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import math, strutils, sugar, tables
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const
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N = 1_000_000_000
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S = sqrt(N.toFloat).int
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var composite: array[3..S, bool]
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for n in countup(3, S, 2):
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if n * n > S: break
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if not composite[n]:
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for k in countup(n * n, S, 2 * n):
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composite[k] = true
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# Prime list. Add a dummy zero to start at index 1.
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let primes = @[0, 2] & collect(newSeq, for n in countup(3, S, 2): (if not composite[n]: n))
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var cache: Table[(Natural, Natural), Natural]
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proc phi(x, a: Natural): Natural =
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if a == 0: return x
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let pair = (x, a)
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if pair in cache: return cache[pair]
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result = phi(x, a - 1) - phi(x div primes[a], a - 1)
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cache[pair] = result
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proc π(n: Natural): Natural =
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if n <= 2: return 0
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let a = π(sqrt(n.toFloat).Natural)
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result = phi(n, a) + a - 1
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var n = 1
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for i in 0..9:
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echo "π(10^$1) = $2".format(i, π(n))
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n *= 10
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@ -0,0 +1,45 @@
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# compile with: nim c -d:danger -t:-march=native --gc:arc
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from std/monotimes import getMonoTime, `-`
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from std/times import inMilliseconds
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from std/math import sqrt
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let masks = [ 1'u8, 2, 4, 8, 16, 32, 64, 128 ] # faster than bit twiddling
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let masksp = cast[ptr[UncheckedArray[byte]]](unsafeAddr(masks[0]))
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proc countPrimes(n: int64): int64 =
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if n < 3:
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return if n < 2: 0 else: 1
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else:
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let rtlmt = n.float64.sqrt.int
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let mxndx = (rtlmt - 1) div 2
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let sz = (mxndx + 8) div 8
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var cmpsts = cast[ptr[UncheckedArray[byte]]](alloc0(sz))
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for i in 1 .. mxndx:
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if (cmpsts[i shr 3] and masksp[i and 7]) != 0: continue
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let sqri = (i + i) * (i + 1)
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if sqri > mxndx: break
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let bp = i + i + 1
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for c in countup(sqri, mxndx, bp):
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let w = c shr 3; cmpsts[w] = cmpsts[w] or masksp[c and 7]
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var pisqrt = 0'i64
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for i in 0 .. mxndx:
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if (cmpsts[i shr 3] and masksp[i and 7]) == 0: pisqrt += 1
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var primes = cast[ptr[UncheckedArray[uint32]]](alloc(sizeof(uint32) * pisqrt.int))
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var j = 0
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for i in 0 .. mxndx:
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if (cmpsts[i shr 3] and masksp[i and 7]) == 0: primes[j] = (i + i + 1).uint32; j += 1
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proc phi(x: int64; a: int): int64 =
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if a <= 1:
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return if a < 1: x else: x - (x shr 1)
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let p = primes[a - 1].int64
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if x <= p: return 1 # very simple one-line optimization that limits exponential growth!
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return phi(x, a - 1) - phi((x.float64 / p.float64).int64, a - 1)
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result = phi(n, pisqrt.int) + pisqrt - 1
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cmpsts.dealloc; primes.dealloc
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let nstrt = getMonoTime()
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var pow = 1'i64
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for i in 0 .. 9: echo "π(10^", i, ") = ", pow.countPrimes; pow *= 10
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let nelpsd = (getMonoTime() - nstrt).inMilliseconds
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echo "This took ", nelpsd, " milliseconds."
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# compile with: nim c -d:danger -t:-march=native --gc:arc
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from std/monotimes import getMonoTime, `-`
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from std/times import inMilliseconds
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from std/math import sqrt
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let masks = [ 1'u8, 2, 4, 8, 16, 32, 64, 128 ] # faster than bit twiddling
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let masksp = cast[ptr[UncheckedArray[byte]]](unsafeAddr(masks[0]))
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const TinyPhiPrimes = [2, 3, 5, 7, 11, 13]
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const TinyPhiCirc = 3 * 5 * 7 * 11 * 13
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const TinyPhiRes = 2 * 4 * 6 * 10 * 12
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const CC = 6
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proc makeTinyPhiLUT(): array[TinyPhiCirc, uint16] =
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for i in 0 .. TinyPhiCirc - 1: result[i] = 1
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for i in 1 .. 6:
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if result[i] == 0: continue
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result[i] = 0; let bp = i + i + 1
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let sqri = (i + i) * (i + 1)
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for c in countup(sqri, TinyPhiCirc - 1, bp): result[c] = 0
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var acc = 0'u16;
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for i in 0 .. TinyPhiCirc - 1: acc += result[i]; result[i] = acc
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const TinyPhiLUT = makeTinyPhiLUT()
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proc tinyPhi(x: int64): int64 {.inline.} =
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let ndx = (x - 1) div 2; let numtot = ndx div TinyPhiCirc.int64
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return numtot * TinyPhiRes.int64 + TinyPhiLUT[(ndx - numtot * TinyPhiCirc.int64).int].int64
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proc countPrimes(n: int64): int64 =
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if n < 169: # below 169 whose sqrt is 13 is where TinyPhi doesn't work...
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if n < 3: return if n < 2: 0 else: 1
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# adjust for the missing "degree" base primes
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if n <= 13: return (n - 1) div 2 + (if n < 9: 1 else: 0)
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return 5 + TinyPhiLUT[(n - 1).int div 2].int64
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let rtlmt = n.float64.sqrt.int
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let mxndx = (rtlmt - 1) div 2
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var cmpsts = cast[ptr[UncheckedArray[byte]]](alloc0((mxndx + 8) div 8))
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for i in 1 .. mxndx:
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if (cmpsts[i shr 3] and masksp[i and 7]) != 0: continue
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let sqri = (i + i) * (i + 1)
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if sqri > mxndx: break
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let bp = i + i + 1
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for c in countup(sqri, mxndx, bp):
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let w = c shr 3; cmpsts[w] = cmpsts[w] or masksp[c and 7]
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var pisqrt = 0'i64
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for i in 0 .. mxndx:
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if (cmpsts[i shr 3] and masksp[i and 7]) == 0: pisqrt += 1
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let primes = cast[ptr[UncheckedArray[uint32]]](alloc(sizeof(uint32) * pisqrt.int))
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var j = 0
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for i in 0 .. mxndx:
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if (cmpsts[i shr 3] and masksp[i and 7]) == 0: primes[j] = (i + i + 1).uint32; j += 1
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var phi = tinyPhi(n)
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proc lvl(m, mbf: int64; mxa: int) = # recurse from bottom left of "tree"...
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for a in CC .. mxa:
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let p = primes[a].int64
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if m < p * p: phi += mbf * (mxa - a + 1).int64; return # rest of level all ones!
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let nm = (m.float64 / p.float64).int64; phi += mbf * tinyPhi(nm)
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if a > CC: lvl(nm, -mbf, a - 1) # split
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# finished level!
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lvl(n, -1, pisqrt.int - 1); result = phi + pisqrt - 1
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cmpsts.dealloc; primes.dealloc
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let strt = getMonoTime()
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var pow = 1'i64
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for i in 0 .. 9: echo "π(10^", i, ") = ", pow.countPrimes; pow *= 10
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let elpsd = (getMonoTime() - strt).inMilliseconds
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echo "This took ", elpsd, " milliseconds."
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@ -0,0 +1,121 @@
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# compile with: nim c -d:danger -t:-march=native --gc:arc
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from std/monotimes import getMonoTime, `-`
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from std/times import inMilliseconds
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from std/math import sqrt
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let masks = [ 1'u8, 2, 4, 8, 16, 32, 64, 128 ] # faster than bit twiddling
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let masksp = cast[ptr[UncheckedArray[byte]]](unsafeAddr(masks[0]))
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# non-recursive Legendre prime counting function for a range `n`...
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# this has O(n^(3/4)/((log n)^2)) time complexity; O(n^(1/2)) space complexity.
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proc countPrimes(n: int64): int64 =
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if n < 3: # can't odd sieve for value less than 3!
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return if n < 2: 0 else: 1
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else:
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proc half(n: int): int {.inline.} = (n - 1) shr 1 # convenience conversion to index
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# dividing using float64 is faster than int64 for some CPU's...
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# precision limits range to maybe 1e16!
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proc divide(nm, d: int64): int {.inline.} = (nm.float64 / d.float64).int
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let rtlmt = n.float64.sqrt.int # precision limits range to maybe 1e16!
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let mxndx = (rtlmt - 1) div 2
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var smalls = # current accumulated counts of odd primes 1 to sqrt range
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cast[ptr[UncheckedArray[uint32]]](alloc(sizeof(uint32) * (mxndx + 1)))
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# initialized for no sieving whatsoever other than odds-only - partial sieved by 2:
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# 0 odd primes to 1; 1 odd prime to 3, etc....
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for i in 0 .. mxndx: smalls[i] = i.uint32
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var roughs = # current odd k-rough numbers up to sqrt of range; k = 2
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cast[ptr[UncheckedArray[uint32]]](alloc(sizeof(uint32) * (mxndx + 1)))
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# initialized to all odd positive numbers 1, 3, 5, ... sqrt range...
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for i in 0 .. mxndx: roughs[i] = (i + i + 1).uint32
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# array of current phi counts for above roughs...
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# these are not strictly `phi`'s since they also include the
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# count of base primes in order to match the above `smalls` definition!
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var larges = # starts as size of counts just as `roughs` so they align!
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cast[ptr[UncheckedArray[int64]]](alloc(sizeof(int64) * (mxndx + 1)))
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# initialized for current roughs after accounting for even prime of two...
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for i in 0 .. mxndx: larges[i] = ((n div (i + i + 1) - 1) div 2).int64
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# cmpsts is a bit-packed boolean array representing
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# odd composite numbers from 1 up to rtlmt used for sieving...
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# initialized as "zeros" meaning all odd positives are potentially prime
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# note that this array starts at (and keeps) 1 to match the algorithm even
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# though 1 is not a prime, as 1 is important in computation of phi...
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var cmpsts = cast[ptr[UncheckedArray[byte]]](alloc0((mxndx + 8) div 8))
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# number of found base primes and current highest used rough index...
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var npc = 0; var mxri = mxndx
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for i in 1 .. mxndx: # start at index for 3; i will never reach mxndx...
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let sqri = (i + i) * (i + 1) # computation of square index!
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if sqri > mxndx: break # stop partial sieving due to square index limit!
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if (cmpsts[i shr 3] and masksp[i and 7]) != 0'u8: continue # if not prime
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# culling the base prime from cmpsts means it will never be found again
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cmpsts[i shr 3] = cmpsts[i shr 3] or masksp[i and 7] # cull base prime
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let bp = i + i + 1 # base prime from index!
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for c in countup(sqri, mxndx, bp): # SoE culling of all bp multiples...
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let w = c shr 3; cmpsts[w] = cmpsts[w] or masksp[c and 7]
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# partial sieving to current base prime is now completed!
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var ri = 0 # to keep track of current used roughs index!
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for k in 0 .. mxri: # processing over current roughs size...
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# q is not necessarily a prime but may be a
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# product of primes not yet culled by partial sieving;
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# this is what saves operations compared to recursive Legendre:
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let q = roughs[k].int; let qi = q shr 1 # index of always odd q!
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# skip over values of `q` already culled in the last partial sieve:
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if (cmpsts[qi shr 3] and masksp[qi and 7]) != 0'u8: continue
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# since `q` cannot be equal to bp due to cull of bp and above skip;
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let d = bp * q # `d` is odd product of some combination of odd primes!
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# the following computation is essential to the algorithm's speed:
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# see above description in the text for how this works:
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larges[ri] = larges[k] -
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(if d <= rtlmt: larges[smalls[d shr 1].int - npc]
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else: smalls[half(divide(n, d.int64))].int64) + npc.int64
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# eliminate rough values that have been culled in partial sieve:
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# note that `larges` and `roughs` indices relate to each other!
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roughs[ri] = q.uint32; ri += 1 # update rough value; advance rough index
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var m = mxndx # adjust `smalls` counts for the newly culled odds...
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# this is faster than recounting over the `cmpsts` array for each loop...
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for k in countdown(((rtlmt div bp) - 1) or 1, bp, 2): # k always odd!
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# `c` is correction from current count to desired count...
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# `e` is end limit index no correction is necessary for current cull...
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let c = smalls[k shr 1] - npc.uint32; let e = (k * bp) shr 1
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while m >= e: smalls[m] -= c; m -= 1 # correct over range down to `e`
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mxri = ri - 1; npc += 1 # set next loop max roughs index; count base prime
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# now `smalls` is a LUT of odd prime accumulated counts for all odd primes;
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# `roughs` is exactly the "k-roughs" up to the sqrt of range with `k` the
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# index of the next prime above the quad root of the range;
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# `larges` is the partial prime counts for each of the `roughs` values...
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# note that `larges` values include the count of the odd base primes!!!
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# `cmpsts` are never used again!
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# the following does the top most "phi tree" calculation:
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result = larges[0] # the answer to here is all valid `phis`
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for i in 1 .. mxri: result -= larges[i] # combined here by subtraction
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# compensate for the included odd base prime counts over subracted above:
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result += ((mxri + 1 + 2 * (npc - 1)) * mxri div 2).int64
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# This loop adds the counts due to the products of the `roughs` primes,
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# of which we only use two different ones at a time, as all the
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# combinations with lower primes than the cube root of the range have
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# already been computed and included with the previous major loop...
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# see text description above for how this works...
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for j in 1 .. mxri: # for all `roughs` (now prime) not including one:
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let p = roughs[j].int64; let m = n div p # `m` is the `p` quotient
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# so that the end limit `e` can be calculated based on `n`/(`p`^2)
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let e = smalls[half((m div p).int)].int - npc
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# following break test equivalent to non-memoization/non-splitting optmization:
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if e <= j: break # stop at about `p` of cube root of range!
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for k in j + 1 .. e: # for all `roughs` greater than `p` to end limit:
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result += smalls[half(divide(m, roughs[k].int64))].int64
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# compensate for all the extra base prime counts just added!
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result -= ((e - j) * (npc + j - 1)).int64
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result += 1 # include the count for the only even prime of two
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smalls.dealloc; roughs.dealloc; larges.dealloc; cmpsts.dealloc
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let strt = getMonoTime()
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var pow = 1'i64
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for i in 0 .. 9: echo "π(10^", i, ") = ", pow.countPrimes; pow *= 10
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let elpsd = (getMonoTime() - strt).inMilliseconds
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echo "This took ", elpsd, " milliseconds."
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