Re: Re: An interesting math problem
- To: mathgroup at smc.vnet.net
- Subject: [mg36185] Re: [mg36125] Re: An interesting math problem
- From: BobHanlon at aol.com
- Date: Mon, 26 Aug 2002 04:16:04 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
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