Section 8.1
Step 1: Let
M
=
O. Output the words "empty set."
Step 2: for
i =
1 to
2
n

1
• replace
M
by
M
+
1;
• write
M
=
101102 •••
IOn
as an ndigit number in base 2;
• for
k
=
1 to
n,
if
10k
=
1, output
ak.
221
Each value of
i
in Step 2 yields one subset of {
aI,
... ,
an}
and, when Step 2 is complete, all subsets
have been output.
17.
(a)
[BB] i.
S
=
1;
S
=
1
+
(3)(2)
= 5;
S
=
5
+
2(22)
=
+
8
=
3.
ii.
S
=
2;
S
=
3
+
2(2)
=
1;
S
=
1
+
1(2)
=
3.
(b) i.
S
=
1;
S
=
1
+
0(5)
=
1;
S
=
1
+
3(5
2
)
=
1
+
75
=
76.
ii.
S
=
3;
S
=
0
+
3(5)=15;
S
=
1
+
15(5)
=
76.
(c)
i.
S
=
4;
S
=
4
+
5( 1)
=
4  5
= 9;
S
=
9
+
6( _1)2
=
+
6
=
3;
S
=
+
(4)( _1)3
=
3  4( 1)
=
+
4
=
1.
ii.
S
=
S
=
6  4( 1)
=
10;
S
=
5
+
1O( 1)
=
5;
S
=
4  5(1)
=
(d) i.
S
=
7;
S
=
7
+
16(3)
=
+
48
=
41;
S
=
41
+
0(3
2
)
=
41;
S
=
41
+
(40)(3
3
)
=
41  40(27)
=
41 
1080
= 1039;
S
= 1039
+
4
)
S
+
17(3
5
)
+
17(243)
+
4131
=
3092.
ii.
S
=
17;
S
=
0
+
17(3)
=
51;
S
= 40
+
51(3)
=
113;
S
=
0
+
113(3)
=
339;
S
=
16
+
339(3)
=
1033;
S
=
+
1033(3)
=
3092.
18. [BB] The first value of Sis
S
=
an.
With
i
=
1 in Step 2, the value of Sis
an1
+
Sx
=
+
anx.
With
i
=
2 in Step 2, the value of Sis
an2
+Sx
=
+
(an1 +anx)x
=
an_2+an_1x+anx2.
With
i
=
3 in Step 2, the value of
Sis an
+
=
an
+
an2X
+
an_1x2
+
anx
3
.
With
i
=
the value of
Sis an n
+
an(n1)X
+
an_(n_2)X
2
+ .
.. +
n
=
ao
+
a1X
+
a2x
2
+
... +
n
as desired.
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 Summer '10
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 Linear Algebra, Graph Theory, Prime number, TEP

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