (mod 4).

{v , v , ..., v
1 5
2n−3 , v2 , v6 , ..., v2n−2 }, if n ≡ 2 (mod 4)
Pf =
{v1 , v5 , ..., v2n−1 , v2 , v6 , ..., v2n },
if n ≡ 1 (mod 4).
81
0
.
International Journal of Trend in Scientific Research and Development(IJTSRD Volume 4 Issue 4 June 2020
Available Online www.ijtsrd.com e-ISSN 2456 - 6470
Clearly, Bfs weakly dominates Pf .
Therefore, f is a MTSDF.
Let n ≡ 2 (mod 4).
v1 v2 v5 v6 . . . v2n−3 v
2n−2
v1
0
1
0 0 ...
0
0
v2
1
0
0
...
0
0
0
0 0 ...
0
0
0
.
A =
.
.
.
.
.
v2n
1
AX = 0 implies X = 0.
Therefore, f is a BMTSDF.
Let n ≡ 1 (mod 4).
v1 v2 v5 v6 . . . v2n−1
v2n
v2
1
0
0
...
0
0
v3
0
1
0 0 ...
0
0
0
0 0 ...
0
0
.
A =
.
.
.
v2n−1
AX = 0 implies X = 0.
Therefore, f is a BMTSDF.
.
.
0
1
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