Is it possible to utilize mathematical operators such as multiplication, division, addition, subtraction, and exponentiation to transform a non-zero numerical value into

Question

Is it possible to utilize mathematical operators such as multiplication, division, addition, subtraction, and exponentiation to transform a non-zero numerical value into

I am currently working with Oracle Siebel software which supports JavaScript expressions using only specific operators such as multiply, divide, subtract, add, and XOR (*, /, -, +, ^). Unfortunately, other operators like ! or ? : are not available for use.

Given the limited set of operators mentioned above, I am trying to figure out if it is possible to convert a number to 1 if it is non-zero, and keep it as 0 if it is already zero. This number can be positive, negative, or zero itself.

For example:

var c = 55;

var d;  // d needs to be set as 1

I attempted using c / c, however, this results in NaN when c is 0. In that scenario, d should remain 0.

The value of c represents a currency amount, meaning it will have up to two decimal places and a maximum of 12 digits before the decimal point.

My objective is to simulate an if condition by converting a number into a Boolean 0 or 1, and then incorporating it into other parts of the expression.

javascript siebel

Answer 1

Answer №1

Consider this expression: n/n^0.

If n is anything other than zero:

 Step    Explanation
------- -------------------------------------------------------------------------------
 n/n^0   Original expression.
 1^0     Any number divided by itself equals 1, so n/n simplifies to 1.
 1       The XOR operation of 1 and 0 results in 1.

If n is zero:

 Step    Explanation
------- -------------------------------------------------------------------------------
 n/n^0   Original expression.
 0/0^0   When n is 0, n/n becomes 0/0.
 NaN^0   Dividing zero by zero is undefined mathematically, resulting in NaN.
 0^0     Prior to any bitwise operation, NaN gets normalized to 0.
         Hence, 0 comes out as the result.
 0       XOR between 0 and 0 gives 0 as the output.

In summary, non-zero values convert to 1 while 0 remains unchanged, taking advantage of JavaScript's behavior where NaN^0 evaluates to 0.

Check out the demonstration below:

[0, 1, 19575, -1].forEach(n => console.log(`${n} turns into ${n/n^0}.`))

'

Answer 2

Consider this expression: n/n^0.

If n is anything other than zero:

 Step    Explanation
------- -------------------------------------------------------------------------------
 n/n^0   Original expression.
 1^0     Any number divided by itself equals 1, so n/n simplifies to 1.
 1       The XOR operation of 1 and 0 results in 1.

If n is zero:

 Step    Explanation
------- -------------------------------------------------------------------------------
 n/n^0   Original expression.
 0/0^0   When n is 0, n/n becomes 0/0.
 NaN^0   Dividing zero by zero is undefined mathematically, resulting in NaN.
 0^0     Prior to any bitwise operation, NaN gets normalized to 0.
         Hence, 0 comes out as the result.
 0       XOR between 0 and 0 gives 0 as the output.

In summary, non-zero values convert to 1 while 0 remains unchanged, taking advantage of JavaScript's behavior where NaN^0 evaluates to 0.

Check out the demonstration below:

[0, 1, 19575, -1].forEach(n => console.log(`${n} turns into ${n/n^0}.`))

'

Answer 3

Answer №2

Dividing c by (c plus a tiny value close to zero)

should do the trick. (The constant a tiny value close to zero is Number.MIN_VALUE, which represents the smallest positive number that can be represented.) If the value of x is 0, then the result will be 0, and if x is any value other than 0 (specifically at least 4.45014771701440252e-308 as mentioned in the question), JavaScript's floating-point math limitation causes the result to always be 1 due to imprecision.

Answer 4

Dividing c by (c plus a tiny value close to zero)

should do the trick. (The constant a tiny value close to zero is Number.MIN_VALUE, which represents the smallest positive number that can be represented.) If the value of x is 0, then the result will be 0, and if x is any value other than 0 (specifically at least 4.45014771701440252e-308 as mentioned in the question), JavaScript's floating-point math limitation causes the result to always be 1 due to imprecision.

Answer 5

Answer №3

((c/c)^c - c) * ((c/c)^c - c) will always yield a result of 1 for both positive and negative numbers, and 0 when c is equal to 0.

Although my equation may be more complex compared to the alternative answer, I believe it offers a more elegant solution that does not rely on predefined constants.

UPDATE: In agreement with @JosephSible's suggestion, an even more concise variation of my formula (in collaboration with @CRice) without using constants can be expressed as:

c/c^c-c

Answer 6

((c/c)^c - c) * ((c/c)^c - c) will always yield a result of 1 for both positive and negative numbers, and 0 when c is equal to 0.

Although my equation may be more complex compared to the alternative answer, I believe it offers a more elegant solution that does not rely on predefined constants.

UPDATE: In agreement with @JosephSible's suggestion, an even more concise variation of my formula (in collaboration with @CRice) without using constants can be expressed as:

c/c^c-c

Answer 7

Answer №4

An intricate solution that is independent of limited precision: When considering the expression x^(2**n), it will result in x+2**n if x is zero, and x-2**n if there is a one in the nth position of x. Therefore, for x=0, (x^(2**n)-x+2**n)/(2**(n+1) will always be equal to 1, but may vary for x != 0. By multiplying all factors of (x^(2**n)-x+2**n)/(2**(n+1)) for every n, then XORing with 1, the desired function can be obtained. However, manual coding for each factor is necessary, particularly when using floating points.

If the operator == is available, then (x==0)^1 can be utilized as well.

Answer 8

An intricate solution that is independent of limited precision: When considering the expression x^(2**n), it will result in x+2**n if x is zero, and x-2**n if there is a one in the nth position of x. Therefore, for x=0, (x^(2**n)-x+2**n)/(2**(n+1) will always be equal to 1, but may vary for x != 0. By multiplying all factors of (x^(2**n)-x+2**n)/(2**(n+1)) for every n, then XORing with 1, the desired function can be obtained. However, manual coding for each factor is necessary, particularly when using floating points.

If the operator == is available, then (x==0)^1 can be utilized as well.

Is it possible to utilize mathematical operators such as multiplication, division, addition, subtraction, and exponentiation to transform a non-zero numerical value into

Answer №1

Answer №2

Answer №3

Answer №4

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