en-us Thu, 02 Sep 2010 12:29:55 -0500 ATM@msgroup.org 60 *All Things Motorcycle* - Puzzles and Games http://www.msgroup.org/forums/mtt/ *All Things Motorcycle* for Guest http://www.msgroup.org/forums/mtt/ http://www.msgroup.org/forums/mtt/rss_32.png *All Things Motorcycle* RSS Feed 144 47 Puzzles and Games - memory testlostinspacehttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=11581&REPLY_ID=112106Puzzles and Games Mon, 29 Mar 2010 00:03:11 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=11581 There are many memory contests out there. How about this one http://www.encyclopedia.com/doc/1G1-206457538.html or this http://www.recordholders.org/en/list/memory.html What else is the mind capable of? Does anyone remember the Ray Bradbury novel F. 451 books in which books are banned? "Rebels" accepted the burning of books but saved the written word by memorizing the great novels in a secret retreat. Phil]]>Puzzles and Games - Great time waster - Circle the Catlostinspacehttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=11218&REPLY_ID=112105Puzzles and Games Sun, 28 Mar 2010 23:30:47 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=11218 This is fun to chase the cat but anyone familar with the Japanese game of GO may find this game fairly elemental. Thanks for the posting. Phil]]>Puzzles and Games - Traffic Rush - iPhone GameWSMedershttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=11415Puzzles and Games Mon, 01 Feb 2010 12:07:23 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=11415 What is interesting though, is that it is quite accurate in it's portrayal of motorcycles.
They are much smaller and harder to see. And since the name of the game is to watch all four ways of traffic and anticipate the timing of the vehicles, you quickly realize how easy it is to miss a motorcycle.
I would love to see global stats on how many accidents involve motorcycles in this game.

http://www.bestfreeiphoneapplicatio...h-iphone-app]]>Puzzles and Games - The trains and tunnels math problemscottrnelsonhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=11211&REPLY_ID=108906Puzzles and Games Fri, 18 Dec 2009 06:42:44 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=11211
quote:
Originally posted by Deseret Rider

Your thought process being a little different both arrived at the same answer I have.
After reading Surely You're Joking, Mr. Feynman about the physicist Richard P. Feynman, and finding out that he did math differently than others, I've stopped trying to conform to "the correct way" of computing things. I wish I had figured that out way back in elementary school.]]>Puzzles and Games - Remember these?Cat-womanhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=9678&REPLY_ID=98248Puzzles and Games Wed, 08 Apr 2009 18:14:52 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9678 I remember playing these on the game machines at the local pizza parlor and my dentists office. oops, guess my age shows there.]]>Puzzles and Games - mazelillyhttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9823Puzzles and Games Sun, 15 Mar 2009 14:31:51 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9823 [88]http://www.fugly.com/media/flashdir...aze_Game.swf]]>Puzzles and Games - Three letter anagramslostinspacehttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9601Puzzles and Games Tue, 03 Feb 2009 23:05:35 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9601
How many groups of three letters can you think of whose letters can be rearranged to make three words?

For example: n-o-w can be own, won, and now. T-e-a can be ate, eat and tea.

So far I've got a list of 11 including the two above.

No using the internet!!

Phil
]]>Puzzles and Games - A Geo/Mapping Gamerioguyhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=9551&REPLY_ID=94528Puzzles and Games Tue, 27 Jan 2009 11:27:01 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9551 This one is quite devious with lots of possibilities for errors. It is a rally our riding group is having in May.

It's allowed to work with others on a solution. The maximum score I've been able to figure out is 1,067 points.

http://motorcycle.meetup.com/47/calendar/9520819/

(Note: The mapping programs I've used grossly overstate the time. You could probably make any 320 mile route planned in 8 hours.)
]]>Puzzles and Games - be smarthotthumzzhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=6912&REPLY_ID=91869Puzzles and Games Tue, 09 Dec 2008 17:17:15 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=6912 Its a crazy game.....but I solved it!]]>Puzzles and Games - great time wasterDaddiohttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=9066&REPLY_ID=90068Puzzles and Games Wed, 29 Oct 2008 10:01:05 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=9066 This looks like what I saw when visiting Rome, Italy. I am grateful I was limited to public transport to get around!]]>Puzzles and Games - into the corporate mindcramersechttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=7022&REPLY_ID=86203Puzzles and Games Fri, 29 Aug 2008 15:46:06 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=7022 Ummm, http://www.snopes.com/history/american/gauge.asp

Steve]]>Puzzles and Games - New WatchMrjavahttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=8171&REPLY_ID=85708Puzzles and Games Fri, 22 Aug 2008 12:44:53 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=8171 Analog / digital watch....
What about checking time with the cell phone?
Ive got about 4 watches (after selling others) that I never wear.
I carry 2 cell phones so I haven't needed a watch in about 5 years]]>Puzzles and Games - Riddle of the surgeon's sonCat-womanhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=8399&REPLY_ID=83748Puzzles and Games Sun, 03 Aug 2008 23:46:46 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=8399 yup, we read gender into professions all to often. I have heard this before, and managed to get it correct then. I feel part of the issue resolves around having no gender-neutral pronoun that does not automatically make a "person" a "thing."


BTW, I have gotten better care via male nurses at times than females.

]]>Puzzles and Games - Miniature golf anyone?bridwell52http://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=8434&REPLY_ID=83669Puzzles and Games Sun, 03 Aug 2008 13:21:55 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=8434 Must be too hot to ride there too.
Thanks
David]]>Puzzles and Games - Awareness testCat-womanhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=6979&REPLY_ID=80915Puzzles and Games Thu, 10 Jul 2008 00:16:45 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=6979 got it...see a good use for targt fixation.]]>Puzzles and Games - Skills: How steady is your hand?Cat-womanhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=7488&REPLY_ID=80914Puzzles and Games Thu, 10 Jul 2008 00:13:18 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=7488 add to the tally:

1 confused dag

2 disturbed bunnies

and 1 wet couch.

I hate those things.]]>Puzzles and Games - Simple logicdegabohhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=7019&REPLY_ID=71167Puzzles and Games Wed, 19 Mar 2008 21:01:42 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=7019 d'oh! I should keep my mouth shut! Or read more slowly. :)

]]>Puzzles and Games - Physics question (Puzzle)Rayvonhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2168&REPLY_ID=70238Puzzles and Games Mon, 03 Mar 2008 17:48:24 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2168 The plane would take off. The wheels would just be spinning 100 mph faster than the plane is travelling. Same physics as taking off from an aircraft carrier. Every knot that the carrier is travelling into the wind reduces the amount of speed the planed needs to become airborne.]]>Puzzles and Games - Red Square Game - Simple, yet frustratingVodkaAndPickleshttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=6766&REPLY_ID=69042Puzzles and Games Fri, 08 Feb 2008 13:41:19 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=6766 33.058 seconds so far]]>Puzzles and Games - So You think you are pretty good at checkers?Destinchttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=5777&REPLY_ID=66921Puzzles and Games Fri, 07 Dec 2007 17:19:37 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=5777 Here is the site


http://www.cs.ualberta.ca/~chinook/]]>Puzzles and Games - Latin Advicesixsigmahttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=6439&REPLY_ID=66389Puzzles and Games Thu, 22 Nov 2007 15:28:51 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=6439
quote:
Originally posted by Baggsy

Very close.

The first two lines are obvious from your translation.

I was going to wait for tomorrow, but someone will just google the first line and get the link to Trin-Trin.

Here's the link if your latin is rusty.

http://www.trinity.utoronto.ca/Abou...ity/History/




My Latin is rusty to the point of non-existence (My Italian Lady gave it a shot)[:D]]]>Puzzles and Games - A very old math problem.elroyhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=6010&REPLY_ID=63105Puzzles and Games Tue, 28 Aug 2007 21:33:11 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=6010 Bugger....
But it is an oldie...not unlike you two?[88]

You guys have now set me a challenge.[}:)]

Sometimes, even I think I might have too much time on my hands.[:I]]]>Puzzles and Games - Logic Problem - Campfire GatheringGeotiggyhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=3131&REPLY_ID=63072Puzzles and Games Tue, 28 Aug 2007 15:19:39 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=3131 These are great for killing time. Keep posting them.

Thanks!!!

]]>Puzzles and Games - Blue ball gamegadgetgeekhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=5915&REPLY_ID=62290Puzzles and Games Wed, 15 Aug 2007 17:46:56 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=5915 damn

Request for URL http://70.85.224.74:80/media/113369...he_blue_ball denied and logged by WebBlocker: denied blocked for Adult/Sexually Explicit.

Now I'm busted.]]>Puzzles and Games - Math Problemjollyrogerhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=5022&REPLY_ID=54455Puzzles and Games Thu, 12 Apr 2007 15:01:06 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=5022
quote:
Originally posted by twc

quote:
Originally posted by jollyroger

I'm not only running the sweepstakes, I am, indeed, currently winning it.

One thing I do when running such events is to disqualify myself from participation. That way there's no conflict of interest. The mathematicians in the crowd might question how you arrived at the payout, but nobody will question your integrity.


Well, I understand that, and thus set it up (it's a bowling sweepstakes) where it's handicapped to where literally anyone in the pool of 60 participants could not only do well, but could win it.
There's no room for my integrity to be challenged; it's set up to be as fair as humanly possible..[8D]]]>Puzzles and Games - Yet Another Physics question (Puzzle)Tomhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=4823&REPLY_ID=52901Puzzles and Games Wed, 07 Mar 2007 08:08:34 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=4823
quote:
Originally posted by scottrnelson

quote:
Originally posted by Tom

One more. This one is easy to calculate (if you remember the formula), but the interesting thing is that the answer just doesn’t feel like it should be right.

Suppose the earth were perfectly round at 25,000 miles (no valley’s or mountains). You lay a rope around it all the way - 25, 000 miles.

You then lay another rope around, but this one is suspended a constant 1 foot above the first rope.

How much longer is the 2nd rope then the first?

You really should just start a new topic for each of these, or the moderator should split them off...

This problem assumes that you have some sort of magic rope that absolutely doesn't stretch or shrink regardless of temperature, humidity, or any other thing that would change its length.

The answer is 6.283185307 feet.

The equation for the circumference of a circle is pi * diameter, so if you increase the diameter by two feet, one foot on this side of the earth and one foot on the opposite side, you've increased the diameter by two feet. It's irrelevant whether the diameter is 8000 miles or one millimeter, you get the same result.


Correct - from a math view point the answer is not surprising at all. But from an intuition viewpoint it just doesn’t seem right to most people - especially when the answer is the same for the earth or a quart jar.]]>Puzzles and Games - Another Physics question (Puzzle)bridwell52http://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=4821&REPLY_ID=52887Puzzles and Games Tue, 06 Mar 2007 18:05:15 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=4821 I think all of you have sprung a leak.
David]]>Puzzles and Games - What's wrong with this proof?banditrvphttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=4554&REPLY_ID=51403Puzzles and Games Tue, 23 Jan 2007 08:29:22 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=4554 A winner!

I always thought it was cute.]]>Puzzles and Games - Son's math problempuddlejumperhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2706&REPLY_ID=50662Puzzles and Games Sat, 06 Jan 2007 00:22:31 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2706 Laughing my butt off when I read these posts!]]>Puzzles and Games - Playing women like an instrument...Jerry Godellhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=3157&REPLY_ID=38865Puzzles and Games Thu, 22 Jun 2006 22:35:13 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=3157 Didn't know playing the piano could be that much fun!]]>Puzzles and Games - Solution - Trying Out MotorcyclesJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=3089Puzzles and Games Thu, 15 Jun 2006 11:50:02 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=3089
The clues are sufficient to solve the problem.

The first pass through the clues results in completing this much of the logic grid:



Then you use observation of the logic grid and the following new information is gleaned (backgrounded in yellow):



Though modest in quantity, a second pass of the clues now reveals more new information:



But that, in turn, allows even more new information to be found via a second observation of the grid:



A third pass through the clues is required and results in a complete solution:





]]>Puzzles and Games - Logic Problem - Trying Out MotorcyclesDeseret Riderhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=3014&REPLY_ID=38190Puzzles and Games Sat, 10 Jun 2006 22:43:13 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=3014 Ah Yeah! Good puzzle Jim-----be back to you on it---IF---I can figure it out. thanks for your effort----give us one of these once in a while ----keeps our minds from going stale---and we appreciate your stimulation.]]>Puzzles and Games - really cool web siteNicolas27http://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2704&REPLY_ID=35103Puzzles and Games Tue, 18 Apr 2006 13:48:30 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2704 yeah, but like I said, had to undo many times. It only shows the finished product, and even the process of me drawing is without all of the undos that I made. I had drawn several pictures before that one and just figured I'd try to draw the bike. I used a pic for reference and just kept trying, erroring, trying, erroring, for several hours until I got what you see.

Anyways, thought it was a neat website and figured I'd share. :) It's pretty addictive though, so be careful, before you know it you'll have wasted hours of your time on it. hehe]]>Puzzles and Games - Video Gameshoward.vhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2692&REPLY_ID=35065Puzzles and Games Mon, 17 Apr 2006 20:37:19 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2692
quote:
Originally posted by jollyroger

quote:
Originally posted by howard.v

Here is a link for the official game website if anybody's interested.
www.touristtrophy-thegame.com

Just as an aside, howard, is that a factory saddle on your rig?
It looks like a Mustang.
Nice bike![8D]


It is a Mustang seat, driver and passenger. I love it. I have had several people ask me if it was an 1100. When I tell them it's a 750, they can't believe it. The retro styling makes it look larger than it is. I think the Mustang seats compliment the look.]]>Puzzles and Games - Simple Math----a problem to solve!Deseret Riderhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2679&REPLY_ID=34941Puzzles and Games Sat, 15 Apr 2006 13:09:56 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2679 Alton----I believe you are right---and my hat is off to you for taking the time to explore the site for information before jumping to conclusions---as I did! Good thinking!]]>Puzzles and Games - Sudoku - 7 - Forced ChainsJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2457Puzzles and Games Fri, 03 Mar 2006 08:50:21 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2457
In fact, forced chains is VERY CLOSE to trial and error and some pundits of the game insist that some puzzles can only be solved using a trial and error method. I am of the opinion, however, that using forced chains IS a form of logic - that is, trial and error seeks to find what works while forced chains seeks to find what does NOT work.

In any event, it is true that some puzzles cannot be solved without resorting to forced chains. For example:



Has viable candidates as follows:



Which can be solved no further than here with the tools I have shown you:



If you, for example, decide to make the top right cell an '8' as follows:



Then pretty soon your puzzle will look like this (Notice that row 2 now contains the impossible situation where two of the cells can only be a '2')



So now you go back to the top right cell (meaning you UNDO ALL OTHER CHANGES MADE SINCE THEN) and make it a '2' instead of an '8' because now you know that the '8' is false.



The puzzle will solve normally thereafter, though that is not necessarily true.

Going back to your first choice ('8' instead of '2' for the top right cell), it is possible that would have left you having to resort to yet another 'forced chains' to try to complete the puzzle. If you DO have to use forced chains a second time then you DO NOT KNOW IF YOUR FIRST CHOICE WAS TRUE OR FALSE. Only if you end up with an impossible puzzle configuration (50% odds) can you know that your choice was FALSE, and only if the puzzle solves completely after making a selection (substantially less than 50% odds) can you know if your choice was TRUE. The forced Chains maneuver is, in other words, MOST VALUABLE TO YOU IF IT PROVES A CHOICE IS FALSE!
]]>Puzzles and Games - Sudoku - 6 - XY-wingJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2423Puzzles and Games Sat, 25 Feb 2006 08:01:12 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2423
First, the logic ...

If there are three cells containing naked pairs which contain only three different digits (X, Y, and Z) in them and they intersect in one, two or three domains (row, column, box), then regardless whether X or Y is true, Z is true in EITHER XZ or YZ. That sounds rather obvious, or convoluted, but as you will see, it is a powerful observation.

XY-wing in ONE domain:


XY-wing in TWO domains:


XY-wing in THREE domains:


For example, if X=4, Y=7, and Z=9 and three cells in any one domain contain 49, 47, and 79, then one of the cells is a 4, one of them is a 7 and one of them is a 9, of course.

In fact, if they are all three contained within a single domain then we have a simple 'triple' (three cells in one domain that contain only three different digits) instead of an 'XY-wing' so it is obvious that each of those cells is actually one of those digits and no other cells in that domain can contain any of those digits.

Here is an example of an 'XY-wing' spread across two domains (I have highlighted all the cells that contain naked pairs):



The pink cell contains the 'XY' (28) digits while one of the blue cell that contains the 'X' value (2) in it is the 'XZ' and the one that contains the 'Y' value (8) in it is the 'YZ' cell. The 'XY' cell is ALWAYS intersected by the other two cells - it is 'in the middle', if you will.

Please note that regardless of which value the 'XY' cell contains (either 2 or 8) then one of the blue cells MUST contain 'Z' (5). THIS is what makes the 'XY-wing' so valuable. For example, if the pink cell is actually a 2 then the blue cell at the top of column 5 MUST BE a 5 but if the pink cell is actually an 8 then the blue cell in row 7 column 6 MUST BE a 5.

Let's look at each of the only two possibilities more closely. Suppose that the pink cell was actually a 2. Then the blue cell in row 1 MUST BE a 5. And that means that none of the dark green cells can also be a 5 so you would cull those candidate cells of 5's.



But if the pink cell was actually an 8 instead of a 2 then the blue cell in column 6 MUST BE a 5. And that means that none of the dark green cells can also be a 5 so you would cull those candidate cells of 5's.



Now we will see the magic of 'XY-wing' patterns. The fact is that you cannot determine whether the pink cell is a 2 or an 8 but you do know that it is one or the other. And that means that the cells colored dark green below CANNOT be a 5 regardless of which value the pink cell contains. That is, the candidate cells that intersect BOTH the 'XZ' and 'YZ' cells CANNOT CONTAIN 'Z' because either the 'XZ' or 'YZ' cell does.



Let's see a couple more examples.

Note that the pink cell is the ONE cell of the three that is intersected by the two others. It is the 'XY' cell. Therefore, the 'Z' digit below is a 1.



Now we find the candidate cells that intersect with BOTH the 'XZ' and 'YZ' cells and we KNOW that they cannot be 1's so we can cull them from the viable candidate lists in those cells.



Here we can cull 2's from any candidate cell that is dark green.



]]>Puzzles and Games - Sudoku - 4 - finishJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2404&REPLY_ID=32808Puzzles and Games Fri, 24 Feb 2006 09:33:54 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2404 [:I] You ain't seen nothin yet. Wait till you try to absorb the next thread about 'colors'. [:I]]]>Puzzles and Games - Sudoku - 5 - ColorsJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2415Puzzles and Games Fri, 24 Feb 2006 09:29:19 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2415
Remember back to our discussion about conjugate pairs. These are instances where you have two, and only two, cells in a domain (row, column or box) that contain the same viable candidate digit. Thus, one of those two cells MUST be that digit.

Well, conjugate pairs often form chains in your puzzle. That is, one of those two cells can be part of more than one conjugate pair.

Let's look at a puzzle that has been solved to the point where we can use this new tool:



Now highlight those cells that contain viable 6 candidates:



We will find a conjugate pair and color one of the cells blue and the other green, then we will extend the chain by connecting the first conjugate pair to another. Note that we will ALWAYS alternate colors - that is, the blue cells always connect to green cells.

Here is the first one:



Here is the next one:



And here is the completed chain:



Now, let me remind you that we have linked conjugate pairs. That means that one of the pairs IS the digit. And, because we have alternated colors, EITHER ALL the blue cells are true OR ALL the green cells are true. That is a profoundly important concept here, as we will soon see. I also want to remind you that sudoku rules insist that there be one and only one of the digits (in this case 6's) in every domain.

So, ANY CELL that intersects with BOTH a blue cell and a green cell CANNOT be a 6! Look at the cell located in row 1 column 5. It intersects with the blue cell in row 1 column 2 and the green cell in row 4 column 5. It CANNOT be a 6 because EITHER the blue cell or the green cell is a 6.



Now let's look at all the cells that contain 9's as viable candidates.



Let's find a conjugate pair and color it.



And then we will complete the entire chain.



Though I can find no candidate cells that intersect both blue and green cells in the result, I do see yet another set of conjugate pairs so I will create a second chain using different colors (pink and amber) for it.



These are DIFFERENT chains. At this point we do not know ANYTHING about how they interact with each other, if at all. So we look for instances of interaction. We find that the pink cell at row 5 column 7 is in the same domain (box) as the blue cell in row 4 column 9 but we also find that NO amber cell intersects with any blue or green cells.

Loosly Coupled Chains



Put your logic hats on now. If there were no intersections between the cells in the first chain and the cells in the second chain then the two chains would be ENTIRELY independent of each other. That is, the state of the blue or green cells would not determine the state of the pink or amber cells. However, we know that if the blue cells are 9's then the pink cells are NOT 9's because of the intersection I pointed out. If, however, the blue cells are NOT 9's that DOES NOT MEAN THAT THE PINK CELLS ARE. In other words, when only one color of the two chains intersects the chains are only loosely coupled.

If, on the other hand, we had found that one or more of the amber cells intersected with a green cell in addition to the already recognized fact that at least one of the pink cells intersects with the blue cells, then that set of chains would be tightly coupled and that, in turn, means that those two chains could be merged and become one chain. That is, all the pink cells could be turned into green cells and all the amber cells could be turned into blue cells.

Because these chains are NOT tightly coupled any cell that intersects the two chains cannot be culled of the 9 candidate. For example, the cell at row 5 column 8 is between a pink and a blue cell. Had the two chains been tightly coupled that would mean that either the pink or the blue cell is a 9, but because they are loosely coupled all we know is that if either the pink or the blue cells is a 9 then the other one is not. Since it is possible that NEITHER the pink nor the blue cells is a 9 then we do not know that the highlighted cell is or is not a 9.

Now, though we have found no candidate cells that can be culled because of intersections with multiple colors, we do see something that is extremely valuable: there are two blue cells in row 4. But that is not possible! By that, I mean, if all the blue cells are 9's then there would be two 9's in row 4 and that cannot be in a sudoku puzzle. In other words, we now KNOW that NO BLUE CELLS are 9 and can remove the 9's from the viable candidate list within each of those cells.

MORE important, and often overlooked, is that we now KNOW that ALL GREEN CELLS are 9!!!

The reason that we colored both chains and left them showing even though we found that they were not tightly coupled should now be apparent. Had it turned out that the blue cells were 9's instead of the green cells then we would have been able to cull 9's from all the pink cells and turned all the amber cells into 9's.



Tightly Coupled Chains

So let's look at a couple of conjugate pair chains that are tightly coupled in order to cement this very powerful idea. Here is a game that has been solved to the point that we can proceed using colors. I have highled all candidate cells that contain 7's and then colored the first chain blue and green:



Now I will color the second chain pink and amber:

Observe that the pink cells intersect with (and only with) green cells while the amber cells intersect with (and only with) blue cells:



We have just discovered that these two chains are Tightly Coupled and can, as a result, merge them. Here we have turned the pink cells into blue cells:



And here we have turned the amber cells into green cells:



In other words, we can treat all the colored cells as if they were part of a single conjugate pair chain. Thus, the three highlighted cells CANNOT be 7's (because they intersect both a blue and a green cell) so we can cull the 7's from them.



Now we have new conjugate pairs that we can add to the chain. Here is the first one and we color it green:



The second one we color blue:



At this point we discover that there are two blue cells in row 8 so now we KNOW that ALL blue cells are NOT 7's and ALL green cells ARE 7's:



But before we start culling we will complete building the chain:



Then we will cull 7's from all the blue cells:



And turn all the dark green cells into 7's (Note that this will leave us with only one cell that has a candidate 7 in it so that it MUST be a 7:



Turning that cell into a 7 resolves ALL 7's in the puzzle:



One more very powerful observation about what you can do with multiple conjugate pair chains - the chains themselves can be treated like conjugate pairs. That is, since EITHER all pink cells or all amber cells are true then if both pink and amber cells intersect with blue cells, for example, then ALL BLUE CELLS MUST BE FALSE - which means that ALL GREEN CELLS MUST BE TRUE (9's). Conjugate pairs and conjugate pair chains are powerful patterns!

Here is an example of two different conjugate chains of viable 6 candidates in a puzzle. In this picture you will see that the amber cells intersect with blue cells:



And here you see that the pink cells also interect with blue cells (or, if you prefer, we see that blue cells intersect with both pink and amber cells - this should make the next concept perfectly clear as we already know that either the pink or amber cells is true):



So now we KNOW that all blue cells MUST be false (so we can cull 6's from their lists of viable candidate digits) and that means that ALL green cells must be true (so we convert them to 6's):



Powerful stuff, colors, no?
]]>Puzzles and Games - An old simple classic one!subvetSSN606http://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2398&REPLY_ID=32802Puzzles and Games Fri, 24 Feb 2006 08:54:26 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2398 Correct! [8D]

Tom]]>Puzzles and Games - Sudoku - 1 - Starting a hard puzzleJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2390&REPLY_ID=32748Puzzles and Games Thu, 23 Feb 2006 14:30:54 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2390 Oh I think you misunderstand the reason people want to program Sudoku solving programs. It is NOT to solve Sudoku puzzles. That is, it is not to input a puzzle and let the program solve it for you.

Rather, it is to learn the 'rules' and patterns necessary for YOU to solve puzzles.

There are no artificial intelligent programs, yet. They all simply follow a set of instructions that you have coded. In other words, you must first recognize that the program MUST be able to recognize a certain pattern and then code it so that it can.

If your program can solve any Sudoku puzzle presented to it then it should be obvious that the programmer FULLY understands the game and can solve those puzzles without the program.

Well, some things that you can do with a program are almost impossible for a mere human to do - essentially in recognizing very complex patterns - because of the HUGE number of possibilities that you must try in order to find those patterns, but with or without the computer if you wrote a program that can solve any Sudoku puzzle, given enough time you could solve that puzzle without the computer.

]]>Puzzles and Games - Sudoku - 3 - X-Wing and SwordfishJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2403Puzzles and Games Thu, 23 Feb 2006 12:25:24 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2403
Here is the puzzle which we were working on in the 'Sudoku Pairs' thread, just before we applied the 'locked candidates' pattern rule. It highlights all cells that contain a 6:



Instead of using that pattern I want to show you yet another way to cull viable candidates using the same data.

Note that the two sets of conjugate pairs shown in blue share the same rows and columns. The conjugate pairs are in different rows. They are NOT, however, seen as conjugate pairs in their respective columns because there are more than two 6's in those columns. In other words, there are candidates that can be culled.

Now let's look even closer at these four cells. There are a total of TWO 6's there (one in each of the two rows.) Further, those TWO 6's are distributed between the two columns. Thus, when one of the cells is found to be a 6 then the cell diagonally across from it is also a 6 (which is apparently why it is called an 'X-wing' pattern.) In other words, those four cells determine where all the 6's are for BOTH rows AND columns.

You will remember from the previous thread that a conjugate pair that is in more than one domain IS a conjugate pair in all domains shared.

So, the conjugate pair (blue cells) in column 7 means you can cull 6's from all other cells in column 7 (dark green) and the conjugate pair in column 9 means that you can cull 6's from all other cells in column 9 as follows:



Now let's look at just the cells that have a 4 in their list of viable candidates:



If you have two conjugate pairs in two shared rows and columns to make an 'X-wing' pattern, then it should not surprise you to realize that if you have THREE cells in Three shared rows and columns that have a particular candidate digit in them that those make another important pattern - a 'Swordfish'. In such a pattern you must now recognize that there will be exactly THREE 4's and one of those 4's will be in each of the shared rows and one of them will be in each of the shared columns.

So, here are the cells that make up a 'Swordfish' pattern in our puzzle.



Yep, I know that two of those cells are not candidate cells as they have already been resolved. All that means is that the swordfish pattern can be made with THREE rows OR THREE columns that contain TWO or THREE shared columns or rows containing the digit being looked at.



This is what makes it difficult to recognize this pattern. But the logic remains valid. There MUST be exactly three 4's distributed amongst these blue cells such that one of them is in each row and each column of the pattern. Thus, ONE of the TWO blue cells in rows 5 and 9 as well as ONE of the THREE blue cells in row 3 must be a 4.

You will note, therefore, that both the three rows and the three columns shared by the pattern can have all 4's in viable candidate lists in the dark green cells removed.



Removing those 4's yields the following puzzle (with those cells containing viable 4 candidates highlighted):



Let's look at another 'X-wing' example at this point. Here is the puzzle with the cells that have 7's as viable candidates. The blue cells form the 'X-wing', the dark green cells show which 7's can be removed from their cells as they are no longer viable.



This results in a puzzle with a new naked singleton (yellow.) We are now very close to having the entire puzzle solved. Honest!




]]>Puzzles and Games - Sudoku - 2 - PairsJames R. Davishttp://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2402Puzzles and Games Thu, 23 Feb 2006 09:56:16 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2402


A naked pair, by itself (that is, the only such pair in its row, column and box) can tell us only one thing - that that cell MUST be one or the other of the digits in the pair. It cannot be used to determine which of those digits the cell actually is and IT CANNOT BE USED TO CULL VIABLE CANDIDATES IN ANY OTHER CELL.



But if you have TWO identical pairs that share a domain (row, column or box) then you know a great deal more. Here you see highlighted two naked pairs that share the same row and the same box. There CANNOT BE MORE THAN TWO SUCH PAIRS IN ANY DOMAIN!! Why? Because one of those cells MUST be a '4' and the other MUST be a '6' and because you can only have ONE '4' or '6' in any domain.

THEREFORE, since we have found out that the '4' and the '6' MUST be in those two cells in the domain, THEY CANNOT BE VIABLE CANDIDATES in any other cell of the same domain. So, though there are no other cells in that row or box that contain '4' or '6' in their list of viable candidates, if there were any you could (and must) eliminate those non-viable candidates from those cells. That, after all, is what the fundamental effort is in solving a Sudoku puzzle. In other words, the pairs shown in blue allow you to eliminate '4' and '6' from all the dark green cells.



What you just learned is the concept of 'scope'. The blue pair of cells share two domains: they are in the same row and they are in the same box. Thus, they effect all the other cells in the shared domains ONLY. The fact that there are 4's and 6's as viable candidates in cells in the last column is not affected by the existence of those two identical pairs as they do not share the column domain (they are not both in column 9.)

But, wait! We have not exhausted the value of pairs yet nor have we even seen the most powerful aspect of pairs. Let us look more closely.

In the two blue cells there are two digits, of course, but far more important is what those two digits are. If you simply ignore that there are 6's in those two cells you have 4's left. TWO of them. That is, a pair of fours, one in each of two different cells. Those two 4's are known as a 'conjugate pair'. What that means is that Either one of them is a 4 OR the other one is a four. That is pure binary. A or B. True or False. The same is true if you ignore the 4's. You are left with a conjugate pair of 6's. Two, and only two, possible locations for the digit 6 IN THAT DOMAIN!

Now there is another fundamental difference between naked pairs and conjugate pairs. A naked pair means that there are two digits in a cell. A conjugate pair means that the same digit is found in only two cells within a domain (even if there are other digits in those cells.)

For example, let's look at all the cells that contain the digit '2':



The blue cells are a conjugate pair of 2's that are in only a single domain (row 3). No other cells in that domain (row 3) can be a '2'. That is, one of those two blue cells IS a '2' while the other one is not. Period.

There is another conjugate pair of 2's shown in blue below. Note that they are part of more than one domain. That is, they are within column 7 as well as in the lower right corner box. Thus, any cell shown as dark green CANNOT be a 2 and thus you are able to eliminate any 2's from their candidate lists.



Finally, there is a third conjugate pair of 2's. Because they share only a single domain (the upper left box), no other cells in that box can be a 2.



So let's see how to use this new knowledge to eliminate viable candidates and resolve cells. Let's look at all the cells that contain 8's. The two blue cells are a conjugate pair of 8's WITHIN the box domain. But note that that pair is also part of row 7. It is NOT a conjugate pair within row 7 because there are three cells that contain 8's in that domain. However, because it IS a conjugate pair in the box we KNOW that one or the other of them IS an 8. THEREFORE, the third (pink) cell in row 7 CANNOT BE AN 8. So, we eliminate the 8 from that cell's list of viable candidates. And, as a result, we will be left with a Singleton 8 in the bottom middle box - in other words, THAT cell IS an 8.



We will convert the blue cell into an 8 and that will, in turn, allow us to remove 8 from the viable candidate lists in the green cells.



After we have done that we are left with a puzzle than has only 36 of its 81 cells left to resolve.



Now let's look at one more way to use conjugate pairs. We will look at all cells that have a 6 as a viable candidate. Note the conjugate pair in column 8 colored blue. It is NOT a conjugate pair in the bottom left box as there are five cells that can contain be a 6 in that box. However, because one of the two blue cells MUST be a 6 (because they are a conjugate pair in column 8) then none of the dark green cells can be a 6. This is called being 'candidate locked'. Specifically, if a conjugate pair is found in a domain then it is a conjugate pair in all domains it is a part of.



That results in a puzzle that looks like this:







]]>Puzzles and Games - A puzzler----for those who like puzzlesnomad danhttp://www.msgroup.org/forums/mtt/topic.asp?whichpage=-1&TOPIC_ID=2374&REPLY_ID=32601Puzzles and Games Sun, 19 Feb 2006 12:25:09 -0500http://www.msgroup.org/forums/mtt/topic.asp?TOPIC_ID=2374 If I could borrow James R’s brain for a while I would go back to school and turn my Masters into a PHD [;^)]]]>