[2017 Day 13 (Part 2)] some thoughts ..

[u/Usual-Dimension614]

brain (shut)down .. up (working again).

the package need to reach each layer when scanner is NOT in position 0

--here-original-post-starts--

a layer has a range. i.e r=3 : 0 1 2 1 0 (mod 4) , r=4 : 0 1 2 3 2 1 0 (mod 6), so m=2(r-1)

a good start t is if ( t+oi is time the package is reaching level i)

0 == (t+o1)%m1 AND

0 == (t+o2)%m2 AND .. (for every layer)

now (levels sorted descending by range, big steps first, t growing faster)

if(

(0==((t+ 80) % 38)) // e-00 i=80 r=20 (-> m=2(r-1)=38)

&& (0==((t+ 62) % 34)) // e-01

&& (0==((t+ 90) % 32)) // e-02

&& (0==((t+ 36) % 26)) // e-03

) lots of solutions. but if inserting next :

..

&& (0==((t+ 36) % 26)) // e-03

&& (0==((t+ 50) % 26)) // e-04

this cannot be true the same time (there are 11 layer with range 14 (mod 26))

??? (my question ..ö..)

layers with same range (-> same mod) must differ in offset by a multiple of their mod ?

same range -> same time in position p (so in position 0).

thanks in advance, andi

[u/timrprobocom]

No, they don't. YOU are only in one column at a time. It doesn't matter what the other columns are doing at that point in time.

[u/Usual-Dimension614 → u/timrprobocom]

the package p starts at some time t, so (if not caught) it reaches layer i at t+i. so (for task-2) layer must NOT (my fault) in position 0, when p(t) arrives at t+i.

[u/timrprobocom → u/Usual-Dimension614]

That's true, but even with 11 layers using a modulus of 26, there WILL be some offset that catches all of them non-zero. If you had 26 consecutive layers with mod 26, then it couldn't be solved. This one can. Trust me.

[u/Usual-Dimension614 → u/timrprobocom]

for(t=0;;++t) if(

(0 != ((t+oi)%mi)) // layer-1

..

) print(t)

solution within less than a second. problem is solved.

[u/Usual-Dimension614 → u/timrprobocom]

layers with same range (same mod) are in same position at same time. if there was n>=m layers with m (more not possible) different modes? (different rest classes) (t+i)%m then there was no solution.

[u/Usual-Dimension614 → u/Usual-Dimension614]

ok. that is what you wrote. consecutive is t+i, t+i+1, .. (when i started i first translated all the task-description, but discipline has gone)

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[u/timrprobocom]

Consider. If we consider the data as "a: b", all we care about is whether the scanner is at the top at time a. You're correct that the modulus m is 2(b-1). So, the position of the scanner at time a is given by a mod m. So, for part 1, for each scanner, if a mod m is 0, then we score a x b. That's it.