Think of a Number. How Do Math Magicians Know What It Is?
quantamagazine.org
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Two numbers are chosen randomly, both are positive integers smaller than 100. Sandy is told the sum of the numbers, while Peter is told the product of the numbers.
Then, this dialog occurs between Sandy and Peter:
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I don't know the numbers.
Sandy: I don't know the numbers.
Peter: I do know the numbers.
What are the numbers?
Source: https://www.reddit.com/r/math/comments/32opae/next_level_che...
> Each sentence is extra data given to the other person.
> When Peter says "I don't know the numbers", means that he doesn't have enough information. For example, if the product of the numbers is 10, it could be (1,10) or (2,5). But if the product is 9801, then Peter would know the answer (99,99). Therefore, his first sentence reveals to Sandy that (99,99) isn't a possible answer. But even this extra data isn't enough for Sandy to know the answer, and she says so. Again, this is extra data for Peter, but , again, is not enough. This go back and forth until suddenly Peter gains enough information to find the answer.
Sandy knows the sum and has now been told that Peter doesn't know the pair. If the sum had been 6, the following pairs are possible: (1+5) (3+3) (4+2), Peter has just ruled out (1x5) and (3x3), so Sandy would be able to narrow it down to (1,5) if the sum had been 6.
So when she tells Peter she can't narrow it down, it tells him that the pair isn't (4,2) either (among many others).
And if Peter's number were 8: (1x8) or (2x4) he'd be able to solve it, but he doesn't so Sandy then knows that (1,8) isn't the solution either.
I think pair (7x11)=77 can't be rule out, because pair (1x77) is also equal 77.
still don't got it...
Can you explain the situation for 3 turns before Peter knows, Sincere thanks.
Let's think about picking two numbers between 1-9.
Peter is given the product 24. He knows there are two possible pairs of numbers between 1-9 which produce a product of 24, (3,8) and (4,6), so he says "I don't know the numbers"
Sandy is given the sum 10. There are many pairs of numbers that produce a sum of 10, [(1,9), (2,8)...]. But she also knows that Peter did not immediately know the answer. If the pair of numbers was (5,5), that would have produced a product of 25. If Peter was given a product of 25, he would have immediately known the answer, since there's only 1 pair of numbers that produces that product.
So Sandy knows the answer isn't (5,5). Similarly, she knows it's not (2,8) or (3,7). The answer could be (1,9) though, since the product of (1,9) is 9, and there's another pair that can produce that product (3,3). If Peter was given the product 9 he wouldn't have immediately known the answer. The answer could also be (4,6), since the product of those is 24, and that can also be achieved with the pair (3,8). So there's only 2 pairs of numbers that add up to 10, and which Peter would not have immediately known based on their product. Sandy knows the answer must be either (1,9) or (4,6). Sandy says "I don't know the numbers".
Peter knows the solution must be either (3,8) or (4,6), and he knows that Sandy did not immediately know the answer. If Sandy had been given the sum 11 though, she should have immediately known the answer. There is only 1 pair of numbers that produces 11, but which does not have a unique product. Yes, the pair (2,9) sums to 11, but the product is unique, and if Peter had been given the product 18 to begin with, he would have immediately known the answer. So because he didn't immediately know the answer, and because that was not enough information for Sandy to say that the pair is (3,8), then Peter knows that the summation of the numbers must not be 11. The only other choice then is (4,6), and so Peter says "I do know the numbers".
Why is not (2,8)? both pair (2, 8) and pair (4, 4) can produce a product of 16.
Thanks anyway!
> Yes, the pair (2,9) sums to 11, but the product is unique, and if Peter had been given the product 18 to begin with, he would have immediately known the answer.
pair (2,9) and pair (3,6) can produce a product of 18, so Peter can't immediately know the answer if he had been given the product 18 to begin with, so the problem can't beed solved.
I think this problem would Caught in an unresolved cycle after unique product and unique sum has been ruled out.
I don't know where i am wrong, Sincere thanks...
That should actually be "if the product of their respective smallest prime factors is over 100". 7x11 can't be ruled out since 1x77 also produces 77, whereas 49x17 and Nx53 can be ruled out.
You can also rule out some of the larger squares, e.g. 25x25 and 64x64, so there's probably still a better phrasing for that
Edit: Can also rule out 1xPrime
I think there is an assumption that both parties are ordering their possible choices in an identical manner, but I am unsure.
After understanding that idea, the rest is just tedious logic/brute-force-search, I believe.
It's also possible that there is ambiguity in the statement or something like that. Hard to say exactly what part isn't connecting with you.
I don't have any guarantees that both of them are sorting each list the same. So how does "I don't have enough information" relay which of those list items to eliminate?
When the person knowing the product says he doesn't know the answer, the sum person now checks the product list for all products with only 1 pair. These pairs can now be removed from the sum and product list. That actually culls the sum list and some entries that previously had 2 or more options now have 1 less. If the entry for the known sum has only 1 pair, he knows the answer. Otherwise he doesn't and now the culling continues with all sums that have only 1 pair. Repeat until you have the solution.
No ordering required. The algorithm can terminate at different rounds depending on the picked number and for some N (like 4) might not have a solution at all! For the given riddle the exact turn the solution was found is given through the conversation.
This is a pretty neat explanation https://alexanderell.is/posts/numbers-game/
So if I told you the product of two numbers (positive integers) was 3, you immediately know 3 is prime and therefore the numbers are 1,3. Therefore either they cannot say they don't know if they were told 3. We can go through every combination and find what possibilities they could be.
Well, it cannot be 1,1 or 1,2 or 1,3 because those produce products of 1, 2 and 3 because they don't have any other pair that can generate them. The same is true of 2,3 which is the only way to get six or 2,4 the only way to get 8. And 3,3 or 3,4 which are the only way to get 9 and 12. And lastly 4,4 is the only way to get to 16. You'll note that this is almost all pairs which is easy to iterate over by making the second number greater than or equal to the first. The only pairs left are 1,4 and 2,2 both of which produce a product of 4. So we, as the sun person, now know the product must be four. Given the product and the sum, we can deduce that it's 1,4 because we were told the sun was 5.
When it goes to 100, it's the same process. Except now I have to iteratively eliminate pairs until there is only one solution.
Before Peter told Sandy he didn't know, only 4 sums could have been caused by a unique pair (198, 3, 2, and 197). Peter telling Sandy he doesn't know lets her rule out lots of pairs, and after doing so, there are 9 sums that would have a single pair remaining that hadn't been ruled out. For example, were the sum 165, Sandy could have concluded that the only possible pairing would be 69 and 96, since the other pairs that add up to that number (e.g., 74 and 91, 80 and 85, etc.) would have unique products that Peter would have known about. That she doesn't know the answer yet therefore tells Peter that 69 and 96 is not a possible pair. Were the product 6624, Peter would now know that the only possible remaining pair was 72 and 92, and he would know the answer. But since he didn't know the answer, now Sandy knows that it can't have been 72 and 92 either.
This crossing out continues until Peter realizes that 70 and 96 was not a viable pair, which lets him realize that the only other way to get 6720 was to have the numbers be 80 and 84, and he declares he knew the answer. [Assuming I got the correct number of rounds]
Round #1: ... (many)
Round #2: ... (many)
Round #3: (1,4), (72,92) and (72,98)
Round #4: (2,3), (80,90)
Round #5: (1,6), (75,96)
Round #6: (72,99)
Round #7: (81,88)
Round #8: (70,99)
Round #9: (77,90)
Round #10: (72,95)
Round #11: (76,90)
Round #12: (70,96)
Round #13: (90,84)
Round #14: (66,98)
Round #15: Solution is "77" and "84"
another variant:
Peter: If I divide by 4, I have remainder x Sandy: If I divide by 4, I have remainder y Peter: If I divide by 25, I have remainder a Sandy: If I divide by 25, I have remainder b Peter and Sandy (in unison): Got it!
Here is some python code that might be more revealing https://www.online-python.com/c5nAfLoIqr
A code review would be greatly appreciated!
I mean, obviously the extremes get eliminated right away, and then the primes, but then I have to track 10,000 number combinations.
You can actually WLOG away all pairs where the second element is larger than the first.
Also, the most common kind of pair I eliminated was actually due to the size constraint (ie, 98 * 99) or pairs of (1, prime) which I didn't actually realize would be a thing until I coded it up.
x + y = s
x * y = p
x = s - y
x = p/y
substitute for x
p/y = s - y
p = sy - y*2
y*2 - sy + p = 0
# use the quadratic formula to solve for y
y = (-b ± √(b²-4ac)) / (2a)
In python:
import numpy as np
x = np.random.randint(1,100)
y = np.random.randint(1,100)
s = x+y
p = xy
a = 1
b = -1s
c = p
y_solved = (-1b + (b*2 - 4ac)*.5)/(2a),(-1b - (b*2 - 4ac)*.5)/(2a)
print(x,y,y_solved)
[0] https://web.archive.org/web/20121228075545/http://devblog.bu...
I started hacking on this protocol using a pencil and paper variation of Diffie-Hellman where instead of a secret key agreement you arrive at the same salary number (as a function of the difference in your secret numbers), but realized you can do a one-sided version of this where the potential employee says to the recruiter "think of your max budget for this role" using a variation the second puzzle in article, though unfortunately the recruiter would likely feel tricked, so I've gone back to working on a DH-like protocol ahead of time because it would be fairer.
Of course I would have liked to have made a blog post out of it, but this article is so close and mine is still half-baked, so if someone beats me to it and wants to leverage the power of being wrong on the internet in this comment thread, we could create a fair protocol that improves the lives of all parties involved in the hiring process.
https://en.wikipedia.org/wiki/Yao%27s_Millionaires%27_proble...
It'd be interesting to see if any of the tricks worked for, say, -7π + ei.
Such a loop of operations containing only multiplication, addition, and subtraction would work on activity complex numbers. Square roots would not.
Spoilers:
First note that, below 16, 11 is the only sum which explains S's first statement: all possible pairs of numbers adding to 11 have non-unique products, accounting for S knowing P would not know the numbers.
Then note that for 18 (9x2), 24 (8x3), and 28 (7x4), all other factor pairs for that product add to a non-11 number below 16. Those non-11 sums are ruled out by S's first statement, so P will know the sum is 11 by her second statement.
Therefore (9,2), (8,3), and (7,4) all look like valid solutions, and it seems likely there are more.
What am I missing?
https://web.archive.org/web/20121228075545/http://devblog.bu...
Any positive integer you take, you end up in a 1-4-2-1 loop. It's not proved yet but there's no number found yet that satisfies otherwise.
Very interesting. What's the use case of this?
Impressing ladies at the bar with your 'deep connection'.