#visualmath — Public Fediverse posts

As in the past problems I've written Python scripts to find the problems and validate their solutions.

Solution to my BP 58: all nonempty boxes contain Nonograms. The boxes on the left, once solved, show grids that represent uppercase letters (the letters A T E F). The boxes on the right, once solved, show digits (8 5 2).

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

#bongardproblem #mathpuzzle #puzzle #visualmath

As in the past problems I've written Python scripts to find the problems and validate their solutions.

Solution to my BP 58: all nonempty boxes contain Nonograms. The boxes on the left, once solved, show grids that represent uppercase letters (the letters A T E F). The boxes on the right, once solved, show digits (8 5 2).

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

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 57: in the left boxes there are mini-Suduku (4x4 or 6x6) with exactly one solution. In the right boxes mini-Sudoku with two solutions.

Left boxes contents:

Box 1 problem:
3 0 1 0
0 1 3 0
0 2 0 3
0 0 0 0
- - - - -
Box 1 solution:
3 4 1 2
2 1 3 4
1 2 4 3
4 3 2 1

Box 2 problem:
0 0 6 0 4 0
0 4 0 0 1 0
2 1 0 5 0 0
0 0 3 0 0 4
6 0 0 0 0 0
0 0 0 0 0 3
- - - - -
Box 2 solution:
1 2 6 3 4 5
3 4 5 6 1 2
2 1 4 5 3 6
5 6 3 1 2 4
6 3 2 4 5 1
4 5 1 2 6 3

Box 3 problem:
0 0 0 4
1 4 0 3
0 3 0 0
0 0 0 2
- - - - -
Box 3 solution:
3 2 1 4
1 4 2 3
2 3 4 1
4 1 3 2

Right boxes contents:

Box 7 problem:
2 0 0 3
0 0 0 0
0 0 3 1
0 3 0 4
- - - - -
Box 7 solution A:
2 1 4 3
3 4 1 2
4 2 3 1
1 3 2 4
- - - - -
Box 7 solution B:
2 4 1 3
3 1 4 2
4 2 3 1
1 3 2 4

Box 8 problem:
0 0 0 0
0 0 0 1
2 1 3 0
4 3 0 0
- - - - -
Box 8 solution A:
1 2 4 3
3 4 2 1
2 1 3 4
4 3 1 2
- - - - -
Box 8 solution B:
1 4 2 3
3 2 4 1
2 1 3 4
4 3 1 2

Box 9 problem:
3 0 1 0 0 6
0 0 4 3 0 0
0 0 0 0 0 0
0 0 0 6 3 0
0 1 5 0 0 3
0 4 0 2 0 0
- - - - -
Box 9 solution A:
3 2 1 5 4 6
5 6 4 3 1 2
4 3 6 1 2 5
1 5 2 6 3 4
2 1 5 4 6 3
6 4 3 2 5 1
- - - - -
Box 9 solution B:
3 2 1 5 4 6
5 6 4 3 2 1
4 3 6 1 5 2
1 5 2 6 3 4
2 1 5 4 6 3
6 4 3 2 1 5

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 57: in the left boxes there are mini-Suduku (4x4 or 6x6) with exactly one solution. In the right boxes mini-Sudoku with two solutions.

Left boxes contents:

Box 1 problem:
3 0 1 0
0 1 3 0
0 2 0 3
0 0 0 0
- - - - -
Box 1 solution:
3 4 1 2
2 1 3 4
1 2 4 3
4 3 2 1

Box 2 problem:
0 0 6 0 4 0
0 4 0 0 1 0
2 1 0 5 0 0
0 0 3 0 0 4
6 0 0 0 0 0
0 0 0 0 0 3
- - - - -
Box 2 solution:
1 2 6 3 4 5
3 4 5 6 1 2
2 1 4 5 3 6
5 6 3 1 2 4
6 3 2 4 5 1
4 5 1 2 6 3

Box 3 problem:
0 0 0 4
1 4 0 3
0 3 0 0
0 0 0 2
- - - - -
Box 3 solution:
3 2 1 4
1 4 2 3
2 3 4 1
4 1 3 2

Right boxes contents:

Box 7 problem:
2 0 0 3
0 0 0 0
0 0 3 1
0 3 0 4
- - - - -
Box 7 solution A:
2 1 4 3
3 4 1 2
4 2 3 1
1 3 2 4
- - - - -
Box 7 solution B:
2 4 1 3
3 1 4 2
4 2 3 1
1 3 2 4

Box 8 problem:
0 0 0 0
0 0 0 1
2 1 3 0
4 3 0 0
- - - - -
Box 8 solution A:
1 2 4 3
3 4 2 1
2 1 3 4
4 3 1 2
- - - - -
Box 8 solution B:
1 4 2 3
3 2 4 1
2 1 3 4
4 3 1 2

Box 9 problem:
3 0 1 0 0 6
0 0 4 3 0 0
0 0 0 0 0 0
0 0 0 6 3 0
0 1 5 0 0 3
0 4 0 2 0 0
- - - - -
Box 9 solution A:
3 2 1 5 4 6
5 6 4 3 1 2
4 3 6 1 2 5
1 5 2 6 3 4
2 1 5 4 6 3
6 4 3 2 5 1
- - - - -
Box 9 solution B:
3 2 1 5 4 6
5 6 4 3 2 1
4 3 6 1 5 2
1 5 2 6 3 4
2 1 5 4 6 3
6 4 3 2 1 5

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 56: in the left boxes there are mini-Suduku (4x4 or 6x6) that have exactly one solution. In the right boxes Sudoku with no solutions.

Left boxes contents:

Box 1, already solved:
2 1 3 4
4 3 1 2
3 4 2 1
1 2 4 3

Box 2:
3 4 0 0
0 2 0 0
0 0 1 0
4 0 0 2
- - - - -
Box 2 solution:
3 4 2 1
1 2 4 3
2 3 1 4
4 1 3 2

Box 3:
0 6 0 0 0 2
0 0 3 0 0 5
4 5 0 0 0 0
0 0 0 0 4 0
2 0 5 0 3 1
6 0 0 0 0 0
- - - - -
Box 3 solution:
5 6 4 3 1 2
1 2 3 4 6 5
4 5 6 1 2 3
3 1 2 5 4 6
2 4 5 6 3 1
6 3 1 2 5 4

Box 4:
0 0 4 0
0 0 1 3
0 3 0 1
1 0 0 0
- - - - -
Box 4 solution:
3 1 4 2
2 4 1 3
4 3 2 1
1 2 3 4

Right boxes contents:

Box 7:
0 3 0 0
2 0 0 4
3 4 0 0
0 0 0 1

Box 8:
4 2 0 0
0 0 0 3
0 3 1 0
0 0 0 2

Box 9:
0 0 4 0 0 0
0 6 0 0 0 0
0 3 0 6 0 5
0 0 0 0 0 0
0 0 2 3 0 0
4 5 6 0 2 1

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 56: in the left boxes there are mini-Suduku (4x4 or 6x6) that have exactly one solution. In the right boxes Sudoku with no solutions.

Left boxes contents:

Box 1, already solved:
2 1 3 4
4 3 1 2
3 4 2 1
1 2 4 3

Box 2:
3 4 0 0
0 2 0 0
0 0 1 0
4 0 0 2
- - - - -
Box 2 solution:
3 4 2 1
1 2 4 3
2 3 1 4
4 1 3 2

Box 3:
0 6 0 0 0 2
0 0 3 0 0 5
4 5 0 0 0 0
0 0 0 0 4 0
2 0 5 0 3 1
6 0 0 0 0 0
- - - - -
Box 3 solution:
5 6 4 3 1 2
1 2 3 4 6 5
4 5 6 1 2 3
3 1 2 5 4 6
2 4 5 6 3 1
6 3 1 2 5 4

Box 4:
0 0 4 0
0 0 1 3
0 3 0 1
1 0 0 0
- - - - -
Box 4 solution:
3 1 4 2
2 4 1 3
4 3 2 1
1 2 3 4

Right boxes contents:

Box 7:
0 3 0 0
2 0 0 4
3 4 0 0
0 0 0 1

Box 8:
4 2 0 0
0 0 0 3
0 3 1 0
0 0 0 2

Box 9:
0 0 4 0 0 0
0 6 0 0 0 0
0 3 0 6 0 5
0 0 0 0 0 0
0 0 2 3 0 0
4 5 6 0 2 1

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 55: the nonempty boxes on the left contain patterns that don't change with time ("Still Life") for the Life cellular automata. In the right boxes there are patterns that generate a large number of active cells (the "acorn" pattern), or an unbounded number of active cells (the Gosper Glider Gun).

See also:
https://conways-game-of-life.fandom.com/wiki/Gosper_Glider_Gun

I'll offer few more Bongard problems based on games.

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 55: the nonempty boxes on the left contain patterns that don't change with time ("Still Life") for the Life cellular automata. In the right boxes there are patterns that generate a large number of active cells (the "acorn" pattern), or an unbounded number of active cells (the Gosper Glider Gun).

See also:
https://conways-game-of-life.fandom.com/wiki/Gosper_Glider_Gun

I'll offer few more Bongard problems based on games.

#bongardproblem #mathpuzzle #puzzle #visualmath

Each box seems to contain something like an IQ quiz. You don't need to solve all the quizzes correctly to find the overall Bongard problem solution.

Detailed box contents:
1: the dots rotate clockwise, circles rotate anticlockwise. Every row is one step ahead, solution=B (second figure from left);
2: a labyrinth with 4 possible paths to the exit, solution=A;
3: a little less easy, it isn't a simple rotation, solution=A;
4: the second column is a XOR between the column 1 and 3, solution=C (A xor is quite common in IQ tests, but sometimes they are broken);
5: scanning the grids by rows, the space between black pixels is a constant, that's different between grids, solution=C (the only one with a constant space between black pixels. Constant=2);
6: adapted from Aaron BP 82. In each sub-box three little crosses are approximately vertices of a equilateral triangle (and the circle and the other crosses are noise). Solution=A;
- - - -
7: Platonic solids, solution=B;
8: you can solve this with the Pick theorem, all figures above the line have a surface area of 10.5, solution=C, same area;
9: Schlegel diagrams of the Platonic solids, solution=C;
10: above the thick horizontal line there are the first four Perfect Numbers represented in Morse code: 6, 28, 496, 8128. Below the thick line there are other numbers, 15000, 335503, 33550336. The next Perfect Number is C.

So the solution to my Bongard problem 54 is: the boxes on the left contain fluid intelligence (gf) quizzes. The right boxes contain crystallized intelligence (gc) quizzes (because you need knowledge to solve them all).

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

#bongardproblem #mathpuzzle #puzzle #visualmath

Each box seems to contain something like an IQ quiz. You don't need to solve all the quizzes correctly to find the overall Bongard problem solution.

Detailed box contents:
1: the dots rotate clockwise, circles rotate anticlockwise. Every row is one step ahead, solution=B (second figure from left);
2: a labyrinth with 4 possible paths to the exit, solution=A;
3: a little less easy, it isn't a simple rotation, solution=A;
4: the second column is a XOR between the column 1 and 3, solution=C (A xor is quite common in IQ tests, but sometimes they are broken);
5: scanning the grids by rows, the space between black pixels is a constant, that's different between grids, solution=C (the only one with a constant space between black pixels. Constant=2);
6: adapted from Aaron BP 82. In each sub-box three little crosses are approximately vertices of a equilateral triangle (and the circle and the other crosses are noise). Solution=A;
- - - -
7: Platonic solids, solution=B;
8: you can solve this with the Pick theorem, all figures above the line have a surface area of 10.5, solution=C, same area;
9: Schlegel diagrams of the Platonic solids, solution=C;
10: above the thick horizontal line there are the first four Perfect Numbers represented in Morse code: 6, 28, 496, 8128. Below the thick line there are other numbers, 15000, 335503, 33550336. The next Perfect Number is C.

So the solution to my Bongard problem 54 is: the boxes on the left contain fluid intelligence (gf) quizzes. The right boxes contain crystallized intelligence (gc) quizzes (because you need knowledge to solve them all).

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

#bongardproblem #mathpuzzle #puzzle #visualmath

Like the precedent, this problem looks a little like an IQ quiz. Despite most IQ quizzes are boring. To solve this you just need some eye to spot a simple pattern. The original problem was a little harder because you had to find the strange figure mixed among the others, while here you just need to understand why the box 7 is different from the others.

Solution to my BP 53: in each of the seven boxes there are (handmade) abstract figures, each one has six independent binary characteristics (outer closed line is a circle or hexagon, mid closed line is a square or pentagon, inner closed line is a triangle or square, background color of the three outer levels. The innermost closed line has always the same filling). For each of the left boxes exactly one characteristic differs compared to the other six boxes. This isn't true for the figure in box 7, that is therefore the most standard of all seven figures.

Detailed box contents:
1: unlike all other six boxes the outer background is white;
2: unlike all other six boxes the mid closed line is a pentagon;
3: the innermost closed line is a square;
4: inside the outer closed line the background is white;
5: inside the mid closed line the background is white;;
6: the outer closed line is an hexagon;
- - - -
7: in this figure all six binary characteristics are the common ones.

If you represent this problem with binary strings you have: 000_001, 000_010, 000_100, 001_000, 010_000, 100_000, 000_000.

#bongardproblem #mathpuzzle #puzzle #visualmath

Like the precedent, this problem looks a little like an IQ quiz. Despite most IQ quizzes are boring. To solve this you just need some eye to spot a simple pattern. The original problem was a little harder because you had to find the strange figure mixed among the others, while here you just need to understand why the box 7 is different from the others.

Solution to my BP 53: in each of the seven boxes there are (handmade) abstract figures, each one has six independent binary characteristics (outer closed line is a circle or hexagon, mid closed line is a square or pentagon, inner closed line is a triangle or square, background color of the three outer levels. The innermost closed line has always the same filling). For each of the left boxes exactly one characteristic differs compared to the other six boxes. This isn't true for the figure in box 7, that is therefore the most standard of all seven figures.

Detailed box contents:
1: unlike all other six boxes the outer background is white;
2: unlike all other six boxes the mid closed line is a pentagon;
3: the innermost closed line is a square;
4: inside the outer closed line the background is white;
5: inside the mid closed line the background is white;;
6: the outer closed line is an hexagon;
- - - -
7: in this figure all six binary characteristics are the common ones.

If you represent this problem with binary strings you have: 000_001, 000_010, 000_100, 001_000, 010_000, 100_000, 000_000.

#bongardproblem #mathpuzzle #puzzle #visualmath

This was quite easy, and it's mostly noise. The next problem is a bit less easy, but still looking like a IQ test.

Solution to my BP 52: in all boxes the small objects are the inputs of a boolean function (0 if stroked, 1 if filled). The output is represented by the filling of the large object. So the boxes on the left represent AND and the boxes on the right represent OR.

#bongardproblem #mathpuzzle #puzzle #visualmath

This was quite easy, and it's mostly noise. The next problem is a bit less easy, but still looking like a IQ test.

Solution to my BP 52: in all boxes the small objects are the inputs of a boolean function (0 if stroked, 1 if filled). The output is represented by the filling of the large object. So the boxes on the left represent AND and the boxes on the right represent OR.

#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

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

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

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

Solution to my BP 50: in the left boxes inside every circle there's an even number of black dots (in the right boxes at least one circle has an odd number of black dots).

#bongardproblem #mathpuzzle #puzzle #visualmath

Solution to my BP 50: in the left boxes inside every circle there's an even number of black dots (in the right boxes at least one circle has an odd number of black dots).

#bongardproblem #mathpuzzle #puzzle #visualmath