whenever you do this max here you have to keep track of for each value of XT which value of XT minus 1 is the one that was chosen as the max so you can work your way back along the chain along those pointers to find the full sequence so details are shown here but essentially very simple it's just like the filtering operations except that now we have for all X T we have to store the Arg max to remember the most compatible value for the proof from the previous time slice so when at the very end we're done at capital T we