sshine
commented
5 years ago The opposite problem of finding the first N for which at least P print jobs are necessary is also interesting. For example, in order to print 7 copies of a status report, you can't go with
- P = 1, since 2² = 4 gives too few copies and 3² = 9 gives too many copies,
- P = 2, since 2² + 1² = 5 gives too few copies and 2² + 2² = 8 gives too many copies,
- P = 3, since 2² + 1² + 1² = 6 gives too few copies and 2² + 2² + 1² = 9 gives too many copies,
But you can go with P = 4 since 2² + 1² + 1² + 1² = 7.
But one problem at a time.
cmccandless
kytrinyx
ErikSchierboom
petertseng
coriolinus
This exercise idea is based on a story from my current work place: My boss'es Ubuntu had a bug that made his printer, when asked to print N copies of something, instead print N² copies. So for our Thursday status meeting, he'd hope that the number of people who show up, for whom he would bring a copy of his status report, was targetable with as few print requests as possible.
For example, if 13 people would show up to the meeting, he would first ask the printer of 3 copies (yielding 9 copies), and then 2 copies (yielding 4 copies), with a final result of 3² + 2² = 13 copies. For some numbers of employees, this was simply not possible with two print jobs, and he'd need to consult the printer a third time.
The exercise is to write a function that computes the groan factor, being the minimal number of print jobs it takes to print N copies of a status report when the printer gives you back N² copies.
Examples:
The suggested exercise name underlines the kafkaesque hope that a specific number of employees N show up on Thursdays such that ∃ a, b ∈ ℕ₀: a² + b² = N for optimal working conditions.