plus two or similar so plus do for the entire simple sequence so it's actually more efficient than doing Huffman each separate symbol so that's we've all read so far where some key assumptions it is assumed that we have a model P of X for which we can do the following track to be numerate all X for Huffman otherwise we can do it build that tree to enumerate everything well you want to enumerate all possible images you might want to possibly encode no you can't build a Huffman tree for that arithmetic coding gets around that by you only need