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Prime factorization
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ORmorni 25 Sep, 2022 @ 2:29pm
Solution thread
29:

This relies on the property that there are few primes below 10:
(once)=(end)7?7?5?5?3?3?3?2?2?2?2?2?2?
(once)=^
# To unary
a0=0aa
a1=1aa
^1=^a
^0=^
# Division by 7
7!aaaaaaa=a7!
7!=(start)111,
^7?=^7!
aaaaaaa7?=7?aaaaaaa
7?=
# Division by 5
5!aaaaa=a5!
5!=(start)101,
^5?=^5!
aaaaa5?=5?aaaaa
5?=
# Division by 3
3!aaa=a3!
3!=(start)11,
^3?=^3!
aaa3?=3?aaa
3?=
# Cleaning up
^=
# Converting back
a=b
1b=b0
0b=1
b=1
# Division by 2
02?=(start)10,
2?=
(end),1=
Last edited by ORmorni; 25 Sep, 2022 @ 3:23pm
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SoMall-dumpling  [developer] 15 Apr, 2023 @ 12:25am 
Thank you for solving my problem！
I'd like to show my original solution here:

(once)=(end)\--*
/0=/
(start)pk/\C$*=
<=(start)p
/\$*=/\C$* #C means an exact division
k/\C=/\C
------------*=*-
++,*=*++,
+*=*+
$$-=-q$$
$-=-q$
$*=(end)*
q=(start)r
rpk=K- #K means the square of the divisor is bigger than the dividend
rp=(start)p
rK=K
pK=K
-k=--
-/\-=-/\
-/\*=-
K=
k?/=?/xy
p?/=?/
?=
/\-=?/\$$-- #? will add 1 to the divisor
p/=/xy
C-=-zC
z=(end)+
C=(end),
(once)1=/xy
xy1=0xyxyxy
xy0=0xyxy
y=(start)k
x\-*=\$*<-
x\-=\------------
-=(end)+
0+=1
1+=+0
+=1
Last edited by SoMall-dumpling; 15 Apr, 2023 @ 12:27am
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