#! /usr/bin/env python
#
def chinese_to_i4 ( n, m, r ):

#*****************************************************************************80
#
## CHINESE_TO_I4 converts a set of Chinese remainders to an equivalent integer.
#
#  Discussion:
#
#    Given a set of N pairwise prime, positive moduluses M(I), and
#    a corresponding set of remainders R(I), this routine finds an
#    integer J such that, for all I,
#
#      J = R(I) mod M(I)
#
#  Licensing:
#
#    This code is distributed under the GNU LGPL license.
#
#  Modified:
#
#    25 May 2015
#
#  Author:
#
#    John Burkardt
#
#  Parameters:
#
#    Input, integer N, the number of moduluses.
#
#    Input, integer M(N), the moduluses.  These should be positive
#    and pairwise prime.
#
#    Input, integer R(N), the Chinese remainder representation of the integer.
#
#    Output, integer J, the corresponding integer.
#
  import numpy as np
  from sys import exit
  from chinese_check import chinese_check
  from congruence import congruence
  from i4vec_product import i4vec_product

  ierror = chinese_check ( n, m )

  if ( ierror != 0 ):
    print ( '' )
    print ( 'CHINESE_TO_I4 - Fatal error!' )
    print ( '  The moduluses are not legal.' )
    exit ( 'CHINESE_TO_I4 - Fatal error!' )
#
#  Set BIG_M.
#
  big_m = i4vec_product ( n, m )
#
#  Solve BIG_M / M(I) * B(I) = 1, mod M(I)
#
  b = np.zeros ( n )

  for i in range ( 0, n ):
    a = big_m // m[i]
    c = 1
    b[i], ierror = congruence ( a, m[i], c )
#
#  Set J = sum ( 1 <= I <= N ) ( R(I) * B(I) * BIG_M / M(I) ) mod M
#
  j = 0
  for i in range ( 0, n ):
    j = ( ( j + r[i] * b[i] * ( big_m // m[i] ) ) % big_m )

  return j

def chinese_to_i4_test ( ):

#*****************************************************************************80
#
## CHINESE_TO_I4_TEST tests CHINESE_TO_I4.
#
#  Licensing:
#
#    This code is distributed under the GNU LGPL license.
#
#  Modified:
#
#    25 May 2015
#
#  Author:
#
#    John Burkardt
#
  import numpy as np
  import platform
  from i4_to_chinese import i4_to_chinese
  from i4vec_print import i4vec_print

  n = 4
  m = np.array ( [ 3, 4, 5, 7 ] )

  print ( '' )
  print ( 'CHINESE_TO_I4_TEST' )
  print ( '  Python version: %s' % ( platform.python_version ( ) ) )
  print ( '  CHINESE_TO_I4 computes an integer with the given' )
  print ( '  Chinese Remainder representation.' )

  i4vec_print ( n, m, '  The moduli:' )

  j = 37

  print ( '' )
  print ( '  The number being analyzed is %d' % ( j ) )

  r = i4_to_chinese ( j, n, m )

  i4vec_print ( n, r, '  The remainders:' )

  j2 = chinese_to_i4 ( n, m, r )

  print ( '' )
  print ( '  The reconstructed number is %d' % ( j2 ) )

  r = i4_to_chinese ( j2, n, m )

  i4vec_print ( n, r, '  The remainders of the reconstructed number:' )
#
#  Terminate.
#
  print ( '' )
  print ( 'CHINESE_TO_I4_TEST' )
  print ( '  Normal end of execution.' )
  return

if ( __name__ == '__main__' ):
  from timestamp import timestamp
  timestamp ( )
  chinese_to_i4_test ( )
  timestamp ( )