Conway's Soldiers

Notation and definitions

Let the target square be labeled x 0 = 1 , and all other squares be labeled x n , where n is the number of squares away (horizontally and vertically, as in manhattan distance) from the target square. If we consider the starting configuration of soldiers, below the thick red line, we can assign a score based on the sum of the values under each soldier, (e.g., x 2 + 2 x 3 etc.) When a soldier jumps over another soldier, there are three cases to consider:

1. A Positive jump: this is when a soldier jumps towards the target square over another soldier. Let the value of the soldier's square be x n , then the total change in score after a positive jump is x n − 2 − x n − 1 − x n = x n − 2 ( 1 − x − x 2 ) .
2. A Neutral jump; this is when a soldier jumps over another soldier but remains an equal distance from the target square after his jump (should this be necessary). In this case the change in score is − x n − 1 .
3. A Negative jump: this is when a soldiers jumps over another into an empty square away from the target square. Here the change in score is x n + 2 − x n + 1 − x n = x n ( x 2 − x − 1 ) .

Choosing a value of x {displaystyle x}

Let us choose a value of x such that the change in score for any type of jump is never positive. This can be done by solving x 2 + x − 1 = 0 yielding x = 5 − 1 2 , − 5 − 1 2 . We choose the positive root, 5 − 1 2 = φ ≈ 0.61803 39887 … , as its absolute value is less than 1, which becomes useful later in the proof. Rearranging φ 2 + φ − 1 = 0 , we can see that:

φ 2 = 1 − φ [and multiplying by φ ;]

φ 3 = φ − φ 2

φ 4 = φ 2 − φ 3

etc...

Summing this to infinity causes all terms on the right hand side to cancel apart from the 1, i.e.,

∑ n = 2 ∞ φ n = 1

This can also be shown with the common ratio, where:

∑ k = 0 ∞ r k = 1 1 − r ( r < 1 )

When r = φ :

∑ k = 2 ∞ φ k = 1 1 − φ − φ − 1 = 1

Solutions

Let us take the first example, where the target square is in the first row above the red line. We now consider the maximum possible initial score, that is when every square has a soldier on it. The sum of the squares on the first row below the red line, is φ + 2 φ 2 + 2 φ 3 + … = φ + 2 ( φ 2 + φ 3 + φ 4 + … ) . [Drawing a diagram helps to visualise this]. In the next line down, every square is one further away from the target square, and so has value φ times the square above it, and so on for all the rows below the line.

Therefore, the total value of all the squares below the line is equal to:

S 1 = ( φ + 2 ( φ 2 + φ 3 + φ 4 … ) ) ( 1 + φ + φ 2 + φ 3 + … ) .

At this stage we observe that ∑ n = 2 ∞ φ n = 1 from above, and thus the above expression simplifies to:

S 1 = ( φ + 2 ) ( 1 + φ + 1 ) = ( φ + 2 ) 2 = 4 + 4 φ + φ 2 = 5 + 3 φ , using φ 2 = 1 − φ for the last step.

Therefore the sum of all the squares below the line when the target square is immediately above the line is 5 + 3 φ .

The next case we consider is when the target square is in the second row above the red line. In this case each square under the red line is one square further away from the target square than in the previous example, so the total score now is found by multiplying the total score we obtained by φ to give:

S 2 = φ ( 5 + 3 φ ) = 5 φ + 3 φ 2 = 5 φ + 3 ( 1 − φ ) = 3 + 2 φ .

Similarly:

S 3 = 2 + φ

S 4 = 1 + φ

S 5 = 1

Thus, we have shown that when the target square is five rows above the red line, the maximum possible original total score of all the soldiers is 1. With a finite number of soldiers the total will be less than one. Therefore, since any type of jump does not increase the total score, and the final score on all the soldiers must be at least 1 ( φ 0 = 1 and the scores on any other soldiers left), the fifth row cannot be reached with a finite number of soldiers originally below the line.

This completes the proof.

Q E D .