9 weights problem

Originally posted by iChopWoodForFree
Each step I will choose the outcome that gives the highest number of possibilities.

*1.abcd < efgh

Possibilities:

ae af ag ah ai be bf bg bh bi ce cf cg ch ci de df dg dh di ei fi gi hi

We know that if any of the odd weights is abc or d then that is the light weight and if it is def or g then that is the heavy weight. This is why there a ...[text shortened]... apology.

Since you were so certain it could be done I found myself compelled to look deeper.

what about abcd=efgh

Originally posted by AThousandYoung
what about abcd=efgh

I covered that at the bottom of my post. Ecxept I mistakenly wrote abcd=defg... Woops LOL.

abcd=efgh gives us 24 possibilities but only 12 pairs and it eliminates weight 'i' which helps cut down on the possibilities in subsequent weighings.

1.abcd=efgh

ab ac ad bc bd cd ef eg eh fg fg gh


These are the 12 pairs but in each pair either weight can be light or heavy giving us 24 possibilities.

2.abgh=cdef

ab gh cd ef

Now four pairs but still eight possibilities.

3.ag<ei (if = then cd is the only remaining pair. If > then the solution given below is the opposite.)

ab gh ef

We know that a,g and f cannot be heavy while b,h and e cannot be light.

4.ae=bf then g is light h is heavy
4.ae<bf then a is light b is heavy
4.ae>bf then f is light e is heavy

Looks like everyone has given up on this.
I first saw it yesterday and I think it is solvable but might take some time.

I may be misunderstanding this problem but

x,x,x,x weighed against x,x,x+e,x-e with x out would be worst start.
Shift left one and carry last as new out.

x,x,x,x weighed against x,x+e,x-e,x with x out.
Shift left one and carry last as new out.

x,x,x,x weighed against x+e,x-e,x,x with x out.
Shift left one and carry last as new out.

x,x,x,x+e weighed against x-e,x,x,x with x out.

Should tell? Is this equivalent to your method ichopwoodforfree?

Originally posted by JerryH
I may be misunderstanding this problem but

x,x,x,x weighed against x,x,x+e,x-e with x out would be worst start.
Shift left one and carry last as new out.

x,x,x,x weighed against x,x+e,x-e,x with x out.
Shift left one and carry last as new out.

x,x,x,x weighed against x+e,x-e,x,x with x out.
Shift left one and carry last as new out.

x,x,x,x+e ...[text shortened]... ainst x-e,x,x,x with x out.

Should tell? Is this equivalent to your method ichopwoodforfree?

That could work!

Originally posted by AThousandYoung
That could work!

Now I'm playing it like a game 🙂

I've solved a few unbalanced starts in four or less but then my weigher ran away 🙁

Anyone up for a Balancing? I'll weigh or play.

The weigher picks where x+e and x-e start in a,b,c,d ^ e,f,g,h _ i and replaces the ^ with <, > or =. Then the player shifts a,b,c,d > e,f,g,h _ i as many places as would like, right or left, but keeps the order. As an example a shift four right: f,g,h,i ^ a,b,c,d _ e. Then the weigher replaces the ^ again: f,g,h,i > a,b,c,d _ e.

So far I have the shift four to the light side rule when starting unbalanced, that hasn't failed in three whole games🙂 Also I'm just blocking out where x+e and x-e could be on each move like: a,b,c,d > e,f,g,h _ i ---> x+e,x+e,x+e,x+e > x,x,x,x _ x-e or x+e,x+e,x+e,x+e > x-e,x-e,x-e,x-e _ x or x,x,x,x > x-e,x-e,x-e,x-e _ x+e. Then the next weighing narrows the possibilities: f,g,h,i > a,b,c,d _ e. ---> Since [f,g,h,i] is greater than [a,b,c,d], i must be x+e and e must be x-e. Solved in two weighings 🙂