Determining the inconsistently weighted object

• Difficulty Level : Medium
• Last Updated : 17 Nov, 2021

Given N objects numbered from 1 to N out of which all are of the same weights except only one object which is not known beforehand. We are also given Q comparisons, in each of which an equal number of objects are placed on both sides of a balance scale, and we are told the heavier side.
The task is to find the inconsistently weighted object or determine if the data is not sufficient enough.
Examples

Input : N = 6
Q = 3
1 2 = 5 6
1 2 3 > 4 5 6
3 4 < 5 6
Output : 4
Explanation: Object 4 is lighter than all other objects.

Input : N = 10
Q = 4
1 2 3 4 < 7 8 9 10
1 = 9
2 3 4 > 1 5 10
6 = 2
Output : Insufficient data

It is told that except only one element, the rest of the elements are of the same weights. So, if we observe carefully, it can be said that:

1. In a ‘=’ comparison, none of the objects on both sides is the inconsistently weighted one.

2. If an object appears on the heavier side in one comparison and on the lighter side in another, then it is not the inconsistently weighted object. This is because, if an object appears on the heavier side then it is of the maximum weight and if it appears on the lighter side then it is of the minimum weight. Since a single element can’t be both maximum and minimum at the same time. So, this case will never occur.

3. The inconsistently weighted object must appear in all of the non-balanced (‘>’ or ‘<‘) comparisons.

We can use the above three observations to narrow down the potential candidates for the inconsistently weighted object. We will consider only those objects which are either on the heavier side or the lighter side; if there is only one such object then it is the required one. If there is no such object, then we will consider all those objects which do not appear in any comparison. If there is only one such object then it is the inconsistently weighted object. If none of these scenarios arises, the data is insufficient.
Below is the implementation of the above approach:

Python3

 # Python program to determine the# inconsistently weighted object # Function to get the difference of two listsdef subt(A, B):    return list(set(A) - set(B)) # Function to get the intersection of two listsdef intersection(A, B):    return list(set(A).intersection(set(B))) # Function to get the intersection of two listsdef union(A, B):    return list(set(A).union(set(B))) # Function to find the inconsistently weighted objectdef inconsistentlyWeightedObject(N, Q, comparisons):    # Objects which appear on the heavier side    heavierObj = [i for i in range(1, N + 1)]         # Objects which appear on the lighter side    lighterObj = [i for i in range(1, N + 1)]    equalObj = [] # Objects which appear in '=' comparisons         # Objects which don't appear in any comparison    objectNotCompared = [i for i in range(1, N + 1)]         for c in comparisons:        objectNotCompared = subt(objectNotCompared, union(c, c))                 if c == '=':            equalObj = union(equalObj, union(c, c))        elif c == '<':            # Removing those objects which do            # not appear on the lighter side            lighterObj = intersection(lighterObj, c)                         # Removing those objects which do            # not appear on the heavier side            heavierObj = intersection(heavierObj, c)        else:            # Removing those objects which do            # not appear on the lighter side            lighterObj = intersection(lighterObj, c)                         # Removing those objects which do            # not appear on the heavier side            heavierObj = intersection(heavierObj, c)         L_iwo = subt(union(heavierObj, lighterObj), equalObj) # Potential candidates     if len(L_iwo) == 1:        return L_iwo    elif not len(L_iwo):        if len(objectNotCompared) == 1:            return objectNotCompared        else:            return 'Insufficient data'    else:        return 'Insufficient data'  # Driver codeN = 6Q = 3comparisons = [ [[1, 2], '=', [5, 6]], [[1, 2, 3], '>', [4, 5, 6]],                                        [[3, 4], '<', [5, 6]] ]print(inconsistentlyWeightedObject(N, Q, comparisons))
Output:
4

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