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July 12

The Non-Zero Difference Puzzle

Here is an interesting puzzle. I am just curious as to how, using logic, can one solve this puzzle. I have solved it using the brute force method, exactly what the author said NOT to do. But how can it be solved analytically?--A Real Kaiser (talk) 00:10, 12 July 2008 (UTC)[reply]

It seems to me like a skewed version of Pascal's triangle. Maybe start there. --Prestidigitator (talk) 05:55, 12 July 2008 (UTC)[reply]

Here's a start: Each entry in the top row is required to be between 1 and 9 (inclusive). The maximum possible difference between any pair of those entries is 8, therefore the maximum possible entry in the second row is 8. Each entry in the second row is between 1 and 8. The maximum possible difference between any pair of them is 7, therefore the maximum possible entry in the third row is 7. Keep applying that logic, and you'll find that the maximum allowable number in the bottom row is 1. So fill the bottom 2 boxes in with 1's.

The row above "1 1" must be either "1 2 1" or "2 1 2". From here, trial-and-error is the best method I can come up with. Try the "2 1 2" and work your way up, and you'll find it doesn't work. A useful rule for reducing the search space is: when you have an N in the Nth row from the bottom, its 2 neighbors above must be an "N+1" and a "1", since N+1 is the maximum allowable number in the (N+1)st row from the bottom. So if you use "2 1 2", both 2's must have "3 1" or "1 3" over them. That means the next row up is "1 3 1 3", "1 3 3 1", "3 1 1 3", or "3 1 3 1", none of which match the "1" in between those 2's. So the second row from the bottom must be "1 2 1". --tcsetattr (talk / contribs) 07:01, 12 July 2008 (UTC)[reply]

Using that method actually doesn't lead to that big of a tree; many possibilities cancel out quickly. Until the last few rows, you can ignore reverses. Also, because of the planted threes, the highest number in the row will not be on the either edge after the bottom row. If you want to see the answer, the spoiler is in the source.

Integration question

Can anyone help with integrating this?

\int {\frac {u^{a}}{1+bu}}\ du

Here, a is a complex number (with real part -1, if that helps; the imaginary part is a parameter), and b is a real number. Mathematica's online integrator was not much help (http://integrals.wolfram.com/index.jsp?expr=(x^a)%2F(1+%2B+b+*+x)&random=false). I suppose it probably doesn't have a closed form integral? Eric. 85.178.19.137 (talk) 18:39, 12 July 2008 (UTC)[reply]

Is the Hypergeometric series form no use at all? just out of curiosity was it for anyhting specific? would the bounds be anything?87.102.86.73 (talk) 21:13, 12 July 2008 (UTC)[reply]

Thanks, I took another look at that and see it wasn't quite as bad as I feared. This integral arose in calculating the location of a vehicle traveling at constant acceleration and turning rate, undergoing drag (which occurs for this year's ICFP contest, www.icfpcontest.org). I was hoping that a sufficiently nice result for the integral would give some insight in optimizing vehicle navigation, but unfortunately not. (The lower bound is zero, and the upper bound is another parameter, so I just left it is an indefinite integral.) Eric. 85.178.58.188 (talk) 09:48, 13 July 2008 (UTC)[reply]

If the integration limits are zero and infinity then put bu= x and use that:

\int _{0}^{\infty }{\frac {x^{-p}}{1+x}}dx={\frac {\pi }{\sin \left(\pi p\right)}}

Count Iblis (talk) 23:07, 12 July 2008 (UTC)[reply]

How do I compute the odds to see if any National Football League franchises are going to the Super Bowl this year?

When's the draft? That should help. Otherwise, how do Las Vegas oddsmakers compute sports odds? Are there people who watch and take into account everything?. ("The quarterback just drank a Gatorade and put on high heels, -there goes another point against his team..." etc.)--Baseball and and and Popcorn Fanatic (talk) 19:30, 12 July 2008 (UTC)[reply]

I think the main factor in working out the odds are how people are betting. If lots of people are betting for a particular team, that suggests they are more likely to win, so you lower the odds, and vice versa. I'm not sure to what extent the bookmakers use knowledge of the sport to try and work out for themselves who's likely to win - quite possibly not at all. --Tango (talk) 21:07, 12 July 2008 (UTC)[reply]

Are you talking about something like ranking of sports team? Because if we rank them, then the top few (the best teams) will go to super bowl. If this is what you talking about, then I did a presentation on it (different ranking schemes) some time ago. I can explain more in detail if you want.--A Real Kaiser (talk) 01:39, 13 July 2008 (UTC)[reply]

What I'm getting at is whether it's horses, football, baseball, etc. sports oddmakers are remarkably accurate.I want to impress my friends and bocome RICH! How do I achieve this skill? —Preceding unsigned comment added by Baseball and and and Popcorn Fanatic (talk • contribs) 06:34, 13 July 2008 (UTC)[reply]

Unfortunately, probabilities alone won't solve this problem. A typical probabilities approach might be to figure out which portion of the time a given team has won in the past, and use that to predict the future expected probability of the event. However, as the make-up of each team changes each year (and even within a year due to injuries, etc.), such a method is rather unreliable here. Instead, you need to basically use knowledge of the sport to figure out who is likely to win games. Things like "pitcher A has a good south-paw pitch, which should be effective against hitter B, but not hitters C or D; therefore, I predict that pitcher A's team will lose". There's a lot more to it than just that, of course, but that's a simple example. The first step, I would imagine, is to get to know the game or sport inside and out. StuRat (talk) 15:19, 13 July 2008 (UTC)[reply]

The reason bookmakers make so much money is because they offer worse odds than are realistic - add together the probabilities of each horse in a race winning and you'll find it adds up to more than 1. There's quite a large margin of error in which they still average a profit. --Tango (talk) 16:58, 13 July 2008 (UTC)[reply]