Balance puzzle


A balance puzzle or weighing puzzle is a logic puzzle about balancing items—often coins—to determine which holds a different value, by using balance scales a limited number of times. These differ from puzzles that assign weights to items, in that only the relative mass of these items is relevant.
KnownGoalMaximum Coins for weighingsNumber of Weighings for coins
Whether target coin is lighter or heavier than othersIdentify coin
Target coin is different from othersIdentify coin
Target coin is different from others, or all coins are the sameIdentify if unique coin exists, and whether it is lighter or heavier

For example, in detecting a dissimilar coin in three weighings, the maximum number of coins that can be analyzed is = 13. Note that with 3 weighs and 13 coins, it is not always possible to determine the identity of the last coin, but merely that the coin is different. In general, with n weighs, you can determine the identity of a coin if you have - 1 or less coins. In the case n = 3, you can truly discover the identity of the different coin out of 12 coins.

Nine-coin problem

A well-known example has up to nine items, say coins, that are identical in weight except one, which is lighter than the others—a counterfeit. The difference is perceptible only by weighing them on scale—but only the coins themselves can be weighed. How can one isolate the counterfeit coin with only two weighings?

Solution

To find a solution, we first consider the maximum number of items from which one can find the lighter one in just one weighing. The maximum number possible is three. To find the lighter one, we can compare any two coins, leaving the third out. If the two coins weigh the same, then the lighter coin must be one of those not on the balance. Otherwise, it is the one indicated as lighter by the balance.
Now, imagine the nine coins in three stacks of three coins each. In one move we can find which of the three stacks is lighter. It then takes only one more move to identify the light coin from within that lighter stack. So in two weighings, we can find a single light coin from a set of.
By extension, it would take only three weighings to find the odd light coin among 27 coins, and four weighings to find it from 81 coins.

Twelve-coin problem

A more complex version has twelve coins, eleven or twelve of which are identical. If one is different, we don't know whether it is heavier or lighter than the others. This time the balance may be used three times to determine if there is a unique coin—and if there is, to isolate it and determine its weight relative to the others. The problem has a simpler variant with three coins in two weighings, and a more complex variant with 39 coins in four weighings.

Solution

This problem has more than one solution. One is easily scalable to a higher number of coins by using base-three numbering: labeling each coin with a different number of three digits in base three, and positioning at the n-th weighings all the coins that are labeled with the n-th digit identical to the label of the plate. Other step-by-step procedures are similar to the following. It is less straightforward for this problem, and the second and third weighings depend on what has happened previously, although that need not be the case.
Given a population of 13 coins in which it is known that 1 of the 13 is different from the rest, it is simple to determine which coin it is with a balance and 3 tests as follows:

With a reference coin

If there is one authentic coin for reference then the suspect coins can be thirteen. Number the coins from 1 to 13 and the authentic coin number 0 and perform these weighings in any order:
If the scales are only off balance once, then it must be one of the coins 1, 2, 3—which only appear in one weighing.
If there is never balance then it must be one of the coins 10–13 that appear in all weighings. Picking out the one counterfeit coin corresponding to each of the 27 outcomes is always possible except when all weighings are balanced, in which case there is no counterfeit coin. If coins 0 and 13 are deleted from these weighings they give one generic solution to the 12-coin problem.
If two coins are counterfeit, this procedure, in general, does not pick either of these, but rather some authentic coin. For instance, if both coins 1 and 2 are counterfeit, either coin 4 or 5 is wrongly picked.

Without a reference coin

In a relaxed variation of this puzzle, one only needs to find the counterfeit coin without necessarily being able to tell its weight relative to the others. In this case, clearly any solution that previously weighed every coin at some point can be adapted to handle one extra coin. This coin is never put on the scales, but if all weighings are balanced it is picked as the counterfeit coin. It is not possible to do any better, since any coin that is put on the scales at some point and picked as the counterfeit coin can then always be assigned weight relative to the others.
A method which weighs the same sets of coins regardless of outcomes lets one either
  1. conclude if they all weigh the same, or find the odd coin and tell if it is lighter or heavier, or
  2. find the odd coin, and, for 12 of them, tell if it is lighter or heavier.
The three possible outcomes of each weighing can be denoted by "\" for the left side being lighter, "/" for the right side being lighter, and "-" for both sides having the same weight. The symbols for the weighings are listed in sequence. For example, "//-" means that the right side is lighter in the first and second weighings, and both sides weigh the same in the third weighing. Three weighings give the following 33 = 27 outcomes. Except for "---", the sets are divided such that each set on the right has a "/" where the set on the left has a "\", and vice versa:

/// \\\
\// /\\
/\/ \/\
//\ \\/
\/- /\-
-\/ -/\
/-\ \-/
\\- //-
-\\ -//
\-\ /-/
/-- \--
-/- -\-
--/ --\
---

As each weighing gives a meaningful result only when the number of coins on the left side is equal to the number on the right side, we disregard the first row, so that each column has the same number of "\" and "/" symbols. The rows are labelled, the order of the coins being irrelevant:

\// A light /\\ A heavy
/\/ B light \/\ B heavy
//\ C light \\/ C heavy
\/- D light /\- D heavy
-\/ E light -/\ E heavy
/-\ F light \-/ F heavy
\\- G light //- G heavy
-\\ H light -// H heavy
\-\ I light /-/ I heavy
/-- J light \-- J heavy
-/- K light -\- K heavy
--/ L light --\ L heavy
--- M either lighter or heavier,
 or all coins weigh the same

Using the pattern of outcomes above, the composition of coins for each weighing can be determined; for example the set "\/- D light" implies that coin D must be on the left side in the first weighing, on the right side in the second, and unused in the third:

1st weighing: left side: ADGI, right side: BCFJ
2nd weighing: left side: BEGH, right side: ACDK
3rd weighing: left side: CFHI, right side: ABEL

The outcomes are then read off the table. For example, if the right side is lighter in the first two weighings and both sides weigh the same in the third, the corresponding code "//- G heavy" implies that coin G is the odd one, and it is heavier than the others.

Generalizations

The generalization of this problem is described in Chudnov.
Let be the -dimensional Euclidean space, be the inner product of vectors
and from For vectors and subsets the operations and are defined, respectively, as ;,, By we shall denote the discrete -cube in ; i.e., the set of all sequences of length over the alphabet The set is the discrete ball of radius with centre at the point Relative weights of objects are given by a vector which defines the configurations of weights of the objects: the th object has standard weight if the weight of the th object is greater by a constant value if . The vector characterizes the types of objects: the standard type, the non-standard type, and it does not contain information about relative weights of non-standard objects.
A weighing is given by a vector the result of a weighing for a situation is The weighing given by a vector has the following interpretation: for a given check the th object participates in the weighing if ; it is put on the left balance pan if and is put on the right pan if For each weighing, both pans should contain the same number of objects: if on some pan the number of objects is smaller than as it should be, then it receives some reference objects. The result of a weighing describes the following cases: the balance if, the left pan outweighs the right one if, and the right pan outweighs the left one if The incompleteness of initial information about the distribution of weights of a group of objects is characterized by the set of admissible distributions of weights of objects which is also called the set of admissible situations, the elements of are called admissible situations.
Each weighing induces the partition of the set by the plane into three parts , and defines the corresponding partition of the set where
Definition 1. A weighing algorithm of length is a sequence where is the function determining the check at each th step, of the algorithm from the results of weighings at the previous steps.
Let be the set of all -syndromes and be the set of situations with the same syndrome ; i.e., ;
Definition 2. A WA is said to:
a) identify the situations in a set if the condition is satisfied for all
b) identify the types of objects in a set if the condition is satisfied for all
It is proved in that for so-called suitable sets an algorithm of identification the types identifies also the situations in
As an example the perfect dynamic algorithms with parameters are constructed in which correspond to the parameters of the perfect ternary Golay code. At the same time, it is established that a static WA with the same parameters does not exist.
Each of these algorithms using 5 weighings finds among 11 coins up to two counterfeit coins which could be heavier or lighter than real coins by the same value. In this case the uncertainty domain contains situations, i.e. the constructed WA lies on the Hamming bound for and in this sense is perfect.
To date it is not known whether there are other perfect WA that identify the situations in for some values of. Moreover, it is not known whether for some there exist solutions for the equation
\sum_ ^t 2^iC_n^i = 3^m
which is, obviously, necessary for the existence of a perfect WA. It is only known that for there are no perfect WA, and for this equation has the unique nontrivial solution which determines the parameters of the constructed perfect WA.

Original parallel weighings puzzle

Konstantin Knop invented this puzzle. There are N indistinguishable coins, one of which is fake. There are two balance scales that can be used in parallel. Each weighing lasts one minute. What is the largest number of coins N for which it is possible to find the fake coin in five minutes?

In literature