No, no, no!
The circuit shows the diode in PARALLEL with the load, but in reverse! That is not in series..
Do the circuit exactly like its drawn, only make sure you get the pins of the 317 right. I think (not sure) they are drawn in the wrong order (compared to where they really are). There is another pic with the elements pictures, and there it should be right.. Check the datasheet for the correct pinout!
About the diode..
Usually they are placed BEFORE a circuit, in series, to protect from reverse polarity.. But when the polarity is right, they drop the voltage, so you need a higher imput voltage..
In order for this circuit to work from 6V up, the diode was put in parallel with the load but in reverse, which is a "dirty" solution to the problem, as it shorts the battery when the polarity is wrong, but at least it doesn't drop the voltage, when the polarity is right.
This means the diode does NOTHING when the polarity is right.. It's just there to protect the LD from accidents.
EDIT: If you're not sure, TEST THE CIRCUIT WITH SOME LEDs! They are cheap and hard to kill. Then power the circuit down, SHORT the capacitor (if it's not yet on the LD) and then connect the LD, with the current set to minimum (pot - max resistance), and slowly turn the pot to achieve the desired current..
You also need a way to measure the current. A 1 Ohm resistor in series with the LD is great for this, as you don't have to disconnect anything - less chance of killing the LD and it doesn't affect the circuit (the LD will still get the same current, only the imput voltage needs to be slightly higher, but it's hardly noticable). We discussed this a couple of pages ago. Measure the voltage drop across the 1 Ohm resistor, then mV = mA..
EDIT EDIT: Ground yourself, then ground youself, and when youre done grounding yourself, ground yourself! Then, before you touch the LD, ground yourself! Then ground yourself some more!