Ladybird riddle

There is exactly one ladybug in each cell of a 5 × 5 5\times 5 square, as shown in the figure. Then, each ladybug simultaneously flies into a randomly selected, adjacent square that shares an edge.

Is it possible that, after all the ladybugs have moved, there is still only one ladybug in each cell?

Yes No

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6 solutions

If a ladybird is originally staying on a red {\color{#D61F06}{\text{red}}} cell, then it must fly into a green {\color{#20A900}{\text{green}}} cell.

If a ladybird is originally staying on a green {\color{#20A900}{\text{green}}} cell, then it must fly into a red {\color{#D61F06}{\text{red}}} cell.

There are 13 13 \space red {\color{#D61F06}{\text{red}}} and 12 green 12\space {\color{#20A900}{\text{green}}} cells, so it is not possible, that after the ladybirds have moved, there will be only ladybird in each cell.

Moderator note:

This is related to round-trip puzzles where on some sort of grid (possibly with squares removed) the goal is to make a closed loop that travels orthoganally and passes through every square. An alternative style to the puzzle picks separate start and end squares. We have an example of this type of problem as one of our 100 Day Summer Challenge series .

The ladybug maneuver is still possible without a possible round trip -- some ladybugs can just switch places, for instance. However, the existence of a round trip does indicate a solution exists, since the every ladybug can just move to the next point on the path from their position.

Note also how the logic here is extremely similar to Jeannine Myer's solution of the first POTW of this set !

no they can it says not from corner to corner, it says edge so they can use edges to move or fly. So its possible and whoever made this answers this wrong.

Ismet Semih Uzuner - 3 years, 10 months ago

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I agree, and keep rereading the question to see if I misinterpreted it, but it definitely says "adjacent square" and "shares an edge." In which case the answer is definitely yes.

Ian McKay - 3 years, 10 months ago

To add on... The number of squares is odd, thus making the proportion of red to green cells uneven.

Jacob Yousif - 3 years, 10 months ago

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If there was a 6×6 square then this scenario was possible. But again in case of rectangles it will be impossible. Please correct me if I am wrong!

Kaushik Chandra - 3 years, 10 months ago

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If the number of cells per rectangle side is even, it is possible, methinks.

Kostis Angelis - 3 years, 10 months ago

If it was a 6 x 5 rectangle then it would be possible.

Wayne Robinson - 3 years, 10 months ago

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Yes. Divide the grid into 2x5 grids and force the insects to wander round them in a clockwise direction.

Other solutions possible.

Malcolm Rich - 3 years, 10 months ago

Nice problem! It took me a few minutes before I realized that there were only two kinds of squares for the purpose of the before/after mapping, and therefore that parity was all that mattered.

Mark Lama - 3 years, 10 months ago

For some, obviously wrong, reason, I thought no movement at all was an option.

Roos Dickson - 3 years, 10 months ago

It's also impossible to draw a closed, non-intersecting loop through all of them. On the squares where the dimensions permit this, there is such possibility of all ladybugs moving one way along such a loop.

Anton Dubovik - 3 years, 10 months ago

The puzzle does not stipulate one ladybird can make only one move to the adjacent square. After moving to the adjacent square, a lady bird can exchange square with the next ladybird. On that scenario all ladybirds can move from their original square, and finally there is one ladybird in each square.

Ananda Moonasingha - 3 years, 9 months ago
Thomas Donnelly
Aug 15, 2017

It is not possible if you have an odd number of squares. In this case, you can see that all the bugs can move clockwise (or counter-clockwise), except for the one in the central square, which must then double-up in an adjacent square.

That proves this particular arrangement does not work, but why is it a proof that no such arrangement exists ?

Maxime Augier - 3 years, 9 months ago
E S
Aug 15, 2017

There is an easier way to intuitively see the answer to this problem by visually splitting the big square into smaller 2x2 or 2x1 squares and moving all bugs in those squares either clockwise/counter clockwise or up/down. By splitting the bigger square into smaller ones so that a circle or half circle(swap zone) can be drawn inside of each them we could see that eventually we will end up with a single spot where a bug have nowhere to go.

Seigneur Pilatte
Aug 17, 2017

I did it the odd way, every move is a vector and the sum of all moves is 0 and it is impossible that 25 perpendicular vectors cancels out.

Excellent idea I really loved it

Sundara Narasimhan - 3 years, 9 months ago
Tom Van Lier
Aug 16, 2017

Every time a ladybug goes left, another one has to go right or vice versa. The same accounts for a ladybug going upwards or downwards.

Since there is an odd number of ladybugs, we can never find a solution with an equal number going right and left, and an equal number going up and down. We can however find a partial balance (horizontally or vertically), but the other direction will always fail to be balanced.

This is indeed neat.

Can you explain why:

Every time a ladybug goes left, another one has to go right or vice versa

Agnishom Chattopadhyay - 3 years, 9 months ago

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To fill the spot the ladybird left open.

Tom Van Lier - 3 years, 9 months ago

Since we know the random movements result in the same configuration (one ladybug per cell) this means the center of mass in the vertical and horizontal directions must be unchanged. Thus the number of shifts in the center of mass in one direction must exactly equal that of the opposite direction.

Alfredo Peguero - 3 years, 9 months ago
Kevin Fagen
Aug 15, 2017

Let all ladybugs that can move in one direction (let's say, to the upward) do so. This would work for a group of 4 rows by 5 columns, or 5 rows by 4 columns. The problem is that whatever row or column is on the end, can't go anywhere, because they have to move to an adjacent square. So as a partial solution, we know that a mass migration of a large group of adjacent ladybugs on this scale, can't be accomplished. The next step I would get to from there would be, is there a smaller group, like a square of 4x4, that could all move in the same direction?

But maybe the partial solution solution is not constructible, although the global solution is?

Agnishom Chattopadhyay - 3 years, 9 months ago

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