# David’s de Bruijn sequence card trick

A few days ago, my friend David asked me if I could help him with a card trick. I said I could, hence this post. I managed to pin David down in front of my camera long enough for him to demonstrate the trick; a full explanation follows this video:

The basic trick is pretty well-known: Card Colm wrote about it for his MAA column a few years ago, and collected a pretty extensive list of people who have written on the subject. David came across it through Persi Diaconis and Ron Graham’s book Magical Mathematics, but hadn’t read the book in detail before we tried to work the trick out for ourselves.

A de Bruijn sequence of length $k$ on an alphabet $A$ is a cyclic sequence which contains all possible strings of length $k$ exactly once. It follows that if you are given a string of $k$ letters, you can tell exactly where it occurs in the sequence. Since it doesn’t matter what the symbols in the alphabet $A$ are as long as they’re all different, we can just refer to a de Bruijn sequence $(k,n)$, containing all subsequences of length $k$ using the numbers $\{0,\dots,n-1\}$.

We decided we were going to assign each card in a standard deck a 6-digit binary string (because $2^6 = 64 \gt 52$) and to arrange them in the order of a $(6,2)$ de Bruijn sequence. Then, if we asked someone to cut the deck, draw out six cards and tell us their colours, we could work out exactly which cards they had in their hand.

There can be more than one de Bruijn sequence for each choice of $k$ and $n$, so we could pick the one that was easiest to construct. For binary sequences, this rule always generates a de Bruijn sequence of length $k$: write $k-1$ zeroes followed by a $1$, then subsequent letters are given by $x_{i+6} := x_i+x_{i+1} \pmod 2$ — add two adjacent numbers to get the one six places along. Because addition modulo $2$ is the same thing as subtraction modulo $2$, the same trick works backwards: $x_i = x_{i+1} + x_{i+6}$.

So the only thing left to do was to assign binary strings to cards. I decided that four digits for face value and two for suit would do, since it lets you split the calculation up into two small steps. So the face values go from $0 = A$, to $12 = K$, and the suits go Diamonds, Clubs, Hearts, Spades. In a hand of six cards drawn from the deck, the sequence of red or black colours gives the value of the last card drawn.

The only problem is that there are twelve strings representing face values greater than $12$, and I don’t think it’s possible to construct a de Bruijn sequence where they all occur in one block. But equally, because there are $64$ six-digit binary strings, any sequence we used would have twelve unused strings at the end. So we had to make one kludge and manually swap “bad” strings in the usable part of the sequence with “good” ones at the end.

Now, decoding binary strings is pretty tedious work, and finding these unusable strings is especially so, so I quickly wrote a Python script to do the calculations for me. Here it is:

And here’s its output:

The Python script constructs the de Bruijn sequence, decodes it into a sequence of 64 cards (some of them virtual), finds bad cards in the deck of 52 and swaps them with good cards in the pile of twelve “castoffs” at the end, making sure the colour is preserved. It turns out there are eight bad cards in the deck that need swapping.

So, we only need to remember eight pairs of binary strings and a simple iterative rule which constructs the sequence in order to perform the trick. Not bad for an afternoon’s work!

References

Universal cycles for combinatorial structures by Chung, Diaconis and Graham

Products of Universal Cycles by Diaconis and Graham

Magical Mathematics: The Mathematical Ideas that Animate Great Magic Tricks by Diaconis and Graham

What’s Black and Red and Red All Over? by Colm Mulcahy

• #### Christian Perfect

Mathematician, koala fan, Aperiodical editor. Usually found paddling in the North Sea, or fiddling with computers.

### 3 Responses to “David’s de Bruijn sequence card trick”

1. Ian Thomson

Christian and David

I shudder to think what “koala fan” might mean.

However, I’m writing to thank you both for this webpage and its resources. I’m a performing magician, and this is a brilliant method for performing what could be an amazing effect. Something similar is described in Alex Stone’s “Fooling Houdini”, but the exact details are left out and from its description I doubt it’s as clean as the script you’ve put together here.
.
However, I doubt that many magicians would be able to follow the theory (unless they’ve learned binary arithmetic and sequence theory in one of their many holidays at Her Mayesty’s Pleasure). As well as being a magic-nerd I’m an ex-scientist, and am in IT, so I managed to grasp most of it with a little effort. But for me this is an advantage, as I can’t imagine any of my magic-nerd brothers wanting to perform it.

Anyway. Can I point out that, in performance (if you’re worried about that) you don’t need to ask for the colour of each card? Just the pattern of one colour would do. Plus if you could discern the colour spread of the six cards without asking any questions at all (invisible markings on the back, for example), that would be a miracle indeed. I’m working on the latter idea.

And that’s it. Thank you again for the work, ingenuity, mental abilities and skill behind putting this together. A 52-card De Bruijn beats all the other ideas I’ve seen for cleanliness, including Colm’s.

Cheers

Ian

2. Rob Stanley

Thanks for the code. I’ve modified it slightly to better suit my needs, and because I think you can take a few more shortcuts.

1. I’ve swapped the order to suit then face value.
2. A face value of 1 is the Ace, 2 the 2, and so on.
3. I’ve included 2x Jokers (counted as black).
4. I’ve rotated the sequence slightly.

This gives me an ordering for the deck of 52 cards of which 6 are in the castoffs. But the replacement rules (only 4, and done by hand!) are easier (also why I need the Jokers in the deck).

1. Impossible Clubs are face value – 13.
3. The ’14 of Diamonds’ is the 8 of Diamonds.
4. The ’15 of Diamonds’ is the Queen of Diamonds.

 #names of cards faces = ['Ace','2','3','4','5','6','7','8','9','10','Jack','Queen','King'] N_faces = faces.__len__() suits = ['Clubs','Spades','Diamonds','Hearts']

 #generate de bruijn sequence def debruijn(): sequence = [0,0,0,0,0,1] for i in range(6,64): sequence.append( (sequence[i-6]+sequence[i-5]) % 2 ) return sequence #get a string of binary digits representing an integer def binarise(n,length): return bin(n)[2:].zfill(length) def debinarise(digits): binary = ''.join([str(x) for x in digits]) return int(binary,2) #decode a six-digit binary string to a card def decode_sequence(seq): number = debinarise(seq[2:]) suit = debinarise(seq[:2]) return suit,number #turn a card into a six-digit binary string def encode_card(card): suit,number = card return ''.join([str(x) for x in binarise(suit,2)+binarise(number,4)]) #display a card's binary encoding and its name def show_card(card): suit,number = card if (suit==1 and (number == 0 or number > 13)): #unknown spades name = "Joker" elif (suit==0 and number > 13): name = "%s of %s" % (faces[number-14], suits[suit]) elif (suit==2 and number == 14): name = "%s of %s" % (faces[7], suits[3]) elif (suit==2 and number == 15): name = "%s of %s" % (faces[11], suits[3]) else: name = "%s of %s" % (faces[number-1], suits[suit]) return '%s: %s' % (encode_card(card), name) def show_cards(cards): for card in cards: print(show_card(card)) def rotate(l,n): return l[n:] + l[:n] sequence = rotate(debruijn(),2) sequence*=2 cards = [decode_sequence(sequence[i:i+6]) for i in range(0,64)] # The deck of cards consists of the first 52 deck = cards[:54] # Get the unused cards from the end of the 64-deck which have usable values castoffs = [(x,y) for (x,y) in cards[54:] if y=N_faces+1 or y == 0] toswap.sort() castoffs.sort() print "Cards:" for i in range(0,54): digits = sequence[i:i+6] card = decode_sequence(digits) print(show_card(card)) print "\nCastoffs:" show_cards(castoffs) 

print "\nTo remember:" print "1. Impossible Clubs are face value - 13." print "2. Impossible Spades are Jokers." print "3. The '14 of Diamonds' is the 8 of Diamonds." print "4. The '15 of Diamonds' is the Queen of Diamonds." show_cards(toswap) 

 Cards: 000100: 4 of Clubs 001000: 8 of Clubs 010000: Joker 100001: Ace of Diamonds 000011: 3 of Clubs 000110: 6 of Clubs 001100: Queen of Clubs 011000: 8 of Spades 110001: Ace of Hearts 100010: 2 of Diamonds 000101: 5 of Clubs 001010: 10 of Clubs 010100: 4 of Spades 101001: 9 of Diamonds 010011: 3 of Spades 100111: 7 of Diamonds 001111: 2 of Clubs 011110: Joker 111101: King of Hearts 111010: 10 of Hearts 110100: 4 of Hearts 101000: 8 of Diamonds 010001: Ace of Spades 100011: 3 of Diamonds 000111: 7 of Clubs 001110: Ace of Clubs 011100: Queen of Spades 111001: 9 of Hearts 110010: 2 of Hearts 100100: 4 of Diamonds 001001: 9 of Clubs 010010: 2 of Spades 100101: 5 of Diamonds 001011: Jack of Clubs 010110: 6 of Spades 101101: King of Diamonds 011011: Jack of Spades 110111: 7 of Hearts 101110: 8 of Hearts 011101: King of Spades 111011: Jack of Hearts 110110: 6 of Hearts 101100: Queen of Diamonds 011001: 9 of Spades 110011: 3 of Hearts 100110: 6 of Diamonds 001101: King of Clubs 011010: 10 of Spades 110101: 5 of Hearts 101010: 10 of Diamonds 010101: 5 of Spades 101011: Jack of Diamonds 010111: 7 of Spades 101111: Queen of Hearts

 Castoffs: 000001: Ace of Clubs 000010: 2 of Clubs 111000: 8 of Hearts 111100: Queen of Hearts 

To remember: 1. Impossible Clubs are face value - 13. 2. Impossible Spades are Jokers. 3. The '14 of Diamonds' is the 8 of Diamonds. 4. The '15 of Diamonds' is the Queen of Diamonds. 001110: Ace of Clubs 001111: 2 of Clubs 010000: Joker 011110: Joker 101110: 8 of Hearts 101111: Queen of Hearts