mathematics of party favors

True story: My wife bought a bunch of party favors for a family gathering. These were rolled up gift wrapping containing some cheap present. She opened on and found a party game labeled “The Mystery Calculator”. She gave it to me.

15927953450_07ec905b2d_o
This is what it is: A small folded cardboard mini-booklet of 6 cards of 4×8 arrays of numbers, ranging from 1 to 63.

Page 1:

\begin{array}{rrrrrrrr}  1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\  17 & 19 & 21 & 23 & 25 & 27 & 29 & 31 \\  33 & 35 & 37 & 39 & 41 & 43 & 45 & 47 \\  49 & 51 & 53 & 55 & 57 & 59 & 61 & 63  \end{array}

Page 2:

\begin{array}{rrrrrrrr}  2 & 3 & 6 & 7 & 10 & 11 & 14 & 15 \\  18 & 19 & 22 & 23 & 26 & 27 & 30 & 31 \\  34 & 35 & 38 & 39 & 42 & 43 & 46 & 47 \\  50 & 51 & 54 & 55 & 58 & 59 & 62 & 63  \end{array}

Page 3:

\begin{array}{rrrrrrrr}  4 & 5 & 6 & 7 & 12 & 13 & 14 & 15 \\  20 & 21 & 22 & 23 & 28 & 29 & 30 & 31 \\  36 & 37 & 38 & 39 & 44 & 45 & 46 & 47 \\  52 & 53 & 54 & 55 & 60 & 61 & 62 & 63  \end{array}

Page 4:

\begin{array}{rrrrrrrr}  8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\  24 & 25 & 26 & 27 & 28 & 29 & 30 & 31 \\  40 & 41 & 42 & 43 & 44 & 45 & 46 & 47 \\  56 & 57 & 58 & 59 & 60 & 61 & 62 & 63  \end{array}

Page 5:

\begin{array}{rrrrrrrr}  16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \\  24 & 25 & 26 & 27 & 28 & 29 & 30 & 31 \\  48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\  56 & 57 & 58 & 59 & 60 & 61 & 62 & 63  \end{array}

Page 6:

\begin{array}{rrrrrrrr}  32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 \\  40 & 41 & 42 & 43 & 44 & 45 & 46 & 47 \\  48 & 49 & 50 & 51 & 52 & 53 & 54 & 55 \\  56 & 57 & 58 & 59 & 60 & 61 & 62 & 63  \end{array}

Here are the rules to the game.

Show the 6 cards to a friend. Ask him or her secretly pick any number from any card. Tell them to look through the 6 cards and find those cards that the number they selected also appears, and give those cards to you. You add the numbers in the upper left-hand corner of these cards. This sum will be the number your friend selected.

I haven’t proven this works but my guess is that this is based on the mathematics of the Hamming code of length 6 (see this previous post and this one).

A curious conjecture about self-reciprocal polynomials

Let p be a polynomial
p(z) = a_0 + a_1z+\dots + a_nz^n\ \ \ \ \ \ a_i\in {\bf C},
and let p^* denote the reciprocal polynomial
p^*(z) = a_n + a_{n-1}z+\dots + a_0z^n=z^np(1/z).
We say p is self-reciprocal if p=p^*. For example,
1+2z+3z^2+2z^3+z^4
and
1+2z+3z^2+3z^3+2z^4+z^5
are self-reciprocal. Suppose
p(z) = a_0 + a_1z+\dots + a_dz^d+a_{d+1}z^{d+1}+a_dz^{d}+\dots +a_{1}z^{2d+1}+a_0z^{2d+2},\ \ \ \ \ \ a_i\in {\bf C},
or
p(z) = a_0 + a_1z+\dots + a_dz^d+a_{d+1}z^{d+1}+a_{d+1}z^{d+2}+a_dz^{d+3}+\dots +a_{1}z^{2d+2}+a_0z^{2d+3},\ \ \ \ \ \ a_i\in {\bf C},

The question is this: for which increasing sequences a_0<a_1<\dots a_{d+1} do the polynomial roots p(z)=0 lie on the unit circle |z|=1?

Some examples:

symmetric-increasing-coeff-plot1
This represents, when d=1 and a_0=1 and a_1=1.1 the largest a_{d+1} which has this roots property is as in the plot.

symmetric-increasing-coeff-plot2
This represents, when d=1 and a_0=1 and a_1=1.2 the largest a_{d+1} which has this roots property is as in the plot.

symmetric-increasing-coeff-plot3
This represents, when d=1 and a_0=1 and a_1=1.3 the largest a_{d+1} which has this roots property is as in the plot.

symmetric-increasing-coeff-plot4
This represents, when d=1 and a_0=1 and a_1=1.4 the largest a_{d+1} which has this roots property is as in the plot.

symmetric-increasing-coeff-plot5
This represents, when d=1 and a_0=1 and a_1=1.5 the largest a_{d+1} which has this roots property is as in the plot.

Conjecture 1: Let s:{\bf Z}_{>0}\to {\bf R}_{>0} be a ”slowly increasing” function.

  1. Odd degree case.
    If g(z)= a_0 + a_1z + \dots + a_dz^d, where a_i=s(i), then the roots of p(z)=g(z)+z^{d+1}g^*(z) all lie on the unit circle.
  2. Even degree case.
    The roots of
    p(z)=a_0 + a_1z + \dots + a_{d-1}z^{d-1} + a_{d}z^{d} + a_{d-1}z^{d+1} + \dots + a_{1}z^{2d-1}+a_0z^{2d}
    all lie on the unit circle.

The next conjecture gives an idea as to how fast a “slowly increasing” function can grow.

Conjecture 2:* Consider the even degree case only. The polynomial
p(z)=a_0 + a_1z + \dots + a_{d-1}z^{d-1} + a_{d}z^{d} + a_{d-1}z^{d+1} + \dots + a_{1}z^{2d-1}+a_0z^{2d}  ,
with a_0=1 and all a_i>0, is symmetric increasing if and only if it can be written as a product of quadratics of the form x^2-2\cos(\theta)x+1, where all but one of the factors satisfy 2\pi/4\leq \theta\leq 4\pi/3. One of the factors can be of the form x^2+\alpha x+1, for some \alpha \geq 0.

* It was pointed out to me by Els Withers that this conjecture is false.