a problem to solve

[mensanator at aol.com]

John Salerno wrote:
> mensanator at aol.com wrote:
>
> > a = [0xf5fdc,0xf6edb,0xbddb7,0x6fddd,0xeb7ed,0xb977f,0xbfed3,0xedef5]
> > b = [0xddb7d,0xfaddb,0xde75f,0xeef7a,0xdd77b,0xdfbce,0xb77dd,0x7ef5d]
> > c = [0xf37bd,0xdfaee,0xddd6f,0xddfb6,0xb9efb,0xb7bbe,0xecfbd,0xb75df]
> > d = [0x77edb,0xbb7ee,0xdf773,0x7bdeb,0x7ddaf,0xdeeeb,0xfb35f,0xbb7dd]
>
> > Once you see that a[0],b[0],c[0],d[0] is not a solution, try a
> > different set of switches. Repeat until you find a set that works.
> > There is, in fact a solution, but I won't reveal it unless you ask.
>
> Using your hex list, I combined the four correct switches with the &
> operator, and the result was 0. Shouldn't it be 1 for the correct answer?

No. First of all, combining them with the & operator would be
the asnswer to having all four lamps lit in the same position.
But you want exactly 3 (in any combination). The correct way
to combine the switches (using my answer of a[7] b[2] c[5] d[3])
is to use the boolean expression I gave you initially:

>>> y = ((c[5] & d[3]) & (a[7] ^ b[2])) | ((a[7] & b[2]) & (c[5] ^ d[3]))

>>> y
1048575

Note the answer is 2**20-1, not 1, the correct answer will have a 1
in EACH of the 20 bit positions (when y is printed in binary).

Is 1048575 the right answer?

>>> import gmpy
>>> print gmpy.digits(y,2)
11111111111111111111

Yes, it is the right answer.

If you calculate y for EVERY possible permutation of switches

a[0] b[0] c[0] d[0]
a[0] b[0] c[0] d[1]
a[0] b[0] c[0] d[2]
a[0] b[0] c[0] d[3]
a[0] b[0] c[0] d[4]
a[0] b[0] c[0] d[5]
a[0] b[0] c[0] d[6]
a[0] b[0] c[0] d[7]
a[0] b[0] c[1] d[0]
.
.
.
a[7] b[7] c[7] d[5]
a[7] b[7] c[7] d[6]
a[7] b[7] c[7] d[7]

you'll find that a[7] b[2] c[5] d[3] is the ONLY solution.