Fractran: Difference between revisions
→Revised Plan: Explain the revised output of N as a set of prime factor powers.
m (→The Plan) |
(→Revised Plan: Explain the revised output of N as a set of prime factor powers.) |
||
Line 879:
To facilitate access without searching the list of primes for N, instead its list is represented by an array of powers, NPPOW, with most entries zero. Thus, NPPOW(i) has the power for PRIME(i) in the factorisation of N, and for N = 2, NPPOW(1) = 1 with all other elements zero. But for FP (the factorisation of P) and FQ (for Q) there is a proper list provided via type FACTORED, whereby FP.PLIST(0) is the count of prime factors and <code>FP.PLIST(1:FP.PLIST(0))</code> fingers the prime numbers that are factors, with FP.PPOW(i) having their corresponding powers. Thus, FP.PLIST(2) has the index of the second prime factor of P (should it have so many), which is PRIME(FP.PLIST(2)), and its power is FP.PPOW(2). Accordingly, to determine if N (as NPPOW) is divisible by one of the fractions FQ, the appropriate elements of NPPOW (that give its powers) must be compared to the corresponding powers in FQ.PPOW, and if ALL of the fingered powers in NPPOW are greater than or equal to those in FQ.PPOW, then a hit! Note that the prime number factorisation of one has no elements in its list, and happily, the ALL operation applied for no tests yields ''true'' as desired, because one divides any number. Thus, in the FRACTRAN op-code system, a fraction P/1 is always a match, and no fractions beyond it will ever be considered.
For output, the value of N will not be shown multiplied out but via a schedule showing the powers of the first few prime numbers that form its factorisation. Rather than staring in puzzlement at opaque monster strings of digits, one can view each separate prime factor's power counting up and down as the calculation proceeds. A simple scan of all the factorisations soon determines the highest prime employed, and this never changes.
===Revised Code===
| |||