Compare four integers, return word based on maximum
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This function should take four integer inputs (a
,b
,c
,d
) and return a binary word based on which values equal the maximum of the four.
The return value will be between 1
and 0xF
.
For example:
a = 6, b = 77, c = 1, d = 4
returns 2
(binary 0010
; only 2nd-least significant bit is set corresponding to b
being sole max value)
a = 4, b = 5, c = 10, d = 10
returns 0xC
(binary 1100
; 3rd- and 4th-least significant bits set corresponding to c
and d
equaling max value)
a = 1, b = 1, c = 1, d = 1
returns 0xF
(binary 1111
; all four bits set because all values equal the max)
Here is a simple implementation:
int getWord(int a, int b, int c, int d)
{
int max = a;
int word = 1;
if (b > max)
{
max = b;
word = 2;
}
else if (b == max)
{
word |= 2;
}
if (c > max)
{
max = c;
word = 4;
}
else if (c == max)
{
word |= 4;
}
if (d > max)
{
word = 8;
}
else if (d == max)
{
word |= 8;
}
return word;
}
return value can be string of 0's and 1's, bool / bit vector, or integer
code-golf bitwise
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|
show 7 more comments
$begingroup$
This function should take four integer inputs (a
,b
,c
,d
) and return a binary word based on which values equal the maximum of the four.
The return value will be between 1
and 0xF
.
For example:
a = 6, b = 77, c = 1, d = 4
returns 2
(binary 0010
; only 2nd-least significant bit is set corresponding to b
being sole max value)
a = 4, b = 5, c = 10, d = 10
returns 0xC
(binary 1100
; 3rd- and 4th-least significant bits set corresponding to c
and d
equaling max value)
a = 1, b = 1, c = 1, d = 1
returns 0xF
(binary 1111
; all four bits set because all values equal the max)
Here is a simple implementation:
int getWord(int a, int b, int c, int d)
{
int max = a;
int word = 1;
if (b > max)
{
max = b;
word = 2;
}
else if (b == max)
{
word |= 2;
}
if (c > max)
{
max = c;
word = 4;
}
else if (c == max)
{
word |= 4;
}
if (d > max)
{
word = 8;
}
else if (d == max)
{
word |= 8;
}
return word;
}
return value can be string of 0's and 1's, bool / bit vector, or integer
code-golf bitwise
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2
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I have a solution in a golfing language, which uses the builtins Reverse, Maximum, Equality-check, Join, Convert from binary to integer, Convert from integer to hexadecimal. Does this means my score is 1 due to the Equality-check? I have the feeling this is too much focused on regular languages, and even for those it's not 100% clear what the scoring for let's say a Maximum-builtin.. :S
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– Kevin Cruijssen
Mar 4 at 7:39
1
$begingroup$
I would suggest you try to: 1. change this question to code-golf, which only care about the number of bytes. 2. or, restrict to some certain language (certain version of compiler/interpreter please), and list all statements and operators allowed, and how to score them.
$endgroup$
– tsh
Mar 4 at 7:43
5
$begingroup$
1 is a better option, IMO. I think this makes a perfectly good code-golf question and I can't see any benefit that would come from restricting the languages available for answers
$endgroup$
– senox13
Mar 4 at 7:54
2
$begingroup$
I updated my question to remove the criteria. Let me know it it's still unclear
$endgroup$
– Mr Anderson
Mar 4 at 8:09
5
$begingroup$
Should I output a decimal number? Or may I output 4 binary digits instead?
$endgroup$
– tsh
Mar 4 at 8:51
|
show 7 more comments
$begingroup$
This function should take four integer inputs (a
,b
,c
,d
) and return a binary word based on which values equal the maximum of the four.
The return value will be between 1
and 0xF
.
For example:
a = 6, b = 77, c = 1, d = 4
returns 2
(binary 0010
; only 2nd-least significant bit is set corresponding to b
being sole max value)
a = 4, b = 5, c = 10, d = 10
returns 0xC
(binary 1100
; 3rd- and 4th-least significant bits set corresponding to c
and d
equaling max value)
a = 1, b = 1, c = 1, d = 1
returns 0xF
(binary 1111
; all four bits set because all values equal the max)
Here is a simple implementation:
int getWord(int a, int b, int c, int d)
{
int max = a;
int word = 1;
if (b > max)
{
max = b;
word = 2;
}
else if (b == max)
{
word |= 2;
}
if (c > max)
{
max = c;
word = 4;
}
else if (c == max)
{
word |= 4;
}
if (d > max)
{
word = 8;
}
else if (d == max)
{
word |= 8;
}
return word;
}
return value can be string of 0's and 1's, bool / bit vector, or integer
code-golf bitwise
$endgroup$
This function should take four integer inputs (a
,b
,c
,d
) and return a binary word based on which values equal the maximum of the four.
The return value will be between 1
and 0xF
.
For example:
a = 6, b = 77, c = 1, d = 4
returns 2
(binary 0010
; only 2nd-least significant bit is set corresponding to b
being sole max value)
a = 4, b = 5, c = 10, d = 10
returns 0xC
(binary 1100
; 3rd- and 4th-least significant bits set corresponding to c
and d
equaling max value)
a = 1, b = 1, c = 1, d = 1
returns 0xF
(binary 1111
; all four bits set because all values equal the max)
Here is a simple implementation:
int getWord(int a, int b, int c, int d)
{
int max = a;
int word = 1;
if (b > max)
{
max = b;
word = 2;
}
else if (b == max)
{
word |= 2;
}
if (c > max)
{
max = c;
word = 4;
}
else if (c == max)
{
word |= 4;
}
if (d > max)
{
word = 8;
}
else if (d == max)
{
word |= 8;
}
return word;
}
return value can be string of 0's and 1's, bool / bit vector, or integer
code-golf bitwise
code-golf bitwise
edited Mar 4 at 19:04
Mr Anderson
asked Mar 4 at 6:50
Mr AndersonMr Anderson
1485
1485
2
$begingroup$
I have a solution in a golfing language, which uses the builtins Reverse, Maximum, Equality-check, Join, Convert from binary to integer, Convert from integer to hexadecimal. Does this means my score is 1 due to the Equality-check? I have the feeling this is too much focused on regular languages, and even for those it's not 100% clear what the scoring for let's say a Maximum-builtin.. :S
$endgroup$
– Kevin Cruijssen
Mar 4 at 7:39
1
$begingroup$
I would suggest you try to: 1. change this question to code-golf, which only care about the number of bytes. 2. or, restrict to some certain language (certain version of compiler/interpreter please), and list all statements and operators allowed, and how to score them.
$endgroup$
– tsh
Mar 4 at 7:43
5
$begingroup$
1 is a better option, IMO. I think this makes a perfectly good code-golf question and I can't see any benefit that would come from restricting the languages available for answers
$endgroup$
– senox13
Mar 4 at 7:54
2
$begingroup$
I updated my question to remove the criteria. Let me know it it's still unclear
$endgroup$
– Mr Anderson
Mar 4 at 8:09
5
$begingroup$
Should I output a decimal number? Or may I output 4 binary digits instead?
$endgroup$
– tsh
Mar 4 at 8:51
|
show 7 more comments
2
$begingroup$
I have a solution in a golfing language, which uses the builtins Reverse, Maximum, Equality-check, Join, Convert from binary to integer, Convert from integer to hexadecimal. Does this means my score is 1 due to the Equality-check? I have the feeling this is too much focused on regular languages, and even for those it's not 100% clear what the scoring for let's say a Maximum-builtin.. :S
$endgroup$
– Kevin Cruijssen
Mar 4 at 7:39
1
$begingroup$
I would suggest you try to: 1. change this question to code-golf, which only care about the number of bytes. 2. or, restrict to some certain language (certain version of compiler/interpreter please), and list all statements and operators allowed, and how to score them.
$endgroup$
– tsh
Mar 4 at 7:43
5
$begingroup$
1 is a better option, IMO. I think this makes a perfectly good code-golf question and I can't see any benefit that would come from restricting the languages available for answers
$endgroup$
– senox13
Mar 4 at 7:54
2
$begingroup$
I updated my question to remove the criteria. Let me know it it's still unclear
$endgroup$
– Mr Anderson
Mar 4 at 8:09
5
$begingroup$
Should I output a decimal number? Or may I output 4 binary digits instead?
$endgroup$
– tsh
Mar 4 at 8:51
2
2
$begingroup$
I have a solution in a golfing language, which uses the builtins Reverse, Maximum, Equality-check, Join, Convert from binary to integer, Convert from integer to hexadecimal. Does this means my score is 1 due to the Equality-check? I have the feeling this is too much focused on regular languages, and even for those it's not 100% clear what the scoring for let's say a Maximum-builtin.. :S
$endgroup$
– Kevin Cruijssen
Mar 4 at 7:39
$begingroup$
I have a solution in a golfing language, which uses the builtins Reverse, Maximum, Equality-check, Join, Convert from binary to integer, Convert from integer to hexadecimal. Does this means my score is 1 due to the Equality-check? I have the feeling this is too much focused on regular languages, and even for those it's not 100% clear what the scoring for let's say a Maximum-builtin.. :S
$endgroup$
– Kevin Cruijssen
Mar 4 at 7:39
1
1
$begingroup$
I would suggest you try to: 1. change this question to code-golf, which only care about the number of bytes. 2. or, restrict to some certain language (certain version of compiler/interpreter please), and list all statements and operators allowed, and how to score them.
$endgroup$
– tsh
Mar 4 at 7:43
$begingroup$
I would suggest you try to: 1. change this question to code-golf, which only care about the number of bytes. 2. or, restrict to some certain language (certain version of compiler/interpreter please), and list all statements and operators allowed, and how to score them.
$endgroup$
– tsh
Mar 4 at 7:43
5
5
$begingroup$
1 is a better option, IMO. I think this makes a perfectly good code-golf question and I can't see any benefit that would come from restricting the languages available for answers
$endgroup$
– senox13
Mar 4 at 7:54
$begingroup$
1 is a better option, IMO. I think this makes a perfectly good code-golf question and I can't see any benefit that would come from restricting the languages available for answers
$endgroup$
– senox13
Mar 4 at 7:54
2
2
$begingroup$
I updated my question to remove the criteria. Let me know it it's still unclear
$endgroup$
– Mr Anderson
Mar 4 at 8:09
$begingroup$
I updated my question to remove the criteria. Let me know it it's still unclear
$endgroup$
– Mr Anderson
Mar 4 at 8:09
5
5
$begingroup$
Should I output a decimal number? Or may I output 4 binary digits instead?
$endgroup$
– tsh
Mar 4 at 8:51
$begingroup$
Should I output a decimal number? Or may I output 4 binary digits instead?
$endgroup$
– tsh
Mar 4 at 8:51
|
show 7 more comments
21 Answers
21
active
oldest
votes
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Jelly, 2 bytes
Takes input as [d,c,b,a]
. Returns a list of Booleans.
Ṁ=
Try it online!
Ṁ
Maximum
=
equal to (implies the other argument is the original argument; vectorises)
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add a comment |
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R, 17 bytes
max(a<-scan())==a
Try it online!
Returns a vector of booleans. Since this output has been confirmed, this is preferable over numerical output, as that one is almost twice longer:
R, 33 bytes
sum(2^which(max(a<-scan())==a)/2)
Try it online!
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add a comment |
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APL (Dyalog Unicode), 4 bytesSBCS
Anonymous tacit prefix function. Takes [a,b,c,d]
as argument. Returns a bit-Boolean array.*
⌈/=⌽
Try it online!
⌈/
Does the maximum of the argument
=
equal (vectorises)
⌽
the reverse of the argument?
* Note that APL stores arrays of Booleans using one bit per value, so this does indeed return a 4-bit word, despite the display form being 0 0 1 0
.
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add a comment |
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Haskell, 20 18 bytes
2 bytes saved thanks to proud haskeller
map=<<(==).maximum
Try it online!
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You can output a string.
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– Adám
Mar 5 at 5:44
1
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Writingmap
instead of(<$>)
would be two bytes shorter!
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– proud haskeller
Mar 7 at 6:23
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@proudhaskeller Good catch. Can't believe that I didn't see that one.
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– Sriotchilism O'Zaic
Mar 7 at 13:40
add a comment |
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Perl 6, 12 bytes
{$_ X==.max}
Try it online!
Anonymous code block that takes a list of integers and returns a list of booleans. If we need to return as a number, it's +4 bytes to wrap the inside of the code block with 2:[...]
.
Explanation:
{ } # Anonymous code block
$_ # With the input
X== # Which values are equal
.max # To the maximum element
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OP now says you don't need to wrap.
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– Adám
Mar 5 at 5:42
add a comment |
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Japt, 5
m¶Urw
Try it!
-4 bytes thanks to @Oliver!
-2 bytes thanks to @Shaggy!
Input is a 4-element array in the following format:
[d, c, b, a]
Output is an array of bits.
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Of course there is ;) There are apparently a lot of shortcuts to learn.
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– dana
Mar 4 at 15:52
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If a boolean array is an acceptable output, this can be 7 bytes
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– Oliver
Mar 4 at 15:56
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@Oliver, 5 bytes ;)
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– Shaggy
Mar 4 at 20:29
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You guys are pretty good :) That's interesting howrw
converts tor("w")
does a reduce by repeatedly getting the max. Same withm¶
getting converted toU.m("===", ...)
. In any case, thanks for the tips!
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– dana
Mar 4 at 23:54
add a comment |
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x86 machine code (MMX/SSE1), 26 bytes (4x int16_t)
x86 machine code (SSE4.1), 28 bytes (4x int32_t or uint32_t)
x86 machine code (SSE2), 24 bytes (4x float32) or 27B to cvt int32
(The last version that converts int32 to float isn't perfectly accurate for large integers that round to the same float. With float input, rounding is the caller's problem and this function works correctly if there are no NaNs, identifying floats that compare == to the max. The integer versions work for all inputs, treating them as signed 2's complement.)
All of these work in 16/32/64-bit mode with the same machine code.
A stack-args calling convention would make it possible to loop over the args twice (finding max and then comparing), possibly giving us a smaller implementation, but I haven't tried that approach.
x86 SIMD has vector->integer bitmap as a single instruction (pmovmskb
or movmskps
or pd), so it was natural for this even though MMX/SSE instructions are at least 3 bytes long. SSSE3 and later instructions are longer than SSE2, and MMX/SSE1 instructions are the shortest. Different versions of pmax*
(packed-integer vertical max) were introduced at different times, with SSE1 (for mmx regs) and SSE2 (for xmm regs) only having signed word (16-bit) and unsigned byte.
(pshufw
and pmaxsw
on MMX registers are new with Katmai Pentium III, so really they require SSE1, not just the MMX CPU feature bit.)
This is callable from C as unsigned max4_mmx(__m64)
with the i386 System V ABI, which passes an __m64
arg in mm0
. (Not x86-64 System V, which passes __m64
in xmm0
!)
line code bytes
num addr
1 global max4_mmx
2 ;; Input 4x int16_t in mm0
3 ;; output: bitmap in EAX
4 ;; clobbers: mm1, mm2
5 max4_mmx:
6 00000000 0F70C8B1 pshufw mm1, mm0, 0b10110001 ; swap adjacent pairs
7 00000004 0FEEC8 pmaxsw mm1, mm0
8
9 00000007 0F70D14E pshufw mm2, mm1, 0b01001110 ; swap high/low halves
10 0000000B 0FEECA pmaxsw mm1, mm2
11
12 0000000E 0F75C8 pcmpeqw mm1, mm0 ; 0 / -1
13 00000011 0F63C9 packsswb mm1, mm1 ; squish word elements to bytes, preserving sign bit
14
15 00000014 0FD7C1 pmovmskb eax, mm1 ; extract the high bit of each byte
16 00000017 240F and al, 0x0F ; zero out the 2nd copy of the bitmap in the high nibble
17 00000019 C3 ret
size = 0x1A = 26 bytes
If there was a pmovmskw
, what would have saved the packsswb
and the and
(3+2 bytes). We don't need and eax, 0x0f
because pmovmskb
on an MMX register already zeros the upper bytes. MMX registers are only 8 bytes wide, so 8-bit AL covers all the possible non-zero bits.
If we knew our inputs were non-negative, we could packsswb mm1, mm0
to produce non-negative signed bytes in the upper 4 bytes of mm1
, avoiding the need for and
after pmovmskb
. Thus 24 bytes.
x86 pack with signed saturation treats the input and output as signed, so it always preserves the sign bit. (https://www.felixcloutier.com/x86/packsswb:packssdw). Fun fact: x86 pack with unsigned saturation still treats the input as signed. This might be why PACKUSDW
wasn't introduced until SSE4.1, while the other 3 combinations of size and signedness existed since MMX/SSE2.
Or with 32-bit integers in an XMM register (and pshufd
instead of pshufw
), every instruction would need one more prefix byte, except for movmskps
replacing the pack/and. But pmaxsd
/ pmaxud
need an extra extra byte...
callable from C as unsigned max4_sse4(__m128i);
with x86-64 System V, or MSVC vectorcall (-Gv
), both of which pass __m128i
/__m128d
/__m128
args in XMM regs starting with xmm0
.
20 global max4_sse4
21 ;; Input 4x int32_t in xmm0
22 ;; output: bitmap in EAX
23 ;; clobbers: xmm1, xmm2
24 max4_sse4:
25 00000020 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
26 00000025 660F383DC8 pmaxsd xmm1, xmm0
27
28 0000002A 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
29 0000002F 660F383DCA pmaxsd xmm1, xmm2
30
31 00000034 660F76C8 pcmpeqd xmm1, xmm0 ; 0 / -1
32
33 00000038 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
34 0000003B C3 ret
size = 0x3C - 0x20 = 28 bytes
Or if we accept input as float
, we can use SSE1 instructions. The float
format can represent a wide range of integer values...
Or if you think that's bending the rules too far, start with a 3-byte 0F 5B C0 cvtdq2ps xmm0, xmm0
to convert, making a 27-byte function that works for all integers that are exactly representable as IEEE binary32 float
, and many combinations of inputs where some of the inputs get rounded to a multiple of 2, 4, 8, or whatever during conversion. (So it's 1 byte smaller than the SSE4.1 version, and works on any x86-64 with just SSE2.)
If any of the float inputs are NaN, note that maxps a,b
exactly implements (a<b) ? a : b
, keeping the element from the 2nd operand on unordered. So it might be possible for this to return with a non-zero bitmap even if the input contains some NaN, depending on where they are.
unsigned max4_sse2(__m128);
37 global max4_sse2
38 ;; Input 4x float32 in xmm0
39 ;; output: bitmap in EAX
40 ;; clobbers: xmm1, xmm2
41 max4_sse2:
42 ; cvtdq2ps xmm0, xmm0
43 00000040 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
44 00000045 0F5FC8 maxps xmm1, xmm0
45
46 00000048 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
47 0000004D 0F5FCA maxps xmm1, xmm2
48
49 00000050 0FC2C800 cmpeqps xmm1, xmm0 ; 0 / -1
50
51 00000054 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
52 00000057 C3 ret
size = 0x58 - 0x40 = 24 bytes
copy-and-shuffle with pshufd
is still our best bet: shufps dst,src,imm8
reads the input for the low half of dst
from dst
. And we need a non-destructive copy-and-shuffle both times, so 3-byte movhlps
and unpckhps
/pd are both out. If we were narrowing to a scalar max, we could use those, but it costs another instruction to broadcast before compare if we don't have the max in all elements already.
Related: SSE4.1 phminposuw
can find the position and value of the minimum uint16_t
in an XMM register. I don't think it's a win to subtract from 65535 to use it for max, but see an SO answer about using it for max of bytes or signed integers.
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add a comment |
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Python 3.8 (pre-release), 67 bytes
Lambda function that takes in 4 integers, bit shifts the boolean result of their comparison to the maximum value with some help from Python 3.8's new assignment operator, and returns the bitwise OR of the results
lambda a,b,c,d:((m:=max(a,b,c,d))==d)<<3|(m==c)<<2|(m==b)<<2|(a==m)
Try it online!
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:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
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– ElPedro
Mar 4 at 23:54
add a comment |
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Java (JDK), 78 bytes
a->{int m=a[0],w=1,p=1;for(int t:a){w=t<m?w:p|(t>m?(m=t)^m:w);p*=2;}return w;}
Try it online!
- Takes the input as an array of
[a,b,c,d]
.
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add a comment |
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05AB1E, 3 2 bytes
ZQ
Input as a list of [d,c,b,a]
, output as a list of boolean.
Try it online or verify all test cases.
Explanation:
Z # Take the maximum of the implicit input-list
Q # Check for each in the (implicit) input-list if it equals this value
# (and output implicitly as result)
$endgroup$
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OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
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JavaScript (ES6), 30 bytes
Takes input as ([d,c,b,a])
. Returns 4 Boolean values.
a=>a.map(x=>x==Math.max(...a))
Try it online!
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1
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
add a comment |
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Ruby, 34 22 bytes
Takes input as an array [d, c, b, a]
and returns an array of 1s and 0s.
->r{r.map{|e|e/r.max}}
Try it online!
$endgroup$
add a comment |
$begingroup$
Python 3, 59 bytes 66 bytes
def f(l):
n=max(a)
for i in 0,1,2,3:a[i]=a[i]==n
return a[::-1]
Try it online!
Takes input as [a,b,c,d]
and outputs a list of booleans.
Edited to be a proper function, then saved 2 bytes by removing brackets around the conditional.
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1
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Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
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– Jonathan Frech
Mar 5 at 9:23
1
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Edited. First time around. Appreciate the heads up!
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– Bsoned
Mar 5 at 17:53
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You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
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– Value Ink
Mar 5 at 23:27
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@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
add a comment |
$begingroup$
1. Python 3.5, 90 bytes
Takes sequence of numbers as parameters. Returns "binary" string
import sys;v=[*map(int,sys.argv[1:])];m=max(v);s=""
for e in v:s=str(int(e==m))+s
print(s)
example:
$ ./script.py 6 77 1 4 77
10010
Explanation
import sys
# convert list of string parameters to list of integers
v=[*map(int,sys.argv[1:])]
# get max
m=max(v)
# init outstring
s=""
# walk through list
for e in v:
# prepend to outstring: int(True)=>1, int(False)=>0
s=str(int(e==m))+s
# print out result
print(s)
$endgroup$
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 26 bytes
n=>n.Select(a=>a==n.Max())
Try it online!
Takes input in format [d,c,b,a]
. All the others down below take input as [a,b,c,d]
C# (Visual C# Interactive Compiler), 35 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max())
Returns an IEnumerable<bool>
.
Try it online!
C# (Visual C# Interactive Compiler), 39 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max()?1:0)
Returns an IEnumerable<int>
, which represent bits.
Try it online!
C# (Visual C# Interactive Compiler), 49 bytes
n=>{for(int i=4;i-->0;Write(n[i]==n.Max()?1:0));}
Prints a binary string to STDOUT.
Try it online!
$endgroup$
$begingroup$
IEnumerable<bool>
is acceptable.
$endgroup$
– Adám
Mar 5 at 5:45
add a comment |
$begingroup$
PHP, 54 bytes
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i-1;echo$r;
or
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i;echo$r/2;
take input from command line arguments. Run with -nr
or try them online.
$endgroup$
add a comment |
$begingroup$
Here's a JS version that outputs as binary
update: Shorter with join, and without the lookup:
JavaScript (Node.js), 42 bytes
a=>a.map(x=>+(x==Math.max(...a))).join('')
Try it online!
Previous, with lookup, 49 bytes
a=>a.map(x=>[0,1][+(x==Math.max(...a))]).join('')
Try it online!
Previous, with reduce, 52 bytes:
a=>a.reduce((y,x)=>y+[0,1][+(x==Math.max(...a))],'')
Try it online!
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
$endgroup$
1
$begingroup$
You can safely remove[0,1][...]
since you're using an index that already is either $0$ or $1$.
$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 51 bytes
x=>{for(int m=x.Max(),i=4;i-->0;)x[i]=x[i]==m?1:0;}
Try it online!
Above is an anonymous function that outputs by modifying an argument. Output is an array of 1's and 0's.
Below is a recursive function that outputs an integer.
C# (Visual C# Interactive Compiler), 60 bytes
int f(intx,int i=3)=>i<0?0:2*f(x,i-1)|(x[i]==x.Max()?1:0);
Try it online!
Both functions take input as a 4-element array.
[d, c, b, a]
$endgroup$
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
add a comment |
$begingroup$
Python 2, 35 bytes
lambda i:[int(x==max(i))for x in i]
Try it online!
Takes input in the format [d,c,b,a] as with the accepted answer from Adám so I guess it is OK.
Alternative for 41 if it's not...
Python 2, 41 bytes
lambda i:[int(x==max(i))for x in i][::-1]
Try it online!
$endgroup$
add a comment |
$begingroup$
Python 3, 42 bytes
f=lambda a:list(map(lambda x:x==max(a),a))
Simply returns a list of whether the element is the max for each element in the input. -2 bytes if you don't count the f=
assignment.
Try it online!
$endgroup$
$begingroup$
f=
doesn't count except in recursive functions
$endgroup$
– ASCII-only
Mar 13 at 3:47
add a comment |
$begingroup$
Batch, 92 bytes
@set m=%1
@set f=@for %%i in (%*)do @
%f%set/a"m=m+(m-=%%i)*(m>>31)
%f%cmd/cset/a!(m-%%i)
Takes arguments as command-line parameters in reverse order. Works by arithmetically calculating the maximum of the parameters by reducing over them and adding on only positive differences from the running maximum, then mapping over each parameter again this time comparing it to the maximum. Conveniently cmd/cset/a
doesn't output a newline, so the results are automatically concatenated together. The %f%
simply saves 5 bytes on what would be a repeated construct.
$endgroup$
add a comment |
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21 Answers
21
active
oldest
votes
21 Answers
21
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Jelly, 2 bytes
Takes input as [d,c,b,a]
. Returns a list of Booleans.
Ṁ=
Try it online!
Ṁ
Maximum
=
equal to (implies the other argument is the original argument; vectorises)
$endgroup$
add a comment |
$begingroup$
Jelly, 2 bytes
Takes input as [d,c,b,a]
. Returns a list of Booleans.
Ṁ=
Try it online!
Ṁ
Maximum
=
equal to (implies the other argument is the original argument; vectorises)
$endgroup$
add a comment |
$begingroup$
Jelly, 2 bytes
Takes input as [d,c,b,a]
. Returns a list of Booleans.
Ṁ=
Try it online!
Ṁ
Maximum
=
equal to (implies the other argument is the original argument; vectorises)
$endgroup$
Jelly, 2 bytes
Takes input as [d,c,b,a]
. Returns a list of Booleans.
Ṁ=
Try it online!
Ṁ
Maximum
=
equal to (implies the other argument is the original argument; vectorises)
answered Mar 4 at 9:58
AdámAdám
28.9k276207
28.9k276207
add a comment |
add a comment |
$begingroup$
R, 17 bytes
max(a<-scan())==a
Try it online!
Returns a vector of booleans. Since this output has been confirmed, this is preferable over numerical output, as that one is almost twice longer:
R, 33 bytes
sum(2^which(max(a<-scan())==a)/2)
Try it online!
$endgroup$
add a comment |
$begingroup$
R, 17 bytes
max(a<-scan())==a
Try it online!
Returns a vector of booleans. Since this output has been confirmed, this is preferable over numerical output, as that one is almost twice longer:
R, 33 bytes
sum(2^which(max(a<-scan())==a)/2)
Try it online!
$endgroup$
add a comment |
$begingroup$
R, 17 bytes
max(a<-scan())==a
Try it online!
Returns a vector of booleans. Since this output has been confirmed, this is preferable over numerical output, as that one is almost twice longer:
R, 33 bytes
sum(2^which(max(a<-scan())==a)/2)
Try it online!
$endgroup$
R, 17 bytes
max(a<-scan())==a
Try it online!
Returns a vector of booleans. Since this output has been confirmed, this is preferable over numerical output, as that one is almost twice longer:
R, 33 bytes
sum(2^which(max(a<-scan())==a)/2)
Try it online!
edited Mar 4 at 21:42
answered Mar 4 at 9:17
Kirill L.Kirill L.
5,9981527
5,9981527
add a comment |
add a comment |
$begingroup$
APL (Dyalog Unicode), 4 bytesSBCS
Anonymous tacit prefix function. Takes [a,b,c,d]
as argument. Returns a bit-Boolean array.*
⌈/=⌽
Try it online!
⌈/
Does the maximum of the argument
=
equal (vectorises)
⌽
the reverse of the argument?
* Note that APL stores arrays of Booleans using one bit per value, so this does indeed return a 4-bit word, despite the display form being 0 0 1 0
.
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 4 bytesSBCS
Anonymous tacit prefix function. Takes [a,b,c,d]
as argument. Returns a bit-Boolean array.*
⌈/=⌽
Try it online!
⌈/
Does the maximum of the argument
=
equal (vectorises)
⌽
the reverse of the argument?
* Note that APL stores arrays of Booleans using one bit per value, so this does indeed return a 4-bit word, despite the display form being 0 0 1 0
.
$endgroup$
add a comment |
$begingroup$
APL (Dyalog Unicode), 4 bytesSBCS
Anonymous tacit prefix function. Takes [a,b,c,d]
as argument. Returns a bit-Boolean array.*
⌈/=⌽
Try it online!
⌈/
Does the maximum of the argument
=
equal (vectorises)
⌽
the reverse of the argument?
* Note that APL stores arrays of Booleans using one bit per value, so this does indeed return a 4-bit word, despite the display form being 0 0 1 0
.
$endgroup$
APL (Dyalog Unicode), 4 bytesSBCS
Anonymous tacit prefix function. Takes [a,b,c,d]
as argument. Returns a bit-Boolean array.*
⌈/=⌽
Try it online!
⌈/
Does the maximum of the argument
=
equal (vectorises)
⌽
the reverse of the argument?
* Note that APL stores arrays of Booleans using one bit per value, so this does indeed return a 4-bit word, despite the display form being 0 0 1 0
.
edited Mar 4 at 8:18
answered Mar 4 at 8:08
AdámAdám
28.9k276207
28.9k276207
add a comment |
add a comment |
$begingroup$
Haskell, 20 18 bytes
2 bytes saved thanks to proud haskeller
map=<<(==).maximum
Try it online!
$endgroup$
$begingroup$
You can output a string.
$endgroup$
– Adám
Mar 5 at 5:44
1
$begingroup$
Writingmap
instead of(<$>)
would be two bytes shorter!
$endgroup$
– proud haskeller
Mar 7 at 6:23
$begingroup$
@proudhaskeller Good catch. Can't believe that I didn't see that one.
$endgroup$
– Sriotchilism O'Zaic
Mar 7 at 13:40
add a comment |
$begingroup$
Haskell, 20 18 bytes
2 bytes saved thanks to proud haskeller
map=<<(==).maximum
Try it online!
$endgroup$
$begingroup$
You can output a string.
$endgroup$
– Adám
Mar 5 at 5:44
1
$begingroup$
Writingmap
instead of(<$>)
would be two bytes shorter!
$endgroup$
– proud haskeller
Mar 7 at 6:23
$begingroup$
@proudhaskeller Good catch. Can't believe that I didn't see that one.
$endgroup$
– Sriotchilism O'Zaic
Mar 7 at 13:40
add a comment |
$begingroup$
Haskell, 20 18 bytes
2 bytes saved thanks to proud haskeller
map=<<(==).maximum
Try it online!
$endgroup$
Haskell, 20 18 bytes
2 bytes saved thanks to proud haskeller
map=<<(==).maximum
Try it online!
edited Mar 7 at 13:40
answered Mar 4 at 15:27
Sriotchilism O'ZaicSriotchilism O'Zaic
35.6k10161369
35.6k10161369
$begingroup$
You can output a string.
$endgroup$
– Adám
Mar 5 at 5:44
1
$begingroup$
Writingmap
instead of(<$>)
would be two bytes shorter!
$endgroup$
– proud haskeller
Mar 7 at 6:23
$begingroup$
@proudhaskeller Good catch. Can't believe that I didn't see that one.
$endgroup$
– Sriotchilism O'Zaic
Mar 7 at 13:40
add a comment |
$begingroup$
You can output a string.
$endgroup$
– Adám
Mar 5 at 5:44
1
$begingroup$
Writingmap
instead of(<$>)
would be two bytes shorter!
$endgroup$
– proud haskeller
Mar 7 at 6:23
$begingroup$
@proudhaskeller Good catch. Can't believe that I didn't see that one.
$endgroup$
– Sriotchilism O'Zaic
Mar 7 at 13:40
$begingroup$
You can output a string.
$endgroup$
– Adám
Mar 5 at 5:44
$begingroup$
You can output a string.
$endgroup$
– Adám
Mar 5 at 5:44
1
1
$begingroup$
Writing
map
instead of (<$>)
would be two bytes shorter!$endgroup$
– proud haskeller
Mar 7 at 6:23
$begingroup$
Writing
map
instead of (<$>)
would be two bytes shorter!$endgroup$
– proud haskeller
Mar 7 at 6:23
$begingroup$
@proudhaskeller Good catch. Can't believe that I didn't see that one.
$endgroup$
– Sriotchilism O'Zaic
Mar 7 at 13:40
$begingroup$
@proudhaskeller Good catch. Can't believe that I didn't see that one.
$endgroup$
– Sriotchilism O'Zaic
Mar 7 at 13:40
add a comment |
$begingroup$
Perl 6, 12 bytes
{$_ X==.max}
Try it online!
Anonymous code block that takes a list of integers and returns a list of booleans. If we need to return as a number, it's +4 bytes to wrap the inside of the code block with 2:[...]
.
Explanation:
{ } # Anonymous code block
$_ # With the input
X== # Which values are equal
.max # To the maximum element
$endgroup$
$begingroup$
OP now says you don't need to wrap.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
Perl 6, 12 bytes
{$_ X==.max}
Try it online!
Anonymous code block that takes a list of integers and returns a list of booleans. If we need to return as a number, it's +4 bytes to wrap the inside of the code block with 2:[...]
.
Explanation:
{ } # Anonymous code block
$_ # With the input
X== # Which values are equal
.max # To the maximum element
$endgroup$
$begingroup$
OP now says you don't need to wrap.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
Perl 6, 12 bytes
{$_ X==.max}
Try it online!
Anonymous code block that takes a list of integers and returns a list of booleans. If we need to return as a number, it's +4 bytes to wrap the inside of the code block with 2:[...]
.
Explanation:
{ } # Anonymous code block
$_ # With the input
X== # Which values are equal
.max # To the maximum element
$endgroup$
Perl 6, 12 bytes
{$_ X==.max}
Try it online!
Anonymous code block that takes a list of integers and returns a list of booleans. If we need to return as a number, it's +4 bytes to wrap the inside of the code block with 2:[...]
.
Explanation:
{ } # Anonymous code block
$_ # With the input
X== # Which values are equal
.max # To the maximum element
answered Mar 4 at 9:25
Jo KingJo King
26.5k364130
26.5k364130
$begingroup$
OP now says you don't need to wrap.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
OP now says you don't need to wrap.
$endgroup$
– Adám
Mar 5 at 5:42
$begingroup$
OP now says you don't need to wrap.
$endgroup$
– Adám
Mar 5 at 5:42
$begingroup$
OP now says you don't need to wrap.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
Japt, 5
m¶Urw
Try it!
-4 bytes thanks to @Oliver!
-2 bytes thanks to @Shaggy!
Input is a 4-element array in the following format:
[d, c, b, a]
Output is an array of bits.
$endgroup$
$begingroup$
Of course there is ;) There are apparently a lot of shortcuts to learn.
$endgroup$
– dana
Mar 4 at 15:52
$begingroup$
If a boolean array is an acceptable output, this can be 7 bytes
$endgroup$
– Oliver
Mar 4 at 15:56
$begingroup$
@Oliver, 5 bytes ;)
$endgroup$
– Shaggy
Mar 4 at 20:29
$begingroup$
You guys are pretty good :) That's interesting howrw
converts tor("w")
does a reduce by repeatedly getting the max. Same withm¶
getting converted toU.m("===", ...)
. In any case, thanks for the tips!
$endgroup$
– dana
Mar 4 at 23:54
add a comment |
$begingroup$
Japt, 5
m¶Urw
Try it!
-4 bytes thanks to @Oliver!
-2 bytes thanks to @Shaggy!
Input is a 4-element array in the following format:
[d, c, b, a]
Output is an array of bits.
$endgroup$
$begingroup$
Of course there is ;) There are apparently a lot of shortcuts to learn.
$endgroup$
– dana
Mar 4 at 15:52
$begingroup$
If a boolean array is an acceptable output, this can be 7 bytes
$endgroup$
– Oliver
Mar 4 at 15:56
$begingroup$
@Oliver, 5 bytes ;)
$endgroup$
– Shaggy
Mar 4 at 20:29
$begingroup$
You guys are pretty good :) That's interesting howrw
converts tor("w")
does a reduce by repeatedly getting the max. Same withm¶
getting converted toU.m("===", ...)
. In any case, thanks for the tips!
$endgroup$
– dana
Mar 4 at 23:54
add a comment |
$begingroup$
Japt, 5
m¶Urw
Try it!
-4 bytes thanks to @Oliver!
-2 bytes thanks to @Shaggy!
Input is a 4-element array in the following format:
[d, c, b, a]
Output is an array of bits.
$endgroup$
Japt, 5
m¶Urw
Try it!
-4 bytes thanks to @Oliver!
-2 bytes thanks to @Shaggy!
Input is a 4-element array in the following format:
[d, c, b, a]
Output is an array of bits.
edited Mar 5 at 3:25
answered Mar 4 at 14:27
danadana
1,921167
1,921167
$begingroup$
Of course there is ;) There are apparently a lot of shortcuts to learn.
$endgroup$
– dana
Mar 4 at 15:52
$begingroup$
If a boolean array is an acceptable output, this can be 7 bytes
$endgroup$
– Oliver
Mar 4 at 15:56
$begingroup$
@Oliver, 5 bytes ;)
$endgroup$
– Shaggy
Mar 4 at 20:29
$begingroup$
You guys are pretty good :) That's interesting howrw
converts tor("w")
does a reduce by repeatedly getting the max. Same withm¶
getting converted toU.m("===", ...)
. In any case, thanks for the tips!
$endgroup$
– dana
Mar 4 at 23:54
add a comment |
$begingroup$
Of course there is ;) There are apparently a lot of shortcuts to learn.
$endgroup$
– dana
Mar 4 at 15:52
$begingroup$
If a boolean array is an acceptable output, this can be 7 bytes
$endgroup$
– Oliver
Mar 4 at 15:56
$begingroup$
@Oliver, 5 bytes ;)
$endgroup$
– Shaggy
Mar 4 at 20:29
$begingroup$
You guys are pretty good :) That's interesting howrw
converts tor("w")
does a reduce by repeatedly getting the max. Same withm¶
getting converted toU.m("===", ...)
. In any case, thanks for the tips!
$endgroup$
– dana
Mar 4 at 23:54
$begingroup$
Of course there is ;) There are apparently a lot of shortcuts to learn.
$endgroup$
– dana
Mar 4 at 15:52
$begingroup$
Of course there is ;) There are apparently a lot of shortcuts to learn.
$endgroup$
– dana
Mar 4 at 15:52
$begingroup$
If a boolean array is an acceptable output, this can be 7 bytes
$endgroup$
– Oliver
Mar 4 at 15:56
$begingroup$
If a boolean array is an acceptable output, this can be 7 bytes
$endgroup$
– Oliver
Mar 4 at 15:56
$begingroup$
@Oliver, 5 bytes ;)
$endgroup$
– Shaggy
Mar 4 at 20:29
$begingroup$
@Oliver, 5 bytes ;)
$endgroup$
– Shaggy
Mar 4 at 20:29
$begingroup$
You guys are pretty good :) That's interesting how
rw
converts to r("w")
does a reduce by repeatedly getting the max. Same with m¶
getting converted to U.m("===", ...)
. In any case, thanks for the tips!$endgroup$
– dana
Mar 4 at 23:54
$begingroup$
You guys are pretty good :) That's interesting how
rw
converts to r("w")
does a reduce by repeatedly getting the max. Same with m¶
getting converted to U.m("===", ...)
. In any case, thanks for the tips!$endgroup$
– dana
Mar 4 at 23:54
add a comment |
$begingroup$
x86 machine code (MMX/SSE1), 26 bytes (4x int16_t)
x86 machine code (SSE4.1), 28 bytes (4x int32_t or uint32_t)
x86 machine code (SSE2), 24 bytes (4x float32) or 27B to cvt int32
(The last version that converts int32 to float isn't perfectly accurate for large integers that round to the same float. With float input, rounding is the caller's problem and this function works correctly if there are no NaNs, identifying floats that compare == to the max. The integer versions work for all inputs, treating them as signed 2's complement.)
All of these work in 16/32/64-bit mode with the same machine code.
A stack-args calling convention would make it possible to loop over the args twice (finding max and then comparing), possibly giving us a smaller implementation, but I haven't tried that approach.
x86 SIMD has vector->integer bitmap as a single instruction (pmovmskb
or movmskps
or pd), so it was natural for this even though MMX/SSE instructions are at least 3 bytes long. SSSE3 and later instructions are longer than SSE2, and MMX/SSE1 instructions are the shortest. Different versions of pmax*
(packed-integer vertical max) were introduced at different times, with SSE1 (for mmx regs) and SSE2 (for xmm regs) only having signed word (16-bit) and unsigned byte.
(pshufw
and pmaxsw
on MMX registers are new with Katmai Pentium III, so really they require SSE1, not just the MMX CPU feature bit.)
This is callable from C as unsigned max4_mmx(__m64)
with the i386 System V ABI, which passes an __m64
arg in mm0
. (Not x86-64 System V, which passes __m64
in xmm0
!)
line code bytes
num addr
1 global max4_mmx
2 ;; Input 4x int16_t in mm0
3 ;; output: bitmap in EAX
4 ;; clobbers: mm1, mm2
5 max4_mmx:
6 00000000 0F70C8B1 pshufw mm1, mm0, 0b10110001 ; swap adjacent pairs
7 00000004 0FEEC8 pmaxsw mm1, mm0
8
9 00000007 0F70D14E pshufw mm2, mm1, 0b01001110 ; swap high/low halves
10 0000000B 0FEECA pmaxsw mm1, mm2
11
12 0000000E 0F75C8 pcmpeqw mm1, mm0 ; 0 / -1
13 00000011 0F63C9 packsswb mm1, mm1 ; squish word elements to bytes, preserving sign bit
14
15 00000014 0FD7C1 pmovmskb eax, mm1 ; extract the high bit of each byte
16 00000017 240F and al, 0x0F ; zero out the 2nd copy of the bitmap in the high nibble
17 00000019 C3 ret
size = 0x1A = 26 bytes
If there was a pmovmskw
, what would have saved the packsswb
and the and
(3+2 bytes). We don't need and eax, 0x0f
because pmovmskb
on an MMX register already zeros the upper bytes. MMX registers are only 8 bytes wide, so 8-bit AL covers all the possible non-zero bits.
If we knew our inputs were non-negative, we could packsswb mm1, mm0
to produce non-negative signed bytes in the upper 4 bytes of mm1
, avoiding the need for and
after pmovmskb
. Thus 24 bytes.
x86 pack with signed saturation treats the input and output as signed, so it always preserves the sign bit. (https://www.felixcloutier.com/x86/packsswb:packssdw). Fun fact: x86 pack with unsigned saturation still treats the input as signed. This might be why PACKUSDW
wasn't introduced until SSE4.1, while the other 3 combinations of size and signedness existed since MMX/SSE2.
Or with 32-bit integers in an XMM register (and pshufd
instead of pshufw
), every instruction would need one more prefix byte, except for movmskps
replacing the pack/and. But pmaxsd
/ pmaxud
need an extra extra byte...
callable from C as unsigned max4_sse4(__m128i);
with x86-64 System V, or MSVC vectorcall (-Gv
), both of which pass __m128i
/__m128d
/__m128
args in XMM regs starting with xmm0
.
20 global max4_sse4
21 ;; Input 4x int32_t in xmm0
22 ;; output: bitmap in EAX
23 ;; clobbers: xmm1, xmm2
24 max4_sse4:
25 00000020 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
26 00000025 660F383DC8 pmaxsd xmm1, xmm0
27
28 0000002A 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
29 0000002F 660F383DCA pmaxsd xmm1, xmm2
30
31 00000034 660F76C8 pcmpeqd xmm1, xmm0 ; 0 / -1
32
33 00000038 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
34 0000003B C3 ret
size = 0x3C - 0x20 = 28 bytes
Or if we accept input as float
, we can use SSE1 instructions. The float
format can represent a wide range of integer values...
Or if you think that's bending the rules too far, start with a 3-byte 0F 5B C0 cvtdq2ps xmm0, xmm0
to convert, making a 27-byte function that works for all integers that are exactly representable as IEEE binary32 float
, and many combinations of inputs where some of the inputs get rounded to a multiple of 2, 4, 8, or whatever during conversion. (So it's 1 byte smaller than the SSE4.1 version, and works on any x86-64 with just SSE2.)
If any of the float inputs are NaN, note that maxps a,b
exactly implements (a<b) ? a : b
, keeping the element from the 2nd operand on unordered. So it might be possible for this to return with a non-zero bitmap even if the input contains some NaN, depending on where they are.
unsigned max4_sse2(__m128);
37 global max4_sse2
38 ;; Input 4x float32 in xmm0
39 ;; output: bitmap in EAX
40 ;; clobbers: xmm1, xmm2
41 max4_sse2:
42 ; cvtdq2ps xmm0, xmm0
43 00000040 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
44 00000045 0F5FC8 maxps xmm1, xmm0
45
46 00000048 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
47 0000004D 0F5FCA maxps xmm1, xmm2
48
49 00000050 0FC2C800 cmpeqps xmm1, xmm0 ; 0 / -1
50
51 00000054 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
52 00000057 C3 ret
size = 0x58 - 0x40 = 24 bytes
copy-and-shuffle with pshufd
is still our best bet: shufps dst,src,imm8
reads the input for the low half of dst
from dst
. And we need a non-destructive copy-and-shuffle both times, so 3-byte movhlps
and unpckhps
/pd are both out. If we were narrowing to a scalar max, we could use those, but it costs another instruction to broadcast before compare if we don't have the max in all elements already.
Related: SSE4.1 phminposuw
can find the position and value of the minimum uint16_t
in an XMM register. I don't think it's a win to subtract from 65535 to use it for max, but see an SO answer about using it for max of bytes or signed integers.
$endgroup$
add a comment |
$begingroup$
x86 machine code (MMX/SSE1), 26 bytes (4x int16_t)
x86 machine code (SSE4.1), 28 bytes (4x int32_t or uint32_t)
x86 machine code (SSE2), 24 bytes (4x float32) or 27B to cvt int32
(The last version that converts int32 to float isn't perfectly accurate for large integers that round to the same float. With float input, rounding is the caller's problem and this function works correctly if there are no NaNs, identifying floats that compare == to the max. The integer versions work for all inputs, treating them as signed 2's complement.)
All of these work in 16/32/64-bit mode with the same machine code.
A stack-args calling convention would make it possible to loop over the args twice (finding max and then comparing), possibly giving us a smaller implementation, but I haven't tried that approach.
x86 SIMD has vector->integer bitmap as a single instruction (pmovmskb
or movmskps
or pd), so it was natural for this even though MMX/SSE instructions are at least 3 bytes long. SSSE3 and later instructions are longer than SSE2, and MMX/SSE1 instructions are the shortest. Different versions of pmax*
(packed-integer vertical max) were introduced at different times, with SSE1 (for mmx regs) and SSE2 (for xmm regs) only having signed word (16-bit) and unsigned byte.
(pshufw
and pmaxsw
on MMX registers are new with Katmai Pentium III, so really they require SSE1, not just the MMX CPU feature bit.)
This is callable from C as unsigned max4_mmx(__m64)
with the i386 System V ABI, which passes an __m64
arg in mm0
. (Not x86-64 System V, which passes __m64
in xmm0
!)
line code bytes
num addr
1 global max4_mmx
2 ;; Input 4x int16_t in mm0
3 ;; output: bitmap in EAX
4 ;; clobbers: mm1, mm2
5 max4_mmx:
6 00000000 0F70C8B1 pshufw mm1, mm0, 0b10110001 ; swap adjacent pairs
7 00000004 0FEEC8 pmaxsw mm1, mm0
8
9 00000007 0F70D14E pshufw mm2, mm1, 0b01001110 ; swap high/low halves
10 0000000B 0FEECA pmaxsw mm1, mm2
11
12 0000000E 0F75C8 pcmpeqw mm1, mm0 ; 0 / -1
13 00000011 0F63C9 packsswb mm1, mm1 ; squish word elements to bytes, preserving sign bit
14
15 00000014 0FD7C1 pmovmskb eax, mm1 ; extract the high bit of each byte
16 00000017 240F and al, 0x0F ; zero out the 2nd copy of the bitmap in the high nibble
17 00000019 C3 ret
size = 0x1A = 26 bytes
If there was a pmovmskw
, what would have saved the packsswb
and the and
(3+2 bytes). We don't need and eax, 0x0f
because pmovmskb
on an MMX register already zeros the upper bytes. MMX registers are only 8 bytes wide, so 8-bit AL covers all the possible non-zero bits.
If we knew our inputs were non-negative, we could packsswb mm1, mm0
to produce non-negative signed bytes in the upper 4 bytes of mm1
, avoiding the need for and
after pmovmskb
. Thus 24 bytes.
x86 pack with signed saturation treats the input and output as signed, so it always preserves the sign bit. (https://www.felixcloutier.com/x86/packsswb:packssdw). Fun fact: x86 pack with unsigned saturation still treats the input as signed. This might be why PACKUSDW
wasn't introduced until SSE4.1, while the other 3 combinations of size and signedness existed since MMX/SSE2.
Or with 32-bit integers in an XMM register (and pshufd
instead of pshufw
), every instruction would need one more prefix byte, except for movmskps
replacing the pack/and. But pmaxsd
/ pmaxud
need an extra extra byte...
callable from C as unsigned max4_sse4(__m128i);
with x86-64 System V, or MSVC vectorcall (-Gv
), both of which pass __m128i
/__m128d
/__m128
args in XMM regs starting with xmm0
.
20 global max4_sse4
21 ;; Input 4x int32_t in xmm0
22 ;; output: bitmap in EAX
23 ;; clobbers: xmm1, xmm2
24 max4_sse4:
25 00000020 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
26 00000025 660F383DC8 pmaxsd xmm1, xmm0
27
28 0000002A 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
29 0000002F 660F383DCA pmaxsd xmm1, xmm2
30
31 00000034 660F76C8 pcmpeqd xmm1, xmm0 ; 0 / -1
32
33 00000038 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
34 0000003B C3 ret
size = 0x3C - 0x20 = 28 bytes
Or if we accept input as float
, we can use SSE1 instructions. The float
format can represent a wide range of integer values...
Or if you think that's bending the rules too far, start with a 3-byte 0F 5B C0 cvtdq2ps xmm0, xmm0
to convert, making a 27-byte function that works for all integers that are exactly representable as IEEE binary32 float
, and many combinations of inputs where some of the inputs get rounded to a multiple of 2, 4, 8, or whatever during conversion. (So it's 1 byte smaller than the SSE4.1 version, and works on any x86-64 with just SSE2.)
If any of the float inputs are NaN, note that maxps a,b
exactly implements (a<b) ? a : b
, keeping the element from the 2nd operand on unordered. So it might be possible for this to return with a non-zero bitmap even if the input contains some NaN, depending on where they are.
unsigned max4_sse2(__m128);
37 global max4_sse2
38 ;; Input 4x float32 in xmm0
39 ;; output: bitmap in EAX
40 ;; clobbers: xmm1, xmm2
41 max4_sse2:
42 ; cvtdq2ps xmm0, xmm0
43 00000040 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
44 00000045 0F5FC8 maxps xmm1, xmm0
45
46 00000048 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
47 0000004D 0F5FCA maxps xmm1, xmm2
48
49 00000050 0FC2C800 cmpeqps xmm1, xmm0 ; 0 / -1
50
51 00000054 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
52 00000057 C3 ret
size = 0x58 - 0x40 = 24 bytes
copy-and-shuffle with pshufd
is still our best bet: shufps dst,src,imm8
reads the input for the low half of dst
from dst
. And we need a non-destructive copy-and-shuffle both times, so 3-byte movhlps
and unpckhps
/pd are both out. If we were narrowing to a scalar max, we could use those, but it costs another instruction to broadcast before compare if we don't have the max in all elements already.
Related: SSE4.1 phminposuw
can find the position and value of the minimum uint16_t
in an XMM register. I don't think it's a win to subtract from 65535 to use it for max, but see an SO answer about using it for max of bytes or signed integers.
$endgroup$
add a comment |
$begingroup$
x86 machine code (MMX/SSE1), 26 bytes (4x int16_t)
x86 machine code (SSE4.1), 28 bytes (4x int32_t or uint32_t)
x86 machine code (SSE2), 24 bytes (4x float32) or 27B to cvt int32
(The last version that converts int32 to float isn't perfectly accurate for large integers that round to the same float. With float input, rounding is the caller's problem and this function works correctly if there are no NaNs, identifying floats that compare == to the max. The integer versions work for all inputs, treating them as signed 2's complement.)
All of these work in 16/32/64-bit mode with the same machine code.
A stack-args calling convention would make it possible to loop over the args twice (finding max and then comparing), possibly giving us a smaller implementation, but I haven't tried that approach.
x86 SIMD has vector->integer bitmap as a single instruction (pmovmskb
or movmskps
or pd), so it was natural for this even though MMX/SSE instructions are at least 3 bytes long. SSSE3 and later instructions are longer than SSE2, and MMX/SSE1 instructions are the shortest. Different versions of pmax*
(packed-integer vertical max) were introduced at different times, with SSE1 (for mmx regs) and SSE2 (for xmm regs) only having signed word (16-bit) and unsigned byte.
(pshufw
and pmaxsw
on MMX registers are new with Katmai Pentium III, so really they require SSE1, not just the MMX CPU feature bit.)
This is callable from C as unsigned max4_mmx(__m64)
with the i386 System V ABI, which passes an __m64
arg in mm0
. (Not x86-64 System V, which passes __m64
in xmm0
!)
line code bytes
num addr
1 global max4_mmx
2 ;; Input 4x int16_t in mm0
3 ;; output: bitmap in EAX
4 ;; clobbers: mm1, mm2
5 max4_mmx:
6 00000000 0F70C8B1 pshufw mm1, mm0, 0b10110001 ; swap adjacent pairs
7 00000004 0FEEC8 pmaxsw mm1, mm0
8
9 00000007 0F70D14E pshufw mm2, mm1, 0b01001110 ; swap high/low halves
10 0000000B 0FEECA pmaxsw mm1, mm2
11
12 0000000E 0F75C8 pcmpeqw mm1, mm0 ; 0 / -1
13 00000011 0F63C9 packsswb mm1, mm1 ; squish word elements to bytes, preserving sign bit
14
15 00000014 0FD7C1 pmovmskb eax, mm1 ; extract the high bit of each byte
16 00000017 240F and al, 0x0F ; zero out the 2nd copy of the bitmap in the high nibble
17 00000019 C3 ret
size = 0x1A = 26 bytes
If there was a pmovmskw
, what would have saved the packsswb
and the and
(3+2 bytes). We don't need and eax, 0x0f
because pmovmskb
on an MMX register already zeros the upper bytes. MMX registers are only 8 bytes wide, so 8-bit AL covers all the possible non-zero bits.
If we knew our inputs were non-negative, we could packsswb mm1, mm0
to produce non-negative signed bytes in the upper 4 bytes of mm1
, avoiding the need for and
after pmovmskb
. Thus 24 bytes.
x86 pack with signed saturation treats the input and output as signed, so it always preserves the sign bit. (https://www.felixcloutier.com/x86/packsswb:packssdw). Fun fact: x86 pack with unsigned saturation still treats the input as signed. This might be why PACKUSDW
wasn't introduced until SSE4.1, while the other 3 combinations of size and signedness existed since MMX/SSE2.
Or with 32-bit integers in an XMM register (and pshufd
instead of pshufw
), every instruction would need one more prefix byte, except for movmskps
replacing the pack/and. But pmaxsd
/ pmaxud
need an extra extra byte...
callable from C as unsigned max4_sse4(__m128i);
with x86-64 System V, or MSVC vectorcall (-Gv
), both of which pass __m128i
/__m128d
/__m128
args in XMM regs starting with xmm0
.
20 global max4_sse4
21 ;; Input 4x int32_t in xmm0
22 ;; output: bitmap in EAX
23 ;; clobbers: xmm1, xmm2
24 max4_sse4:
25 00000020 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
26 00000025 660F383DC8 pmaxsd xmm1, xmm0
27
28 0000002A 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
29 0000002F 660F383DCA pmaxsd xmm1, xmm2
30
31 00000034 660F76C8 pcmpeqd xmm1, xmm0 ; 0 / -1
32
33 00000038 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
34 0000003B C3 ret
size = 0x3C - 0x20 = 28 bytes
Or if we accept input as float
, we can use SSE1 instructions. The float
format can represent a wide range of integer values...
Or if you think that's bending the rules too far, start with a 3-byte 0F 5B C0 cvtdq2ps xmm0, xmm0
to convert, making a 27-byte function that works for all integers that are exactly representable as IEEE binary32 float
, and many combinations of inputs where some of the inputs get rounded to a multiple of 2, 4, 8, or whatever during conversion. (So it's 1 byte smaller than the SSE4.1 version, and works on any x86-64 with just SSE2.)
If any of the float inputs are NaN, note that maxps a,b
exactly implements (a<b) ? a : b
, keeping the element from the 2nd operand on unordered. So it might be possible for this to return with a non-zero bitmap even if the input contains some NaN, depending on where they are.
unsigned max4_sse2(__m128);
37 global max4_sse2
38 ;; Input 4x float32 in xmm0
39 ;; output: bitmap in EAX
40 ;; clobbers: xmm1, xmm2
41 max4_sse2:
42 ; cvtdq2ps xmm0, xmm0
43 00000040 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
44 00000045 0F5FC8 maxps xmm1, xmm0
45
46 00000048 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
47 0000004D 0F5FCA maxps xmm1, xmm2
48
49 00000050 0FC2C800 cmpeqps xmm1, xmm0 ; 0 / -1
50
51 00000054 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
52 00000057 C3 ret
size = 0x58 - 0x40 = 24 bytes
copy-and-shuffle with pshufd
is still our best bet: shufps dst,src,imm8
reads the input for the low half of dst
from dst
. And we need a non-destructive copy-and-shuffle both times, so 3-byte movhlps
and unpckhps
/pd are both out. If we were narrowing to a scalar max, we could use those, but it costs another instruction to broadcast before compare if we don't have the max in all elements already.
Related: SSE4.1 phminposuw
can find the position and value of the minimum uint16_t
in an XMM register. I don't think it's a win to subtract from 65535 to use it for max, but see an SO answer about using it for max of bytes or signed integers.
$endgroup$
x86 machine code (MMX/SSE1), 26 bytes (4x int16_t)
x86 machine code (SSE4.1), 28 bytes (4x int32_t or uint32_t)
x86 machine code (SSE2), 24 bytes (4x float32) or 27B to cvt int32
(The last version that converts int32 to float isn't perfectly accurate for large integers that round to the same float. With float input, rounding is the caller's problem and this function works correctly if there are no NaNs, identifying floats that compare == to the max. The integer versions work for all inputs, treating them as signed 2's complement.)
All of these work in 16/32/64-bit mode with the same machine code.
A stack-args calling convention would make it possible to loop over the args twice (finding max and then comparing), possibly giving us a smaller implementation, but I haven't tried that approach.
x86 SIMD has vector->integer bitmap as a single instruction (pmovmskb
or movmskps
or pd), so it was natural for this even though MMX/SSE instructions are at least 3 bytes long. SSSE3 and later instructions are longer than SSE2, and MMX/SSE1 instructions are the shortest. Different versions of pmax*
(packed-integer vertical max) were introduced at different times, with SSE1 (for mmx regs) and SSE2 (for xmm regs) only having signed word (16-bit) and unsigned byte.
(pshufw
and pmaxsw
on MMX registers are new with Katmai Pentium III, so really they require SSE1, not just the MMX CPU feature bit.)
This is callable from C as unsigned max4_mmx(__m64)
with the i386 System V ABI, which passes an __m64
arg in mm0
. (Not x86-64 System V, which passes __m64
in xmm0
!)
line code bytes
num addr
1 global max4_mmx
2 ;; Input 4x int16_t in mm0
3 ;; output: bitmap in EAX
4 ;; clobbers: mm1, mm2
5 max4_mmx:
6 00000000 0F70C8B1 pshufw mm1, mm0, 0b10110001 ; swap adjacent pairs
7 00000004 0FEEC8 pmaxsw mm1, mm0
8
9 00000007 0F70D14E pshufw mm2, mm1, 0b01001110 ; swap high/low halves
10 0000000B 0FEECA pmaxsw mm1, mm2
11
12 0000000E 0F75C8 pcmpeqw mm1, mm0 ; 0 / -1
13 00000011 0F63C9 packsswb mm1, mm1 ; squish word elements to bytes, preserving sign bit
14
15 00000014 0FD7C1 pmovmskb eax, mm1 ; extract the high bit of each byte
16 00000017 240F and al, 0x0F ; zero out the 2nd copy of the bitmap in the high nibble
17 00000019 C3 ret
size = 0x1A = 26 bytes
If there was a pmovmskw
, what would have saved the packsswb
and the and
(3+2 bytes). We don't need and eax, 0x0f
because pmovmskb
on an MMX register already zeros the upper bytes. MMX registers are only 8 bytes wide, so 8-bit AL covers all the possible non-zero bits.
If we knew our inputs were non-negative, we could packsswb mm1, mm0
to produce non-negative signed bytes in the upper 4 bytes of mm1
, avoiding the need for and
after pmovmskb
. Thus 24 bytes.
x86 pack with signed saturation treats the input and output as signed, so it always preserves the sign bit. (https://www.felixcloutier.com/x86/packsswb:packssdw). Fun fact: x86 pack with unsigned saturation still treats the input as signed. This might be why PACKUSDW
wasn't introduced until SSE4.1, while the other 3 combinations of size and signedness existed since MMX/SSE2.
Or with 32-bit integers in an XMM register (and pshufd
instead of pshufw
), every instruction would need one more prefix byte, except for movmskps
replacing the pack/and. But pmaxsd
/ pmaxud
need an extra extra byte...
callable from C as unsigned max4_sse4(__m128i);
with x86-64 System V, or MSVC vectorcall (-Gv
), both of which pass __m128i
/__m128d
/__m128
args in XMM regs starting with xmm0
.
20 global max4_sse4
21 ;; Input 4x int32_t in xmm0
22 ;; output: bitmap in EAX
23 ;; clobbers: xmm1, xmm2
24 max4_sse4:
25 00000020 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
26 00000025 660F383DC8 pmaxsd xmm1, xmm0
27
28 0000002A 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
29 0000002F 660F383DCA pmaxsd xmm1, xmm2
30
31 00000034 660F76C8 pcmpeqd xmm1, xmm0 ; 0 / -1
32
33 00000038 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
34 0000003B C3 ret
size = 0x3C - 0x20 = 28 bytes
Or if we accept input as float
, we can use SSE1 instructions. The float
format can represent a wide range of integer values...
Or if you think that's bending the rules too far, start with a 3-byte 0F 5B C0 cvtdq2ps xmm0, xmm0
to convert, making a 27-byte function that works for all integers that are exactly representable as IEEE binary32 float
, and many combinations of inputs where some of the inputs get rounded to a multiple of 2, 4, 8, or whatever during conversion. (So it's 1 byte smaller than the SSE4.1 version, and works on any x86-64 with just SSE2.)
If any of the float inputs are NaN, note that maxps a,b
exactly implements (a<b) ? a : b
, keeping the element from the 2nd operand on unordered. So it might be possible for this to return with a non-zero bitmap even if the input contains some NaN, depending on where they are.
unsigned max4_sse2(__m128);
37 global max4_sse2
38 ;; Input 4x float32 in xmm0
39 ;; output: bitmap in EAX
40 ;; clobbers: xmm1, xmm2
41 max4_sse2:
42 ; cvtdq2ps xmm0, xmm0
43 00000040 660F70C8B1 pshufd xmm1, xmm0, 0b10110001 ; swap adjacent pairs
44 00000045 0F5FC8 maxps xmm1, xmm0
45
46 00000048 660F70D14E pshufd xmm2, xmm1, 0b01001110 ; swap high/low halves
47 0000004D 0F5FCA maxps xmm1, xmm2
48
49 00000050 0FC2C800 cmpeqps xmm1, xmm0 ; 0 / -1
50
51 00000054 0F50C1 movmskps eax, xmm1 ; extract the high bit of each dword
52 00000057 C3 ret
size = 0x58 - 0x40 = 24 bytes
copy-and-shuffle with pshufd
is still our best bet: shufps dst,src,imm8
reads the input for the low half of dst
from dst
. And we need a non-destructive copy-and-shuffle both times, so 3-byte movhlps
and unpckhps
/pd are both out. If we were narrowing to a scalar max, we could use those, but it costs another instruction to broadcast before compare if we don't have the max in all elements already.
Related: SSE4.1 phminposuw
can find the position and value of the minimum uint16_t
in an XMM register. I don't think it's a win to subtract from 65535 to use it for max, but see an SO answer about using it for max of bytes or signed integers.
edited Mar 23 at 4:46
answered Mar 5 at 9:27
Peter CordesPeter Cordes
2,1521018
2,1521018
add a comment |
add a comment |
$begingroup$
Python 3.8 (pre-release), 67 bytes
Lambda function that takes in 4 integers, bit shifts the boolean result of their comparison to the maximum value with some help from Python 3.8's new assignment operator, and returns the bitwise OR of the results
lambda a,b,c,d:((m:=max(a,b,c,d))==d)<<3|(m==c)<<2|(m==b)<<2|(a==m)
Try it online!
$endgroup$
$begingroup$
:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
$endgroup$
– ElPedro
Mar 4 at 23:54
add a comment |
$begingroup$
Python 3.8 (pre-release), 67 bytes
Lambda function that takes in 4 integers, bit shifts the boolean result of their comparison to the maximum value with some help from Python 3.8's new assignment operator, and returns the bitwise OR of the results
lambda a,b,c,d:((m:=max(a,b,c,d))==d)<<3|(m==c)<<2|(m==b)<<2|(a==m)
Try it online!
$endgroup$
$begingroup$
:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
$endgroup$
– ElPedro
Mar 4 at 23:54
add a comment |
$begingroup$
Python 3.8 (pre-release), 67 bytes
Lambda function that takes in 4 integers, bit shifts the boolean result of their comparison to the maximum value with some help from Python 3.8's new assignment operator, and returns the bitwise OR of the results
lambda a,b,c,d:((m:=max(a,b,c,d))==d)<<3|(m==c)<<2|(m==b)<<2|(a==m)
Try it online!
$endgroup$
Python 3.8 (pre-release), 67 bytes
Lambda function that takes in 4 integers, bit shifts the boolean result of their comparison to the maximum value with some help from Python 3.8's new assignment operator, and returns the bitwise OR of the results
lambda a,b,c,d:((m:=max(a,b,c,d))==d)<<3|(m==c)<<2|(m==b)<<2|(a==m)
Try it online!
answered Mar 4 at 22:01
senox13senox13
1315
1315
$begingroup$
:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
$endgroup$
– ElPedro
Mar 4 at 23:54
add a comment |
$begingroup$
:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
$endgroup$
– ElPedro
Mar 4 at 23:54
$begingroup$
:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
$endgroup$
– ElPedro
Mar 4 at 23:54
$begingroup$
:= reminds me of the old days where that was the assignment operator with Lotus Notes formula. Guess I'll have to take a look at 3.8 for old times sake :)
$endgroup$
– ElPedro
Mar 4 at 23:54
add a comment |
$begingroup$
Java (JDK), 78 bytes
a->{int m=a[0],w=1,p=1;for(int t:a){w=t<m?w:p|(t>m?(m=t)^m:w);p*=2;}return w;}
Try it online!
- Takes the input as an array of
[a,b,c,d]
.
$endgroup$
add a comment |
$begingroup$
Java (JDK), 78 bytes
a->{int m=a[0],w=1,p=1;for(int t:a){w=t<m?w:p|(t>m?(m=t)^m:w);p*=2;}return w;}
Try it online!
- Takes the input as an array of
[a,b,c,d]
.
$endgroup$
add a comment |
$begingroup$
Java (JDK), 78 bytes
a->{int m=a[0],w=1,p=1;for(int t:a){w=t<m?w:p|(t>m?(m=t)^m:w);p*=2;}return w;}
Try it online!
- Takes the input as an array of
[a,b,c,d]
.
$endgroup$
Java (JDK), 78 bytes
a->{int m=a[0],w=1,p=1;for(int t:a){w=t<m?w:p|(t>m?(m=t)^m:w);p*=2;}return w;}
Try it online!
- Takes the input as an array of
[a,b,c,d]
.
edited Mar 4 at 22:45
answered Mar 4 at 11:10
Olivier GrégoireOlivier Grégoire
9,39511944
9,39511944
add a comment |
add a comment |
$begingroup$
05AB1E, 3 2 bytes
ZQ
Input as a list of [d,c,b,a]
, output as a list of boolean.
Try it online or verify all test cases.
Explanation:
Z # Take the maximum of the implicit input-list
Q # Check for each in the (implicit) input-list if it equals this value
# (and output implicitly as result)
$endgroup$
$begingroup$
OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
05AB1E, 3 2 bytes
ZQ
Input as a list of [d,c,b,a]
, output as a list of boolean.
Try it online or verify all test cases.
Explanation:
Z # Take the maximum of the implicit input-list
Q # Check for each in the (implicit) input-list if it equals this value
# (and output implicitly as result)
$endgroup$
$begingroup$
OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
05AB1E, 3 2 bytes
ZQ
Input as a list of [d,c,b,a]
, output as a list of boolean.
Try it online or verify all test cases.
Explanation:
Z # Take the maximum of the implicit input-list
Q # Check for each in the (implicit) input-list if it equals this value
# (and output implicitly as result)
$endgroup$
05AB1E, 3 2 bytes
ZQ
Input as a list of [d,c,b,a]
, output as a list of boolean.
Try it online or verify all test cases.
Explanation:
Z # Take the maximum of the implicit input-list
Q # Check for each in the (implicit) input-list if it equals this value
# (and output implicitly as result)
edited Mar 5 at 7:21
answered Mar 4 at 9:23
Kevin CruijssenKevin Cruijssen
42.3k570217
42.3k570217
$begingroup$
OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
$begingroup$
OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
$begingroup$
OP has been updated — you don't need to convert to hex.
$endgroup$
– Adám
Mar 5 at 5:42
add a comment |
$begingroup$
JavaScript (ES6), 30 bytes
Takes input as ([d,c,b,a])
. Returns 4 Boolean values.
a=>a.map(x=>x==Math.max(...a))
Try it online!
$endgroup$
1
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
add a comment |
$begingroup$
JavaScript (ES6), 30 bytes
Takes input as ([d,c,b,a])
. Returns 4 Boolean values.
a=>a.map(x=>x==Math.max(...a))
Try it online!
$endgroup$
1
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
add a comment |
$begingroup$
JavaScript (ES6), 30 bytes
Takes input as ([d,c,b,a])
. Returns 4 Boolean values.
a=>a.map(x=>x==Math.max(...a))
Try it online!
$endgroup$
JavaScript (ES6), 30 bytes
Takes input as ([d,c,b,a])
. Returns 4 Boolean values.
a=>a.map(x=>x==Math.max(...a))
Try it online!
edited Mar 5 at 10:20
answered Mar 4 at 8:39
ArnauldArnauld
80.4k797333
80.4k797333
1
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
add a comment |
1
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
1
1
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
$begingroup$
OP has clarified that you can indeed return 4 Boolean values.
$endgroup$
– Adám
Mar 5 at 5:41
add a comment |
$begingroup$
Ruby, 34 22 bytes
Takes input as an array [d, c, b, a]
and returns an array of 1s and 0s.
->r{r.map{|e|e/r.max}}
Try it online!
$endgroup$
add a comment |
$begingroup$
Ruby, 34 22 bytes
Takes input as an array [d, c, b, a]
and returns an array of 1s and 0s.
->r{r.map{|e|e/r.max}}
Try it online!
$endgroup$
add a comment |
$begingroup$
Ruby, 34 22 bytes
Takes input as an array [d, c, b, a]
and returns an array of 1s and 0s.
->r{r.map{|e|e/r.max}}
Try it online!
$endgroup$
Ruby, 34 22 bytes
Takes input as an array [d, c, b, a]
and returns an array of 1s and 0s.
->r{r.map{|e|e/r.max}}
Try it online!
edited Mar 5 at 15:52
answered Mar 5 at 0:00
iamnotmaynardiamnotmaynard
93349
93349
add a comment |
add a comment |
$begingroup$
Python 3, 59 bytes 66 bytes
def f(l):
n=max(a)
for i in 0,1,2,3:a[i]=a[i]==n
return a[::-1]
Try it online!
Takes input as [a,b,c,d]
and outputs a list of booleans.
Edited to be a proper function, then saved 2 bytes by removing brackets around the conditional.
$endgroup$
1
$begingroup$
Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
$endgroup$
– Jonathan Frech
Mar 5 at 9:23
1
$begingroup$
Edited. First time around. Appreciate the heads up!
$endgroup$
– Bsoned
Mar 5 at 17:53
$begingroup$
You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
$endgroup$
– Value Ink
Mar 5 at 23:27
$begingroup$
@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
add a comment |
$begingroup$
Python 3, 59 bytes 66 bytes
def f(l):
n=max(a)
for i in 0,1,2,3:a[i]=a[i]==n
return a[::-1]
Try it online!
Takes input as [a,b,c,d]
and outputs a list of booleans.
Edited to be a proper function, then saved 2 bytes by removing brackets around the conditional.
$endgroup$
1
$begingroup$
Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
$endgroup$
– Jonathan Frech
Mar 5 at 9:23
1
$begingroup$
Edited. First time around. Appreciate the heads up!
$endgroup$
– Bsoned
Mar 5 at 17:53
$begingroup$
You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
$endgroup$
– Value Ink
Mar 5 at 23:27
$begingroup$
@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
add a comment |
$begingroup$
Python 3, 59 bytes 66 bytes
def f(l):
n=max(a)
for i in 0,1,2,3:a[i]=a[i]==n
return a[::-1]
Try it online!
Takes input as [a,b,c,d]
and outputs a list of booleans.
Edited to be a proper function, then saved 2 bytes by removing brackets around the conditional.
$endgroup$
Python 3, 59 bytes 66 bytes
def f(l):
n=max(a)
for i in 0,1,2,3:a[i]=a[i]==n
return a[::-1]
Try it online!
Takes input as [a,b,c,d]
and outputs a list of booleans.
Edited to be a proper function, then saved 2 bytes by removing brackets around the conditional.
edited Mar 5 at 17:58
answered Mar 5 at 6:23
BsonedBsoned
214
214
1
$begingroup$
Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
$endgroup$
– Jonathan Frech
Mar 5 at 9:23
1
$begingroup$
Edited. First time around. Appreciate the heads up!
$endgroup$
– Bsoned
Mar 5 at 17:53
$begingroup$
You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
$endgroup$
– Value Ink
Mar 5 at 23:27
$begingroup$
@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
add a comment |
1
$begingroup$
Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
$endgroup$
– Jonathan Frech
Mar 5 at 9:23
1
$begingroup$
Edited. First time around. Appreciate the heads up!
$endgroup$
– Bsoned
Mar 5 at 17:53
$begingroup$
You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
$endgroup$
– Value Ink
Mar 5 at 23:27
$begingroup$
@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
1
1
$begingroup$
Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
$endgroup$
– Jonathan Frech
Mar 5 at 9:23
$begingroup$
Hello and welcome to PPCG. As it stands, your answer has the form of a snippet, which is disallowed. Please fix your answer to conform with our I/O consensus, i.e. make it a function or full program.
$endgroup$
– Jonathan Frech
Mar 5 at 9:23
1
1
$begingroup$
Edited. First time around. Appreciate the heads up!
$endgroup$
– Bsoned
Mar 5 at 17:53
$begingroup$
Edited. First time around. Appreciate the heads up!
$endgroup$
– Bsoned
Mar 5 at 17:53
$begingroup$
You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
$endgroup$
– Value Ink
Mar 5 at 23:27
$begingroup$
You can reduce it to 37 bytes using this list comprehension within a lambda. Welcome to PPCG, and enjoy your stay!
$endgroup$
– Value Ink
Mar 5 at 23:27
$begingroup$
@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
$begingroup$
@ValueInk Removing the superfluous whitespace saves yet another byte.
$endgroup$
– Jonathan Frech
Mar 6 at 16:32
add a comment |
$begingroup$
1. Python 3.5, 90 bytes
Takes sequence of numbers as parameters. Returns "binary" string
import sys;v=[*map(int,sys.argv[1:])];m=max(v);s=""
for e in v:s=str(int(e==m))+s
print(s)
example:
$ ./script.py 6 77 1 4 77
10010
Explanation
import sys
# convert list of string parameters to list of integers
v=[*map(int,sys.argv[1:])]
# get max
m=max(v)
# init outstring
s=""
# walk through list
for e in v:
# prepend to outstring: int(True)=>1, int(False)=>0
s=str(int(e==m))+s
# print out result
print(s)
$endgroup$
add a comment |
$begingroup$
1. Python 3.5, 90 bytes
Takes sequence of numbers as parameters. Returns "binary" string
import sys;v=[*map(int,sys.argv[1:])];m=max(v);s=""
for e in v:s=str(int(e==m))+s
print(s)
example:
$ ./script.py 6 77 1 4 77
10010
Explanation
import sys
# convert list of string parameters to list of integers
v=[*map(int,sys.argv[1:])]
# get max
m=max(v)
# init outstring
s=""
# walk through list
for e in v:
# prepend to outstring: int(True)=>1, int(False)=>0
s=str(int(e==m))+s
# print out result
print(s)
$endgroup$
add a comment |
$begingroup$
1. Python 3.5, 90 bytes
Takes sequence of numbers as parameters. Returns "binary" string
import sys;v=[*map(int,sys.argv[1:])];m=max(v);s=""
for e in v:s=str(int(e==m))+s
print(s)
example:
$ ./script.py 6 77 1 4 77
10010
Explanation
import sys
# convert list of string parameters to list of integers
v=[*map(int,sys.argv[1:])]
# get max
m=max(v)
# init outstring
s=""
# walk through list
for e in v:
# prepend to outstring: int(True)=>1, int(False)=>0
s=str(int(e==m))+s
# print out result
print(s)
$endgroup$
1. Python 3.5, 90 bytes
Takes sequence of numbers as parameters. Returns "binary" string
import sys;v=[*map(int,sys.argv[1:])];m=max(v);s=""
for e in v:s=str(int(e==m))+s
print(s)
example:
$ ./script.py 6 77 1 4 77
10010
Explanation
import sys
# convert list of string parameters to list of integers
v=[*map(int,sys.argv[1:])]
# get max
m=max(v)
# init outstring
s=""
# walk through list
for e in v:
# prepend to outstring: int(True)=>1, int(False)=>0
s=str(int(e==m))+s
# print out result
print(s)
edited Mar 7 at 7:27
answered Mar 6 at 20:31
ReneRene
1413
1413
add a comment |
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 26 bytes
n=>n.Select(a=>a==n.Max())
Try it online!
Takes input in format [d,c,b,a]
. All the others down below take input as [a,b,c,d]
C# (Visual C# Interactive Compiler), 35 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max())
Returns an IEnumerable<bool>
.
Try it online!
C# (Visual C# Interactive Compiler), 39 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max()?1:0)
Returns an IEnumerable<int>
, which represent bits.
Try it online!
C# (Visual C# Interactive Compiler), 49 bytes
n=>{for(int i=4;i-->0;Write(n[i]==n.Max()?1:0));}
Prints a binary string to STDOUT.
Try it online!
$endgroup$
$begingroup$
IEnumerable<bool>
is acceptable.
$endgroup$
– Adám
Mar 5 at 5:45
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 26 bytes
n=>n.Select(a=>a==n.Max())
Try it online!
Takes input in format [d,c,b,a]
. All the others down below take input as [a,b,c,d]
C# (Visual C# Interactive Compiler), 35 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max())
Returns an IEnumerable<bool>
.
Try it online!
C# (Visual C# Interactive Compiler), 39 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max()?1:0)
Returns an IEnumerable<int>
, which represent bits.
Try it online!
C# (Visual C# Interactive Compiler), 49 bytes
n=>{for(int i=4;i-->0;Write(n[i]==n.Max()?1:0));}
Prints a binary string to STDOUT.
Try it online!
$endgroup$
$begingroup$
IEnumerable<bool>
is acceptable.
$endgroup$
– Adám
Mar 5 at 5:45
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 26 bytes
n=>n.Select(a=>a==n.Max())
Try it online!
Takes input in format [d,c,b,a]
. All the others down below take input as [a,b,c,d]
C# (Visual C# Interactive Compiler), 35 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max())
Returns an IEnumerable<bool>
.
Try it online!
C# (Visual C# Interactive Compiler), 39 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max()?1:0)
Returns an IEnumerable<int>
, which represent bits.
Try it online!
C# (Visual C# Interactive Compiler), 49 bytes
n=>{for(int i=4;i-->0;Write(n[i]==n.Max()?1:0));}
Prints a binary string to STDOUT.
Try it online!
$endgroup$
C# (Visual C# Interactive Compiler), 26 bytes
n=>n.Select(a=>a==n.Max())
Try it online!
Takes input in format [d,c,b,a]
. All the others down below take input as [a,b,c,d]
C# (Visual C# Interactive Compiler), 35 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max())
Returns an IEnumerable<bool>
.
Try it online!
C# (Visual C# Interactive Compiler), 39 bytes
n=>n.Select((a,b)=>n[3-b]==n.Max()?1:0)
Returns an IEnumerable<int>
, which represent bits.
Try it online!
C# (Visual C# Interactive Compiler), 49 bytes
n=>{for(int i=4;i-->0;Write(n[i]==n.Max()?1:0));}
Prints a binary string to STDOUT.
Try it online!
edited Mar 13 at 3:38
answered Mar 4 at 16:34
Embodiment of IgnoranceEmbodiment of Ignorance
2,778127
2,778127
$begingroup$
IEnumerable<bool>
is acceptable.
$endgroup$
– Adám
Mar 5 at 5:45
add a comment |
$begingroup$
IEnumerable<bool>
is acceptable.
$endgroup$
– Adám
Mar 5 at 5:45
$begingroup$
IEnumerable<bool>
is acceptable.$endgroup$
– Adám
Mar 5 at 5:45
$begingroup$
IEnumerable<bool>
is acceptable.$endgroup$
– Adám
Mar 5 at 5:45
add a comment |
$begingroup$
PHP, 54 bytes
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i-1;echo$r;
or
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i;echo$r/2;
take input from command line arguments. Run with -nr
or try them online.
$endgroup$
add a comment |
$begingroup$
PHP, 54 bytes
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i-1;echo$r;
or
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i;echo$r/2;
take input from command line arguments. Run with -nr
or try them online.
$endgroup$
add a comment |
$begingroup$
PHP, 54 bytes
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i-1;echo$r;
or
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i;echo$r/2;
take input from command line arguments. Run with -nr
or try them online.
$endgroup$
PHP, 54 bytes
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i-1;echo$r;
or
while($i<4)$argv[++$i]<max($argv)||$r+=1<<$i;echo$r/2;
take input from command line arguments. Run with -nr
or try them online.
answered Mar 4 at 16:12
TitusTitus
13.4k11240
13.4k11240
add a comment |
add a comment |
$begingroup$
Here's a JS version that outputs as binary
update: Shorter with join, and without the lookup:
JavaScript (Node.js), 42 bytes
a=>a.map(x=>+(x==Math.max(...a))).join('')
Try it online!
Previous, with lookup, 49 bytes
a=>a.map(x=>[0,1][+(x==Math.max(...a))]).join('')
Try it online!
Previous, with reduce, 52 bytes:
a=>a.reduce((y,x)=>y+[0,1][+(x==Math.max(...a))],'')
Try it online!
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
$endgroup$
1
$begingroup$
You can safely remove[0,1][...]
since you're using an index that already is either $0$ or $1$.
$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
add a comment |
$begingroup$
Here's a JS version that outputs as binary
update: Shorter with join, and without the lookup:
JavaScript (Node.js), 42 bytes
a=>a.map(x=>+(x==Math.max(...a))).join('')
Try it online!
Previous, with lookup, 49 bytes
a=>a.map(x=>[0,1][+(x==Math.max(...a))]).join('')
Try it online!
Previous, with reduce, 52 bytes:
a=>a.reduce((y,x)=>y+[0,1][+(x==Math.max(...a))],'')
Try it online!
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
$endgroup$
1
$begingroup$
You can safely remove[0,1][...]
since you're using an index that already is either $0$ or $1$.
$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
add a comment |
$begingroup$
Here's a JS version that outputs as binary
update: Shorter with join, and without the lookup:
JavaScript (Node.js), 42 bytes
a=>a.map(x=>+(x==Math.max(...a))).join('')
Try it online!
Previous, with lookup, 49 bytes
a=>a.map(x=>[0,1][+(x==Math.max(...a))]).join('')
Try it online!
Previous, with reduce, 52 bytes:
a=>a.reduce((y,x)=>y+[0,1][+(x==Math.max(...a))],'')
Try it online!
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
$endgroup$
Here's a JS version that outputs as binary
update: Shorter with join, and without the lookup:
JavaScript (Node.js), 42 bytes
a=>a.map(x=>+(x==Math.max(...a))).join('')
Try it online!
Previous, with lookup, 49 bytes
a=>a.map(x=>[0,1][+(x==Math.max(...a))]).join('')
Try it online!
Previous, with reduce, 52 bytes:
a=>a.reduce((y,x)=>y+[0,1][+(x==Math.max(...a))],'')
Try it online!
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
fa=>a.map(x=>+(x==Math.max(...a))).join('')
console.log(f([ 4, 1,77, 6])) // 0010
console.log(f([10,10, 5, 4])) // 1100
console.log(f([ 1, 1, 1, 1])) // 1111
edited Mar 4 at 16:45
answered Mar 4 at 15:18
PureferretPureferret
535726
535726
1
$begingroup$
You can safely remove[0,1][...]
since you're using an index that already is either $0$ or $1$.
$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
add a comment |
1
$begingroup$
You can safely remove[0,1][...]
since you're using an index that already is either $0$ or $1$.
$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
1
1
$begingroup$
You can safely remove
[0,1][...]
since you're using an index that already is either $0$ or $1$.$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
You can safely remove
[0,1][...]
since you're using an index that already is either $0$ or $1$.$endgroup$
– Arnauld
Mar 4 at 16:38
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
$begingroup$
@Arnauld seems obvious now. Thanks!
$endgroup$
– Pureferret
Mar 4 at 16:54
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 51 bytes
x=>{for(int m=x.Max(),i=4;i-->0;)x[i]=x[i]==m?1:0;}
Try it online!
Above is an anonymous function that outputs by modifying an argument. Output is an array of 1's and 0's.
Below is a recursive function that outputs an integer.
C# (Visual C# Interactive Compiler), 60 bytes
int f(intx,int i=3)=>i<0?0:2*f(x,i-1)|(x[i]==x.Max()?1:0);
Try it online!
Both functions take input as a 4-element array.
[d, c, b, a]
$endgroup$
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 51 bytes
x=>{for(int m=x.Max(),i=4;i-->0;)x[i]=x[i]==m?1:0;}
Try it online!
Above is an anonymous function that outputs by modifying an argument. Output is an array of 1's and 0's.
Below is a recursive function that outputs an integer.
C# (Visual C# Interactive Compiler), 60 bytes
int f(intx,int i=3)=>i<0?0:2*f(x,i-1)|(x[i]==x.Max()?1:0);
Try it online!
Both functions take input as a 4-element array.
[d, c, b, a]
$endgroup$
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
add a comment |
$begingroup$
C# (Visual C# Interactive Compiler), 51 bytes
x=>{for(int m=x.Max(),i=4;i-->0;)x[i]=x[i]==m?1:0;}
Try it online!
Above is an anonymous function that outputs by modifying an argument. Output is an array of 1's and 0's.
Below is a recursive function that outputs an integer.
C# (Visual C# Interactive Compiler), 60 bytes
int f(intx,int i=3)=>i<0?0:2*f(x,i-1)|(x[i]==x.Max()?1:0);
Try it online!
Both functions take input as a 4-element array.
[d, c, b, a]
$endgroup$
C# (Visual C# Interactive Compiler), 51 bytes
x=>{for(int m=x.Max(),i=4;i-->0;)x[i]=x[i]==m?1:0;}
Try it online!
Above is an anonymous function that outputs by modifying an argument. Output is an array of 1's and 0's.
Below is a recursive function that outputs an integer.
C# (Visual C# Interactive Compiler), 60 bytes
int f(intx,int i=3)=>i<0?0:2*f(x,i-1)|(x[i]==x.Max()?1:0);
Try it online!
Both functions take input as a 4-element array.
[d, c, b, a]
edited Mar 4 at 17:12
answered Mar 4 at 13:56
danadana
1,921167
1,921167
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
add a comment |
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
You don't need to output an integer.
$endgroup$
– Adám
Mar 5 at 5:43
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
$begingroup$
@Adam - thanks :) I realized this after posting while I was working on my other answer. Before I had a chance to change, there was another C# answer that used a lot of the good tricks.
$endgroup$
– dana
Mar 5 at 19:11
add a comment |
$begingroup$
Python 2, 35 bytes
lambda i:[int(x==max(i))for x in i]
Try it online!
Takes input in the format [d,c,b,a] as with the accepted answer from Adám so I guess it is OK.
Alternative for 41 if it's not...
Python 2, 41 bytes
lambda i:[int(x==max(i))for x in i][::-1]
Try it online!
$endgroup$
add a comment |
$begingroup$
Python 2, 35 bytes
lambda i:[int(x==max(i))for x in i]
Try it online!
Takes input in the format [d,c,b,a] as with the accepted answer from Adám so I guess it is OK.
Alternative for 41 if it's not...
Python 2, 41 bytes
lambda i:[int(x==max(i))for x in i][::-1]
Try it online!
$endgroup$
add a comment |
$begingroup$
Python 2, 35 bytes
lambda i:[int(x==max(i))for x in i]
Try it online!
Takes input in the format [d,c,b,a] as with the accepted answer from Adám so I guess it is OK.
Alternative for 41 if it's not...
Python 2, 41 bytes
lambda i:[int(x==max(i))for x in i][::-1]
Try it online!
$endgroup$
Python 2, 35 bytes
lambda i:[int(x==max(i))for x in i]
Try it online!
Takes input in the format [d,c,b,a] as with the accepted answer from Adám so I guess it is OK.
Alternative for 41 if it's not...
Python 2, 41 bytes
lambda i:[int(x==max(i))for x in i][::-1]
Try it online!
edited Mar 5 at 0:03
answered Mar 4 at 21:41
ElPedroElPedro
3,5531023
3,5531023
add a comment |
add a comment |
$begingroup$
Python 3, 42 bytes
f=lambda a:list(map(lambda x:x==max(a),a))
Simply returns a list of whether the element is the max for each element in the input. -2 bytes if you don't count the f=
assignment.
Try it online!
$endgroup$
$begingroup$
f=
doesn't count except in recursive functions
$endgroup$
– ASCII-only
Mar 13 at 3:47
add a comment |
$begingroup$
Python 3, 42 bytes
f=lambda a:list(map(lambda x:x==max(a),a))
Simply returns a list of whether the element is the max for each element in the input. -2 bytes if you don't count the f=
assignment.
Try it online!
$endgroup$
$begingroup$
f=
doesn't count except in recursive functions
$endgroup$
– ASCII-only
Mar 13 at 3:47
add a comment |
$begingroup$
Python 3, 42 bytes
f=lambda a:list(map(lambda x:x==max(a),a))
Simply returns a list of whether the element is the max for each element in the input. -2 bytes if you don't count the f=
assignment.
Try it online!
$endgroup$
Python 3, 42 bytes
f=lambda a:list(map(lambda x:x==max(a),a))
Simply returns a list of whether the element is the max for each element in the input. -2 bytes if you don't count the f=
assignment.
Try it online!
answered Mar 7 at 18:07
ThePlasmaRailgunThePlasmaRailgun
1789
1789
$begingroup$
f=
doesn't count except in recursive functions
$endgroup$
– ASCII-only
Mar 13 at 3:47
add a comment |
$begingroup$
f=
doesn't count except in recursive functions
$endgroup$
– ASCII-only
Mar 13 at 3:47
$begingroup$
f=
doesn't count except in recursive functions$endgroup$
– ASCII-only
Mar 13 at 3:47
$begingroup$
f=
doesn't count except in recursive functions$endgroup$
– ASCII-only
Mar 13 at 3:47
add a comment |
$begingroup$
Batch, 92 bytes
@set m=%1
@set f=@for %%i in (%*)do @
%f%set/a"m=m+(m-=%%i)*(m>>31)
%f%cmd/cset/a!(m-%%i)
Takes arguments as command-line parameters in reverse order. Works by arithmetically calculating the maximum of the parameters by reducing over them and adding on only positive differences from the running maximum, then mapping over each parameter again this time comparing it to the maximum. Conveniently cmd/cset/a
doesn't output a newline, so the results are automatically concatenated together. The %f%
simply saves 5 bytes on what would be a repeated construct.
$endgroup$
add a comment |
$begingroup$
Batch, 92 bytes
@set m=%1
@set f=@for %%i in (%*)do @
%f%set/a"m=m+(m-=%%i)*(m>>31)
%f%cmd/cset/a!(m-%%i)
Takes arguments as command-line parameters in reverse order. Works by arithmetically calculating the maximum of the parameters by reducing over them and adding on only positive differences from the running maximum, then mapping over each parameter again this time comparing it to the maximum. Conveniently cmd/cset/a
doesn't output a newline, so the results are automatically concatenated together. The %f%
simply saves 5 bytes on what would be a repeated construct.
$endgroup$
add a comment |
$begingroup$
Batch, 92 bytes
@set m=%1
@set f=@for %%i in (%*)do @
%f%set/a"m=m+(m-=%%i)*(m>>31)
%f%cmd/cset/a!(m-%%i)
Takes arguments as command-line parameters in reverse order. Works by arithmetically calculating the maximum of the parameters by reducing over them and adding on only positive differences from the running maximum, then mapping over each parameter again this time comparing it to the maximum. Conveniently cmd/cset/a
doesn't output a newline, so the results are automatically concatenated together. The %f%
simply saves 5 bytes on what would be a repeated construct.
$endgroup$
Batch, 92 bytes
@set m=%1
@set f=@for %%i in (%*)do @
%f%set/a"m=m+(m-=%%i)*(m>>31)
%f%cmd/cset/a!(m-%%i)
Takes arguments as command-line parameters in reverse order. Works by arithmetically calculating the maximum of the parameters by reducing over them and adding on only positive differences from the running maximum, then mapping over each parameter again this time comparing it to the maximum. Conveniently cmd/cset/a
doesn't output a newline, so the results are automatically concatenated together. The %f%
simply saves 5 bytes on what would be a repeated construct.
answered Mar 13 at 21:29
NeilNeil
82.6k745179
82.6k745179
add a comment |
add a comment |
If this is an answer to a challenge…
…Be sure to follow the challenge specification. However, please refrain from exploiting obvious loopholes. Answers abusing any of the standard loopholes are considered invalid. If you think a specification is unclear or underspecified, comment on the question instead.
…Try to optimize your score. For instance, answers to code-golf challenges should attempt to be as short as possible. You can always include a readable version of the code in addition to the competitive one.
Explanations of your answer make it more interesting to read and are very much encouraged.…Include a short header which indicates the language(s) of your code and its score, as defined by the challenge.
More generally…
…Please make sure to answer the question and provide sufficient detail.
…Avoid asking for help, clarification or responding to other answers (use comments instead).
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2
$begingroup$
I have a solution in a golfing language, which uses the builtins Reverse, Maximum, Equality-check, Join, Convert from binary to integer, Convert from integer to hexadecimal. Does this means my score is 1 due to the Equality-check? I have the feeling this is too much focused on regular languages, and even for those it's not 100% clear what the scoring for let's say a Maximum-builtin.. :S
$endgroup$
– Kevin Cruijssen
Mar 4 at 7:39
1
$begingroup$
I would suggest you try to: 1. change this question to code-golf, which only care about the number of bytes. 2. or, restrict to some certain language (certain version of compiler/interpreter please), and list all statements and operators allowed, and how to score them.
$endgroup$
– tsh
Mar 4 at 7:43
5
$begingroup$
1 is a better option, IMO. I think this makes a perfectly good code-golf question and I can't see any benefit that would come from restricting the languages available for answers
$endgroup$
– senox13
Mar 4 at 7:54
2
$begingroup$
I updated my question to remove the criteria. Let me know it it's still unclear
$endgroup$
– Mr Anderson
Mar 4 at 8:09
5
$begingroup$
Should I output a decimal number? Or may I output 4 binary digits instead?
$endgroup$
– tsh
Mar 4 at 8:51