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        riddles >> cs >> Base 13 subtraction
        
(Message started by: minniekp on Aug 17^th, 2007, 1:59am)

Title: Base 13 subtraction
Post by minniekp on Aug 17^th, 2007, 1:59am

How do you subtract two numbers in base 13 given below:

(A023AC5B)
-(129B5321) to the base 13

This was asked in Adobe written test.

Thanks

Title: Re: Base 13 subtraction
Post by inexorable on Aug 17^th, 2007, 2:54am

8A55593A
can't it be answered by just doing by hand?
the carry will be 13 as its base 13

Title: Re: Base 13 subtraction
Post by towr on Aug 17^th, 2007, 3:12am

Simple way is to convert both to integers, subtract, and return to base 13.

Another way is to turn it into an addition, for the number you want to subtract, change every digit to 13- that digit. Add the two numbers base-13 and ermm, add 1 extra for good measure? And mind the overflow.
There's some details to bare in mind (for which I don't have time now); but basically adapt what you'd do for binary.

Title: Re: Base 13 subtraction
Post by Grimbal on Aug 17^th, 2007, 5:13am

OK, here is how it's done. Remember A=10, B=11, etc. Start from the right.

B-1 = A.
5-2 = 3
C-3 = 12-3 = 9
A-5 = 10-5 = 5

so far so good.

3-B = 3-11 = uh... it gets negative
so add 13 to 3 and remember the carry
16-11 = 5

2-1(carry)-9 = ...
add 13 to 2 = 15, remember carry
15-1-9 = 5

0-1-2 = ...
add 13 to 0 = 13, remember carry
13-1-2 = 10 = A

A-1-1 = 8

put it together: 8A55593A

Title: Re: Base 13 subtraction
Post by towr on Aug 17^th, 2007, 7:17am

Here's my approach worked out:


 A023AC5B
-129B5321
--------- (modulo 10000000)
 A023AC5B
+BA3179AB (12's/C's complement -> 129B5321+BA3179AB = CCCCCCCC)
+       1 (because CCCCCCCC + 1 = 10000000)
---------
18A55593A
--------- mod 10000000
 8A55593A

Title: Re: Base 13 subtraction
Post by Grimbal on Aug 17^th, 2007, 7:43am

Of course, when you add BA3179AB+1, in fact you add BA3179AB.CCC...
Because CCCCCCCC.CCC... = 0 (mod 100000000)
;)