Lecture 5

Cn ln pn now we look at the two terms assuming that we

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Unformatted text preview: hose that end with Plutonium. C(n) = L(n) + P(n) Now we look at the two terms, assuming that we add one new piece to the end. 15 Given are all chains of length n-1. There are again C(n-1) of those. To which of them can we add a piece of lead? To ALL of them. Therefore, 16 Given are all chains of length n-1. There are again C(n-1) of those. To which of them can we add a piece of Plutonium? Only to those that END with a piece of LEAD!! There are L(n-1) of those. Therefore, L(n) = C(n-1) P(n) = L(n-1) 17 18 C(n) = L(n) + P(n) = C(n-1) + L(n-1) = C(n-1) + C(n-2) Back to our example: This is the Fibonacci formula. However, we have slightly different base cases here. C(3) = C(2) + C(1) = 3 + 2 = 5. That exactly what we had before. s C(1) = 2 C(2) = 3 P or L PL of LP or LL 19 20...
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