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For an odds calculator for a board game, I need to calculate how many rounds a battle will last on average. Because there is a possibility that both sides in the battle will miss, a battle can theoretically last forever. Therefore I cannot traverse all branches, but need to calculate a mathematical limit. By verifying with a simulator, I have found that the following function correctly approximates the average number of rounds left:
// LIMIT could be any number, the larger it is, the more accurate the result.
const LIMIT = 100;
// r is the number of rounds left if at least 1 of the sides hit
// x is the chance that both sides miss and the round count gets increased,
// but the battle state stays the same.
function approximateLimitForNumberOfRounds(r: number, x: number) {
let approx = r / (1 - x);
// n -> infinity
for (let n = 1; n < LIMIT; n++) {
approx += x ** n;
}
return approx;
}
How can I modify this function to exactly calculate the number of rounds left, instead of approximating it? (noting that since x is a chance, it is contained in (0, 1) or 0 < x < 1).
We can note that approx takes on the following values:
r / (1 - x) # I refer to this as 'a' below
a + x
a + x + x^2
a + x + x^2 + x^3
a + x + x^2 + ... + x^n
Thus, we can simplify the mathematical expression to be:
a + (the sum of x^k from k = 1 to k = n)
Next, we must note that the sequence x + x^2 + x^3 ... forms a geometric sequence with first term x and common ratio x. Since x is bounded by 0 < x < 1, this will have a limiting sum, namely:
x + x^2 + x^3 + ... x^inf = x/(1-x)
(this obviously fails when x = 1, as well as in the original function where r / (1 - x) is taken, but in that case, you will simply have the sum as infinity and approx would escape to infinity if it were not undefined; so I am assuming that x != 1 in the following calculations and x = 1 can be / has been dealt with separately).
Now, since we have both a single expression for x + x^2 + ... to infinity, and a single expression for approx that includes x + x^2 + ... then we can write approx using both of these two facts:
approx = r / (1 - x) + x / (1 - x)
approx = (r + x) / (1 - x)
And there you go! That is the mathematical equivalent of the logic you've outlined in your question, compressed to a single statement (which I believe is correct :)).
Let's say you have a function that takes both x and y, real numbers that are integers, as arguments.
What would you put inside that function, using only mathematical operators, so that no two given sequences of arguments could ever return the same value, be it any kind of value?
Example of a function that fails at doing this:
function myfunction(x,y){
return x * y;
}
// myfunction(2,6) and myfunction(3,4) will both return 12
// myfunction(2,6) and myfunction(6,2) also both return 12.
As already noted in comments, at the level of JavaScript numbers such a function can't exist, simply because assuming that we're working with integer-valued IEEE 754 binary64 floats there are more possible input pairs than possible output values.
But to the mathematical question of whether there is a simple, injective function from pairs of integers to a single integer, the answer is yes. Here's one such function that uses only addition and multiplication, so should fit the questioner's "using only mathematical operators" constraint.
First we map each of the inputs from the domain of integers to the domain of nonnegative integers. The polynomial map x ↦ 2*x*x + x will do that for us, and maps distinct values to distinct values. (Sketch of proof: if 2*x*x + x == 2*y*y + y for some integers x and y, then rearranging and factoring gives (x - y) * (2*x + 2*y + 1) == 0; the second factor can never be zero for integers x and y, so the first factor must be zero and x == y.)
Second, given a pair of nonnegative integers (a, b), we map that pair to a single (nonnegative) integer using (a, b) ↦ (a + b)*(a + b) + a. It's easy to see that this, too, is injective: given the value of (a + b)*(a + b) + a, I can recover the value of a + b by taking the integer square root, and from there recover a and b.
Here's some Python code demonstrating the above:
def encode_pair(x, y):
""" Encode a pair of integers as a single (nonnegative) integer. """
a = 2*x*x + x
b = 2*y*y + y
return (a + b)*(a + b) + a
We can easily check that there are no repetitions for small x and y: here we take all pairs (x, y) with -500 <= x < 500 and -500 <= y < 500, and find the set containing encode_pair(x, y) for each combination. If all goes well, we should end up with a set with exactly 1 million entries, one per input combination.
>>> all_outputs = {encode_pair(x, y) for x in range(-500, 500) for y in range(-500, 500)}
>>> len(all_outputs)
1000000
>>> min(all_outputs)
0
But perhaps a more convincing way to establish the injectivity is to give an explicit inverse, showing that the original (x, y) can be recovered from the output. Here's that inverse function. It makes use of Python's integer square root operation math.isqrt, which is available only for Python >= 3.8, but is easy to implement yourself if you need it.
from math import isqrt
def decode_pair(n):
""" Decode an integer produced by encode_pair. """
a_plus_b = isqrt(n)
a = n - a_plus_b*a_plus_b
b = a_plus_b - a
c = isqrt(8*a + 1)
d = isqrt(8*b + 1)
return ((2 - c%4) * c - 1) // 4, ((2 - d%4) * d - 1) // 4
Example usage:
>>> encode_pair(3, 7)
15897
>>> decode_pair(15897)
(3, 7)
Depending on what you allow as a "mathematical operator" (which isn't really a particularly well-defined term), there are tighter functions possible. Here's a variant of the above that provides not just an injection but a bijection: every integer appears as the encoding of some pair of integers. It extends the set of mathematical operators used to include subtraction, division and absolute value. (Note that all divisions appearing in encode_pair are exact integer divisions, without any remainder.)
def encode_pair(x, y):
""" Encode a pair of integers as a single integer.
This gives a bijective map Z x Z -> Z.
"""
ax = (abs(2 * x + 1) - 1) // 2 # x if x >= 0, -1-x if x < 0
sx = (ax - x) // (2 * ax + 1) # 0 if x >= 0, 1 if x < 0
ay = (abs(2 * y + 1) - 1) // 2 # y if y >= 0, -1-y if y < 0
sy = (ay - y) // (2 * ay + 1) # 0 if y >= 0, 1 if y < 0
xy = (ax + ay + 1) * (ax + ay) // 2 + ax # encode ax and ay as xy
an = 2 * xy + sx # encode xy and sx as an
n = an - (2 * an + 1) * sy # encode an and sy as n
return n
def decode_pair(n):
""" Inverse of encode_pair. """
# decode an and sy from n
an = (abs(2 * n + 1) - 1) // 2
sy = (an - n) // (2 * an + 1)
# decode xy and sx from an
sx = an % 2
xy = an // 2
# decode ax and ay from xy
ax_plus_ay = (isqrt(8 * xy + 1) - 1) // 2
ax = xy - ax_plus_ay * (ax_plus_ay + 1) // 2
ay = ax_plus_ay - ax
# recover x from ax and sx, and y from ay and sy
x = ax - (1 + 2 * ax) * sx
y = ay - (1 + 2 * ay) * sy
return x, y
And now every integer appears as the encoding of exactly one pair, so we can start with an arbitrary integer, decode it to a pair, and re-encode to recover the same integer:
>>> n = -12345
>>> decode_pair(n)
(67, -44)
>>> encode_pair(67, -44)
-12345
The encode_pair function above is deliberately quite verbose, in order to explain all the steps involved. But the code and the algebra can be simplified: here's exactly the same computation expressed more compactly.
def encode_pair_cryptic(x, y):
""" Encode a pair of integers as a single integer.
This gives a bijective map Z x Z -> Z.
"""
c = abs(2 * x + 1)
d = abs(2 * y + 1)
e = (2 * y + 1) * ((c + d)**2 * c + 2 * (c - d) * c - 4 * x - 2)
return (e - 2 * c * d) // (4 * c * d)
encode_pair_cryptic gives exactly the same results as encode_pair. I'll give one example, and leave the reader to figure out the equivalence.
>>> encode_pair(47, -53)
-9995
>>> encode_pair_cryptic(47, -53)
-9995
I'm no math wiz but found this question kinda fun so I gave it a shot. This is by no means scalable to large number since I'm using prime numbers as exponents and gets out of control really quick. But tested up to 90,000 combinations and found no duplicates.
The code below has a couple extra functions generateValues() and hasDuplicates() that is just there to run and test multiple values coming from the output of myFunction()
BigNumber.config({ EXPONENTIAL_AT: 10 })
// This function is just to generate the array of prime numbers
function getPrimeArray(num) {
const array = [];
let isPrime;
let i = 2;
while (array.length < num + 1) {
for (let j = 2; (isPrime = i === j || i % j !== 0) && j <= i / 2; j++) {}
isPrime && array.push(i);
i++;
}
return array;
}
function myFunction(a, b) {
const primes = getPrimeArray(Math.max(a, b));
// Using the prime array, primes[a]^primes[b]
return BigNumber(primes[a]).pow(primes[b]).toString();
}
function generateValues(upTo) {
const results = [];
for (let i = 1; i < upTo + 1; i++) {
for (let j = 1; j < upTo + 1; j++) {
console.log(`${i},${j}`)
results.push(myFunction(i,j));
}
}
return results.sort();
}
function hasDuplicates(arr) {
return new Set(arr).size !== arr.length;
}
const values = generateValues(50)
console.log(`Checked ${values.length} values; duplicates: ${hasDuplicates(values)}`)
<script src="https://cdnjs.cloudflare.com/ajax/libs/bignumber.js/8.0.2/bignumber.min.js"></script>
Explanation of what's going on:
Using the example of myFunction(1,3)
And the array of primes [2, 3, 5, 7]
This would take the 2nd and 4th items, 3 and 7 which would result in 3^7=2187
Using 300 as the max generated 90,000 combinations with no duplicates (However it took quite some time.) I tried using a max of 500 but the fan on my laptop sounded like a jet engine taking off so gave up on it.
If x and y are some fixed size integers (eg 8 bits) then what you want is possible if the return of f has at least as many bits as the sum of the number of bits of x an y (ie 16 in the example) and not otherwise.
In the 8 bit example f(x,y) = (x<<8)+y would do. This is because if g(z) = ((z>>8), z&255) then g(f(x,y)) = (x,y). The impossibility comes from the pigeon hole principle: if we want (in the example) to map the pairs (x,y) (of which there 2^16) 1-1 to some integer type, then we must have at least 2^16 values of this type.
function myfunction(x,y){
x = 1/x;
y = 1/y;
let yLength = ("" + y).length
for(let i = 0; i < yLength; i++){
x*=10;
}
return (x + y)
}
console.log(myfunction(2,12))
console.log(myfunction(21,2))
Based on your question and you comments, I understood the following:
You want to pass 2 real numbers into a function. The function should use mathematical operators to generate a new result.
Your question is, if there is any kind of mathematical equation/function you could use, that would ALWAYS deliver a unique result.
If that's so, then the answer is no. You can make your function as complicated as possible and get a result(c) using the two numbers (a & b).
In this case I would look for another combination which could give me the result(c) using the same equation/function. Therefore I would use the system of linear equation to solve this mathematical issue.
In general, a system with fewer equations than unknowns has infinitely many solutions, but it may have no solution. Such a system is known as an underdetermined system.
In our case we would have one equation which gives us one result and two unknowns, therefore it would have infinitely many solutions because we already have a solution, so there is no way for the system to have no solutions at all.
More about this topic.
Edit:
I just recognized that some of us understood the domain of the function in a different way. I was thinking about real numbers (R) but it seems many assumed you talk about integers (Z) only.
Well I guess
real integers
wasnt clear enough, at least for me.
So if we would use integers only, I have no idea if that is possible to always have different results. Some users suggested a topic about that here I am also interested to take a look into that too.
I'd like to calculate the mathematical logarithm "by hand"...
... where stands for the logarithmBase and stands for the value.
Some examples (See Log calculator):
The base 2 logarithm of 10 is 3.3219280949
The base 5 logarithm of 15 is 1.6826061945
...
Hoever - I do not want to use a already implemented function call like Math.ceil, Math.log, Math.abs, ..., because I want a clean native solution that just deals with +-*/ and some loops.
This is the code I got so far:
function myLog(base, x) {
let result = 0;
do {
x /= base;
result ++;
} while (x >= base)
return result;
}
let x = 10,
base = 2;
let result = myLog(base, x)
console.log(result)
But it doesn't seems like the above method is the right way to calculate the logarithm to base N - so any help how to fix this code would be really appreciated.
Thanks a million in advance jonas.
You could use a recursive approach:
const log = (base, n, depth = 20, curr = 64, precision = curr / 2) =>
depth <= 0 || base ** curr === n
? curr
: log(base, n, depth - 1, base ** curr > n ? curr - precision : curr + precision, precision / 2);
Usable as:
log(2, 4) // 2
log(2, 10) // 3.32196044921875
You can influence the precision by changing depth, and you can change the range of accepted values (currently ~180) with curr
How it works:
If we already reached the wanted depth or if we already found an accurate value:
depth <= 0 || base ** curr === n
Then it just returns curr and is done. Otherwise it checks if the logarithm we want to find is lower or higher than the current one:
base ** curr > n
It will then continue searching for a value recursively by
1) lowering depth by one
2) increasing / decreasing curr by the current precision
3) lower precision
If you hate functional programming, here is an imperative version:
function log(base, n, depth = 20) {
let curr = 64, precision = curr / 2;
while(depth-- > 0 && base ** curr !== n) {
if(base ** curr > n) {
curr -= precision;
} else {
curr += precision;
}
precision /= 2;
}
return curr;
}
By the way, the algorithm i used is called "logarithmic search" commonly known as "binary search".
First method: with a table of constants.
First normalize the argument to a number between 1 and 2 (this is achieved by multiplying or dividing by 2 as many times as necessary - keep a count of these operations). For efficiency, if the values can span many orders of magnitude, instead of equal factors you can use a squared sequence, 2, 4, 16, 256..., followed by a dichotomic search when you have bracketed the value.
F.i. if the exponents 16=2^4 works but not 256=2^8, you try 2^6, then one of 2^5 and 2^7 depending on outcome. If the final exponent is 2^d, the linear search takes O(d) operations and the geometric/dichotomic search only O(log d). To avoid divisions, it is advisable to keep a table of negative powers.
After normalization, you need to refine the mantissa. Compare the value to √2, and if larger multiply by 1/√2. This brings the value between 1 and √2. Then compare to √√2 and so on. As you go, you add the weights 1/2, 1/4, ... to the exponent when a comparison returns greater.
In the end, the exponent is the base 2 logarithm.
Example: lg 27
27 = 2^4 x 1.6875
1.6875 > √2 = 1.4142 ==> 27 = 2^4.5 x 1.1933
1.1933 > √√2 = 1.1892 ==> 27 = 2^4.75 x 1.0034
1.0034 < √√√2 = 1.0905 ==> 27 = 2^4.75 x 1.0034
...
The true value is 4.7549.
Note that you can work with other bases, in particular e. In some contexts, base 2 allows shortcuts, this is why I used it. Of course, the square roots should be tabulated.
Second method: with a Taylor series.
After the normalization step, you can use the standard series
log(1 + x) = x - x²/2 + x³/3 - ...
which converges for |x| < 1. (Caution: we now have natural logarithms.)
As convergence is too slow for values close to 1, it is advisable to use the above method to reduce to the range [1, √2). Then every new term brings a new bit of accuracy.
Alternatively, you can use the series for log((1 + x)/(1 - x)), which gives a good convergence speed even for the argument 2. See https://fr.wikipedia.org/wiki/Logarithme_naturel#D%C3%A9veloppement_en_s%C3%A9rie
Example: with x = 1.6875, y = 0.2558 and
2 x (0.2558 + 0.2558³/3 + 0.2558^5/5) = 0.5232
lg 27 ~ 4 + 0.5232 / ln 2 = 4.7548
I am having a asp application and in that amount column is there. I need to find out how many thousands and hundreds and tens are there in that amount
For example
if i am having amount as 3660 means
1000's - 3
100's - 6
10's - 6
like this i need
Can any body help me
The simple answer is to divide the number by 1000 whatever is the quotient that is the number of 1000's in the amount. Then divide the remainder with the 100's the quotient will be the number of 100's. And then again divide the remainder with 10, the quotient will be the number of 10's
Something like this:
quotient = 3660 / 1000; //This will give you 3
remainder = 3660 % 1000; //This will give you 660
Then,
quotient1 = remainder/ 100; //This will give you 6
remainder1 = remainder % 100; //This will give you 60
And finally
quotient2 = remainder1 / 10; //This will give you 6
Is it not easier to use type coercion and change the data type to string?
Then you can easily check the value by checking the value at selected index position,
var number = 1234;
number2 = new String(number);
var thousands = number2[0];
var hundreds = number2[1];
and so on....
It may not be usable in what you're doing, it was for me :)
If the "javascript" tag is the correct one, then you've already gotten some answers. If the "asp-classic" tag is actually the correct one, then chances are your scripting language is VBScript, not Javascript.
Did you just pick multiples of 10 as an example, or is that the actual multiple you're looking for? Because if it's the latter, then all you need to do is split the number into digits — that's what the base 10 number system means, after all.
Function SplitNum(theNum)
dim L, i, s, n
n = CStr(theNum)
L = Len(n)
s = ""
for i = 1 to 3
if s <> "" then s = "," & s
if i >= L then
s = "0" & s
else
s = Left(Right(n,i+1),1) & s
end if
next
if L > 4 then s = left(n,L-4) & s
SplitNum = s
End Function
If your actual divisors are something other than multiples of 10, you'll need to do arithmetic. The integer division operator in VBScript is \. (Integer division is basically the "opposite" of the modulus function.)
Function GetMultiples(theNum)
dim q, r
q = theNum \ 1000 & ","
r = theNum Mod 1000
q = q & r \ 100 & ","
r = r Mod 100
q = q & r \ 10
GetMultiples = q
End Function
Try this out...
Here is a fiddle that demonstrates how to use the output..
http://jsfiddle.net/Villarrealized/L3AxZ/1/
function getMultiplesOfTen(number){
number = parseInt(number);
if(typeof(number)!=="number") return number;
var result = {};
(function breakDown(num){
if(isNaN(num))return num;//if it's invalid return
if(num<=0)return false;
num = num.toFixed(0);//get rid of decimals
var divisor = Math.pow(10,num.length-1),//ex. when num = 300, divisor = 100
quotient = Math.floor(num/divisor);
result[divisor]=quotient;//add it to our object
breakDown(num % divisor);//break down the remainder
})(number);
//return result as an object
return result;
}
This function will return an object with the multiple of ten as the key and the number as the value
ex. getMultiplesOfTen(150)=={100:1,10:5} == 1 multiple of 100 and 5 multiples of 10.
Let's say we have the number 7354. To find the thousands:
variable a = 7354 / 1000
variable b = a % 10
The number which is stored in variable b now is the number if the thousands.
To find the hundreds:
variable c = 7354 / 100
variable d = a % 10
The number which is stored in variable d now is the number if the hundreds.
To find the tens:
variable e = 7354 / 10
variable f = a % 10
The number which is stored in variable f now is the number if the tens.
To find the ones:
7354 % 10
This works for every number in the place of 7354, even for bigger numbers than 7354.
The first digit in the quotient of 1,592÷64
1
is in the Choose... ones tens hundreds thousands place.
Decide where the first digit of the quotient should be placed. Do not complete the division.
The first digit of the quotient for 2,370÷24
2,370÷24
should be in the Choose... ones tens hundreds thousands place
2021 version here:
Cast your number as a string, spread it and reverse it like so:
x = 1234
x = [...x.toString()].reverse() // [4, 3, 2, 1]
thous = x[3] // 1
hunds = x[2] // 2
tens = x[1] // 3
units = x[0] // 4
y = [...x.toString()].reverse()[3] // 1 because 1000 has 3 zeros
I suppose you can use some fancy system of powers of 10 to get those indexes. So let's do exactly that and get #Villarrealized 15 lines of 2013 code condensed into just a few lines of 2021 code:
function placeValues(someNumber = 1234) {
x = [...someNumber.toString()].reverse().reduce((p, c, i) => {
p[10**i] = c;
return p;
}, {});
return x; // {1000:1, 100:2, 10:3, 1:4}
}
Maybe i am just not that good enough in math, but I am having a problem in converting a number into pure alphabetical Bijective Hexavigesimal just like how Microsoft Excel/OpenOffice Calc do it.
Here is a version of my code but did not give me the output i needed:
var toHexvg = function(a){
var x='';
var let="_abcdefghijklmnopqrstuvwxyz";
var len=let.length;
var b=a;
var cnt=0;
var y = Array();
do{
a=(a-(a%len))/len;
cnt++;
}while(a!=0)
a=b;
var vnt=0;
do{
b+=Math.pow((len),vnt)*Math.floor(a/Math.pow((len),vnt+1));
vnt++;
}while(vnt!=cnt)
var c=b;
do{
y.unshift( c%len );
c=(c-(c%len))/len;
}while(c!=0)
for(var i in y)x+=let[y[i]];
return x;
}
The best output of my efforts can get is: a b c d ... y z ba bb bc - though not the actual code above. The intended output is suppose to be a b c ... y z aa ab ac ... zz aaa aab aac ... zzzzz aaaaaa aaaaab, you get the picture.
Basically, my problem is more on doing the ''math'' rather than the function. Ultimately my question is: How to do the Math in Hexavigesimal conversion, till a [supposed] infinity, just like Microsoft Excel.
And if possible, a source code, thank you in advance.
Okay, here's my attempt, assuming you want the sequence to be start with "a" (representing 0) and going:
a, b, c, ..., y, z, aa, ab, ac, ..., zy, zz, aaa, aab, ...
This works and hopefully makes some sense. The funky line is there because it mathematically makes more sense for 0 to be represented by the empty string and then "a" would be 1, etc.
alpha = "abcdefghijklmnopqrstuvwxyz";
function hex(a) {
// First figure out how many digits there are.
a += 1; // This line is funky
c = 0;
var x = 1;
while (a >= x) {
c++;
a -= x;
x *= 26;
}
// Now you can do normal base conversion.
var s = "";
for (var i = 0; i < c; i++) {
s = alpha.charAt(a % 26) + s;
a = Math.floor(a/26);
}
return s;
}
However, if you're planning to simply print them out in order, there are far more efficient methods. For example, using recursion and/or prefixes and stuff.
Although #user826788 has already posted a working code (which is even a third quicker), I'll post my own work, that I did before finding the posts here (as i didnt know the word "hexavigesimal"). However it also includes the function for the other way round. Note that I use a = 1 as I use it to convert the starting list element from
aa) first
ab) second
to
<ol type="a" start="27">
<li>first</li>
<li>second</li>
</ol>
:
function linum2int(input) {
input = input.replace(/[^A-Za-z]/, '');
output = 0;
for (i = 0; i < input.length; i++) {
output = output * 26 + parseInt(input.substr(i, 1), 26 + 10) - 9;
}
console.log('linum', output);
return output;
}
function int2linum(input) {
var zeros = 0;
var next = input;
var generation = 0;
while (next >= 27) {
next = (next - 1) / 26 - (next - 1) % 26 / 26;
zeros += next * Math.pow(27, generation);
generation++;
}
output = (input + zeros).toString(27).replace(/./g, function ($0) {
return '_abcdefghijklmnopqrstuvwxyz'.charAt(parseInt($0, 27));
});
return output;
}
linum2int("aa"); // 27
int2linum(27); // "aa"
You could accomplish this with recursion, like this:
const toBijective = n => (n > 26 ? toBijective(Math.floor((n - 1) / 26)) : "") + ((n % 26 || 26) + 9).toString(36);
// Parsing is not recursive
const parseBijective = str => str.split("").reverse().reduce((acc, x, i) => acc + ((parseInt(x, 36) - 9) * (26 ** i)), 0);
toBijective(1) // "a"
toBijective(27) // "aa"
toBijective(703) // "aaa"
toBijective(18279) // "aaaa"
toBijective(127341046141) // "overflow"
parseBijective("Overflow") // 127341046141
I don't understand how to work it out from a formula, but I fooled around with it for a while and came up with the following algorithm to literally count up to the requested column number:
var getAlpha = (function() {
var alphas = [null, "a"],
highest = [1];
return function(decNum) {
if (alphas[decNum])
return alphas[decNum];
var d,
next,
carry,
i = alphas.length;
for(; i <= decNum; i++) {
next = "";
carry = true;
for(d = 0; d < highest.length; d++){
if (carry) {
if (highest[d] === 26) {
highest[d] = 1;
} else {
highest[d]++;
carry = false;
}
}
next = String.fromCharCode(
highest[d] + 96)
+ next;
}
if (carry) {
highest.push(1);
next = "a" + next;
}
alphas[i] = next;
}
return alphas[decNum];
};
})();
alert(getAlpha(27)); // "aa"
alert(getAlpha(100000)); // "eqxd"
Demo: http://jsfiddle.net/6SE2f/1/
The highest array holds the current highest number with an array element per "digit" (element 0 is the least significant "digit").
When I started the above it seemed a good idea to cache each value once calculated, to save time if the same value was requested again, but in practice (with Chrome) it only took about 3 seconds to calculate the 1,000,000th value (bdwgn) and about 20 seconds to calculate the 10,000,000th value (uvxxk). With the caching removed it took about 14 seconds to the 10,000,000th value.
Just finished writing this code earlier tonight, and I found this question while on a quest to figure out what to name the damn thing. Here it is (in case anybody feels like using it):
/**
* Convert an integer to bijective hexavigesimal notation (alphabetic base-26).
*
* #param {Number} int - A positive integer above zero
* #return {String} The number's value expressed in uppercased bijective base-26
*/
function bijectiveBase26(int){
const sequence = "ABCDEFGHIJKLMNOPQRSTUVWXYZ";
const length = sequence.length;
if(int <= 0) return int;
if(int <= length) return sequence[int - 1];
let index = (int % length) || length;
let result = [sequence[index - 1]];
while((int = Math.floor((int - 1) / length)) > 0){
index = (int % length) || length;
result.push(sequence[index - 1]);
}
return result.reverse().join("")
}
I had to solve this same problem today for work. My solution is written in Elixir and uses recursion, but I explain the thinking in plain English.
Here are some example transformations:
0 -> "A", 1 -> "B", 2 -> "C", 3 -> "D", ..
25 -> "Z", 26 -> "AA", 27 -> "AB", ...
At first glance it might seem like a normal 26-base counting system
but unfortunately it is not so simple.
The "problem" becomes clear when you realize:
A = 0
AA = 26
This is at odds with a normal counting system, where "0" does not behave
as "1" when it is in a decimal place other than then unit.
To understand the algorithm, consider a simpler but equivalent base-2 system:
A = 0
B = 1
AA = 2
AB = 3
BA = 4
BB = 5
AAA = 6
In a normal binary counting system we can determine the "value" of decimal places by
taking increasing powers of 2 (1, 2, 4, 8, 16) and the value of a binary number is
calculated by multiplying each digit by that digit place's value.
e.g. 10101 = 1 * (2 ^ 4) + 0 * (2 ^ 3) + 1 * (2 ^ 2) + 0 * (2 ^ 1) + 1 * (2 ^ 0) = 21
In our more complicated AB system, we can see by inspection that the decimal place values are:
1, 2, 6, 14, 30, 62
The pattern reveals itself to be (previous_unit_place_value + 1) * 2.
As such, to get the next lower unit place value, we divide by 2 and subtract 1.
This can be extended to a base-26 system. Simply divide by 26 and subtract 1.
Now a formula for transforming a normal base-10 number to special base-26 is apparent.
Say the input is x.
Create an accumulator list l.
If x is less than 26, set l = [x | l] and go to step 5. Otherwise, continue.
Divide x by 2. The floored result is d and the remainder is r.
Push the remainder as head on an accumulator list. i.e. l = [r | l]
Go to step 2 with with (d - 1) as input, e.g. x = d - 1
Convert """ all elements of l to their corresponding chars. 0 -> A, etc.
So, finally, here is my answer, written in Elixir:
defmodule BijectiveHexavigesimal do
def to_az_string(number, base \\ 26) do
number
|> to_list(base)
|> Enum.map(&to_char/1)
|> to_string()
end
def to_09_integer(string, base \\ 26) do
string
|> String.to_charlist()
|> Enum.reverse()
|> Enum.reduce({0, nil}, fn
char, {_total, nil} ->
{to_integer(char), 1}
char, {total, previous_place_value} ->
char_value = to_integer(char + 1)
place_value = previous_place_value * base
new_total = total + char_value * place_value
{new_total, place_value}
end)
|> elem(0)
end
def to_list(number, base, acc \\ []) do
if number < base do
[number | acc]
else
to_list(div(number, base) - 1, base, [rem(number, base) | acc])
end
end
defp to_char(x), do: x + 65
end
You use it simply as BijectiveHexavigesimal.to_az_string(420). It also accepts on optional "base" arg.
I know the OP asked about Javascript but I wanted to provide an Elixir solution for posterity.
I have published these functions in npm package here:
https://www.npmjs.com/package/#gkucmierz/utils
Converting bijective numeration to number both ways (also BigInt version is included).
https://github.com/gkucmierz/utils/blob/main/src/bijective-numeration.mjs