Gold nugget storage
6
$begingroup$
Given a positive integer, write it as the sum of numbers, where each of them is in $kt$. 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
code-golf
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
$endgroup$
|
show 5 more comments
6
$begingroup$
Given a positive integer, write it as the sum of numbers, where each of them is in $kt$. 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
code-golf
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
$endgroup$
4
$begingroup$
Could you give slightly more detail/put into words how the input results in the output?
$endgroup$
– Quintec
Nov 11 '18 at 16:45
2
$begingroup$
If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
$endgroup$
– trichoplax
Nov 11 '18 at 16:52
4
$begingroup$
Is it basically a sort of change-making problem ? With the coins denominations being in the set1,9,81 × 1...64?
$endgroup$
– digEmAll
Nov 11 '18 at 17:01
2
$begingroup$
Are you trying to minimize the number of items or number of stacks?
$endgroup$
– fəˈnɛtɪk
Nov 11 '18 at 17:02
3
$begingroup$
I suggest adding5274 = 64*81 + 10*9as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get64*81 + 1*81 + 9*1(as mine did)
$endgroup$
– Black Owl Kai
Nov 11 '18 at 21:03
|
show 5 more comments
6
6
6
1
$begingroup$
Given a positive integer, write it as the sum of numbers, where each of them is in $kt$. 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
code-golf
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
$endgroup$
Given a positive integer, write it as the sum of numbers, where each of them is in $kt$. 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
code-golf
code-golf
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
edited Nov 11 '18 at 23:40
l4m2
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
asked Nov 11 '18 at 16:34
l4m2l4m2
4,7361735
4,7361735
4
$begingroup$
Could you give slightly more detail/put into words how the input results in the output?
$endgroup$
– Quintec
Nov 11 '18 at 16:45
2
$begingroup$
If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
$endgroup$
– trichoplax
Nov 11 '18 at 16:52
4
$begingroup$
Is it basically a sort of change-making problem ? With the coins denominations being in the set1,9,81 × 1...64?
$endgroup$
– digEmAll
Nov 11 '18 at 17:01
2
$begingroup$
Are you trying to minimize the number of items or number of stacks?
$endgroup$
– fəˈnɛtɪk
Nov 11 '18 at 17:02
3
$begingroup$
I suggest adding5274 = 64*81 + 10*9as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get64*81 + 1*81 + 9*1(as mine did)
$endgroup$
– Black Owl Kai
Nov 11 '18 at 21:03
|
show 5 more comments
4
$begingroup$
Could you give slightly more detail/put into words how the input results in the output?
$endgroup$
– Quintec
Nov 11 '18 at 16:45
2
$begingroup$
If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
$endgroup$
– trichoplax
Nov 11 '18 at 16:52
4
$begingroup$
Is it basically a sort of change-making problem ? With the coins denominations being in the set1,9,81 × 1...64?
$endgroup$
– digEmAll
Nov 11 '18 at 17:01
2
$begingroup$
Are you trying to minimize the number of items or number of stacks?
$endgroup$
– fəˈnɛtɪk
Nov 11 '18 at 17:02
3
$begingroup$
I suggest adding5274 = 64*81 + 10*9as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get64*81 + 1*81 + 9*1(as mine did)
$endgroup$
– Black Owl Kai
Nov 11 '18 at 21:03
4
4
$begingroup$
Could you give slightly more detail/put into words how the input results in the output?
$endgroup$
– Quintec
Nov 11 '18 at 16:45
$begingroup$
Could you give slightly more detail/put into words how the input results in the output?
$endgroup$
– Quintec
Nov 11 '18 at 16:45
2
2
$begingroup$
If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
$endgroup$
– trichoplax
Nov 11 '18 at 16:52
$begingroup$
If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
$endgroup$
– trichoplax
Nov 11 '18 at 16:52
4
4
$begingroup$
Is it basically a sort of change-making problem ? With the coins denominations being in the set
1,9,81 × 1...64 ?$endgroup$
– digEmAll
Nov 11 '18 at 17:01
$begingroup$
Is it basically a sort of change-making problem ? With the coins denominations being in the set
1,9,81 × 1...64 ?$endgroup$
– digEmAll
Nov 11 '18 at 17:01
2
2
$begingroup$
Are you trying to minimize the number of items or number of stacks?
$endgroup$
– fəˈnɛtɪk
Nov 11 '18 at 17:02
$begingroup$
Are you trying to minimize the number of items or number of stacks?
$endgroup$
– fəˈnɛtɪk
Nov 11 '18 at 17:02
3
3
$begingroup$
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)$endgroup$
– Black Owl Kai
Nov 11 '18 at 21:03
$begingroup$
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)$endgroup$
– Black Owl Kai
Nov 11 '18 at 21:03
|
show 5 more comments
5 Answers
5
active
oldest
votes
3
$begingroup$
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
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
$endgroup$
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
add a comment |
2
$begingroup$
Perl 6, 47 bytes
+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)
Try it online!
A greedy algorithm seems to work.
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
$endgroup$
add a comment |
2
$begingroup$
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!
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
$endgroup$
$begingroup$
@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
I see. Butf=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184)(56 bytes) should work, right?
$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
$begingroup$
@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
add a comment |
2
$begingroup$
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.
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
$endgroup$
1
$begingroup$
Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version5184can beŽK≠. In addition, theSafter91can 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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
add a comment |
0
$begingroup$
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
$endgroup$
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
3
$begingroup$
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
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
$endgroup$
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
add a comment |
3
$begingroup$
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
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
$endgroup$
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
add a comment |
3
3
3
$begingroup$
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
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
$endgroup$
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
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
edited Nov 11 '18 at 20:15
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
answered Nov 11 '18 at 18:34
Erik the OutgolferErik the Outgolfer
31.8k429103
31.8k429103
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
add a comment |
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
Since the output in testable area is trivial, can't quite check?
$endgroup$
– l4m2
Nov 11 '18 at 23:35
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
$begingroup$
@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.
$endgroup$
– Erik the Outgolfer
Nov 12 '18 at 8:37
add a comment |
2
$begingroup$
Perl 6, 47 bytes
+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)
Try it online!
A greedy algorithm seems to work.
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
$endgroup$
add a comment |
2
$begingroup$
Perl 6, 47 bytes
+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)
Try it online!
A greedy algorithm seems to work.
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
$endgroup$
add a comment |
2
2
2
$begingroup$
Perl 6, 47 bytes
+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)
Try it online!
A greedy algorithm seems to work.
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
$endgroup$
Perl 6, 47 bytes
+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)
Try it online!
A greedy algorithm seems to work.
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
answered Nov 11 '18 at 23:36
nwellnhofnwellnhof
6,73511126
6,73511126
add a comment |
add a comment |
2
$begingroup$
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!
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
$endgroup$
$begingroup$
@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
I see. Butf=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184)(56 bytes) should work, right?
$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
$begingroup$
@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
add a comment |
2
$begingroup$
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!
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
$endgroup$
$begingroup$
@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
I see. Butf=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184)(56 bytes) should work, right?
$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
$begingroup$
@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
add a comment |
2
2
2
$begingroup$
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!
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
$endgroup$
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!
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
edited Nov 11 '18 at 23:37
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
answered Nov 11 '18 at 18:04
ArnauldArnauld
75.9k693320
75.9k693320
$begingroup$
@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
I see. Butf=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184)(56 bytes) should work, right?
$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
$begingroup$
@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
add a comment |
$begingroup$
@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
I see. Butf=n=>n&&1+f(n<5184?n>64&&n%(n<576?9:n<648?72:81):n-5184)(56 bytes) should work, right?
$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
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@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
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@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
@nwellnhof This would fail for several values (576, 632, 633, ...)
$endgroup$
– Arnauld
Nov 11 '18 at 21:57
$begingroup$
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?$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
$begingroup$
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?$endgroup$
– nwellnhof
Nov 11 '18 at 22:59
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@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
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@nwellnhof Yes, it does. :)
$endgroup$
– Arnauld
Nov 11 '18 at 23:37
add a comment |
2
$begingroup$
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.
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
$endgroup$
1
$begingroup$
Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version5184can beŽK≠. In addition, theSafter91can 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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
add a comment |
2
$begingroup$
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.
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
$endgroup$
1
$begingroup$
Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version5184can beŽK≠. In addition, theSafter91can 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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
add a comment |
2
2
2
$begingroup$
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.
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
$endgroup$
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.
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
edited Nov 12 '18 at 7:56
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
answered Nov 11 '18 at 20:34
Black Owl KaiBlack Owl Kai
68011
68011
1
$begingroup$
Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version5184can beŽK≠. In addition, theSafter91can 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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
add a comment |
1
$begingroup$
Not sure why you opted to choose the legacy version, since your answer also works in the new Elixir rewrite, but in the new version5184can beŽK≠. In addition, theSafter91can 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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
1
1
$begingroup$
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.$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
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.$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 7:31
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
@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!
$endgroup$
– Black Owl Kai
Nov 12 '18 at 7:56
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
$begingroup$
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.
$endgroup$
– Kevin Cruijssen
Nov 12 '18 at 8:00
add a comment |
0
$begingroup$
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
$endgroup$
add a comment |
0
$begingroup$
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
$endgroup$
add a comment |
0
0
0
$begingroup$
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
$endgroup$
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
answered Nov 12 '18 at 11:01
alephalphaalephalpha
21.3k32991
21.3k32991
add a comment |
add a comment |
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4
$begingroup$
Could you give slightly more detail/put into words how the input results in the output?
$endgroup$
– Quintec
Nov 11 '18 at 16:45
2
$begingroup$
If something requires clarification in the comments, it would be useful to edit the challenge to include that clarification
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– trichoplax
Nov 11 '18 at 16:52
4
$begingroup$
Is it basically a sort of change-making problem ? With the coins denominations being in the set
1,9,81 × 1...64?$endgroup$
– digEmAll
Nov 11 '18 at 17:01
2
$begingroup$
Are you trying to minimize the number of items or number of stacks?
$endgroup$
– fəˈnɛtɪk
Nov 11 '18 at 17:02
3
$begingroup$
I suggest adding
5274 = 64*81 + 10*9as a test case, this checks if answers take as many blocks as possible if there are more than 576 nuggets and get64*81 + 1*81 + 9*1(as mine did)$endgroup$
– Black Owl Kai
Nov 11 '18 at 21:03