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d'alar'cop
5:05 PM
@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.).
1 hour later…
mdc32
6:28 PM
@d'alar'cop someone found the solution with 8... i don't know if you saw it or not
d'alar'cop
yeah I saw it
Joel.. he's always good
d'alar'cop
6:40 PM
@mdc32 what do you think about it?
mdc32
7:16 PM
@d'alar'cop I think it works
I'm not too happy about it but there's nothing we can do
d'alar'cop
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
mdc32
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
d'alar'cop
yes it would.. but there's surely a lot of symmety and so forth
mdc32
yeah like his 4b and 4c are nearly identical
d'alar'cop
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..
mdc32
7:30 PM
should we try it?
d'alar'cop
yes
but I can't do it right now :p
mdc32
same here, i have to do homework
d'alar'cop
ok, well we probably have at least some days..
Joel is probably working on it
mdc32
he definitely is
d'alar'cop
yes, well there's surely a pattern emerging between the 2 solutions..
luckily he used the same kind of notation..
mdc32
7:36 PM
yeah which is useful
d'alar'cop
yes it all reads like one solution really
mdc32
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
d'alar'cop
good idea...
do you feel like it should be possible?
mdc32
yes
we just need a pattern between the 4 in 3 and the 6 in 4
d'alar'cop
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
I am pretty sure it is :) —
Lembik
21 mins ago
mdc32
7:49 PM
yeah so i think 8 in 5 is possible
then 10 in 6 should be possible, because 12 in 7 is
d'alar'cop
yep
let's think about it as finding the heavy ones
that might help... and it's sort of how Joel framed his answer
mdc32
i never thought of it like that
d'alar'cop
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
mdc32
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
d'alar'cop
yep
d'alar'cop
8:05 PM
ok I'll see you later mate
all the best :)
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Coin Weighing with a Single Device
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