Assuming that $$U$$ represents a Universal Turing Machine and $$\epsilon$$ is the empty string:

$$K_U(x) = \min_{|p|} \{p: U \circ p(\epsilon) = x\}$$

where $$x \neq x' \implies p \neq p'$$ so $$K_U(\cdot)$$ is injective.

## Proof that $$K_U(\cdot)$$ is not a computable search procedure:

If $$K_U(\cdot)$$ is computable, then we would be able to determine whether an arbitrary program $$g$$ halts on the empty string $$\epsilon$$, as we would be able to evaluate:

$$U \circ g(\epsilon)$$

Furthermore, given that $$g$$ is arbitrary and $$g(\epsilon)$$ represents an arbitrary string we would also be able to evaluate:

$$U \circ f \circ g(\epsilon)$$

for an arbitrary program $$f$$ that composes with $$g$$. It follows that we would be able to solve the halting problem. QED.