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Ugo [173]
3 years ago
12

If after a run of arc consistency during the backtracking search we end up with the filtered domains of *all* of the not yet ass

igned variables being empty, this means the CSP has no solution. True or false?
Computers and Technology
1 answer:
Irina-Kira [14]3 years ago
3 0

Answer:

True

Explanation:

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MrMuchimi

Answer:

EA9_{16} = 3753

CB2_{16} = 3250

(1011 1110 1101 1011 1010)_2 = 781754

(1010 1000 1011 1000 1110 1101)_2 = 11057389

(1011 1110 1101 1011 1010)_2 = BEDBA

(1010 1000 1011 1000 1110 1101)_2 =  A8B8ED

74510_8= 221416

67210_8 = 203212

Explanation:

Solving (a): To base 10

(i)\ EA9_{16

We simply multiply each digit by a base of 16 to the power of their position.

i.e.

EA9_{16} = E * 16^2 + A * 16^1 + 9 * 16^0

EA9_{16} = E * 256 + A * 16 + 9 * 1

In hexadecimal

A = 10; E = 14

So:

EA9_{16} = 14 * 256 + 10 * 16 + 9 * 1

EA9_{16} = 3753

(ii)\ CB2_{16}

This gives:

CB2_{16} = C * 16^2 + B * 16^1 + 2 * 16^0

CB2_{16} = C * 256 + B * 16 + 2 * 1

In hexadecimal

C = 12; B =11

So:

CB2_{16} = 12 * 256 + 11 * 16 + 2 * 1

CB2_{16} = 3250

Solving (b): To base 10

(i)\ (1011 1110 1101 1011 1010)_2

We simply multiply each digit by a base of 2 to the power of their position.

i.e.

(1011 1110 1101 1011 1010)_2 = 1 * 2^{19} + 0 * 2^{18} + 1 * 2^{17} + 1 * 2^{16} +1 * 2^{15} + 1 * 2^{14} + 1 * 2^{13} + 0 * 2^{12} + 1 * 2^{11} + 1 * 2^{10} + 0 * 2^9 + 1 * 2^8 +1 * 2^7 + 0 * 2^6 + 1 * 2^5 + 1 * 2^4 + 1 * 2^3 + 0 * 2^2 + 1 * 2^1 + 0 * 2^0

(1011 1110 1101 1011 1010)_2 = 781754

(ii)\ (1010 1000 1011 1000 1110 1101)_2

(1010 1000 1011 1000 1110 1101)_2 = 1 * 2^{23} + 0 * 2^{22} + 1 * 2^{21} + 0 * 2^{20} +1 * 2^{19} + 0 * 2^{18} + 0 * 2^{17} + 0 * 2^{16} + 1 * 2^{15} + 0 * 2^{14} + 1 * 2^{13} + 1 * 2^{12} +1 * 2^{11} + 0 * 2^{10} + 0 * 2^9 + 0 * 2^8 + 1 * 2^7 + 1 * 2^6 + 1 * 2^5 + 0 * 2^4 + 1*2^3 + 1 * 2^2 + 0 * 2^1 + 1 * 2^0

(1010 1000 1011 1000 1110 1101)_2 = 11057389

Solving (c): To base 16

i.\ (1011 1110 1101 1011 1010)_2

First, convert to base 10

In (b)

(1011 1110 1101 1011 1010)_2 = 781754

Next, is to divide 781754 by 16 and keep track of the remainder

781754/16\ |\ 48859\ R\ 10

48859/16\ |\ 3053\ R\ 11

3053/16\ |\ 190\ R\ 13

190/16\ |\ 11\ R\ 14

11/16\ |\ 0\ R\ 11

Write out the remainder from bottom to top

(11)(14)(13)(11)(10)

In hexadecimal

A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.

(11)(14)(13)(11)(10)=BEDBA

So:

(1011 1110 1101 1011 1010)_2 = BEDBA

ii.\ (1010 1000 1011 1000 1110 1101)_2

In b

(1010 1000 1011 1000 1110 1101)_2 = 11057389

Next, is to divide 11057389 by 16 and keep track of the remainder

11057389/16\ |\ 691086\ R\ 13

691086/16\ |\ 43192\ R\ 14

43192/16\ |\ 2699\ R\ 8

2699/16\ |\ 168\ R\ 11

168/16\ |\ 10\ R\ 8

10/16\ |\ 0\ R\ 10

Write out the remainder from bottom to top

(10)8(11)8(14)(13)

In hexadecimal

A = 10; B = 11; C = 12; D = 13; E = 14; F = 15.

(10)8(11)8(14)(13) = A8B8ED

So:

(1010 1000 1011 1000 1110 1101)_2 =  A8B8ED

Solving (d): To octal

(i.)\ 74510

Divide 74510 by 8 and keep track of the remainder

74510/8\ |\ 9313\ R\ 6

9313/8\ |\ 1164\ R\ 1

1164/8\ |\ 145\ R\ 4

145/8\ |\ 18\ R\ 1

18/8\ |\ 2\ R\ 2

2/8\ |\ 0\ R\ 2

Write out the remainder from bottom to top

74510_8= 221416

(ii.)\ 67210

Divide 67210 by 8 and keep track of the remainder

67210/8\ |\ 8401\ R\ 2

8401/8\ |\ 1050\ R\ 1

1050/8\ |\ 131\ R\ 2

131/8\ |\ 16\ R\ 3

16/8\ |\ 2\ R\ 0

2/8\ |\ 0\ R\ 2

Write out the remainder from bottom to top

67210_8 = 203212

8 0
3 years ago
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