5 ms·

Put it in the B configuration. There are 12 boys, which can be thought of as 24 halves bundled in pairs: a half on the outside of the circle, and a half on the

Put it in the B configuration. There are 12 boys, which can be thought of as 24 halves bundled in pairs: a half on the outside of the circle, and a half on the inside of the circle. (Conveniently, also, each bundled pair of halves includes one half with a flag and one half without a flag). Some of these halves are more substantial looking than others, mind you. Rotate it to the A configuration: There are still 24 halves, bundled in pairs. But now we count it as 13 boys instead of 12. Why? Because in the bottom left, two of the halves that got paired together are both flag-and-head halves, so even though that was just two halves in our original count, it feels like the two of them bundled together should count as two boys instead of just one now. (Correspondingly, in the top right, two of the halves that get bundled together in the A configuration are both non-flag halves. But the result still seems sufficient to call one full boy instead of zero boys.)

thehappypm 5y ago
Which boy is gone? :)
- lowbloodsugar 5y ago
  My guess: The problem is one of discrete values. You are asking about "a boy", but there are no "a boy"s. There are only fractions of boys. The total sum of "boy fractions" is the same, but the number of pairs of "boy fractions" that meet the distinction of "a boy" changes. "A boy" does not vanish because there were never "a boy"s in the first place. One might have better luck with "boy heads", and you can see that at roughly 5 oclock there is a "boy head" that turns into a "boy arm". So part of the trick is that some of the "boy fractions" change in your interpretation. Without the trick, you would reasonably say, "Wait, there is still a boy head there! So there are still 13 boys!"
  - pis0m0jad0 5y ago
    I also couldn't help but notice this after reading your comment, but in the image it seems pretty clearly that the A configuration is pretty incoherent. The 5 o'clock boy you mentioned, for example, clearly has a sleeve for the right side of his face, the boy at 2 o'clock is also missing a large chunk of his head, and of course the two boys at 8 o'clock overlap, but, now that I look again, in an incoherent way. To me it seems that the answer to where the boy goes is simply that "the A configuration is invalid, but slightly so that we might not notice"
    - crdrost 5y ago
      Something like that, yeah. If done right, the idea is that in the one configuration you have thirteen 24/25ths of a boy, so thirteen 96%-boys. You hide this as each one being slightly skinny. In the other configuration you have twelve 26/25ths of a boy, so twelve 104%-boys. You hide this as each one being slightly fat. The danger is that if you make the drawing too detailed and gorgeous, someone will be able to look at a little detail like eyes or so, if one of these 4% slices contains an eye then you may end up with boys with 1 or 3 eyes, something that any looker would say spoils the illusion. You have a couple options there. - People will accept a "dead zone" where the two moving parts interact, you can try to locate an eye inside the "dead zone" so that you don't trigger this W-T-F moment. - Use cartoonishness/ambiguity. So maybe this dot means "eye" on this cartoon but "freckle" on that cartoon, similarly by locating the exact eye on the border between the two, maybe half a line goes from being "long eyelashes" to being "the middle of a winking eye" or so.
- quartz 5y ago
  The inside boy at 8 o'clock. In position A he's there, in position B he's gone. Think of each position as a discrete state rather than thinking of it as "moving boys".
  - larrydag 5y ago
    Also the 2 o'clock boy has no flag in A then he has a flag in B.
- Igelau 5y ago
  Jack, in the red shirt.
  - kazinator 5y ago
    Waldo, in the striped sweater.
- mannykannot 5y ago
  It is clearer if you think of it as partial pictures of 24 boys, 12 on the inside and 12 on the outside. In either position, each picture on the inside lines up with one on the outside, and vice-versa (each position is a bijection.) In position B, each of the pairs form a picture of one boy, leading us to count 12. The same is true for position A, except for the one case where each part-picture is well more than 50% (and, in particular, there are two complete heads.) Consequently, this cannot be seen as a picture of one boy, and we count it as two. In this view, the answer to your question is 'one of each pair, except in one case in position A only.'

thehappypm 5y ago

Which boy is gone? :)

lowbloodsugar 5y ago
My guess: The problem is one of discrete values. You are asking about "a boy", but there are no "a boy"s. There are only fractions of boys. The total sum of "boy fractions" is the same, but the number of pairs of "boy fractions" that meet the distinction of "a boy" changes. "A boy" does not vanish because there were never "a boy"s in the first place. One might have better luck with "boy heads", and you can see that at roughly 5 oclock there is a "boy head" that turns into a "boy arm". So part of the trick is that some of the "boy fractions" change in your interpretation. Without the trick, you would reasonably say, "Wait, there is still a boy head there! So there are still 13 boys!"
- pis0m0jad0 5y ago
  I also couldn't help but notice this after reading your comment, but in the image it seems pretty clearly that the A configuration is pretty incoherent. The 5 o'clock boy you mentioned, for example, clearly has a sleeve for the right side of his face, the boy at 2 o'clock is also missing a large chunk of his head, and of course the two boys at 8 o'clock overlap, but, now that I look again, in an incoherent way. To me it seems that the answer to where the boy goes is simply that "the A configuration is invalid, but slightly so that we might not notice"
  - crdrost 5y ago
    Something like that, yeah. If done right, the idea is that in the one configuration you have thirteen 24/25ths of a boy, so thirteen 96%-boys. You hide this as each one being slightly skinny. In the other configuration you have twelve 26/25ths of a boy, so twelve 104%-boys. You hide this as each one being slightly fat. The danger is that if you make the drawing too detailed and gorgeous, someone will be able to look at a little detail like eyes or so, if one of these 4% slices contains an eye then you may end up with boys with 1 or 3 eyes, something that any looker would say spoils the illusion. You have a couple options there. - People will accept a "dead zone" where the two moving parts interact, you can try to locate an eye inside the "dead zone" so that you don't trigger this W-T-F moment. - Use cartoonishness/ambiguity. So maybe this dot means "eye" on this cartoon but "freckle" on that cartoon, similarly by locating the exact eye on the border between the two, maybe half a line goes from being "long eyelashes" to being "the middle of a winking eye" or so.
quartz 5y ago
The inside boy at 8 o'clock. In position A he's there, in position B he's gone. Think of each position as a discrete state rather than thinking of it as "moving boys".
- larrydag 5y ago
  Also the 2 o'clock boy has no flag in A then he has a flag in B.
Igelau 5y ago
Jack, in the red shirt.
- kazinator 5y ago
  Waldo, in the striped sweater.
mannykannot 5y ago
It is clearer if you think of it as partial pictures of 24 boys, 12 on the inside and 12 on the outside. In either position, each picture on the inside lines up with one on the outside, and vice-versa (each position is a bijection.) In position B, each of the pairs form a picture of one boy, leading us to count 12. The same is true for position A, except for the one case where each part-picture is well more than 50% (and, in particular, there are two complete heads.) Consequently, this cannot be seen as a picture of one boy, and we count it as two. In this view, the answer to your question is 'one of each pair, except in one case in position A only.'