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    <p>Starting from point 1, the best solution up until a point j, must include the new end-point j in the last line segment, so the problem becomes where do I have to place the last split to minimize the cost of this last line-segment? </p> <p>Fortunately the cost is calculated in terms of subproblems of the same problem you are trying to solve, and fortunately you've already solved these smaller subproblems by the time you've moved to the next point j. So at the new point j you are trying to find an optimal point i, between points 1 and j, to split off a new line segment that includes j, and minimizes the cost: optimal_cost_up_to(i) + cost_of_split + cost_of_lsq_fit(i+1,j). Now the confusing part is that at any point it might seem like you are just finding a single split, but in reality all the previous splits are determined by optimal_cost_up_to(i), and since you've already calculated the optimal cost for all these points leading up to j, then you just need to memoize the answers so that the algorithm doesn't have to recalculate these costs each time it advances a point.</p> <p>In python I'd probably use a dictionary to store the results, although for this dynamic programming algorithm an array might be better...</p> <p>anyways...</p> <pre><code> def optimalSolution(points,split_cost) solutions = {0:{'cost':0,'splits':[]}} for j in range(1,len(points)): best_split = None best_cost = lsqFitCost(points,0,j) for i in range(0,j): cost = solutions[i]['cost'] + split_cost + lsqFitCost(points,i+1,j) if cost &lt; best_cost: best_cost = cost best_split = i if best_split != None: solution[j] = {'cost':best_cost,'splits':solution[best_split]['splits']+[best_split]} else: solution[j] = {'cost':best_cost,'splits':[]} return solutions </code></pre> <p>it's not pretty, and I haven't checked it (there might be an error involving the case where no split is the best cost), but hopefully it'll help get you on the right path? Note that lsqFitCost has to do a lot of work on each iteration, but for a least squares line fit like this you can make this cost a lot less by maintaining running sums used in the calculation... you should Google least squares line fitting for more info. This could make lsqFitCost constant so the overall time would be O(N^2).</p>
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    1. POSegmented least squares algorithm, don't understand this dynamic programming concept at all
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