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#5
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The easy proof is the commutative property of multiplication, and my calculus is too rusty to do it for real.
Here's my best attempt:Given x, an odd number of containers in scoring position, let y = the optimal stack height (which you have proofed) y = x/2 In the case of an odd number of containers, y will be a complex fraction (where x = 11, y = 5 1/2). Since we cannot have 1/2 a container, we must round y either up or down, by adding 0.5 or subtracting 0.5. We can calculate 2 scores based on (1) adding another container to the stack and (2) not adding another container to the stack. s1 = (y+0.5) (y-0.5) s2 = (y-0.5) (y+0.5) Given the commutative property of multiplication (which states that A*B=B*A), we prove that s1 = s2. |
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