author | Christian Urban <christian.urban@kcl.ac.uk> |
Wed, 25 Nov 2020 17:28:48 +0000 | |
changeset 369 | 569aea2d052b |
parent 345 | 40657f9a4e4a |
permissions | -rw-r--r-- |
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// Preliminary Part about the 3n+1 conjecture |
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//============================================ |
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object CW6a { |
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//(1) Complete the collatz function below. It should |
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// recursively calculate the number of steps needed |
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// until the collatz series reaches the number 1. |
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// If needed, you can use an auxiliary function that |
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// performs the recursion. The function should expect |
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// arguments in the range of 1 to 1 Million. |
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def collatz(n: Long) : Long = ??? |
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//(2) Complete the collatz_max function below. It should |
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// calculate how many steps are needed for each number |
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// from 1 up to a bound and then calculate the maximum number of |
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// steps and the corresponding number that needs that many |
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// steps. Again, you should expect bounds in the range of 1 |
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// up to 1 Million. The first component of the pair is |
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// the maximum number of steps and the second is the |
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// corresponding number. |
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def collatz_max(bnd: Long) : (Long, Long) = ??? |
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//(3) Implement a function that calculates the last_odd |
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// number in a collatz series. For this implement an |
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// is_pow_of_two function which tests whether a number |
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// is a power of two. The function is_hard calculates |
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// whether 3n + 1 is a power of two. Again you can |
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// assume the input ranges between 1 and 1 Million, |
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// and also assume that the input of last_odd will not |
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// be a power of 2. |
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def is_pow_of_two(n: Long) : Boolean = ??? |
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def is_hard(n: Long) : Boolean = ??? |
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def last_odd(n: Long) : Long = ??? |
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} |
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