Well, it is implied that since you can’t remove 1 coin prom either P1 or P2 in (2), you simply weigh P1,P2!

If however we want bloody N=3 included to the general strategu, here is an alternative for case (4):

Proceed as in (2) w/o removing coins from P1,P2. If P1,P2 don’t balance, add the coin from P3 to P1 (which can then be devided exactly), and proceed.

Alkis

]]>Since I have not seen a clear cut and complete solution to this coin-weighing problem and since I liked it a lot, here is my contribution (quite late, but good things are durable!)

It can be done for all N>2 and there are more than one methods, depending on N. Here, I tried to forulate a general strategy, that I believe it’s optimum for k-redeem (if I undestood it well):

Divide N by 2. You get 3 parts: P1,P2 containing N1 coins, equal to the quotient, and P3 containing 0/1 coins, equal to the reminder.

1) N2=0 and N1 is even: Weigh P1,P2 against each other. Divide the part that is heavier (say P1) into 2 parts and weigh them against each other: 1a) If they balance (i.e all coins in P1 are good), P2 contains a *lighter* coin. 1b) If they don’t balance, P1 contains a *heavier* coin.

2) N2=0 and N1 is odd: Remove 1 coin from both P1 and P2, setting N1 to even and creating P3 with N2=2. Weigh P1,P2: 2a) If they balance, P3 contains the fake coin. Weigh two coins from P1 (good) against P3. If P1 is heavier, P3 contains a *lighter* coin, otherwise P3 contains a *heavier* coin. 2b) If they don’t balance, proceed as in (1) (ignoring N2).

3) N2=1 and N1 is even: Proceed as in (2) w/o having to remove coins from P1,P2.

4) N2=1 and N1 is odd: Proceed as in (2), with P3 containing now N2=3 coins.

Voila! I hope you enjoyed it!

Alkis

]]>For all N=2k+1, place a coin in each tray. The fake coin sits in the only tray, where the two trays opposite of it are level. If the fake coin is at the lowest elevation of all coins, then it is heavier; otherwise it will be at the highest elevation, and thus lighter.

For all N=2k (k!=1), place a coin in each tray. The fake coin is either the coin highest in elevation or lowest. Switch the coin lowest in elevation with one of the two coins beside the one highest in elevation. If the scale does not shift in orientation, the coin highest is the fake (and lighter than the others). If the scale does shift, then the coin lowest is fake (and heavier).

Does not work for N=[0,2]. May require instrumentation more precise than the human eye.

]]>It is clear that N/3 (more precisely Ceiling(N/3)) is possible. Given 3k coins, divide into 3 groups of k.

First weighing: compare groups A and B.

Second weighing: compare groups B and C.

Divide the N coins equally so that we have two bundles of N/2 coins each.

Assume that the fake coin is lighter.

Now take the two bundles on the balance and take the bundle which is heavier(as according to our assumption this bundle will contain all the real coins).

Now take this particular bundle and divide it as N/4 coins equally and weigh it on a balance.

If they are equal, our assumption was correct otherwise the fake coin is heavier.

]]>Your solution doesn’t work for several small N.

]]>As in the classical problem, if W weighings (of given sets of coins) provide enough information solve the problem, there are at most (3^W – 3)/2 possible outcomes of those weighings, which together with the factor of S ambiguity, has to be enough information to fully determine the false coin. In other words, the maximum number of coins that can be handled with W weighings is at most S*(3^W – 3)*(1/2) coins. For example, in 2 weighings we can reduce 15 coins down to one of 5 possibilities and determine light/heavy status of the false coin. The information bound may be attainable in all cases by following the classical (S=1) weighing schedule until ambiguity is reduced to S or fewer coins, but I have not tried to check whether this is really true.

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