Gold nugget storage









up vote
6
down vote

favorite
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Given a positive integer, write it as the sum of numbers, where each of them is in $kin1,9,81,tin1,2,3,...,64$. How many numbers at least are used? Shortest code win.



Samples:



Input Output Method
1 1 1
2 1 2
64 1 64
65 2 64+1
72 1 72
343 2 342+1
576 1 576
577 2 576+1
5184 1 5184
46656 9 5184+5184+5184+5184+5184+5184+5184+5184+5184
5274 2 5184+90


enter image description here










share|improve this question



















  • 4




    Could you give slightly more detail/put into words how the input results in the output?
    – Quintec
    Nov 11 at 16:45






  • 2




    If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
    – trichoplax
    Nov 11 at 16:52






  • 4




    Is it basically a sort of change-making problem ? With the coins denominations being in the set 1,9,81 × 1...64 ?
    – digEmAll
    Nov 11 at 17:01







  • 2




    Are you trying to minimize the number of items or number of stacks?
    – fəˈnɛtɪk
    Nov 11 at 17:02






  • 3




    I suggest adding 5274 = 64*81 + 10*9 as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get 64*81 + 1*81 + 9*1 (as mine did)
    – Black Owl Kai
    Nov 11 at 21:03















up vote
6
down vote

favorite
1












Given a positive integer, write it as the sum of numbers, where each of them is in $kin1,9,81,tin1,2,3,...,64$. How many numbers at least are used? Shortest code win.



Samples:



Input Output Method
1 1 1
2 1 2
64 1 64
65 2 64+1
72 1 72
343 2 342+1
576 1 576
577 2 576+1
5184 1 5184
46656 9 5184+5184+5184+5184+5184+5184+5184+5184+5184
5274 2 5184+90


enter image description here










share|improve this question



















  • 4




    Could you give slightly more detail/put into words how the input results in the output?
    – Quintec
    Nov 11 at 16:45






  • 2




    If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
    – trichoplax
    Nov 11 at 16:52






  • 4




    Is it basically a sort of change-making problem ? With the coins denominations being in the set 1,9,81 × 1...64 ?
    – digEmAll
    Nov 11 at 17:01







  • 2




    Are you trying to minimize the number of items or number of stacks?
    – fəˈnɛtɪk
    Nov 11 at 17:02






  • 3




    I suggest adding 5274 = 64*81 + 10*9 as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get 64*81 + 1*81 + 9*1 (as mine did)
    – Black Owl Kai
    Nov 11 at 21:03













up vote
6
down vote

favorite
1









up vote
6
down vote

favorite
1






1





Given a positive integer, write it as the sum of numbers, where each of them is in $kin1,9,81,tin1,2,3,...,64$. How many numbers at least are used? Shortest code win.



Samples:



Input Output Method
1 1 1
2 1 2
64 1 64
65 2 64+1
72 1 72
343 2 342+1
576 1 576
577 2 576+1
5184 1 5184
46656 9 5184+5184+5184+5184+5184+5184+5184+5184+5184
5274 2 5184+90


enter image description here










share|improve this question















Given a positive integer, write it as the sum of numbers, where each of them is in $kin1,9,81,tin1,2,3,...,64$. How many numbers at least are used? Shortest code win.



Samples:



Input Output Method
1 1 1
2 1 2
64 1 64
65 2 64+1
72 1 72
343 2 342+1
576 1 576
577 2 576+1
5184 1 5184
46656 9 5184+5184+5184+5184+5184+5184+5184+5184+5184
5274 2 5184+90


enter image description here







code-golf






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 11 at 23:40

























asked Nov 11 at 16:34









l4m2

4,6081634




4,6081634







  • 4




    Could you give slightly more detail/put into words how the input results in the output?
    – Quintec
    Nov 11 at 16:45






  • 2




    If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
    – trichoplax
    Nov 11 at 16:52






  • 4




    Is it basically a sort of change-making problem ? With the coins denominations being in the set 1,9,81 × 1...64 ?
    – digEmAll
    Nov 11 at 17:01







  • 2




    Are you trying to minimize the number of items or number of stacks?
    – fəˈnɛtɪk
    Nov 11 at 17:02






  • 3




    I suggest adding 5274 = 64*81 + 10*9 as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get 64*81 + 1*81 + 9*1 (as mine did)
    – Black Owl Kai
    Nov 11 at 21:03













  • 4




    Could you give slightly more detail/put into words how the input results in the output?
    – Quintec
    Nov 11 at 16:45






  • 2




    If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
    – trichoplax
    Nov 11 at 16:52






  • 4




    Is it basically a sort of change-making problem ? With the coins denominations being in the set 1,9,81 × 1...64 ?
    – digEmAll
    Nov 11 at 17:01







  • 2




    Are you trying to minimize the number of items or number of stacks?
    – fəˈnɛtɪk
    Nov 11 at 17:02






  • 3




    I suggest adding 5274 = 64*81 + 10*9 as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get 64*81 + 1*81 + 9*1 (as mine did)
    – Black Owl Kai
    Nov 11 at 21:03








4




4




Could you give slightly more detail/put into words how the input results in the output?
– Quintec
Nov 11 at 16:45




Could you give slightly more detail/put into words how the input results in the output?
– Quintec
Nov 11 at 16:45




2




2




If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
– trichoplax
Nov 11 at 16:52




If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
– trichoplax
Nov 11 at 16:52




4




4




Is it basically a sort of change-making problem ? With the coins denominations being in the set 1,9,81 × 1...64 ?
– digEmAll
Nov 11 at 17:01





Is it basically a sort of change-making problem ? With the coins denominations being in the set 1,9,81 × 1...64 ?
– digEmAll
Nov 11 at 17:01





2




2




Are you trying to minimize the number of items or number of stacks?
– fəˈnɛtɪk
Nov 11 at 17:02




Are you trying to minimize the number of items or number of stacks?
– fəˈnɛtɪk
Nov 11 at 17:02




3




3




I suggest adding 5274 = 64*81 + 10*9 as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get 64*81 + 1*81 + 9*1 (as mine did)
– Black Owl Kai
Nov 11 at 21:03





I suggest adding 5274 = 64*81 + 10*9 as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get 64*81 + 1*81 + 9*1 (as mine did)
– Black Owl Kai
Nov 11 at 21:03











5 Answers
5






active

oldest

votes

















up vote
3
down vote














Jelly, 17 bytes



64R×9;Ɗ⁺ff€¥@ŒṗẈṂ


Try it online!



-1 thanks to Jonathan Allan.



Explanation (you can't test for inputs larger than 58 over TIO):



64R×9;Ɗ⁺ff€¥@ŒṗẈṂ Arguments: x
64R [1..64]
×9;Ɗ Multiply by 9, prepend to original list
⁺ Do the above once more
Œṗ Positive integer partitions of x
¥@ Call with reversed arguments (x = partitions, y = flattened outer product)
f€ For each partition in x, keep the elements that are in y
f Keep the elements of x that have remained intact after the above
Ẉ Lengths of the remaining partitions
Ṃ Minimum





share|improve this answer






















  • Since the output in testable area is trivial, can't quite check?
    – l4m2
    Nov 11 at 23:35










  • @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
    – Erik the Outgolfer
    Nov 12 at 8:37

















up vote
2
down vote














Perl 6, 47 bytes





+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)


Try it online!



A greedy algorithm seems to work.






share|improve this answer



























    up vote
    2
    down vote













    JavaScript (ES6), 72 66 57 56 bytes



    Saved 1 byte thanks to @nwellnhof





    f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n>719?81:72):n-5184)


    Try it online!






    share|improve this answer






















    • @nwellnhof This would fail for several values (576, 632, 633, ...)
      – Arnauld
      Nov 11 at 21:57










    • I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
      – nwellnhof
      Nov 11 at 22:59










    • @nwellnhof Yes, it does. :)
      – Arnauld
      Nov 11 at 23:37

















    up vote
    2
    down vote














    05AB1E, 30 27 bytes



    ŽK≠‰`91vDy64*›i1sy9*%]64/îO


    Try it online! or verify all test cases



    -3 bytes thanks to Kevin Cruijssen



    This is my first 05AB1E submission, so I am sure that this can be optimized.






    share|improve this answer


















    • 1




      Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
      – Kevin Cruijssen
      Nov 12 at 7:31











    • @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
      – Black Owl Kai
      Nov 12 at 7:56










    • Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
      – Kevin Cruijssen
      Nov 12 at 8:00

















    up vote
    0
    down vote














    Wolfram Language (Mathematica), 66 bytes



    Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&


    Try it online!






    share|improve this answer




















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      5 Answers
      5






      active

      oldest

      votes








      5 Answers
      5






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      3
      down vote














      Jelly, 17 bytes



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ


      Try it online!



      -1 thanks to Jonathan Allan.



      Explanation (you can't test for inputs larger than 58 over TIO):



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ Arguments: x
      64R [1..64]
      ×9;Ɗ Multiply by 9, prepend to original list
      ⁺ Do the above once more
      Œṗ Positive integer partitions of x
      ¥@ Call with reversed arguments (x = partitions, y = flattened outer product)
      f€ For each partition in x, keep the elements that are in y
      f Keep the elements of x that have remained intact after the above
      Ẉ Lengths of the remaining partitions
      Ṃ Minimum





      share|improve this answer






















      • Since the output in testable area is trivial, can't quite check?
        – l4m2
        Nov 11 at 23:35










      • @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
        – Erik the Outgolfer
        Nov 12 at 8:37














      up vote
      3
      down vote














      Jelly, 17 bytes



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ


      Try it online!



      -1 thanks to Jonathan Allan.



      Explanation (you can't test for inputs larger than 58 over TIO):



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ Arguments: x
      64R [1..64]
      ×9;Ɗ Multiply by 9, prepend to original list
      ⁺ Do the above once more
      Œṗ Positive integer partitions of x
      ¥@ Call with reversed arguments (x = partitions, y = flattened outer product)
      f€ For each partition in x, keep the elements that are in y
      f Keep the elements of x that have remained intact after the above
      Ẉ Lengths of the remaining partitions
      Ṃ Minimum





      share|improve this answer






















      • Since the output in testable area is trivial, can't quite check?
        – l4m2
        Nov 11 at 23:35










      • @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
        – Erik the Outgolfer
        Nov 12 at 8:37












      up vote
      3
      down vote










      up vote
      3
      down vote










      Jelly, 17 bytes



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ


      Try it online!



      -1 thanks to Jonathan Allan.



      Explanation (you can't test for inputs larger than 58 over TIO):



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ Arguments: x
      64R [1..64]
      ×9;Ɗ Multiply by 9, prepend to original list
      ⁺ Do the above once more
      Œṗ Positive integer partitions of x
      ¥@ Call with reversed arguments (x = partitions, y = flattened outer product)
      f€ For each partition in x, keep the elements that are in y
      f Keep the elements of x that have remained intact after the above
      Ẉ Lengths of the remaining partitions
      Ṃ Minimum





      share|improve this answer















      Jelly, 17 bytes



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ


      Try it online!



      -1 thanks to Jonathan Allan.



      Explanation (you can't test for inputs larger than 58 over TIO):



      64R×9;Ɗ⁺ff€¥@ŒṗẈṂ Arguments: x
      64R [1..64]
      ×9;Ɗ Multiply by 9, prepend to original list
      ⁺ Do the above once more
      Œṗ Positive integer partitions of x
      ¥@ Call with reversed arguments (x = partitions, y = flattened outer product)
      f€ For each partition in x, keep the elements that are in y
      f Keep the elements of x that have remained intact after the above
      Ẉ Lengths of the remaining partitions
      Ṃ Minimum






      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited Nov 11 at 20:15

























      answered Nov 11 at 18:34









      Erik the Outgolfer

      31.1k429102




      31.1k429102











      • Since the output in testable area is trivial, can't quite check?
        – l4m2
        Nov 11 at 23:35










      • @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
        – Erik the Outgolfer
        Nov 12 at 8:37
















      • Since the output in testable area is trivial, can't quite check?
        – l4m2
        Nov 11 at 23:35










      • @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
        – Erik the Outgolfer
        Nov 12 at 8:37















      Since the output in testable area is trivial, can't quite check?
      – l4m2
      Nov 11 at 23:35




      Since the output in testable area is trivial, can't quite check?
      – l4m2
      Nov 11 at 23:35












      @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
      – Erik the Outgolfer
      Nov 12 at 8:37




      @l4m2 You can't do so over TIO, but you might be able to do so if you install Jelly locally. That's why I've added the explanation.
      – Erik the Outgolfer
      Nov 12 at 8:37










      up vote
      2
      down vote














      Perl 6, 47 bytes





      +($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)


      Try it online!



      A greedy algorithm seems to work.






      share|improve this answer
























        up vote
        2
        down vote














        Perl 6, 47 bytes





        +($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)


        Try it online!



        A greedy algorithm seems to work.






        share|improve this answer






















          up vote
          2
          down vote










          up vote
          2
          down vote










          Perl 6, 47 bytes





          +($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)


          Try it online!



          A greedy algorithm seems to work.






          share|improve this answer













          Perl 6, 47 bytes





          +($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)


          Try it online!



          A greedy algorithm seems to work.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 11 at 23:36









          nwellnhof

          6,4431125




          6,4431125




















              up vote
              2
              down vote













              JavaScript (ES6), 72 66 57 56 bytes



              Saved 1 byte thanks to @nwellnhof





              f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n>719?81:72):n-5184)


              Try it online!






              share|improve this answer






















              • @nwellnhof This would fail for several values (576, 632, 633, ...)
                – Arnauld
                Nov 11 at 21:57










              • I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
                – nwellnhof
                Nov 11 at 22:59










              • @nwellnhof Yes, it does. :)
                – Arnauld
                Nov 11 at 23:37














              up vote
              2
              down vote













              JavaScript (ES6), 72 66 57 56 bytes



              Saved 1 byte thanks to @nwellnhof





              f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n>719?81:72):n-5184)


              Try it online!






              share|improve this answer






















              • @nwellnhof This would fail for several values (576, 632, 633, ...)
                – Arnauld
                Nov 11 at 21:57










              • I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
                – nwellnhof
                Nov 11 at 22:59










              • @nwellnhof Yes, it does. :)
                – Arnauld
                Nov 11 at 23:37












              up vote
              2
              down vote










              up vote
              2
              down vote









              JavaScript (ES6), 72 66 57 56 bytes



              Saved 1 byte thanks to @nwellnhof





              f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n>719?81:72):n-5184)


              Try it online!






              share|improve this answer














              JavaScript (ES6), 72 66 57 56 bytes



              Saved 1 byte thanks to @nwellnhof





              f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n>719?81:72):n-5184)


              Try it online!







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Nov 11 at 23:37

























              answered Nov 11 at 18:04









              Arnauld

              71.5k688299




              71.5k688299











              • @nwellnhof This would fail for several values (576, 632, 633, ...)
                – Arnauld
                Nov 11 at 21:57










              • I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
                – nwellnhof
                Nov 11 at 22:59










              • @nwellnhof Yes, it does. :)
                – Arnauld
                Nov 11 at 23:37
















              • @nwellnhof This would fail for several values (576, 632, 633, ...)
                – Arnauld
                Nov 11 at 21:57










              • I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
                – nwellnhof
                Nov 11 at 22:59










              • @nwellnhof Yes, it does. :)
                – Arnauld
                Nov 11 at 23:37















              @nwellnhof This would fail for several values (576, 632, 633, ...)
              – Arnauld
              Nov 11 at 21:57




              @nwellnhof This would fail for several values (576, 632, 633, ...)
              – Arnauld
              Nov 11 at 21:57












              I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
              – nwellnhof
              Nov 11 at 22:59




              I see. But f=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184) (56 bytes) should work, right?
              – nwellnhof
              Nov 11 at 22:59












              @nwellnhof Yes, it does. :)
              – Arnauld
              Nov 11 at 23:37




              @nwellnhof Yes, it does. :)
              – Arnauld
              Nov 11 at 23:37










              up vote
              2
              down vote














              05AB1E, 30 27 bytes



              ŽK≠‰`91vDy64*›i1sy9*%]64/îO


              Try it online! or verify all test cases



              -3 bytes thanks to Kevin Cruijssen



              This is my first 05AB1E submission, so I am sure that this can be optimized.






              share|improve this answer


















              • 1




                Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
                – Kevin Cruijssen
                Nov 12 at 7:31











              • @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
                – Black Owl Kai
                Nov 12 at 7:56










              • Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
                – Kevin Cruijssen
                Nov 12 at 8:00














              up vote
              2
              down vote














              05AB1E, 30 27 bytes



              ŽK≠‰`91vDy64*›i1sy9*%]64/îO


              Try it online! or verify all test cases



              -3 bytes thanks to Kevin Cruijssen



              This is my first 05AB1E submission, so I am sure that this can be optimized.






              share|improve this answer


















              • 1




                Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
                – Kevin Cruijssen
                Nov 12 at 7:31











              • @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
                – Black Owl Kai
                Nov 12 at 7:56










              • Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
                – Kevin Cruijssen
                Nov 12 at 8:00












              up vote
              2
              down vote










              up vote
              2
              down vote










              05AB1E, 30 27 bytes



              ŽK≠‰`91vDy64*›i1sy9*%]64/îO


              Try it online! or verify all test cases



              -3 bytes thanks to Kevin Cruijssen



              This is my first 05AB1E submission, so I am sure that this can be optimized.






              share|improve this answer















              05AB1E, 30 27 bytes



              ŽK≠‰`91vDy64*›i1sy9*%]64/îO


              Try it online! or verify all test cases



              -3 bytes thanks to Kevin Cruijssen



              This is my first 05AB1E submission, so I am sure that this can be optimized.







              share|improve this answer














              share|improve this answer



              share|improve this answer








              edited Nov 12 at 7:56

























              answered Nov 11 at 20:34









              Black Owl Kai

              5717




              5717







              • 1




                Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
                – Kevin Cruijssen
                Nov 12 at 7:31











              • @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
                – Black Owl Kai
                Nov 12 at 7:56










              • Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
                – Kevin Cruijssen
                Nov 12 at 8:00












              • 1




                Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
                – Kevin Cruijssen
                Nov 12 at 7:31











              • @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
                – Black Owl Kai
                Nov 12 at 7:56










              • Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
                – Kevin Cruijssen
                Nov 12 at 8:00







              1




              1




              Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
              – Kevin Cruijssen
              Nov 12 at 7:31





              Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version 5184 can be ŽK≠. In addition, the S after 91 can be removed since it's done implicitly; and }} can be ] (although you'd still need }} in your test suite on TIO). Try it online or verify all test cases.
              – Kevin Cruijssen
              Nov 12 at 7:31













              @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
              – Black Owl Kai
              Nov 12 at 7:56




              @KevinCruijssen I used the legacy version because tio.run/#05ab1e redirects there and because I didn't know about multiple versions.Thanks for the tips!
              – Black Owl Kai
              Nov 12 at 7:56












              Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
              – Kevin Cruijssen
              Nov 12 at 8:00




              Ah, I see. The legacy version is written in Python, and was used for the past few years. A few months back the Elixir rewrite was released as new version, with loads of new features. Some builtins have changed slightly, so I do still use the legacy version sometimes, but I mostly use the new version now. :) PS: If you haven't seen it yet: Tips for golfing in 05AB1E might be interesting to read through. And feel free to ask anything in the 05AB1E chat if you need help.
              – Kevin Cruijssen
              Nov 12 at 8:00










              up vote
              0
              down vote














              Wolfram Language (Mathematica), 66 bytes



              Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&


              Try it online!






              share|improve this answer
























                up vote
                0
                down vote














                Wolfram Language (Mathematica), 66 bytes



                Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&


                Try it online!






                share|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote










                  Wolfram Language (Mathematica), 66 bytes



                  Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&


                  Try it online!






                  share|improve this answer













                  Wolfram Language (Mathematica), 66 bytes



                  Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&


                  Try it online!







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 12 at 11:01









                  alephalpha

                  21k32888




                  21k32888



























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