This column’s topics:
Combinatorics in The New York Times.
Recent work by Neil J. A. Sloane and David Cutler addresses the “pancake-cutting” problem: If you cut a pancake $n$ times, how many pieces can you get? Of course, since this is mathematics, the pancake is infinitely large and the knife infinitely long. For a straight-edged knife the answer has been known for some time—$n$ cuts produce at most $(n^2+n+2)/2$ pieces. The new results examine “exotic” knives, made up of a finite number of straight edges. The only rule is that the whole cutting surface is connected. Siobhan Roberts reported on the work for The New York Times.
One of Sloane and Cutler’s exotic knives is shaped like a long-legged ${\sf A}$, with two blades going off to infinity and one strut between them. A single ${\sf A}$-knife divides the planar pancake into three parts, while two (once positioned and affinely adjusted) can yield 14 separate pieces.
According to Sloane and Cutler’s preprint, the ${\sf A}$-knife research began last summer, when Sloane spoke to undergraduates at Rutgers about the pancake-slicing problem. Cutler was in attendance, along with two other students, Edward Xiong and Jonathan Pei. Inspired by Sloane’s talk, the three students solved a version of the problem in which the knife is a tripod: three infinite blades radiating from one point. They proved that $n$ tripods can cut the plane into at most $(9n^2-5n+2)/2$ pieces.
Later, Cutler and Sloane experimentally studied ${\sf A}$-knives; their results exactly matched Xiong, Pei and Cutler’s formula for tripods. “This was not a coincidence!” they write. They went on to prove that for a given number of cuts, tripods and ${\sf A}$-knives always achieve the same maximum number of pancake pieces. Their paper also generalizes Xiong, Pei and Cutler’s result to $k$-pods, where the knife has $k$ infinite blades instead of 3. With $n$ cuts of a $k$-pod, one can create $\binom{n}{2}k^2 + n(k-1) +1$ pancake pieces.
The algebra of melody.
Last summer, the University of Waterloo’s Olga Ibragimova and Chrystopher Nehaniv published “Algebraic Applications in Investigation of Musical Symmetry,” a study on the tonal structure of melody. (February 19, Phys.org posted a press release, whose headline “Why some tunes stick: Mathematical symmetry helps explain catchy melodies” is not completely accurate: Ibragimova and Nehaniv’s article does not mention stickiness or catchiness.)
In this preliminary study, the authors do not consider the length of notes nor their exact pitches. Instead, each note is represented by a number locating its pitch class in the middle octave on a piano (C$=1$, C#$=2$, D$=3$, … , B$=12$). Mathematically, they think of an $n$-note melody as a function $m \colon L \mapsto C$. Here, $L$ is the space of locations 
The following familiar 8-note melody, with its numerical encoding, will be used to illustrate the musical transformations considered by Ibragimova and Nehaniv. Click “Play” on the sound-bar to hear it.
The authors consider four operations that transform a melody without changing what they call its “inherent properties.” These are transposition, translation, inversion, and retrograde.
- Transposition shifts all the notes in a melody up or down by a specified number of steps.
In this transposition, the notes in the sample melody are shifted up by 5 steps: $[1~5~1~12~3~6~3~1]$ becomes $[6~10~6~5~8~11~8~6]$. Notice that the addition is done in modular “clock arithmetic” where $12+5=5$. - Translation shifts all the notes in a melody backwards or forwards by a specified number of steps.
This translation shifts the notes in the sample melody back by 3 steps: $[1~5~1~12~3~6~3~1]$ becomes $[12~3~6~3~1~1~5~1]$. For melodies of length $n$ this translation is done modulo $n$, so that notes pushed off one end reappear at the other. - Inversion flips a melody upside-down. In this study, inversion moves pitch-class $k$ to pitch-class $13-k$.
The rule $k \rightarrow 13-k$ changes our sample melody $[1~5~1~12~3~6~3~1]$ to its inversion $[12~8~12~1~10~7~10~12]$. - Retrograde plays the original melody backwards.
Under the retrograde transformation, the sample melody $[1~5~1~12~3~6~3~1]$ becomes $[1~3~6~3~12~1~5~1]$.
Since each melody $m$ is a map from the locations 
Each of our four operations is a permutation of $C^n$ which preserves two things: the size of the jumps between notes, and their relative position in the sequence. These must be the “inherent properties” that the authors have in mind. If you compose two of these transformations, for example transposing and then reflecting, the new melody will retain its inherent properties. The four types of operations, and all their possible combinations, form a group ${\bf G}$, which we can think of as the group of musical permutations of $C^n$.
Ibragimova and Nehaniv point out that translation and retrograde keep the same notes, but rearrange the locations $L$, while transposition and inversion preserve $L$ and change the pitch-classes $C$. They explain that ${\bf G}$ is structurally the direct product of $G_L$ and $G_C$,$${\bf G} = G_L \times G_C,$$where $G_L$ is the group of symmetries of 
Moreover, $G_L$ is the “dihedral” group $D_n$ of symmetries of an $n$-gon (see the illustration below), and $G_C$ is $D_{12}$.
Given a transformation $g$ in ${\bf G}$, Ibragimova and Nehaniv construct a length-8 melody which is symmetric with respect to $g$. They choose $g = (R^2, r^3)$, where $R$ is translation forward by one position and $r$ is transposition up by one semitone. This means they want a melody which, when translated two steps to the right and transposed up three semitones, sounds the same as it did to start. They are free to choose the first two notes, which they take to be 1 (C) and 3 (D); their construction yields $[1~ 3~ 4~ 6~ 7~ 9~ 10~ 12]$.
—Tony Phillips, Stony Brook University