G.A. Puzzle #1

Introducing this library, I want to kick off with an exciting puzzle. I haven't made a video or a picture of it yet, sorry about that, but here goes Standard Notation:

Air Altar Puzzle:
Gopper at [2,1].
Orbs: A[-3,-7], B[-1, 7] and C[7,2].


A cuppa to the one who solves it first!

Theme:

Spoiler:

Is the shortest 14 ticks?

icy001 wrote:

Is the shortest 14 ticks?

Succumb! Conform!

https://gopacademy.forumotion.com/t14-notation-in-gop

What?

I'm asking if the shortest solution is 14 ticks! Or else is it 15 ticks?

Edit: Or 13? :o

I've got a solution... that is a) exceptionally unorthodox, b) probably not even that quick, c) utterly useless if you had other orbs to score and d) quite possibly doesn't even work.

Oh well, here it is:
(Using ctrl-running here)
Also since Timo and I use different ways to describe tick counting I'm just going to describe it by the positions.

-Run OFF, click on orb at [-3,7], switch run ON and hold to the point where the orb is at [0,4] and player is at [0,3]
-Hold orb at [7,2] until orb is at [4,2]
-Run ON, click orb at [-1,7] until orb is at [-1,5] and player is at [0,5] ::::: This is where I'm not sure it works. alternatively could just tap is and then run to that spot but i think that would be slower.
-Run OFF, click orb that is now at [4,2], player will walk to [1,5] and tap the orb
-switch to repel, repel tap orb now at [0,4]
-Run ON, run to [0,7], tap orb now at [-1,5], Click again to score.

And there we have it!

Multiple errors, but the funniest one is that you can't step on (0, 7) on the air altar :D

Didn't even notice this was on the air altar. shows how much attention I pay.

Not sure if it's the fastest, but Lightgrace and I came up with the following solution:

1w1n(1) + 0(1) + C(2) + B(2) + C(1) + 2e1s(1) + pre.2s(1) + A(1) + 2n(1) + 0(1) + A[2s](3) + B(1) = 16 ticks

Am I missing a prototick before B(1) at the end?

Also, a funnier solution:

2e1n(1) + 0(1) + B(2) + C(1) + 2e(1) + 0(1) + Rep.C(1) + 0(1) + Att.A(1) + 2w(1) + 0(1) + A[1ws3](4) + B(1) = 17 ticks

Again, probably missing protocik before B(1) at the end?
♥

Aimee wrote:

Not sure if it's the fastest, but Lightgrace and I came up with the following solution:

1w1n(1) + 0(1) + C(2) + B(2) + C(1) + 2e1s(1) + pre.2s(1) + A(1) + 2n(1) + 0(1) + A[2s](3) + B(1) = 16 ticks

Am I missing a prototick before B(1) at the end?

Also, a funnier solution:

2e1n(1) + 0(1) + B(2) + C(1) + 2e(1) + 0(1) + Rep.C(1) + 0(1) + Att.A(1) + 2w(1) + 0(1) + A[1ws3](4) + B(1) = 17 ticks

Again, probably missing protocik before B(1) at the end?

2e1n(1) + 0(1) + B(2) + C(1) + 2e(1) + 0(1) + Rep.C(2) + 0(1) + Att.A(1) + 2w(1) + 0(1) + A[1ws3](3) + B(1) = 17 ticks
Nevermind optimal tick-efficiency, this is correct because it's awesome. I want that orbdrag hold of A to be 3 ticks, am I wrong?

Not sure. I tried to count the orbdrag, but it's pretty hard. Also, if it would be 3 ticks (and running 2 of it), wouldn't there be a prototick before you tap (B) at the end? That's why I did 4 ticks and no prototick.

♥

Aimee wrote:

Not sure. I tried to count the orbdrag, but it's pretty hard. Also, if it would be 3 ticks (and running 2 of it), wouldn't there be a prototick before you tap (B) at the end? That's why I did 4 ticks and no prototick.

♥

You start running 2 ticks right after finishing the prototick on it. So you stand on (3,-1) for one tick while holding the orb before tapping B. That's how I think it works. =)

I had actually designed this puzzle on paper earlier, and I misplaced an orb at (-1,7) instead of (0,7). >_>