function for writing out a base 7 word, like binary counter style
I need a function similar to the ones explained here...
JS function for writing out a word, binary counter style
...but using base 7 (or others) to generate (count) letters up from A to G Like so...
---a
---b
---c
---d
---e
---f
---g
--aa
--ab
--ac
--ad
--ae
--af
etc. up to gggg
Is there a simple way to change one 开发者_C百科of those functions to make this happen?
p.s. This is pretty cool...
They used...var iterations = Math.pow(2,str.length)
and Math.pow(string.length,2)
They both work but only because the string length was 4 and the base was 2
inadvertently correcti.e 4^2=16 2^4=16
If any other string length was used one of them becomes wrong.
i.e. 2^10=1024 10^2=100
Maybe this is gives a first start: Working Demo
It basically just takes a number and converts it to a base 7 number:
var map = 'abcdefg';
var n = 13
var out = '';
if(n == 0) out = map.charAt(0) + out;
while (n > 0) {
out = map.charAt(n % 7) + out;
n = Math.floor(n / 7);
}
// gives "bg" for 13
This would imply maintaining a normal counter in decimal system and convert every number.
Note that aa
in this system does not exit, as a
maps to 0
.
function change_num(num,base) {
var lets = 'abcdefghijklmnopqrstuvwxyz';
lets = lets.split('');
if (num / base < 1) {
return lets[num];
}
if (num / (Math.pow(base,2)) < 1) {
return lets[Math.floor(num/base)] + lets[(num % base)];
}
if (num / (Math.pow(base,3)) < 1) {
var numreturn;
numreturn = lets[Math.floor(num/(Math.pow(base,2)))];
numreturn += lets[Math.floor((num%Math.pow(base,2))/base)];
numreturn += lets[(num % base)];
return numreturn;
}
if (num / (Math.pow(base,4)) < 1) {
var numreturn;
numreturn = lets[Math.floor(num/Math.pow(base,3))];
numreturn += lets[Math.floor(num%Math.pow(base,3)/(Math.pow(base,2)))];
numreturn += lets[Math.floor((num%Math.pow(base,2))/base)];
numreturn += lets[(num % base)];
return numreturn;
}
if (num / (Math.pow(base,5)) < 1) {
var numreturn;
numreturn = lets[Math.floor(num/Math.pow(base,4))];
numreturn += lets[Math.floor(num%Math.pow(base,4)/(Math.pow(base,3)))];
numreturn += lets[Math.floor(num%Math.pow(base,3)/(Math.pow(base,2)))];
numreturn += lets[Math.floor((num%Math.pow(base,2))/base)];
numreturn += lets[(num % base)];
return numreturn;
}
}
Works for any base. Couldn't be bothered to write a recursive function so did it the hard way, you can probably infer what I'm doing.
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