Answer
See explanation
Work Step by Step
Below is a straightforward algorithmic approach to verify **injectivity** (one‐to‐one property) of a function \(f\) whose domain and codomain are both finite. We assume:
- The domain is a finite set \(X = \{x_1, x_2, \dots, x_m\}\).
- The codomain is some finite set \(Y\) (we do not necessarily need to list out all elements of \(Y\), only to store images \(f(x)\)).
- We have access to a “black box” procedure \(\mathrm{Compute}(f,x)\) that returns \(f(x)\) for any \(x \in X\).
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## Idea of the Algorithm
A function \(f\) is injective if **no two distinct inputs** share the same output. Equivalently, as you iterate through the domain, you must never see the same function value more than once.
### Pseudocode
```
Algorithm IsOneToOne(f, X):
1. Create an empty set S to store images of f.
2. For each element x in X (the domain):
a. y ← Compute(f, x) // Get the image of x under f
b. If y is in S:
return "f is NOT one-to-one" // Found a duplicate image
Else:
Add y to S
3. // If we finish the loop without finding duplicates:
return "f IS one-to-one"
```
### Explanation
1. **Initialize a set \(S\)** to keep track of values already encountered.
2. **Loop** over each element \(x\) in the domain \(X\).
3. **Compute** \(f(x)\).
4. **Check** whether \(f(x)\) has already appeared in \(S\).
- If yes, we have found two different inputs \(x_1\neq x_2\) with \(f(x_1)=f(x_2)\), so \(f\) is not injective.
- If no, insert \(f(x)\) into \(S\) and move on.
5. **If** we finish processing every \(x\) without finding any duplicate images, then \(f\) is injective.
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## Complexity
- You make **one pass** through all elements of the domain.
- Each pass involves:
1. **Computing** \(f(x)\) once, which we assume is handled by an existing procedure.
2. **Checking** membership in \(S\) and inserting if not found.
- If you use an efficient data structure (like a hash set) for \(S\), membership check and insertion are effectively \(O(1)\) on average, making the overall algorithm \(O(m)\), where \(m = |X|\).
- If you used a simpler data structure (like a list), membership check is \(O(m)\) each time, and overall worst‐case time could be \(O(m^2)\).
Either way, for a finite domain, this procedure is finite and guarantees to detect or refute injectivity.
