Counting primes in Scrabble racks - Page 2

Brian-E · 2008-03-07, 12:27

Sorry that I'm so dim, but...
what is that hex number and how does it answer the original question?
Appreciating your help in advance, guys.

ckdo · 2008-03-07, 12:47

Quote:

Originally Posted by retina

Take a standard English set of Scrabble tiles [...]

Count the number of unique racks you can draw, from a full bag, for draws of 1 tile through 50 tiles. Factorise, and post the largest prime found among all draw sizes.

e.g.

Code:

 1 tile draw - 27 unique racks = (3*3*3)
 2 tile draw - ?? unique racks = (a*b*...*c)
...
50 tile draw - ?? unique racks = (x*y*...*z)

And post the largest number among a...z (there is only one unique answer).

The number I posted is the largest prime factor you come across when you factorise the number of unique racks of 1 to 50 tiles, respectively.

Brian-E · 2008-03-07, 14:21

Quote:

Originally Posted by ckdo

The number I posted is the largest prime factor you come across when you factorise the number of unique racks of 1 to 50 tiles, respectively.

OK.
But how did you calculate it? Could you give any pointer to your method?

ckdo · 2008-03-07, 15:45

Warning! Major spoiler ahead!

I split the problem like this: Drawing [B]n[/B] tiles from the entire set is equal to drawing [B]a[/B] tiles from the [N,L,B,P,V,K] subset, [B]b[/B] tiles from the [R,S,C,F,W,J] subset, [B]c[/B] tiles from the [T,D,M,H,Y,X] subset and [B]d[/B] tiles from the [E,A,I,O,U,G,Blank,Q,Z] subset such that [B]a+b+c+d=n[/B] with [B]0<=[a,b,c]<=17[/B] and [B]0<=d<=49[/B]. The subsets were chosen like this because the distribution of the letters in the first three is identical: 6,4,2,2,2,1 from left to right. Saves computing stuff twice. Splitting the problem further didn't seem necessary. Now you only need to make two tables how many possibilities there are to draw 0..17 and 0..49 tiles from sets with elements distributed as given (once) and then for all [B]n=1..50[/B] find fitting [B]a,b,c[/B] values, multiply the combinations for the four subsets and add them up leaving you with the number of racks of [B]n[/B] tiles. Then you "only" need to factor those numbers (up to 15 digits in length) and remember which factor was the largest you see. Which hopefully is 219163709706077, the number of unique racks of 36 tiles. [B]qed.[/B]

retina · 2008-03-07, 16:00

There is also away to compute the answer without needing any tables and also it only uses addition, never needs any multiplications. It can also extended without limit to any set of coefficients (tiles) given in any order. Actually the big clue is this word, "coefficients"

I also discovered there is a super set distribution with 200 tiles, the numbers are a bit larger up to 22 digits maximum, so the same thing, what is the largest prime from all factorised numbers of draws up to 100 tiles?

Brian-E · 2008-03-07, 16:52

I find ckdo's method ingenious, although I can't quite see why computing the tables for the (very clever) subsets requires less work than simply computing the whole table in one go drawing from all 100 tiles.
Therefore I'm interested to think further and try to discover retina's method with the hint provided. I guess it will be needed to solve the bigger superset problem.

Edit: Forget that "although...", sorry. Of course the division to subsets makes it easier because there are fewer combinations of numbers from each group.

ckdo · 2008-03-07, 17:09

The 64 bit border is one I'm not tempted to cross anywhen soon.

retina · 2008-03-07, 17:14

Quote:

Originally Posted by ckdo

The 64 bit border is one I'm not tempted to cross anywhen soon.

Who's afraid of the big bad wolf ...

nibble4bits · 2008-03-15, 06:14

Arbitary precision arithmatic is your friend! Have fun waiting 6 months for your calculation that would take 2 months with less-than-64-bit numbers. :)

2008-03-07, 16:52	#17
Brian-E "Brian" Jul 2007 The Netherlands CC5₁₆ Posts	I find ckdo's method ingenious, although I can't quite see why computing the tables for the (very clever) subsets requires less work than simply computing the whole table in one go drawing from all 100 tiles. Therefore I'm interested to think further and try to discover retina's method with the hint provided. I guess it will be needed to solve the bigger superset problem. Edit: Forget that "although...", sorry. Of course the division to subsets makes it easier because there are fewer combinations of numbers from each group. Last fiddled with by Brian-E on 2008-03-07 at 17:00

Similar Threads
Thread	Thread Starter	Forum	Replies	Last Post
Path Counting	henryzz	Puzzles	13	2014-09-17 11:21
SCRABBLE	Raman	Hobbies	9	2013-12-11 21:56
Counting on Fingers	Orgasmic Troll	Lounge	3	2005-12-31 22:22
Counting Cubes	Numbers	Puzzles	6	2005-09-03 00:26
counting bricks	michael	Puzzles	8	2004-01-14 16:27

2008-03-07, 12:27	#12
Brian-E "Brian" Jul 2007 The Netherlands 7·467 Posts	Sorry that I'm so dim, but... what is that hex number and how does it answer the original question? Appreciating your help in advance, guys.

2008-03-07, 15:45	#15
ckdo Dec 2007 Cleves, Germany 2·5·53 Posts	Warning! Major spoiler ahead! I split the problem like this: Drawing [B]n[/B] tiles from the entire set is equal to drawing [B]a[/B] tiles from the [N,L,B,P,V,K] subset, [B]b[/B] tiles from the [R,S,C,F,W,J] subset, [B]c[/B] tiles from the [T,D,M,H,Y,X] subset and [B]d[/B] tiles from the [E,A,I,O,U,G,Blank,Q,Z] subset such that [B]a+b+c+d=n[/B] with [B]0<=[a,b,c]<=17[/B] and [B]0<=d<=49[/B]. The subsets were chosen like this because the distribution of the letters in the first three is identical: 6,4,2,2,2,1 from left to right. Saves computing stuff twice. Splitting the problem further didn't seem necessary. Now you only need to make two tables how many possibilities there are to draw 0..17 and 0..49 tiles from sets with elements distributed as given (once) and then for all [B]n=1..50[/B] find fitting [B]a,b,c[/B] values, multiply the combinations for the four subsets and add them up leaving you with the number of racks of [B]n[/B] tiles. Then you "only" need to factor those numbers (up to 15 digits in length) and remember which factor was the largest you see. Which hopefully is 219163709706077, the number of unique racks of 36 tiles. [B]qed.[/B]

2008-03-07, 16:00	#16
retina Undefined "The unspeakable one" Jun 2006 My evil lair 2²·1,549 Posts	There is also away to compute the answer without needing any tables and also it only uses addition, never needs any multiplications. It can also extended without limit to any set of coefficients (tiles) given in any order. Actually the big clue is this word, "coefficients" I also discovered there is a super set distribution with 200 tiles, the numbers are a bit larger up to 22 digits maximum, so the same thing, what is the largest prime from all factorised numbers of draws up to 100 tiles?

2008-03-07, 17:09	#18
ckdo Dec 2007 Cleves, Germany 2×5×53 Posts	The 64 bit border is one I'm not tempted to cross anywhen soon.

2008-03-15, 06:14	#20
nibble4bits Nov 2005 182₁₀ Posts	Arbitary precision arithmatic is your friend! Have fun waiting 6 months for your calculation that would take 2 months with less-than-64-bit numbers. :)