The input is a single positive integer n
The output isn with its most significant bit set to 0.
1 -> 0
2 -> 0
10 -> 2
16 -> 0
100 -> 36
267 -> 11
350 -> 94
500 -> 244
For example: 350 in binary is 101011110. Setting its most significant bit (i.e. the leftmost 1 bit) to 0 turns it into 001011110 which is equivalent to the decimal integer 94, the output. This is OEIS A053645.
i<=n with n/i for -1 byte. This isn't my golf, someone else tried to edit it into your post but I rolled it back because edits for golfing posts are not accepted according to our community rules.
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lambda n:n^2**len(bin(n))/8
Unfortunately, this does not work for 1:
lambda n:int(bin(n)[3:],2)
.²óo-
Removing the most significant bit from an integer N is equivalent to finding the distance from N to the highest integer power of 2 lower than N.
Thus, I used the formula N - 2floor(log2N):
.² - Logarithm with base 2.ó - Floor to an integer.o - 2 raised to the power of the result above.- - Difference.
b¦C also works... doesn't it? Convert to binary, MSB is always at index 1, remove MSB, convert back.
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Nov 15, 2017 at 12:01
1!
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Nov 15, 2017 at 12:15
b¦C works for 1 in the new 05AB1E version. Although I like your .²óo- more tbh. :)
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Sep 22, 2022 at 14:05
BḊḄ
BḊḄ Main Link
B Convert to binary
Ḋ Dequeue; remove the first element
Ḅ Convert from binary
Ḅ and Ḋ two-byte codepoints? This would change the overall size to 5 bytes.
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Nov 15, 2017 at 12:30
main(i){scanf("%d",&i);return i&~(1<<31-__builtin_clz(i));}
This gcc answer uses only integer bitwise and arithmetic operations. No logarithms here! It may have issues with an input of 0, and is totally non-portable.
It's my first answer on this site, so I'd love feedback and improvements. I sure had fun with learning bitwise expressions.
main, a function is a valid submission. Changing this into a function and taking input as an argument to said function saves 18 bytes.
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Nov 17, 2017 at 16:16
n->n^1<<(int)Math.log2(n) will work and is likely shorter than 38 bytes. It was my second (yet untested) idea, if the highestOneBit one didn't work appropriately. Out of curiosity, what was your solution
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Nov 15, 2017 at 20:54
n->n^1<<(int)(Math.log(n)/Math.log(2)) because Math.log2 doesn't exist in Java. ;P Only Math.log, Math.log10 and Math.loglp are available.
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Nov 16, 2017 at 7:57
Math.log2 doesn't exist indeed... My bad. See? One nice method (highestOneBit) exists but not another one (Math.log2). Java is weird ;-)
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Nov 16, 2017 at 8:41
B0T(XB
Saved two bytes thanks to Cinaski. Switching to assignment indexing instead of reference indexing was 2 bytes shorter :)
% Grab input implicitly: 267
B % Convert to binary: [1 0 0 0 0 1 0 1 1]
0T( % Set the first value to 0: [0 0 0 0 0 1 0 1 1]
XB % Convert to decimal: 11
4L rather than [2J]. Another fun 6 bytes: tZlcW- (only works in MATLAB, not in TIO/Octave)
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Nov 15, 2017 at 9:19
ḋtḋ
Explanation:
-- implicit input, e.g. 350
ḋ -- convert number to list of binary digits (TNum -> [TNum]): [1,0,1,0,1,1,1,1,0]
t -- remove first element: [0,1,0,1,1,1,1,0]
ḋ -- convert list of binary digits to number ([TNum] -> TNum): 94
lambda n:n-2**len(bin(n))/8
lambda n:n-2**len(bin(n))/8 # Lambda Function: takes `n` as an argument
lambda n: # Declaration of Lambda Function
len(bin(n)) # Number of bits + 2
2** # 2 ** this ^
/8 # Divide by 8 because of the extra characters in the binary representation
n- # Subtract this from the original
2**len(bin(n))/8 can also be spelled 1<<len(bin(n))-3, and then it will work in both 2 and 3 (no bytes saved/added).
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-8 bytes thanks to caird coinheringaahing. I typed that from memory. :o
lambda n:int('0'+bin(n)[3:],2)
function(x)x-2^(log2(x)%/%1)
Easiest to calculate the most significant bit via 2 ^ floor(log2(x)) rather than carry out base conversions, which are quite verbose in R
Byte code:
0F BD C8 0F B3 C8 C3Disassembly:
bsr ecx, eax
btr eax, ecx
retaccepts and returns the value in the eax register.
Perform a reverse scan for the first set bit, and then reset that bit.
Saved 2 bytes thanks to ovs
a=>a^1<<Math.log2(a)
a=>'0b'+a.toString`2`.slice`1`^0
.slice`1`^0 when .slice(1)^0 would work just as well, haha
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Nov 15, 2017 at 0:19
}.&.#:
Pretty simple.
}.&.#:
#: convert to list of binary digits
&. apply right function, then left, then the inverse of right
}. behead
Tacit prefix function.
2⊥1↓2∘⊥⍣¯1
2∘⊥… decode from base-2…
…⍣¯1 negative one time (i.e. encode in base-2)
1↓ drop the first bit
2⊥ decode from base-2
-7 Bytes thanks to Ventero. -2 Bytes thanks to historicrat.
->n{/./=~'%b'%n;$'.to_i 2}
->n{n.to_s(2)[1..-1].to_i 2}
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-1 is not required: ->n{n.to_s(2)[1..].to_i 2}
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Built-in in gcc used.
f(c){return c^1<<31-__builtin_clz(c);}
31- with ~ should save two bytes.
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Excel, 20 bytes
=A1-2^INT(LOG(A1,2))
-5 bytes thanks to @IanM_Matrix1
=BIN2DEC(MID(DEC2BIN(A1),2,99))
Nothing interesting.
BASE(A1,2) instead of DEC2BIN(A1)
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Nov 24, 2020 at 13:45
BASE.
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(You can omit destination register on add when same as source)
clz x1,x0
add x1,1
lsl x0,x1
lsr x0,x1
ret
shr/shl/ret and wants instead something like lsr/lsl/bx lr.
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a^2sl
Explanation:
l Log base 2 of input.
s Cast ^ to integer (this is the position of the most significant bit.)
^2 Raise 2 to ^ (get the value of said bit)
a Subtract ^ from input
./-l
o@i
. Duplicate an implicit zero at the bottom of the stack. Does nothing.
/ Switch to Ordinal mode, move SE.
i Read all input as a string.
l Convert to lower case (does nothing, because the input doesn't contain letters).
i Try reading all input again, pushes an empty string.
/ Switch to Cardinal mode, move W.
. Duplicate. Since we're in Cardinal mode, this tries to duplicate an integer.
To get an integer, the empty string is discarded implicitly and the input is
converted to the integer value it represents. Therefore, at the end of this,
we get two copies of the integer value that was input.
l Clear lower bits. This sets all bits except the MSB to zero.
- Subtract. By subtracting the MSB from the input, we set it to zero. We could
also use XOR here.
/ Switch to Ordinal, move NW (and immediately reflect to SW).
o Implicitly convert the result to a string and print it.
/ Switch to Ordinal, move S.
@ Terminate the program.
#-2^Floor@Log2@#&
This is my first Mathematica answer, feel free to tell me what have I screwed up.
-4 bytes thanks to @HyperNeutrino!
So as it turns out, someone made a similar program before, and sent it to the OEIS. However, keep in mind that the floor of a logarithm is basically defined as the number of digits of a number. This is just a coincidence, or rather a task simple enough that many people will get the same answer.
() is apparently unnecessary
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#~BitClear~-1& which would be 14 bytes. Seems like the natural extension of the syntax. Maybe in version 12.
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Nov 14, 2017 at 20:07
^2p¢ÊÉ
^2p¢ÊÉ
¢ Get binary form of input
Ê Get length of that
É Subtract 1
2p Raise 2 to the power of that
^ XOR with the input
1 can fail: 4 bytes¢Ån2
Explanation: get input binary (¢), slice off first char (Å), parse as binary back to a number (n2).
{2b()b}
Explanation:
{ } Block: 267
2b Binary: [1 0 0 0 0 1 0 1 1]
( Pop: [0 0 0 0 1 0 1 1] 1
) Increment: [0 0 0 0 1 0 1 1] 2
b Base convert: 11
Reuse the MSB (which is always 1) to avoid having to delete it; the equivalent without that trick would be {2b1>2b} or {2b(;2b}.
^(^1|\1\1)*1
Input and output in unary (the test suite includes conversion from and to decimal for convenience).
This is quite easy to do in unary. All we want to do is delete the largest power of 2 from the input. We can match a power of 2 with some forward references. It's actually easier to match values of the form 2n-1, so we'll do that and match one 1 separately:
^(^1|\1\1)*1
The group 1 either matches a single 1 at the beginning to kick things off, or it matches twice what it did on the last iteration. So it matches 1, then 2, then 4 and so on. Since these get added up, we're always one short of a power of 2, which we fix with the 1 at the end.
Due the trailing linefeed, the match is simply removed from the input.
10obviously gives0:D \$\endgroup\$