Day 7b: The Sum of Its Parts - Part 2
As you're about to begin construction, four of the Elves offer to help. "The sun will set soon; it'll go faster if we work together." Now, you need to account for multiple people working on steps simultaneously. If multiple steps are available, workers should still begin them in alphabetical order.
Each step takes 60
seconds plus an amount corresponding to its letter: A=1
, B=2
, C=3
, and so on. So, step A
takes 60+1=61
seconds, while step Z
takes 60+26=86
seconds. No time is required between steps.
To simplify things for the example, however, suppose you only have help from one Elf (a total of two workers) and that each step takes 60
fewer seconds (so that step A
takes 1
second and step Z
takes 26
seconds). Then, using the same instructions as above, this is how each second would be spent:
Second Worker 1 Worker 2 Done
0 C .
1 C .
2 C .
3 A F C
4 B F CA
5 B F CA
6 D F CAB
7 D F CAB
8 D F CAB
9 D . CABF
10 E . CABFD
11 E . CABFD
12 E . CABFD
13 E . CABFD
14 E . CABFD
15 . . CABFDE
Each row represents one second of time. The Second column identifies how many seconds have passed as of the beginning of that second. Each worker column shows the step that worker is currently doing (or .
if they are idle). The Done column shows completed steps.
Note that the order of the steps has changed; this is because steps now take time to finish and multiple workers can begin multiple steps simultaneously.
In this example, it would take 15
seconds for two workers to complete these steps.
With 5 workers and the 60+ second step durations described above, how long will it take to complete all of the steps?
Builds off of Day 7 part 1 by maintaining an array of workers
.
const makeGetTimeNeeded =
(minSeconds: number) =>
(node: string) =>
node.charCodeAt(0) - 'A'.charCodeAt(0) + 1 + minSeconds;
const getTimeToDoParallelWork = ({graph, minSeconds, numOfWorkers}: ISimulation) => {
const queue = getOrderedToplogicalSortQueue(graph);
const getTimeNeeded = makeGetTimeNeeded(minSeconds);
const workers: {node: string; timeNeeded: number}[] = [];
let timeLapsed = 0;
// keep looping even if the queue is empty to increment the timeLapsed
while (!queue.isEmpty() || workers.length > 0) {
// free up any workers that are done, and update remaining nodes
// still in queue to tell 'em the job is done
for (let i = 0; i < workers.length; i++) {
if (workers[i].timeNeeded <= timeLapsed) {
queue.values.forEach(v => v.mustBeAfter.delete(workers[i].node));
workers.splice(i, 1);
i--; // stay at current index if we remove a worker
}
}
queue.heapify();
// if there are available workers and jobs that can be started
while (
workers.length < numOfWorkers &&
!queue.isEmpty() &&
queue.peek()!.mustBeAfter.size === 0
) {
const { node } = queue.dequeue()!;
workers.push({node, timeNeeded: timeLapsed + getTimeNeeded(node)});
}
timeLapsed++;
}
return timeLapsed - 1;
};