Can the number be split into powers of 2?

Clash Royale CLAN TAG#URR8PPP

up vote
6
down vote

favorite

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

So we have $$4281$$ that can be split 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 split 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 8-1-02-4)
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 11 mins ago

Luis Mendo

73.1k885285

asked 1 hour ago

Charlie

6,8371979

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  56 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
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just leaving this here as tip for everyone. Here are three possible ways to check if a number is a power of 2: 1) Check if log2(n) doesn't contain decimal digits after the comma. 2) Check if n AND (n-1) == 0. 3) Create a list of square-nrs and check if n is in that list.
  – Kevin Cruijssen
  37 mins ago
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

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

So we have $$4281$$ that can be split 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 split 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 8-1-02-4)
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 11 mins ago

Luis Mendo

73.1k885285

asked 1 hour ago

Charlie

6,8371979

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  56 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
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just leaving this here as tip for everyone. Here are three possible ways to check if a number is a power of 2: 1) Check if log2(n) doesn't contain decimal digits after the comma. 2) Check if n AND (n-1) == 0. 3) Create a list of square-nrs and check if n is in that list.
  – Kevin Cruijssen
  37 mins ago
  
  

  
  
  
  

add a comment | 

up vote
6
down vote

favorite

up vote
6
down vote

favorite

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

So we have $$4281$$ that can be split 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 split 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 8-1-02-4)
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 11 mins ago

Luis Mendo

73.1k885285

asked 1 hour ago

Charlie

6,8371979

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

So we have $$4281$$ that can be split 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 split 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 8-1-02-4)
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 11 mins ago

Luis Mendo

73.1k885285

asked 1 hour ago

Charlie

6,8371979

share|improve this question

edited 11 mins ago

Luis Mendo

73.1k885285

asked 1 hour ago

Charlie

6,8371979

share|improve this question

share|improve this question

edited 11 mins ago

Luis Mendo

73.1k885285

edited 11 mins ago

Luis Mendo

73.1k885285

edited 11 mins ago

Luis Mendo

73.1k885285

73.1k885285

asked 1 hour ago

Charlie

6,8371979

asked 1 hour ago

Charlie

6,8371979

asked 1 hour ago

Charlie

6,8371979

6,8371979

• 
  
  
  

  
  

  
  
  
  
  
  

  
  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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  56 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
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just leaving this here as tip for everyone. Here are three possible ways to check if a number is a power of 2: 1) Check if log2(n) doesn't contain decimal digits after the comma. 2) Check if n AND (n-1) == 0. 3) Create a list of square-nrs and check if n is in that list.
  – Kevin Cruijssen
  37 mins ago
  
  

  
  
  
  

add a 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
  1 hour 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
  1 hour ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
  – Shieru Asakoto
  56 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
  53 mins ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Just leaving this here as tip for everyone. Here are three possible ways to check if a number is a power of 2: 1) Check if log2(n) doesn't contain decimal digits after the comma. 2) Check if n AND (n-1) == 0. 3) Create a list of square-nrs and check if n is in that list.
  – Kevin Cruijssen
  37 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
1 hour 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
1 hour 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
1 hour 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
1 hour ago

2

2

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

Suggested test case: 101 (falsy because of the 0) ... or should this still be true (1 - 01)?
– Shieru Asakoto
56 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
53 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
53 mins ago

1

1

Just leaving this here as tip for everyone. Here are three possible ways to check if a number is a power of 2: 1) Check if log2(n) doesn't contain decimal digits after the comma. 2) Check if n AND (n-1) == 0. 3) Create a list of square-nrs and check if n is in that list.
– Kevin Cruijssen
37 mins ago

Just leaving this here as tip for everyone. Here are three possible ways to check if a number is a power of 2: 1) Check if log2(n) doesn't contain decimal digits after the comma. 2) Check if n AND (n-1) == 0. 3) Create a list of square-nrs and check if n is in that list.
– Kevin Cruijssen
37 mins ago

add a comment | 

6 Answers
6

active

oldest

votes

up vote
3
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|improve this answer

answered 56 mins ago

Kevin Cruijssen

31.6k553171

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
  – Mr. Xcoder
  33 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

JavaScript (Node.js), ₆₉₆₄ 58 bytes

f=(x,m=10,q=!(x%m&x%m-1|!x))=>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.

-11 bytes by golfing the power-of-2 check.

share|improve this answer

edited 44 mins ago

answered 1 hour ago

Bubbler

3,937541

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 1 hour ago

TFeld

12.4k2834

add a comment | 

up vote
2
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 1 hour ago

ovs

17.7k21058

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 75 69 bytes

-6 bytes thanks @Arnauld. At most 32-bit support

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

Try it online!

Input as a string.

share|improve this answer

edited 46 mins ago

answered 1 hour ago

Shieru Asakoto

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
1
down vote

Jelly, 9 bytes

ŒṖḌl2ĊƑ€Ẹ

Check out the test suite!

How?

ŒṖḌl2ĊƑ€Ẹ Full program. N = integer input.
ŒṖ All possible partitions of the digits of N.
 Ḍ Undecimal (i.e. join to numbers).
 l2 Log2. Note: returns -(-inf+nanj) for 0, so it doesn't fail.
 ĊƑ€ For each, check if the logarithm equals its ceil.
 Ẹ Any. Return 0 if there are no truthy elements, 1 otherwise.

share|improve this answer

edited 2 mins ago

answered 10 mins ago

Mr. Xcoder

30.8k758195

add a comment | 

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6 Answers
6

active

oldest

votes

6 Answers
6

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
3
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|improve this answer

answered 56 mins ago

Kevin Cruijssen

31.6k553171

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
  – Mr. Xcoder
  33 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
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|improve this answer

answered 56 mins ago

Kevin Cruijssen

31.6k553171

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
  – Mr. Xcoder
  33 mins ago
  
  

  
  
  
  

add a comment | 

up vote
3
down vote

up vote
3
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|improve this answer

answered 56 mins ago

Kevin Cruijssen

31.6k553171

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|improve this answer

answered 56 mins ago

Kevin Cruijssen

31.6k553171

share|improve this answer

share|improve this answer

answered 56 mins ago

Kevin Cruijssen

31.6k553171

answered 56 mins ago

Kevin Cruijssen

31.6k553171

answered 56 mins ago

Kevin Cruijssen

31.6k553171

31.6k553171

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
  – Mr. Xcoder
  33 mins ago
  
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
  – Mr. Xcoder
  33 mins ago
  
  

  
  
  
  

1

1

Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
– Mr. Xcoder
33 mins ago

Nice, my solution was .œ.²1%O0å (9 bytes as well). Mine failed 0, however.
– Mr. Xcoder
33 mins ago

add a comment | 

up vote
3
down vote

JavaScript (Node.js), ₆₉₆₄ 58 bytes

f=(x,m=10,q=!(x%m&x%m-1|!x))=>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.

-11 bytes by golfing the power-of-2 check.

share|improve this answer

edited 44 mins ago

answered 1 hour ago

Bubbler

3,937541

add a comment | 

up vote
3
down vote

JavaScript (Node.js), ₆₉₆₄ 58 bytes

f=(x,m=10,q=!(x%m&x%m-1|!x))=>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.

-11 bytes by golfing the power-of-2 check.

share|improve this answer

edited 44 mins ago

answered 1 hour ago

Bubbler

3,937541

add a comment | 

up vote
3
down vote

up vote
3
down vote

JavaScript (Node.js), ₆₉₆₄ 58 bytes

f=(x,m=10,q=!(x%m&x%m-1|!x))=>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.

-11 bytes by golfing the power-of-2 check.

share|improve this answer

edited 44 mins ago

answered 1 hour ago

Bubbler

3,937541

JavaScript (Node.js), ₆₉₆₄ 58 bytes

f=(x,m=10,q=!(x%m&x%m-1|!x))=>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.

-11 bytes by golfing the power-of-2 check.

share|improve this answer

edited 44 mins ago

answered 1 hour ago

Bubbler

3,937541

share|improve this answer

share|improve this answer

edited 44 mins ago

edited 44 mins ago

edited 44 mins ago

answered 1 hour ago

Bubbler

3,937541

answered 1 hour ago

Bubbler

3,937541

answered 1 hour ago

Bubbler

3,937541

3,937541

add a comment | 

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 1 hour 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 1 hour 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 1 hour 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 1 hour ago

TFeld

12.4k2834

share|improve this answer

share|improve this answer

answered 1 hour ago

TFeld

12.4k2834

answered 1 hour ago

TFeld

12.4k2834

answered 1 hour ago

TFeld

12.4k2834

12.4k2834

add a comment | 

add a comment | 

up vote
2
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 1 hour ago

ovs

17.7k21058

add a comment | 

up vote
2
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 1 hour ago

ovs

17.7k21058

add a comment | 

up vote
2
down vote

up vote
2
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 1 hour ago

ovs

17.7k21058

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 1 hour ago

ovs

17.7k21058

share|improve this answer

share|improve this answer

answered 1 hour ago

ovs

17.7k21058

answered 1 hour ago

ovs

17.7k21058

answered 1 hour ago

ovs

17.7k21058

17.7k21058

add a comment | 

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 75 69 bytes

-6 bytes thanks @Arnauld. At most 32-bit support

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

Try it online!

Input as a string.

share|improve this answer

edited 46 mins ago

answered 1 hour ago

Shieru Asakoto

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
2
down vote

JavaScript (Node.js), 75 69 bytes

-6 bytes thanks @Arnauld. At most 32-bit support

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

Try it online!

Input as a string.

share|improve this answer

edited 46 mins ago

answered 1 hour ago

Shieru Asakoto

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

JavaScript (Node.js), 75 69 bytes

-6 bytes thanks @Arnauld. At most 32-bit support

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

Try it online!

Input as a string.

share|improve this answer

edited 46 mins ago

answered 1 hour ago

Shieru Asakoto

1,980312

JavaScript (Node.js), 75 69 bytes

-6 bytes thanks @Arnauld. At most 32-bit support

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

Try it online!

Input as a string.

share|improve this answer

edited 46 mins ago

answered 1 hour ago

Shieru Asakoto

1,980312

share|improve this answer

share|improve this answer

edited 46 mins ago

edited 46 mins ago

edited 46 mins ago

answered 1 hour ago

Shieru Asakoto

1,980312

answered 1 hour ago

Shieru Asakoto

1,980312

answered 1 hour ago

Shieru Asakoto

1,980312

1,980312

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  

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

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

add a comment | 

up vote
1
down vote

Jelly, 9 bytes

ŒṖḌl2ĊƑ€Ẹ

Check out the test suite!

How?

ŒṖḌl2ĊƑ€Ẹ Full program. N = integer input.
ŒṖ All possible partitions of the digits of N.
 Ḍ Undecimal (i.e. join to numbers).
 l2 Log2. Note: returns -(-inf+nanj) for 0, so it doesn't fail.
 ĊƑ€ For each, check if the logarithm equals its ceil.
 Ẹ Any. Return 0 if there are no truthy elements, 1 otherwise.

share|improve this answer

edited 2 mins ago

answered 10 mins ago

Mr. Xcoder

30.8k758195

add a comment | 

up vote
1
down vote

Jelly, 9 bytes

ŒṖḌl2ĊƑ€Ẹ

Check out the test suite!

How?

ŒṖḌl2ĊƑ€Ẹ Full program. N = integer input.
ŒṖ All possible partitions of the digits of N.
 Ḍ Undecimal (i.e. join to numbers).
 l2 Log2. Note: returns -(-inf+nanj) for 0, so it doesn't fail.
 ĊƑ€ For each, check if the logarithm equals its ceil.
 Ẹ Any. Return 0 if there are no truthy elements, 1 otherwise.

share|improve this answer

edited 2 mins ago

answered 10 mins ago

Mr. Xcoder

30.8k758195

add a comment | 

up vote
1
down vote

up vote
1
down vote

Jelly, 9 bytes

ŒṖḌl2ĊƑ€Ẹ

Check out the test suite!

How?

ŒṖḌl2ĊƑ€Ẹ Full program. N = integer input.
ŒṖ All possible partitions of the digits of N.
 Ḍ Undecimal (i.e. join to numbers).
 l2 Log2. Note: returns -(-inf+nanj) for 0, so it doesn't fail.
 ĊƑ€ For each, check if the logarithm equals its ceil.
 Ẹ Any. Return 0 if there are no truthy elements, 1 otherwise.

share|improve this answer

edited 2 mins ago

answered 10 mins ago

Mr. Xcoder

30.8k758195

Jelly, 9 bytes

ŒṖḌl2ĊƑ€Ẹ

Check out the test suite!

How?

ŒṖḌl2ĊƑ€Ẹ Full program. N = integer input.
ŒṖ All possible partitions of the digits of N.
 Ḍ Undecimal (i.e. join to numbers).
 l2 Log2. Note: returns -(-inf+nanj) for 0, so it doesn't fail.
 ĊƑ€ For each, check if the logarithm equals its ceil.
 Ẹ Any. Return 0 if there are no truthy elements, 1 otherwise.

share|improve this answer

edited 2 mins ago

answered 10 mins ago

Mr. Xcoder

30.8k758195

share|improve this answer

share|improve this answer

edited 2 mins ago

edited 2 mins ago

edited 2 mins ago

answered 10 mins ago

Mr. Xcoder

30.8k758195

answered 10 mins ago

Mr. Xcoder

30.8k758195

answered 10 mins ago

Mr. Xcoder

30.8k758195

30.8k758195

add a comment | 

add a comment | 

 

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