Let $\mathbf{UNARY}-\mathbf{SSUM}$ be the subset sum problem in which all numbers are represented in unary. Why does the $\mathbf{NP}\mathbf{-}$ completeness proof for $\mathbf{SUBSET}\mathbf{-}\mathbf{SUM}$ fail to show $\mathbf{UNARY}\mathbf{-}\mathbf{SSUM}$ is $\mathbf{NP}\mathbf{-}$ complete? Show that $\mathbf{UNARY}\mathbf{-}\mathbf{SSUM}\mathbf{\in }\mathbf{P}$

Another subsets sum challenge $\mathrm{UNARY}-\mathrm{SSUM}$is one in which all integers are expressed in unary.

The amount of the input and the number of steps necessary for that input may be used to determine the function's difficulty.

• This length of something like the inputs generally equal to the value if somehow the input is in unary. The quantity of squares needed to represent the input is $\mathrm{x}$when the input value is $\mathrm{x}.$
• If the input is binary, the input value $x$ must be represented by $\log \mathrm{x}$ squares. As a result, if the binary input size is $N$ , the unary input size is $2^N$ .
• This same lowering proof takes more often than exponential time reducing $3\mathrm{SAT}\mathrm{to}\mathrm{UNARY}-\mathrm{SSUM}$, implying that perhaps the volume of both the input is expanding exponentially.

Since this simplification isn't really exponential time, the Nanoparticles demonstration for $\mathrm{SUBSET}-\mathrm{SUM}\mathrm{fails}\mathrm{to}\mathrm{establish}\mathrm{UNARY}-\mathrm{SSUM}\mathrm{is}\mathrm{NP}-\mathrm{complete}$. In unary, the input size is $2^{\mathrm{N}}$, which requires exponential time.

• As a result, the $\mathrm{NP}-$completeness argument of $\mathrm{SUBSET}-\mathrm{SUM}$fails to demonstrate that $\mathrm{SUBSET}-\mathrm{SUM}$is also $\mathrm{NP}-$complete.