Re: fast modular reduction (proof?)

[danisch@ira.uka.de (Hadmut Danisch)]

-----BEGIN PGP SIGNED MESSAGE-----


> In the following pseudocode, B is the radix in which the numbers are 
> represented (2^32 for a 32-bit machine), n is the length of modulus in 
> blocks, U is B^(n+1) mod the modulus, X is the number to be reduced, k+1 
> is the length of X, and Y is the result.
> 
> 1. Y = X
> 2. For i from k down to n+1, repeat steps 3 and 4
> 3.    Y = Y - Y[i] * B^i + Y[i] * U * B^(i-n-1)
> 4.    If Y >= B^i, then Y = Y - B^i + U * B^(i-n-1)




To do a proof I rewrite the algorithm:



n = len(modulus)                   // modulus < B^n

Y = X                              // obviously Y = X mod modulus


K =  B ^ (n+1) - U                 // U = B ^ (n+1) mod modulus,
                                   // therefore K = 0 mod modulus
                                   // furthermore K > 0


for (i=len(Y)-1 ; i>n  ;  i--)
 {

   F = B ^ (i-n-1) * K             // F > 0
                                   // F = 0 mod modulus

   Y -=  Y[i] * F                  // Y shrinking, but
                                   // Y still the same mod modulus


   if ( Y >= B^i )
      Y -= F                       // again shrinking,
                                   // still the same mod modulus

 }


This shows that Y was shrinking, but is still equal to X mod modulus.





To see whether Y really shrinks enough:

  Y = sum(i=0..len(Y)-1)  Y[i] * B^i


In the step Y = Y - Y[i] * B^i the highest block of Y is deleted (what
could be done fast by reducing the length of Y). 

  Now  Y < B^i





Afterwards the same value mod modulus is added to keep Y constant: 

  Y = Y + Y[i] * U * B^(i-n-1).


  Y[i]<B   ->  Y[i]+1  <= B
  
  U < modulus < B^n , therefore U < B^n

  ->    (Y[i]+1) * U < B * B^n  = B^(n+1)

  ->    Y[i] * U  < B^(n+1) - U

  ->    Y[i] * U *  B^ (i-n-1 )  < F




  Therefore after doing the addition

    Y < B^i + F


  Check of the last step:

  0 <= U < B^n therefore 

  B^n  <  B^(n+1) - U  <= B^(n+1)

  Therefore in every loop


  B^(i-1) < F <= B^i


  ->   Y-F  < B^i



Partial Correctness:






Y = X                              [ Y = X mod 
                                     Y < B^len(X) ]


K =  B ^ (n+1) - U                 [ K = 0 mod
                                     
                                     B^n < K <= B^(n+1) ]                
                                   
                                   


for (i=len(X)-1 ; i>n  ;  i--)
 {
        [ Y = X mod  ,  Y < B^(i+1) ]


   F = B ^ (i-n-1) * K 
   
        [ F = 0 mod  ,   B^(i-1) < F <= B^i  ]

        [ 0 <= Y[i] < B ]

        [ Y[i] * F  = 0 mod ,
          0 <= Y[i] * F  < B^(i+1) ]


        [ Y >= Y[i] * B^i  ->  Y >= Y[i] * F ]

   Y -=  Y[i] * F

        [ Y = X mod ,
          Y < B^i + F  (reason see above) ,
          Y >= 0 ]
          



   if ( Y >= B^i )
      Y -= F                       // again shrinking,
                                   // still the same mod modulus



      [ Y = X mod ,
        Y >= 0    ,
        Y < B^i ]

 }



Last i was n+1, therefore

  Y = X mod ,  Y >= 0 , Y < B^(n+1)


This is not enough,  Y < B^n is requested. The loop can't be done once
more because i-n-1 would become negative.

k+1 was the length of X, and  n the length of the modulus. You walk
down from k to n+1 . In every loop you remove one block of the
number. This means you have to do len(X)-len(modulus) loops. In the
pseudocode you do only len(X)-len(modulus)-1 loops.



One loop seems to be missing. This may be a
result of confusion whether your Y starts with Y[0] or Y[1]. 


I do understand the algorithm as:

n = len(modulus)
U = B^n  mod modulus
K = B^n - U                         // = 0 mod modulus,  0 < K < B^n
Y = X

for(i=len(X)-1 ; i>= n ; i--)       // squeeze Block i in Number Y
 {                                  // Y < B ^ (i+1)

   F = B ^ ( i-n )  * K             // F = 0 mod modulus

   Y -= Y[i] * F                    // subtract  Y[i] * B^i, now Y < B ^ i
                                    // add the equivalent Y[i] * B^(i-n)*U <= F
                                    // now Y < B^i + F

   if ( Y >= B[i] )
     Y -= F                         // now Y < B^i

 }

Last i was n, therefore Y < B^n , Y = X mod modulus , 
but perhaps still Y >= modulus.

Ok, algorithm understood and agreed (after modifying the loop counter).

Any more agreements or disagreements?

Hadmut

-----BEGIN PGP SIGNATURE-----
Version: 2.6.2

iQCVAwUBME8QImc1jG5vDiNxAQFREgQAiMWfgy3bZtPckCNToAvKP7A+JevlnrqG
wJT6/111656nAZHDF5Htr2vwPEmiR522IvPKMG2MnOanMIRn2bgBYG3GECA3zlDo
ZnXNT1OXdtZzv848WPryLVjSJrrVTX/PUN9d6LaTqLBP5pZXwAYkO5kLU6/WP6yL
tF+fniXDZvk=
=Ev0S
-----END PGP SIGNATURE-----