IDENTIFYING ALL FAKE COINS :
We can number the coins $1...36$ , then choose 4 out of these. That is $36\times35\times34\times33$ Possibilities, but the order is not important, hence we have to Divide by $1\times2\times3\times4$, which gives us:
$$3\times35\times17\times33$$
These 4 Coins can be either heavier or lighter , hence 2 Cases.
Hence Total Possibilities : $2\times3\times35\times17\times33$
Each Weighting give 3 Possibilities : left heavy or right heavy or both Equal.
When we have $N$ Weightings , we have $3^N$ Possibilities.
$$3^N \approx 2\times3\times35\times17\times33 $$
We get N = 11 which is the theoretical Necessity.
$3^{11} = 177147$
$2\times3\times35\times17\times33 = 117810$
IDENTIFYING WHETHER FAKE COINS ARE HEAVIER OR LIGHTER :
We can get this in fewer Weightings because we want to extract lesser information.
Number the Coins $1...36$
STARTING STAGE :
Compare 1 with 2.
If Equal, Compare 1+2 with 3+4.
If Equal, Compare 1+2+3+4 with 5+6+7+8.
If Equal, we have 8 Coins which are genuine.
ENDING STAGE :
Compare these 8 with the other 8. We will get heavier (Answer) or lighter (Answer) or Equal (unknown).
When unknown : We can then put these 16 against other 16 to get the Answer.
If still Equal, then the last 4 are the fake Coins. We can Check with 4 genuine Coins to get the Answer.
We have used 6 Weightings in Worst Case.
If we had InEqual Weightings in STARTING STAGE itself, then we have a set of Coins (having maybe 2 or 4 or 8 Coins) where at least 1 is a fake Coin.
The ENDING STAGE will change in this Situation.
We may have the Set (with X = 2 or 4 or 8 Coins) containing at least 1 fake Coin.
Take X=2 Case :
Compare the lighter Coin with 3 , 4 , 5 , 6 , 7 , 8 , 9 : If all Equal , then the heavier Coin was the fake.
If lighter Coin is always lighter , then that is the fake Coin.
Other wise , we can make a Pile with heavier Coins & a Pile with lighter Coins. Which ever Pile has more than 4 Coins is genuine.
We have used 8 Weightings in Worst Case.
Take X=4 Case :
Take X=8 Case :
Hence We can get the Answer in 6 Weightings in Worst Case.
While trying to update that , I got a new Solution which , I figure , is easier :
Compare 1 & 2 : If InEqual , we put these into heavier Pile & lighter Pile.
Compare lighter Coin with 3 & put in Correct Pile.
Compare lighter Coin with 4 & put in Correct Pile.
Compare lighter Coin with 5 & put in Correct Pile.
(X) If all Equal till now , we have 5 genuine Coins. Compare with 5 others to get the Answer. If Equal , these are also genuine , hence put these together. Compare with 10 other Coins. If Equal , these are also genuine , hence put these together to get 20 genuine Coins. Compare with 16 of these with the other 16 to get the Answer.
Worst Case : 7 Weightings.
(Y) If not all Equal till now , at this Point , we will have a Pile with 1..4 heavier Coins & a Pile with 4..1 lighter Coins , with 5 Coins in total.
Compare Some lighter Coin with 6 , 7 , 8 , 9. Put each Coin into the Correct Pile. We will have a Pile with at least 5 Coins, with the other Pile having at most 4 Coins.
The larger Pile is genuine , the other is fake.
Worst Case : 8 Weightings.