help solving string question

Durring my last semester, we were given a bonus quiz to write a function the received and numberical input and produced the revise as an output.

example a=123

b=321

we were not allowed to use any string variable or function. Any ideas now that the semester is over Id like to know what you think.

My original thought was to divide the input but the got harder with larger number as the function should receive any number

[Joseph Pauwels]

boy it is a good thing this was not an English class. that spelling and miss used work are horrible. I am glad yall knew what I meant

Answer 1

[Star Strider]

This works, but obviously only for integers:

a = 123;
% a = fix(rand*1E6)   % Test Integer

La = fix(log10(a));
x = a;
for k1 = La:-1:0
    d(k1+1) = fix(x/(10^k1));
    x = rem(x,10^k1);
end
v10 = 10.^(La:-1:0)';
Flipped_a = d*v10

The Flipped_a variable is the result. I tested it on other random integers as well.

[Joseph Pauwels]

that was the point, only on integers. fix and rem arent string variables right?

[Star Strider]

The fix and rem functions are not string functions.

• The fix function rounds toward zero.
• The rem function (similar to mod) returns the remainder after division.

[Wesley Ooms]

fix, log10, rand, and rem are functions. The way i read the question is that no function is allowed.

Answer 2

[Carlos]

a=[1 2 3 4 5];
>> b=zeros(1,length(a));
count=1;
l=length(a);
while(count<=l)
b(count)=a(l+1-count);
count=count+1;
end

[Joseph Pauwels]

Thanks Carlos, that i could do. but not what i meant 12345 not 1 2 3 4 5.

Answer 3

[Wesley Ooms]

totally not optimized, but this will do the trick:

 clear a b

 a=32385

 i=1
 while floor(a/10)
 b(i)=a-floor(a/10)*10

 i=i+1   
 a =floor(a/10)

 end
 b(i)=a
 d=1
 e=0
 for c=0:numel(b)-1

    e=e+d*b(end-c)   
    d=d*10
 end

[Wesley Ooms]

ok, i see i used 2 functions (floor and numel) you can get around the numel by placing the part of the for loop in the while loop, but i don't kno how to get around floor. I think Carlos answer is not what your teacher meant.

[Wesley Ooms]

you can get around floor by using this:

clear a b

 a=132385

 i=1
 c=0
 d=1
 e=0

 while a
  c=-1:(a/10)
  a1=c(end)  
 b(i)=a-a1*10
 a=a1
 i=i+1   

 end

 for c=0:i-2

    e=e+d*b(end-c)     
    c=c+1

    d=d*10;
 end

[Joseph Pauwels]

it doesn't spit out an answer, it just keeps running

here my code

function [b] = reverse( a ) %UNTITLED2 Summary of this function goes here % Detailed explanation goes here

i=1; c=0; d=1; e=0;

while a c=-1:(a/1); a1=c(end); b(i)=a-a1*10; a=a1; i=i+1; end for c=0:i-2; e=e+d*b(end-c); c=c+1; d=d*10; end end

[Wesley Ooms]

that is because you

do c=-1:(a/1); instead of c=-1:(a/10);

the following must work: (but i admit it is not optimized for speed)

    function b=swapnumber(a)
    b = 0;
    while a
        c = -1:a/10;
        c = c(end); 
        b = (b+a-c*10)*10;
        a = c;
    end
    b = b/10;

[Wesley Ooms]

the following code also works but is faster for large numbers since it predetermines the size of the number

    function b=swapnumber(a)
    b=0;d=0;while a>1;a=.1*a;d=d+1;end
    for i=0:d;c=0:a*10;c=c(end);b=b+c*10^i;a=a*10-c;end

Answer 4

[Sagar Damle]

I think the code which I am going to put here is the standard code to reverse a number.(This code is used in C language,of course,according to its own syntax!)Also,it is easy to understand.Remember this code,I think it is very helpful!

a = 126986;
 b = a;    % Save value of "a" in new variable "b". 
 reverse = 0;
 while b > 0      % OR  while b ~= 0  (Both 'while' statements are same.)
     r = rem(b,10);
     reverse = reverse * 10 + r;
     b = floor(b/10);
 end
 a
 reverse

Answer 5

[Joshua Amuga]

IVP :=ode({y''(x)+4*y'(x)+3*y,y(0)=3,y'(0)=4},y(x)); Undefined function or variable 'IVP'. this is what i got