Generate triangular signal

Question

Task:

Given sample index, x, calculate sample value f(x) of triangular wave, with period of 4 samples and amplitude 1. Offset can be negative and sample value could be either {0, 1, -1}.

Test cases:

   -5 -> -1
   -4 -> 0
   -3 -> 1
   -2 -> 0
   -1 -> -1
    0 -> 0
    1 -> 1
    2 -> 0
    3 -> -1
    4 -> 0
    5 -> 1

Personally I know two approaches in C - the first is using lookup table, the second is using conditional instructions. For brownie points, could you impress me with a pure "math" approach? (I mean a pure functional approach, e.g. not using conditional instructions or using memory for LUT.) But this is not an restriction. If you can't, or your language does not support it - just post any solution

What do you mean by "offset can be negative"? Also this is basically just a trigonometric function, so I'd be surprised if it isn't a dupe of something. — FryAmTheEggman, May 16, 2017 at 1:08
@JungHwanMin This has much more relaxed rules, so it's not really a dupe (though it asks for the same thing). — user45941, May 16, 2017 at 1:09

JungHwan Min · Accepted Answer · 2017-05-16 01:18:49Z

12

Mathematica, 8 bytes

Im[I^#]&

Explanation

Im[I^#]&
   I^#    (* Raise the imaginary unit to the input power *)
Im[   ]   (* Take the imaginary part *)

edited May 16, 2017 at 1:18

answered May 16, 2017 at 1:12

JungHwan Min

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3

\$\begingroup\$ Ohh, beautiful approach. How did I not see this? :D \$\endgroup\$
– hyper-neutrino ♦
May 16, 2017 at 1:19
\$\begingroup\$ cant see algo how to generate imaginary unit in C.. =( im assembler man ^^ using builtins in exotic languages is not in my favor ) however is an answer.. \$\endgroup\$
– user69099
May 17, 2017 at 23:54

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Community · Accepted Answer · 2020-06-17 09:04:33Z

7

TI-Basic, 7 5 4 bytes

sin(90Ans

(Degree mode) -1 byte from @immibis from my old answer.

Old answer

imag(i^Ans

Pure-math approach on a calculator. :)

Just for fun, here's another pure-math (ish) solution for 9 bytes (In radian mode), or 8 bytes (Degree mode)

2/πsin^-1sin(πAns/2 # Radians
90^-1sin^-1sin(90Ans # Degrees

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 16, 2017 at 2:43

pizzapants184

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\$\begingroup\$ yeah, but you just forget an imag() implementation.. but usin others code is fine, though.. good answer :) \$\endgroup\$
– user69099
May 18, 2017 at 0:03
2

\$\begingroup\$ @xakepp35 I don't understand. imag() is a valid function on TI-BASIC. \$\endgroup\$
– JungHwan Min
May 18, 2017 at 13:52
\$\begingroup\$ What's wrong with sin(90Ans? Why do you need the extra 90-1sin-1? \$\endgroup\$
– user253751
May 19, 2017 at 0:47
\$\begingroup\$ @immibis The 90^-1sin^-1 makes it a triangle wave for all values, but sin(90Ans works for what the question asks. \$\endgroup\$
– pizzapants184
May 19, 2017 at 3:46

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xnor · Accepted Answer · 2017-05-16 02:24:51Z

6

Python 2, 20 bytes

lambda n:n%2-n%4/3*2

Try it online!

I'm running a brute-force search for shorter arithmetic or bitwise expressions, I'll see if anything turns up. This one I hand-found.

answered May 16, 2017 at 2:24

xnor

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2

\$\begingroup\$ Brute force expression search? Nice! \$\endgroup\$
– Graviton
May 16, 2017 at 2:31

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Dennis · Accepted Answer · 2017-05-16 05:33:23Z

Julia 0.5, 12 bytes

!n=(2-n&3)%2

I like this approach because it's unlikely to be the shortest in any other language.

Try it online!

How it works

Julia's operator precedence is a bit unusual: unlike most other languages, bitwise operators have the same precedence as their arithmetic counterparts, so & (bitwise multiplication) has the same precedence as *.

First, n&3 takes the input modulo 4, with positive sign.

The result – 0, 1, 2, or 3 – is then subtracted from 2, yielding 2, 1, 0, or -1.

Finally, we take the signed remainder of the division by 2, returning 0, 1, 0, or -1.

Dennis · Accepted Answer · 2017-05-16 01:13:57Z

4

Jelly, 3 bytes

ı*Ċ

Try it online!

How it works

ı*Ċ  Main link. Argument: n

ı*   Elevate i, the imaginary unit, to the n-th power.
  Ċ  Take the imaginary part of the result.

answered May 16, 2017 at 1:13

Dennis

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Community · Accepted Answer · 2020-06-17 09:04:33Z

4

dc, 13

Not sure if you count the modulo % operator as "pure math":

?1+d*v4%1-2%p

Try it online. Note that dc uses _ instead of - to indicate negative numbers.

Explanation

?              # read input
 1+            # add 1
   d*v         # duplicate, multiply, square root (poor-mans abs())
      4%       # mod 4
        1-     # subtract 1
          2%   # mod 2
            p  # print

Note that dc's % mod operator is the standard "CPU" version that maps negative values to negative values.

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 16, 2017 at 1:59

Digital Trauma

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\$\begingroup\$ Can you just do abs((x+1)%4)-1 instead? \$\endgroup\$
– Magic Octopus Urn
May 16, 2017 at 18:02

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Community · Accepted Answer · 2020-06-17 09:04:33Z

brainfuck, 136 bytes

>,>++++<[>->+<[>]>[<+>-]<<[<]>-]>>>>-[>+<-----]>--[-<+>>>+>>>+>>>+<<<<<<<<]<->>>>>>->--[>+<++++++]>++<<<<<<<<<<[[->>>+<<<]>>>-]>[.[-]]>.

Try it online!

There's probably a more trivial answer, but this essentially uses a table of values. Although brainfuck takes input as ASCII characters with positive values from 0 to 127, it still works as if it were able to accept negative values (to test, replace the , with n amount of - characters).

How it works

>,                                   take input (X)
>++++<                               take second input for modulo (4)
[>->+<[>]>[<+>-]<<[<]>-]             calculate X mod 4
>>>>-[>+<-----]>--                   create initial '1' character
[-<+>>>+>>>+>>>+<<<<<<<<]            duplicate '1' four times as 1,1,1,1
<->>>>>>->--[>+<++++++]>++<<<<<<<<<< change 1,1,1,1 to 0,1,0,-1 
[[->>>+<<<]>>>-]>[.[-]]>.            move to the right X%4 * 3 times, then print the following two characters ( 0, 1, 0,-1)

hyper-neutrino · Accepted Answer · 2017-05-16 02:31:05Z

1

Python, 26 24 21 bytes

lambda x:(1j**x).imag

-2 bytes thanks to ValueInk for realizing that the mathematical method is actually longer than the trivial approach :P
-3 bytes thanks to Dennis for pointing out that I don't need the int(...), thus making this shorter :)

edited May 16, 2017 at 2:31

answered May 16, 2017 at 1:34

hyper-neutrino♦

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\$\begingroup\$ lambda x:[0,1,0,-1][x%4] is actually shorter than your int-coerced answer lol \$\endgroup\$
– Value Ink
May 16, 2017 at 1:44
\$\begingroup\$ @ValueInk Oh...um this is embarrasing lol \$\endgroup\$
– hyper-neutrino ♦
May 16, 2017 at 1:52
2

\$\begingroup\$ Why would you use int() in the first place though? \$\endgroup\$
– Dennis
May 16, 2017 at 1:59
\$\begingroup\$ @Dennis Because .imag gives a floating-point value and I'm not sure if that's allowed by the specs. It doesn't matter now :) \$\endgroup\$
– hyper-neutrino ♦
May 16, 2017 at 2:19
\$\begingroup\$ If floats aren't allowed, JavaScript will not be able to compete. \$\endgroup\$
– Dennis
May 16, 2017 at 2:28

| Show 1 more comment

Jonathan Allan · Accepted Answer · 2017-05-16 03:01:01Z

1

Python, 20 bytes

lambda n:n%2*(2-n%4)

An unnamed function which returns the result.

Try it online!

answered May 16, 2017 at 3:01

Jonathan Allan

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

Mathematica, 18 bytes

#~JacobiSymbol~46&

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 16, 2017 at 2:03

ZaMoC

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5

\$\begingroup\$ This doesn't quite work: inputs 9 and 11 both give 1 as output, for example. The period of this function is 184, not 4. \$\endgroup\$
– Greg Martin
May 16, 2017 at 4:22
1

\$\begingroup\$ However, JacobiSymbol[-4,#]& does work, and just costs one more byte. Nice idea! \$\endgroup\$
– Greg Martin
May 16, 2017 at 6:15
\$\begingroup\$ again cant see an algrithm( just builtines written by others and couple of short code.. ah, all as usual. fine answer though \$\endgroup\$
– user69099
May 17, 2017 at 23:56

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alephalpha · Accepted Answer · 2017-05-16 03:53:50Z

1

Pari/GP, 12 bytes

n->imag(I^n)

Try it online!

answered May 16, 2017 at 3:53

alephalpha

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Community · Accepted Answer · 2020-06-17 09:04:33Z

1

PHP, 20 bytes

<?=2<=>($argn&3?:2);

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 16, 2017 at 5:44

user63956

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Ørjan Johansen · Accepted Answer · 2017-05-16 17:18:37Z

1

Haskell, 19 bytes

Port of Dennis's Julia solution, just because he said it wouldn't be shortest in any other language. (Someone might still prove me wrong that it's shortest in Haskell.)

f n=rem(2-n`mod`4)2

Try it online!

Haskell has two different remainder functions, one (rem) works like the Julia one, while the other (mod) gives a positive result even when the first argument is negative, and so is suitable for translating &3. (Haskell's actual &, called .&., alas requires an import Data.Bits.)

answered May 16, 2017 at 17:18

Ørjan Johansen

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\$\begingroup\$ As far as I can tell, a port of Dennis's Julia solution is optimal for JavaScript too (14 bytes). Shows that even Dennis is fallible! \$\endgroup\$
– Neil
May 16, 2017 at 22:02

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user41805 · Accepted Answer · 2017-05-16 17:43:53Z

1

Octave, 22 bytes

@(x)round(sin(x*pi/2))

Try it online!

answered May 16, 2017 at 17:43

user41805

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Community · Accepted Answer · 2020-06-17 09:04:33Z

0

Ruby, 20 bytes

Simple and clean.

->x{[0,1,0,-1][x%4]}

edited Jun 17, 2020 at 9:04

CommunityBot

1

answered May 16, 2017 at 1:45

Value Ink

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algmyr · Accepted Answer · 2017-05-16 02:52:50Z

0

C99, 27 bytes

Assuming you want the wave to be centered at the origin:

f(n){return cpow(1i,n)/1i;}

otherwise f(n){return cpow(1i,n);} will do. I originally had a cimag in there, but apparently trying to return an int from a _Complex int yields the real part, so I used that. It makes sense, but it'snnothing I would have predicted. Behavior is the same in gcc and clang

answered May 16, 2017 at 2:52

algmyr

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\$\begingroup\$ cpow is undefined xD \$\endgroup\$
– user69099
May 17, 2017 at 23:59
\$\begingroup\$ some #includes omitted, does not compile))))) \$\endgroup\$
– user69099
May 18, 2017 at 0:00
\$\begingroup\$ but +1 just for C \$\endgroup\$
– user69099
May 18, 2017 at 0:00
1

\$\begingroup\$ @xakepp35 That's actually not the fault of the includes, it's a linker problem. Compile with -std=c99 -lm and it should work. It works fine for me with both gcc and clang without any includes. Well, by fine I mean there are no errors, but a large number of warnings. \$\endgroup\$
– algmyr
May 18, 2017 at 0:06

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Magic Octopus Urn · Accepted Answer · 2017-05-16 18:04:28Z

0

05AB1E, 5 bytes

4%<Ä<

Try it online!

The output is reversed, but from what I understood this is allowed:

+1 byte to multiply output by -1 using (.

   -5 -> 1
   -4 -> 0
   -3 -> -1
   -2 -> 0
   -1 -> 1
    0 -> 0
    1 -> -1
    2 -> 0
    3 -> 1
    4 -> 0
    5 -> -1

4%    # Amplitude of 4...
  <   # Period of 1...
   Ä  # Absolute value...
    < # Period of 1 centered at 0...

edited May 16, 2017 at 18:04

answered May 16, 2017 at 17:09

Magic Octopus Urn

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\$\begingroup\$ what I understood this is allowed \$\endgroup\$
– user69099
May 18, 2017 at 0:03
\$\begingroup\$ test cases are not passed ;-) \$\endgroup\$
– user69099
May 18, 2017 at 0:04
\$\begingroup\$ but +1 nice try \$\endgroup\$
– user69099
May 18, 2017 at 0:04

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Maria · Accepted Answer · 2017-05-16 18:56:12Z

0

Pyth - 7 bytes (possibly 6)

ta2%tQ4

Try it

If the phase of the wave isn't important, 6 bytes:

ta2%Q4

Try it

Explanation:

ta2%tQ4
     Q    # The input
    t     # Subtract 1 to get the phase right (might not be necessary)
   %  4   # Take mod 4
 a2       # Absolute value of the result - 2
t         # Subtract 1 so the result is in [-1,0,1]

edited May 16, 2017 at 18:56

answered May 16, 2017 at 17:48

Maria

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Robert Benson · Accepted Answer · 2017-05-16 21:13:45Z

0

AWK, 26 bytes

{$0=(sqrt(($0%4)^2)-2)%2}1

Try it online!

This is an alternate approach using trig functions without the modulus operator.

{$0=int(sin($0*atan2(0,-1)/2))}1

Try it online !

edited May 16, 2017 at 21:13

answered May 16, 2017 at 20:46

Robert Benson

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Etheryte · Accepted Answer · 2017-05-17 00:50:26Z

0

Javascript ES6, 18 17 bytes

n=>n&1&&(++n&2)-1

Firstly, check if the input is even or odd and return 0 for all even values. For all odd inputs, increment and bitwise with 0b10 to remove any bits that don't interest us, then return the answer with an offset.

const f = n=>n&1&&(++n&2)-1;

for (let i = -5; i < 6; i++) {
  document.body.appendChild(document.createElement('pre')).innerHTML = `f(${i}) => ${f(i)}`;
}

edited May 17, 2017 at 0:50

answered May 16, 2017 at 20:55

Etheryte

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1

\$\begingroup\$ Save a byte by replacing ? :0 with && \$\endgroup\$
– Steve Bennett
May 16, 2017 at 22:28
\$\begingroup\$ @SteveBennett Thanks, great idea! \$\endgroup\$
– Etheryte
May 17, 2017 at 0:50

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kuilin · Accepted Answer · 2017-05-17 03:58:49Z

0

JavaScript, 15 bytes

n=>n&3&&2-(n&3)

Bitwise and 3 is equivalent to modulo 4 except without the weird rule on JavaScript's modulos of negative numbers. I did polynomial regression on the first four points at first but then realized I was being dumb because (1, 1), (2, 0), and (3, -1) is just 2-n.

a=n=>n&1&&2-(n&3);
console.log([a(-5), a(-4), a(-3), a(-2), a(-1), a(0), a(1), a(2), a(3), a(4), a(5)])

answered May 17, 2017 at 3:58

kuilin

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\$\begingroup\$ very well! +1 for the answer \$\endgroup\$
– user69099
May 18, 2017 at 0:05

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Giuseppe · Accepted Answer · 2017-10-02 14:20:38Z

0

R, 19 bytes

function(n)Im(1i^n)

Try it online!

A port of JungHwan Min's Mathematica answer.

answered Oct 2, 2017 at 14:20

Giuseppe

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Stack Exchange Network

Generate triangular signal

Task:

Test cases:

22 Answers 22

Mathematica, 8 bytes

Explanation

TI-Basic, 7 5 4 bytes

Python 2, 20 bytes

Julia 0.5, 12 bytes

How it works

Jelly, 3 bytes

How it works

dc, 13

Explanation

brainfuck, 136 bytes

How it works

Python, 26 24 21 bytes

Python, 20 bytes

Mathematica, 18 bytes

Pari/GP, 12 bytes

PHP, 20 bytes

Haskell, 19 bytes

Octave, 22 bytes

Ruby, 20 bytes

C99, 27 bytes

05AB1E, 5 bytes

Pyth - 7 bytes (possibly 6)

AWK, 26 bytes

Javascript ES6, 18 17 bytes

JavaScript, 15 bytes

R, 19 bytes

Your Answer

Linked

Hot Network Questions

Generate triangular signal

Task:

Test cases:

22 Answers 22

Mathematica, 8 bytes

Explanation

TI-Basic, 7 5 4 bytes

Python 2, 20 bytes

Julia 0.5, 12 bytes

How it works

Jelly, 3 bytes

How it works

dc, 13

Explanation

brainfuck, 136 bytes

How it works

Python, 26 24 21 bytes

Python, 20 bytes

Mathematica, 18 bytes

Pari/GP, 12 bytes

PHP, 20 bytes

Haskell, 19 bytes

Octave, 22 bytes

Ruby, 20 bytes

C99, 27 bytes

05AB1E, 5 bytes

Pyth - 7 bytes (possibly 6)

AWK, 26 bytes

Javascript ES6, 18 17 bytes

JavaScript, 15 bytes

R, 19 bytes

Your Answer

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