@mdc32 and it's true that there is no solution using that system. the assumption that the weighing patterns (i.e. which coins you choose to weigh) is dynamic (i.e. changes depending on previous results.).

@d'alar'cop someone found the solution with 8... i don't know if you saw it or not

yeah I saw it

Joel.. he's always good

@mdc32 what do you think about it?

@d'alar'cop I think it works

I'm not too happy about it but there's nothing we can do

well the last ditch is to solve the 12 coins..

to show that there is a 7 solution or that 8 in the best

@mdc32 so we learned that the weighing need to be dynamic

that makes brute-forcing very impractical

yeah, but even still i feel like it might be too big

i mean for 6 in 4 he has 4a 4b 4c and 4d all with 4 outcomes

i think 12 in 7 might have like 64 or even 128 outcomes which would be a lot to keep track of

yes it would.. but there's surely a lot of symmety and so forth

yeah like his 4b and 4c are nearly identical

yes, and as in all the other times... many cases are discounted... like when things go well..

we can probably extrapolate a lot from the current one for 6..

should we try it?

yes

but I can't do it right now :p

same here, i have to do homework

ok, well we probably have at least some days..

Joel is probably working on it

he definitely is

yes, well there's surely a pattern emerging between the 2 solutions..

luckily he used the same kind of notation..

yeah which is useful

yes it all reads like one solution really

well if it keeps going as it has (4 in 3, 6 in 4) then we could get 8 in 5, or 10 in 6, or eventually 12 in 7

i might try 8 in 5 to see if theres a pattern

good idea...

do you feel like it should be possible?

yes

we just need a pattern between the 4 in 3 and the 6 in 4

the author implies that 7 is possible

the problem is that 4 in 3 is special..

because it's not dynamic

that's why the code I was using worked..

for 4 in 3

yeah so i think 8 in 5 is possible

then 10 in 6 should be possible, because 12 in 7 is

yep

let's think about it as finding the heavy ones

that might help... and it's sort of how Joel framed his answer

i never thought of it like that

call the heavy ones odd... and we want to find the odd ones

can we use 0 and 1? 10 and 20 irritates me :p

yeah 0 and 1 is fine

it doesn't matter if theres two tens adding up to 20 because you know how many you're weighing

yep

ok I'll see you later mate

all the best :)