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By the very nature of this solution, its best-golfed form rounds arbitrarily – sometimes up, sometimes down. As such, it's more complicated to explain why it works, so I suggest reading the explanation of the 104 byte always-round-down solution below first. (This is oddly appropriate, given the order of operations in regexes that use right-to-left evaluated variable-length lookbehind.)

By the very nature of this solution, its best-golfed form rounds arbitrarily – sometimes up, sometimes down. As such, it's more complicated to explain why it works, so I suggest reading the explanation of the 99 byte always-round-down solution below first. (This is oddly appropriate, given the order of operations in regexes that use right-to-left evaluated variable-length lookbehind.)

(?=(x*)(1円x?))(x*)(?^=((?=(2円$|))x(?^=((?^=(7円?x))(^10円|3円))*5円(x*)(?^3=9円(x*))))*1円(?^7=3円)|^)

 # tail = N = input number
(?=
 (x*) # 1円 = floor(N / 2)
 (1円x?) # 2円 = ceil(N / 2)
)
(x*) # 3円 = guess for floor(sqrt(N*N/2))
(?^= # Atomic lookinto: tail = N
 (
 (?=(2円$|)) # 5円 = 2円 if tail == 2,円 0 otherwise;
 # Note that if tail==2,円 it means this is the last iteration
 # if N is odd – but if N is even, this iteration will fail to
 # match, as the last iteration will have already finished.
 x # tail -= 1
 (?^= # Atomic lookinto: tail = N
 # if 10円 is set: tail -= 10円; 7円 += 1;
 # X = floor((tail - 5円) / 3円); tail -= 3円*X; 7円 += X;
 # Note that in practice, 0 ≤ X ≤ 1
 (
 (?^=
 ( # 7円 = sum of the following:
 7円? # previous value of 7円 if set, 0 otherwise
 x # 1
 )
 )
 (
 ^10円 # On the first iteration, if 10円 is set, tail -= 10円
 |
 3円 # Assert tail ≥ 3円; tail -= 3円
 )
 )* # Match the above as many times as possible, minimum 0
 5円 # tail -= 5円
 (x*) # 9円 = tail = remainder = dividend % 3円
 (?^3= # Atomic lookinto: tail = 3円
 9円 # tail -= 9円
 (x*) # 10円 = tail = 3円 - 9円
 )
 )
 )*
 1円 # Assert tail ≥ 1,円 forcing the loop above to iterate 2円 times
 (?^7= # Lookinto: tail = 7円
 3円 # Assert 3円 ≤ tail, i.e. 3円 ≤ 7円
 )
|
 ^ # or N == 0, in which case the output = the input; this needs to be
 # a special case because "(?^7=...)" cannot match if 7円 is unset.
)

(?=(x*)(1円x?))(x*)(?^=((?=(2円$|))x(?^=((?^=(7円?x))(^10円|3円))*5円(x*)(?^3=9円(x*))))*1円(?^7=3円)|$)

 # tail = N = input number
(?=
 (x*) # 1円 = floor(N / 2)
 (1円x?) # 2円 = ceil(N / 2)
)
(x*) # 3円 = guess for floor(sqrt(N*N/2))
(?^= # Atomic lookinto: tail = N
 (
 (?=(2円$|)) # 5円 = 2円 if tail == 2,円 0 otherwise;
 # Note that if tail==2,円 it means this is the last iteration
 # if N is odd – but if N is even, this iteration will fail to
 # match, as the last iteration will have already finished.
 x # tail -= 1
 (?^= # Atomic lookinto: tail = N
 # if 10円 is set: tail -= 10円; 7円 += 1;
 # X = floor((tail - 5円) / 3円); tail -= 3円*X; 7円 += X;
 # Note that in practice, 0 ≤ X ≤ 1
 (
 (?^=
 ( # 7円 = sum of the following:
 7円? # previous value of 7円 if set, 0 otherwise
 x # 1
 )
 )
 (
 ^10円 # On the first iteration, if 10円 is set, tail -= 10円
 |
 3円 # Assert tail ≥ 3円; tail -= 3円
 )
 )* # Match the above as many times as possible, minimum 0
 5円 # tail -= 5円
 (x*) # 9円 = tail = remainder = dividend % 3円
 (?^3= # Atomic lookinto: tail = 3円
 9円 # tail -= 9円
 (x*) # 10円 = tail = 3円 - 9円
 )
 )
 )*
 1円 # Assert tail ≥ 1,円 forcing the loop above to iterate 2円 times
 (?^7= # Lookinto: tail = 7円
 3円 # Assert 3円 ≤ tail, i.e. 3円 ≤ 7円
 )
|
 $ # or N == 0, in which case the output = the input; this needs to be
 # a special case because "(?^7=...)" cannot match if 7円 is unset.
)

 # tail = N = input number
(?=
 (x*) # 1円 = floor(N / 2)
 (1円x?) # 2円 = ceil(N / 2)
)
(x*) # 3円 = guess for floor(sqrt(N*N/2))
(?^= # Atomic lookinto: tail = N
 (
 (?=(2円$|)) # 5円 = 2円 if tail == 2,円 0 otherwise;
 # Note that if tail==2,円 it means this is the last iteration
 # if N is odd – but if N is even, this iteration will fail to
 # match, as the last iteration will have already finished.
 x # tail -= 1
 (?^= # Atomic lookinto: tail = N
 # if 10円 is set: tail -= 10円; 7円 += 1;
 # X = floor(tail / 3円); tail -= 3円*X; 7円 += X;
 # Note that in practice, 0 ≤ X ≤ 1
 (
 (?^=
 ( # 7円 = sum of the following:
 7円? # previous value of 7円 if set, 0 otherwise
 x # 1
 )
 )
 (
 ^10円 # On the first iteration, if 10円 is set, tail -= 10円
 |
 3円 # Assert tail ≥ 3円; tail -= 3円
 )
 )* # Match the above as many times as possible, minimum 0
 5円 # tail -= 5円
 (x*) # 9円 = tail = remainder = dividend % 3円
 (?^3= # Atomic lookinto: tail = 3円
 9円 # tail -= 9円
 (x*) # 10円 = tail = 3円 - 9円
 )
 )
 )*
 1円 # Assert tail ≥ 1,円 forcing the loop above to iterate 2円 times
 (?^7= # Lookinto: tail = 7円
 3円 # Assert 3円 ≤ tail, i.e. 3円 ≤ 7円
 )
|
 ^ # or N == 0, in which case the output = the input; this needs to be
 # a special case because "(?^7=...)" cannot match if 7円 is unset.
)

 # tail = N = input number
(?=
 (x*) # 1円 = floor(N / 2)
 (1円x?) # 2円 = ceil(N / 2)
)
(x*) # 3円 = guess for floor(sqrt(N*N/2))
(?^= # Atomic lookinto: tail = N
 (
 (?=(2円$|)) # 5円 = 2円 if tail == 2,円 0 otherwise;
 # Note that if tail==2,円 it means this is the last iteration
 # if N is odd – but if N is even, this iteration will fail to
 # match, as the last iteration will have already finished.
 x # tail -= 1
 (?^= # Atomic lookinto: tail = N
 # if 10円 is set: tail -= 10円; 7円 += 1;
 # X = floor((tail - 5円) / 3円); tail -= 3円*X; 7円 += X;
 # Note that in practice, 0 ≤ X ≤ 1
 (
 (?^=
 ( # 7円 = sum of the following:
 7円? # previous value of 7円 if set, 0 otherwise
 x # 1
 )
 )
 (
 ^10円 # On the first iteration, if 10円 is set, tail -= 10円
 |
 3円 # Assert tail ≥ 3円; tail -= 3円
 )
 )* # Match the above as many times as possible, minimum 0
 5円 # tail -= 5円
 (x*) # 9円 = tail = remainder = dividend % 3円
 (?^3= # Atomic lookinto: tail = 3円
 9円 # tail -= 9円
 (x*) # 10円 = tail = 3円 - 9円
 )
 )
 )*
 1円 # Assert tail ≥ 1,円 forcing the loop above to iterate 2円 times
 (?^7= # Lookinto: tail = 7円
 3円 # Assert 3円 ≤ tail, i.e. 3円 ≤ 7円
 )
|
 ^ # or N == 0, in which case the output = the input; this needs to be
 # a special case because "(?^7=...)" cannot match if 7円 is unset.
)