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solve 58 in uiua (attempt sieve)

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Fredrik Robertsen
2025-09-15 21:53:04 +02:00
parent 5abc6a896e
commit bb04a2eb61
1 changed files with 19 additions and 73 deletions
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@@ -1,66 +1,12 @@
# rotate 90 deg # project euler 58
r ← ⍉≡⇌
# create spiral numbers with n rows and cols p ← =1⧻°/×
# spiral ? n Sieve ← ▽⊸≡p+2⇡
s ← (
+1⇡ⁿ2 # create n^2 numbers
⊸(
⊸(-1×2√/↥) # t*2 - 1, where t = sqrt(max(range(n^2))) = n
⌈÷2↙ # create [1 1 2 2 3 ...] until sum is n^2
)
⊜□˜▽+1°⊏ # successively take 1, 1, 2, 2, 3, ... until all n^2 numbers taken
⇌/◇(˜⊂r)⍜⊢⍚¤ # combine all taken with rotating array to form spiral
)
# memoized recursive version of s Primes ← Sieve 37417
σ ← |1 memo(
⨬(⊂⊸(++1⇡∩/↥⊃△(⊢⊢)) r σ-1
| ⋅(¤[1]) # base-case
)⊸=1
)
# generates each diagonal of spiral array
Σ ← +1\+ +¤[2 4 6]×8 ⇡⌊÷2
# get numbers on diagonal
d ← ⊜°¤ ↥⊸r⊞=. °⊏
# is_prime(n) ? n # is_prime(n) ? n
p ← =1⧻°/× # p ← ⨬(=|1)⊸=¯1⊸(⍣⊢¯1▽⤚(=0≡◿) Primes ¤)
┌─╴test
p ↚ ⊜(⊜⋕⊸≠@\s)⊸≠@\n
⍤⤙≍ s 7 p $ 37 36 35 34 33 32 31
$ 38 17 16 15 14 13 30
$ 39 18 5 4 3 12 29
$ 40 19 6 1 2 11 28
$ 41 20 7 8 9 10 27
$ 42 21 22 23 24 25 26
$ 43 44 45 46 47 48 49
└─╴
Sol! ← +1×2⧻⍢(+2⟜(÷⊃⧻/+≡p ^0)|⋅(>0.1)) ⊙1
# Sol!(d s) 3 # too slow
# (σ -1×2) == s
# Sol!(d σ -1×2) 3 # too slow (too much memory usage)
# Sol!(⊂˜↯1÷2-1⟜(♭ Σ)) 3 # still too slow
# okay... let's do it mathematically from scratch:
# primes on diagonals/number of elements on diagonals
# Chance ← ÷-1×2⟜(
# ⊃((+5+∩×₄⊃×₂ⁿ₂)⇡
# | (+3+⊃×₆(×4ⁿ2))⇡
# | (+7+⊃×₁₀(×4ⁿ2))⇡
# ) ⌊÷2
# /+≡p⊂⊂
# )
#
# ⍢(+2|>0.1 Chance) 3
# that solved it in 4 minutes (with p memoized)... too slow!
NewP ← ( NewP ← (
# get the next three prime candidates (lower right is odd square) # get the next three prime candidates (lower right is odd square)
@@ -71,18 +17,18 @@ NewP ← (
++∩₃p # count how many are prime (0-3) ++∩₃p # count how many are prime (0-3)
) )
# initial values are: Sol ← (
# - 5: the amount of numbers so far; 1, 3, 5, 7, 9 (counter for diagonal) # initial values are:
# - NewP.: = 3, since 3, 5, 7 are prime (counter for primes on diagonal) # - 5: the amount of numbers so far; 1, 3, 5, 7, 9 (counter for diagonal)
# - 0: current NewP generation. NewP 0 is 3 (see above) # - NewP.: = 3, since 3, 5, 7 are prime (counter for primes on diagonal)
5 NewP. 0 # - 0: current NewP generation. NewP 0 is 3 (see above)
⍢(+4 ⊙+ ⊙⊙(⊸NewP+1) # add 4 for new corners, generate NewP(n+1) 5 NewP. 0
| >0.1 ÷ # check prime ratio ⍢(+4 ⊙+ ⊙⊙(⊸NewP+1) # add 4 for new corners, generate NewP(n+1)
| >0.1 ÷ # check prime ratio
)
# discard number of primes and diagonal values,
# scale generation up to side length
# newp gen 0 -> 0*2 + 3 = side length 3
# newp gen 1 -> 1*2 + 3 = side length 5, ...
⋅⋅(+3×2)
) )
# discard number of primes and diagonal values,
# scale generation up to side length
# newp gen 0 -> 0*2 + 3 = side length 3
# newp gen 1 -> 1*2 + 3 = side length 5, ...
⋅⋅(+3×2)
# that's a 240x speedup, from 4 minutes to 1 second. we can go faster...