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MathGroup Archive 2002

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RE: Re: Re: Re: An interesting math problem

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36385] RE: [mg36349] Re: [mg36185] Re: [mg36125] Re: An interesting math problem
  • From: "DrBob" <drbob at bigfoot.com>
  • Date: Wed, 4 Sep 2002 02:56:42 -0400 (EDT)
  • Reply-to: <drbob at bigfoot.com>
  • Sender: owner-wri-mathgroup at wolfram.com

By taking sums and differences of the counterweights 1, 3, ... 3^(k-1)
you can precisely express all integers up to the integer whose base 3
notation is composed of k ones.

By doubling those counterweights, you can precisely express all the EVEN
integers up to TWICE that limit, or -1 + 3^k.  If you know that unknowns
are limited to the integers from 1 to 3^k, this allows you to precisely
weight the even numbers and bracket the odd numbers.  Hence
counterweights 2, 6, ... 2*3^(k-1) are sufficient for unknowns up to
3^k.

Notice that this is an efficient coding scheme:  Each counterweight is
multiplied by -1, 0, or 1, so there can be at most 3^k distinct
counterweight sums.  More than half are negative or zero, however; the
positive sums and differences number at most (3^k - 1)/2.  Since we need
to enumerate (and CAN enumerate with even weights) only the even numbers
and because we get the maximum unknown 3^k by elimination, we "cover" up
to twice as many as we can enumerate, plus one.  Twice (3^k - 1)/2 plus
one is 3^k, so we're getting the maximum coverage possible for a given
number of counterweights.

Bobby Treat

-----Original Message-----
From: Julio Vera [mailto:jvera at adinet.com.uy] 
To: mathgroup at smc.vnet.net
Subject: [mg36385] [mg36349] Re: [mg36185] Re: [mg36125] Re: An interesting math
problem


I liked this very much.

For n=4 there are 3^n=81 possible combinations, but only 40 positive
ones -
the rest would be 0 and each of the positive ones multiplied by -1. As
Bob
stated, this is obtained by:

(3^n-1)/2=40

For these 40 positive integers to match 1 to 40 correspondingly, the
biggest
should be 40 (also, the smallest should be 1). So we have:

a+b+c+d=40 or 1+b+c+d=40 (we state that a<b<c<d)

This could be a particular case (for a=3 and n=4) of:

In[1]:= Sum[a^i , {i, 0, n - 1}] == (a^n-1)/(a-1)
Out[1]= True

With 5 counterweights, we would be able to determine the weight of 121
elements

In[]:= n = 5; a = 3; (a^n - 1)/(a - 1)
Out[]= 121

And the 5 counterweights would be the 5 elements of the corresponding
sum.

In[]:=
Table[a^h, {h, 0, n - 1}]

Out[]=
{1, 3, 9, 27, 81}

I did not understand what Bryan said about 2,4,10,28 being a solution.

At first, I thought b could be either 2 or 3. b, the second of the 40
positive combinations, could be

b=2 or b-a=2, meaining b=3

But, if b=2, we would have 2 of the 40 positive combinations that would
equal
1:

a=1 and b-a=1

This means we would not be able to get all 40 values with the 40
positive
combinations. so b must be 3.

Maybe the same reasoning would lead to 9 and 27 being also unique.
{1,3,9,27} could be the most "efficient" solution (of a+b+c+d=40), in
the
sense that it would not generate any repetition. And the most efficient
solution would be the only solution, since the repetitions are not
acceptable.

I did not understand, either, what Bob said about 34 being the largest
possible
weight.

I am very curious about these two issues (a second solution, and 34), so
I
would very much appreciate if you could elaborate on them.

Thanks very much,

Julio
----- Original Message -----
From: <BobHanlon at aol.com>
To: mathgroup at smc.vnet.net
Subject: [mg36385] [mg36349] [mg36185] Re: [mg36125] Re: An interesting math
problem


>
> In a message dated 8/22/02 6:23:12 AM, timreh719 at yahoo.com.tw writes:
>
> >I'm sorry for that my question is not clear,I have correct below.
> >
> >timreh719 at yahoo.com.tw (bryan) wrote in message
> news:<ajvp7h$ibk$1 at smc.vnet.net>...
> >> Hi All:
> >>    I have a very interesting math problem:If I have a scales,and I
> >> have 40 things that their mass range from 1~40 which each is a
nature
> >> number,and now I can only make 4 counterweights to measure out each
> >> mass of those things.Question:What mass should the counterweights
> >> be???
> >> The answer is that 1,3,9,27   and I wnat to use mathematica to
solve
> >> this problem.
> >>     In fact,I think that this physical problem has various
> >> answer,ex.2,4,10,28
> >> this way also work,because if I have a thing which weight 3 , and I
> >> can measure out by comparing 2<3<4 . But,If I want to solve this
math
> >> problem:
> >> {x|x=k1*a+k2*b+k3*c+k4*d}={1,2,3,4,,,,,,40} where a,b,c,d is nature
> numbers.
> >> and {k1,k2,k3,k4}={1,0,-1}
> >> How to solve it ??
> >> Thank you very much in advance and hope mail to me your method and
> >> mathematica solving method.  appreciate any idea sharing
> >>                                                        sincerely
> >> bryan
> >
>
> Just use brute force.
>
> Needs["DiscreteMath`Combinatorica`"];
>
> var = {a, b, c, d}; n = Length[var];
>
> s = Outer[Times, var, {-1, 0, 1} ];
>
> f = Flatten[Outer[Plus, Sequence@@s]];
>
> Since the length of f is just 3^n then the range of numbers
> to be covered should be {-(3^n-1)/2, (3^n-1)/2}.
> Consequently, the largest of the weights can not exceed
> (3^n-1)/2 - (1+2+...+(n-1)) or
>
> ((3^n-1) - n(n-1))/2
>
> 34
>
> Thread[var->#]& /@
>
>   (First /@ Select[{var,f} /. Thread[var->#]& /@
>
>           KSubsets[Range[((3^n-1) - n(n-1))/2], n],
>
>         Sort[#[[2]]] == Range[-(3^n-1)/2,(3^n-1)/2]&])
>
> {{a -> 1, b -> 3, c -> 9, d -> 27}}
>
>
> Bob Hanlon
> Chantilly, VA  USA
>







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