## Jane’s Homework

### December 12, 2017

Today’s exercise is from user Jane, who needs homework help:

Consider all partitionings of a list of positive integers into two partitions. For each partitioning, compute the sums of the two partitions, then compute the least common multiple of the two sums. Report the maximum of all possible least common multiples. For example, given the list (2 3 4 6), the possible partitionings, the associated sums, and the least common multiples, are:

() (2 3 4 6) - 0 15 - 0 (2) (3 4 6) - 2 13 - 26 (3) (2 4 6) - 3 12 - 12 (2 3) (4 6) - 5 10 - 10 (4) (2 3 6) - 4 11 - 44 (2 4) (3 6) - 6 9 - 18 (3 4) (2 6) - 7 8 - 56 (2 3 4) (6) - 9 6 - 18 (6) (2 3 4) - 6 9 - 18 (2 6) (3 4) - 8 7 - 56 (3 6) (2 4) - 9 6 - 18 (2 3 6) (4) - 11 4 - 44 (4 6) (2 3) - 10 5 - 10 (2 4 6) (3) - 12 3 - 12 (3 4 6) (2) - 13 2 - 26 (2 3 4 6) () - 15 0 - 0Thus, the maximum least common multiple is 56.

Jane writes that she wants to be able to recursively find all partitions given a list, and she thinks she will have to use `lambda`

in her program.

Your task is to write a program that finds the maximum least common multiple of the part-sums of all possible 2-partitionings of a list of positive integers. When you are finished, you are welcome to read or run a suggested solution, or to post your own solution or discuss the exercise in the comments below.

Two methods to partition, one recursive and the other using the powerset function from the itertools documentation.

Here’s another way: iteratively generate combinations but bail out when the combination sum gets too large (this works for any value of limit and it’s straightforward to generalize to multisets):

Here’s a Haskell version that doesn’t use explicit recursion. The part2s

function uses replicateM (the monadic version of replicate, for lists) to

generate all length 4 (in our example) combinations of Left and Right, the two

constructors for Either. For each combination we zip it with the original list,

using the “reverse application operator”, to produce a list of Eithers which is

then partitioned into two lists.

A couple of solutions in Racket.

Here’s a solution in C.

The code iterates over the power set of the input list. The approach is based on Loopless Gray binary generation (Algorithm L in TAOCP section 7.2.1.1), which calculates a single element index to be added or removed on each iteration. Partition sums are re-calculated with an addition or subtraction of the updated element. LCM is then calculated on these updated sums.

Output:

I forgot to free the memory I allocated in the calc_janes_homework function.

maskandfocusshould both be free’d before the function returns.Here’s a solution in racket.