1007 14Hamiltonian2_10

1007 14Hamiltonian2_10 - ENGG1007 Foundations of Computer...

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1 ENGG1007 Foundations of Computer Science Hamiltonian Graphs 2 Professor Francis Chin, Dr SM Yiu November 15/17, 2010 Chapter 9.5
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2 ENGG1007 FCS Rotating Drum First bit Third bit Second bit
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3 ENGG1007 FCS Rotating Drum Positioning revisited The 2 n positions of the drum can be determined by the contacts of n rings. The innermost ring is for the first bit, …and the outermost ring for the last bit. 111 000 110 001 101 010 100 011 First bit Second bit Third bit Errors might easily be introduced at the position transition. For example: 001 to 010, we might have 001 ____ 010 Perfect transition 001 010 000 Gap introduces error 001 010 011 Overlap introduces error
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4 ENGG1007 FCS Rotating Drum Positioning revisited 111 000 110 001 101 010 100 011 100 000 101 001 111 011 110 010 First bit Second bit Third bit Errors might easily be introduced at the transition from one position to the next if the change of the bit positions is more than one bit. EG. Transition between 111 and 000. Any bit representation is possible. Thus the bit representations of two adjacent positions should not differ too much (with only 1 bit transition). 100 101 000
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5 ENGG1007 FCS Gray Code Consider bit strings {000,001,010,011,100,101,110,111} Can you permute them so that adjacent bit strings defer by one bit only? Transform to G(V,E), where V = set of bit strings, E = { (x,y) if x and y differ by 1 bit } 000 001 010 100 101 110 111 011 First bit Second bit Third bit It is easy to see that G is a hypercube. The solution for the rotating drum positioning problem is to find a Hamiltonian cycle on hypercube.
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6 ENGG1007 FCS Hamiltonian Circuit on Hypercube The HC on hypercube can be constructed by induction. Q 2 01 10 11 00 001 000 100 010 110 111 011 101 Q 3 0000 1000 0001 1001 0100 1100 0101 1101 0010 1010 0011 1011 0110 1110 0111 1111 Q 4 Assume (0 x 1 x n-1 , 1 x 1 ... x n-1 ) is an edge in HC in Q n . Hypercube Q n+1 is constructed by making 2 half-cube Q n , The HC for Q n+1 : follow the HC in the “0”-half-cube until 0 x 1 ... x n-1 0, then jump to 1 x 1 ... x n-1 0 follow the HC in the “1”-half-cube until 1 x 1 ... x n-1 1, then jump to 0 x 1 ... x n-1 1 and follow the remaining HC in the “0”-half-cube.
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1007 14Hamiltonian2_10 - ENGG1007 Foundations of Computer...

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