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We give an informal effective construction of the j p(i)'s in successive stages s = 1, 2, 3, . . . . At the beginning of stage s > 0, j p(s) will be completely undefined and j p(0) will be defined only on the initial segment {0, . . . , ns - 1}, where ns denotes the least number not yet in domain ( j p(0)). We initialize j p(0)(0) to p(0) and go to stage 1. |
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Begin ( Stage s )
Set j p(0)(ns) =1, and for each  , set j p(s)(x) = j p(0)(x).
Let j = M(p(0), j p(0)[ns + 1]).
If M(p(s), j p(s)[ns + 1]) = j, then go to step 1, else go to step 2.
Step 1.
 Let  be the least x not yet in domain ( j p(0)).
For l = 0, . . ., a, set  .
Set  .
Repeat
Set j p(s)(y) = j p(0)(y) = 0 and reset y = y + 1.
until, if ever, at least one of the following two conditions holds.
Condition 1A: M(p(s),  .
Condition 1B:  .
Step 1a. If Condition 1A holds then go to step 2.
Step 1b. If Condition 1B, but not 1A, holds then:
Let l be the least convergence point discovered in Condition 1B.
Set  .
For each  , set  if  .
Go to stage s + 1.
Step 2.
Let  be the least x not yet in domain( j p(s)).
For each  not yet in domain ( j p(0)), set j p(0)(x) = j p(s)(x).
Repeat
Set  and rest 
until, if ever, at least one of the following two conditions holds.
Condition 2A:  .
Condition 2B:  .
Step 2a. If Condition 2A holds then go to stage s + 1.
Step 2b. If Condition 2B, but not 2A, holds then go to Step 1. |
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