What is the difference between bottom-up and top-down

Question

The bottom-up approach  to dynamic programming  consists in first looking at the  smaller  subproblems  and then solve the larger subproblems using the solution to the smaller problems   The top-down consists in solving the problem in a  natural manner  and check if you have calculated the solution to the subproblem before   I m a little confused  What is the difference between these two

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Simply saying top down approach uses recursion for calling Sub problems again and again  where as bottom up approach use the single without calling any one and hence it is more efficient

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Top down and bottom up DP are two different ways of solving the same problems   Consider a memoized  top down  vs dynamic  bottom up  programming solution to computing fibonacci numbers      fib cache       def memo fib n     global fib cache   if n    0 or n    1       return 1   if n in fib cache       return fib cache n    ret   memo fib n - 1    memo fib n - 2    fib cache n    ret   return ret  def dp fib n      partial answers    1  1     while len partial answers   lt   n       partial answers append partial answers -1    partial answers -2      return partial answers n   print memo fib 5   dp fib 5    I personally find memoization much more natural   You can take a recursive function and memoize it by a mechanical process  first lookup answer in cache and return it if possible  otherwise compute it recursively and then before returning  you save the calculation in the cache for future use   whereas doing bottom up dynamic programming requires you to encode an order in which solutions are calculated  such that no  big problem  is computed before the smaller problem that it depends on

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A key feature of dynamic programming is the presence of overlapping subproblems  That is  the problem that you are trying to solve can be broken into subproblems  and many of those subproblems share subsubproblems  It is like  Divide and conquer   but you end up doing the same thing many  many times  An example that I have used since 2003 when teaching or explaining these matters  you can compute Fibonacci numbers recursively   def fib n     if n  lt  2      return n   return fib n-1    fib n-2    Use your favorite language and try running it for fib 50   It will take a very  very long time  Roughly as much time as fib 50  itself  However  a lot of unnecessary work is being done  fib 50  will call fib 49  and fib 48   but then both of those will end up calling fib 47   even though the value is the same  In fact  fib 47  will be computed three times  by a direct call from fib 49   by a direct call from fib 48   and also by a direct call from another fib 48   the one that was spawned by the computation of fib 49     So you see  we have overlapping subproblems   Great news  there is no need to compute the same value many times  Once you compute it once  cache the result  and the next time use the cached value  This is the essence of dynamic programming  You can call it  top-down    memoization   or whatever else you want  This approach is very intuitive and very easy to implement  Just write a recursive solution first  test it on small tests  add memoization  caching of already computed values   and --- bingo  --- you are done   Usually you can also write an equivalent iterative program that works from the bottom up  without recursion  In this case this would be the more natural approach  loop from 1 to 50 computing all the Fibonacci numbers as you go   fib 0    0 fib 1    1 for i in range 48     fib i 2    fib i    fib i 1    In any interesting scenario the bottom-up solution is usually more difficult to understand  However  once you do understand it  usually you d get a much clearer big picture of how the algorithm works  In practice  when solving nontrivial problems  I recommend first writing the top-down approach and testing it on small examples  Then write the bottom-up solution and compare the two to make sure you are getting the same thing  Ideally  compare the two solutions automatically  Write a small routine that would generate lots of tests  ideally -- all small tests up to certain size --- and validate that both solutions give the same result  After that use the bottom-up solution in production  but keep the top-bottom code  commented out  This will make it easier for other developers to understand what it is that you are doing  bottom-up code can be quite incomprehensible  even you wrote it and even if you know exactly what you are doing   In many applications the bottom-up approach is slightly faster because of the overhead of recursive calls  Stack overflow can also be an issue in certain problems  and note that this can very much depend on the input data  In some cases you may not be able to write a test causing a stack overflow if you don t understand dynamic programming well enough  but some day this may still happen   Now  there are problems where the top-down approach is the only feasible solution because the problem space is so big that it is not possible to solve all subproblems  However  the  caching  still works in reasonable time because your input only needs a fraction of the subproblems to be solved --- but it is too tricky to explicitly define  which subproblems you need to solve  and hence to write a bottom-up solution  On the other hand  there are situations when you know you will need to solve all subproblems  In this case go on and use bottom-up    I would personally use top-bottom for Paragraph optimization a k a the Word wrap optimization problem  look up the Knuth-Plass line-breaking algorithms  at least TeX uses it  and some software by Adobe Systems uses a similar approach   I would use bottom-up for the Fast Fourier Transform

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Lets take fibonacci series as an example  1 1 2 3 5 8 13 21      first number  1 Second number  1 Third Number  2   Another way to put it    Bottom first  number  1 Top  Eighth  number on the given sequence  21   In case of first five fibonacci number   Bottom first  number  1 Top  fifth  number  5    Now lets take a look of recursive Fibonacci series algorithm as an example  public int rcursive int n        if   n    1      n    2             return 1        else           return rcursive n - 1    rcursive n - 2             Now if we execute this program with following commands  rcursive 5     if we closely look into the algorithm  in-order to generate fifth number it requires 3rd and 4th numbers  So my recursion actually start from top 5  and then goes all the way to bottom lower numbers  This approach is actually top-down approach    To avoid doing same calculation multiple times we use Dynamic Programming techniques  We store previously computed value and reuse it  This technique is called memoization  There are more to Dynamic programming other then memoization which is not needed to discuss current problem    Top-Down  Lets rewrite our original algorithm and add memoized techniques     public int memoized int n  int   memo        if  n  lt   2            return 1        else if  memo n     -1            return memo n         else           memo n    memoized n - 1  memo    memoized n - 2  memo             return memo n       And we execute this method like following     int n   5      int   memo   new int n   1       Arrays fill memo  -1       memoized n  memo     This solution is still top-down as algorithm start from top value and go to bottom each step to get our top value   Bottom-Up  But  question is  can we start from bottom  like from first fibonacci number then walk our way to up  Lets rewrite it using this techniques   public int dp int n        int   output   new int n   1       output 1    1      output 2    1      for  int i   3  i  lt   n  i              output i    output i - 1    output i - 2             return output n       Now if we look into this algorithm it actually start from lower values then go to top  If i need 5th fibonacci number i am actually calculating 1st  then second then third all the way to up 5th number  This techniques actually called bottom-up techniques    Last two  algorithms full-fill dynamic programming requirements  But one is top-down and another one is bottom-up  Both algorithm has similar space and time complexity

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Dynamic Programming is often called Memoization   1 Memoization is the top-down technique start solving the given problem by breaking it down  and dynamic programming is a bottom-up technique start solving from the trivial sub-problem  up towards the given problem   2 DP finds the solution by starting from the base case s  and works its way upwards  DP solves all the sub-problems  because it does it bottom-up     Unlike Memoization  which solves only the needed sub-problems    DP has the potential to transform exponential-time brute-force solutions into polynomial-time algorithms  DP may be much more efficient because its iterative      On the contrary  Memoization must pay for the  often significant  overhead due to recursion    To be more simple  Memoization uses the top-down approach to solve the problem i e  it begin with core main  problem then breaks it into sub-problems and solve these sub-problems similarly  In this approach same sub-problem can occur multiple times and consume more CPU cycle  hence increase the time complexity  Whereas in Dynamic programming same sub-problem will not be solved multiple times but the prior result will be used to optimize the solution

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Following is the DP based solution for Edit Distance problem which is top down  I hope it will also help in understanding the world of Dynamic Programming   public int minDistance String word1  String word2     Standard dynamic programming puzzle           int m   word2 length                int n   word1 length           if m    0     Cannot miss the corner cases                   return n          if n    0              return m          int     DP   new int n   1  m   1            for int j  1   j  lt   m  j                  DP 0  j    j                    for int i  1   i  lt   n  i                  DP i  0    i                     for int i  1   i  lt   n  i                  for int j  1   j  lt   m  j                      if word1 charAt i - 1     word2 charAt j - 1                       DP i  j    DP i-1  j-1                   else                 DP i  j    Math min Math min DP i-1  j   DP i  j-1    DP i-1  j-1     1     Main idea is this                                   return DP n  m       You can think of its recursive implementation at your home  It s quite good and challenging if you haven t solved something like this before

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rev4  A very eloquent comment by user Sammaron has noted that  perhaps  this answer previously confused top-down and bottom-up  While originally this answer  rev3  and other answers said that  bottom-up is memoization    assume the subproblems    it may be the inverse  that is   top-down  may be  assume the subproblems  and  bottom-up  may be  compose the subproblems    Previously  I have read on memoization being a different kind of dynamic programming as opposed to a subtype of dynamic programming  I was quoting that viewpoint despite not subscribing to it  I have rewritten this answer to be agnostic of the terminology until proper references can be found in the literature  I have also converted this answer to a community wiki  Please prefer academic sources  List of references   Web  1 2   Literature  5    Recap  Dynamic programming is all about ordering your computations in a way that avoids recalculating duplicate work  You have a main problem  the root of your tree of subproblems   and subproblems  subtrees   The subproblems typically repeat and overlap   For example  consider your favorite example of Fibonnaci  This is the full tree of subproblems  if we did a naive recursive call   TOP of the tree fib 4   fib 3                          fib 2    fib 2             fib 1        fib 1               fib 0     fib 1    fib 0    fib 1        fib 1               fib 0      fib 1    fib 0  BOTTOM of the tree    In some other rare problems  this tree could be infinite in some branches  representing non-termination  and thus the bottom of the tree may be infinitely large  Furthermore  in some problems you might not know what the full tree looks like ahead of time  Thus  you might need a strategy algorithm to decide which subproblems to reveal      Memoization  Tabulation  There are at least two main techniques of dynamic programming which are not mutually exclusive    Memoization - This is a laissez-faire approach  You assume that you have already computed all subproblems and that you have no idea what the optimal evaluation order is  Typically  you would perform a recursive call  or some iterative equivalent  from the root  and either hope you will get close to the optimal evaluation order  or obtain a proof that you will help you arrive at the optimal evaluation order  You would ensure that the recursive call never recomputes a subproblem because you cache the results  and thus duplicate sub-trees are not recomputed    example  If you are calculating the Fibonacci sequence fib 100   you would just call this  and it would call fib 100  fib 99  fib 98   which would call fib 99  fib 98  fib 97      etc     which would call fib 2  fib 1  fib 0  1 0 1  Then it would finally resolve fib 3  fib 2  fib 1   but it doesn t need to recalculate fib 2   because we cached it  This starts at the top of the tree and evaluates the subproblems from the leaves subtrees back up towards the root   Tabulation - You can also think of dynamic programming as a  table-filling  algorithm  though usually multidimensional  this  table  may have non-Euclidean geometry in very rare cases    This is like memoization but more active  and involves one additional step  You must pick  ahead of time  the exact order in which you will do your computations  This should not imply that the order must be static  but that you have much more flexibility than memoization    example  If you are performing fibonacci  you might choose to calculate the numbers in this order  fib 2  fib 3  fib 4     caching every value so you can compute the next ones more easily  You can also think of it as filling up a table  another form of caching   I personally do not hear the word  tabulation  a lot  but it s a very decent term  Some people consider this  dynamic programming   Before running the algorithm  the programmer considers the whole tree  then writes an algorithm to evaluate the subproblems in a particular order towards the root  generally filling in a table   footnote  Sometimes the  table  is not a rectangular table with grid-like connectivity  per se  Rather  it may have a more complicated structure  such as a tree  or a structure specific to the problem domain  e g  cities within flying distance on a map   or even a trellis diagram  which  while grid-like  does not have a up-down-left-right connectivity structure  etc  For example  user3290797 linked a dynamic programming example of finding the maximum independent set in a tree  which corresponds to filling in the blanks in a tree      At it s most general  in a  dynamic programming  paradigm  I would say the programmer considers the whole tree  then writes an algorithm that implements a strategy for evaluating subproblems which can optimize whatever properties you want  usually a combination of time-complexity and space-complexity   Your strategy must start somewhere  with some particular subproblem  and perhaps may adapt itself based on the results of those evaluations  In the general sense of  dynamic programming   you might try to cache these subproblems  and more generally  try avoid revisiting subproblems with a subtle distinction perhaps being the case of graphs in various data structures  Very often  these data structures are at their core like arrays or tables  Solutions to subproblems can be thrown away if we don t need them anymore     Previously  this answer made a statement about the top-down vs bottom-up terminology  there are clearly two main approaches called Memoization and Tabulation that may be in bijection with those terms  though not entirely   The general term most people use is still  Dynamic Programming  and some people say  Memoization  to refer to that particular subtype of  Dynamic Programming   This answer declines to say which is top-down and bottom-up until the community can find proper references in academic papers  Ultimately  it is important to understand the distinction rather than the terminology      Pros and cons  Ease of coding  Memoization is very easy to code  you can generally  write a  memoizer  annotation or wrapper function that automatically does it for you   and should be your first line of approach  The downside of tabulation is that you have to come up with an ordering     this is actually only easy if you are writing the function yourself  and or coding in an impure non-functional programming language    for example if someone already wrote a precompiled fib function  it necessarily makes recursive calls to itself  and you can t magically memoize the function without ensuring those recursive calls call your new memoized function  and not the original unmemoized function    Recursiveness  Note that both top-down and bottom-up can be implemented with recursion or iterative table-filling  though it may not be natural   Practical concerns  With memoization  if the tree is very deep  e g  fib 10 6    you will run out of stack space  because each delayed computation must be put on the stack  and you will have 10 6 of them   Optimality  Either approach may not be time-optimal if the order you happen  or try to  visit subproblems is not optimal  specifically if there is more than one way to calculate a subproblem  normally caching would resolve this  but it s theoretically possible that caching might not in some exotic cases   Memoization will usually add on your time-complexity to your space-complexity  e g  with tabulation you have more liberty to throw away calculations  like using tabulation with Fib lets you use O 1  space  but memoization with Fib uses O N  stack space    Advanced optimizations  If you are also doing a extremely complicated problems  you might have no choice but to do tabulation  or at least take a more active role in steering the memoization where you want it to go   Also if you are in a situation where optimization is absolutely critical and you must optimize  tabulation will allow you to do optimizations which memoization would not otherwise let you do in a sane way  In my humble opinion  in normal software engineering  neither of these two cases ever come up  so I would just use memoization   a function which caches its answers   unless something  such as stack space  makes tabulation necessary    though technically to avoid a stack blowout you can 1  increase the stack size limit in languages which allow it  or 2  eat a constant factor of extra work to virtualize your stack  ick   or 3  program in continuation-passing style  which in effect also virtualizes your stack  not sure the complexity of this  but basically you will effectively take the deferred call chain from the stack of size N and de-facto stick it in N successively nested thunk functions    though in some languages without tail-call optimization you may have to trampoline things to avoid a stack blowout      More complicated examples  Here we list examples of particular interest  that are not just general DP problems  but interestingly distinguish memoization and tabulation  For example  one formulation might be much easier than the other  or there may be an optimization which basically requires tabulation    the algorithm to calculate edit-distance 4   interesting as a non-trivial example of a two-dimensional table-filling algorithm

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Dynamic programming problems can be solved using either bottom-up or top-down approaches  Generally  the bottom-up approach uses the tabulation technique  while the top-down approach uses the recursion  with memorization  technique  But you can also have bottom-up and top-down approaches using recursion as shown below  Bottom-Up  Start with the base condition and pass the value calculated until now recursively  Generally  these are tail recursions  int n   5  fibBottomUp 1  1  2  n    private int fibBottomUp int i  int j  int count  int n        if  count  gt  n  return 1      if  count    n  return i   j      return fibBottomUp j  i   j  count   1  n      Top-Down  Start with the final condition and recursively get the result of its sub-problems  int n   5  fibTopDown n    private int fibTopDown int n        if  n  lt   1  return 1      return fibTopDown n - 1    fibTopDown n - 2

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