Custom Density Forum Comp

Ideas, info, suggestions and locations
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DMarden
Posts: 13
Joined: Sun Aug 21, 2011 1:43 pm

Custom Density Forum Comp

Post by DMarden »

LMSComp.jpg
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Welcome to the first forum-based elimination competition!

The Rules
This is a multiple-round elimination competition. It will begin with a qualifying round in which no one will be eliminated. The qualifying round will last one week in order to gather many participants. Following that, each round will last 3 days. These rounds will have a different grid size each time (decided by me) and contestants will need to achieve as many mines as they can before the round ends. Depending upon the number of entrants, either 1 or 2 people with the lowest mine count will be eliminated each round until one player is left standing.

If you are new to Minesweeper, please ensure that you use one of the official versions that allow saved replays. You can download a copy of either Minesweeper X, Clone, Arbiter, or ViennaSweeper HERE

To Qualify
Achieve 20 mines or more on an 8x8 grid.

Entry Deadline
Please submit a link to your replay to me via PM by 8pm (GMT) Monday, September 12th.

Current Contestant Lineup
1. qqwref
2.
3.
4.
5.
6.
7.
8.
Last edited by DMarden on Wed Sep 07, 2011 11:21 am, edited 1 time in total.
qqwref
Posts: 125
Joined: Thu Sep 23, 2010 4:17 pm

Re: Custom Density Forum Comp

Post by qqwref »

This sounds like a fun competition; I like the element trying to beat the other results when a higher number of mines requires more luck. My only big suggestion is that you impose a limit on the number of mines (obviously we don't want people solving e.g. 8x8 63 mines). I'm not sure what this limit should be.
NF player. Best scores 1-10-39.
DMarden
Posts: 13
Joined: Sun Aug 21, 2011 1:43 pm

Re: Custom Density Forum Comp

Post by DMarden »

qqwref wrote:This sounds like a fun competition; I like the element trying to beat the other results when a higher number of mines requires more luck. My only big suggestion is that you impose a limit on the number of mines (obviously we don't want people solving e.g. 8x8 63 mines). I'm not sure what this limit should be.
The limit will be (x-1)(y-1) so the limit for 8x8 would be 49 mines.
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