Gold nugget storage
up vote
6
down vote
favorite
1
Given a positive integer, write it as the sum of numbers, where each of them is in ${kt|kin{1,9,81},tin{1,2,3,...,64}}$. 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
code-golf
asked Nov 11 at 16:34
l4m2
4,5231634
|
show 5 more comments
up vote
6
down vote
favorite
1
Given a positive integer, write it as the sum of numbers, where each of them is in ${kt|kin{1,9,81},tin{1,2,3,...,64}}$. 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
code-golf
asked Nov 11 at 16:34
l4m2
4,5231634
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 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)
– Black Owl Kai
Nov 11 at 21:03
|
show 5 more comments
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 ${kt|kin{1,9,81},tin{1,2,3,...,64}}$. 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
code-golf
asked Nov 11 at 16:34
l4m2
4,5231634
Given a positive integer, write it as the sum of numbers, where each of them is in ${kt|kin{1,9,81},tin{1,2,3,...,64}}$. 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
code-golf
code-golf
asked Nov 11 at 16:34
l4m2
4,5231634
asked Nov 11 at 16:34
l4m2
4,5231634
edited Nov 11 at 23:40
asked Nov 11 at 16:34
l4m2
4,5231634
asked Nov 11 at 16:34
l4m2
4,5231634
asked Nov 11 at 16:34
l4m2
4,5231634
4,5231634
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 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)
– Black Owl Kai
Nov 11 at 21:03
|
show 5 more comments
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 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)
– 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
|
show 5 more comments
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
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
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
add a comment |
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.
answered Nov 11 at 23:36
nwellnhof
6,3231125
add a comment |
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!
answered Nov 11 at 18:04
Arnauld
70.8k688298
@nwellnhof This would fail for several values (576, 632, 633, ...)
– Arnauld
Nov 11 at 21:57
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?
– nwellnhof
Nov 11 at 22:59
@nwellnhof Yes, it does. :)
– Arnauld
Nov 11 at 23:37
add a comment |
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.
answered Nov 11 at 20:34
Black Owl Kai
5617
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 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.
– 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
add a comment |
up vote
0
down vote
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 at 11:01
alephalpha
21k32888
add a comment |
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
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
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
add a comment |
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
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
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
add a comment |
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
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
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 at 18:34
Erik the Outgolfer
31k429102
edited Nov 11 at 20:15
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
answered Nov 11 at 18:34
Erik the Outgolfer
31k429102
31k429102
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
add a comment |
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
add a comment |
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.
answered Nov 11 at 23:36
nwellnhof
6,3231125
add a comment |
up vote
2
down vote
Perl 6, 47 bytes
{+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)}
Try it online!
A greedy algorithm seems to work.
answered Nov 11 at 23:36
nwellnhof
6,3231125
add a comment |
up vote
2
down vote
up vote
2
down vote
Perl 6, 47 bytes
{+($_,(*X-(1,9,81 X*^65)).grep(*>=0).min...^0)}
Try it online!
A greedy algorithm seems to work.
answered Nov 11 at 23:36
nwellnhof
6,3231125
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 at 23:36
nwellnhof
6,3231125
answered Nov 11 at 23:36
nwellnhof
6,3231125
answered Nov 11 at 23:36
nwellnhof
6,3231125
answered Nov 11 at 23:36
nwellnhof
6,3231125
6,3231125
add a comment |
add a comment |
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!
answered Nov 11 at 18:04
Arnauld
70.8k688298
@nwellnhof This would fail for several values (576, 632, 633, ...)
– Arnauld
Nov 11 at 21:57
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?
– nwellnhof
Nov 11 at 22:59
@nwellnhof Yes, it does. :)
– Arnauld
Nov 11 at 23:37
add a comment |
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!
answered Nov 11 at 18:04
Arnauld
70.8k688298
@nwellnhof This would fail for several values (576, 632, 633, ...)
– Arnauld
Nov 11 at 21:57
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?
– nwellnhof
Nov 11 at 22:59
@nwellnhof Yes, it does. :)
– Arnauld
Nov 11 at 23:37
add a comment |
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!
answered Nov 11 at 18:04
Arnauld
70.8k688298
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 at 18:04
Arnauld
70.8k688298
edited Nov 11 at 23:37
answered Nov 11 at 18:04
Arnauld
70.8k688298
answered Nov 11 at 18:04
Arnauld
70.8k688298
answered Nov 11 at 18:04
Arnauld
70.8k688298
70.8k688298
@nwellnhof This would fail for several values (576, 632, 633, ...)
– Arnauld
Nov 11 at 21:57
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?
– nwellnhof
Nov 11 at 22:59
@nwellnhof Yes, it does. :)
– Arnauld
Nov 11 at 23:37
add a comment |
@nwellnhof This would fail for several values (576, 632, 633, ...)
– Arnauld
Nov 11 at 21:57
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?
– 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
add a comment |
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.
answered Nov 11 at 20:34
Black Owl Kai
5617
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 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.
– 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
add a comment |
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.
answered Nov 11 at 20:34
Black Owl Kai
5617
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 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.
– 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
add a comment |
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.
answered Nov 11 at 20:34
Black Owl Kai
5617
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 at 20:34
Black Owl Kai
5617
edited Nov 12 at 7:56
answered Nov 11 at 20:34
Black Owl Kai
5617
answered Nov 11 at 20:34
Black Owl Kai
5617
answered Nov 11 at 20:34
Black Owl Kai
5617
5617
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 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.
– 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
add a comment |
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 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.
– 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
add a comment |
up vote
0
down vote
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 at 11:01
alephalpha
21k32888
add a comment |
up vote
0
down vote
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 at 11:01
alephalpha
21k32888
add a comment |
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!
answered Nov 12 at 11:01
alephalpha
21k32888
Wolfram Language (Mathematica), 66 bytes
Min[Length/@IntegerPartitions[#,All,Union[#,9#,81#]&@Range@64,#]]&
Try it online!
answered Nov 12 at 11:01
alephalpha
21k32888
answered Nov 12 at 11:01
alephalpha
21k32888
answered Nov 12 at 11:01
alephalpha
21k32888
answered Nov 12 at 11:01
alephalpha
21k32888
21k32888
add a comment |
add a comment |
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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*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)– Black Owl Kai
Nov 11 at 21:03