The Card Trick That Cannot Be Explained? Top Answer Update

5 Intermediate and Advanced Card Tricks Every Magician Should Learn

triumph

Triumph card trick explained

Triumph is one of the most famous tricks ever invented. It is loved because it is so beautifully simple and direct, but also incredibly effective.

The basic triumph routine is that a playing card is chosen and then lost back in the deck. Half of the cards are then revealed and shuffled into the remaining face-down deck, creating a chaotic jumble of cards facing different directions. After making a magical gesture, the magician spreads the cards out to show that they are now all magically facing the same direction, except for one card – the spectator’s chosen card.

history of triumph

The basic idea behind Triumph can be attributed to a mechanical trick deck effect called Inverto, published by Theodore DeLand in 1914, according to Genii Magazine’s Magicpedia. Over the next three decades, this unique effect inspired many other “reverse card” and “slop shuffle” effects.

Finally, in 1946, one of the greatest magicians of all time, Dai Vernon (affectionately known as “The Professor”) published his version of a “reverse card” effect in Stars of Magic – Volume 2. This effect not only spawned the Triumph name, but also introduced the Triumph Shuffle – a special strip-out shuffle in which the playing cards are secretly rearranged under the guise of ripple shuffling and slicing the deck.

Because the Professor’s version is still regularly used by professional magicians around the world, and inspired many of the myriad variations available today, magicians often credit Dai Vernon with inventing Triumph.

Where can I learn Triumph?

As mentioned above, Triumph is a very popular card trick that has been further developed by numerous other professional magicians over the years. Here you can learn the original Triumph routine as well as some of our other favorite variations.

The ambitious card

Ambitious card trick explained

The Ambitious Card is an amazing magic trick that should be in every wizard’s repertoire. There’s a reason it’s taught in virtually every beginner’s magic book such as The Royal Road to Card Magic, Mark Wilson’s Complete Course in Magic, and Roberto Giobbi’s Card College.

In its simplest form, an Ambitious Card routine consists of a selected card (which is also usually signed) appearing repeatedly at the top of the deck after being placed in the middle of the deck or elsewhere. Most of these routines generally involve several stages that get progressively more impossible before finally ending with a surprise ending where the chosen card completely disappears from the deck and reappears in an impossible place like an envelope or wallet (usually a special card-to-wallet prop). ).

While beginners can certainly pull off the ambitious card trick, many intermediate to advanced magicians love this effect for the versatility it offers. As with the classic cup and ball magic trick, there is a seemingly endless amount of variation and sleight of hand that can be performed. Much like a jazz musician, many magicians will even improvise certain parts of an Ambitious Card routine based on the audience they are performing for.

History of the ambitious card

While some attribute the first iteration of the Ambitious Card to a book published in 1886 by French magician Alberti entitled Recueil de Tours de Physique Amusante (translated a year later by Professor Hoffmann in Drawing Room Conjuring), others believe that the The Ambitious Card first appeared in Jean Nicolas Ponsin’s 1853 magic book Nouvelle Magie Blanche Dévoilée. And to muddy the waters even further, some historians believe that Ponsin’s routine was even inspired by an effect explained more than 130 years earlier in Richard Neve’s The Merry Companion; or pleasures for geniuses.

While its exact origins may be a bit unclear, the Ambitious Card is another effect that once again has Dai Vernon and the legendary Harry Houdini to thank for cementing its place in magic history.

In 1922, Houdini, who was in Chicago exposing spiritualists and false psychics at the Majestic Theater, attended a local meeting of the Society of American Magicians. During this event, a 27-year-old Vernon, who had heard Houdini’s hilarious claim that he could figure out any magic trick or illusion if he saw it three times, decided to test his dexterity on the world’s most famous magician.

Houdini was reluctant to let this then completely unknown young magician show him a card trick. However, according to legend, he eventually accepted the challenge and was fooled not just three, but seven times by Vernon’s card trick – later revealed to be a signed card version of the Ambitious Card.

From there, the Ambitious Card became known as “Trick That Fooled Houdini” and Vernon, now dubbed “The Man Who Fooled Houdini”, became one of the most influential magicians of all time – inspiring everyone from Ricky Jay to David Copperfield.

Where can I learn the Ambitious Map?

As well as the spellbooks mentioned above, here are some other great resources to help you learn different variations of the ambitious map and hopefully motivate you to create your own unique version!

Any Card to Any Number (ACAAN)

Each card explained by each number

The Any Card at Any Number plot, often referred to simply as ACAAN, is a card magic classic in which a freely chosen playing card appears in a position in the deck that correlates to a number also chosen by a spectator ( i.e. the ace of spades is 27 cards from top to bottom). These types of effects often utilize a memorized deck, as taught in Simon Aronson’s “Bound to Please” or Juan Tamariz’s “Mnemonica,” and sometimes even use a special playing card or trick deck.

Modified versions where the playing card is forced (better known as Card at Any Number or CAAN) and other variations on the ACAAN concept are also very popular among professional magicians.

History of every card at every number

While the Conjuring Archive believes the first effect, “selected card appearing at any number in the deck,” may have appeared as early as 1786, and masters of card magic such as T. Nelson Downs, Dai Vernon, and Ed Marlo also developed similar stylistic effects, Much of ACAAN’s mainstream popularity can be attributed to David Berglas.

In the 1950s, Berglas, an acclaimed British magician and mentalist described by world-renowned mentalist Derren Brown as “one of the greatest magicians alive” created what many magicians call the “Holy Grail of card magic”. This card trick, later dubbed The Berglas Effect by Jon Racherbaumer in his 1984 book At the Table, is the cleanest and most direct version of the Any Card at Any Number plot ever created. With the Berglas effect, every playing card can be found at every named number in an ordinary deck.

Where can I learn every card at every number?

While the Berglas effect is as close to real magic as you can get, it’s also an incredibly difficult feat. In fact, it takes more than 60 pages of Richard Kaufman’s aptly titled book, The Berglas Effect, to unveil the secret behind this famous card trick. Thankfully, there are a variety of other amazing ACAAN versions you can learn:

Monte with three cards

Three Card Monte explained

Three Card Monte, sometimes known as the three-card trick or follow the lady, is a card trick that began as a street scam used by scammers to harass innocent bystanders in cities around the world.

The game of Three Card Monte is pretty simple. Three playing cards are used, one of which is the “money card” (usually a queen of hearts or queen of spades) which the player is attempting to follow along with two other face down playing cards. It is designed to look incredibly simple so that the victims or brands are tricked into wagering money. However, scammers naturally cheat to ensure they always win. Since this usually requires the use of the same type of sleight of hand as magicians, the three-card monte is very popular in the magic community – particularly among card technicians who specialize in gambling demonstrations.

Story of Three Card Monte

Although it is very likely that it has been around much longer, it is believed that the three-card monte became a popular trick among scammers in the mid-19th century. In his 1861 book Les Tricheries Des Grecs Dévoilées (later translated by Professor Hoffmann into an English book entitled Card Sharping Exposed), Jean Eugène Robert-Houdin mentioned the use of “Les Trois Cartes” (or literally “The Three Cards”) “).

The first detailed description of the three-card trick eventually appeared in Professor Hoffmann’s seminal grimoire, Modern Magic. In fact, it wasn’t books about magic, but books about gamblers and cheaters that introduced the precise details of the three-card monte techniques into the mainstream. In fact, an entire section on the Three Card Monte and the use of bent and marked cards appeared in John Nevil Maskelyne’s 1894 book Sharps and Flats: A Complete Revelation of the Secrets of Cheat at Games of Chance and Skill.

As you can see, magicians have always had a keen interest in the three-card monte. So if S.W. Erdnase’s The Expert at the Card Table offered magicians the first truly detailed behind-the-scenes look at the mysteries of Three Card Monte, including the use of crimps and bent corners. The infamous scam quickly became a mainstay in the magical community.

Where can I learn the three card monte?

Beyond traditional street cheating, many variations of the Three Card Monte have been developed, such as the Two Card Monte (or Hand Sandwich) made famous by David Blaine, as well as those that use special gimmick cards to perform impossible magic feats. Here you can learn some of the most popular versions of the Three Card Monte performed by professional magicians.

Ace Montage

Ace assembly explained

The Ace Montage is probably the easiest card magic trick to describe on this list. It’s as simple as it sounds. Four aces (or other quadruplets) are shown as separate before traveling invisibly and magically rejoining themselves into a stack (usually by following a “leader ace” or other “leader card”).

History of the Ace Assembly

Although he did not claim to have created it, the idea of ​​an ace gathering was described by the French magician Jean Nicholas Ponsin in his 1853 book Nouvelle Magie Blanche Dévoilée. This description was later reused in English by Professor Louis Hoffmann in his iconic spell text Modern Magic in 1876. While Johann Nepomuk Hofzinser may have developed an ace assembly as early as 1857, it was not published until 1910 when Ottokar Fischer authored his book J.N. Hofzinser Kartenkunste (which was later translated into an English book called Hofzinser’s Card Conjuring by H.H. Sharpe).

While not credited with creating the Ace Assembly, Hofzinser’s version, used as the finale to his famous The Power of Belief routine, was eventually developed by Mac MacDonald into MacDonald’s Aces – a gaffed-card variant that remains one of the most popular to this day most popular Ace Assembly card tricks on the market.

The Ace Assembly is a right of passage for most magicians and over the years many professional magicians have laid claim to its continued evolution from Ken Krenzel’s Progressive Aces (aka Succession Aces) to Lin Searle’s Ultimate Aces (or Technicolor Assembly). , Jazz Aces by Peter Kane, and variations by Bill Miesel, Ed Marlo, and Larry Jennings with no deck cards known as Open Travelers (or Invisible Palm Aces).

Where can I learn the Ace Assembly?

You could probably spend a lifetime studying the Ace Assembly. But to get you started, here are some of our favorite resources (many of which include the incredible variations listed above).

ALSO SEE: THE BEST Card Magicians!

SEE ALSO: SIMPLE card tricks

ALSO: BEST card magicians!

ALSO: SIMPLE card tricks

Card magic tricks were once called “poetry of magic” by the great card magician Johann Hofzinser. Yes, they can be terrible when your little brother asks you to count seven cards in three decks, but they can also be great. This is the range of poetry in card magic. Check out the two links above if you want to learn a few simple ones or see what great card mages can do in their hands. The following list highlights ten (and a few honorable mentions) of what we think are the greatest and most amazing classics of the genre. Enjoy…

by Roland Sarlot

Ripping and destroying things like newspapers, napkins, bills, and cards, and then reassembling them into one piece is a classic magic trick. A great trick for a smaller audience or just a few people. While not the most dynamic rendition of this trick, Guy Hollingworth does his amazing take on a torn and restored playing card. He calls it “The Reformation” and it ups the ante with the use of a SIGNED CARD. He tears the card into FOUR PIECES and then puts them back together piece by piece. No glue, no tape, no camera tricks! Clever and oh so very difficult to make.

A staple of any magician, the Disappearing Deck uses expert dexterity with deception to amaze the contestant. In this case, an unsuspecting newscaster on the morning show of “Grand Prix World Champion of Magic” Shawn (aka Steve, haha) wows Farquhar, who with a snap of his fingers makes the deck THAT’S IN THE HANDS OF THE PARTICIPANT disappear. It is incredibly replaced with a clear plastic square the same size as the deck! Watch the news anchor freak out.

In this classic trick, four aces appear to disappear before your eyes and then reappear in one spot. On the surface, McDonald’s Aces (or Grandpa’s Aces in Copperfield’s case) seems like any other four aces card trick, but Copperfield brings a story to the trick centered on learning magic from his Grandpa. What really sets this trick apart is how Copperfield handles the cards with a poetic tenderness to heavy music that creates a powerful sleight of hand. He masters this trick with his trademark finesse, right down to how the aces first appear from the deck and then miraculously disappear before your very eyes. Breathtaking!

First, let’s say that there are endless variations of this repetitive card trick, and you don’t always have to show just five cards. 6 card and 10 card are also standard versions. This little trick packs A LOT of MAGIC in a very short amount of time. One of the great things about this magic trick (which is very rare among card magic tricks) is that it can be performed standing up, allowing more than one audience at a table to experience it. In this performance, Wayne Dobson performs an outstanding version of this classic trick as an opener in front of the Queen of England! A great trick and a fabulous performer!!!

10. Cards Across There are so many great card tricks out there and it’s such a tough decision to rate these tricks, but we didn’t want to skip this trick. This is performed by the Las Vegas Magician Mac King and is a standard piece from his repertoire. (By the way, next time you’re in Nevada, he has one of the best magic shows in Las Vegas, and by far the best value. Period!) As with most tricks on this site, the variations are endless, but here Mac has three magically disappear and reappear. A great trick and a great presentation! (Be sure to watch his bonus 30-second finale, but don’t blink…)

9. Card in Box The highly respected and creative magician among magicians, Tommy Wonder, seats unsuspecting guests at a table and shows his guests a small box and then pulls out a deck of cards. (Skip to minute 3 of the video.) After some shuffling and announcing that a card is missing, the magician asks a guest to choose a card and turn it over from a neatly laid out deck. Lo and behold, the card chosen is the same as the one in… you guessed it. In the box! Not only the same card, but also THE SIGNED CARD. Really stunning!

8. Out of This World All-round magician Michael Ammar takes a normal deck of cards, lets one contestant pass it, and performs magic with his mighty impressive powers in this incredible trick. (Skip to second 25 in the video.) While using the contestant’s power of impression to correctly place face down cards in red and black decks, the cards match the decks incredibly, “save two,” the two cards that the magician has written down BEFORE the trick has even started. It’s pretty awesome.

7. Oil and Water The great Argentinian sleight of hand René Lavand not only performs this amazing trick, but ONE HANDED!!! Although he doesn’t speak English, the magic speaks for itself. He sits at a table with his closest friends (or bitter enemies) when he pulls out a deck, cuts the deck, and then chooses six cards; three black and three red. He shuffles the six cards face down in a red and black pattern, then turns them over to show they are actually in groups of red or black and not in a pattern. He keeps repeating the trick, just slower for us at home. Last time he does the whole deck trick and the whole deck is color aligned perfectly. Unbelievable! Check out René on the list of the best card magicians of all time.

6. Sam the Bellhop An ancient storyline in card magic, but Bill Malone (master teacher and glitterati performer) delights guests with a delightful tale of Sam the Bellhop and the 654 Club. (Skip to second 20 of the video to begin.) As the story continues, Malone and the guests shuffle the cards MANY times, but each facedown top card matches the numbers in the story, with repeat counts of 6, 5, and 4 A fun and super amazing trick.

5. Card on the Ceiling Michael Ammar is at it again… (Skip to minute 2:30 in the video above.) It begins with an unsuspecting spectator selecting a card and signing it. He has the card placed back in the deck and ties the deck together with rubber bands, locking in the chosen card. The audience’s attention is drawn to the ceiling before the magician throws the deck at the ceiling, causing the signed card to stick…to…the…CEILING! It somehow escaped the rubber banded deck AND somehow stuck to the ceiling! WHAT?!?!?

4. Chicago opener “The Magician’s Magician” Daryl demonstrates this classic by shuffling a deck and turning it over, then asking a spectator to choose a card (any card, really!). The chosen card is shown to the audience but not the magician, then placed back in the deck face down. Daryl explains that a card changes color. Sure enough, when you spread the cards face down, a card in the red deck has a blue back. And the odd card? The same that was selected but originally had a red back!!! Then the trick is repeated for a very surprising ending. Watch carefully!

3. Invisible Deck In this episode of The Ellen Show, a viewer performs an amazing card trick. It all starts when the magician asks Ellen to shuffle an imaginary, some would say invisible, deck of cards. Ellen chooses an imaginary card and returns it facedown to the face-up INVISIBLE deck. The magician, magically holding a real box of cards (what can I say!?!?) asks Ellen which card she chose. Miraculously, the chosen card lies face down in the middle of the deck. A true marvel and a true card magic classic presented directly by Don Alan 60 years ago!

2. Card in Impossible Place One of the greatest card magicians of all time, Jimmy Grippo, shuffles a deck and has a contestant choose a card from the deck and show everyone before returning it and shuffling the deck again. (Start at 4:10 in the video.) Through an honest magic act, Jimmy declares that the right card will fly up his sleeve. And like a kind of magician, he pulls the chosen card out of his jacket pocket. The trick is repeated with a signed card, with equally amazing results. Check out Jimmy on the list of the greatest card magicians! This is one of Jimmy’s “map in an impossible place” example. SOOOooo, are you ready for the best magic card trick ever? DRUM ROLL PLEASE…

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A card power is one of the greatest tools at a magician’s disposal, and they are a crucial component of hundreds (probably thousands) of effects. Once you learn a great card power, you can amaze ANYONE with ANY deck of cards.

Because of this, it’s important that you choose a GOOD power to use. But which power is the best?

This is where we come in. In this blog post, we will go through the top 5 powers you can learn today. We also tell you exactly where to go if you want to learn them.

Let’s get right into it…

Choice #3: The Dribble Power/The Waterfall Power

This one is a firm favorite of many wizards. It looks very fair but allows you to force any card you want.

The waterfall force (or trickle force) looks like this:

You take the cards in one hand and dribble them into the other hand (hence the name “waterfall squad”) and instruct your spectator to tell you “stop” at any time.

When they do, stop and they look at the map they stopped at. It seems like a perfectly fair choice to them, but you’ve actually just forced their card of choice on them.

To learn the waterfall power/the dribble power, see this:

Choice #2: One-Way Force Deck

This is the best way to go if you absolutely CANNOT afford to miss out. For example, when you’re performing your first stage show, you might feel nervous about getting every part of your performance right. So, to feel less stressed, you could opt for the “one-way force deck” to make sure your force card ALWAYS hits.

This method obviously has disadvantages, which we will talk about in a moment.

But first, let’s talk about what the one-way force deck actually IS.

The One Way Force Deck is a normal deck of cards, except for one very important fact. ALL cards are the same card. For example, every single one of the 52 cards in your deck could be the King of Hearts.

Of course, this makes it VERY easy to force the map you want; All you do is give your spectator the option to choose a card as usual. You can make this seem like true free choice (because in a way it IS free choice – free choice of card, except all cards are the same), but at the end of the day, your spectator has the card they wanted impose on them.

I’ve seen the one way force deck used in many different tricks and presentations with varying degrees of success. For example, it’s often good to use the one-way force deck as part of an “ACAAN” (any card, any number) routine.

Now let’s talk about the disadvantages of the one-way force deck.

Here’s the big deal: your audience can’t look at the deck. This should go without saying, but whatever you do, DO NOT let your audience see that all cards are created equal. Unfortunately, this means you can’t hand the deck over for review at the end of the trick, or show the audience that it’s a normal deck at the beginning of the trick.

However, there is a simple solution to this…

A deck switch.

A deck switch is exactly what it sounds like – when you swap out one deck for another. If you’re doing a deck switch, you can do the one-way power with the one-way deck, and then once the card you want has been chosen, switch decks so you have a regular deck. With THIS deck, you can dish out stuff for scrutiny and have your audience examine every card in the deck – you’ve got nothing to hide! There are many different ways to change decks, but the best place to start is definitely this book:

The art of changing decks

The Art of Switching Decks by Roberto Giobbi and Hermetic Press

Choice #1: The classic power

In my personal opinion, this is THE best power you will ever find for a number of reasons.

First, you can do it with ANY deck. Second, it looks EXACTLY like a normal selection. Your audience should never assume that they have chosen a power card.

But first, what is the classical power?

The classic power is a power where you allow the spectator to choose a card like they normally would. They spread across the cards and they remove the card they want (or think they want – actually it’s the card you were trying to force them to take).

One of the problems with many powers is that they require the magician to use a selection process that seems a bit odd. Why make the viewer choose a card when you’re dribbling through the cards unless you’re doing something fun? These are the kind of questions that your audience will subconsciously pick up on.

But the classic power is incredible because your audience can’t tell when they’re hitting a real card and when you’re forcing a card on them.

I like to exercise this power, even if I don’t have to. If I do a trick that requires my spectator to draw a card, I still try, even if I don’t have to use the classic power! If I miss, it doesn’t matter because I didn’t have to force a map anyway. But doing it right gives me important practice and experience in successfully performing the classical force.

How do you learn classical power?

First of all you need a good teacher. Aaron Fisher is one of the top card magicians in the world and an expert on the classic power. In the following training he will show you exactly how to perform the classic strength step by step.

Listen:

Aaron Fisher – Classic Force

This training is only available for CC members. Not a CC member yet? Click this link to get your first month for just $1!

Conclusion: Find the power that works best for you

In truth, the best power for you is the one that feels best and that you can do most safely. If a power works well for you, use it! It doesn’t matter if it’s not “ideal” – if it works for you, then it IS the ideal power.

But where do you find all these many powers to choose from?

After all, we’ve only just scratched the surface with the forces listed in this post. There are many more powers for you to explore.

But where do you find them all?

Lucky for you, we have the perfect resource.

The training plan below is a “deep dive” into the many powers and ways to force things. Watching this will unlock many more powers so you can find the one that works best for YOU.

Listen:

This training is only available for CC members. Not a CC member yet? Click this link to get your first month for just $1!

We hope you found this post on best card forces useful! If you have any suggestions for other forces that you think belong on this list, don’t hesitate to let us know in the comments section. If you found this article useful, share it with your friends!

Force Magic contains active synthetic pyrethroids (Prallethrin dan Permethrin). The active synthetic pyrethroids have the ability to kill insects but are harmless to mammals (humans) as the substances are easily excreted from our bodies through feces, urine and the respiratory tract.

IN PROGRESS

The magician lays out 21 cards face up in three columns and seven rows. He asks a person in the audience to memorize one of the cards. Then the spectator is instructed to indicate the column containing the chosen card. The magician collects the cards column by column and then lays them out row by row again in the same sort of grid; three columns, seven rows. Again, the viewer is prompted to view a column. The process is repeated a third time, after which the magician collects the cards and recites the magic word: “Abracadabra!”. He then deals the cards and spells out a letter “Abracadabra” for each card. As he pronounces the last “a”, he deals the chosen card!

How does this work? Why does it work? Does it work for any number of cards? How many columns and rows does the trick need? How many times does the spectator have to point to a column for the trick to work? All these questions, believe it or not, are answered in this post.

How it works

Also known as The Abracadabra Card, The Eleventh Card, Sim Sala Bim, Lucky Guess, and 49er Fool’s Gold (where there are 49 cards in a 7×7 grid), this trick is very simple. The magician need only collect the cards column by column, taking care to maintain the relative order of the cards in the column shown and inserting the column shown between the others. In other words, the given column should end in the middle, similar to the famous bologna or whatever people want to use as the middle layer of a sandwich. One of the easiest ways to maintain the order of cards in a column is to let each row overlap the previous one a bit, but keep the columns apart so each column can be picked up quickly and in style. This way the magician can collect the cards quickly and the spectator will have a harder time realizing the fact that their pillar is jammed. Another important part of the trick is that the cards must be laid out row by row. By repeating this process several times, depending on the number of cards, the chosen card is “magically” drawn into the middle, i. H. the 11th position when using 21 cards.

A mathematical point of view

By counting the cards in rows – calling the first card number zero – as we lay them out on the grid, we can assign a number to the selected card. This number is called the “row-by-row position” and is denoted by p. Each time we place the indicated column between the others, the chosen card will get a new row-by-row position. If r is the number of rows, c is the number of columns, and y is the row (zero being the top row) containing the chosen card, we can derive the formula for the new row-by-row position. Since the chosen card is in row y in the column shown, and we are using zero-based counting, there are y cards above it in that column. Also, we collect this column after collecting half of the columns (i.e. r(c/2) cards). Therefore, y + r(c / 2) cards must be before the chosen card after we collect them all. Since we’re using zero-based counting and the cards are dealt in rows, this is the new value of p.

p = y + r(c / 2)

(1)

Note that r(c / 2) does not equal cr / 2 since we are dealing with integer division. In some cases the two are the same, but we can no longer take that for granted in the integer domain, where truncation complicates things a bit. For example, imagine nine cards laid out in three columns and three rows, so c = 3 and r = 3. In this case, c/2 = 1. We get this by truncating at the decimal point, so it becomes 1, 5 to 1. Therefore, r(c / 2) = 3. However, cr = 9 and cr / 2 = 4.

The variable y is redundant because we can easily calculate which row the card is in given the row-by-row position and the number of cards in a row, i.e. the number of columns.

y = p / c

(2)

Note that this is also integer division. Let’s do a sanity check with these nine cards; c = 3 and r = 3. The ranked positions of the first three cards are 0, 1, and 2. Integer division by 3 gives y = 0 for all of these cards. So far, so good . The sequential positions of the next three cards – 3, 4, or 5 – divided by 3 equals 1. Good. 6, 7, or 8 divided by 3 equals 2. The formula works as expected.

The recursive formula

Now we can finally create a recursive formula to calculate the new row-by-row position of the chosen card simply by replacing y in (1) with the right-hand side of (2).

p i = p i – 1 / c + r (c / 2)

(3)

The formula above gives us a way to calculate the new row-by-row position p i given the old position p i – 1. The 21 trick works if and only if the recursion eventually reaches pk = cr/2 for some positive integer k, provided c and r are odd and 0 ≤ p0 < cr. A reader who wasn't paying attention asks you might wonder why c and r have to be odd. The answer is simple, otherwise the grid will not have a center position. There seem to be several ways to prove that the 21 trick works. Below are some methods that seem plausible. Go backwards from p k = cr / 2 and prove that it is possible to reach any position row by row. While this sounds quite combinatorial and aesthetically pleasing, upon closer inspection it appears quite complex. "Solve" the recursion and calculate a non-recursive function. Show that the limit of this function is cr / 2. This might be impossible due to the integer divisions and is beyond the scope of this entry. Show that the chosen card can never move up if it is above the middle row and that it can never move down if it is below the middle row. Show that it always changes lines. Show that it never happens in the middle row. Show that when it reaches the middle row, its next row-by-row position is cr/2. This method is clear, to the point, and seems practical. Show that | p i - cr / 2 | < | p i - 1 - cr / 2 |, preferably by treating p 0 < cr / 2 and p 0 > cr / 2 as two different cases. This method seems to delve deep into the intricacies of mathematics.

Without claiming the accuracy and/or plausibility of the other three methods, this entry uses the third method to prove that the card chosen will indeed arrive at cr/2 sooner or later.

Why it works

As outlined above, the proof consists of four parts. The first part shows that the chosen card cannot move in the wrong direction, the second shows that the card always moves, the third shows that the movement of the card is limited by the middle row and the fourth shows that the card moves to the middle column if it is in the middle row. The first two parts together prove that the chosen card always moves towards the middle row. Together with the third part, they show that the card is indeed always moved to the middle row. All four parts together mean that the selected card is always moved to the center of the grid.

direction of movement

Since we’re trying to prove something based on which row the chosen card is in, we’d better construct a formula that tells us how the y-value changes. That’s pretty easy; Just replace the p in (2) with the right side of (1) to get a new recursive formula.

y i = (y i – 1 + r(c / 2)) / c

(4)

[Not done.]

constant movement

Consider a square lattice, i.e. when r = c. Now let’s see what happens if the chosen card in a step i of the trick is in the same row as in the previous step i – 1.

1: y i – 1 = y i 2: y i – 1 = (y i – 1 + r(r / 2)) / r 3: y i – 1 = (r(r / 2)) / r 4: y i – 1 = r / 2

If the card doesn’t change rows, it appears that it’s already in the middle row. The omission of y i – 1 in line 3 may seem a bit cryptic. This obviously has something to do with integer division. Since we are dividing by r, an integer greater than y i – 1, and r is dividing r(r / 2), the variable y i – 1 can be omitted without changing the result. It is actually the bit that is “scraped off” by integer division. For example (8 + 3) / 4 = 2, which is equal to 8 / 4. The 3 disappears completely because it is smaller than 4.

If r < c, then y i - 1 < c (obviously y i - 1 < r, or the chosen card would not be in the grid), and in this case we can perform the same kind of omission of y i - 1 when dividing by c . 1: y i - 1 = y i 2: y i - 1 = (y i - 1 + r(c / 2)) / c 3: y i - 1 = (r(c / 2)) / c 4: y i - 1 = r / 2 Line 4 needs some explanation here. It's not entirely obvious that r(c / 2) divided by c equals r / 2. To even begin to understand this division, we must remember that this is integer division and both r and c are odd. Hence c = 2m + 1 and r = 2n + 1 for some integers m and n. Now we can express the division in a way that gives us an actual result. (r(c / 2)) / c = (2n + 1)((2m + 1) / 2) / (2m + 1) = (2n + 1)m / (2m + 1) = (2mn + m) / (2m + 1) Well, n(2m + 1) + (m - n) = 2mn + m, and by dividing each side by 2m + 1, we see that n = (2mn + m) / (2m + 1). Note that this only works because n(2m + 1) is obviously divisible by 2m + 1 and m - n is positive (as implied by r < c) and therefore equals zero when divided by 2m + 1 . [I'm currently working on the proof for r > c.]

Limited Movement

Remembering that c and r are odd, we move on to the next part of the proof, where we just need to use a little algebra. Let’s consider the case when the chosen card is above (or in) the middle row.

1: y i – 1 ≤ r / 2 2: y i – 1 + r(c / 2) ≤ r / 2 + r(c / 2) 3: (y i – 1 + r(c / 2)) / c ≤ ( r / 2 + r(c / 2)) / c 4: y i ≤ (r / 2 + r(c / 2)) / c 5: y i ≤ ((2n + 1) / 2 + ( 2n + 1)( (2m + 1) / 2)) / (2m + 1) 6: y i ≤ (n + (2n + 1)m) / (2m + 1) 7: y i ≤ (2mn + m + n ) / (2m + 1) 8: y i ≤ ((4mn + 2m + 2n + 1) / (2m + 1)) / 2 9: y i ≤ (2n + 1) / 2 10: y i ≤ r / 2

So if the card is at any point above or in the middle row, the next step won’t move it below the middle row. Although most of the algebra above is straightforward, the extra 1 introduced in line 8 may require justification. The following is a simplified version of introducing the 1; x = (2x + 1) / 2. In this case, if we consider that we are dealing with integers, we can clearly see that the extra one divided by two equals zero, and so introducing it doesn’t really change anything . Other than that, the proof wouldn’t work.

The proof when the chosen card is below the middle row is very similar; we can simply change the “less than or equal to” character to a “greater than or equal to” character. The combination of these two pieces of evidence says more than the fact that the card’s movement is limited by the middle row; it also proves that if the card is already in the middle row, it cannot leave it.

Final centering

If the chosen card is in the middle row, its row-by-row position is p i divided by the number of columns equal to r / 2. Combining this with (3), we see that p i + 1 = r / 2 + r( c/2). ). Again, the key is that both c and r must be odd for the trick to work. The rest is algebra.

r / 2 + r(c / 2) = (2n + 1) / 2 + (2n + 1)((2m + 1) / 2) = n + (2n + 1)m = 2mn + m + n

cr / 2 = (2m + 1)(2n + 1) / 2 = (4mn + 2m + 2n + 1) / 2 = 2mn + m + n

Thus p i + 1 = cr / 2 = p k , which means that if the chosen card is in the middle row, in the next step it will be moved to the middle of that row. This is an excellent example of something that is perfectly obvious when considered practically, but can be expressed in such abstract terms that one forgets the point.

number of steps

The number of steps required to reach the center of the grid appears to be log c cr rounded up to the nearest whole number. This is because the position a selected card is moved to depends solely on its y value (which row it is in). Thus, regardless of which of the c positions in this row it occupies, the card moves to a specific position in the grid. Therefore we can draw the grid as a tree by drawing an edge between each position in a row and the position to which the map will move from that row. This produces a c-ary tree when c = r, but otherwise the tree is only approximately c-ary.

The map always moves towards the root – the center of the grid – and so the maximum number of steps is equal to the height of the tree, which is log c cr for a c-nary tree. Since the height of a tree cannot be a real value, the log value must be rounded up.

Closing remarks

Hopefully the readers of this entry now know more about the 21 trick (and the sanity of the average h2g2 researcher) than they ever hoped to know. In any case, they can sleep a little better at night, knowing that the 21 trick won’t fail if done correctly, and that it can be performed even with any odd number of cards (as long as each card is clearly identifiable) and odd column and line counts.