laimancode's blog

fun proof problem
By laimancode, 10 years ago, In English

hi everyone! my first post

Problem: a wooden rod is separated into pieces of length L1, L2, ..., Ln. Cutting a rod of length L into two pieces costs L. But we want to cut the rod into n-1 pieces. There is an optimal order of cuts that produces the minimum cost. (can be calculated by dynamic programming)

Is the minimum cost always smallest when the pieces are in order from smallest to greatest? I think it is, but do not know how to prove.

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10 years ago, # |
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No. If we can rearrange pieces arbitrary (and I read from your last question that it is), than the optimal configuration can be reached with Huffman tree. Let's solve problem in reverse: we have pieces with length L1 .. Ln and we want to bring them together in any order paying La + Lb for connecting segments with lengths La and Lb.

It can be proved that the optimal algorithm is the following: select two minimal pieces (two arbitrary ones if many minimums exist) and connect them together. Furthermore, this is the only optimal algorithm.

The following example shows that it is not always possible to connect rods and leave them sorted.

3 3 4 5 7
(3 3) 4 5 7
(3 3) (4 5) 7

after the second step we should connect (3 3) and 7, but they are separated by 4 and 5 and the order is broken.

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    10 years ago, # ^ |
    ← Rev. 6   Vote: I like it 0 Vote: I do not like it

    what if u combine the connected pieces to get the sum, and sort at each stage? then it will look like this.

    • 3 7 4 3 5, sort and get 3 3 4 5 7
    • 6 9 7, sort and get 6 7 9
    • 13 9, sort and get 9 13
    • 22, single piece so not possible to combine further

    and answer will be sum of numbers at each stage, (6+9) + (13) + (22) = 50.

    EDIT: complexity analysis below.

    • first step:
    • second step:
    • third step:
    • ith step:
    • last step:

    overall:

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      10 years ago, # ^ |
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      Yes, you can sort them at each stage, but after this operation original pieces won't be sorted. 6 7 9 is 3 3 7 4 5 in original order, and, obviously, this is not a sorted array :)

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      10 years ago, # ^ |
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      Are you saying that after the sort, you combine element x with element x+1 (for even x)? If so, I think this is wrong.

      For the input {1, 2, 100, 101}, your algorithm would give 408, while the correct answer is 313.

      Your algorithm's pairings: ((1 2) (100 101))

      Correct pairings: (((1 2) 100) 101)

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        10 years ago, # ^ |
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        No, he's not saying that. UTFG on Huffman code for more.

        He's saying that after the sort, we combine elements 1 and 2. No more than that. You're just misunderstanding ifsmirnov's example notation — braces denote already merged pieces; he just didn't write the pieces in sorted order. For your example, it's

        1 2 100 101
(1 2) 100 101
101 ((1 2) 100)
(101 ((1 2) 100)

        for his, it's 

        3 3 4 5 7
4 5 (3 3) 7
(3 3) 7 (4 5)
(4 5) ((3 3) 7)
((4 5) ((3 3) 7))
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          10 years ago, # ^ |
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          I've read about Huffman codes, no need to be rude.

          Did you read the edit in JuanMata's comment ("complexity analysis below")? If you sort all segments in each step, complexity is O(N^2 log N). However, he said that complexity is O(N log N). Furthermore, he says that the second step takes O(N/2 log (N/2)) time, indicating that each step in his algorithm reduces N to N/2. Your interpretation would reduce N to N-1, which doesn't seem to be what he intended.

          Obviously this can be sped up by using a priority queue, but that's not really what's being commented on here. I was simply correcting JuanMata's algorithm.

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            10 years ago, # ^ |
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            I wasn't being rude (at least not from the point of view I'm used to), just direct :D

            Oh. Sorry then. I assumed that the replies were transitive in this case, and it kind of made sense for me there.

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              10 years ago, # ^ |
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              OK, no offense taken, I may have misinterpreted your response.

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            10 years ago, # ^ |
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            Xellos tried to be very confusing below, so now i think its my turn! :D

            i wrongly interpreted ifsmirnov's correct approach, and ended up giving a wrong solution. then aquamongoose correctly interpreted my wrong interpretation of ifsmirnov, and corrected my solution. later Xellos wrongly interpreted aquamongoose (thinking that i correctly interpreted ifsmirnov and that aquamongoose wrongly interpreted me), but ended up giving the same solution as both of us.

            is it too confusing? if not, is it atleast more confusing than Xellos's comment? :D

            finally after all the confusion due to wrong interpretations (not to mention comments like this! :P), luckily we have all come to a consensus about the final solution, which i have posted below.

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10 years ago, # |
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i misinterpreted ifsmirnov's approach, and so my previous comment is wrong (as shown by aquamongoose above).

the correct solution (what ifsmirnov really meant) is to find the minimum two numbers at each stage, and combine only these two into their sum.
for the same example as above, the array at each stage would look like:

  • step 0: 3 3 4 5 7 (input)
  • step 1: 6 4 5 7
  • step 2: 6 9 7
  • step 3: 13 9
  • step 4: 22

therefore minimum answer is 6+9+13+22 = 50.
complexity:
EDIT: complexity can be bettered to using Xellos's suggestion to use a map or priority_queue to store the values.

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    10 years ago, # ^ |
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    As shown by my comment which is right, aquamongoose's comment which disproved ifsmirnov's comment is wrong, so ifsmirnov's comment and you comment which responded to that one are right. And this comment is right, but not because it's according to aquamongoose's comment, but because it just says the same as everyone. Is it confusing enough, or should I try more? :D

    Also to the complexity: Use the set<>, Luke! What you're doing is removing 2 elements from the set and adding their sum. There's no need for O(N2) here.

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      10 years ago, # ^ |
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      Set is bad, set doesn't allow equal elements to be in it. Use multiset or priority_queue, which suits perfectly to such kind of tasks.

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        10 years ago, # ^ |
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        Details. That was not my point.

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      10 years ago, # ^ |
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      try to confuse more :D

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