这个挑战是应该在GOLF CPU中编写的一系列最少操作问题中的第一个。您可以在这里找到下一个

数字的分区N是加在一起的数字列表N。一个主要的分区是加起来素数的列表N。

对于这个挑战，您将获得一个整数N ≥ 2。您需要为生成最短的主分区N。如果有多个可能的分区，则可以打印其中任何一个。

例子：

9: [2, 7]
12: [5, 7]
95: [89, 3, 3]
337: [337]
1023749: [1023733, 13, 3]
20831531: [20831323, 197, 11]

您的程序应使用GOLF CPU编写。对于输入/输出，您可以使用STDIO或寄存器。该列表可以按任何顺序排列，如果您使用的是STDOUT，则可以用空格或逗号分隔（无需括号）。显然，不允许对解决方案进行硬编码，也不得对头几个质数进行硬编码。

这是最少的操作问题，因此以最少的周期数解决上面的示例的答案就成功了！

159,326,251个周期

输入为n，输出为r，s和t（忽略零）。

# Input in register n
# Outputs in registers r, s, t
# (I use the return value as a debug parameter)

# hardcoded case n=2
cmp c, n, 2
jz skip_n2, c
  mov r, 2
  halt 0
skip_n2:
# hardcoded case n=4
cmp c, n, 4
jz skip_n4, c
  mov r, 2
  mov s, 2
  halt 0
skip_n4:

# Sieve of Eratosthenes
mov i, 1
sieve_loop:
  add i, i, 2
  lb a, i
  jnz sieve_loop, a

  mulu j, k, i, i
  geu c, j, n
  jnz end_sieve_loop, c

  sieve_inner_loop:
    sb j, 1
    add j, j, i
    lequ c, j, n
    jnz sieve_inner_loop, c

  jmp sieve_loop

end_sieve_loop:

lb a, n

# if n is even, skip to search
and c, n, 1
jz search, c

# if n is prime, the partition is simply [n]
jnz not_prime, a
  mov r, n
  halt 1
not_prime:

# if n is odd, check n-2
sub i, n, 2
lb a, i

jnz sub_3, a
# if n-2 is prime, the partition is [2, n-2]
mov r, 2
mov s, i
halt 2

sub_3:
# otherwise the partition is [3] + partition(n-3)
mov t, 3
sub n, n, 3

search:
mov i, 1
sub n, n, 1

search_loop:
  add i, i, 2
  sub n, n, 2
  lb a, i
  jnz search_loop, a
  lb a, n
  jnz search_loop, a
  mov r, i
  mov s, n
  halt 3

测试用例：

robert@unity:~/golf-cpu$ ./assemble.py partition.golf
robert@unity:~/golf-cpu$ ./golf.py -p r,s,t partition.bin n=9
2, 7, 0
Execution terminated after 51 cycles with exit code 2.
robert@unity:~/golf-cpu$ ./golf.py -p r,s,t partition.bin n=12
5, 7, 0
Execution terminated after 77 cycles with exit code 3.
robert@unity:~/golf-cpu$ ./golf.py -p r,s,t partition.bin n=95
3, 89, 3
Execution terminated after 302 cycles with exit code 3.
robert@unity:~/golf-cpu$ ./golf.py -p r,s,t partition.bin n=337
337, 0, 0
Execution terminated after 1122 cycles with exit code 1.
robert@unity:~/golf-cpu$ ./golf.py -p r,s,t partition.bin n=1023749
13, 1023733, 3
Execution terminated after 6654139 cycles with exit code 3.
robert@unity:~/golf-cpu$ ./golf.py -p r,s,t partition.bin n=20831531
229, 20831299, 3
Execution terminated after 152670560 cycles with exit code 3.
robert@unity:~/golf-cpu$ 

示例的总循环数：477,918,603

更新1：更新为使用Lemoine的猜想。

更新2：更新为使用Eratosthenes筛子，而不是天真地找到素数。

运行：

python3 assemble.py 52489-prime-partitions.golf
python3 golf.py 52489-prime-partitions.bin x=<INPUT>

示例运行：

$ python3 golf.py 52489-prime-partitions.bin x=10233
5
5
10223
Execution terminated after 194500 cycles with exit code 0.

循环计数，例如输入：

Input    Cycles
9        191
12       282
95       1,666
337      5,792
1023749  21,429,225
20831531 456,481,447

我们认为(N+1)*8堆的第一个字节是包含N+164位值的数组。（由于堆的大小受到限制，因此仅适用于N < 2^57）。项的值i*8表示是否i为质数：

Value Description
-1    Not a prime
0     Unknown
1     The largest prime found
n > 1 This is a prime and the next prime is n

当我们完成数组的构建后，它将看起来像[-1, -1, 3, 5, -1, 7, -1, 11, -1, -1, -1, 13, ...]。

我们使用Eratosthenes筛子构建阵列。

接下来，程序以伪代码执行以下操作：

if is_prime(x):
    print x
else:
    if is_even(x):
        for p in primes:
            if is_prime(x - p):
                print p, x - p
                exit
    else:
        if is_prime(x - 2):
            print 2, x - 2
        else:
            for p in primes:
                if is_prime(x - 2 * p):
                    print p, p, 2 * p
                    exit

由于Lemoine的猜想和Goldbach的弱猜想，可以保证此方法有效。Lemoine的猜想尚未得到证明，但对于2 ^ 57以下的数字可能是正确的。

    call build_primes

    mov q, x
    call is_prime

    jnz print_prime, a

    and b, x, 1
    jz find_pair, b

    # Check if x - 2 is a prime
    sub q, x, 2
    call is_prime
    jnz print_prime_odd2, a

# Input: x, b
find_pair:
    mov p, 2
find_pair_loop:
    mov d, p
    jz find_pair_even, b

    add d, d, p

find_pair_even:
    sub q, x, d

    call is_prime
    jnz print_prime2_or_3, a

    shl i, p, 3
    lw p, i
    jmp find_pair_loop

print_prime2_or_3:
    jz print_prime2, b

    mov x, p
    call write_int_ln

print_prime2:
    mov x, p
    call write_int_ln

    mov x, q
    call print_prime

print_prime_odd2:
    mov p, 2
    call print_prime2

print_prime:
    call write_int_ln
    halt 0

# Input: x
# Memory layout: [-1, -1, 3, 5, -1, 7, -1, 11, ...]
# x: max integer
# p: current prime
# y: pointer to last found prime
# i: current integer
build_primes:
    sw 0, -1
    sw 8, -1
    sw 16, 1
    mov y, 16

    mov p, 2

build_primes_outer:
    mulu i, r, p, p
    jnz build_primes_final, r

    geu a, i, x
    jnz build_primes_final, a

build_primes_inner:
    shl m, i, 3
    sw m, -1

    add i, i, p

    geu a, i, x
    jz build_primes_inner, a

build_primes_next:
    inc p
    shl m, p, 3
    lw a, m
    jnz build_primes_next, a

    sw y, p
    mov y, m
    sw y, 1

    jmp build_primes_outer

build_primes_final:
    inc p
    geu a, p, x
    jnz build_primes_ret, a

    shl m, p, 3
    lw a, m
    jnz build_primes_final, a

    sw y, p
    mov y, m
    sw y, 1

    jmp build_primes_final

build_primes_ret:
    ret

# Input: q
# Output: a
is_prime:
    shl m, q, 3
    lw a, m
    neq a, a, -1
    ret a

write_int:
    divu x, m, x, 10
    jz write_int_done, x
    call write_int
write_int_done:
    add m, m, ord("0")
    sw -1, m
    ret

write_int_ln:
    call write_int
    mov m, ord("\n")
    sw -1, m
    ret