Killing by numbers: the mathematics of warfare

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Greece:  Leuctra (371 BC)

The city-states of Greece have constantly fought for supremacy in the region. A brief period of peace has come to an end, as the currently dominant Sparta has openly challenged the city of Thebes’ political position. For refusing to abolish the Boeotian Confederacy, a political union of city-states which Thebes heads, Sparta has declared war. The city of Thebes and her allies have amassed 7,200 foot-soldiers (called hoplites) and, lead by Epaminondas, they look over the fields of Leuctra, where entrenched in their camp rest 9,600 Spartan hoplites.

King Cleombrotus I, the Spartan king, is in a rather dour mood. Despite recent military success, there have been bad omens on the march to Leuctra, not least of which was that the animals brought for sacrifice to the gods were eaten by wolves. However, having sought council from his various generals, all of whom are certain of a Spartan victory, he calls on you, a strange traveller who bestows surprising insight into the future.

One-on-one combat and Lanchester’s linear law

As with all mathematical modeling, if we hope to obtain any meaningful results, we must understand the basic mechanics of the process we wish to simulate. In our case, we require better knowledge of ancient Greek warfare. The Greek city-state armies consisted primarily of hoplites. These men fought in the notorious phalanx formation: densely packed rows of men such that a wall of spears and shields faced the opposing army. Each army would march forward, thegreek_hoplite bravest and most experienced men at the front so that the advancing troops wouldn’t attempt to slow down or flee. Upon collision of the armies, shields shook, spears snapped, bones broke and soldiers tumbled. Eventually, after sustaining sufficient losses, one side’s formation would collapse, causing a mass retreat, and inevitably defeat.

A simple model for this style of combat, first devised by Lanchester, assumes that fighting between the forces occurs one-on-one. This means that every man fights only his opposite number, and all troops not fighting wait their turn to fill in the ranks at the front. Allowing troop numbers to be continuous in time $t$, the equations for the rate of change of the Spartan troops, $S(t)$, and the Theban troops, $T(t)$, can be written as

\frac{dS}{dt}=-K_T N,  \tag{1}\\
\frac{dT}{dt}=-K_S N. \tag{2}

Here, $N$ is the total number of troops fighting at any one time for either side. Given that Cleombrotus will form his phalanxes twelve rows deep—fairly standard for Greek hoplites—the front line will be $N = 9600/12 = 800$ men wide. To avoid being outflanked, Cleombrotus suspects that the Thebans will match this 800 wide frontline with a nine row phalanx.  A characteristic of battles in ancient Greece was that casualty numbers remained relatively low: we shall assume that once either army lacks the numbers to maintain six rows of 800 ($T<4800$ or $S<4800$), panic will set in and the army will flee.

The values $K_T$ and $K_S$ determine the fighting strength of each opposing army. If $K_T = 1$, it means that per unit time every engaged Theban kills a Spartan, while if $K_T =0$, no $T$ ever successfully kills an $S$. These Lanchester attrition rates do not have to be a constant: they could be time dependent (troops get fatigued as the battle continues) or dependent on $S$ and $T$ (being outnumbered may dampen your hopes, or alternatively make you fight harder).  However, for simplicity, we will take both $K_T$ and $K_S$ to be constant.

So, the question: how much better must the Thebans be at fighting than their Spartan counterparts to win?

Our coupled system is very easy to solve. Dividing (1) by (2), we get that

\frac{dS}{dT}= \frac{K_T}{K_S}. \tag{3}

Integrating the above, and using the initial values of S and T we get that

K_S (S(t)-9600) = K_T (T(t)-7200). \tag{4}

Notice that the above equation is a linear relationship between the number of troops and their combat efficiency (known as Lanchester’s linear law). This is due to our one-on-one combat assumption. To ensure a Theban victory, it would be required that, for some time $t^*$, $S(t^*) = 4800$ and $T(t^*)>4800$.  Substituting $S(t) =4800$ into equation (4), re-arranging for $T$ and then setting $T(t)>4800$, we find that the Thebans are victorious if

K_T > \frac{4800}{2400} K_S = 2 K_S.


A graph showing how the battle plays out for various $K_T/K_S$. The dotted line is at 4,800, at which point the army flees and is defeated.


Relaying this information to Cleombrotus, you can’t help but notice his confidence: Spartans are the most revered warriors of their time, and despite his allies not fielding the highest quality troops, there’s nothing to suggest that they are two times ‘weaker’ than Thebe’s Boeotian allies.

Ranged combat: aimed fire and Lanchester’s square law

Although the bulk of Greek combat was fought between hoplites, both armies did field some lighter troops (commonly non-Greek mercenaries), known as peltasts. They carried javelins and slingshots which could be thrown at the opposing army. They were generally used to harass flanks, and apart from a few special cases, did not play a vital role. Still, let’s see what happens when we pitch Sparta’s 500 peltasts against Thebe’s 1,000.

The model we could use for such a scenario is known as Lanchester’s aimed fire model. The equations for the rate of change of the Spartan peltasts, $P(t)$, and the Theban peltasts, $Q(t)$, are given by:

\frac{dP}{dt}=-\alpha_Q Q,  \tag{5}\\
\frac{dQ}{dt}=-\alpha_P P. \tag{6}

This is because combat between javelin throwers is no longer one-to-one. Rather, all units are firing at the opposition at the same time. Hence, the death rate for $P$ is equal to the number of $Q$ firing at them, multiplied by the attrition rate $\alpha_Q$. Once again the variable $\alpha_Q$ does not need to be constant. A natural choice could be that it is proportional to $P$, the number of targets there are for $Q$ to shoot (a model used for unaimed fire). Like before, however, we shall keep them constant.

Dividing (5) by (6), we get that

\frac{dP}{dQ}= \frac{\alpha_Q}{\alpha_P} \frac{Q}{P}

Solving via seperation of variables, and using initial data, we find that

\alpha_P ( P(t)^2 – 500^2) = \alpha_Q  (Q(t)^2 – 1000^2)

The linear relation between numbers and efficiency is replaced with a quadratic relation. This result is the famous Lanchester’s square law. From this equation we see that quantity is more important than quality: for a draw in the above engagement, the Spartan peltasts, who are outnumbered two to one, would have to be four times as effective ($\alpha_P / \alpha_Q > 4$). These equations were derived in 1916, during World War I, which may explain some of the military thinking of the time.


The day of the battle


A comparison of what the Spartans expected (top image) and what actually happened. The red symbolizes the elite troops of either army: it was tradition to put your best troops on your right flank to stop your army drifting to the right, a habit caused by the shield being strapped to the left arm. Epaminondas placed his on the left flank so that the elite Spartans could be defeated before his inferior numbers became a problem.

After breakfast, and a splash of wine, Cleombrotus and his army march onto the flat planes of Leuctra. The king and his fiercest warriors take the position of honour on the right flank. In the distance, the Thebans can be seen through a haze of dust as they approach. A skirmish begins between both armies’ peltasts, while the hoplites close the gap between their front lines. As the Thebans get closer, Cleombrotus notices something strange: breaking convention, the Thebans have stacked a 50 man deep column on the Spartan right, with the rest of their army spread thin along the rest of the line. Furthermore, this weaker section of the army is approaching slower than the head of the main column. The 50 deep Theban mass crashes into Cleombrotus’ position. At first the depth of the column did little to help: fighting at the front is fierce and both sides take casualties, obeying the linear law of one-on-one combat. However, as the battle continues, it becomes clear that the 50 deep column cannot be beaten by a mere 12 rows of Spartans. Watching the Spartan elite slowly diminish, and eventually flee, Sparta’s allies decide to follow suit and run from the battle as well, despite still facing inferior numbers in front of them. King Cleombrotus is killed, as are many Spartans, and Sparta’s dominance over Greece comes to a bloody end.


A note on the model

The Lanchester equations are a very simple example of a mathematical model of populations. Similar equations are used to model many other predator-prey interactions (the classic example being foxes and rabbits). You can of course have more than two different ‘species’, and the species do not necessarily have to be living organisms.

The equations seem far too simple to have any true reflection on warfare. The most glaring weakness is the Lanchester attrition rates. Boiling an army’s ability to fight down to a single coefficient, with no spatial or time dependence, is a rather hard assumption to swallow. Furthermore, the model requires both armies to be homogeneous (ie for all the troops to be treated as the same). This was part of the brilliance of Epaminondas’ plan: he did not see 9,600 identical enemy hoplites, but a small force of Spartans and a larger force of less enthusiastic allies. By “crushing the head of the serpent”, Epaminondas ensured his inferior numbers never became a problem. As JK Anderson puts it, a sizeable part of both armies were “probably little more than spectators”.

Despite the model’s weaknesses, the square law does reveal interesting features about the aimed fire model over one-on-one combat. A smaller army engaged in aimed-fire can defeat a larger army by breaking the larger force into two, and dealing with each smaller chunk separately. Such a tactic would fail to work in combat obeying the linear law: had Sparta’s allies not fled the field, maybe the history books would be a little different!


Taking two forces with equal Lanchester attrition rates fighting under the aimed fire model, we see that a smaller force of 750 (red $R$) can successfully defeat a large force of 1,000 (blue $B$) if they fight the blue army in two separate battles (right-hand side), while suffering a heavy defeat if they meet the full blue army (left-hand side).

Further reading:

On the history: There is plenty available online discussing both the battle itself and the events leading up to it. “Military Theory and Practice in the age of Xenophon” by JK Anderson is the definitive book regarding the military side of things.

On the maths: A cool post applying Lanchester’s equations to dog fighting in World War II is available here

[Pictures: Banner:  Wikimedia Commons CC BY 2.5;  Hoplites: Public domain. Other images: Chalkdust.]



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Alex Doak is a PhD student at UCL researching fluid mechanics.
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