Can the number be splitted into powers of 2?

Clash Royale CLAN TAG#URR8PPP

up vote
4
down vote

favorite

Yesterday while playing with my kid I noticed the number in his toy train:

So we have $$4281$$ that can be splitted into $$4-2-8-1$$ or $$2^2-2^1-2^3-2^0$$

So simple challenge: given a non-negative number as input, return consistent truthy and falsey values that represent whether or not the string representation of the number (in base 10 and without leading zeroes) can be somehow splitted into numbers that are powers of 2.

Examples:

4281 truthy (4-2-8-1)
164 truthy (16-4 or 1-64)
8192 truthy (the number itself is a power of 2)
81024 truthy (8-1024 or 8-1-024)
101 truthy (1-01)
0 falsey (0 cannot be represented as 2^x for any x)
1 truthy
3 falsey
234789 falsey
256323 falsey (we have 256 and 32 but then 3)
8132 truthy (8-1-32)

Tests for very large numbers (not really necessary to be handled by your code):
81024256641116 truthy (8-1024-256-64-1-1-16)
64512819237913 falsey

This is code-golf, so may the shortest code for each language win!

code-golf string number decision-problem

share|improve this question

edited 2 mins ago

asked 59 mins ago

Charlie

6,8271979

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any limit to how many digits must be supported? Is it OK if it works in theory (assume much memory/time) for large numbers? The reason why I ask: A possible solution is to create all possible "splits" (don't remember the word). 164 -> "1,6,4", "1, 64", "16,4", "164". The number of "splits" can get quite large for large numbers.
  – Stewie Griffin
  38 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StewieGriffin initially I thought about limiting the input number to the range of a standard int type (4 bytes), but actually I do not mind if your code does not support very large numbers. Just state in your answer the limitations of your code.
  – Charlie
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  6 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can 81024 only be 8-1024, or is 8-1-024 also valid? EDIT: Basically the same question whether leading 0s are allowed as for @ShieruAsakoto suggested test case above..
  – Kevin Cruijssen
  4 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShieruAsakoto I've been testing the 101 case with the current answers and they all return true, because it can be splitted into 1-01 that are both powers of 2, so I'll consider that case to be truthy.
  – Charlie
  3 mins ago
  

  
  
  
  

 | 
show 1 more comment

up vote
4
down vote

favorite

Yesterday while playing with my kid I noticed the number in his toy train:

So we have $$4281$$ that can be splitted into $$4-2-8-1$$ or $$2^2-2^1-2^3-2^0$$

So simple challenge: given a non-negative number as input, return consistent truthy and falsey values that represent whether or not the string representation of the number (in base 10 and without leading zeroes) can be somehow splitted into numbers that are powers of 2.

Examples:

4281 truthy (4-2-8-1)
164 truthy (16-4 or 1-64)
8192 truthy (the number itself is a power of 2)
81024 truthy (8-1024 or 8-1-024)
101 truthy (1-01)
0 falsey (0 cannot be represented as 2^x for any x)
1 truthy
3 falsey
234789 falsey
256323 falsey (we have 256 and 32 but then 3)
8132 truthy (8-1-32)

Tests for very large numbers (not really necessary to be handled by your code):
81024256641116 truthy (8-1024-256-64-1-1-16)
64512819237913 falsey

This is code-golf, so may the shortest code for each language win!

code-golf string number decision-problem

share|improve this question

edited 2 mins ago

asked 59 mins ago

Charlie

6,8271979

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any limit to how many digits must be supported? Is it OK if it works in theory (assume much memory/time) for large numbers? The reason why I ask: A possible solution is to create all possible "splits" (don't remember the word). 164 -> "1,6,4", "1, 64", "16,4", "164". The number of "splits" can get quite large for large numbers.
  – Stewie Griffin
  38 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StewieGriffin initially I thought about limiting the input number to the range of a standard int type (4 bytes), but actually I do not mind if your code does not support very large numbers. Just state in your answer the limitations of your code.
  – Charlie
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  6 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can 81024 only be 8-1024, or is 8-1-024 also valid? EDIT: Basically the same question whether leading 0s are allowed as for @ShieruAsakoto suggested test case above..
  – Kevin Cruijssen
  4 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShieruAsakoto I've been testing the 101 case with the current answers and they all return true, because it can be splitted into 1-01 that are both powers of 2, so I'll consider that case to be truthy.
  – Charlie
  3 mins ago
  

  
  
  
  

 | 
show 1 more comment

up vote
4
down vote

favorite

up vote
4
down vote

favorite

Yesterday while playing with my kid I noticed the number in his toy train:

So we have $$4281$$ that can be splitted into $$4-2-8-1$$ or $$2^2-2^1-2^3-2^0$$

So simple challenge: given a non-negative number as input, return consistent truthy and falsey values that represent whether or not the string representation of the number (in base 10 and without leading zeroes) can be somehow splitted into numbers that are powers of 2.

Examples:

4281 truthy (4-2-8-1)
164 truthy (16-4 or 1-64)
8192 truthy (the number itself is a power of 2)
81024 truthy (8-1024 or 8-1-024)
101 truthy (1-01)
0 falsey (0 cannot be represented as 2^x for any x)
1 truthy
3 falsey
234789 falsey
256323 falsey (we have 256 and 32 but then 3)
8132 truthy (8-1-32)

Tests for very large numbers (not really necessary to be handled by your code):
81024256641116 truthy (8-1024-256-64-1-1-16)
64512819237913 falsey

This is code-golf, so may the shortest code for each language win!

code-golf string number decision-problem

share|improve this question

edited 2 mins ago

asked 59 mins ago

Charlie

6,8271979

Yesterday while playing with my kid I noticed the number in his toy train:

So we have $$4281$$ that can be splitted into $$4-2-8-1$$ or $$2^2-2^1-2^3-2^0$$

So simple challenge: given a non-negative number as input, return consistent truthy and falsey values that represent whether or not the string representation of the number (in base 10 and without leading zeroes) can be somehow splitted into numbers that are powers of 2.

Examples:

4281 truthy (4-2-8-1)
164 truthy (16-4 or 1-64)
8192 truthy (the number itself is a power of 2)
81024 truthy (8-1024 or 8-1-024)
101 truthy (1-01)
0 falsey (0 cannot be represented as 2^x for any x)
1 truthy
3 falsey
234789 falsey
256323 falsey (we have 256 and 32 but then 3)
8132 truthy (8-1-32)

Tests for very large numbers (not really necessary to be handled by your code):
81024256641116 truthy (8-1024-256-64-1-1-16)
64512819237913 falsey

This is code-golf, so may the shortest code for each language win!

code-golf string number decision-problem

code-golf string number decision-problem

share|improve this question

edited 2 mins ago

asked 59 mins ago

Charlie

6,8271979

share|improve this question

edited 2 mins ago

asked 59 mins ago

Charlie

6,8271979

share|improve this question

share|improve this question

edited 2 mins ago

edited 2 mins ago

edited 2 mins ago

asked 59 mins ago

Charlie

6,8271979

asked 59 mins ago

Charlie

6,8271979

asked 59 mins ago

Charlie

6,8271979

6,8271979

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any limit to how many digits must be supported? Is it OK if it works in theory (assume much memory/time) for large numbers? The reason why I ask: A possible solution is to create all possible "splits" (don't remember the word). 164 -> "1,6,4", "1, 64", "16,4", "164". The number of "splits" can get quite large for large numbers.
  – Stewie Griffin
  38 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StewieGriffin initially I thought about limiting the input number to the range of a standard int type (4 bytes), but actually I do not mind if your code does not support very large numbers. Just state in your answer the limitations of your code.
  – Charlie
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  6 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can 81024 only be 8-1024, or is 8-1-024 also valid? EDIT: Basically the same question whether leading 0s are allowed as for @ShieruAsakoto suggested test case above..
  – Kevin Cruijssen
  4 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShieruAsakoto I've been testing the 101 case with the current answers and they all return true, because it can be splitted into 1-01 that are both powers of 2, so I'll consider that case to be truthy.
  – Charlie
  3 mins ago
  

  
  
  
  

 | 
show 1 more comment

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Any limit to how many digits must be supported? Is it OK if it works in theory (assume much memory/time) for large numbers? The reason why I ask: A possible solution is to create all possible "splits" (don't remember the word). 164 -> "1,6,4", "1, 64", "16,4", "164". The number of "splits" can get quite large for large numbers.
  – Stewie Griffin
  38 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @StewieGriffin initially I thought about limiting the input number to the range of a standard int type (4 bytes), but actually I do not mind if your code does not support very large numbers. Just state in your answer the limitations of your code.
  – Charlie
  20 mins ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  6 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Can 81024 only be 8-1024, or is 8-1-024 also valid? EDIT: Basically the same question whether leading 0s are allowed as for @ShieruAsakoto suggested test case above..
  – Kevin Cruijssen
  4 mins ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @ShieruAsakoto I've been testing the 101 case with the current answers and they all return true, because it can be splitted into 1-01 that are both powers of 2, so I'll consider that case to be truthy.
  – Charlie
  3 mins ago
  

  
  
  
  

Any limit to how many digits must be supported? Is it OK if it works in theory (assume much memory/time) for large numbers? The reason why I ask: A possible solution is to create all possible "splits" (don't remember the word). 164 -> "1,6,4", "1, 64", "16,4", "164". The number of "splits" can get quite large for large numbers.
– Stewie Griffin
38 mins ago

Any limit to how many digits must be supported? Is it OK if it works in theory (assume much memory/time) for large numbers? The reason why I ask: A possible solution is to create all possible "splits" (don't remember the word). 164 -> "1,6,4", "1, 64", "16,4", "164". The number of "splits" can get quite large for large numbers.
– Stewie Griffin
38 mins ago

@StewieGriffin initially I thought about limiting the input number to the range of a standard int type (4 bytes), but actually I do not mind if your code does not support very large numbers. Just state in your answer the limitations of your code.
– Charlie
20 mins ago

@StewieGriffin initially I thought about limiting the input number to the range of a standard int type (4 bytes), but actually I do not mind if your code does not support very large numbers. Just state in your answer the limitations of your code.
– Charlie
20 mins ago

2

2

Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
– Shieru Asakoto
6 mins ago

Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
– Shieru Asakoto
6 mins ago

Can 81024 only be 8-1024, or is 8-1-024 also valid? EDIT: Basically the same question whether leading 0s are allowed as for @ShieruAsakoto suggested test case above..
– Kevin Cruijssen
4 mins ago

Can 81024 only be 8-1024, or is 8-1-024 also valid? EDIT: Basically the same question whether leading 0s are allowed as for @ShieruAsakoto suggested test case above..
– Kevin Cruijssen
4 mins ago

1

1

@ShieruAsakoto I've been testing the 101 case with the current answers and they all return true, because it can be splitted into 1-01 that are both powers of 2, so I'll consider that case to be truthy.
– Charlie
3 mins ago

@ShieruAsakoto I've been testing the 101 case with the current answers and they all return true, because it can be splitted into 1-01 that are both powers of 2, so I'll consider that case to be truthy.
– Charlie
3 mins ago

 | 
show 1 more comment

5 Answers
5

active

oldest

votes

up vote
2
down vote

Python 2, 85 bytes

f=lambda n:bin(int(n)).count('1')==1or any(f(n[:i])*f(n[i:])for i in range(1,len(n)))

Try it online!

share|improve this answer

answered 52 mins ago

TFeld

12.4k2834

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 75 70 bytes

-5 bytes thanks @Arnauld. At most 32-bit support

f=x=>+x?[...x].some((_,i)=>!((y=x.slice(0,++i))&~-y)&f(x.slice(i))):!x

Try it online!

Input as a string.

share|improve this answer

edited 4 mins ago

answered 32 mins ago

Shieru Asakoto

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Oh dang I forgot the bitwise trick again! Thanks!
  – Shieru Asakoto
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
  – Arnauld
  2 mins ago
  

  
  
  
  

add a comment | 

up vote
1
down vote

Python 2, 72 bytes

f=lambda n,m=10:n>=m/10and(n%m&~-(n%m)<1and(n/m<1or f(n/m))or f(n,m*10))

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.6k21058

add a comment | 

up vote
1
down vote

JavaScript (Node.js), ₆₉ 64 bytes

f=(x,m=10,q=!(x%m&x%m-1|x%m<m/10))=>x<m?q:q&&f(x/m|0)||f(x,10*m)

Try it online!

Input as number. The logic part is quite convoluted, so no idea how to untangle it and get rid of q.

-5 bytes by golfing the power-of-2 check. Also rejects leading zeroes, e.g. 101 where 1/01 is an invalid partition.

share|improve this answer

edited 2 mins ago

answered 15 mins ago

Bubbler

3,917541

add a comment | 

up vote
0
down vote

05AB1E, 9 bytes

Ýos.œåPOĀ

Try it online or verify all test cases. (NOTE: The т in the header is 100 to only get the first 100 square numbers, instead of the first input amount of square numbers. It works in the second case as well, but is pretty inefficient and might time-out on TIO.)

Explanation:

Ý # Create a list in the range [0,n], where n is the (implicit) input
 # (or 100 in the TIO)
 # i.e. 81024 → [0,1,2,3,...,81024]
 o # Square each
 # → [1,2,4,8,...,451..216 (nr with 24391 digits)]
 s # Swap to take the input
 .œ # Create each possible partition of this input
 # i.e. 81024 → [["8","1","0","2","4"],["8","1","0","24"],...,["8102","4"],["81024"]]
 å # Check for each if it's in the list of squares
 # → [[1,1,0,1,1],[1,1,0,0],...,[0,1],[0]]
 P # Check for each inner list whether all are truthy
 # → [0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0]
 O # Take the sum
 # → 2
 Ā # Truthify (and output implicitly)
 # → 1 (truthy)

share

answered 5 mins ago

Kevin Cruijssen

31.5k553171

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["\$", "\$"]]);
);
);
, "mathjax-editing");

StackExchange.ifUsing("editor", function ()
StackExchange.using("externalEditor", function ()
StackExchange.using("snippets", function ()
StackExchange.snippets.init();
);
);
, "code-snippets");

StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "200"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);

else
createEditor();

);

function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
convertImagesToLinks: false,
noModals: false,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);



);

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f173833%2fcan-the-number-be-splitted-into-powers-of-2%23new-answer', 'question_page');

);

Post as a guest

Name Email

5 Answers
5

active

oldest

votes

5 Answers
5

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Python 2, 85 bytes

f=lambda n:bin(int(n)).count('1')==1or any(f(n[:i])*f(n[i:])for i in range(1,len(n)))

Try it online!

share|improve this answer

answered 52 mins ago

TFeld

12.4k2834

add a comment | 

up vote
2
down vote

Python 2, 85 bytes

f=lambda n:bin(int(n)).count('1')==1or any(f(n[:i])*f(n[i:])for i in range(1,len(n)))

Try it online!

share|improve this answer

answered 52 mins ago

TFeld

12.4k2834

add a comment | 

up vote
2
down vote

up vote
2
down vote

Python 2, 85 bytes

f=lambda n:bin(int(n)).count('1')==1or any(f(n[:i])*f(n[i:])for i in range(1,len(n)))

Try it online!

share|improve this answer

answered 52 mins ago

TFeld

12.4k2834

Python 2, 85 bytes

f=lambda n:bin(int(n)).count('1')==1or any(f(n[:i])*f(n[i:])for i in range(1,len(n)))

Try it online!

share|improve this answer

answered 52 mins ago

TFeld

12.4k2834

share|improve this answer

share|improve this answer

answered 52 mins ago

TFeld

12.4k2834

answered 52 mins ago

TFeld

12.4k2834

answered 52 mins ago

TFeld

12.4k2834

12.4k2834

add a comment | 

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 75 70 bytes

-5 bytes thanks @Arnauld. At most 32-bit support

f=x=>+x?[...x].some((_,i)=>!((y=x.slice(0,++i))&~-y)&f(x.slice(i))):!x

Try it online!

Input as a string.

share|improve this answer

edited 4 mins ago

answered 32 mins ago

Shieru Asakoto

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Oh dang I forgot the bitwise trick again! Thanks!
  – Shieru Asakoto
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
  – Arnauld
  2 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 75 70 bytes

-5 bytes thanks @Arnauld. At most 32-bit support

f=x=>+x?[...x].some((_,i)=>!((y=x.slice(0,++i))&~-y)&f(x.slice(i))):!x

Try it online!

Input as a string.

share|improve this answer

edited 4 mins ago

answered 32 mins ago

Shieru Asakoto

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Oh dang I forgot the bitwise trick again! Thanks!
  – Shieru Asakoto
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
  – Arnauld
  2 mins ago
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

JavaScript (Node.js), 75 70 bytes

-5 bytes thanks @Arnauld. At most 32-bit support

f=x=>+x?[...x].some((_,i)=>!((y=x.slice(0,++i))&~-y)&f(x.slice(i))):!x

Try it online!

Input as a string.

share|improve this answer

edited 4 mins ago

answered 32 mins ago

Shieru Asakoto

1,980312

JavaScript (Node.js), 75 70 bytes

-5 bytes thanks @Arnauld. At most 32-bit support

f=x=>+x?[...x].some((_,i)=>!((y=x.slice(0,++i))&~-y)&f(x.slice(i))):!x

Try it online!

Input as a string.

share|improve this answer

edited 4 mins ago

answered 32 mins ago

Shieru Asakoto

1,980312

share|improve this answer

share|improve this answer

edited 4 mins ago

edited 4 mins ago

edited 4 mins ago

answered 32 mins ago

Shieru Asakoto

1,980312

answered 32 mins ago

Shieru Asakoto

1,980312

answered 32 mins ago

Shieru Asakoto

1,980312

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Oh dang I forgot the bitwise trick again! Thanks!
  – Shieru Asakoto
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
  – Arnauld
  2 mins ago
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Arnauld Oh dang I forgot the bitwise trick again! Thanks!
  – Shieru Asakoto
  18 mins ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
  – Arnauld
  2 mins ago
  

  
  
  
  

@Arnauld Oh dang I forgot the bitwise trick again! Thanks!
– Shieru Asakoto
18 mins ago

@Arnauld Oh dang I forgot the bitwise trick again! Thanks!
– Shieru Asakoto
18 mins ago

Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
– Arnauld
2 mins ago

Using a ternary saves 1 byte: (y=x.slice(0,++i))&~-y?0:f(x.slice(i)).
– Arnauld
2 mins ago

add a comment | 

up vote
1
down vote

Python 2, 72 bytes

f=lambda n,m=10:n>=m/10and(n%m&~-(n%m)<1and(n/m<1or f(n/m))or f(n,m*10))

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.6k21058

add a comment | 

up vote
1
down vote

Python 2, 72 bytes

f=lambda n,m=10:n>=m/10and(n%m&~-(n%m)<1and(n/m<1or f(n/m))or f(n,m*10))

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.6k21058

add a comment | 

up vote
1
down vote

up vote
1
down vote

Python 2, 72 bytes

f=lambda n,m=10:n>=m/10and(n%m&~-(n%m)<1and(n/m<1or f(n/m))or f(n,m*10))

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.6k21058

Python 2, 72 bytes

f=lambda n,m=10:n>=m/10and(n%m&~-(n%m)<1and(n/m<1or f(n/m))or f(n,m*10))

Try it online!

share|improve this answer

answered 12 mins ago

ovs

17.6k21058

share|improve this answer

share|improve this answer

answered 12 mins ago

ovs

17.6k21058

answered 12 mins ago

ovs

17.6k21058

answered 12 mins ago

ovs

17.6k21058

17.6k21058

add a comment | 

add a comment | 

up vote
1
down vote

JavaScript (Node.js), ₆₉ 64 bytes

f=(x,m=10,q=!(x%m&x%m-1|x%m<m/10))=>x<m?q:q&&f(x/m|0)||f(x,10*m)

Try it online!

Input as number. The logic part is quite convoluted, so no idea how to untangle it and get rid of q.

-5 bytes by golfing the power-of-2 check. Also rejects leading zeroes, e.g. 101 where 1/01 is an invalid partition.

share|improve this answer

edited 2 mins ago

answered 15 mins ago

Bubbler

3,917541

add a comment | 

up vote
1
down vote

JavaScript (Node.js), ₆₉ 64 bytes

f=(x,m=10,q=!(x%m&x%m-1|x%m<m/10))=>x<m?q:q&&f(x/m|0)||f(x,10*m)

Try it online!

Input as number. The logic part is quite convoluted, so no idea how to untangle it and get rid of q.

-5 bytes by golfing the power-of-2 check. Also rejects leading zeroes, e.g. 101 where 1/01 is an invalid partition.

share|improve this answer

edited 2 mins ago

answered 15 mins ago

Bubbler

3,917541

add a comment | 

up vote
1
down vote

up vote
1
down vote

JavaScript (Node.js), ₆₉ 64 bytes

f=(x,m=10,q=!(x%m&x%m-1|x%m<m/10))=>x<m?q:q&&f(x/m|0)||f(x,10*m)

Try it online!

Input as number. The logic part is quite convoluted, so no idea how to untangle it and get rid of q.

-5 bytes by golfing the power-of-2 check. Also rejects leading zeroes, e.g. 101 where 1/01 is an invalid partition.

share|improve this answer

edited 2 mins ago

answered 15 mins ago

Bubbler

3,917541

JavaScript (Node.js), ₆₉ 64 bytes

f=(x,m=10,q=!(x%m&x%m-1|x%m<m/10))=>x<m?q:q&&f(x/m|0)||f(x,10*m)

Try it online!

Input as number. The logic part is quite convoluted, so no idea how to untangle it and get rid of q.

-5 bytes by golfing the power-of-2 check. Also rejects leading zeroes, e.g. 101 where 1/01 is an invalid partition.

share|improve this answer

edited 2 mins ago

answered 15 mins ago

Bubbler

3,917541

share|improve this answer

share|improve this answer

edited 2 mins ago

edited 2 mins ago

edited 2 mins ago

answered 15 mins ago

Bubbler

3,917541

answered 15 mins ago

Bubbler

3,917541

answered 15 mins ago

Bubbler

3,917541

3,917541

add a comment | 

add a comment | 

up vote
0
down vote

05AB1E, 9 bytes

Ýos.œåPOĀ

Try it online or verify all test cases. (NOTE: The т in the header is 100 to only get the first 100 square numbers, instead of the first input amount of square numbers. It works in the second case as well, but is pretty inefficient and might time-out on TIO.)

Explanation:

Ý # Create a list in the range [0,n], where n is the (implicit) input
 # (or 100 in the TIO)
 # i.e. 81024 → [0,1,2,3,...,81024]
 o # Square each
 # → [1,2,4,8,...,451..216 (nr with 24391 digits)]
 s # Swap to take the input
 .œ # Create each possible partition of this input
 # i.e. 81024 → [["8","1","0","2","4"],["8","1","0","24"],...,["8102","4"],["81024"]]
 å # Check for each if it's in the list of squares
 # → [[1,1,0,1,1],[1,1,0,0],...,[0,1],[0]]
 P # Check for each inner list whether all are truthy
 # → [0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0]
 O # Take the sum
 # → 2
 Ā # Truthify (and output implicitly)
 # → 1 (truthy)

share

answered 5 mins ago

Kevin Cruijssen

31.5k553171

add a comment | 

up vote
0
down vote

05AB1E, 9 bytes

Ýos.œåPOĀ

Try it online or verify all test cases. (NOTE: The т in the header is 100 to only get the first 100 square numbers, instead of the first input amount of square numbers. It works in the second case as well, but is pretty inefficient and might time-out on TIO.)

Explanation:

Ý # Create a list in the range [0,n], where n is the (implicit) input
 # (or 100 in the TIO)
 # i.e. 81024 → [0,1,2,3,...,81024]
 o # Square each
 # → [1,2,4,8,...,451..216 (nr with 24391 digits)]
 s # Swap to take the input
 .œ # Create each possible partition of this input
 # i.e. 81024 → [["8","1","0","2","4"],["8","1","0","24"],...,["8102","4"],["81024"]]
 å # Check for each if it's in the list of squares
 # → [[1,1,0,1,1],[1,1,0,0],...,[0,1],[0]]
 P # Check for each inner list whether all are truthy
 # → [0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0]
 O # Take the sum
 # → 2
 Ā # Truthify (and output implicitly)
 # → 1 (truthy)

share

answered 5 mins ago

Kevin Cruijssen

31.5k553171

add a comment | 

up vote
0
down vote

up vote
0
down vote

05AB1E, 9 bytes

Ýos.œåPOĀ

Try it online or verify all test cases. (NOTE: The т in the header is 100 to only get the first 100 square numbers, instead of the first input amount of square numbers. It works in the second case as well, but is pretty inefficient and might time-out on TIO.)

Explanation:

Ý # Create a list in the range [0,n], where n is the (implicit) input
 # (or 100 in the TIO)
 # i.e. 81024 → [0,1,2,3,...,81024]
 o # Square each
 # → [1,2,4,8,...,451..216 (nr with 24391 digits)]
 s # Swap to take the input
 .œ # Create each possible partition of this input
 # i.e. 81024 → [["8","1","0","2","4"],["8","1","0","24"],...,["8102","4"],["81024"]]
 å # Check for each if it's in the list of squares
 # → [[1,1,0,1,1],[1,1,0,0],...,[0,1],[0]]
 P # Check for each inner list whether all are truthy
 # → [0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0]
 O # Take the sum
 # → 2
 Ā # Truthify (and output implicitly)
 # → 1 (truthy)

share

answered 5 mins ago

Kevin Cruijssen

31.5k553171

05AB1E, 9 bytes

Ýos.œåPOĀ

Try it online or verify all test cases. (NOTE: The т in the header is 100 to only get the first 100 square numbers, instead of the first input amount of square numbers. It works in the second case as well, but is pretty inefficient and might time-out on TIO.)

Explanation:

Ý # Create a list in the range [0,n], where n is the (implicit) input
 # (or 100 in the TIO)
 # i.e. 81024 → [0,1,2,3,...,81024]
 o # Square each
 # → [1,2,4,8,...,451..216 (nr with 24391 digits)]
 s # Swap to take the input
 .œ # Create each possible partition of this input
 # i.e. 81024 → [["8","1","0","2","4"],["8","1","0","24"],...,["8102","4"],["81024"]]
 å # Check for each if it's in the list of squares
 # → [[1,1,0,1,1],[1,1,0,0],...,[0,1],[0]]
 P # Check for each inner list whether all are truthy
 # → [0,0,1,0,0,0,0,1,0,0,0,0,0,0,0,0]
 O # Take the sum
 # → 2
 Ā # Truthify (and output implicitly)
 # → 1 (truthy)

share

answered 5 mins ago

Kevin Cruijssen

31.5k553171

share

share

answered 5 mins ago

Kevin Cruijssen

31.5k553171

answered 5 mins ago

Kevin Cruijssen

31.5k553171

answered 5 mins ago

Kevin Cruijssen

31.5k553171

31.5k553171

add a comment | 

add a comment | 

 

draft saved

draft discarded

 

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fcodegolf.stackexchange.com%2fquestions%2f173833%2fcan-the-number-be-splitted-into-powers-of-2%23new-answer', 'question_page');

);

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Sign up or log in

StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name Email

Name Email

Name

Email

Name Email

Name

Email