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I have
let f = x => x % 4 === 0 ? 0 : 4 - x % 4
But that's a piece of garbage function. Help.
x is never going to be negative.
Here's a sort of Table of Truth, or something.
x x % 4 4 - (x % 4) f(x)
0 0 4 0
1 1 3 3
2 2 2 2
3 3 1 1
4 0 4 0
5 1 3 3
6 2 2 2
7 3 1 1
8 0 4 0
9 1 3 3
I'm trying to find some correlations here, but it's late and I don't think my brain is working correctly. zzz
What I'm seeing in the f(x) column is a sort of reverse modulus, whereby the outputs cycle from 032103210... instead of 01230123...
I'm sensing some use of Math.max or Math.min in combination with Math.abs might help … There's probably an x * -1 in there somewhere too …
Can you help me write f such that it doesn't suck so badly ?
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The modulo operator, denoted by %, is an arithmetic operator. The modulo division operator produces the remainder of an integer division. produces the remainder when x is divided by y.
Modular arithmetic is a system of arithmetic for integers, which considers the remainder. In modular arithmetic, numbers "wrap around" upon reaching a given fixed quantity (this given quantity is known as the modulus) to leave a remainder.
The easiest way to understand modular arithmetic is to think of it as finding the remainder of a number upon division by another number. For example, since both 15 and -9 leave the same remainder 3 when divided by 12, we say that
An intuitive usage of modular arithmetic is with a 12-hour clock. If it is 10:00 now, then in 5 hours the clock will show 3:00 instead of 15:00. 3 is the remainder of 15 with a modulus of 12. N N. Two integers N N. In such a case, we say that a ≡ b ( m o d N). a \equiv b\pmod N. a≡ b (mod N).
Modular division is totally different from modular addition, subtraction and multiplication. It also does not exist always. (a / b) mod m is not equal to ( (a mod m) / (b mod m)) mod m. The modular inverse of a mod m exists only if a and m are relatively prime i.e. gcd (a, m) = 1. if (a x b) mod m = 1 then b is modular inverse of a.
N N. In such a case, we say that a ≡ b ( m o d N). a \equiv b\pmod N. a≡ b (mod N). The easiest way to understand modular arithmetic is to think of it as finding the remainder of a number upon division by another number.
Moving Redu's use of bitwise operators a bit ahead:
f(x) = -x & 3
Table of Truths™
x x -x 3 -x&3 -x&3
- ---- ----- ---- ---- ----
0 0000 0000 0011 0000 0
1 0001 -0001 0011 0011 3
2 0010 -0010 0011 0010 2
3 0011 -0011 0011 0001 1
4 0100 -0100 0011 0000 0
5 0101 -0101 0011 0011 3
6 0110 -0110 0011 0010 2
7 0111 -0111 0011 0001 1
8 1000 -1000 0011 0000 0
9 1001 -1001 0011 0011 3
var i,
f = x => -x & 3;
for (i = 0; i < 20; i++) {
console.log(i, f(i));
}.as-console-wrapper { max-height: 100% !important; top: 0; }Original solution with negative value and a negative offset of 3 then modulo and add the same offset again.
f(x) = (-x - 3) % 4 + 3
var i,
f = x => (-x - 3) % 4 + 3;
for (i = 0; i < 20; i++) {
console.log(i, f(i));
}.as-console-wrapper { max-height: 100% !important; top: 0; }I thought you don't want to use modulo. Then here is your code.
var f = x => 2 * (x & 2 ? ~x & 1 : x & 1) + (x & 1)
x x % 4 4 - (x % 4) f(x)
0 0 4 0
1 1 3 3
2 2 2 2
3 3 1 1
4 0 4 0
5 1 3 3
6 2 2 2
7 3 1 1
8 0 4 0
9 1 3 3
Explanation: I just had to recall the back old days of truth tables which helped me to solve out this problem. So now we have inputs and output. Since we work in modulo 4 we are interested only in the last two bits.
Input Output
0 : 0 0 0 0
1 : 0 1 1 1
2 : 1 0 1 0
3 : 1 1 0 1
So if we look, the output 2^1 digit is XOR of input 2^0 and 2^1 hence 2 * (x & 2 ? ~x & 1 : x & 1) and the output 2^0 digit is in fact input 2^0 digit. Which is (x & 1) hence... var f = x => 2 * (x & 2 ? ~x & 1 : x & 1) + (x & 1)
Note: (foo XOR bar = foo ? !bar : bar)
u v
y z z w
x x & 2 ? ~x y & 1 x & 1 2 * z w + v f(x)
-- ------ ------ --- ------- ------ ------ ----- ----- ----
0 0000 0000 F -0001 0001 0000 0 0 0
1 0001 0000 F -0010 0000 0001 2 3 3
2 0010 0010 T -0011 0001 0000 2 2 2
3 0011 0010 T -0100 0000 0001 0 1 1
4 0100 0000 F -0101 0001 0000 0 0 0
5 0101 0000 F -0110 0000 0001 2 3 3
7 0110 0010 T -0111 0001 0000 2 2 2
8 0111 0010 T -1000 0000 0001 0 1 1
9 1000 0000 F -1001 0001 0000 0 0 0
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