《Inferring Decision Trees Using the Minimum Description Length Principle*》
Information And Computation 80, 227-248(1989)

My difficulty is: how to get 18.170 bits when computing Coding Decision Tree Costs?
here are some relevant part in this article.

---------------the decision tree is:--------------

3rd page of article( page 229 on the top-right corner)

-----------------------------formula for computing cost-------------
10th page of article ( page 236 on the top-left corner)
The total cost for this procedure is thus:
$$L(n,k,b)=lo{g}_2(b+1)+lo{g}_2[C_n^k]$$

-----------------------the sequence relevant to the above decision tree-------------------------------
13th page of article ( page 239 on the top-right corner)

---

All above are trying to encode the whole decision tree,
and then send the decision tree from sender to receiver.

My difficulty is: how to get 18.170 bits mentioned above?

---

My understanding is:
Outlook: 2 bits
Humidity lg(3)bits
Windy lg(2)bit (not mentioned in article ,I just guess)

---

the whole sequence in this paper is :
1 Outlook 1 Humidity 0 N 0 P 0 P 1 Windy 0 N 0 P
then,
there are 8digits above, the following I guess may be wrong:

3 decicion nodes cost:3 bits
5 leaves cost: 5 bits

---

2+lg(3)+log(2)+3+5+?=18.17
?=5.585,
but how to get 5.585?
Although the following picture(encoding exception) has the number 5.585,
I guess it cannot be used as the explanation of above 5.585.

Could you tell me how to get “18.170 bits” mentioned in this article ?
Thanks very much!

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