Editing Shannon–Fano coding (section)

===Example===

[[Image:ShannonCodeAlg.svg|right|thumb|300px|Shannon–Fano Algorithm]]

We continue with the previous example.

:{| class="wikitable" style="text-align: center;"
! Symbol
! A
! B
! C
! D
! E
|-
! Count
| 15
| 7
| 6
| 6
| 5
|-
! Probabilities
| 0.385
| 0.179
| 0.154
| 0.154
| 0.128
|}

All symbols are sorted by frequency, from left to right (shown in Figure a). Putting the dividing line between symbols B and C results in a total of 22 in the left group and a total of 17 in the right group. This minimizes the difference in totals between the two groups.

With this division, A and B will each have a code that starts with a 0 bit, and the C, D, and E codes will all start with a 1, as shown in Figure b. Subsequently, the left half of the tree gets a new division between A and B, which puts A on a leaf with code 00 and B on a leaf with code 01.

After four division procedures, a tree of codes results. In the final tree, the three symbols with the highest frequencies have all been assigned 2-bit codes, and two symbols with lower counts have 3-bit codes as shown table below:

:{| class="wikitable" style="text-align: center;"
! Symbol
! A
! B
! C
! D
! E
|-
! Probabilities
| 0.385
| 0.179
| 0.154
| 0.154
| 0.128
|-
! First division
| colspan="2" | 0
| colspan="3" | 1
|-
! Second division
| rowspan="2" style="vertical-align:top"; | 0
| rowspan="2" style="vertical-align:top"; | 1
| rowspan="2" style="vertical-align:top"; | 0
| colspan="2" | 1
|-
! Third division
| 0
| 1
|-
! Codewords
| 00
| 01
| 10
| 110
| 111
|}

This results in lengths of 2 bits for A, B and C and per 3 bits for D and E, giving an average length of

:<math display="block">\frac{2\,\text{bits}\cdot(15+7+6) + 3\,\text{bits} \cdot (6+5)}{39\, \text{symbols}} \approx 2.28\,\text{bits per symbol.}</math>

We see that Fano's method, with an average length of 2.28, has outperformed Shannon's method, with an average length of 2.62.