[superprismatic]

A string of decimal digits is produced

as follows. The 10 decimal digits are

split into 2 disjoint subsets A and B.

The elements in each subset are

arranged in some order and each has

a pointer to its first element. A

fair coin is now flipped. If heads

comes up, the element under the

pointer in subset A is chosen and the

pointer is moved to the next element

of A. Similarly, if tails comes up,

the element under the pointer in B

is chosen and its pointer advanced.

The pointers advance circularly, so

a pointer goes from the end of its

subset to the beginning of it. This

continues until a digit string of

the desired length is obtained.

For example, suppose A={9,1,2} and

B={5,3,8,6,7,4,0}. Then the sequence

of flips H,H,T,H,T,H,T,T,T,H will

produce the digit string 9,1,5,2,3,

9,8,6,7,1.

What would the ordered subsets be if

the decimal digit string is 4,3,0,1,

8,9,7,6,5,2,4,3,1,0,8,7,9,6,4,0,7,5,

6,2,3 ?

[Guest]

Good afternoon (for me at least) brain stretch.

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groups of 4 and 6

[Guest]

Sounds like a good project for a coding class...I like it.

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A = {4,0,7,6}; B = {3,1,8,9,5,2} Sequence = H,T,H,T,T,T,H,H,T,T,H,T,T,H,T,H,T,H,H,H,H,T,H,T,T

[Guest]

Another good puzzle

[Guest]

  Reveal hidden contents

A = (4,0,7,6)

B = (3,1,8,9,5,2)

Can be solved by hand.

Look at the frequency of numbers between two of the same kind.

For example, between the second and third 4's, there is 3,1,0,8,7,9,6.

What is missing here is 2 and 5. Therefore 2 and 5 cannot be in the same group as 4.

On the other hand between the two 2's, you have:

4,3,1,0,8,7,9,6,4,0,7,5,6

The 4's, 7's, 6's, and 0's all repeat before the 2 finishes it's cycle.

Therefore 4, 7, 6, and 0 cannot be in the same group as 2.

So far we have (4,7,6,0) and (2,5);

Between the second and third 1's, we see a 0 repeating, so 1 must be in the opposite group of 0.

Following similar logic, the answer can be determined fairly quickly.

[superprismatic]

  On 9/15/2009 at 10:37 PM, mmiguel1 said:

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A = (4,0,7,6)

B = (3,1,8,9,5,2)

Can be solved by hand.

Look at the frequency of numbers between two of the same kind.

For example, between the second and third 4's, there is 3,1,0,8,7,9,6.

What is missing here is 2 and 5. Therefore 2 and 5 cannot be in the same group as 4.

On the other hand between the two 2's, you have:

4,3,1,0,8,7,9,6,4,0,7,5,6

The 4's, 7's, 6's, and 0's all repeat before the 2 finishes it's cycle.

Therefore 4, 7, 6, and 0 cannot be in the same group as 2.

So far we have (4,7,6,0) and (2,5);

Between the second and third 1's, we see a 0 repeating, so 1 must be in the opposite group of 0.

Following similar logic, the answer can be determined fairly quickly.

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Very nice explanation of a good solution method!

[Guest]

  Reveal hidden contents

4076,318952

[superprismatic]

  On 9/17/2009 at 11:11 PM, Dawid said:

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4076,318952

Nice going!