Disappearing Bicyclist – Sam Loyd (1906) | Hacker News Reader

Disappearing Bicyclist – Sam Loyd (1906)

geogebra.org

98 pointsalizauf64 comments

You can isolate the discrepancy simply by considering just the bottom left quadrant.

When you flip configurations, an extra boy is shifted into the sector, forming the boy pair, so you can now count three boys in that sector instead of two. What is shifted out is just a fraction of a leg, so there is a net gain of one boy.

The remainder of the circle is constructed so that there appears is no net change in the number of boys there. The fraction of a leg which is shifted in replaces a fraction of a leg, and the boy which is shifted out is replaced by a boy.

But note that this is a paradox: you can't shift a leg into one end of a register, such that a person is shifted out of the other end, and yet have the person count in that register remain the same! That's like shifting a 0 into a bit register, such that 1 is shifted out the other end, yet the parity remains the same.

The subterfuge is that in configuration B, there is a Siamese twin body with two heads (A + 2 clockwise). The casual observer glosses over this, because the heads are fused together quite well, and counts them as one boy.

When the configuration is switched to A, this twin head is separated. There are no more Siamese heads sharing a body, and that separation is what keeps the apparent head count the same, even though a head is shifted out in compensation for a leg shifted in.

Another noteworthy feature is that when the boy pair is formed, it is also Siamese twins, sharing one leg. So in the A configuration we have Siamese twins sharing a leg, which gets counted as two people, but in the B configuration, we count Siamese sharing an entire body as one person, because the heads are so well fused as to almost look like one.

If you consistently count all instances of Siamese twins as either two people, or else as one person, then the count is consistently 13 or consistently 12.

The paradox here rests in a kind of "visual equivocation fallacy". The equivocation fallacy is that, during the course of counting boys around the circle in the two configurations, the observer's definition of whether a Siamese twin is one person or two changes, due to visual deception. Well-fused heads count as one person, but joining at the hip and sharing a leg counts as two people.

Much appreciate the care you took to write this but multiple readings and flipping configurations later I'm as lost as I was when I first counted the discrepancy. Can it be described in a simple sentence?

In one sentence:

Two-headed body at A+2, in B config, wrongly counted as one boy, corresponds with hip-joined Siamese twins at A+7, in A config.

Maybe this can help. Look at this cropped image:

https://i.imgur.com/egAouqW.png

It looks like three boys; but there are four heads here!

The top boy has about 1/3 of a head coming from the inner disc.

The bottom boy has about 1/3 of a head coming from the outer disc.

The middle boy has a 2/3rds portion from each disc: basically two fused heads.

This is in the B position. When the inner circle rotates clockwise to A, these pieces are redistributed. The top boy still gets enough of a fractional head from the (cropped away) previous boy. (Not quite. More precisely, in position A, the top boy gets no head material at all from the previous boy, yet has enough head material from the outer disc that he still has a complete head.)

The top boy's 1/3 of a head goes to the middle boy, where it combines with the outer 2/3rds to make one more or less normal head now: no more double head here. The middle boy's previous inner 2/3rds moves to the bottom boy, where it also makes a normal head.

So, the double head being gone in configuration A, we now count 3 heads rather than 4.

Since 4 heads became 3: an extra head shifted out of here in the direction of rotation, and that's what supplies the extra head for the other boy in the bottom left, who now gets joined at the hip with a twin, sharing a leg with him.

Looking at the cropped image again, consider what would happen if the middle double-headed figure did not have a big 2/3rds of a head from the inner disc, but only 1/3rd, making just one normal head. Upon rotating to the A position, the bottom boy would now get that inner 1/3rd and therefore would not have a complete head!

I believe that there is justification in relating this to the Tangram Paradox:

https://mathworld.wolfram.com/TangramParadox.html

Here, a subterfuge involving subtle fractional differences is also taking place. Very similar, corresponding body shapes actually have a different area, and that difference corresponds to the extra piece.

In the bicycle wheel problem, we ignore that two heads are two individuals if they have the same body and are fused at the head, yet we do not ignore that two heads are two individuals if they share only a hip and leg.

In the Tangram paradox, we cannot ignore the glaringly obvious extra area of the small rectangle forming the foot, but find it hard to perceive the extra area of the larger body.

Hard to see extra area versus hard to see two heads that are partially merged.

Spent another 30 mins on all these generous explanations, some of which made some sort of sense, some of the time.

I've effectively given up, and can only conclude the following:

State A has under 13 boys and state B has over 12 boys and we round up or down visually. This explanation doesn't satisfy me at all but is enough for me to move on, defeated.

See if this helps: https://news.ycombinator.com/item?id=28650420

Including the follow-up child comment.

I've reproduced the effect using equally spaced, identical geometric figures, linearly arranged.

Consider that in state A there is a spot with two boys. Consider that in state B no such spot exists. Why is that?

Broadly, in state A from that position, the boys are on the outside AND inside. As you progress around the circle, they transition from inside to outside more or less smoothly. By rotating the most inside boy to be aligned with the most outside boy, you haven't mismatched any of the other boys enough to invalidate them as part of your count. Consider that in state B you discount the boy part in the position where in state A you counted 2 boys.

If that doesn't do it for you, can you describe what helped and where I added or maintained confusion?

This was my attempt:

https://news.ycombinator.com/item?id=28646279

Here's a picture with side-by-side comparison and things numbered: https://i.imgur.com/zw6n7C7.png

EDIT: I just left this without comment because I had to go eat dinner with family, but notice how in the first picture there are four faces that are right along the split. But in the second picture, there are only three.

Bravo for the explanation. These types of visual games are often convincing because they are reliant on the observer just glancing for a few seconds and not prolonged study.

OK, so that's a good explanation as far as the head counting is concerned. But the extra head has a body. Well 3/4 of a body: the Siamese twin has his own torso, two arms and one leg. Where is that from?

the 3/4 of a body come from to produce the Siamese twin, who has his own torso, two arms and a leg?

This is kind of like, imagine we have a slide rule with two moving parts:

    . . , | | | |
    | | | | ; ' '

Here I have six vertical features, placed side by side. Now I slide the rows relative to each other:

      . . , | | | |
    | | | | ; ' '

wee, now I have seven vertical features.

Let's do it with circles:

https://i.imgur.com/9tuNkbT.png

Just by chopping and sliding, I turned 10 circles into 11. Apparently. But of course, the 11 objects are not really circles; they don't have the full area.

Here it is with stick figures.

https://i.imgur.com/MbrCthN.png

We can do the "head examination" here too, but it's actually irrelevant and "wrong-headed" because that situation is a consequence of what I did, rather than the basis.

We can ask, where did the extra head come from for the new stick figure on the far right? And here it is actually plain to see: the previous four figures have different slices of their head removed, which precisely add up to a complete head. There is nothing special about the head; the same holds for all the entire figure.

Sam Loyd's bicycle wheel drawing conceals the principle by having the boy figures in different positions, and around a wheel rather than a linear slider.

For anyone else that missed it – there is a green slider at the top to rotate the inner disk.

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.)

Which boy is gone? :)

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!"

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"

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.

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".

Also the 2 o'clock boy has no flag in A then he has a flag in B.

Jack, in the red shirt.

Waldo, in the striped sweater.

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.'

Part of the trick is that the boys are split into two parts and progress from more inside to more outside along the wheel. Since the unit "a boy" is based on real world experience where boys can be different sizes and especially since how much space a boy takes up in an image can vary depending on angle, we mentally accept that this mashup of different amounts of body parts adds up to "one boy" rather than thinking "No, that's just 85 percent of a boy and over there we have 105 percent of a boy!"

As someone else noted, we accept in one position that two partial boys each count as a whole boy and then in another position we accept that they each count as just part of a whole boy when paired differently.

(Edited in hopes of improving clarity.)

To add to this, the boy at 3 o'clock, whose head is split in two equal pieces, is the key to the sleight of hand.

Does his head stay or does it move?

If it moves, the boy at 2 o'clock should count as 2 heads in position B. If it stays, then he should count as 2 heads in position B, since the head below him moves up.

But because it's ambiguous, we count it as both moving and not moving.

Yeah the size of the boy half goes from say 5% to 95% around the circle, on both sides. In the 12 boys configuration the halves are paired correctly to add up to 100% sized boys. But in the other, there are 12 boys smaller than 100%, and then at 8 o'clock two 95% pieces appear together so we round them up to two. Neat illusion.

Looked at it differently, we go between two states:

  1+12, 2+11, 3+10, ..., 10+3, 11+2, 12+1
  (12 pairs, each adding up to 13)

and

  0+12, 1+11, 2+10, ..., 10+2, 11+1, 12+0
  (13 pairs, each adding up to 12)

Unfortunately this doesn't work with banknotes ... unless you can find lots of people who are willing to accept a banknote that appears to have been torn into two pieces and then stuck together again with sticky tape, with the line of the tear being a weird curve that just happens to cross both serial numbers in roughly the same place.

It does work with chocolate, though...

http://mathandmultimedia.com/2014/07/22/explanation-infinite...

‘Nobody’ looks at banknotes. Also, historically, banknotes were torn and repaired more often. Because of that, you can just cut a 1/10th width strip out of 9 banknotes and glue them together to make a 9/10 width tenth banknote.

Examples: https://books.google.com/books?id=e7QzAQAAMAAJ&pg=PA318&lpg=... (1804)

https://books.google.com/books?id=osjhDwAAQBAJ&pg=PA114&lpg=... (1850s)

A more tricky recent variant replaces the cut-out part with a fake part: https://bc.ctvnews.ca/can-you-spot-the-fake-splice-and-tape-... (why do these criminals take the effort and risk of creating a fake fiver? I would discard the remains of the fiver, and spend all effort on improving the technique for transplanting the hologram to the fake 100)

Of course, this works better with small denominations, if only because people expect larger denominations to look newer.

The risk of getting caught also is fairly large, I think, but a good criminal can feign innocence, claiming to have gotten the note elsewhere.

If I had to guess they were using the five ask a test to a) see if it would pass a cursory inspection, which it appears it didn't, and b) refine the technique. People are generally more skeptical of larger denominations (especially 100 in the US which you can sometimes get a bit of grief over when trying to use), but if someone were to notice the 5 is counterfeit somehow it's not quite as suspicious just to pay with a different (legitimate) note.

Regarding that story about the Canadian notes, it seems like an unnecessary risk to me to go spend the altered $5 bill with the foil. It sounds like you can turn $5 in to $100, and only have to risk getting the $100 in to circulation. Or you can turn $5 in to $105, but then you have to risk getting 2 bills in to circulation. It's significantly more risk for a very tiny increase, no?

That sounds like something from Martin Gardener, I definitely heard of this trick!

Yes, I believe he has a Dr. Matrix story in which Dr. Matrix attempts to do this and gets arrested.

Edit: "Sing Sing" in The Magic Numbers of Dr. Matrix.

> In desperation he did a foolish thing. He tried to make some twenty-dollar bills. His method was bizarre and surprising. With a paper cutter he sliced each of fourteen bills into two parts, cutting them neatly along the broken vertical lines on each of the schematic bills shown on the left side of Figure 4. [...]

> Unfortunately—or rather, fortunately—the United States government places duplicate serial numbers at opposite comers of every bill, and most of the numerologist's new bills therefore bore pairs of serial numbers that did not match. True, Dr. Matrix's method of making new bills was not exactly counterfeiting—he merely "rearranged" the parts of genuine bills. Nevertheless, the Treasury Department took a dim view of his work and it was not long until he found himself firmly confined within the matrix of cells at Sing Sing.

(note that although Gardner is best known as a nonfiction writer, the Dr. Matrix stories are fictional)

There's a good Numberphile video explaining this illusion: https://www.youtube.com/watch?v=cE44nr4d3iY

(Hint: look at the heads)

This one includes flags as an easy hint.

The flags don't really "do" anything. In the one configuration, every flag is held by a boy. In the other configuration one boy has no flag, and there is a severed hand holding a flag.

That's actually wrong for the other configuration. In one, every boy has a flag. In the other, there's one boy with no flag, and 12 boys with flags. The appearance of a boy with no flags is an insight.

None of the flags straddle the boundary between the two discs, so it is obvious that no trick is present affecting their apparent count: there are always 13.

The diagram's inner disc contains one complete torso with arms and hands, which holds two flags in either configuration, so "every boy has a flag" is only true if you actually mean "has at least one flag", not if you mean "has one flag".

Since the number of flags is not affected by configuration, one boy not having a flag is unsurprising. There are always 13 flags, so if there are apparently 13 boys, and one is still holding two flags, then someone must necessarily not have a flag.

(In the A configuration, there are actually two boys with no flags. One clutches his head with both hands, holding no flag, and the other, next to the A, is flanked by a floating, detached hand which holds a flag. Therefore, that boy has no hand, and no flag. This severed does not seem essential, either; I believe the picture could be somehow repaired so that the detached hand is reattached. Then in the A configuration would be exactly one boy with no flag, as you say.)

> it is obvious that no trick is present affecting their apparent count

I see you found the insight from the hint: flags are governed by different rules than boys.

I was pretty confused until I noticed the bottom right head turning into an arm.

This is the most elegant explanation.

Is this just a visual version of the missing dollar riddle? https://en.wikipedia.org/wiki/Missing_dollar_riddle

This was my grandpa's go to riddle for his grandkids (but with Francs). :-)

No... the missing dollar riddle asks people to confuse cash inflows and outflows by changing the subjective interpretation of a middleman, who starts out being an outflow (he/she is "part of the business") and then is treated like an inflow (he/she is "one of the money-havers"), it's essentially a linguistic puzzle in an accounting context, one person can be referred to in two different ways.

This one is closer to the missing square puzzle, https://en.wikipedia.org/wiki/Missing_square_puzzle .

This is a calculus puzzle, you have approximately 13 boys "missing a tiny slice" of themselves versus 12 boys with "an extra little slice" of themselves... it's more clever than just "we take the slices and reassemble them", it's "the slices slowly sweep across the boys' body so that when we slide the circles we essentially accumulate a whole 'second boy' inside the internal circle." Sort of like how the missing square puzzle has a clever way to have two almost-parallel lines which are not parallel hiding a long skinny rhombus containing the extra area, but the rearrangement exposes it in a much more visually arresting form as a whole missing square.

The boy who occupies the 4-5 o'clockish space in Configuration A appears to have a fist where half his face should be. I nominate him as the vanisher since there's no fist-face in Configuration B.

in a sense, there are only two boys, an inside boy and an outside boy, and where those two boys are together (at around 8 o'clock) you actually don't see 100% of either of them, yet you count it as two boys

in the A position, look at around 8 oclock where the two boys next to each other.

look at the "inside" boy, and then move clockwise and you just see just his foot on the inside, continue and you see a leg, going all the way around you can see almost all of the "inside boy" emerge.

in a like manner, look at the outside boy of the two boys together, and proceeding counter-clockwise you see the foot, the leg, etc as almost all of the outside boy emerges.

it reminds me of a trick where you take a stack of dollar bills and slice a thin strip from each one, with the slice taken on each successive bill moving across, and tape each bill together again.

At the end, tape all the little strips together and you have an extra bill!

Yeah, I don't understand why moving the two boys side-by-side is considered a trick. It's much harder to understand how to be misled than to discover the solution.

See also the Missing Square puzzle: https://psychology.wikia.org/wiki/Missing_square_puzzle

Don't read the text if you want to figure it out yourself.

I de-mystified this by considering the boys as a spiral. Move the wheel, the spiral increases by one node.

In summary, this puzzle is puzzling because we are given a wheel and keep thinking 'circle' instead of, "is it actually a wheel, or not?"

A similar puzzle is Magic Microbes (ctrl-f it) from this page:

www.karlsims.com/puzzles.html

Reminds me of the infinite chocolate trick, https://www.youtube.com/watch?v=qnpugKVitl0

> Where does he go?

1906. Typhoid probably got him.

Can only count 13 boys whether from point A or B..what am I missing?

Moving the arrow on the center disk of the two physical disks would rotate the insides of the circle and realign the person parts.

Took me a second too.

Thanks. Doesn't help me understand it (nor do any of the above comments) but does help me see it.

The little line at the top is actually an interactive slider control so you can rotate the inner disk from point A to point B.

Definitely not an intuitive interface.

Counting clockwise from A, 5th bicyclist head disappears into hand.

i learnt about this sort of trick from martin gardner (who possibly coined the term "geometric vanishes" to describe them). he was how i first encountered sam loyd too.

Simple, the "B" boys each average 1/13 bigger.

Ah, an early compression algorithm.

Anybody want to figure out the year?

The puzzle was originally conceived in 1906.

Love me some Sam Loyd, there is no greater chess puzzler in my mind.

It's the less-racist remix of his 1896 "Get Off The Earth" puzzle, which contains twelve or thirteen "chinamen" in the exact same poses around a globe.

http://www.marianotomatis.it/blog.php?post=blog/20110715&sec...

Whoever currently owns his name is still selling it, except now it refers to "warriors". https://www.samloyd.com/product/get-off-the-earth-card/

Video demo of the "Get Off the Earth" which is a little jarring. It looks to be the same as a bike one. The video has lots of other versions/variations on these puzzles:

https://youtu.be/KdwJQbxLFHI?t=27

Added above. Thanks!

Reply on news.ycombinator.com