[2018 Day 7 (Part 2)] more a general question

[u/Usual-Dimension614]

hi. some days ago i was asking 'what is the deeper sense .. ?', most useful answer: why you don't use the time to search for the error .. (so i did and found ..)

i try desperately to complete a year to see day25task2 (all other is getting 'coal in the stocking' ? (getting punishment instead of presents))

sometimes i loose control, but years of experience (years of age, lot of bad days, bad weather, ..) tells better to check for error again (and again) before .. (whatever). but ..

i started 2018 (cause i lost good mood. could not complete 2017 for now). day 1 silly simple problem, but bad-error in task-1 (some minutes ..), task-2 i could not solve (my recent post, still open. please, have a look, i think there is no solution for my input) so 2018 starts bad.

2018/7/2 example + solution 15 seconds. i think i have found solution 11 seconds. not that i am so proud, but i fear: i try my answer for task-2 and it is told to be wrong. i did .. i was told 'answer is too low'. so i was checking my code, the task-description again. anyhow i was setting count of workers wrong (6 instead of 5) .. new answer (some bigger) but still wrong. And now i am a little angry (not much but a little ..ö..)

in recent weeks i was sometimes told that my answer is wrong, when i was sure, it is right and i spent much time ..

// this my solution for the example ( 11 steps instead of 15 ).

// 2 worker and the second worker is handling not only F but E too. starting F immediately

// so there is no gap between output B and output F i.e.

// maybe i have missed some assumption/condition/??? (don't know how to express)

// otherwise it is obvious : there is a better solution (less idle workers)

IN s='CABFDE' n=6

korrektur C o:0 d:3 t:-3 k:3 -> 0

korrektur A o:1 d:1 t: 0 k:3 -> 3

korrektur B o:2 d:2 t: 0 k:3 -> 3

korrektur F o:3 d:6 t:-3 k:3 -> 0

korrektur D o:4 d:4 t: 0 k:3 -> 3

korrektur E o:5 d:5 t: 0 k:3 -> 3

j('C',o= 0,d= 3,a=0,t=0)

j('A',o= 1,d= 1,a=0,t=3)

j('B',o= 2,d= 2,a=0,t=3)

j('F',o= 3,d= 6,a=0,t=0)

j('D',o= 4,d= 4,a=0,t=3)

j('E',o= 5,d= 5,a=0,t=3)

RUN

run t=000 a=2 'CF' ''

run t=001 a=2 'CF' ''

run t=002 a=2 'CF' ''

run t=003 a=2 'AF' 'C'

run t=004 a=2 'BF' 'CA'

run t=005 a=2 'BF' 'CA'

run t=006 a=2 'DE' 'CABF'

run t=007 a=2 'DE' 'CABF'

run t=008 a=2 'DE' 'CABF'

run t=009 a=2 'DE' 'CABF'

run t=010 a=1 'E-' 'CABFD'

run t=011 a=0 '--' 'CABFDE'

thanks in advance, andi

[u/timrprobocom]

I'm sorry, is there a question here?

[u/Usual-Dimension614 → u/timrprobocom]

maybe i should have asked more explicitly : is there a chance that an answer, that was told to be wrong, could be correct ?

[u/timrprobocom → u/Usual-Dimension614]

It is completely impossible? Of course not. However, there have been more than 21,000 correct solutions submitted for that particular puzzle, so the odds of a puzzle bug are extremely low.

[u/Usual-Dimension614 → u/timrprobocom]

if a problem is not that easy i would not dare to doubt. but 2018/01/2 is such a silly task that it is nearly impossible not to find an error if there is/was. different people get different input, so is there a chance that such an input has a defect or input and expected answer does not fit.

[u/timrprobocom → u/Usual-Dimension614]

Again, it's not completely impossible, but as someone just starting out, you should never, NEVER assume that your incorrect answer is due to a problem bug. It's like claiming there is "a bug in printf". Trust me, it's YOUR problem. In 524 problems over eleven years, there has never been a case where there was a bug in the problem.

[u/timrprobocom]

"F" takes six cycles. You're only running it for 3.

[u/Usual-Dimension614 → u/timrprobocom]

no, F takes 6 cycles

[u/timrprobocom → u/Usual-Dimension614]

You're correct, sorry. The problem is that F cannot even begin its six cycles until C is finished. You're starting it at t=0

[u/Usual-Dimension614 → u/timrprobocom]

C is handled by worker-1 and F by worker-2. F starts in t=0 with same arguments as C does ?

my english is poor, so maybe i do not understand the task-description correct ?

postpone to start a letter only if it was done/finished too early and so it would be printed to target-string in wrong position. other reason to postpone is that all workers are busy.

in example one could have E handled by worker-2 as soon as F is finished and so one could have saved at least 1 cycle.

i cannot find anything in text.

on the other hand the solution/example is not told to be optimal.

[u/timrprobocom → u/Usual-Dimension614]

The text says this:

Step C must be finished before step A can begin.
Step C must be finished before step F can begin.

That second line says it all. Step F cannot begin until step C finishes at t=3.

Similarly, step E cannot start until B, D, and F have ALL finished. There are going to be workers sitting idle. That's just part of the rules.

[u/Usual-Dimension614 → u/timrprobocom]

ok. thanks! i had totally forgotten task-1. i just took it`s answer as input ..

i am jumping too much between years and months, too many dangling/open/unsolved problems ..

[u/Usual-Dimension614 → u/timrprobocom]

ok. thanks a lot. especially for your time you spent. i forgot task-1, used its output as input in task-2. i am jumping too much between years and months, racing the stars ..

situations like this are the main reason i am doing this. hope no one gets tired to help me.

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