1 // Core Part 1 about the 3n+1 conjecture |
1 // Core Part 1 about the 3n+1 conjecture |
2 //============================================ |
2 //============================================ |
3 |
3 |
4 object C1 { |
4 object C1 { |
5 |
5 |
6 //(1) Complete the collatz function below. It should |
6 // ADD YOUR CODE BELOW |
7 // recursively calculate the number of steps needed |
7 //====================== |
8 // until the collatz series reaches the number 1. |
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9 // If needed, you can use an auxiliary function that |
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10 // performs the recursion. The function should expect |
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11 // arguments in the range of 1 to 1 Million. |
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12 |
8 |
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9 |
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10 //(1) |
13 def collatz(n: Long) : Long = ??? |
11 def collatz(n: Long) : Long = ??? |
14 |
12 |
15 |
13 |
16 //(2) Complete the collatz_max function below. It should |
14 //(2) |
17 // calculate how many steps are needed for each number |
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18 // from 1 up to a bound and then calculate the maximum number of |
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19 // steps and the corresponding number that needs that many |
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20 // steps. Again, you should expect bounds in the range of 1 |
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21 // up to 1 Million. The first component of the pair is |
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22 // the maximum number of steps and the second is the |
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23 // corresponding number. |
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24 |
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25 def collatz_max(bnd: Long) : (Long, Long) = ??? |
15 def collatz_max(bnd: Long) : (Long, Long) = ??? |
26 |
16 |
27 //(3) Implement a function that calculates the last_odd |
17 //(3) |
28 // number in a collatz series. For this implement an |
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29 // is_pow_of_two function which tests whether a number |
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30 // is a power of two. The function is_hard calculates |
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31 // whether 3n + 1 is a power of two. Again you can |
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32 // assume the input ranges between 1 and 1 Million, |
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33 // and also assume that the input of last_odd will not |
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34 // be a power of 2. |
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35 |
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36 def is_pow_of_two(n: Long) : Boolean = ??? |
18 def is_pow_of_two(n: Long) : Boolean = ??? |
37 |
19 |
38 def is_hard(n: Long) : Boolean = ??? |
20 def is_hard(n: Long) : Boolean = ??? |
39 |
21 |
40 def last_odd(n: Long) : Long = ??? |
22 def last_odd(n: Long) : Long = ??? |