Editing Post correspondence problem (section)

===Example 1===
Consider the following two lists:
{{col-begin|width=auto}}
{{col-break}}
{| class="wikitable" style="text-align:center;background:#55FF83;font-style:italic;"
|-style="font-weight:bold;"
|width=40 | α<sub>1</sub>
|width=40 | α<sub>2</sub>
|width=40 | α<sub>3</sub>
|-
|a
|ab
|bba
|}
{{col-break|gap=5px|align=center}}
{| class="wikitable" style="text-align:center;background:#87E6FF;font-style:italic;"
|-style="font-weight:bold;"
|width=40 | β<sub>1</sub>
|width=40 | β<sub>2</sub>
|width=40 | β<sub>3</sub>
|-
|baa
|aa
|bb
|}
{{col-end}}
A solution to this problem would be the sequence (3, 2, 3, 1), because

: <math>\alpha_3 \alpha_2 \alpha_3 \alpha_1 = bba \cdot ab \cdot bba \cdot a = bbaabbbaa = bb \cdot aa \cdot bb \cdot baa = \beta_{3} \beta_{2} \beta_{3} \beta_{1}.</math>

Furthermore, since (3, 2, 3, 1) is a solution, so are all of its "repetitions", such as (3, 2, 3, 1, 3, 2, 3, 1), etc.; that is, when a solution exists, there are infinitely many solutions of this repetitive kind.

However, if the two lists had consisted of only <math>\alpha_2, \alpha_3</math> and <math>\beta_{2}, \beta_{3}</math> from those sets, then there would have been no solution (the last letter of any such α string is not the same as the letter before it, whereas β only constructs pairs of the same letter).

A convenient way to view an instance of a Post correspondence problem is as a collection of blocks of the form
{{col-begin|width=auto}}  
{{col-break}}
{| class="wikitable" style="text-align:center;font-style:italic;font-weight:bold;"
| bgcolor="#55FF83" width="40" | α<sub>i</sub>
|-
| bgcolor="#87E6FF" width="40" | β<sub>i</sub>
|}
{{col-end}}
there being an unlimited supply of each type of block. Thus the above example is viewed as
{{col-begin|width=auto}}
{{col-break|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | a
|-
| bgcolor="#87E6FF" width="40" | baa
|}
''i = 1''
{{col-break|gap=5px|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | ab
|-
| bgcolor="#87E6FF" width="40" | aa
|}
''i = 2''
{{col-break|gap=5px|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | bba
|-
| bgcolor="#87E6FF" width="40" | bb
|}
''i = 3''
{{col-end}}
where the solver has an endless supply of each of these three block types. A solution corresponds to some way of laying blocks next to each other so that the string in the top cells corresponds to the string in the bottom cells. Then the solution to the above example corresponds to:
{{col-begin|width=auto}}
{{col-break|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | bba
|-
| bgcolor="#87E6FF" width="40" | bb
|}
''i<sub>1</sub> = 3''
{{col-break|gap=5px|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | ab
|-
| bgcolor="#87E6FF" width="40" | aa
|}
''i<sub>2</sub> = 2''
{{col-break|gap=5px|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | bba
|-
| bgcolor="#87E6FF" width="40" | bb
|}
''i<sub>3</sub> = 3''
{{col-break|gap=5px|align=center}}
{| class="wikitable" style="font-style:italic;text-align:center;"
| bgcolor="#55FF83" width="40" | a
|-
| bgcolor="#87E6FF" width="40" | baa
|}
''i<sub>4</sub> = 1''
{{col-end}}