Posts for Dacicus

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Posted: 8/29/2005 6:04 AM
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Dacicus
Editor, Experienced Forum User, Published Author, Player (67)
Joined: 6/22/2005
Posts: 1041
schneelocke wrote:
it wasn't displayed for long enough to read the URL.
You could stop the playback to read the URL. I also think that the subtitles could be distracting, though. Maybe the URL to the movie description page could be placed on a screen after the one with the URL to the Nesvideos site.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/29/2005 3:30 AM
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Dacicus
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Joined: 6/22/2005
Posts: 1041
Enhasa wrote:
No, not really, I was just referring to the length of the undertaking.
Oh. Then I totally agree with you.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/29/2005 3:27 AM
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Dacicus
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Joined: 6/22/2005
Posts: 1041
Enhasa wrote:
Simple, the normal requirements would be for mere mortals, while saving every lemming would be for those seeking more challenge.
It depends on what your goal is. Mine is to get through the game as quickly as possible, which means that, in some levels, I'm having trouble killing the unnecessary lemmings quickly enough.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/28/2005 10:53 PM
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Dacicus
Editor, Experienced Forum User, Published Author, Player (67)
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Posts: 1041
Enhasa wrote:
Wow, I really want to see Lemmings runs more than can be adequately described, but I never figured anyone would be crazy enough to try them until we reached a point where almost all the easier runs were already completed.
Oh. Is Lemmings really considered to be that hard? I mean, some of the later levels seem nearly impossible when I look at them superficially, but the thing that bugs me most is finding a way to improve an early level after I've recorded several levels past it.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/28/2005 9:52 PM
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Dacicus
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Posts: 1041
Enhansa wrote:
As for saving every single lemming, it simply can't be done in some levels. The best I can do on the SNES version is losing 2 in classic and 1 in polar. I think it is impossible to do any better, but I could be wrong.
I've been working on Lemmings for the NES, and I second that. For me, it's quite logical that if you could save every lemming in every level, they would require you to do so. However, they don't. Anyway, I'm starting to get bored of improving the same levels over and over again, so I might put it off for a while.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/28/2005 6:14 PM
(Edited: 9/19/2005 5:05 AM)
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Dacicus
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Joined: 6/22/2005
Posts: 1041
I just remembered a little trick that I found while playing through this game. It basically allows you to always hit the knights and the lizards. I found it useful against the knights because they can block your normal hits with their shields, but not this one. You jump, then do a crouching stab while in the air. It works best if you jump when you're pretty close to the enemy and continue moving toward them while stabbing in the air. It shouldn't be hard to perform with frame advance. I can upload a movie of this if you want to see it. I'm not quite sure if it would be any faster than normal fighting. In fact, it would be quite pointless if you have a way to get past the enemies without fighting them.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/28/2005 5:14 PM
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Dacicus
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Posts: 1041
adelikat wrote:
In recording this .fcm I stumbled on to something very interesting. All other times I've tried this I ended up somewhere else in the palace, typically at the beginning screen but stuck in the wall, making it useless. In this case I ended up in palace 1! Notice when you watch it. The color scheme is still blue like the 2nd palace, but I've been tranported to the 1st!
I've seen this trick on one or more sites. For example, here. It's also on David Wonn's site.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/26/2005 10:18 PM
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Dacicus
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Posts: 1041
What functions would you be allowed to use for the programs? Using the functions we have used so far, I don't think it would be possible to find a "highest" number. You could always add a ! or ? to the end to make it even higher, so there would need to be a limit on the unary operators that you could use.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/26/2005 7:45 AM
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Dacicus
Editor, Experienced Forum User, Published Author, Player (67)
Joined: 6/22/2005
Posts: 1041
What!? wrote:
Zelda 1 swordless!
You need the sword to hurt Ganon, unless there's some new glitch about which I don't know. The best you could do without the sword is just get to Ganon's room.
adelikat wrote:
any chess game! then chess would be "solved" and I would hold the secret!
What's a "perfect" chess game, though? No piece losses? No checks received? Both? I'm not quite sure either of those would be possible, unless one of the players would play horribly.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/26/2005 7:22 AM
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Dacicus
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So I guess the scope of this topic has moved from figuring out the expressions by hand to writing a program that can produce them for you?
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/26/2005 7:15 AM
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Dacicus
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Posts: 1041
Randil wrote:
The average is NOT N*(S+1/2) - R.
Isn't the second parenthesis supposed to be before the division?
Randil wrote:
What you must do is subtract R from every possible outcome of N*(S+1/2)
I don't quite understand this. Do you mean to subtract R from every possible average of NdS (what you wrote, I think) or from every possible result of NdS? I'm inclined to think that you meant every possible result, but please correct me if I'm wrong. In any case, there are S^N possible results when you roll N dice with S sides, so that could get quite complicated to program. Well, maybe not with loops.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/25/2005 5:04 AM
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Dacicus
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Joined: 6/22/2005
Posts: 1041
If someone could figure out 145 with only three 4s, it would be easy.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/24/2005 9:37 PM
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Dacicus
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Posts: 1041
Well, you could also think of it as signifying the nth triangular number, which is commonly written as T_n. So just imagine that the ? is the T_ part and whatever follows is the subscript.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/24/2005 9:27 PM
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Dacicus
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Posts: 1041
We're using ? to represent the sum from 1 to whatever. In math, it's usually denoted with a capital sigma, but it's easier to type as ? for this little game.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/24/2005 6:21 AM
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Dacicus
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Posts: 1041
250 = (4? × √4)? + 4? × 4 251 = [(√4)? × 4 + 4?]? - √4 252 = (4? + √4) × (4 + √4)? 253 = {√4 × [4 + 4 + (√4)?]}? 254 = 4 ^ 4 - 4 / √4 255 = 4 ^ 4 - 4 / 4 256 = 4 × 4 × 4 × 4 257 = 4 ^ 4 + 4 / 4 258 = 4 ^ 4 + 4 / √4 259 = [(√4)? × 4 + 4?]? + [(√4)?]? 260 = 4 ^ 4 + √4 + √4 261 = 4 ^ 4 + √4 + (√4)? 262 = 4 ^ 4 + 4 + √4 263 = 4 ^ 4 + 4 + (√4)? 264 = 4 ^ 4 + 4 + 4 265 = 4 ^ 4 + (√4)? ^ √4 266 = 4 ^ 4 + (√4 + √4)?
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/24/2005 3:36 AM
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Dacicus
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Posts: 1041
216 = (4? × √4)? + (√4)? × √4 217 = (4? × √4)? + 4 + (√4)? 218 = (4? × √4)? + 4 + 4 219 = (4? × √4)? + (√4)? × (√4)? 220 = (4? × √4)? + (√4 + √4)? 221 = {(√4)? × [4 + (√4)?]}? - 4? 222 = (4? × √4)? + 4 × (√4)? 223 = [(4 +√4)?]? - 4 - 4 224 = [(4 +√4)?]? - (√4)? - 4 225 = [(√4)? × 4 + (√4)?] ^ √4 226 = [(4 +√4)?]? - (√4)? - √4 227 = [(4 +√4)?]? - √4 - √4 228 = ({[(√4)?]?}?)? - √4 - 4 / 4 229 = [(4 +√4)?]? - 4 / √4 230 = [(4 +√4)?]? - 4 / 4
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 11:37 PM
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Dacicus
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175 = (4?)? + {(√4)? × [√4 + (√4)?]}?
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 11:15 PM
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Dacicus
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159 = (4!)? / √4 + (√4)? × (√4)? 160 = (4!)? / √4 + (√4 + √4)? 161 = {(√4)? × [(√4)?]?}? - (√4 + √4)? 162 = {(√4)? × [(√4)?]?}? - (√4)? × (√4)? 163 = {(√4)? × [(√4)?]?}? - 4 - 4 164 = {(√4)? × [(√4)?]?}? - 4 - (√4)? 165 = {(√4)? × [(√4)?]?}? - 4 - √4 166 = {(√4)? × [(√4)?]?}? - √4 - (√4)? 167 = {(√4)? × [(√4)?]?}? - √4 - √4 168 = (4? × √4)? - {[(√4)?]?}? × √4 169 = {(√4)? × [(√4)?]?}? - 4 / √4 170 = {(√4)? × [(√4)?]?}? - 4 / 4 171 = {(√4)? × [(√4)?]?}? - 4 + 4 172 = {(√4)? × [(√4)?]?}? + 4 / 4 173 = {(√4)? × [(√4)?]?}? + 4 / √4
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 11:00 PM
(Edited: 8/23/2005 11:16 PM)
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Dacicus
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Actually, you wrote 13? incorrectly in your list, so everything after it is also wrong. 13? = 78 + 13 = 91
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 10:49 PM
(Edited: 8/23/2005 10:58 PM)
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Dacicus
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Posts: 1041
There's a formula for the triangular numbers, which is what you're writing out. It's: n? = n × (n + 1) / 2. 139 = ((4 + 4) × √4)? + (√4)? 141 = (4 × 4)? + (√4)? + √4 142 = (4 × 4)? + 4 + √4 143 = (4 × 4)? + (√4)? + 4 144 = (4 × 4)? + 4 + 4 145 = (4 × 4)? + (√4)? × (√4)? 146 = (4 × 4)? + (√4 + √4)?
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 10:37 PM
(Edited: 8/23/2005 10:39 PM)
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Dacicus
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134 = 44 × (√4)? + √4 135 = 44 × (√4)? + (√4)? 136 = (4 + 4 + 4 + 4)?
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 10:27 PM
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Dacicus
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Posts: 1041
128 = (4 + 4) × 4 × 4
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 10:06 PM
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Dacicus
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As far as possible, I guess. 112 = 4? × 4? + (√4)? × 4
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 9:54 PM
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Dacicus
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103 = 4? × 4? + ((√4)?)? - (√4)? 104 = 4? × 4? + √4 × √4 105 = 4? × 4? + √4 + (√4)? 106 = 4? × 4? + 4 + √4 107 = 4? × 4? + 4 + (√4)? 108 = 4? × 4? + 4 + 4 109 = 4? × 4? + (√4)? × (√4)? 110 = 4? × 4? + (√4 + √4)? 31 has been completed. I think it's on the first page. EDIT: If it's not, then 4? × 4? / 4 + ((√4)?)? works.
Current Projects: TAS: Wizards & Warriors III.
Posted: 8/23/2005 9:37 PM
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Dacicus
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Posts: 1041
73 = ((√4)?) ^ 4 - 4 - 4 74 = (4 + √4)? × 4 - 4? 75 = (4?) × (4?) × (√4)? / 4 76 = ((√4)?) ^ 4 - √4 - (√4)? 77 = ((√4)?) ^ 4 - √4 - √4 78 = (√4)? × (√4)? ^ (√4)? - (√4)? 79 = ((√4)?) ^ 4 + √4 - 4 80 = (44 - 4) × √4 81 = ((√4)?) ^ 4 + 4 - 4 82 = ((√4)?) ^ 4 + 4 / 4 83 = ((√4)?) ^ 4 + 4 - √4 84 = 44 × √4 - 4 85 = 44 × √4 - (√4)? 86 = 44 × √4 - √4 87 = (4 + √4)? × 4 + (√4)? 88 = 44 × 4 / √4 89 = ((√4)?) ^ 4 + 4 + 4 90 = 44 × √4 + √4 91 = 44 × √4 + (√4)? 92 = 44 × √4 + 4 93 = (4?) × (4?) - 4 - (√4)? 94 = (4?) × (4?) - 4 - √4 95 = (4?) × (4?) - √4 - (√4)? 96 = (4?) × (4?) - √4 - √4 97 = (4?) × (4?) + (√4)? - ((√4)?)? 98 = (4?) × (4?) - ((√4)?)? / (√4)? 99 = (4?) × (4?) - 4 / 4 100 = ((√4 + √4)?) × ((√4 + √4)?)
Current Projects: TAS: Wizards & Warriors III.
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