quote:from kacinpa:

Do you suppose they would reproduce at a different rate if the grain were other than Quadrotriticale?

I would speculate that so long as they are well fed, the rate would remain the same regardless. I am not sure if they would like to be in a cooler of whatever worms Klingons consider good food, but I once again speculate an ample supply of edible grains would satisfy them regardless of type.]]>

AMAZING, SIMPLE HOME REMEDIES:

1. AVOID CUTTING YOURSELF WHEN SLICING VEGETABLES BY GETTING SOMEONE

ELSE TO HOLD THE VEGETABLES WHILE YOU CHOP.

2. AVOID ARGUMENTS WITH THE FEMALES ABOUT LIFTING THE TOILET SEAT BY

USING THE SINK.

3. FOR HIGH BLOOD PRESSURE SUFFERERS ~ SIMPLY CUT YOURSELF AND BLEED

FOR A FEW MINUTES, THUS REDUCING THE PRESSURE ON YOUR VEINS. REMEMBER TO

USE A TIMER.

4. A MOUSE TRAP PLACED ON TOP OF YOUR ALARM CLOCK WILL PREVENT YOU

FROM ROLLING OVER AND GOING BACK TO SLEEP AFTER YOU HIT THE SNOOZE BUTTON.

5. IF YOU HAVE A BAD COUGH, TAKE A LARGE DOSE OF LAXATIVES. THEN

YOU'LL BE AFRAID TO COUGH.

6. YOU NEED ONLY TWO TOOLS IN LIFE - WD-40 AND DUCT TAPE. IF IT DOESN'T

MOVE AND SHOULD, USE THE WD-40. IF IT SHOULDN'T MOVE AND DOES,

USE THE DUCT TAPE.

7. IF YOU CAN'T FIX IT WITH A HAMMER, YOU'VE GOT AN ELECTRICAL PROBLEM.

THOUGHTS for the day:

SOME PEOPLE ARE LIKE SLINKIES - NOT REALLY GOOD FOR ANYTHING BUT

THEY BRING A SMILE TO YOUR FACE WHEN THEY'RE PUSHED DOWN THE STAIRS.

NEVER, UNDER ANY CIRCUMSTANCES, TAKE A LAXATIVE AND SLEEPING PILLS

ON THE SAME NIGHT

]]>

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]]>

quote:After readingOriginally posted by Deseret Rider

Your thought process being a little different both arrived at the same answer I have.

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

]]>

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.)

]]>

Steve]]>

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]]>

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

]]>

Thanks

David]]>

1 confused dag

2 disturbed bunnies

and 1 wet couch.

I hate those things.]]>

]]>

http://www.cs.ualberta.ca/~chinook/]]>

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]]]>

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]]]>

Thanks!!!

]]>

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.]]>

quote:Originally posted by twcquote: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]]]>

quote:Originally posted by scottrnelsonquote: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.]]>

David]]>

I always thought it was cute.]]>

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:

]]>

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]]>

quote:Originally posted by jollyrogerquote: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.]]>

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.

]]>

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,

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.

]]>

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.

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

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

Because these chains are NOT

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:

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.

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

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?

]]>

Tom]]>

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.

]]>

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

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!

]]>

A naked pair, by itself (that is, the only such pair in its row, column

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:

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