In computer science, a predicate is called an invariant to a sequence of operations provided that: if the predicate is true before starting the sequence, then it is true at the end of the sequence.
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Although computer programs are typically mainly specified in terms of what they change, it's equally important to know or specify the invariants of a program. This is especially useful when reasoning about the program. The theory of optimizing compilers, the methodology of design by contract, and formal methods for determining program correctness, all pay close attention to invariants in computer programs.
Programmers often make use of assertions in their code to make invariants explicit. Some object oriented programming languages have a special syntax for specifying class invariants.
The MU puzzle is a good example of a logical problem where determining an invariant is useful. The puzzle is as follows:
Is it possible to convert MI into MU using these four transformation rules only?
One could spend many hours applying these transformation rules to strings. However, it might be quicker to find a predicate that's invariant to all rules, and makes getting to MU impossible. Logically looking at the puzzle, the only way to get rid of any I's is to have three consecutive I's in the string. This makes the following invariant interesting to consider:
This is an invariant to the problem if for each of the transformation rules the following holds: if the invariant held before applying the rule, it will also hold after applying it. If we look at the net effect of applying the rules on the number of I's and U's we can see this actually is the case for all rules:
| Rule | #I's | #U's | Effect on invariant |
|---|---|---|---|
| 1 | +0 | +1 | Number of I's is unchanged. If the invariant held, it still does. |
| 2 | ×2 | ×2 | If n is not a multiple of 3, then 2×n isn't either. The invariant still holds. |
| 3 | −3 | +1 | If n is not a multiple of 3, n−3 isn't either. The invariant still holds. |
| 4 | +0 | −2 | Number of I's is unchanged. If the invariant held, it still does. |
The table above shows clearly that the invariant holds for each of the possible transformation rules, which basically means that whichever rule we pick, at whatever state, if the number of I's was not a multiple of three before applying the rule, it won't be afterwards either.
Given that there is a single I in the starting string MI, and one is not a multiple of three, it's impossible to go from MI to MU as zero is a multiple of three.