How many bit strings of length 28
- Have at least one consecutive 000?
- Doesn't have consecutive 000's?
I'm using a TI-nspire calculator. Can I do it with the nCr function?
I tried to do it but I did not find a way.
Thank you.
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Let $a_n$ be a bit string of length n without 000, then it can be
$a_{(n-3)}$ with 100 added at end,
or $a_{(n-2)}$ with 10 added at end,
or $a_{(n-1)}$ with 1 added at end.
So $a_n = a_{(n-1)} +a_{(n-2)} + a_{(n-3)}$
starting with $a_0 = 1, a_1=2, a_2 = 4$
The ending of any successful chain can be categorised as 1(111,101,011,001) 10(110,010) or 100.
1 can be added to any successful chain of length (n-1) no matter what it ended with.
10 can be added to any successful chain of length (n-2) no matter what it ended with.
100 can be added to any successful chain of length (n-3) no matter what it ended with.
true blue anil
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i'm beginning with discrete mathematics, could you give me a little bit more advice on what you wrote. Thank – Pierre-Luc Bolduc Jul 13 '15 at 04:03
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Further explanation added. – true blue anil Jul 13 '15 at 06:28
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thank you, i did a recurence and i found 29 249 425 without consecutive 000 and to have at least one consecutive 000 i have to do 2^28 - 29 249 425. Does it seams to be good? – Pierre-Luc Bolduc Jul 13 '15 at 19:57
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You're welcome ! Look up Tribonacci numbers on OEIS to check your figures. – true blue anil Jul 14 '15 at 04:07
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Thank :) and could you explain me how did you find a0 =1 a1=1 a2=4 – Pierre-Luc Bolduc Jul 14 '15 at 15:22
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In recurrence relations, you may need a few starting values. a1, by the way, was given as 2 [1 and 0] similarly, a2 is 4 [00,01,10,11] You should be able to check that a0+a1+a2 = 7 is correct for a 3 bit string (all but 000 are successful chains) – true blue anil Jul 14 '15 at 17:43