Ponder This Challenge - June 2026 - The Superhero Team Movies

A certain comic book publisher holds the copyright for a certain superhero team, and wishes to create a movie franchise based on them. To minimize superhero fatigue with the audience, the movie franchise is planned ahead with the goal of having the minimum number of action sequences which cover all the possible superhero combinations.

Each superhero movie is built in the same manner: First, one hero appears and participates in an action sequence. Next, another hero joins them and the next action sequence contains both heroes, and so on. A movie need not contain all the heroes on the team.

As an example, assume the superhero team contains the heroes Dinolady (D), Vacuum Cleaner (V), and MathMan (M). A possible list of movies is:

The first movie provides action sequences starring D solo, and D and V together. Mathematically speaking, {D} and {D,V}.

The second movie includes the scenes {V} and {M,V}.

The third movie has the scenes {M}, {D, M} and the grand finale of {D, M, V}.

In this example, we managed to cover all the possible superhero combinations without using any such combination twice, using a total of 7 combinations. In general this is not possible. With 4 superheroes there are 15 combinations total, but we will need at least 17 combinations spread across 6 films to cover them all (since we have 6 films and 4 superheroes, we'll have two solo action sequences starring a hero that already had a solo action sequence).

In order to write solutions compactly, we can avoid spaces and linebreaks and use "0" to signify a new movie begins. With these conventions, the above solution becomes "0DV0VM0MDV".

Your goal: Find an optimal solution for the case of $n=6$ heroes. Give your solution in the compacted format.

A bonus "*" will be given for finding an optimal solution for the case of $n=10$ heroes.

Solution

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  Solution

  The optimal solution strings are of length 102 for n=6 and 1898 for n=10 (this includes the "0" in the strings).

  Seeing this is simple combinatorics. For $n=6$, there are ${6\choose 3}=20$ subsets of 3 superheroes that needs to be covered. Each movie has at most one size-3 set of superheroes, so a minimum of 20 movies is required. In each of those movies, we have at least 3 superheroes, so we already have a string of size 80: 20 "0" and 60 letters per movie.

  There is a total of ${6\choose 4}=15$ size-4 subsets and again, each movie can contain only one, and each one requires adding an additional hero to the movie. So we need 15 more letters. The same goes for ${6\choose 5}=6$ and ${6\choose 6}=1$ for size 5 and 6 sets. So in total, the minimum length of the relevant string is $20+20\cdot 3+15+6+1=102$. Applied to $n=10$ the same reasoning yields 1898.

  To find the movies, one can consider the directed graph of subsets of ${1,2,\ldots,n}$ where we have an edge $A\to B$ is $B$ is obtained by adding a new element to $A$. This is a layared graph (each layer corresponding to some set cardinality) where every two adjacent layers form a bipartite graph. One can use the Hopcroft-Karp algorithm to obtain maximum matching between layers; the obtained paths are the movies (when a path does not start from the first layer, we can add the heroes up to the starting point in an arbitrary order).

  Specific concrete solutions:

  • 0FDB0FEB0FDC0FEC0FED0FDAB0FEAB0FCAD0ECAF0EDAF0FBCD0EBCF0DBEF0DCEF0FABCD0EABCF0DABEF0CADEF0BCDEF0ABCDEF

  • 0JHFDB0JIFDB0JHGDB0JIGDB0JIHDB0JHFEB0JIFEB0JHGEB0JIGEB0JIHEB0JHGFB0JIGFB0JIHFB0JIHGB0JHFDC0JIFDC0JHGDC0JIGDC0JIHDC0JHFEC0JIFEC0JHGEC0JIGEC0JIHEC0JHGFC0JIGFC0JIHFC0JIHGC0JHFED0JIFED0JHGED0JIGED0JIHED0JHGFD0JIGFD0JIHFD0JIHGD0JHGFE0JIGFE0JIHFE0JIHGE0JIHGF0JHFDAB0JIFDAB0JHGDAB0JIGDAB0JIHDAB0JHFEAB0JIFEAB0JHGEAB0JIGEAB0JIHEAB0JHGFAB0JIGFAB0JIHFAB0JIHGAB0JHFCAD0JIFCAD0JHGCAD0JIGCAD0JIHCAD0JHECAF0JIECAF0JGECAH0IGECAJ0IHECAJ0JGFCAH0IGFCAJ0IHFCAJ0IHGCAJ0JHEDAF0JIEDAF0JGEDAH0IGEDAJ0IHEDAJ0JGFDAH0IGFDAJ0IHFDAJ0IHGDAJ0JGFEAH0IGFEAJ0IHFEAJ0IHGEAJ0IHGFAJ0JHFBCD0JIFBCD0JHGBCD0JIGBCD0JIHBCD0JHEBCF0JIEBCF0JGEBCH0IGEBCJ0IHEBCJ0JGFBCH0IGFBCJ0IHFBCJ0IHGBCJ0JHDBEF0JIDBEF0JGDBEH0IGDBEJ0IHDBEJ0JFDBGH0IFDBGJ0HFDBIJ0HGDBIJ0JFEBGH0IFEBGJ0HFEBIJ0HGEBIJ0HGFBIJ0JHDCEF0JIDCEF0JGDCEH0IGDCEJ0IHDCEJ0JFDCGH0IFDCGJ0HFDCIJ0HGDCIJ0JFECGH0IFECGJ0HFECIJ0HGECIJ0HGFCIJ0JFEDGH0IFEDGJ0HFEDIJ0HGEDIJ0HGFDIJ0HGFEIJ0JHFABCD0JIFABCD0JHGABCD0JIGABCD0JIHABCD0JHEABCF0JIEABCF0JGEABCH0IGEABCJ0IHEABCJ0JGFABCH0IGFABCJ0IHFABCJ0IHGABCJ0JHDABEF0JIDABEF0JGDABEH0IGDABEJ0IHDABEJ0JFDABGH0IFDABGJ0HFDABIJ0HGDABIJ0JFEABGH0IFEABGJ0HFEABIJ0HGEABIJ0HGFABIJ0JHCADEF0JICADEF0JGCADEH0IGCADEJ0IHCADEJ0JFCADGH0IFCADGJ0HFCADIJ0HGCADIJ0JECAFGH0IECAFGJ0HECAFIJ0GECAHIJ0GFCAHIJ0JEDAFGH0IEDAFGJ0HEDAFIJ0GEDAHIJ0GFDAHIJ0GFEAHIJ0JHBCDEF0JIBCDEF0JGBCDEH0IGBCDEJ0IHBCDEJ0JFBCDGH0IFBCDGJ0HFBCDIJ0HGBCDIJ0JEBCFGH0IEBCFGJ0HEBCFIJ0GEBCHIJ0GFBCHIJ0JDBEFGH0IDBEFGJ0HDBEFIJ0GDBEHIJ0FDBGHIJ0FEBGHIJ0JDCEFGH0IDCEFGJ0HDCEFIJ0GDCEHIJ0FDCGHIJ0FECGHIJ0FEDGHIJ0JHABCDEF0JIABCDEF0JGABCDEH0IGABCDEJ0IHABCDEJ0JFABCDGH0IFABCDGJ0HFABCDIJ0HGABCDIJ0JEABCFGH0IEABCFGJ0HEABCFIJ0GEABCHIJ0GFABCHIJ0JDABEFGH0IDABEFGJ0HDABEFIJ0GDABEHIJ0FDABGHIJ0FEABGHIJ0JCADEFGH0ICADEFGJ0HCADEFIJ0GCADEHIJ0FCADGHIJ0ECAFGHIJ0EDAFGHIJ0JBCDEFGH0IBCDEFGJ0HBCDEFIJ0GBCDEHIJ0FBCDGHIJ0EBCFGHIJ0DBEFGHIJ0DCEFGHIJ0JABCDEFGH0IABCDEFGJ0HABCDEFIJ0GABCDEHIJ0FABCDGHIJ0EABCFGHIJ0DABEFGHIJ0CADEFGHIJ0BCDEFGHIJ0ABCDEFGHIJ

Solvers

• *King Pig (2/6/2026 3:51 PM IDT)
• *Daniel Chong Jyh Tar (2/6/2026 4:07 PM IDT)
• *Alper Halbutogullari (2/6/2026 4:37 PM IDT)
• *Stéphane Higueret (2/6/2026 5:19 PM IDT)
• *Bertram Felgenhauer (2/6/2026 6:41 PM IDT)
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• *Jack Saleeby (3/6/2026 6:21 AM IDT)
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• *Dawson Yao (3/6/2026 8:28 AM IDT)
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• *Jason Shaw (4/6/2026 11:48 PM IDT)
• *Bert Dobbelaere (5/6/2026 12:20 AM IDT)
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• *Stephen Ebert (5/6/2026 2:41 AM IDT)
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