- 21 Dec '06 00:07 / 2 edits

It's a puzzle .. not a statement of fact my friend .. take a deep breath .. there, now don't you feel better?*Originally posted by dmnelson84***http://www.4degreez.com/misc/personality_disorder_test.mv**

Take that test. It's from another thread. My guess is that you'll score high in the Narcissistic field.

Now if you had called me sadistic that would be a different matter

EDIT: low Narcissism i'm afraid but moderate Schizoid and borderline .. go figure .. - 21 Dec '06 00:38

You'd have to play (1800-1357)/32 games to reach that rating (don't forget to round up to get the real figure).*Originally posted by idioms***I am rated 1800 on RHP, I was rated 1357. What is the minimum number of games I could have played to get this rating given that the highest rating of my opponent was 2455?**

BONUS: How many different combinations of ratings are possible to achieve this minimum number of games?

There are lots of different combinations possible to reach your goal, so long as whenever you beat a new opponent they're 400 points higher than you. - 21 Dec '06 01:28

Yep that's right, the formula you gave is the*Originally posted by Irax***You'd have to play (1800-1357)/32 games to reach that rating (don't forget to round up to get the real figure).**

There are lots of different combinations possible to reach your goal, so long as whenever you beat a new opponent they're 400 points higher than you.*lower bound*of this problem. What if the highest rating of my opponents was also 1800? would the formula still be valid?

There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455 - 21 Dec '06 02:00

For the different combinations, I guess you'd have 2455-1357+400 different ratings possible for your first game, then 2455-1357+400-32 different ratings for your second, so that's*Originally posted by idioms***Yep that's right, the formula you gave is the***lower bound*of this problem. What if the highest rating of my opponents was also 1800? would the formula still be valid?

There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455

1498+32n possibilities for each game where n will run from 0 to 13 (For each of the 14 games).

So multiply them together to get the permatations.

(1498-32*0) * (1498-32*1) * (1498-32*2) * .... * (1498-32*13)

I could calculate a figure at this point, but there must be a better way to express this mathematically, it's been too long since I've done this kind of stuff, my memory fails me right now! - 21 Dec '06 02:04

The formula wouldn't hold if the max rating of your opponent was 1800, and since the formula to calculate rating change isn't linear, I wouldn't know how to rearrange everything to create a new formula, but what I can say is that it'll be no more than ((1800-1357)/32) * 2 games, because you'll be gaining at least 16 rating points each game, until you exceeded 1800.*Originally posted by idioms***Yep that's right, the formula you gave is the***lower bound*of this problem. What if the highest rating of my opponents was also 1800? would the formula still be valid?

There certainly are lots of combinations but the number is finite as your opponents rating has an upper bound of 2455 - 21 Dec '06 04:36 / 1 edit

Exactly!**Originally posted by Irax***For the different combinations, I guess you'd have 2455-1357+400 different ratings possible for your first game, then 2455-1357+400-32 different ratings for your second, so that's**

1498+32n possibilities for each game where n will run from 0 to 13 (For each of the 14 games).

So multiply them together to get the permatations.

(1498-32*0) * (1498-32*1) * ...[text shortened]... ically, it's been too long since I've done this kind of stuff, my memory fails me right now!

Unfortunately there isn't an easy way of expressing this without continued fractions since 32 isn't a divisor of 1498

*Technically this gives a rating of 1805 but that's close enough .. I think it may even be 1803 because the "expected win" factor kicks in sometime in the 1700's but then the calculation becomes horrid - 21 Dec '06 12:45

Is there a way to show this calculation mathematically?*Originally posted by XanthosNZ***After 19 wins against an 1800 player your rating would have gone from 1357 to 1807 (assuming rating changes are rounded to the nearest integer).**

It'd be fairly straight forward to implement this as a computer program to work out the final answer, but I'm curious to know if maths has the language to describe it. - 21 Dec '06 17:14 / 1 edit

It's an iterative computation where:*Originally posted by Irax***Is there a way to show this calculation mathematically?**

It'd be fairly straight forward to implement this as a computer program to work out the final answer, but I'm curious to know if maths has the language to describe it.

t(n+1) = round(n + 32 * (1 - 1/(10^((X-n)/400)+1)))

Where X is the opponent's rating and the function round rounds to the nearest integer.

It is (I think) theoretically possible to turn this from an iterative calculation into a non-iterative one (for a given start value) but the math is beyond me. - 21 Dec '06 17:55 / 1 edit

Find an appropriate matrice M in where the iterative formula is coded, and multiply this to itself a number of times, M^n, and find wich n that satisfy the criteria. To find the M, however, is beyond my level of math. I flunked that exam, I'm afraid.*Originally posted by XanthosNZ***It's an iterative computation where:**

t(n+1) = round(n + 32 * (1 - 1/(10^((X-n)/400)+1)))

Where X is the opponent's rating and the function round rounds to the nearest integer.

It is (I think) theoretically possible to turn this from an iterative calculation into a non-iterative one (for a given start value) but the math is beyond me.

For XanthosNZ's iterative formula it suffice with an 2x2-matrice.

A matrice where the fibonacci series is coded was demonstrated in the class so you could actually see which n satisfied the equation f(n) is greater than a given number. M is a 3x3-matrice. - 22 Dec '06 10:04

Does that work for iterative series that don't have polynomial equations?*Originally posted by FabianFnas***Find an appropriate matrice M in where the iterative formula is coded, and multiply this to itself a number of times, M^n, and find wich n that satisfy the criteria. To find the M, however, is beyond my level of math. I flunked that exam, I'm afraid.**

For XanthosNZ's iterative formula it suffice with an 2x2-matrice.

A matrice where the fibonacci seri ...[text shortened]... ly see which n satisfied the equation f(n) is greater than a given number. M is a 3x3-matrice. - 22 Dec '06 10:21

I'm not sure, I flunked that exam I'm afraid, but if you want me to I could try to find my notes from the class?*Originally posted by XanthosNZ***Does that work for iterative series that don't have polynomial equations?**

However, if fibonacci serie is not polynomial, then it works for non-polynomials too. - 22 Dec '06 10:47

For fibonacci the iterative series is:*Originally posted by FabianFnas***I'm not sure, I flunked that exam I'm afraid, but if you want me to I could try to find my notes from the class?**

However, if fibonacci serie is not polynomial, then it works for non-polynomials too.

t(n+2) = t(n) + t(n+1)

which is linear (and therefore a polynomial).