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  1. PO
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    <pre><code>import heapq import itertools import operator def increasing(fn, left, right): """ Given two never decreasing iterators produce another iterator resulting from passing the value from left and right to fn. This iterator should also be never decreasing. """ # Imagine an infinite 2D-grid. # Each column corresponds to an entry from right # Each row corresponds to an entry from left # Each cell correspond to apply fn to those two values # If the number of columns were finite, then we could easily solve # this problem by keeping track of our current position in each column # in each iteration, we'd take the smallest value report it, and then # move down in that column. This works because the values must increase # as we move down the column. That means the current set of values # under consideration must include the lowest value not yet reported # To extend this to infinite columns, at any point we always track a finite # number of columns. The last column current tracked is always in the top row # if it moves down from the top row, we add a new column which starts at the top row # because the values are increasing as we move to the right, we know that # this last column is always lower then any columns that come after it # Due to infinities, we need to keep track of all # items we've ever seen. So we put them in this list # The list contains the first part of the incoming iterators that # we have explored left_items = [next(left)] right_items = [next(right)] # we use a heap data structure, it allows us to efficiently # find the lowest of all value under consideration heap = [] def add_value(left_index, right_index): """ Add the value result from combining the indexed attributes from the two iterators. Assumes that the values have already been copied into the lists """ value = fn( left_items[left_index], right_items[right_index] ) # the value on the heap has the index and value. # since the value is first, low values will be "first" on the heap heapq.heappush( heap, (value, left_index, right_index) ) # we know that every other value must be larger then # this one. add_value(0,0) # I assume the incoming iterators are infinite while True: # fetch the lowest of all values under consideration value, left_index, right_index = heapq.heappop(heap) # produce it yield value # add moving down the column if left_index + 1 == len(left_items): left_items.append(next(left)) add_value(left_index+1, right_index) # if this was the first row in this column, add another column if left_index == 0: right_items.append( next(right) ) add_value(0, right_index+1) def fib(): a = 1 b = 1 while True: yield a a,b = b,a+b r = increasing(operator.add, fib(), itertools.count() ) for x in range(100): print next(r) </code></pre>
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    1. PO"sorted 1-d iterator" based on "2-d iterator" (Cartesian product of iterators)
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    1. PTAnswer
    1. USClaudiu
    1. USWinston Ewert
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    1. This table or related slice is empty.
    1. This table or related slice is empty.
    1. PO"sorted 1-d iterator" based on "2-d iterator" (Cartesian product of iterators)
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    1. VO
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      1. This table or related slice is empty.
      1. VTUpMod
    2. VO
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      1. This table or related slice is empty.
      1. VTUpMod
    3. VO
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    1. COThis is the same as my answer, only with code, so it's better.
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      1. USRyan Thompson
    2. COThis doesn't always work because you don't know that when you generate the next column the first value will be >= the previous column's first value. e.g. the first grid of outputs can look like this: `[[1,666,5],[666,666,666],[666,666,666],[666,666,666]]`. you only know `op(a,b)` > `op(a,c)` if `b>c`, you don't know that `op(g,i)`>`op(h,i)` if `g>h`. all you know is `op(a,b)>=a` and `op(a,b)>=b`, or `min(a,b)<=op(a,b)`, which is what I was getting at in my answer. so you have to generate columns until the leftmost input into the op is >= the value you want to yield.
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      1. USClaudiu
    3. COit is a great solution if that were true, but until then it is misleading.. unless you return some value of `N` like the OP suggested - i think that's probably the best solution (yours + returning `N`)
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      1. USClaudiu
 

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