Knight and Calculator Keypad Riddle: Find valid 10 digit numbers by the knight moves on calculator keypad. Visit every digit once only. Solve in 15 mins.
Knight and Calculator Keypad Riddle
A simple calculator can display ten digits 0 to 9 and so has 10 number keys as in the figure shown.
Press the buttons of the Keypad using the moves of the chess knight with each number digit key visited only once.
In this way, how many ten-digit numbers can you form?
The chess knight moves two steps straight and then one step perpendicularly across in the shape of an L. The knight in the figure starts its move from key 3 and can move on to three keys, 0, 4 or 8 by its two straight and one across, L shape movement.
Time to solve: 15 minutes.
Solution to the knight and calculator keypad riddle
To try possible moves, starting from each of the ten keys is time-consuming and tedious. Instead,
Conclusion 1:We'll analyze the nature of the keys regarding freedom of movement of the keys.
Freedom of movement of a key represents the number of keys the knight can jump to from the key being examined.
Let's show the calculator again for convenience.
Freedom of movement must be the key idea in solving this awkward puzzle. It represents directly the limitations of the knight's moves from a key.
For example, from number key 3 the knight can move to three positions, but starting from key 0 it can move only to 3 and 5. Degree of freedom of movement is 3 for the number key 3, but 2 for number key 0.
Analyze further to discover a key pattern related to freedom of movement of the knight.
Number keys 0, 1, 2, 4, 6, 7, 8 and 9 all have degree of freedom 2 and key 3 has degree of freedom 3. But the key pattern is,
Key pattern 1: From number key 5 the knight can move only to 0. It follows, number key 5 can be reached only from the key 0.
Make the important conclusion based on key pattern 1,
Conclusion 2: Key 5 must be the end point or the starting point of a valid sequence of jumps of the knight over the ten number keys.
This fixes our strategy,
Strategy: We'll explore only one of the two possibilities, the ten-digit sequence starting with '50'.
Reasoning: Whatever numbers we get by starting with '50', we'll get a valid number by just reversing each valid ten-digit number got by this strategy.
Time to explore the knight's moves with knowledge gained till now.
First, the knight moves from digit 5 to digit 0.
Where can it jump across from 0? Possibility is only 1. It can jump only to digit 3 not visited earlier. Number formed till now is '503'.
From digit 3, the knight can jump only onto digits 4 or 8, not visited earlier.
Let's take the digit 4 jump. Number formed '5034'.
From digit 4, you can take only one possible jump onto digit 9. Number formed, '50349'.
Again, from digit 9, the knight can jump onto only the digit 2. Number formed '503492'.
Repeat the moves this way and make the knight jump to each of the ten digits once. As a result, form the number '5034927618'. It is the first valid number and its reverse '8167294305' is the second valid number.
More possible valid ten-digit numbers
Go back to where you chose digit 4 instead of digit 8 and formed the number at that stage '5034'. You got two valid numbers from the choice.
Now you will explore instead, the jump onto digit 8 from digit 3, forming the number '5038'.
In the same way as before, easy to form the third valid number '5038167294', and its reverse, '4927618305'.
The only four possible valid ten-digit numbers that satisfy the knight jump conditions are,
'5034927618', '8167294305', '5038167294', '4927618305'.
Consider that discovering the two reversed numbers starting from digit 4 and digit 8 won't have been easy at all.
That's why instead of starting from any number digit key, we have started from the CRITICAL KEY of digit 5 thus restricting number of possibilities that resulted in easy discoveries.
This is systematic problem solving based on discovery and use of key patterns.
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