10 May 2012

Comparing dose distributions: The gamma test

In my last post I discussed dose distribution comparison with dose difference and distance-to-agreement (DTA) tests. Another widely used and closely related method for comparing dose distributions is the gamma test.

The gamma test was first introduced by Low et al. in 1998 as a single metric that combined features of both dose difference and DTA, while performing robustly in the regions where those are prone to failure. Conceptually, gamma is very similar to dose difference and DTA, but combines them into an abstract metric resembling a distance (Eq. 1). In this way both dose difference and DTA are taken into account for every point compared (rather than either-or as previously discussed).

Eq. 1

Eq. 2

In the above equations I have used somewhat different notation than Low et al. in an attempt to make things slightly clearer.

If we wish to compare two dose distributions, e.g. a measured versus a calculated distribution, we will have a dose, Da(ra), in the first distribution at point ra, and a dose, Db(rb), at the corresponding point rb in the second distribution. The DTA condition is fulfilled when Da(ra) = Db(rb+r), where r is an arbitrary point a distance |r| away from rb. This condition defines an isodose contour in distribution b around point rb. Away from this contour the DTA, dDTA, is undefined. DTA is used with a threshold passing value, δDTA, e.g. 3mm. A DTA smaller than the threshold is considered passing for a simple DTA test. For gamma, δDTA is used to normalize the DTA value, such that a normal “passing” value would then be unity.

Dose difference is simply the difference of the two doses at the corresponding points: |Da(ra) = Db(rb)|. As with DTA, a pass/fail threshold, δDD, is used in the simple dose difference test, but is used to normalize the result in the gamma equation, such that the normal "passing" value would be unity.

We now have two components: normalized DTA and normalized dose difference. By squaring these values, adding, and taking the square root, we have a distance-like metric, Γ, shown in Eq. 1. Because DTA is only defined for values of r, such that Da(ra) = Db(rb+r), Γ is only defined when that condition is met (geometrically located along the DTA isodose contour).

Finally, the actual gamma index, γ, is determined by finding the minimum value of Γ by varying r. This essentially means traveling along the isodose contour and finding the point at which DTA is smallest.

The convention is for passing γ to be ≤ 1 and failing to be > 1. You will notice that a point yielding normalized DTA = 1 and normalized dose difference = 1 would now fail, since the corresponding γ would be √2.

What γ provides is a single value to evaluate, versus using separate tests and then considering both. As with DTA, γ presents challenges in efficient implementation (clearly Eq.'s 1 and 2 are not hand solvable).

Your comments (especially corrections) are appreciated.


Further reading:

  • D. A. Low, W. B. Harms, S. Mutic, and J. A. Purdy, A technique for the quantitative evaluation of dose distributions, Med. Phys. 25, 656 (1998); http://dx.doi.org/10.1118/1.598248


  1. The convention is for passing γ to be ≤ 1 and failing to be > 1. You will notice that a point yielding normalized DTA = 1 and normalized dose difference = 1 would now fail, since the corresponding γ would be √2.



    1. You could look at it to mean that the gamma index is a slightly stricter criterion. But it's also important to keep in mind that all of these tests have a level of arbitrariness. Why pass at 1 and not 1.1?

    2. Actually, the gamma index is a looser criterion than the traditional DTA/dose difference criterion. Any point that will pass either DTA or %DD will always pass a gamma test (as long as it is implemented correctly and accurately). Also, a point that would fail both DTA and %DD may STILL pass the gamma test.

      For example, see http://snag.gy/0zOSQ.jpg, which illustrates an example like the one you proposed above. The green line is gamma, and it points to the closest approach of the red dose distribution. At that point, the normalized distance (x-axis) is 1, and the normalized dose difference (y-axis) is 1, but gamma is 1.4 (like you stated). However, note that this situation would not be DTA=1 and %DD=1 under our old criteria, since DTA must be strictly horizontal, and %DD must be strictly vertical.

      The crux is that gamma index at a point is the minimum value taken out of of all the GAMMA values calculated at that point. See http://snag.gy/DXfGP.jpg to see an illustration. Also look at http://snag.gy/5U5XU.jpg to see a situation where both DTA and %DD fail, but gamma passes.

    3. I am trying to figure out why DTA or %DD fails in figure http://snag.gy/5U5XU.jpg. To me it looks like DTA <1 and %DD <1 this point should pass. Plz clarify. Thx

  2. In paragraph 4 when you say |Da(ra) = Db(rb)|, I suppose that was a typo since it makes more sense for the = to be -, right?

  3. Nice write up. Clears up some some stuff I glazed over in the original paper.

  4. I need to compare two dose distribution, one is calculated from TG-43 formalism and other is hand calculated manually using same position and angle( r and theta). I can calculate dose difference between two but not sure how d^2 dta( ra,rb+r). Both dose distribution are calculated based on same position(r) and angle(theta) so how do I calculate d^2 data(ra,rb+r).

    Thanks in advance

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