#visualmath — Public Fediverse posts

My solution BP 72: in the left boxes some regular (Archimedean) tilings in the Poincare disk model. In the right boxes, other hyperbolic tiling. An alternative simpler solution notes that in the right boxes the center of the circle is a vertex.

On the right there are tilings created by applying different Conway operators to the regular tiling {5, 4}. Images adapted from "Visual Illusions in the Hyperbolic Plane", Tuan Dung Do and Craig S. Kaplan, Bridges 2026 Conference Proceedings:
https://archive.bridgesmathart.org/2026/bridges2026-93.html

Perhaps those nice conference proceedings could be used to create some other Bongard Problems.

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 72: in the left boxes some regular (Archimedean) tilings in the Poincare disk model. In the right boxes, other hyperbolic tiling. An alternative simpler solution notes that in the right boxes the center of the circle is a vertex.

On the right there are tilings created by applying different Conway operators to the regular tiling {5, 4}. Images adapted from "Visual Illusions in the Hyperbolic Plane", Tuan Dung Do and Craig S. Kaplan, Bridges 2026 Conference Proceedings:
https://archive.bridgesmathart.org/2026/bridges2026-93.html

Perhaps those nice conference proceedings could be used to create some other Bongard Problems.

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution of the sub-Bongard problems:
1: Like my BP61, domino-tillable or not;
2: On the left letters with an even number of strokes;
3: Like my BP62, on the left there's at least an internal node with an even number of children;
4: On the left even permutations;
5: On the left an even number of white zones;
6: On the right all graph cycles are odd;
------
7: Platonic solids vs regular filled polygons;
8: On the left at least one circle passes through the center of another;
9: On the left convex point sets;
10: Unkots vs trifoil knots (not easy);
11: Few Cube nets vs few Octahedron nets;
12: On the left two rectangles cross each other in 4 points.

So my solution BP 71: on the left there are parity-related Bongard problems.

(To keep creation time reasonable, some sub-problems are adapted from Aaron BPs and other BPs).

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 70: every box contains the representation of a planar graph (and every box on the left has a corresponding box on the right with the same graph, to make this problem simpler to solve). The graphs on the left are embedded (using various algorithms) in ways that show their symmetries and regularity more, compared to ones in the right boxes. So the embeddings on the left are generally of "higher quality".

See also:
https://spupyrev.github.io/planar-vibe/gallery.html
https://spupyrev.github.io/planar-vibe/

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 69: the left boxes show all the 6 fundamental solutions to the 7 queens problem (that has 40 solutions in total).

The n-queens puzzle is the problem of placing n chess queens on an n*n chessboard so that no two queens threaten each other; thus, a solution requires that no two queens share the same row, column, or diagonal.

Below a visual solution too.

See also:
https://en.wikipedia.org/wiki/Eight_queens_puzzle

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 68: In the left boxes every edge belongs to at most one cycle (the graph is a cactus graph).

From Wikipedia:
"In graph theory, a cactus (sometimes called a cactus tree) is a connected graph in which any two simple cycles have at most one vertex in common. Equivalently, it is a connected graph in which every edge belongs to at most one simple cycle, or (for nontrivial cacti) in which every block (maximal subgraph without a cut-vertex) is an edge or a cycle."

Below a visual solution too.

See also:
https://en.wikipedia.org/wiki/Cactus_graph

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 67: all boxes show Alpha Shapes (a generalization of Convex Hulls) of a random (some uniform, some normal) distribution of (9-30+) 2D points. I've used the nice "alphahull" package in R Studio. In the left boxes the alpha parameter is lower (2.0 - 5), while in the right boxes it's higher (5.5+) or extremely large (100, so for such datasets it becomes a proper Convex Hull).

See also:
https://en.wikipedia.org/wiki/Alpha_shape

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 66: in the left boxes the total number of runs (sub-sequences of equal color) is odd.

I think this rule is easy and quick to verify, but less intuitive to find.

This problem is simpler to understand as a 4th of this sequence by Bongard, sequences like this help with less intuitive concepts:
https://oebp.org/present.php?bp=88
https://oebp.org/present.php?bp=89
https://oebp.org/present.php?bp=90

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 65: in the left boxes the largest (stroked) object contains the center of the box.

The auxiliary lines that could help finding the solution are the diagonals of each box. If you draw them, the solution becomes much simpler to spot. Below I put the visual solution too.

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution BP 64: this BP is based on the concept of strictly Y-monotone polygons. In some boxes there are more than one figure, so in the left boxes there's a strictly Y-monotone 'ensemble' of objects.

Below I put a visual solution too.

See also:
https://en.wikipedia.org/wiki/Monotone_polygon

#bongardproblem #mathpuzzle #puzzle #visualmath

My solution: in the left boxes the support for the smallest enclosing circle is two extreme objects, in the right boxes three objects.

So those minimal circles touch two dots on the left, and three on the right. It's a computational geometry concept. Below I put a visual solution.

See also:
https://en.wikipedia.org/wiki/Smallest-circle_problem

#bongardproblem #mathpuzzle #puzzle #visualmath

This was a simple warming-up problem, mostly based on perception.

My solution: in the left boxes every tree node either has zero children (it's a leaf) or an odd number of children. I attach the visual solution too.

#bongardproblem #mathpuzzle #puzzle #visualmath

In this Bongard problem the first box is a hint, it shows "domino tiles", that is two cells aligned. It shows that you can cover it fully, without leaving empty spaces and without getting outside the border. The other 11 boxes should be solved in a similar way.

So my solution is: in the left boxes you can cover the grids with domino tiles (domino tileable). Unlike the grids in the right boxes.

Below I put an image that shows a solved choice of tilings for the left boxes. How can you be sure the ones on the right can't be covered by tiles? Every 2-tile covers exactly 1 cyan and 1 white cell. So if a grid is solvable, it should have the same number of white an cyan cells. The ones on the right don't have this equal number. This proof strategy is based on parity.

Regarding the solution given by Sean Reid, the two solutions are similar but not equivalent. Below I put a small image with a grid that can be domino-tiled, but doesn't allow a Hamiltonian cycle.

See also:
https://en.wikipedia.org/wiki/Domino_tiling
https://en.wikipedia.org/wiki/Parity_(mathematics)

For fun see also "Domino tilings beyond 2D":
https://arxiv.org/abs/2507.22625

#bongardproblem #mathpuzzle #puzzle #visualmath

This problem was easy enough.

Solution to my BP 60: the left boxes contain both artificial and natural Floyd-Steinberg dithered images. The right boxes contain the logarithm of the modulus of the 2D FFT (dithered and converted to binary). And in this Bongard problem each right box corresponds to the ones on the left.

The left box contents:
1: human eye;
2: small white centered square;
3: uniform distribution of about 990 white pixels (it's the negative of box 7 of my BP n. 32);
4: blue noise;
5: Laves tiling [4.8.8], dual of one of the Archimedean tilings (negative of box 12 of my BP n. 7);
6: Aperiodic tilings of Penrose. Thin and thick rhombuses (negative of box 2 of my BP n.6).

#bongardproblem #mathpuzzle #puzzle #visualmath

Detailed contents of all boxes:
1: the first matrix gets attached before the second one to produce the 4x4 matrices of the left quadrants;
2: starting from the cross of the left quadrants, going down, you end at the crosses of the right quadrants;
3: in the right quadrants a group of dots replaces fractally each dot of the left quadrants;
4: the map is a 90 degrees clockwise rotation;
5: each quadrant has a 12x12 grid. The right quadrants have 3x the number of black spaces of the left quadrants;
6: the map is to move the left object until it touches the right one;
- - - -
7: if the vertical axis represents the map, it's a randomization of the object(s) of the left quadrants;
8: the distance between the segments of the top quadrants gets mapped to the thickness of the vertical segments of the bottom quadrants;
9: the number of extrema of the wavy lines of the top quadrants mapped to a sequence of integers;
10: the circles in bottom quadrants as alternative way to represent nesting, but two leaves are represented by a circle;
11: the two bottom quadrants represent with a staircase/segment the discrete/continue nature of the upper quadrants contents;
12: the arrows in the bottom quadrants point to the direction of increase of sides of the polygons above.

I am not sure what the global solution to Aaron BP 146 is. My guess is: the boxes on the right show more abstract/conceptual map functions, while the boxes on the left show straightforward map functions that are simpler to define precisely.

#bongardproblem #mathpuzzle #puzzle #visualmath

Solutions of all sub-problems:
1: concave objects / stars;
2: concave objects / C shaped objects;
3: generic filled quadrilaterals / filled squares;
4: generic polygons also with intersecting sides / regular star polygons;
5: objects with some holes / convex polygons with 6 holes;
6: generic objects / C shaped objects;
- - - -
7: objects with bilateral symmetry / objects with rotational symmetry;
8: concave objects / regular (convex) filled polygons;
9: empty regular polygons / C shaped objects;
10: non-square rhombuses / rectangles (oblongs and squares);
11: rod with 5 spaces and 2 up pegs / rod with 5 spaces and and 3 up pegs;
12: regular empty polygons / regular filled polygons.

So my solution to Aaron David Fairbanks BP 77 is: in sub-problems of the left boxes the rule that identifies the objects on the right is a specialization (denotes a subset) of the objects on the left. In the right boxes the objects on the right aren't a subset of the objects on the left (so if you want to identify both exactly you need two different rules).

#bongardproblem #mathpuzzle #puzzle #visualmath

Still easy enough.

Solution to Aaron David Fairbanks n. 58: like the precedent Bongard problem, in the upper part of every box there are little tables that define one or more boolean functions with one or two arguments, plus a symbol to denote them. The little black squares represent true values. In the boxes on the left the algebraic expression below (with variables) simplifies to the constant true. In the right boxes the expression can't be simplified to a true constant.

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 51: like my precedent Bongard problem, all boxes contain Venn diagrams that represent error correction codes Hamming(7,4), some of them have one extra parity bit (Hamming(8,4)). The boxes on the left have no errors, so all bits are correct. The boxes on the right have one or more wrong bits. The "???? ????" means that an error can't be corrected in an unambiguous way.

A note:
>Hamming codes have a minimum distance of 3, which means that the decoder can detect and correct a single error, but it cannot distinguish a double bit error of some codeword from a single bit error of a different codeword. Thus, some double-bit errors will be incorrectly decoded as if they were single bit errors and therefore go undetected, unless no correction is attempted. To remedy this shortcoming, Hamming codes can be extended by an extra parity bit. This way, it is possible to increase the minimum distance of the Hamming code to 4, which allows the decoder to distinguish between single bit errors and two-bit errors. Thus the decoder can detect and correct a single error and at the same time detect (but not correct) a double error.<

See also:
https://en.wikipedia.org/wiki/Hamming(7,4)
https://en.wikipedia.org/wiki/Hamming_code

Detailed boxes contents:
1: 10 0101 0100101;
2: 6 0110 1100110;
3: 14 0111 00011110;
4: 15 1111 1111111;
5: 9 1001 0011001;
6: 2 0100 10011001;
- - - -
7: 15 1111 1111111 (1 bit switched off);
8: 0 0000 0000000 (1 bit switched on);
9: 3 1100 0111100 (2 bit switched on);
10: 11 1101 10101010 (1 bit switched on);
11: 4 0010 01010101 (1 bit switched off);
12: 13 1011 01100110 (2 bit switched off).

#bongardproblem #mathpuzzle #puzzle #visualmath

One of the most laborious Bongard problems to create (despite I've used ideas and data from two different Twitter/X messages, a Mathstodon message, and a Wikipedia page). Solving it is much faster.

Solution to my BP 38: the boxes represent resistor nets, where all resistors are 1 Ohm. In the boxes on the left between the left and right terminals there are rational approximations, of various precision (like 22/7 and 377/120), of Pi Ohms. The circuits in the right boxes have other total resistance values.

With twelve 1 Ohm resistors the best you can do is 22/7 Ohms.

Ideas from:
https://mathstodon.xyz/@davidphys1/115479833494488141
https://x.com/shapoco/status/1984484578296545646
https://www.wolframcloud.com/obj/81d1cb66-9e1e-4121-ba54-721881fead2b

```
# Example solution for the box 3:
para = lambda *s: 1.0 / sum(1.0 / x for x in s)
ser = lambda *s: sum(s)
a = ser(para(1, 1), 1)
b = para(1, ser(1, 1))
c = ser(b, 1)
d = para(1, 1)
x = ser(1, 1)
r1 = a
r2 = 1
r3 = 1
ra = r2 + r3 + (r2 * r3) / r1
rb = r1 + r3 + (r1 * r3) / r2
rc = r1 + r2 + (r1 * r2) / r3
p = ser(para(rb, c), para(ra, d))
result = ser(x, para(rc, p))
print(result) # 3.1415929
```

See also:
https://en.wikipedia.org/wiki/Y-%CE%94_transform

#bongardproblem #mathpuzzle #puzzle #visualmath