#Converting a Word to a Number

20 messages · Page 1 of 1 (latest)

normal vaultBOT
strange sage
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do you know how to rewrite numbers in diff bases

potent ore
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its just simple

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A is 1 Z is 26

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AA is 27

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after aa we have ab then ac .....................

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then ba bb bc bd be .......

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............

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.......

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.............

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za zb zc ................zz

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counting all we have 26*26

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then number of zz is 26*26+27

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same for all

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next series also 26* 26* 26+703

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👍

delicate yew
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Since you seem interested in this stuff try to recreate this function, btw its 1 y= function

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granite atlas
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letter2number(A) = 1, letter2number(B) = 2, ..., letter2number(Z) = 26

f(NULL) = 0
f(SEQ:CHAR) = letter2number(CHAR)+26*f(SEQ)

dusk compass
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ive got the code if you want it @pale edge

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Let word be $a_1a_2a_3...a_n$ \
Let the position of those letters be $p_1$, $p_2$, $p_3$, ..., $p_n$ \
So, the decimal is $(p_1)(26^{n-1}) + (p_2)(26^{n-2}) + (p_3)(26^{n-3}) + ... + (p_{n-1})(26^{1}) + (p_n)(26^0)$ \ \

Viceversa it is a bit harder, but let N be the decimal number. Doing the same as we did but reversed, we know that $\lfloor log{26} N \rfloor = n-1$ where $\lfloor x \rfloor$ is the floor function. \
Now to get $p_1$ you do $\lfloor \frac{N}{n-1} \rfloor$ \
Then, you substract $N - (p_1)(26^{n-1})$ and repeat the process until you get $p_n$