[R] expand.grid game

Tue Jan 12 21:19:13 CET 2010

How trivial is probably subjective, I don't think it is much above trivial.  I would not have been surprised to see this question on an exam in my undergraduate (300 or junior level) probability course (the hard part was remembering the details from that class from over 20 years ago).  My favorite test question of all time came from that course: "You have a deck of poker cards with the 3's removed (and jokers), you deal yourself 5 cards at random, what is the probability of getting a straight (not including straight flushes)?"

This problem is simpler.  Just think of the 8 places in the number as urns, and the 17 1's as balls to be put into the urns.  One ball has to go in the first urn, so you have 16 left, there are choose(16+8-1,8-1) ways to distribute 16 undistinguishable balls among 8 distinguishable urns. But that includes some solutions with more than 9 balls in an urn which violates the digits restriction, so subtract off the illegal counts.  If we place 10 balls in the first urn, then we have 7 remaining balls to distribute between the 8 urns or choose( 7+8-1, 7), If we place 1 ball in the first urn and 10 balls in one of the 7 other urns (7*), then there are choose( 6+8-1, 7 ) ways to distribute the remaining 6 balls in the 8 urns.  Not too complicated once you remember (or look up) the formula for urns and balls.  

-- 
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.snow at imail.org
801.408.8111

> -----Original Message-----
> From: baptiste auguie [mailto:baptiste.auguie at googlemail.com]
> Sent: Tuesday, January 12, 2010 12:20 PM
> To: Greg Snow
> Cc: r-help
> Subject: Re: [R] expand.grid game
> 
> Nice --- am I missing something or was this closed form solution not
> entirely trivial to find?
> 
> I ought to compile the various clever solutions given in this thread
> someday, it's fascinating!
> 
> Thanks,
> 
> baptiste
> 
> 2010/1/12 Greg Snow <Greg.Snow at imail.org>:
> > This also has a closed form solution:
> >
> >> choose(16+8-1,7) - choose(7+8-1, 7) - 7*choose(6+8-1,7)
> > [1] 229713
> >
> >
> > --
> > Gregory (Greg) L. Snow Ph.D.
> > Statistical Data Center
> > Intermountain Healthcare
> > greg.snow at imail.org
> > 801.408.8111
> >
> >
> >> -----Original Message-----
> >> From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-
> >> project.org] On Behalf Of Brian Diggs
> >> Sent: Thursday, December 31, 2009 3:08 PM
> >> To: baptiste auguie; David Winsemius
> >> Cc: r-help
> >> Subject: Re: [R] expand.grid game
> >>
> >> baptiste auguie wrote:
> >> > 2009/12/19 David Winsemius <dwinsemius at comcast.net>:
> >> >> On Dec 19, 2009, at 9:06 AM, baptiste auguie wrote:
> >> >>
> >> >>> Dear list,
> >> >>>
> >> >>> In a little numbers game, I've hit a performance snag and I'm
> not
> >> sure
> >> >>> how to code this in C.
> >> >>>
> >> >>> The game is the following: how many 8-digit numbers have the sum
> of
> >> >>> their digits equal to 17?
> >> >> And are you considering the number "00000089" to be in the
> >> acceptable set?
> >> >> Or is the range of possible numbers in 10000079:98000000 ?
> >> >>
> >> >
> >> > The latter, the first digit should not be 0. But if you have an
> >> > interesting solution for the other case, let me know anyway.
> >> >
> >> > I should also stress that this is only for entertainment and
> >> curiosity's sake.
> >> >
> >> > baptiste
> >> >
> >>
> >> I realize I'm late coming to this, but I was reading it in my post-
> >> vacation catch-up and it sounded interesting so I thought I'd give
> it a
> >> shot.
> >>
> >> After coding a couple of solutions that were exponential in time
> (for
> >> the number of digits), I rearranged things and came up with
> something
> >> that is linear in time (for the number of digits) and gives the
> count
> >> of numbers for all sums at once:
> >>
> >> library(plyr)
> >> nsum3 <- function(digits) {
> >>   digits <- as.integer(digits)[[1L]]
> >>   if (digits==1) {
> >>     rep(1,9)
> >>   } else {
> >>     dm1 <- nsum3(digits-1)
> >>     Reduce("+", llply(0:9, function(x) {c(rep(0,x),dm1,rep(0,9-
> x))}))
> >>   }
> >> }
> >>
> >> nsums <- llply(1:8, nsum3)
> >> nsums[[5]][17]
> >> # [1] 3675
> >> nsums[[8]][17]
> >> # [1] 229713
> >>
> >> The whole thing runs in well under a second on my machine (a several
> >> years old dual core Windows machine).  In the results of nsum3, the
> i-
> >> th element is the number of "numbers" whose digits sum to i.  The
> basic
> >> idea is recursion on the number of digits; if n_{t,d} is the number
> of
> >> d-digit "numbers" that sum to t, then n_{t,d} = \sum_{i\in(0,9)}
> n_{t-
> >> i,d-1}. (Adding the digit i to each of those "numbers" makes their
> sum
> >> t and increases the digits to d).  When digits==1, then 0 isn't a
> valid
> >> choice and that also implies the sum of digits can't be 0, which
> fits
> >> well with the 1 indexing of arrays.
> >>
> >> --
> >> Brian Diggs, Ph.D.
> >> Senior Research Associate, Department of Surgery, Oregon Health &
> >> Science University
> >>
> >> ______________________________________________
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> >