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As a programmer, you might be aware that Alfred’s algorithm can fail even when there is a set of coins with value exactly $c$ . For example, if Alfred has one coin worth \3$, one coin worth$$4$, and two coins worth$$5$, and the moose costs$$12$, then Alfred will add both of the$$5$coins to ​$S$and then give up, since adding any other coin would cause the value of the coins in ​$S$to exceed ​$$12$. Of course, Alfred could instead combine one$$3$coin, one$$4$coin, and one$$5$ coin to reach the total.