 Show Posts
|
|
Pages: [1] 2 3 ... 23
|
|
2
|
Playing the Game / Showcase / Extended Level Playlist
|
on: October 13, 2011, 11:16:06 PM
|
We have lots of great levels on the forum. Unfortunately, they are in the wrong order for a new player. The older, simpler puzzles are buried deep, and the puzzles on the front page of the 'custom puzzle levels' forum often require a considerable amount of experience to solve. As a result, few new players enter the community.
I suggest we collect together a list of our best puzzles, in the approximate order in which they should be tackled by a player who has just finished the built-in levels.
What I have so far, in order:
Kevan's Random-to-random matching (lygofog)*^
MagiMaster's Passing Trains series (mahytoh, celalal, pidugos, fysoxih, domedyg )*
*needs updating to the new physics model
^needs an exploit fixed
|
|
|
|
|
7
|
Playing the Game / Custom Puzzle Levels / Re: colcolpicle's puzzles
|
on: October 30, 2010, 04:22:18 AM
|
Cross the streams kibomef:feropym. The timing is rather tricky on this one. The [level]half furnace draw[/level] on the left side needed to replace the traditional version because that version was 2 ticks slow no matter what I tried.
Here's an interesting problem I heard recently. it's not rubicon related but it is math which is kind of related.
How many numbers are there between 1000 and 10000 where each digit is at least one more than the previous digit. For clarification, the first number that has this property is 1234. Once you figure that out, can you figure out a formula for this problem between 10^n to 10^(n+1) for any n?
126. Formula is  item n+2 in row 10 of Pascal's Triangle, or 0 if n>9 or n<-1.
Quick explanation for people with little math background:
The number in each cell in Pascal's Triangle is the number of unique paths from the vertex to that point. I have a scheme where every number with this property corresponds to exactly one path and vice versa.
To generate the number for a path, start at the vertex with a counter at zero. At each step in the path, increment the counter, then write a digit with the value of the counter if the step was down-right. You will always end up with the same number of digits because you perform a constant number of rightwards steps, and you will end up with digits in increasing order because the counter only increments.
To generate the path for a number, start at the vertex with a counter at 1 and a marker before the first digit. For each step in the path, if the next digit is equal to the counter, step down-right and move the marker one digit; otherwise step down-left. Either way, increment the counter (and stop if the counter goes past 9). The digit 'after' the last one is larger than 9 for this. The path will end in the right spot because it will always have one down-right step per digit and 9 steps total (because of the counter).
The reason why it's item n+2 is that the formula given asks about n+1 digit numbers, and the first item in each row corresponds to zero-digit numbers.
|
|
|
|
|
8
|
Playing the Game / Custom Puzzle Levels / Re: Newt's Puzzles
|
on: September 14, 2010, 05:19:58 PM
|
NEW PUZZLEDNA Strand: bilexyf jylebyfEDIT: Forgot to lock the inside of the machine to prevent tampering...
From a transcription puzzle to a full-speed member of the Sliding Glass Floor family? Ouch. I think this one has occasionally has bursts that are impossible to handle. |
|
|
|
|
|
Rubicon Forum
