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POCombine two lists of indices to create a 1-1 mapping between two other lists
 

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  1. POCombine two lists of indices to create a 1-1 mapping between two other lists
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    <p>We'll start with the example. </p> <p>I have two lists:</p> <pre><code>segments = [16, 19, 22, 26] labels = [horse, cow, mule] </code></pre> <p>These two lists has a prefers (in order) some of the elements in the other list as shown in the tables below. Note that a segment can think that it should have a certain label, while the corresponding label doesn't think it should have that segment.</p> <pre><code> | Preferred Label | Preferred Segment | (in order, L -&gt; R) | (in order, L -&gt; R) ---+------------------- ------+------------------- 16 | horse, cow, mule horse | 26, 19, 22, 16 19 | horse, mule cow | 16, 22, 19 22 | mule, cow, horse mule | 26, 22, 19 26 | mule </code></pre> <p>which can be expressed in the form of two index lists:</p> <pre><code>// We have 4 segments, each have a ranked list (length 0-3) of preferred label labelIndicesForSegments = [[0, 1, 2], [0, 2], [2, 1, 0], [2]] // We have 3 labels, each have a ranked list (length 0-4) of preferred segment segmentIndicesForLabels = [[3, 1, 2, 0], [0, 2, 1], [3, 2, 1]] </code></pre> <p>Now, how do I create the "optimal 1 to 1 mapping" between segments and labels? I.e., the segments gets a label as much to the left as possible in its table, and vice versa. The algorithm should force pairs even if a segment and a label doesn't have each other in their preference list (this, however, should be avoided as far as possible).</p> <p>I do not know what would be the best answer to the example above. Some answers could be</p> <pre><code>1: [(16, horse), (19, mule), (22, cow)] 2: [(16, cow), (19, horse), (26, mule)] 3: [(19, horse), (22, cow), (26, mule)] </code></pre> <p>But how do I choose what segment will be without a matching label? And how do I compute which one is the most optimal? </p> <p>As you can see pretty much everything can vary. Segments and labels doesn't have to be of equal length and the list of indices doesn't have to be of equal length. While algorithm speed always matters I can say that in this case I won't be working with more than 10-20 elements in either segments or labels.</p> <p>My problem is that I have no entry point on how to solve this. Most likely there is an algorithm to solve this but I don't know it's name. I'm also implementing this in Java, in case you want to make a concrete non-pseudocode example ;)</p>
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      1. POCombine two lists of indices to create a 1-1 mapping between two other lists
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    1. COIt's not clear what you need to accomplish. Please post the expected answer for your sample with the explanation.
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    2. COI would create a table holding the sum of the preferences and then select for every row the element with higher preference sum available. Hope this help!
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    3. COIf there is a one to one correspondence between the segments and the labels, then can you explain how segmentIndicesForLabels first element would include the segment 3? Element 3 in labelIndicesForSegments only contains the index of 2. So, shouldn't element 2 in segmentIndicesForLabels contain only 2 and 1? Likewise, element 1 in segmentIndicesForLabels should not contain element 1 since 1 does not appear in labelIndicesForSegments for that element. If this is not right, then could you edit the question and explain the problem better.
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      1. POCombine two lists of indices to create a 1-1 mapping between two other lists
      1. USBob Bryan
 

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