The Colorings Function for a Graph
Definition For the graph 𝐺, let 𝑪𝒐𝒍𝑮 (𝒌) be the number of distinct colorings of the
graph 𝐺 using at most 𝑘 colors.
𝐻
𝐶𝑜𝑙𝐻 (3) = 6
𝐽
𝐶𝑜𝑙𝐽 (3) = 12
NOTE: We are only counting distinct colorings. In general, this is much easier than trying to
count the number of non-isomorphic colorings.
Definition For the graph 𝐺, let 𝑪𝒐𝒍𝑮 (𝒌) be the number of distinct colorings of the
graph 𝐺 using at most 𝑘 colors.
Recall that 𝐾𝑛 denotes the complete graph on 𝑛 vertices.
Let 𝑁𝑛 denote the “null graph” on 𝑛 vertices (i.e. the graph with no edges.)
What is 𝑪𝒐𝒍𝑵𝒏 (𝒌)?
What is 𝑪𝒐𝒍𝑲𝒏 (𝒌)?
Definition For the graph 𝐺, let 𝑪𝒐𝒍𝑮 (𝒌) be the number of distinct colorings of the
graph 𝐺 using at most 𝑘 colors.
Recall that 𝐾𝑛 denotes the complete graph on 𝑛 vertices.
Let 𝑁𝑛 denote the “null graph” on 𝑛 vertices (i.e. the graph with no edges.)
𝑪𝒐𝒍𝑵𝒏 (𝒌) = 𝒌𝒏
𝑪𝒐𝒍𝑲𝒏 (𝒌) = 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐) ⋯ (𝒌 − 𝒏 + 𝟏)
Recall that 𝐿𝑛 denotes the line graph with 𝑛 vertices.
Claim:
𝑪𝒐𝒍𝑳𝒏 (𝒌) = 𝒌(𝒌 − 𝟏)𝒏−𝟏
Example 1a: What is 𝐶𝑜𝑙𝐺 (𝑘) for the graph below?
𝐺
Example 1a: What is 𝐶𝑜𝑙𝐺 (𝑘) for the graph below?
𝑣1
𝐺
𝑣2
𝑣3
𝑣4
1. There are 𝑘 possible colors for 𝑣1 .
2. There are then (𝑘 − 1) possibilities for 𝑣2 .
3. Since 𝑣3 is connected to 𝑣1 and 𝑣2 , there are (𝑘 − 2) possibilities for 𝑣3 .
4. Finally, there are (𝑘 − 2) possibilities left for 𝑣4 .
Thus the multiplication principle implies:
𝑪𝒐𝒍𝑮 (𝒌) = 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)𝟐
Example 1b: What is 𝐶𝑜𝑙𝐺 (𝑘) for the graph below?
𝑣1
𝐺
𝑣2
𝑣3
𝑣4
There are 𝑘 possible colors for 𝑣1 .
There are then (𝑘 − 1) possibilities both for 𝑣2 and for 𝑣3 .
Why is the answer not
𝑪𝒐𝒍𝑮 (𝒌) = 𝒌(𝒌 − 𝟏)𝟐 (𝒌 − 𝟑) ?
Important Point: If you reason correctly, you may use any vertex as a starting
point. The first vertex always has 𝑘 colors available. Some choices of order will
lead to easier reasoning than others.
Example 1b: What is 𝐶𝑜𝑙𝐺 (𝑘) for the graph below?
𝑣1
𝐺
𝑣2
𝑣3
𝑣4
There are 𝑘 possible colors for 𝑣1 .
There are then (𝑘 − 1) possibilities for 𝑣2 .
Once those first two colors are chosen, the correct approach:
a. There are (𝑘 − 2) ways for 𝑣3 to be different from 𝑣1 and 𝑣2 .
In this case, there are (𝑘 − 3) possibilities for 𝑣4
So there are 𝑘(𝑘 − 1)(𝑘 − 2)(𝑘 − 3) colorings with 𝑣2 and 𝑣3 different.
b. There is only one way for 𝑣3 to be the same color as 𝑣2 . In this case, there
are (𝑘 − 2) possibilities for 𝑣4 .
So there are 𝑘(𝑘 − 1)1(𝑘 − 2) colorings with 𝑣2 and 𝑣3 identical.
This gives us a result of
𝑪𝒐𝒍𝑮 (𝒌) = 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)(𝒌 − 𝟑) + 𝒌(𝒌 − 𝟏) ∙ 𝟏 ∙ (𝒌 − 𝟐)
= 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)[(𝒌 − 𝟑) + 𝟏]
= 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)𝟐
This is the same answer as before.
Example 2:
𝑣2
𝑣1
𝑣1
Example 3:
𝑣2
𝑣3
𝑣4
𝑣5
Example 2:
𝑣2
𝑣1
# of colorings with
𝑣1 and 𝑣2 same
# of colorings with
𝑣1 and 𝑣2 different
So 𝐶𝑜𝑙𝐺 (𝑘) = 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)𝟑 + 𝒌 ∙ 𝟏 ∙ (𝒌 − 𝟏)𝟑
NOTE: Each term in the colored expressions above and below list the number of possibilities in
order, i.e. as (# 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑜𝑟 𝑣1 )(# 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑜𝑟 𝑣2 ) ⋯ (# 𝑝𝑜𝑠𝑠𝑖𝑏𝑙𝑒 𝑓𝑜𝑟 𝑣5 ).
𝑣1
Example 3:
𝑣2
𝑣3
𝑣4
𝑪𝒐𝒍𝑯 (𝒌)
𝑣5
= 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)𝟑 + 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)𝟏(𝒌 − 𝟏) + 𝒌(𝒌 − 𝟏)𝟏(𝒌 − 𝟏)(𝒌 − 𝟐)
𝒗𝟑 different from 𝒗𝟐
𝒗𝟒 different from 𝒗𝟑
𝒗𝟑 different from 𝒗𝟐
𝒗𝟒 equal to 𝒗𝟑
= 𝒌(𝒌 − 𝟏)(𝒌 − 𝟐)𝟑 + 𝟐𝒌(𝒌 − 𝟏)𝟐 (𝒌 − 𝟐)
𝒗𝟑 equal to 𝒗𝟐