A day in the park with two daughters. Problem solved - with a little help from my friends.

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While Eddie was away at summer camp I took the two girls to Victory Park on a Sunday. Zoriana had the camera half the time. We bicycled there with Marianna in the child seat, arriving at the boat rental right at 10:00 when they opened up. I rowed while Zoriana took these pictures. First is the fisherman, who patiently wave us away from their floats and tackle as we go by.

Marianna is content to be a passenger for the time being, but she is itching to participate in rowing when she gets bigger.

Victory Park is in the middle of the city. Real estate developers recognized an ideal location for a high-rise apartments – right next to a park where you could take the kids. The recently built apartments, shops and restaurants are all upscale.

We rented the boat for an hour. That included a timeout for ice cream.

Frogs and turtles are part of the attraction. The turtles are used to people and hard to startle. Zoriana was able to snap a series of pictures of this guy before he finally slipped away into the water. There was also a beautiful grey heron whose picture we didn’t get.

I have written before about how beautifully maintained the playground equipment is, and the respect with which it is treated. Here you see Zoriana making friends with every little girl she meets. Marianna is exploring all of the playthings. I tried to capture as well the amount of traffic this place gets.

No matter how hard I tried to keep track of the kids, they are often out of sight for a four and five minutes at a time. But since every parent there faces the same problem and nobody else is worried, neither am I.

Here is a second series of photographs from the playground. Kids crawling all over the place, on top of one another, flirting with one another and so on.

This has been an easy blog to write.

Subscribers Notch Johnson and Camilla Dietrich figured out the solutions to Eddie’s math problem. For my own understanding I summarize what they determined in writing.

On six externally identical weights, the masses of which are natural numbers from 1 to 6 g, Petryk pasted plates with the inscriptions "1 g", "2 g", ..., "6 g". How can Andriyuk, a high school student, determine whether the plates are correctly pasted after two weighings on the shawl scales? (balance scales)

This problem statement confirms that
(1)   The masses are correct – 1 thru 6
(2)   The labels are correct – 1 thru 6
Therefore if any labels are incorrectly pasted there must be at least two that are mislabeled.  If the label for 1 is on the 2g weight is, then the label for 2 cannot be on the 2g weight, so a second ball must also be mislabeled.

There are three ways to use five balls that should balance.  Leaving out odd weights one at a time: 1, 3 and 5.

1.     4&6 vs 2,3&5     10g per side
2.     5&4 vs 1,2 & 6.   9g per side
3.     6&2 vs 1,3&4   8g per side.

Use two of these weighings.  Note that no two of them put the same two balls on the same side, such that the labels could be reversed.  Viz: if 2 and 3 were swapped in #1, they would be found out by either #2 or #3 above. 

Therefore, if any two of these weighings balance, the labels are correct.

Further rumination indicates that this is probably all there is to it. There are 6! = 720 permutations as to how the weights could be mislabeled. 

My intuition tells me that leaving only one out of each of two weighings is the key. You could have a wild permutation, such as labeling them all in reverse (6=1, 5=2, etc) that would work for one weighing, but that it would be found out in the second weighing.  However, I don't know how to prove it.

Thanks to you all for your work on this. Thanks also for your many comments on Chat GPT and other artificial intelligence. It makes me confident that I am not missing any significant tricks.

That’s the news from Lake WeBeGone, where the lazy summer days, which seem to have just started, are halfway over. Alas.

Graham