Trades

Last modified: April 22, 2026

As in CenturionDEX v2, swaps in CenturionDEX v3 are governed locally by a constant-product rule. The essential difference is that, in CenturionDEX v3, the active liquidity is not constant over the entire price axis: it may change whenever the current price reaches an initialized tick.

Consequently, a swap must be executed interval by interval, where each interval is bounded by two consecutive initialized ticks. Within each such interval, the liquidity parameter is constant, so the trade is computed exactly as a constant-product step. If the requested trade would move the price beyond the current interval, only the maximal feasible part of the trade is executed there; then the boundary tick is crossed, the tick-dependent state variables are updated, and the remaining part of the trade continues in the next interval.

Let $\phi \in [0,1)$ denote the trading-fee rate, and let $p$ be the current price of token $X$ in units of token $Y$ . Let $i_c$ be the current tick index, so that

i_c=\left\lfloor \log_{1.0001}(p)\right\rfloor.

Let $\mathcal I \subset \mathbb Z$ be the set of initialized tick indexes. Define

i_\ell=\max\{i\in\mathcal I \mid i\le i_c\}, \qquad i_u=\min\{i\in\mathcal I \mid i> i_c\}.

The corresponding boundary prices are

p_\ell = 1.0001^{i_\ell}, \qquad p_u = 1.0001^{i_u}.

By construction, the current price lies in the active interval

p \in [p_\ell,p_u),

and there is no initialized tick strictly between $i_\ell$ and $i_u$ . Hence, the active liquidity is constant throughout this interval. Denote that local liquidity parameter by $L$ .

Within the interval $[p_\ell,p_u)$ , the virtual reserves are

x_v=\frac{L}{\sqrt{p}}, \qquad y_v=L\sqrt{p}.

The corresponding real balances available inside the current interval are

x_r=L\left(\frac{1}{\sqrt{p}}-\frac{1}{\sqrt{p_u}}\right), \qquad y_r=L\left(\sqrt{p}-\sqrt{p_\ell}\right).

The maximal real balances attainable inside the same interval are

x_{\max}=L\left(\frac{1}{\sqrt{p_\ell}}-\frac{1}{\sqrt{p_u}}\right), \qquad y_{\max}=L\left(\sqrt{p_u}-\sqrt{p_\ell}\right).

The quantities $x_r$ and $y_r$ represent, respectively, the currently available in-range balances of tokens $X$ and $Y$ . The differences

x_{\max}-x_r=L\left(\frac{1}{\sqrt{p_\ell}}-\frac{1}{\sqrt{p}}\right), \qquad y_{\max}-y_r=L\left(\sqrt{p_u}-\sqrt{p}\right)

measure how much additional effective input of tokens $X$ and $Y$ , respectively, can still be absorbed before the price reaches one of the interval boundaries.

We now analyze the four canonical swap modes.

Case 1: exact output in token $X$

Suppose the trader wants to receive an amount $\Delta x_{\mathrm{out}}>0$ of token $X$ . This trade moves the price upward.

\Delta x_{\mathrm{out}} \le x_r,

then the entire trade can be executed inside the current interval. The new virtual balance of token $X$ is

x_v' = x_v - \Delta x_{\mathrm{out}},

and therefore the updated price is

p'=\left(\frac{L}{x_v-\Delta x_{\mathrm{out}}}\right)^2.

The corresponding effective input in token $Y$ is

\Delta y_{\mathrm{eff}} = y_v'-y_v = L\sqrt{p'}-L\sqrt{p},

so the gross input paid by the trader is

\Delta y_{\mathrm{in}}=\frac{\Delta y_{\mathrm{eff}}}{1-\phi},

and the fee collected in token $Y$ is

\phi\,\Delta y_{\mathrm{in}}.

If instead

\Delta x_{\mathrm{out}} > x_r,

then the current interval does not contain enough token $X$ to satisfy the full request. In that case, the maximal feasible first step uses exactly the amount $x_r$ , so the price moves all the way to the upper boundary $p_u$ . Indeed,

x_v-x_r =\frac{L}{\sqrt{p}}-L\left(\frac{1}{\sqrt{p}}-\frac{1}{\sqrt{p_u}}\right) =\frac{L}{\sqrt{p_u}}.

Hence, after this partial execution, the updated price is precisely $p_u$ . The effective amount of token $Y$ absorbed in this first step is

L\left(\sqrt{p_u}-\sqrt{p}\right),

so the corresponding gross input is

\frac{L\left(\sqrt{p_u}-\sqrt{p}\right)}{1-\phi}.

The remaining unsatisfied output request is

\Delta x_{\mathrm{out}}-x_r.

At this point, the protocol crosses the upper initialized tick $i_u$ , updates the current tick index, the total active liquidity, and the fee-outside variables associated with $i_u$ , and then continues the trade in the next interval to the right.

Case 2: exact input in token $X$

Suppose now that the trader sends a gross amount $\Delta x_{\mathrm{in}}>0$ of token $X$ and wants to receive token $Y$ . This trade moves the price downward.

Since the fee is charged on the input token, the effective amount that participates in the swap is

\Delta x_{\mathrm{eff}}=(1-\phi)\Delta x_{\mathrm{in}}.

x_r+\Delta x_{\mathrm{eff}} \le x_{\max},

equivalently,

\Delta x_{\mathrm{eff}} \le x_{\max}-x_r,

then the trade is completed within the current interval. The new virtual balance of token $X$ is

x_v' = x_v + \Delta x_{\mathrm{eff}},

and the updated price becomes

p'=\left(\frac{L}{x_v+\Delta x_{\mathrm{eff}}}\right)^2.

The output amount of token $Y$ is then

\Delta y_{\mathrm{out}} = y_v-y_v' = L\sqrt{p}-L\sqrt{p'} = L\left(\sqrt{p}-\sqrt{p'}\right).

If instead

x_r+\Delta x_{\mathrm{eff}} > x_{\max},

then the current interval can absorb only the effective amount

\delta x_{\mathrm{eff}} = x_{\max}-x_r = L\left(\frac{1}{\sqrt{p_\ell}}-\frac{1}{\sqrt{p}}\right).

The corresponding gross input spent in this first step is

\delta x_{\mathrm{in}}=\frac{\delta x_{\mathrm{eff}}}{1-\phi},

and the fee collected is

\phi\,\delta x_{\mathrm{in}} = \frac{\phi}{1-\phi}\,\delta x_{\mathrm{eff}}.

After this partial execution, the virtual balance of token $X$ becomes

x_v+\delta x_{\mathrm{eff}} =\frac{L}{\sqrt{p}}+L\left(\frac{1}{\sqrt{p_\ell}}-\frac{1}{\sqrt{p}}\right) =\frac{L}{\sqrt{p_\ell}},

so the price reaches the lower boundary $p_\ell$ , as expected.

The remaining gross input that must still be processed is

\Delta x_{\mathrm{in}}-\delta x_{\mathrm{in}} = \Delta x_{\mathrm{in}}-\frac{x_{\max}-x_r}{1-\phi}.

The protocol then crosses the lower initialized tick $i_\ell$ , updates the current tick index, the total active liquidity, and the fee-outside variables associated with $i_\ell$ , and continues the remaining part of the trade in the next interval to the left.

Case 3: exact output in token $Y$

Suppose that the trader wants to receive an amount $\Delta y_{\mathrm{out}}>0$ of token $Y$ . This trade moves the price downward.

\Delta y_{\mathrm{out}} \le y_r,

then the entire trade can be executed inside the current interval. The new virtual balance of token $Y$ is

y_v' = y_v - \Delta y_{\mathrm{out}},

and therefore the updated price is

p'=\left(\frac{y_v-\Delta y_{\mathrm{out}}}{L}\right)^2.

The corresponding effective input in token $X$ is

\Delta x_{\mathrm{eff}} = x_v'-x_v = \frac{L}{\sqrt{p'}}-\frac{L}{\sqrt{p}},

so the gross input paid by the trader is

\Delta x_{\mathrm{in}}=\frac{\Delta x_{\mathrm{eff}}}{1-\phi},

and the fee collected in token $X$ is

\phi\,\Delta x_{\mathrm{in}}.

If instead

\Delta y_{\mathrm{out}} > y_r,

then the current interval does not contain enough token $Y$ to satisfy the full request. The maximal feasible first step consumes exactly the amount $y_r$ , so the price moves to the lower boundary $p_\ell$ . Indeed,

y_v-y_r =L\sqrt{p}-L\left(\sqrt{p}-\sqrt{p_\ell}\right) =L\sqrt{p_\ell}.

Thus the updated price is precisely $p_\ell$ . The effective amount of token $X$ required in this first step is

L\left(\frac{1}{\sqrt{p_\ell}}-\frac{1}{\sqrt{p}}\right),

and the corresponding gross input is

\frac{L\left(\frac{1}{\sqrt{p_\ell}}-\frac{1}{\sqrt{p}}\right)}{1-\phi}.

The remaining unsatisfied output request is

\Delta y_{\mathrm{out}}-y_r.

Case 4: exact input in token $Y$

Finally, suppose that the trader sends a gross amount $\Delta y_{\mathrm{in}}>0$ of token $Y$ and wants to receive token $X$ . This trade moves the price upward.

The effective amount that participates in the swap is

\Delta y_{\mathrm{eff}}=(1-\phi)\Delta y_{\mathrm{in}}.

y_r+\Delta y_{\mathrm{eff}} \le y_{\max},

equivalently,

\Delta y_{\mathrm{eff}} \le y_{\max}-y_r,

then the trade is completed within the current interval. The new virtual balance of token $Y$ is

y_v' = y_v + \Delta y_{\mathrm{eff}},

so the updated price becomes

p'=\left(\frac{y_v+\Delta y_{\mathrm{eff}}}{L}\right)^2.

The output amount of token $X$ is

\Delta x_{\mathrm{out}} = x_v-x_v' = \frac{L}{\sqrt{p}}-\frac{L}{\sqrt{p'}}.

If instead

y_r+\Delta y_{\mathrm{eff}} > y_{\max},

then the current interval can absorb only the effective amount

\delta y_{\mathrm{eff}} = y_{\max}-y_r = L\left(\sqrt{p_u}-\sqrt{p}\right).

The corresponding gross input spent in this first step is

\delta y_{\mathrm{in}}=\frac{\delta y_{\mathrm{eff}}}{1-\phi},

and the fee collected is

\phi\,\delta y_{\mathrm{in}} = \frac{\phi}{1-\phi}\,\delta y_{\mathrm{eff}}.

After this partial execution, the virtual balance of token $Y$ becomes

y_v+\delta y_{\mathrm{eff}} =L\sqrt{p}+L\left(\sqrt{p_u}-\sqrt{p}\right) =L\sqrt{p_u},

so the price reaches the upper boundary $p_u$ .

The remaining gross input that must still be processed is

\Delta y_{\mathrm{in}}-\delta y_{\mathrm{in}} = \Delta y_{\mathrm{in}}-\frac{y_{\max}-y_r}{1-\phi}.

The protocol then crosses the upper initialized tick $i_u$ , updates the current tick index, the total active liquidity, and the fee-outside variables associated with $i_u$ , and continues the remaining part of the trade in the next interval to the right.

Interval-by-interval execution

The previous four cases describe one local step of the CenturionDEX v3 swap algorithm. In each case, one of two things happens:

the trade is completed inside the current interval $[p_\ell,p_u)$ ; or
the trade consumes all liquidity available in the relevant direction inside that interval, reaches one of its boundaries, crosses the corresponding initialized tick, updates the tick-dependent state, and then continues in the adjacent interval.

Thus, a full trade is generally a concatenation of such local constant-product steps across consecutive initialized-tick intervals.

A final remark is important. In the discussion above, whenever the current interval is exhausted, we implicitly assume that there exists a next initialized tick in the direction of the trade. This is usually the relevant regime, but it need not always hold. If no further initialized tick exists in the required direction, then no liquidity is available beyond the boundary just reached. In that situation, the swap algorithm stops, and the order is only partially filled.

Worked Example

We now present a fully explicit example illustrating how the CenturionDEX v3 protocol operates at the level of ticks, active liquidity, fee growth, and position accounting.

Consider a CenturionDEX v3 pool with tokens

X=\mathrm{CTN}, \qquad Y=\mathrm{USDC},

trading fee

\phi=0.003,

and tick spacing equal to $1$ . Suppose the current price is

p_0=2500 \quad \text{USDC/CTN}.

Assume that exactly two liquidity providers have opened positions.

The first provider chooses a price interval whose lower and upper tick indexes are

i_1=76137, \qquad i_3=80151,

so the corresponding tick prices are

t_1 = 1.0001^{76137}\approx 2024.988581, \qquad t_3 = 1.0001^{80151}\approx 3025.099528.

Since the current price $p_0=2500$ lies inside the interval $[t_1,t_3]$ , this provider must deposit both tokens. Suppose that they choose to deposit $4$ CTN. Then, from the in-range formula for the real balance of token $X$ ,

x(p)=L\left(\frac{1}{\sqrt{p}}-\frac{1}{\sqrt{p_b}}\right),

we obtain

4 = L_1\left(\frac{1}{\sqrt{2500}}-\frac{1}{\sqrt{t_3}}\right),

and therefore

L_1 = \frac{4}{\dfrac{1}{50}-\dfrac{1}{\sqrt{3025.099528}}} \approx 2199.638149.

The corresponding USDC deposit is

y_1 = L_1\left(\sqrt{2500}-\sqrt{t_1}\right) \approx 2199.638149\,(50-\sqrt{2024.988581}) \approx 10998.469836.

Hence, the first provider opens the position

P_1=[t_1,t_3], \qquad L_1\approx 2199.638149,

with initial deposit

4 \text{ CTN} \qquad\text{and}\qquad 10998.469836 \text{ USDC}.

Now consider a second provider. Let their lower and upper tick indexes be

i_2=79029, \qquad i_4=81891,

so that

t_2 = 1.0001^{79029}\approx 2704.047169, \qquad t_4 = 1.0001^{81891}\approx 3600.005212.

Since the current price $p_0=2500$ satisfies $p_0<t_2$ , this second position is opened below its active interval. Therefore, the provider needs to deposit only token $X$ . Suppose they deposit $6$ CTN. Then, from the below-range formula,

x=L\left(\frac{1}{\sqrt{p_a}}-\frac{1}{\sqrt{p_b}}\right),

we obtain

6 = L_2\left(\frac{1}{\sqrt{t_2}}-\frac{1}{\sqrt{t_4}}\right),

and thus

L_2 = \frac{6}{\dfrac{1}{\sqrt{2704.047169}}-\dfrac{1}{\sqrt{3600.005212}}} \approx 2340.142070.

Hence, the second provider opens the position

P_2=[t_2,t_4], \qquad L_2\approx 2340.142070,

with initial deposit

6 \text{ CTN} \qquad\text{and}\qquad 0 \text{ USDC}.

At the initial price $p_0$ , we have

t_1 < p_0 < t_2,

so the current active interval is $[t_1,t_2]$ , and only the first position is active there. Therefore, the active liquidity at the initial state is

L_{\mathrm{tot}}=L_1.

Trade 1

Suppose now that a trader wants to buy $0.8$ CTN from the pool. Since the trader is receiving token $X$ , this is the exact-output-in- $X$ case.

At the initial price $p_0=2500$ , the virtual reserves of the active interval $[t_1,t_2]$ are

x_v=\frac{L_1}{\sqrt{p_0}} =\frac{2199.638149}{50} \approx 43.992763,

and

y_v=L_1\sqrt{p_0} =2199.638149\cdot 50 \approx 109981.907438.

The requested output amount is

\Delta x_{\mathrm{out}}^{(1)}=0.8.

Since the input fee is charged on token $Y$ , the required gross USDC input is given by the local exact-output formula

\Delta y_{\mathrm{in}}^{(1)} = \frac{\Delta x_{\mathrm{out}}^{(1)}\,y_v} {(1-\phi)\bigl(x_v-\Delta x_{\mathrm{out}}^{(1)}\bigr)}.

Substituting the numerical values, we obtain

\Delta y_{\mathrm{in}}^{(1)} = \frac{0.8\cdot 109981.907438} {0.997\,(43.992763-0.8)} \approx 2043.172761.

The trading fee charged in this swap is therefore

\phi\,\Delta y_{\mathrm{in}}^{(1)} = 0.003\cdot 2043.172761 \approx 6.129518.

After the trade, the virtual balance of token $X$ becomes

x_v' = x_v-\Delta x_{\mathrm{out}}^{(1)} \approx 43.992763-0.8 = 43.192763,

and hence the updated price is

p_1 = \left(\frac{L_1}{x_v'}\right)^2 = \left(\frac{2199.638149}{43.192763}\right)^2 \approx 2593.465733.

Thus, after Trade 1, the current price is

p_1\approx 2593.465733,

which still lies inside the interval $[t_1,t_2]$ .

Because only token- $Y$ fees were generated, the global fee-growth variable in token $Y$ becomes

f_g^Y = \frac{6.129518}{L_1} \approx \frac{6.129518}{2199.638149} \approx 0.002786603.

At this stage, all token- $X$ fee-growth variables remain zero.

Trade 2

Suppose now that another trader deposits

4000 \text{ USDC}

into the pool in order to buy CTN. This is the exact-input-in- $Y$ case.

At price $p_1$ , the real balance of token $Y$ available inside the current interval $[t_1,t_2]$ is

y_r = L_1\left(\sqrt{p_1}-\sqrt{t_1}\right) \approx 13035.513079.

The maximum real balance of token $Y$ that can be reached inside the same interval is

y_{\max} = L_1\left(\sqrt{t_2}-\sqrt{t_1}\right) \approx 15398.743781.

Since the fee is charged on the input token, the effective amount of USDC that would participate in the swap if the trade were executed entirely in the current interval is

(1-\phi)\cdot 4000 = 0.997\cdot 4000 = 3988.

Therefore,

y_r + (1-\phi)\cdot 4000 \approx 13035.513079 + 3988 = 17023.513079 > y_{\max}=15398.743781.

So the trade cannot be completed entirely inside the interval $[t_1,t_2]$ .

First step of Trade 2: from $p_1$ to $t_2$

The largest effective amount of USDC that the current interval can absorb is

y_{\max}-y_r \approx 15398.743781 - 13035.513079 = 2363.230702.

Hence, the gross USDC amount used in the first step is

b_1 = \frac{y_{\max}-y_r}{1-\phi} = \frac{2363.230702}{0.997} \approx 2370.341727.

The fee charged in this first step is

0.003\,b_1 \approx 0.003\cdot 2370.341727 \approx 7.111025.

The corresponding CTN output of this first step is

a_1 = L_1\left(\frac{1}{\sqrt{p_1}}-\frac{1}{\sqrt{t_2}}\right) \approx 0.892398.

As expected, after this first step the price reaches the upper boundary of the current interval:

p=t_2\approx 2704.047169.

At that moment, the protocol crosses the initialized tick $i_2$ . Since position $P_2$ begins at $t_2$ , the active liquidity is updated from

L_{\mathrm{tot}}=L_1

L_{\mathrm{tot}}=L_1+L_2 \approx 4539.780218.

Also, because tick $i_2$ has just been crossed from left to right, the outside-fee variable at that tick is updated to the current global fee growth:

f_o^Y(i_2)\leftarrow f_g^Y.

Before the crossing, the global fee growth is

f_g^Y = \frac{6.129518+7.111025}{L_1} \approx 0.006019419.

Therefore,

f_o^Y(i_2)\approx 0.006019419.

Second step of Trade 2: from $t_2$ to $p_3$

After the first step, the remaining gross USDC input is

b_2 = 4000-b_1 \approx 4000-2370.341727 = 1629.658273.

Its effective amount is

(1-\phi)b_2 = 0.997\cdot 1629.658273 \approx 1624.769298.

Now the trade continues in the interval $[t_2,t_3]$ , whose active liquidity is

L_1+L_2 \approx 4539.780218.

The maximum real USDC balance available in that interval is

y_{\max}^{(2)} = (L_1+L_2)\left(\sqrt{t_3}-\sqrt{t_2}\right) \approx 13621.389191.

Since

1624.769298 < 13621.389191,

the remaining part of the trade is fully completed inside $[t_2,t_3]$ .

At the start of this second step, the virtual balance of token $Y$ is

y_v=(L_1+L_2)\sqrt{t_2}.

After adding the effective USDC input, the new virtual balance is

y_v' = (L_1+L_2)\sqrt{t_2} + (1-\phi)b_2.

Therefore, the updated price is

p_3 = \left(\frac{y_v'}{L_1+L_2}\right)^2 = \left(\sqrt{t_2}+\frac{(1-\phi)b_2}{L_1+L_2}\right)^2 \approx 2741.396770.

The CTN output of the second step is

a_2 = (L_1+L_2)\left(\frac{1}{\sqrt{t_2}}-\frac{1}{\sqrt{p_3}}\right) \approx 0.596759.

Hence, the total CTN received in Trade 2 is

a_1+a_2 \approx 0.892398+0.596759 = 1.489157 \text{ CTN}.

The fee charged in the second step is

0.003\,b_2 \approx 0.003\cdot 1629.658273 \approx 4.888975.

Thus, after Trade 2, the global fee growth in token $Y$ becomes

f_g^Y = 0.006019419 + \frac{4.888975}{L_1+L_2} \approx 0.006019419 + 0.001076919 = 0.007096338.

At the end of the two trades, the current price is

p_3 \approx 2741.396770,

which lies in the interval

[t_2,t_3).

Table 5-4. Summary of price movements and fee-variable updates in Trades 1 and 2

Throughout this example, only token- $Y$ fees are generated, so all token- $X$ fee-growth variables remain equal to zero.

Step	Price movement	Fee collected (USDC)	$f_g^Y$	$f_o^Y(i_2)$	$f_a^Y(i_2)$	$f_b^Y(i_2)$
Initial state	$p_0\in[t_1,t_2)$	$0$	$0$	$0$	$0$	$0$
Trade 1	$p_0 \to p_1$	$6.129518$	$0.002786603$	$0$	$0$	$0.002786603$
Trade 2, step 1	$p_1 \to t_2$	$7.111025$	$0.006019419$	$0$	$0$	$0.006019419$
Tick cross	$t_2$	$0$	$0.006019419$	$0.006019419$	$0$	$0.006019419$
Trade 2, step 2	$t_2 \to p_3$	$4.888975$	$0.007096338$	$0.006019419$	$0.001076919$	$0.006019419$

Fees collected by the two providers

Since the final price satisfies

t_2 < p_3 < t_3,

position $P_1=[t_1,t_3]$ is active at the final price, and so is position $P_2=[t_2,t_4]$ .

Because all fees in the example were collected in token $Y$ , we compute only the token- $Y$ fees earned by each provider.

For the first provider, the fee growth inside the interval $[t_1,t_3]$ is

f_{\mathrm{in},1}^Y = f_g^Y - f_b^Y(i_1) - f_a^Y(i_3).

Since $t_1<p_3<t_3$ , and neither $i_1$ nor $i_3$ was crossed during the two trades, we have

f_b^Y(i_1)=0, \qquad f_a^Y(i_3)=0.

Hence,

f_{\mathrm{in},1}^Y=f_g^Y\approx 0.007096338,

and therefore the total token- $Y$ fees earned by the first provider are

F_1^Y = L_1\,f_{\mathrm{in},1}^Y \approx 2199.638149\cdot 0.007096338 \approx 15.609375 \text{ USDC}.

For the second provider, the fee growth inside $[t_2,t_4]$ is

f_{\mathrm{in},2}^Y = f_g^Y - f_b^Y(i_2) - f_a^Y(i_4).

At the final price we have $t_2<p_3<t_4$ , so

f_b^Y(i_2)=f_o^Y(i_2)\approx 0.006019419, \qquad f_a^Y(i_4)=0.

Therefore,

f_{\mathrm{in},2}^Y = 0.007096338 - 0.006019419 = 0.001076919,

and the second provider has earned

F_2^Y = L_2\,f_{\mathrm{in},2}^Y \approx 2340.142070\cdot 0.001076919 \approx 2.520143 \text{ USDC}.

Observe that

F_1^Y + F_2^Y \approx 15.609375 + 2.520143 = 18.129518 \text{ USDC},

which coincides with the total fees collected in the two trades:

6.129518 + 7.111025 + 4.888975 = 18.129518.

Impermanent loss of the first provider

We now compare the current value of each position with the value that the provider would have had by simply holding the originally deposited assets.

For the first provider, the initial deposit was

4 \text{ CTN} \qquad\text{and}\qquad 10998.469836 \text{ USDC}.

If those assets had been held outside the pool, then at the final price $p_3$ their value would be

W_1(p_3) = 4p_3 + 10998.469836 \approx 4\cdot 2741.396770 + 10998.469836 \approx 21964.056918 \text{ USDC}.

On the other hand, the actual balances of the first position at price $p_3$ are

x_1(p_3) = L_1\left(\frac{1}{\sqrt{p_3}}-\frac{1}{\sqrt{t_3}}\right) \approx 2.018457 \text{ CTN},

and

y_1(p_3) = L_1\left(\sqrt{p_3}-\sqrt{t_1}\right) \approx 16185.985532 \text{ USDC}.

Therefore, the value of the position itself, excluding fees, is

V_1(p_3) = x_1(p_3)\,p_3 + y_1(p_3) \approx 2.018457\cdot 2741.396770 + 16185.985532 \approx 21719.377409 \text{ USDC}.

The impermanent-loss amount is thus

W_1(p_3)-V_1(p_3) \approx 21964.056918 - 21719.377409 \approx 244.679509 \text{ USDC}.

Equivalently, the relative impermanent loss is

\frac{V_1(p_3)}{W_1(p_3)}-1 \approx -0.011140,

that is, approximately $-1.1140\%$ .

If the uncollected fees are added back, the total marked-to-market value becomes

V_1(p_3)+F_1^Y \approx 21719.377409 + 15.609375 = 21734.986784 \text{ USDC}.

Thus, even after including fees, the first provider remains below the value of passively holding the initial assets by approximately

21964.056918 - 21734.986784 = 229.070134 \text{ USDC}.

Impermanent loss of the second provider

The second provider initially deposited only token $X$ , namely

6 \text{ CTN}.

Therefore, if they had simply held their initial assets outside the pool, their value at price $p_3$ would be

W_2(p_3)=6p_3 \approx 6\cdot 2741.396770 \approx 16448.380622 \text{ USDC}.

At the final price $p_3$ , which lies inside the interval $[t_2,t_4]$ , the real balances of the second position are

x_2(p_3) = L_2\left(\frac{1}{\sqrt{p_3}}-\frac{1}{\sqrt{t_4}}\right) \approx 5.692386 \text{ CTN},

and

y_2(p_3) = L_2\left(\sqrt{p_3}-\sqrt{t_2}\right) \approx 837.527546 \text{ USDC}.

Hence, the value of the position itself, excluding fees, is

V_2(p_3) = x_2(p_3)\,p_3 + y_2(p_3) \approx 5.692386\cdot 2741.396770 + 837.527546 \approx 16442.616292 \text{ USDC}.

Therefore, the impermanent-loss amount is

W_2(p_3)-V_2(p_3) \approx 16448.380622 - 16442.616292 \approx 5.764330 \text{ USDC},

and the corresponding relative impermanent loss is

\frac{V_2(p_3)}{W_2(p_3)}-1 \approx -0.000350,

that is, approximately $-0.0350\%$ .

If the earned fees are included, then the total marked-to-market value becomes

V_2(p_3)+F_2^Y \approx 16442.616292 + 2.520143 = 16445.136435 \text{ USDC}.

Hence, even after including fees, the second provider remains below the hold benchmark by approximately

16448.380622 - 16445.136435 = 3.244187 \text{ USDC}.

Final interpretation

This worked example illustrates four distinct mechanisms of the CenturionDEX v3 protocol.

First, liquidity is organized interval by interval. Initially, only the first position is active because the current price lies in $[t_1,t_2)$ . After Trade 2 crosses tick $t_2$ , the second position becomes active and the local liquidity jumps from $L_1$ to $L_1+L_2$ .

Second, fees are accumulated globally per unit liquidity and are then decomposed relative to initialized ticks through the outside-fee variables. In this example, because both trades were CTN-buying trades paid in USDC, all fees were collected in token $Y$ , and tick $i_2$ played the key role in separating the fee share of the second provider from the fee share accrued before their range became active.

Third, the two providers experience very different economic outcomes. The first provider, whose position was active from the beginning, captured the majority of the fees, but also experienced a materially larger impermanent loss. The second provider earned fewer fees, yet their impermanent loss remained small because the final price stayed close to the lower boundary of their active range.

Finally, the example shows that fee income and impermanent loss are distinct phenomena. The fees generated by trading may partially offset the mark-to-market divergence between the LP position and the passive-hold benchmark, but they do not automatically eliminate it.

Trades

Case 1: exact output in token XXX

Case 2: exact input in token XXX

Case 3: exact output in token YYY

Case 4: exact input in token YYY

Interval-by-interval execution

Worked Example

Trade 1

Trade 2

First step of Trade 2: from p1p_1p1​ to t2t_2t2​

Second step of Trade 2: from t2t_2t2​ to p3p_3p3​

Table 5-4. Summary of price movements and fee-variable updates in Trades 1 and 2

Fees collected by the two providers

Impermanent loss of the first provider

Impermanent loss of the second provider

Final interpretation

Case 1: exact output in token $X$

Case 2: exact input in token $X$

Case 3: exact output in token $Y$

Case 4: exact input in token $Y$

First step of Trade 2: from $p_1$ to $t_2$

Second step of Trade 2: from $t_2$ to $p_3$