Fractional base M to N conversion

Function [seqY, remX] = baseNtobaseM(seqX, N, M, K) converts fractional seqX in base N to fractional seqY upto K fractional points

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Title : Converting a fractional sequence in given base to another base

Author : Amir R. Forouzan

Email : [e-mail]

Company : University of Isfahan

Created on : 3 Feb. 2025

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Function : baseNtobaseM

Purpose : Converts a fractional number in base N to a number in base M up to K fractional points.

Inputs :

- seqX : Vector representing the input number in base N. Each element seqX(k)

represents the k-th fractional point of X and is a non-negative integer < N.

- N : The base of the input number seqX.

- M : The base to which the input number seqX is to be converted.

- K : Number of fractional points to which the output number seqY is to be computed.

Outputs :

- seqY : Vector representing the output number in base M. Each element seqY(k)

represents the k-th fractional point of Y and is a non-negative integer < M.

- remX : Vector representing the remainder in base N. Each element remX(k)

represents the k-th fractional point of R and is a non-negative integer < N.

Description:

seqX represents X = (0.seqX(1) seqX(2) seqX(3) ...) in base N.

seqY represents Y = (0.seqY(1) seqY(2) seqY(3) ...) in base M.

remX represents R = (0.remX(1) remX(2) remX(3) ...) in base N.

The function converts seqX to seqY and calculates remX such that X = Y + R / N^K.

Example :

Converting X = 0.5 in base 2 to base 3 up to 4 fractional points.

X in base 2 is represented by seqX = [1] and N = 2, M = 3, K = 4.

[seqY, remX] = baseNtobaseM([1], 2, 3, 4)

Results:

seqY =

1

1

1

1

remX =

1

Thus Y = 1/3 + 1/9 + 1/27 + 1/81 and R = 1/2.

We see that Y + R / 81 = 0.5 as expected.

Code:

Y = 1/3 + 1/9 + 1/27 + 1/81;

R = 1/2;

result = Y + R / 81

result =

0.5000