Posts for Flip

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Posted: 3/18/2018 7:33 PM
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Flip
Experienced Forum User, Published Author, Skilled player (1535)
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Regretfully I haven't been able to produce that yet, but in other news there's also a level 4 skip available, see his Latest run. That one is far easier to reproduce thankfully, so this now means we skip 3/4 of the levels in a TAS :)
Posted: 1/20/2018 8:09 AM
(Edited: 1/20/2018 8:19 AM)
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Flip
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Wouldn't be 6 inputs if it was 7...
Posted: 1/20/2018 8:06 AM
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Flip
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Ah yes, here's the 1 frame saved http://dehacked.2y.net/microstorage.php/info/458408262/Disney%27s%20Aladdin%20%28U%29%20%5B%21%5D.bk2 Just did it hastily to try to not be beaten.
Posted: 1/19/2018 8:16 PM
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Flip
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Very nice, let's see what everybody's opinion is. It takes ~378131, that's 1h45m + to get to the credits, that's mega boring.
Posted: 1/16/2018 5:32 PM
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Flip
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A Magnifying Glass could also work for the icon.
Posted: 12/30/2017 8:23 AM
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Flip
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See if your Anti-virus software has deleted any .dll files from Bizhawk. I know Norton loves to do that, despite me telling it over and over again that they're trusted files; so I keep on having to restore them.
Posted: 12/9/2017 8:58 AM
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Flip
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It's just using Pythagoras over and over again. Measure each diagonal between 2 axes, and use that value to work out the next diagonal. Repeat until the distance along all dimensions is covered.
Posted: 12/3/2017 12:19 PM
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Flip
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.|.||..| .||..|.| .|||..|| ..|..... .|||.||. .||.|||| .|||.|.. .||..|.| ...|.|.
Posted: 11/26/2017 9:08 AM
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Flip
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Nice catch. Here's another mathematical limerick. (12 + 144 + 20 + 3x4^0.5 ) / 7 + 5x11 = 9^2 A dozen a gross and a score plus three times the square root of four divided by seven plus five times eleven is nine squared and not a bit more.
Posted: 11/10/2017 8:55 PM
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Flip
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Riddler Express Since we're dealing with hexagons here, sections form nice equilateral triangles. The shaded area in question is one third of its triangle, where the triangle occupies one sixth of the main hexagon, thus the shaded part has area 1/3 x 1/6 = 1/18. Riddler Classic This seems relatively easy, just split it up into a bunch of semicircles. Thus A=A1+A2=(a1-a2+a3)+(a4-a5+a6) =π/2 ( (10+2)2 - (10-2)2 + 22 ) + π/2 ( (5+2)2 - (5-2)2 + 22 ) =π/2 ( 122 - 82 + 22 + 72 - 32 + 22) = π/2 (144-64+4+49-9+4) = π/2 (128) = 64π
Posted: 11/3/2017 7:33 PM
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Flip
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Riddler Classic A typical path can be described as 2211221222111 etc. Given n leaps of size 2, this covers total length 2n, thus leaving the remaining 20-2n leaps for size 1, for a total of n+(20-2n)=20-n leaps total. Thus a path with a given amount n of 2-leaps can be ordered in (20-n)Cn ways. Since we can have a maximum total of 10 leaps of size 2, this gives a grant total T=Sum{n=0,10} (20-n)Cn = 20C0+19C1+18C2+....+10C10 =1+19+153+680+1820+3003+3003+1716+495+55+1 =10946 But for jumping with more than two potential leap sizes, dunno.
Posted: 10/28/2017 8:22 AM
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Flip
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FractalFusion wrote:
The result of racing ostwx isn't necessarily twx; it could be wtx, stx, ost, xot, ... (This would of course decide the second and third places.)
Yes, those letters were just chosen for convenience. Similarly we could have the outcome of ejoty......tej or w/e. The problem would be symmetrical either way, so the algorithm would still work. For example, as a random selection process, use 2 pangrams. First, the actual speed order which we hope to derive: "Pack my box with five dozen liquor jugs": P<A<C<K<M<Y<B<O<X<W<I<T<H<F<V<E<D<N<L<Q<U<R<J<G<S And then a random race order, say "Sphinx of black quartz, judge my vow" sphin........pmihs.......ihs xofbl.........boxfl........xfl ackqu.......ackqu.......kqu rtjdg.........tdrjg.........rjg emyvw......mywve....wve Race the winners sluge.......elugs......ugs Identifying s as the overall winner, which is correct. Next, use the info from this most recent race to narrow down the potential 2nd/3rd place from the original races. ih. ... ..u .jg ... ihujg.......ihujg.......ujg. Identifying j/g as 3rd/2nd, which is also correct. So this does in fact work in 7 races.
Posted: 10/27/2017 5:39 PM
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Flip
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Riddler Express. First, race 5 packs of 5, take note of the top 3 places. Thus abcde...............cde fghij................hij klmno..............mno pqrst...............rst uvwxy.............wxy That'll be 5 races. The first two columns are discarded since we now know at least 3 horses are faster than them. So next, race the winners. ejoty............oty That identifies y as the overall winner, but we still need positions 2/3. We can discard cde/hij, since we know both e and j are lower than 3rd, and so cd/hi must be too. Likewise t is at most second, so anything it beat can be at most 3rd. O can be at most 3rd, so anything before cannot be in top 3. Thus this narrows down the selection from our initial winners down to ... ... ..o .st wx. Which conveniently is 5 remaining, so we can just race 'em. ostwx..........twx Which gives you the 2nd/3rd place in 7 races, deriving the top three: wxy
Posted: 10/21/2017 9:55 PM
(Edited: 10/22/2017 3:26 PM)
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Flip
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My attempt at Express...failed. Just do a cyclic shuffle of your current solution, and take note of how many you get right each time. 4 blocks gives 4!/4=6 unique cyclic orders. Assuming correct order 1234, here are 6 columns for the potential guesses, with the amount of correct blocks noted after. 1234..4.......1243..2.......1324..2 2341..0.......2431..1.......3241..1 3412..0.......4312..0.......2413..0 4123..0.......3124..1.......4132..1 1342..1.......1423..1.......1432..2 3421..0.......4231..2.......4321..0 4213..1.......2314..1.......3214..2 2134..2.......3142..0.......4132..1 If you get order {4000} you win, if you get the unique cyclic order {2021} then you've identified which order you have, and since the cyclic order does not repeat itself then you can also identify which specific setup you must have, thus guaranteeing the solution with at most 4 wrong guesses. For the other 4 sadly, the cyclic orders {2101} are not unique, which means we still need further moves to tell these 3 sequences apart. Right now, there isn't any easy move to identify which of the 4 sequences you have in only one move, it'll probably be doable in two, requiring 6 failed moves thus 7 total. Edit: Failure realised.
Posted: 6/13/2017 10:05 AM
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RAM Watch no longer seems to remember which columns you want to be displayed, as they reset each time you reload Bizhawk.
Posted: 6/8/2017 7:59 PM
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In RAM Watch, when there's a long list of addresses, the scroll function doesn't appear to work properly. More specifically, when moving a given address a long way up or down, the scroll height doesn't want to change from its initial location. EG Say I've added another address to the list, which by default appears at the bottom. I want to shove it in the middle of this long list, by clicking the Move Up button a few times. Since it's a long list, I'll need to scroll upwards to see if I've moved it far enough into the correct position. So I check, and it appears that I haven't, so I need to keep moving it. But by clicking Move Up again the list will instantly scroll all the way back down to the bottom so I can no longer see where the actual address is now. I have to keep scrolling back up to view it each time. The only way to stop the list auto scrolling back to its initial location each time, is to unselect and reselect the address.
Posted: 4/1/2017 9:16 AM
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That was very smurf. I voted smurf.
Posted: 2/6/2017 2:52 PM
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Flip
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Upon further investigation, it's not just the Skeleton Arm + Paste which creates that extra icon, it's basically any combination which does not remove an object from your cursor as the active one. EG combining Hook + Ramrod will remove either one from your mouse, and create the Gaff in the inventory. Using Paste on Rubber Plug will remove the object from the mouse and create the Sticky Rubber Plug in your inventory. Whereas using Paste on Skeleton Arm, or Skeleton Arm on Paste, will keep your active item on your cursor and also create the glitched icon in the corner. Other objects which have similar effects include poking holes in the Jar Lid, cutting out the Portrait Face, filling the Pie Pan full of Shaving Cream, etc.
Posted: 2/5/2017 10:20 AM
(Edited: 2/5/2017 8:41 PM)
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Flip
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The gold tooth glitch extends far beyond what's shown in the video. If after popping the first bubble, you give Blondebeard some new gum he'll decline as intended. If you give him the chewed gum he'll accept for some reason, and create a second bubble. However, it won't allow two of them to exist at the same time, but you can get around that. Simply take the first tooth, offer some chewed gum, and then place your current one into the chewed gum. That'll change it to another object, and thus it won't be deleted from your inventory upon looting the second. There are then 2 which exist, have a choice of which one you want to keep, to take outside. Try to leave and Blondbeard will stop you, and force you to hand over your Tooth-in-Gum. Simply loot the 2nd one now, and you can take that one outside without hesitation. Useless sadly, as it appears to be a separate object from the tooth you loot from the puddle, because Cutthroat Bill won't accept it as a valid treasure. The other way is to loot the 2nd gold tooth, hand that back to Blondebeard, and then walk out with the Tooth-in-Gum. This then becomes really buggy, as you can then try to inhale the Balloons, and then chew the Tooth-in-Gum which tries to start a sequence, which differs depending on your location. The effects can be as follows: -Asking cabana boy for oil -Opening all stage lights -Examining Blood Island sticker on trunk -Using Gum machine -Refusing to play caber toss (By using it on the playing field, it's the same effect as if you tried to manually pick up a caber) -Asking Haggis for another haircut (Softlocks, no way to leave) Doing it anywhere else has either has no effect, or just crashes the game. It'll complain about "Local script 2009 not being room __"; and so this also has 0 potential for saving time.
Posted: 1/22/2017 5:43 PM
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Upon further investigation, that's very easy to reproduce. Just run against the wall, and jump for a single frame to go straight through it. Nothing to do with invincibility frames or needing the bat to displace you. No applications anywhere it seems, as this is the only corner with a height gap of 24, and even then it just makes you fall through into the previous zone. Thus this cannot be used as a skip.
Posted: 1/20/2017 6:27 PM
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Interesting. I know that you can jump and nudge yourself into a few corners, but even if you manage to fall through it's not much use as level progression generally happens upwards. I was certainly wondering if anything like that could occur in Sultan's dungeon. I myself have found other improvements in Genie's level, and have already improved level 01 by a tiny bit too.
Posted: 1/6/2017 3:01 PM
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Nice discovery. Applications at 0:49, can just jump up directly onto zip line. 7:31, can jump and whip without landing. Near ~8:10, any chance of jumping over the wall, so you don't need to go down and up through chute?
Posted: 1/2/2017 6:21 PM
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It's nothing to do with losing all your lives, it's about failing that level 3 times so it passes you out of pity. Losing lives in earlier stages doesn't help, plus it would mean you would actually get down to 0 and have to sit through a Continue screen.
Posted: 1/1/2017 9:05 AM
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Yes vote, glad to see all the wrinkles ironed out, and good job in creating it in such a short time.
Posted: 12/23/2016 5:05 PM
(Edited: 12/24/2016 6:25 PM)
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Awesome. You can do 1 better, you can fall off the ledge, go left, and then trigger the pole while facing right instead. You can then aim your trajectory left to the next section instead, up here: Plus there's an error in the current run (Edit: As was clearly stated on previous page anyway). For some reason it collects the flute at 2:31, which is optional and bad in a speedrun because it's what causes the rope at 2:56 to take the detour for apples/health. Not a concern now though, since we don't go anywhere near it. Good discovery.
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