I'm not actually sure if it's advantageous to annihilate the diagonal first, as you would need to "manufacture" more oscillated antineutrinos, which would take more moves.. But with it in the way, it takes longer to move the two manufactured antineutrinos.
I got 65 or so by not annihilating the diagonal.
Edit: I got 53 moves by annihilating about 6 of the diagonal, so maybe it's faster.
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!
Man, that's a pretty dastardly level to try and get perfect on. I'm down to 43, though. I'd guess my strategy is similar to your method, using the right-most two columns pretty exclusively.
When we're ready (and I doubt 43 is perfect, still), I can always plug it into the 'solver'. Though I get the feeling a level *this* intense may well crash the solver.
testtubegames wrote:Man, that's a pretty dastardly level to try and get perfect on. I'm down to 43, though. I'd guess my strategy is similar to your method, using the right-most two columns pretty exclusively.
When we're ready (and I doubt 43 is perfect, still), I can always plug it into the 'solver'. Though I get the feeling a level *this* intense may well crash the solver.
Edit: 42!
What's the link for the solver?
This is one complex level!
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))
I'm not actually sure if it's advantageous to annihilate the diagonal first, as you would need to "manufacture" more oscillated antineutrinos, which would take more moves.. But with it in the way, it takes longer to move the two manufactured antineutrinos.
I got 65 or so by not annihilating the diagonal.
Edit: I got 53 moves by annihilating about 6 of the diagonal, so maybe it's faster.
Could you plug in this level?
Please?
Binomial Theorem: ((a+b)^n)= sum k=0->k=n((n!(a^(n-k))(b^k))/(k!(n-k)!))
Yikes, just as I feared, this one broke the solver!
As a background, the solver works by, step by step, compiling all the possible states of the game in one (growing) list. So, at step 0 it has one state (the starting state). At step 1 it has 3 states (the starting state and the two possible moves you could make). At step 2 it has more... and so on. Normally it works fine, but memory is the upper limit. The list can only grow so long before my computer starts to hang.
This made it to Step 22... where, with 24 million states in the queue, it froze. Moral of the story: this is a tough level.
Now I suppose I *could* replace a bunch of the irrelevant antineutrinos with crates, so as to cut down on the possible moves... but even then, I can't imagine the solver could handle the level. The number of states tends to grow pretty exponentially... and we're only halfway to our expected win-state (around 40, perhaps?). Looks like we're going to have to do this one by hand...
testtubegames wrote:Yikes, just as I feared, this one broke the solver!
As a background, the solver works by, step by step, compiling all the possible states of the game in one (growing) list. So, at step 0 it has one state (the starting state). At step 1 it has 3 states (the starting state and the two possible moves you could make). At step 2 it has more... and so on. Normally it works fine, but memory is the upper limit. The list can only grow so long before my computer starts to hang.
This made it to Step 22... where, with 24 million states in the queue, it froze. Moral of the story: this is a tough level.
Now I suppose I *could* replace a bunch of the irrelevant antineutrinos with crates, so as to cut down on the possible moves... but even then, I can't imagine the solver could handle the level. The number of states tends to grow pretty exponentially... and we're only halfway to our expected win-state (around 40, perhaps?). Looks like we're going to have to do this one by hand...
Have you tried deleting unused gamestates, and checking for duplicate states? Might help a bit.
$1 = 100¢ = (10¢)^2 = ($0.10)^2 = $0.01 = 1¢ [1]
Always check your units or you will have no money!