Subset sum problem: Difference between revisions
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main = nullset weights</lang> |
main = nullset weights</lang> |
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The name of the function that takes the weighted set is <code>nullset</code>. It manipulates a partial result represented as a list of pairs, each containing a subset of weighted words and the sum of their weights. Here is a rough translation: |
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* <code>=><></code> fold right combinator with the empty list as the vacuuous case |
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* <code>~&ng?r\~&r</code> If the partial result contains a zero sum, return it. |
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* <code>^TnK2hS\~&r</code> Concatenate the partial result with the new list of subsets (computed as follows) and delete duplicate sums. |
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* <code>^C/~&lmPlNCX</code> Cons a singleton subset containing the next word to the partial results. |
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* <code>*D</code> Distribute the next word in the set to the partial results and do the following to each. |
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* <code>sum@lmPrnPX</code> Add the weight of the new word to the existing sum. |
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* <code>~&lrmPC</code> Cons the new word to the list of existing ones. |
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* <code>~&nZFihmPB+</code> To conclude, search for a result with a zero sum, if any, and return its associated subset of weighted words. |
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output: |
output: |
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<pre> |
<pre> |
Revision as of 19:03, 1 January 2012
Implement a function/procedure/method/subroutine that takes a set/array/list/stream/table/collection of words with integer weights, and identifies a non-empty subset of them whose weights sum to zero (cf. the Dropbox Diet candidate screening exercise and the Subset sum problem Wikipedia article).
For example, for this set of weighted words, one solution would be the set of words {elysee, efferent, deploy, departure, centipede, bonnet, balm, archbishop}, because their respective weights of -326, 54, 44, 952, -658, 452, 397, and -915 sum to zero.
word | weight |
---|---|
alliance | -624 |
archbishop | -915 |
balm | 397 |
bonnet | 452 |
brute | 870 |
centipede | -658 |
cobol | 362 |
covariate | 590 |
departure | 952 |
deploy | 44 |
diophantine | 645 |
efferent | 54 |
elysee | -326 |
eradicate | 376 |
escritoire | 856 |
exorcism | -983 |
fiat | 170 |
filmy | -874 |
flatworm | 503 |
gestapo | 915 |
infra | -847 |
isis | -982 |
lindholm | 999 |
markham | 475 |
mincemeat | -880 |
moresby | 756 |
mycenae | 183 |
plugging | -266 |
smokescreen | 423 |
speakeasy | -745 |
vein | 813 |
Another solution would be the set of words {flatworm, gestapo, infra, isis, lindholm, plugging, smokescreen, speakeasy}, because their respective weights of 503, 915, -847, -982, 999, -266, 423, and -745 also sum to zero.
You may assume the weights range from -1000 to 1000. If there are multiple solutions, only one needs to be found. Use any algorithm you want and demonstrate it on a set of at least 30 weighted words with the results shown in a human readable form. Note that an implementation that depends on enumerating all possible subsets is likely to be infeasible.
Ursala
This solution scans the set sequentially while maintaining a record of all distinct sums obtainable by words encountered thus far, and stops when a zero sum is found. <lang Ursala>#import std
- import int
weights =
{
'alliance': -624, 'archbishop': -915, 'balm': 397, 'bonnet': 452, 'brute': 870, 'centipede': -658, 'cobol': 362, 'covariate': 590, 'departure': 952, 'deploy': 44, 'diophantine': 645, 'efferent': 54, 'elysee': -326, 'eradicate': 376, 'escritoire': 856, 'exorcism': -983, 'fiat': 170, 'filmy': -874, 'flatworm': 503, 'gestapo': 915, 'infra': -847, 'isis': -982, 'lindholm': 999, 'markham': 475, 'mincemeat': -880, 'moresby': 756, 'mycenae': 183, 'plugging': -266, 'smokescreen': 423, 'speakeasy': -745, 'vein': 813}
nullset = ~&nZFihmPB+ =><> ~&ng?r\~&r ^TnK2hS\~&r ^C/~&lmPlNCX *D ^A/sum@lmPrnPX ~&lrmPC
- cast %zm
main = nullset weights</lang>
The name of the function that takes the weighted set is nullset
. It manipulates a partial result represented as a list of pairs, each containing a subset of weighted words and the sum of their weights. Here is a rough translation:
=><>
fold right combinator with the empty list as the vacuuous case~&ng?r\~&r
If the partial result contains a zero sum, return it.^TnK2hS\~&r
Concatenate the partial result with the new list of subsets (computed as follows) and delete duplicate sums.^C/~&lmPlNCX
Cons a singleton subset containing the next word to the partial results.*D
Distribute the next word in the set to the partial results and do the following to each.sum@lmPrnPX
Add the weight of the new word to the existing sum.~&lrmPC
Cons the new word to the list of existing ones.~&nZFihmPB+
To conclude, search for a result with a zero sum, if any, and return its associated subset of weighted words.
output:
< 'flatworm': 503, 'gestapo': 915, 'infra': -847, 'isis': -982, 'lindholm': 999, 'plugging': -266, 'smokescreen': 423, 'speakeasy': -745>