Mixing
Maximize the number of factorials in your solution to 6 5 4 3 = 1
Clash Royale CLAN TAG#URR8PPP
up vote
9
down vote
favorite
Using the same rules as Make 6 5 4 3 = 1, but maximize the number of factorials. You may not take the factorial of 1 or 2.
This is harder than it looks. A great answer has six factorials, an amazing answer has seven factorials, and an outstanding answer has eight or more factorials.
Edit: Shoot, I should have blocked factorials of numbers bigger than twenty like I did in my program.
mathematics formation-of-numbers number-theory
asked Aug 28 at 14:08
Neil G
19419
 |Â
show 8 more comments
up vote
9
down vote
favorite
Using the same rules as Make 6 5 4 3 = 1, but maximize the number of factorials. You may not take the factorial of 1 or 2.
This is harder than it looks. A great answer has six factorials, an amazing answer has seven factorials, and an outstanding answer has eight or more factorials.
Edit: Shoot, I should have blocked factorials of numbers bigger than twenty like I did in my program.
mathematics formation-of-numbers number-theory
asked Aug 28 at 14:08
Neil G
19419
I'm guessing we shouldn't take the factorial of 2 either
– Bennett Bernardoni
Aug 28 at 14:11
@BennettBernardoni Thanks, yes.
– Neil G
Aug 28 at 14:11
Is the answer really different than wolfram42's answer to the other puzzle? (3 factorials)
– xhienne
Aug 28 at 14:17
@xhienne That was my opinion but if he found 8 that is probably precise enough for another question.
– Saeïdryl
Aug 28 at 14:18
1
@Saeïdryl I understand that you can not use (6 - 5)! nor (6 - 4)!
– xhienne
Aug 28 at 14:21
 |Â
show 8 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
Using the same rules as Make 6 5 4 3 = 1, but maximize the number of factorials. You may not take the factorial of 1 or 2.
This is harder than it looks. A great answer has six factorials, an amazing answer has seven factorials, and an outstanding answer has eight or more factorials.
Edit: Shoot, I should have blocked factorials of numbers bigger than twenty like I did in my program.
mathematics formation-of-numbers number-theory
asked Aug 28 at 14:08
Neil G
19419
Using the same rules as Make 6 5 4 3 = 1, but maximize the number of factorials. You may not take the factorial of 1 or 2.
This is harder than it looks. A great answer has six factorials, an amazing answer has seven factorials, and an outstanding answer has eight or more factorials.
Edit: Shoot, I should have blocked factorials of numbers bigger than twenty like I did in my program.
mathematics formation-of-numbers number-theory
asked Aug 28 at 14:08
Neil G
19419
edited Aug 28 at 14:31
asked Aug 28 at 14:08
Neil G
19419
asked Aug 28 at 14:08
Neil G
19419
asked Aug 28 at 14:08
Neil G
19419
19419
I'm guessing we shouldn't take the factorial of 2 either
– Bennett Bernardoni
Aug 28 at 14:11
@BennettBernardoni Thanks, yes.
– Neil G
Aug 28 at 14:11
Is the answer really different than wolfram42's answer to the other puzzle? (3 factorials)
– xhienne
Aug 28 at 14:17
@xhienne That was my opinion but if he found 8 that is probably precise enough for another question.
– Saeïdryl
Aug 28 at 14:18
1
@Saeïdryl I understand that you can not use (6 - 5)! nor (6 - 4)!
– xhienne
Aug 28 at 14:21
 |Â
show 8 more comments
I'm guessing we shouldn't take the factorial of 2 either
– Bennett Bernardoni
Aug 28 at 14:11
@BennettBernardoni Thanks, yes.
– Neil G
Aug 28 at 14:11
Is the answer really different than wolfram42's answer to the other puzzle? (3 factorials)
– xhienne
Aug 28 at 14:17
@xhienne That was my opinion but if he found 8 that is probably precise enough for another question.
– Saeïdryl
Aug 28 at 14:18
1
@Saeïdryl I understand that you can not use (6 - 5)! nor (6 - 4)!
– xhienne
Aug 28 at 14:21
I'm guessing we shouldn't take the factorial of 2 either
– Bennett Bernardoni
Aug 28 at 14:11
I'm guessing we shouldn't take the factorial of 2 either
– Bennett Bernardoni
Aug 28 at 14:11
@BennettBernardoni Thanks, yes.
– Neil G
Aug 28 at 14:11
@BennettBernardoni Thanks, yes.
– Neil G
Aug 28 at 14:11
Is the answer really different than wolfram42's answer to the other puzzle? (3 factorials)
– xhienne
Aug 28 at 14:17
Is the answer really different than wolfram42's answer to the other puzzle? (3 factorials)
– xhienne
Aug 28 at 14:17
@xhienne That was my opinion but if he found 8 that is probably precise enough for another question.
– Saeïdryl
Aug 28 at 14:18
@xhienne That was my opinion but if he found 8 that is probably precise enough for another question.
– Saeïdryl
Aug 28 at 14:18
1
1
@Saeïdryl I understand that you can not use (6 - 5)! nor (6 - 4)!
– xhienne
Aug 28 at 14:21
@Saeïdryl I understand that you can not use (6 - 5)! nor (6 - 4)!
– xhienne
Aug 28 at 14:21
 |Â
show 8 more comments
6 Answers
6
active
oldest
votes
up vote
10
down vote
accepted
I'll attempt with the rule "No factorials on numbers greater than 20". Here is a solution with 7. Some formatting help is appreciated
$$frac6!(5!div4!)!! div 3!!$$
Method:
I started with the observation that 6! / 5! = 6, Likewise 5!/4! = 5. 3! = 6 so 6! = 3!! I then built up from there. So 6! = (6!/(5! / 4!))!
Figured out the missing factorial:
$$left(frac6!(5!div4!)!! - 3!!right)!$$ works since 0! = 1 by definition.
edited Aug 28 at 19:01
Community♦
1
answered Aug 28 at 16:43
wolfram42
2,185119
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
3
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
1
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
 |Â
show 1 more comment
up vote
43
down vote
The answer is
Infinite factorials even without factorials of 1 or 2. By this equation $$6!!!ldots * (5-4) div 3!!!!ldots = 1$$
Notice that the 3 has one more factorial. This comes from the fact that $3!=6$ and we can keep adding factorials to both.
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
1
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
7
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
add a comment |Â
up vote
11
down vote
How about
Infinitely many factorials?
A basic example would be : $frac6 * (5-4) 3! = 1$
But you could add some factorials in the lot : $frac6! * (5-4) 3!! = 1 $
Or even : $frac6!!!!!!!times (5-4) (3!)!!!!!!! = 1 $
Really, as long as there is one more factorial behind the 3, we're good!
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
answered Aug 28 at 14:29
Keelhaul
7,1132473
2
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
1
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
add a comment |Â
up vote
6
down vote
Starting from wolfram42's answer to that other puzzle, the maximum number of factorials is:
infinite since you can add any equal number of
!to the numerator and denominator of(6!) / (5 * 4! * 3!)
Well, since I'm no mathematician, I may say "infinite plus three" actually ;-)
Note: that's also true for any answer to the other puzzle that involved a division, like JonMark Perry's two solutions
answered Aug 28 at 14:29
xhienne
2,854526
add a comment |Â
up vote
2
down vote
Here's the general programming solution to these kinds of problems. It also prints out for every result, how many solutions there are, which is useful in coming up with these kinds of problem:
from enum import IntEnum
from fractions import Fraction
from memoized import memoized
import math
NUMBERS = [6, 5, 4, 3]
RESULT = Fraction(1)
class Levels(IntEnum):
ADDITION = 0
MULTIPLICATION = 1
NEGATION = 2
POWER = 3
FACTORIAL = 4
BARE = 5
class Result:
def __init__(self, value, way, level):
"""
* value is the value achieved.
* way is string representation of an expression.
* level is the outermost operator applied to the expression, which
determines how this expression should be braketed when it's part of a
larger expression.
"""
assert isinstance(value, Fraction)
assert isinstance(level, Levels)
self.value = value
self.way = way
self.level = level
def bracketed(self, for_level):
if for_level > self.level:
return f"(self.way)"
return self.way
def combine(left_rs, right_rs):
for l in left_rs:
for r in right_rs:
if r.level not in (Levels.NEGATION, Levels.ADDITION):
# We block a - (-b), and a + (-b) since these can be written as
# a + b, and a - b. We also block a + (b + c), and a - (b + c)
# since these can be written as a + b + c, and a - b - c.
yield Result(l.value + r.value,
f"l.bracketed(Levels.ADDITION) + r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
yield Result(l.value - r.value,
f"l.bracketed(Levels.ADDITION) - r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
if r.level != Levels.MULTIPLICATION:
# We block a * (b * c) since this can be written as a * b * c.
# We block a * (b / c) since this can be written as a * b / c.
yield Result(l.value * r.value,
f"l.bracketed(Levels.MULTIPLICATION) * r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
if r.value != 0 and (r.level != Levels.MULTIPLICATION):
# We block a / (b / c) since this can be written as a / b * c.
# We block a / (b * c) since this can be written as a / b / c.
yield Result(l.value / r.value,
f"l.bracketed(Levels.MULTIPLICATION) / r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
new_value = None
power_okay = l.value >= 0 or r.value.denominator % 2 == 1
if l.value == 1:
new_value = Fraction(1)
elif l.value == -1:
if power_okay:
new_value = Fraction(-1
if r.value.numerator % 2 == 1
else 1)
elif l.value == 0:
if r.value > 0:
new_value = Fraction(0)
else:
# Only whole number powers are allowed for numbers other than
# -1, 0, 1.
if (power_okay
and r.value.denominator == 1
and -8 < r.value.numerator < 8):
new_value = l.value ** r.value
if new_value is not None:
# The left side is bracketed when it's a power because power is
# right associative.
yield Result(new_value,
f"l.bracketed(Levels.POWER + 1) ^ r.bracketed(Levels.POWER)",
Levels.POWER)
if (l.level == Levels.BARE
and r.level == Levels.BARE and len(r.way) == 1):
# We concatenate 123 as 1(23), but not as (12)3.
yield Result(l.value * 10 + r.value,
f"l.wayr.way",
Levels.BARE)
def negate(rs):
for r in rs:
yield r
if r.level > Levels.MULTIPLICATION:
# We block -(a + b) since this can be written -a - b.
# We block -(a - b) since this can be written -a + b.
# We block -(a * b) since this can be written -a * b.
# We block -(a / b) since this can be written -a / b.
yield Result(-r.value,
f"-r.bracketed(Levels.NEGATION)",
Levels.NEGATION)
def factorialize(rs):
for r in rs:
yield r
x = 0
while (r.value.denominator == 1
and (r.value.numerator == 0 or (3 <= r.value.numerator <= 20))):
# We block fractional and negative factorials.
# We block 1 and 2 factorial.
# We block gigantic factorials.
r = Result(Fraction(math.factorial(r.value.numerator)),
f"r.bracketed(Levels.FACTORIAL + 1)!",
Levels.FACTORIAL)
yield r
# memoization makes the recursive solution into a dynamic programming solution.
@memoized(hashable=False)
def do(digits):
"""
Given a list of digits, produce a list of Result objects.
"""
l = len(digits)
if l == 1:
# Return a Result object for a single digit.
new_rs = [Result(Fraction(digits[0]), str(digits[0]), Levels.BARE)]
else:
# The list of results for a list of digits is the total results for every contiguous partition.
new_rs =
for i in range(1, l):
new_rs.extend(list(combine(do(digits[:i]),
do(digits[i:]))))
return list(factorialize(negate(new_rs)))
rs = do(NUMBERS)
vs =
for r in rs:
vs.setdefault(r.value, ).append(r)
print(f"Solutions to NUMBERS = RESULT")
for r in sorted(vs[RESULT], key=lambda r: len(r.way)):
print(r.way)
print()
print(f" value solutions")
for v, l in sorted(vs.items(), key=lambda vl: (len(vl[1]), len(str(vl[0])))):
print(f"str(v):>30 len(l):<12")
prints
Solutions to [6, 5, 4, 3] = 1
65 - 4 ^ 3
(6 - 5) ^ 43
65 + (-4) ^ 3
-(-6 + 5) ^ 43
-6 - 5 + 4 * 3
6 + 5 - 4 - 3!
6 - 5 * (4 - 3)
6 - 5 / (4 - 3)
6 - 5 ^ (4 - 3)
6 / (5 + 4 - 3)
6 - 5! / 4 / 3!
(6 - 5) ^ 4 ^ 3
(6 - 5) ^ (-43)
...
(((-6 + 5 + 4)!)! - (3!)!)!
((6! / (5! / 4!)!)! - (3!)!)!
((-(-6) ^ (5 - 4))! - (3!)!)!
answered Aug 28 at 18:25
Neil G
19419
1
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
1
So what does this output?
– aschepler
Aug 30 at 2:08
@aschepler Added
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
add a comment |Â
up vote
1
down vote
9 factorials, following the rules (answer below)
Some tricks to use
First trick:
$0!=1$ (by definition). We will use the fact that $forall xinmathbbR:0cdot x=0 Rightarrow (0cdot x)!=1$.
Second trick:
Since we cannot use $1!$ or $2!$ and we cannot use $n!$ for $n>20$, we have to look the other way: think smaller. Lucky for us, we have $x!=Gamma(x+1)$ where $Gamma$ is the gamma function and this holds for $xinmathbbR$.
Building our answer
First step (using the first trick):
We need zero. That's pretty easy, we have $6!-(3!)!=6!-6!=0$.
Second step (using the second trick):
We need many factorials but we can only use $4$ and $5$ because we used the others in the first step. To get those, we aren't going to worry about the outcome because we will multiply by $0$ anyway. We only have to be careful not to get to close to $0$ or $1$ (we shall stay at least $0.01$ distance away from them) and not to exceed $20$.
Calculations in the second step:
The first factorial $left(frac-54right)!approx -4.9$.
The second factorial $left(left(frac-54right)!right)!approx 0.50$.
The third factorial $left(left(left(frac-54right)!right)!right)!approx 0.89$.
The fourth factorial $left(left(left(left(frac-54right)!right)!right)!right)!approx 0.96$.
The fifth factorial $left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$.
The sixth factorial $left(left(left(left(left(left(frac-54right)!right)!right)!right)!right)!right)!approx 0.993$.
Therefore, we can use 5 factorials here, because with the sixth step we get to close to $1$.
Answer
$$x=6!-(3!)!=0$$
$$y=left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$$
$$textitfinal answer=(xcdot y)!=0!=1$$
answered Aug 29 at 15:08
JJJ
1505
add a comment |Â
6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
10
down vote
accepted
I'll attempt with the rule "No factorials on numbers greater than 20". Here is a solution with 7. Some formatting help is appreciated
$$frac6!(5!div4!)!! div 3!!$$
Method:
I started with the observation that 6! / 5! = 6, Likewise 5!/4! = 5. 3! = 6 so 6! = 3!! I then built up from there. So 6! = (6!/(5! / 4!))!
Figured out the missing factorial:
$$left(frac6!(5!div4!)!! - 3!!right)!$$ works since 0! = 1 by definition.
edited Aug 28 at 19:01
Community♦
1
answered Aug 28 at 16:43
wolfram42
2,185119
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
3
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
1
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
 |Â
show 1 more comment
up vote
10
down vote
accepted
I'll attempt with the rule "No factorials on numbers greater than 20". Here is a solution with 7. Some formatting help is appreciated
$$frac6!(5!div4!)!! div 3!!$$
Method:
I started with the observation that 6! / 5! = 6, Likewise 5!/4! = 5. 3! = 6 so 6! = 3!! I then built up from there. So 6! = (6!/(5! / 4!))!
Figured out the missing factorial:
$$left(frac6!(5!div4!)!! - 3!!right)!$$ works since 0! = 1 by definition.
edited Aug 28 at 19:01
Community♦
1
answered Aug 28 at 16:43
wolfram42
2,185119
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
3
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
1
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
 |Â
show 1 more comment
up vote
10
down vote
accepted
up vote
10
down vote
accepted
I'll attempt with the rule "No factorials on numbers greater than 20". Here is a solution with 7. Some formatting help is appreciated
$$frac6!(5!div4!)!! div 3!!$$
Method:
I started with the observation that 6! / 5! = 6, Likewise 5!/4! = 5. 3! = 6 so 6! = 3!! I then built up from there. So 6! = (6!/(5! / 4!))!
Figured out the missing factorial:
$$left(frac6!(5!div4!)!! - 3!!right)!$$ works since 0! = 1 by definition.
edited Aug 28 at 19:01
Community♦
1
answered Aug 28 at 16:43
wolfram42
2,185119
I'll attempt with the rule "No factorials on numbers greater than 20". Here is a solution with 7. Some formatting help is appreciated
$$frac6!(5!div4!)!! div 3!!$$
Method:
I started with the observation that 6! / 5! = 6, Likewise 5!/4! = 5. 3! = 6 so 6! = 3!! I then built up from there. So 6! = (6!/(5! / 4!))!
Figured out the missing factorial:
$$left(frac6!(5!div4!)!! - 3!!right)!$$ works since 0! = 1 by definition.
edited Aug 28 at 19:01
Community♦
1
answered Aug 28 at 16:43
wolfram42
2,185119
edited Aug 28 at 19:01
Community♦
1
edited Aug 28 at 19:01
Community♦
1
edited Aug 28 at 19:01
Community♦
1
1
answered Aug 28 at 16:43
wolfram42
2,185119
answered Aug 28 at 16:43
wolfram42
2,185119
answered Aug 28 at 16:43
wolfram42
2,185119
2,185119
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
3
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
1
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
 |Â
show 1 more comment
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
3
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
1
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Missing one pair of parentheses I think, but yes amazing! How did you do it?
– Neil G
Aug 28 at 16:44
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Actually i made a quick edit and I confused myself. I'll readjust in a minute
– wolfram42
Aug 28 at 16:46
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
Nice, you're pretty close to 8
– Neil G
Aug 28 at 16:57
3
3
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
Presumably by $n!!$ you mean $(n!)!$, not the double factorial operation $prod_k in mathbb N, n - 2 k > 0 (n - 2 k)$?
– wchargin
Aug 28 at 22:44
1
1
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
I assumed that the double factorial operation was not allowed since it would be easy to get more than 8 that way.
– wolfram42
Aug 29 at 14:07
 |Â
show 1 more comment
up vote
43
down vote
The answer is
Infinite factorials even without factorials of 1 or 2. By this equation $$6!!!ldots * (5-4) div 3!!!!ldots = 1$$
Notice that the 3 has one more factorial. This comes from the fact that $3!=6$ and we can keep adding factorials to both.
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
1
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
7
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
add a comment |Â
up vote
43
down vote
The answer is
Infinite factorials even without factorials of 1 or 2. By this equation $$6!!!ldots * (5-4) div 3!!!!ldots = 1$$
Notice that the 3 has one more factorial. This comes from the fact that $3!=6$ and we can keep adding factorials to both.
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
1
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
7
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
add a comment |Â
up vote
43
down vote
up vote
43
down vote
The answer is
Infinite factorials even without factorials of 1 or 2. By this equation $$6!!!ldots * (5-4) div 3!!!!ldots = 1$$
Notice that the 3 has one more factorial. This comes from the fact that $3!=6$ and we can keep adding factorials to both.
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
The answer is
Infinite factorials even without factorials of 1 or 2. By this equation $$6!!!ldots * (5-4) div 3!!!!ldots = 1$$
Notice that the 3 has one more factorial. This comes from the fact that $3!=6$ and we can keep adding factorials to both.
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
answered Aug 28 at 14:26
Bennett Bernardoni
2,073415
2,073415
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
1
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
7
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
add a comment |Â
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
1
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
7
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
Wow, that's brilliant
– Neil G
Aug 28 at 14:28
1
1
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
(I know this is the best answer, but I hope you don't mind that I marked the other answer as accepted since it was the one I was shooting for.)
– Neil G
Aug 28 at 16:46
7
7
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@NeilG Unless this was secretly a riddle, you should accept the best possible answer. It turns out that one of the responders was more clever than you - reward that!
– Jeremy Weirich
Aug 29 at 12:16
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
@jeremy well I upvoted him ?
– Neil G
Aug 29 at 12:24
add a comment |Â
up vote
11
down vote
How about
Infinitely many factorials?
A basic example would be : $frac6 * (5-4) 3! = 1$
But you could add some factorials in the lot : $frac6! * (5-4) 3!! = 1 $
Or even : $frac6!!!!!!!times (5-4) (3!)!!!!!!! = 1 $
Really, as long as there is one more factorial behind the 3, we're good!
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
answered Aug 28 at 14:29
Keelhaul
7,1132473
2
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
1
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
add a comment |Â
up vote
11
down vote
How about
Infinitely many factorials?
A basic example would be : $frac6 * (5-4) 3! = 1$
But you could add some factorials in the lot : $frac6! * (5-4) 3!! = 1 $
Or even : $frac6!!!!!!!times (5-4) (3!)!!!!!!! = 1 $
Really, as long as there is one more factorial behind the 3, we're good!
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
answered Aug 28 at 14:29
Keelhaul
7,1132473
2
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
1
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
add a comment |Â
up vote
11
down vote
up vote
11
down vote
How about
Infinitely many factorials?
A basic example would be : $frac6 * (5-4) 3! = 1$
But you could add some factorials in the lot : $frac6! * (5-4) 3!! = 1 $
Or even : $frac6!!!!!!!times (5-4) (3!)!!!!!!! = 1 $
Really, as long as there is one more factorial behind the 3, we're good!
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
answered Aug 28 at 14:29
Keelhaul
7,1132473
How about
Infinitely many factorials?
A basic example would be : $frac6 * (5-4) 3! = 1$
But you could add some factorials in the lot : $frac6! * (5-4) 3!! = 1 $
Or even : $frac6!!!!!!!times (5-4) (3!)!!!!!!! = 1 $
Really, as long as there is one more factorial behind the 3, we're good!
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
answered Aug 28 at 14:29
Keelhaul
7,1132473
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
edited Aug 30 at 2:18
JonMark Perry
13.8k42667
13.8k42667
answered Aug 28 at 14:29
Keelhaul
7,1132473
answered Aug 28 at 14:29
Keelhaul
7,1132473
answered Aug 28 at 14:29
Keelhaul
7,1132473
7,1132473
2
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
1
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
add a comment |Â
2
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
1
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
2
2
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
Damn you, @Bennett Bernardoni ! :p
– Keelhaul
Aug 28 at 14:29
1
1
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
Great answer! And congratz on the $7k$ rep! :D
– user477343
Aug 29 at 22:01
add a comment |Â
up vote
6
down vote
Starting from wolfram42's answer to that other puzzle, the maximum number of factorials is:
infinite since you can add any equal number of
!to the numerator and denominator of(6!) / (5 * 4! * 3!)
Well, since I'm no mathematician, I may say "infinite plus three" actually ;-)
Note: that's also true for any answer to the other puzzle that involved a division, like JonMark Perry's two solutions
answered Aug 28 at 14:29
xhienne
2,854526
add a comment |Â
up vote
6
down vote
Starting from wolfram42's answer to that other puzzle, the maximum number of factorials is:
infinite since you can add any equal number of
!to the numerator and denominator of(6!) / (5 * 4! * 3!)
Well, since I'm no mathematician, I may say "infinite plus three" actually ;-)
Note: that's also true for any answer to the other puzzle that involved a division, like JonMark Perry's two solutions
answered Aug 28 at 14:29
xhienne
2,854526
add a comment |Â
up vote
6
down vote
up vote
6
down vote
Starting from wolfram42's answer to that other puzzle, the maximum number of factorials is:
infinite since you can add any equal number of
!to the numerator and denominator of(6!) / (5 * 4! * 3!)
Well, since I'm no mathematician, I may say "infinite plus three" actually ;-)
Note: that's also true for any answer to the other puzzle that involved a division, like JonMark Perry's two solutions
answered Aug 28 at 14:29
xhienne
2,854526
Starting from wolfram42's answer to that other puzzle, the maximum number of factorials is:
infinite since you can add any equal number of
!to the numerator and denominator of(6!) / (5 * 4! * 3!)
Well, since I'm no mathematician, I may say "infinite plus three" actually ;-)
Note: that's also true for any answer to the other puzzle that involved a division, like JonMark Perry's two solutions
answered Aug 28 at 14:29
xhienne
2,854526
edited Aug 28 at 14:38
answered Aug 28 at 14:29
xhienne
2,854526
answered Aug 28 at 14:29
xhienne
2,854526
answered Aug 28 at 14:29
xhienne
2,854526
2,854526
add a comment |Â
add a comment |Â
up vote
2
down vote
Here's the general programming solution to these kinds of problems. It also prints out for every result, how many solutions there are, which is useful in coming up with these kinds of problem:
from enum import IntEnum
from fractions import Fraction
from memoized import memoized
import math
NUMBERS = [6, 5, 4, 3]
RESULT = Fraction(1)
class Levels(IntEnum):
ADDITION = 0
MULTIPLICATION = 1
NEGATION = 2
POWER = 3
FACTORIAL = 4
BARE = 5
class Result:
def __init__(self, value, way, level):
"""
* value is the value achieved.
* way is string representation of an expression.
* level is the outermost operator applied to the expression, which
determines how this expression should be braketed when it's part of a
larger expression.
"""
assert isinstance(value, Fraction)
assert isinstance(level, Levels)
self.value = value
self.way = way
self.level = level
def bracketed(self, for_level):
if for_level > self.level:
return f"(self.way)"
return self.way
def combine(left_rs, right_rs):
for l in left_rs:
for r in right_rs:
if r.level not in (Levels.NEGATION, Levels.ADDITION):
# We block a - (-b), and a + (-b) since these can be written as
# a + b, and a - b. We also block a + (b + c), and a - (b + c)
# since these can be written as a + b + c, and a - b - c.
yield Result(l.value + r.value,
f"l.bracketed(Levels.ADDITION) + r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
yield Result(l.value - r.value,
f"l.bracketed(Levels.ADDITION) - r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
if r.level != Levels.MULTIPLICATION:
# We block a * (b * c) since this can be written as a * b * c.
# We block a * (b / c) since this can be written as a * b / c.
yield Result(l.value * r.value,
f"l.bracketed(Levels.MULTIPLICATION) * r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
if r.value != 0 and (r.level != Levels.MULTIPLICATION):
# We block a / (b / c) since this can be written as a / b * c.
# We block a / (b * c) since this can be written as a / b / c.
yield Result(l.value / r.value,
f"l.bracketed(Levels.MULTIPLICATION) / r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
new_value = None
power_okay = l.value >= 0 or r.value.denominator % 2 == 1
if l.value == 1:
new_value = Fraction(1)
elif l.value == -1:
if power_okay:
new_value = Fraction(-1
if r.value.numerator % 2 == 1
else 1)
elif l.value == 0:
if r.value > 0:
new_value = Fraction(0)
else:
# Only whole number powers are allowed for numbers other than
# -1, 0, 1.
if (power_okay
and r.value.denominator == 1
and -8 < r.value.numerator < 8):
new_value = l.value ** r.value
if new_value is not None:
# The left side is bracketed when it's a power because power is
# right associative.
yield Result(new_value,
f"l.bracketed(Levels.POWER + 1) ^ r.bracketed(Levels.POWER)",
Levels.POWER)
if (l.level == Levels.BARE
and r.level == Levels.BARE and len(r.way) == 1):
# We concatenate 123 as 1(23), but not as (12)3.
yield Result(l.value * 10 + r.value,
f"l.wayr.way",
Levels.BARE)
def negate(rs):
for r in rs:
yield r
if r.level > Levels.MULTIPLICATION:
# We block -(a + b) since this can be written -a - b.
# We block -(a - b) since this can be written -a + b.
# We block -(a * b) since this can be written -a * b.
# We block -(a / b) since this can be written -a / b.
yield Result(-r.value,
f"-r.bracketed(Levels.NEGATION)",
Levels.NEGATION)
def factorialize(rs):
for r in rs:
yield r
x = 0
while (r.value.denominator == 1
and (r.value.numerator == 0 or (3 <= r.value.numerator <= 20))):
# We block fractional and negative factorials.
# We block 1 and 2 factorial.
# We block gigantic factorials.
r = Result(Fraction(math.factorial(r.value.numerator)),
f"r.bracketed(Levels.FACTORIAL + 1)!",
Levels.FACTORIAL)
yield r
# memoization makes the recursive solution into a dynamic programming solution.
@memoized(hashable=False)
def do(digits):
"""
Given a list of digits, produce a list of Result objects.
"""
l = len(digits)
if l == 1:
# Return a Result object for a single digit.
new_rs = [Result(Fraction(digits[0]), str(digits[0]), Levels.BARE)]
else:
# The list of results for a list of digits is the total results for every contiguous partition.
new_rs =
for i in range(1, l):
new_rs.extend(list(combine(do(digits[:i]),
do(digits[i:]))))
return list(factorialize(negate(new_rs)))
rs = do(NUMBERS)
vs =
for r in rs:
vs.setdefault(r.value, ).append(r)
print(f"Solutions to NUMBERS = RESULT")
for r in sorted(vs[RESULT], key=lambda r: len(r.way)):
print(r.way)
print()
print(f" value solutions")
for v, l in sorted(vs.items(), key=lambda vl: (len(vl[1]), len(str(vl[0])))):
print(f"str(v):>30 len(l):<12")
prints
Solutions to [6, 5, 4, 3] = 1
65 - 4 ^ 3
(6 - 5) ^ 43
65 + (-4) ^ 3
-(-6 + 5) ^ 43
-6 - 5 + 4 * 3
6 + 5 - 4 - 3!
6 - 5 * (4 - 3)
6 - 5 / (4 - 3)
6 - 5 ^ (4 - 3)
6 / (5 + 4 - 3)
6 - 5! / 4 / 3!
(6 - 5) ^ 4 ^ 3
(6 - 5) ^ (-43)
...
(((-6 + 5 + 4)!)! - (3!)!)!
((6! / (5! / 4!)!)! - (3!)!)!
((-(-6) ^ (5 - 4))! - (3!)!)!
answered Aug 28 at 18:25
Neil G
19419
1
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
1
So what does this output?
– aschepler
Aug 30 at 2:08
@aschepler Added
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
add a comment |Â
up vote
2
down vote
Here's the general programming solution to these kinds of problems. It also prints out for every result, how many solutions there are, which is useful in coming up with these kinds of problem:
from enum import IntEnum
from fractions import Fraction
from memoized import memoized
import math
NUMBERS = [6, 5, 4, 3]
RESULT = Fraction(1)
class Levels(IntEnum):
ADDITION = 0
MULTIPLICATION = 1
NEGATION = 2
POWER = 3
FACTORIAL = 4
BARE = 5
class Result:
def __init__(self, value, way, level):
"""
* value is the value achieved.
* way is string representation of an expression.
* level is the outermost operator applied to the expression, which
determines how this expression should be braketed when it's part of a
larger expression.
"""
assert isinstance(value, Fraction)
assert isinstance(level, Levels)
self.value = value
self.way = way
self.level = level
def bracketed(self, for_level):
if for_level > self.level:
return f"(self.way)"
return self.way
def combine(left_rs, right_rs):
for l in left_rs:
for r in right_rs:
if r.level not in (Levels.NEGATION, Levels.ADDITION):
# We block a - (-b), and a + (-b) since these can be written as
# a + b, and a - b. We also block a + (b + c), and a - (b + c)
# since these can be written as a + b + c, and a - b - c.
yield Result(l.value + r.value,
f"l.bracketed(Levels.ADDITION) + r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
yield Result(l.value - r.value,
f"l.bracketed(Levels.ADDITION) - r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
if r.level != Levels.MULTIPLICATION:
# We block a * (b * c) since this can be written as a * b * c.
# We block a * (b / c) since this can be written as a * b / c.
yield Result(l.value * r.value,
f"l.bracketed(Levels.MULTIPLICATION) * r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
if r.value != 0 and (r.level != Levels.MULTIPLICATION):
# We block a / (b / c) since this can be written as a / b * c.
# We block a / (b * c) since this can be written as a / b / c.
yield Result(l.value / r.value,
f"l.bracketed(Levels.MULTIPLICATION) / r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
new_value = None
power_okay = l.value >= 0 or r.value.denominator % 2 == 1
if l.value == 1:
new_value = Fraction(1)
elif l.value == -1:
if power_okay:
new_value = Fraction(-1
if r.value.numerator % 2 == 1
else 1)
elif l.value == 0:
if r.value > 0:
new_value = Fraction(0)
else:
# Only whole number powers are allowed for numbers other than
# -1, 0, 1.
if (power_okay
and r.value.denominator == 1
and -8 < r.value.numerator < 8):
new_value = l.value ** r.value
if new_value is not None:
# The left side is bracketed when it's a power because power is
# right associative.
yield Result(new_value,
f"l.bracketed(Levels.POWER + 1) ^ r.bracketed(Levels.POWER)",
Levels.POWER)
if (l.level == Levels.BARE
and r.level == Levels.BARE and len(r.way) == 1):
# We concatenate 123 as 1(23), but not as (12)3.
yield Result(l.value * 10 + r.value,
f"l.wayr.way",
Levels.BARE)
def negate(rs):
for r in rs:
yield r
if r.level > Levels.MULTIPLICATION:
# We block -(a + b) since this can be written -a - b.
# We block -(a - b) since this can be written -a + b.
# We block -(a * b) since this can be written -a * b.
# We block -(a / b) since this can be written -a / b.
yield Result(-r.value,
f"-r.bracketed(Levels.NEGATION)",
Levels.NEGATION)
def factorialize(rs):
for r in rs:
yield r
x = 0
while (r.value.denominator == 1
and (r.value.numerator == 0 or (3 <= r.value.numerator <= 20))):
# We block fractional and negative factorials.
# We block 1 and 2 factorial.
# We block gigantic factorials.
r = Result(Fraction(math.factorial(r.value.numerator)),
f"r.bracketed(Levels.FACTORIAL + 1)!",
Levels.FACTORIAL)
yield r
# memoization makes the recursive solution into a dynamic programming solution.
@memoized(hashable=False)
def do(digits):
"""
Given a list of digits, produce a list of Result objects.
"""
l = len(digits)
if l == 1:
# Return a Result object for a single digit.
new_rs = [Result(Fraction(digits[0]), str(digits[0]), Levels.BARE)]
else:
# The list of results for a list of digits is the total results for every contiguous partition.
new_rs =
for i in range(1, l):
new_rs.extend(list(combine(do(digits[:i]),
do(digits[i:]))))
return list(factorialize(negate(new_rs)))
rs = do(NUMBERS)
vs =
for r in rs:
vs.setdefault(r.value, ).append(r)
print(f"Solutions to NUMBERS = RESULT")
for r in sorted(vs[RESULT], key=lambda r: len(r.way)):
print(r.way)
print()
print(f" value solutions")
for v, l in sorted(vs.items(), key=lambda vl: (len(vl[1]), len(str(vl[0])))):
print(f"str(v):>30 len(l):<12")
prints
Solutions to [6, 5, 4, 3] = 1
65 - 4 ^ 3
(6 - 5) ^ 43
65 + (-4) ^ 3
-(-6 + 5) ^ 43
-6 - 5 + 4 * 3
6 + 5 - 4 - 3!
6 - 5 * (4 - 3)
6 - 5 / (4 - 3)
6 - 5 ^ (4 - 3)
6 / (5 + 4 - 3)
6 - 5! / 4 / 3!
(6 - 5) ^ 4 ^ 3
(6 - 5) ^ (-43)
...
(((-6 + 5 + 4)!)! - (3!)!)!
((6! / (5! / 4!)!)! - (3!)!)!
((-(-6) ^ (5 - 4))! - (3!)!)!
answered Aug 28 at 18:25
Neil G
19419
1
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
1
So what does this output?
– aschepler
Aug 30 at 2:08
@aschepler Added
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Here's the general programming solution to these kinds of problems. It also prints out for every result, how many solutions there are, which is useful in coming up with these kinds of problem:
from enum import IntEnum
from fractions import Fraction
from memoized import memoized
import math
NUMBERS = [6, 5, 4, 3]
RESULT = Fraction(1)
class Levels(IntEnum):
ADDITION = 0
MULTIPLICATION = 1
NEGATION = 2
POWER = 3
FACTORIAL = 4
BARE = 5
class Result:
def __init__(self, value, way, level):
"""
* value is the value achieved.
* way is string representation of an expression.
* level is the outermost operator applied to the expression, which
determines how this expression should be braketed when it's part of a
larger expression.
"""
assert isinstance(value, Fraction)
assert isinstance(level, Levels)
self.value = value
self.way = way
self.level = level
def bracketed(self, for_level):
if for_level > self.level:
return f"(self.way)"
return self.way
def combine(left_rs, right_rs):
for l in left_rs:
for r in right_rs:
if r.level not in (Levels.NEGATION, Levels.ADDITION):
# We block a - (-b), and a + (-b) since these can be written as
# a + b, and a - b. We also block a + (b + c), and a - (b + c)
# since these can be written as a + b + c, and a - b - c.
yield Result(l.value + r.value,
f"l.bracketed(Levels.ADDITION) + r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
yield Result(l.value - r.value,
f"l.bracketed(Levels.ADDITION) - r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
if r.level != Levels.MULTIPLICATION:
# We block a * (b * c) since this can be written as a * b * c.
# We block a * (b / c) since this can be written as a * b / c.
yield Result(l.value * r.value,
f"l.bracketed(Levels.MULTIPLICATION) * r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
if r.value != 0 and (r.level != Levels.MULTIPLICATION):
# We block a / (b / c) since this can be written as a / b * c.
# We block a / (b * c) since this can be written as a / b / c.
yield Result(l.value / r.value,
f"l.bracketed(Levels.MULTIPLICATION) / r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
new_value = None
power_okay = l.value >= 0 or r.value.denominator % 2 == 1
if l.value == 1:
new_value = Fraction(1)
elif l.value == -1:
if power_okay:
new_value = Fraction(-1
if r.value.numerator % 2 == 1
else 1)
elif l.value == 0:
if r.value > 0:
new_value = Fraction(0)
else:
# Only whole number powers are allowed for numbers other than
# -1, 0, 1.
if (power_okay
and r.value.denominator == 1
and -8 < r.value.numerator < 8):
new_value = l.value ** r.value
if new_value is not None:
# The left side is bracketed when it's a power because power is
# right associative.
yield Result(new_value,
f"l.bracketed(Levels.POWER + 1) ^ r.bracketed(Levels.POWER)",
Levels.POWER)
if (l.level == Levels.BARE
and r.level == Levels.BARE and len(r.way) == 1):
# We concatenate 123 as 1(23), but not as (12)3.
yield Result(l.value * 10 + r.value,
f"l.wayr.way",
Levels.BARE)
def negate(rs):
for r in rs:
yield r
if r.level > Levels.MULTIPLICATION:
# We block -(a + b) since this can be written -a - b.
# We block -(a - b) since this can be written -a + b.
# We block -(a * b) since this can be written -a * b.
# We block -(a / b) since this can be written -a / b.
yield Result(-r.value,
f"-r.bracketed(Levels.NEGATION)",
Levels.NEGATION)
def factorialize(rs):
for r in rs:
yield r
x = 0
while (r.value.denominator == 1
and (r.value.numerator == 0 or (3 <= r.value.numerator <= 20))):
# We block fractional and negative factorials.
# We block 1 and 2 factorial.
# We block gigantic factorials.
r = Result(Fraction(math.factorial(r.value.numerator)),
f"r.bracketed(Levels.FACTORIAL + 1)!",
Levels.FACTORIAL)
yield r
# memoization makes the recursive solution into a dynamic programming solution.
@memoized(hashable=False)
def do(digits):
"""
Given a list of digits, produce a list of Result objects.
"""
l = len(digits)
if l == 1:
# Return a Result object for a single digit.
new_rs = [Result(Fraction(digits[0]), str(digits[0]), Levels.BARE)]
else:
# The list of results for a list of digits is the total results for every contiguous partition.
new_rs =
for i in range(1, l):
new_rs.extend(list(combine(do(digits[:i]),
do(digits[i:]))))
return list(factorialize(negate(new_rs)))
rs = do(NUMBERS)
vs =
for r in rs:
vs.setdefault(r.value, ).append(r)
print(f"Solutions to NUMBERS = RESULT")
for r in sorted(vs[RESULT], key=lambda r: len(r.way)):
print(r.way)
print()
print(f" value solutions")
for v, l in sorted(vs.items(), key=lambda vl: (len(vl[1]), len(str(vl[0])))):
print(f"str(v):>30 len(l):<12")
prints
Solutions to [6, 5, 4, 3] = 1
65 - 4 ^ 3
(6 - 5) ^ 43
65 + (-4) ^ 3
-(-6 + 5) ^ 43
-6 - 5 + 4 * 3
6 + 5 - 4 - 3!
6 - 5 * (4 - 3)
6 - 5 / (4 - 3)
6 - 5 ^ (4 - 3)
6 / (5 + 4 - 3)
6 - 5! / 4 / 3!
(6 - 5) ^ 4 ^ 3
(6 - 5) ^ (-43)
...
(((-6 + 5 + 4)!)! - (3!)!)!
((6! / (5! / 4!)!)! - (3!)!)!
((-(-6) ^ (5 - 4))! - (3!)!)!
answered Aug 28 at 18:25
Neil G
19419
Here's the general programming solution to these kinds of problems. It also prints out for every result, how many solutions there are, which is useful in coming up with these kinds of problem:
from enum import IntEnum
from fractions import Fraction
from memoized import memoized
import math
NUMBERS = [6, 5, 4, 3]
RESULT = Fraction(1)
class Levels(IntEnum):
ADDITION = 0
MULTIPLICATION = 1
NEGATION = 2
POWER = 3
FACTORIAL = 4
BARE = 5
class Result:
def __init__(self, value, way, level):
"""
* value is the value achieved.
* way is string representation of an expression.
* level is the outermost operator applied to the expression, which
determines how this expression should be braketed when it's part of a
larger expression.
"""
assert isinstance(value, Fraction)
assert isinstance(level, Levels)
self.value = value
self.way = way
self.level = level
def bracketed(self, for_level):
if for_level > self.level:
return f"(self.way)"
return self.way
def combine(left_rs, right_rs):
for l in left_rs:
for r in right_rs:
if r.level not in (Levels.NEGATION, Levels.ADDITION):
# We block a - (-b), and a + (-b) since these can be written as
# a + b, and a - b. We also block a + (b + c), and a - (b + c)
# since these can be written as a + b + c, and a - b - c.
yield Result(l.value + r.value,
f"l.bracketed(Levels.ADDITION) + r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
yield Result(l.value - r.value,
f"l.bracketed(Levels.ADDITION) - r.bracketed(Levels.ADDITION + 1)",
Levels.ADDITION)
if r.level != Levels.MULTIPLICATION:
# We block a * (b * c) since this can be written as a * b * c.
# We block a * (b / c) since this can be written as a * b / c.
yield Result(l.value * r.value,
f"l.bracketed(Levels.MULTIPLICATION) * r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
if r.value != 0 and (r.level != Levels.MULTIPLICATION):
# We block a / (b / c) since this can be written as a / b * c.
# We block a / (b * c) since this can be written as a / b / c.
yield Result(l.value / r.value,
f"l.bracketed(Levels.MULTIPLICATION) / r.bracketed(Levels.MULTIPLICATION + 1)",
Levels.MULTIPLICATION)
new_value = None
power_okay = l.value >= 0 or r.value.denominator % 2 == 1
if l.value == 1:
new_value = Fraction(1)
elif l.value == -1:
if power_okay:
new_value = Fraction(-1
if r.value.numerator % 2 == 1
else 1)
elif l.value == 0:
if r.value > 0:
new_value = Fraction(0)
else:
# Only whole number powers are allowed for numbers other than
# -1, 0, 1.
if (power_okay
and r.value.denominator == 1
and -8 < r.value.numerator < 8):
new_value = l.value ** r.value
if new_value is not None:
# The left side is bracketed when it's a power because power is
# right associative.
yield Result(new_value,
f"l.bracketed(Levels.POWER + 1) ^ r.bracketed(Levels.POWER)",
Levels.POWER)
if (l.level == Levels.BARE
and r.level == Levels.BARE and len(r.way) == 1):
# We concatenate 123 as 1(23), but not as (12)3.
yield Result(l.value * 10 + r.value,
f"l.wayr.way",
Levels.BARE)
def negate(rs):
for r in rs:
yield r
if r.level > Levels.MULTIPLICATION:
# We block -(a + b) since this can be written -a - b.
# We block -(a - b) since this can be written -a + b.
# We block -(a * b) since this can be written -a * b.
# We block -(a / b) since this can be written -a / b.
yield Result(-r.value,
f"-r.bracketed(Levels.NEGATION)",
Levels.NEGATION)
def factorialize(rs):
for r in rs:
yield r
x = 0
while (r.value.denominator == 1
and (r.value.numerator == 0 or (3 <= r.value.numerator <= 20))):
# We block fractional and negative factorials.
# We block 1 and 2 factorial.
# We block gigantic factorials.
r = Result(Fraction(math.factorial(r.value.numerator)),
f"r.bracketed(Levels.FACTORIAL + 1)!",
Levels.FACTORIAL)
yield r
# memoization makes the recursive solution into a dynamic programming solution.
@memoized(hashable=False)
def do(digits):
"""
Given a list of digits, produce a list of Result objects.
"""
l = len(digits)
if l == 1:
# Return a Result object for a single digit.
new_rs = [Result(Fraction(digits[0]), str(digits[0]), Levels.BARE)]
else:
# The list of results for a list of digits is the total results for every contiguous partition.
new_rs =
for i in range(1, l):
new_rs.extend(list(combine(do(digits[:i]),
do(digits[i:]))))
return list(factorialize(negate(new_rs)))
rs = do(NUMBERS)
vs =
for r in rs:
vs.setdefault(r.value, ).append(r)
print(f"Solutions to NUMBERS = RESULT")
for r in sorted(vs[RESULT], key=lambda r: len(r.way)):
print(r.way)
print()
print(f" value solutions")
for v, l in sorted(vs.items(), key=lambda vl: (len(vl[1]), len(str(vl[0])))):
print(f"str(v):>30 len(l):<12")
prints
Solutions to [6, 5, 4, 3] = 1
65 - 4 ^ 3
(6 - 5) ^ 43
65 + (-4) ^ 3
-(-6 + 5) ^ 43
-6 - 5 + 4 * 3
6 + 5 - 4 - 3!
6 - 5 * (4 - 3)
6 - 5 / (4 - 3)
6 - 5 ^ (4 - 3)
6 / (5 + 4 - 3)
6 - 5! / 4 / 3!
(6 - 5) ^ 4 ^ 3
(6 - 5) ^ (-43)
...
(((-6 + 5 + 4)!)! - (3!)!)!
((6! / (5! / 4!)!)! - (3!)!)!
((-(-6) ^ (5 - 4))! - (3!)!)!
answered Aug 28 at 18:25
Neil G
19419
edited Aug 30 at 6:51
answered Aug 28 at 18:25
Neil G
19419
answered Aug 28 at 18:25
Neil G
19419
answered Aug 28 at 18:25
Neil G
19419
19419
1
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
1
So what does this output?
– aschepler
Aug 30 at 2:08
@aschepler Added
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
add a comment |Â
1
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
1
So what does this output?
– aschepler
Aug 30 at 2:08
@aschepler Added
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
1
1
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
This code just bounced off my head. It would be great if will add some explanations in comments.
– haccks
Aug 29 at 9:33
1
1
So what does this output?
– aschepler
Aug 30 at 2:08
So what does this output?
– aschepler
Aug 30 at 2:08
@aschepler Added
– Neil G
Aug 30 at 5:09
@aschepler Added
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
@haccks I added some comments.
– Neil G
Aug 30 at 5:09
add a comment |Â
up vote
1
down vote
9 factorials, following the rules (answer below)
Some tricks to use
First trick:
$0!=1$ (by definition). We will use the fact that $forall xinmathbbR:0cdot x=0 Rightarrow (0cdot x)!=1$.
Second trick:
Since we cannot use $1!$ or $2!$ and we cannot use $n!$ for $n>20$, we have to look the other way: think smaller. Lucky for us, we have $x!=Gamma(x+1)$ where $Gamma$ is the gamma function and this holds for $xinmathbbR$.
Building our answer
First step (using the first trick):
We need zero. That's pretty easy, we have $6!-(3!)!=6!-6!=0$.
Second step (using the second trick):
We need many factorials but we can only use $4$ and $5$ because we used the others in the first step. To get those, we aren't going to worry about the outcome because we will multiply by $0$ anyway. We only have to be careful not to get to close to $0$ or $1$ (we shall stay at least $0.01$ distance away from them) and not to exceed $20$.
Calculations in the second step:
The first factorial $left(frac-54right)!approx -4.9$.
The second factorial $left(left(frac-54right)!right)!approx 0.50$.
The third factorial $left(left(left(frac-54right)!right)!right)!approx 0.89$.
The fourth factorial $left(left(left(left(frac-54right)!right)!right)!right)!approx 0.96$.
The fifth factorial $left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$.
The sixth factorial $left(left(left(left(left(left(frac-54right)!right)!right)!right)!right)!right)!approx 0.993$.
Therefore, we can use 5 factorials here, because with the sixth step we get to close to $1$.
Answer
$$x=6!-(3!)!=0$$
$$y=left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$$
$$textitfinal answer=(xcdot y)!=0!=1$$
answered Aug 29 at 15:08
JJJ
1505
add a comment |Â
up vote
1
down vote
9 factorials, following the rules (answer below)
Some tricks to use
First trick:
$0!=1$ (by definition). We will use the fact that $forall xinmathbbR:0cdot x=0 Rightarrow (0cdot x)!=1$.
Second trick:
Since we cannot use $1!$ or $2!$ and we cannot use $n!$ for $n>20$, we have to look the other way: think smaller. Lucky for us, we have $x!=Gamma(x+1)$ where $Gamma$ is the gamma function and this holds for $xinmathbbR$.
Building our answer
First step (using the first trick):
We need zero. That's pretty easy, we have $6!-(3!)!=6!-6!=0$.
Second step (using the second trick):
We need many factorials but we can only use $4$ and $5$ because we used the others in the first step. To get those, we aren't going to worry about the outcome because we will multiply by $0$ anyway. We only have to be careful not to get to close to $0$ or $1$ (we shall stay at least $0.01$ distance away from them) and not to exceed $20$.
Calculations in the second step:
The first factorial $left(frac-54right)!approx -4.9$.
The second factorial $left(left(frac-54right)!right)!approx 0.50$.
The third factorial $left(left(left(frac-54right)!right)!right)!approx 0.89$.
The fourth factorial $left(left(left(left(frac-54right)!right)!right)!right)!approx 0.96$.
The fifth factorial $left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$.
The sixth factorial $left(left(left(left(left(left(frac-54right)!right)!right)!right)!right)!right)!approx 0.993$.
Therefore, we can use 5 factorials here, because with the sixth step we get to close to $1$.
Answer
$$x=6!-(3!)!=0$$
$$y=left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$$
$$textitfinal answer=(xcdot y)!=0!=1$$
answered Aug 29 at 15:08
JJJ
1505
add a comment |Â
up vote
1
down vote
up vote
1
down vote
9 factorials, following the rules (answer below)
Some tricks to use
First trick:
$0!=1$ (by definition). We will use the fact that $forall xinmathbbR:0cdot x=0 Rightarrow (0cdot x)!=1$.
Second trick:
Since we cannot use $1!$ or $2!$ and we cannot use $n!$ for $n>20$, we have to look the other way: think smaller. Lucky for us, we have $x!=Gamma(x+1)$ where $Gamma$ is the gamma function and this holds for $xinmathbbR$.
Building our answer
First step (using the first trick):
We need zero. That's pretty easy, we have $6!-(3!)!=6!-6!=0$.
Second step (using the second trick):
We need many factorials but we can only use $4$ and $5$ because we used the others in the first step. To get those, we aren't going to worry about the outcome because we will multiply by $0$ anyway. We only have to be careful not to get to close to $0$ or $1$ (we shall stay at least $0.01$ distance away from them) and not to exceed $20$.
Calculations in the second step:
The first factorial $left(frac-54right)!approx -4.9$.
The second factorial $left(left(frac-54right)!right)!approx 0.50$.
The third factorial $left(left(left(frac-54right)!right)!right)!approx 0.89$.
The fourth factorial $left(left(left(left(frac-54right)!right)!right)!right)!approx 0.96$.
The fifth factorial $left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$.
The sixth factorial $left(left(left(left(left(left(frac-54right)!right)!right)!right)!right)!right)!approx 0.993$.
Therefore, we can use 5 factorials here, because with the sixth step we get to close to $1$.
Answer
$$x=6!-(3!)!=0$$
$$y=left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$$
$$textitfinal answer=(xcdot y)!=0!=1$$
answered Aug 29 at 15:08
JJJ
1505
9 factorials, following the rules (answer below)
Some tricks to use
First trick:
$0!=1$ (by definition). We will use the fact that $forall xinmathbbR:0cdot x=0 Rightarrow (0cdot x)!=1$.
Second trick:
Since we cannot use $1!$ or $2!$ and we cannot use $n!$ for $n>20$, we have to look the other way: think smaller. Lucky for us, we have $x!=Gamma(x+1)$ where $Gamma$ is the gamma function and this holds for $xinmathbbR$.
Building our answer
First step (using the first trick):
We need zero. That's pretty easy, we have $6!-(3!)!=6!-6!=0$.
Second step (using the second trick):
We need many factorials but we can only use $4$ and $5$ because we used the others in the first step. To get those, we aren't going to worry about the outcome because we will multiply by $0$ anyway. We only have to be careful not to get to close to $0$ or $1$ (we shall stay at least $0.01$ distance away from them) and not to exceed $20$.
Calculations in the second step:
The first factorial $left(frac-54right)!approx -4.9$.
The second factorial $left(left(frac-54right)!right)!approx 0.50$.
The third factorial $left(left(left(frac-54right)!right)!right)!approx 0.89$.
The fourth factorial $left(left(left(left(frac-54right)!right)!right)!right)!approx 0.96$.
The fifth factorial $left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$.
The sixth factorial $left(left(left(left(left(left(frac-54right)!right)!right)!right)!right)!right)!approx 0.993$.
Therefore, we can use 5 factorials here, because with the sixth step we get to close to $1$.
Answer
$$x=6!-(3!)!=0$$
$$y=left(left(left(left(left(frac-54right)!right)!right)!right)!right)!approx 0.983$$
$$textitfinal answer=(xcdot y)!=0!=1$$
answered Aug 29 at 15:08
JJJ
1505
answered Aug 29 at 15:08
JJJ
1505
answered Aug 29 at 15:08
JJJ
1505
answered Aug 29 at 15:08
JJJ
1505
1505
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fpuzzling.stackexchange.com%2fquestions%2f70838%2fmaximize-the-number-of-factorials-in-your-solution-to-6-5-4-3-1%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
I'm guessing we shouldn't take the factorial of 2 either
– Bennett Bernardoni
Aug 28 at 14:11
@BennettBernardoni Thanks, yes.
– Neil G
Aug 28 at 14:11
Is the answer really different than wolfram42's answer to the other puzzle? (3 factorials)
– xhienne
Aug 28 at 14:17
@xhienne That was my opinion but if he found 8 that is probably precise enough for another question.
– Saeïdryl
Aug 28 at 14:18
1
@Saeïdryl I understand that you can not use (6 - 5)! nor (6 - 4)!
– xhienne
Aug 28 at 14:21