Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Four-colour theorem
and the computer assisted proof of Appel and Haken from 1976
Ville Kivioja
University of Jyväskylä
Graduate student seminar 14. October 2016
1 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
First known source: A letter from August de Morgan to Sir
William Rowan Hamilton, year 1852:
”A student of mine asked me today to give him a reason for the
fact which I did not know was a fact and do not yet. He says that
if a figure be anyhow divided, and the compartments differently
coloured, so that figures with any portion of common boundary line
are differently coloured, then four colours may be wanted but no
more.”
2 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
First known source: A letter from August de Morgan to Sir
William Rowan Hamilton, year 1852:
”A student of mine asked me today to give him a reason for the
fact which I did not know was a fact and do not yet. He says that
if a figure be anyhow divided, and the compartments differently
coloured, so that figures with any portion of common boundary line
are differently coloured, then four colours may be wanted but no
more.”
Francis Guthrie was the name of this student, as was revealed
later.
2 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
3 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
De Morgan / Guthrie: ”. . . if a figure be anyhow divided, and the
compartments differently coloured, so that figures with any portion
of common boundary line are differently coloured, then four colours
may be wanted but no more.”
4 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
De Morgan / Guthrie: ”. . . if a figure be anyhow divided, and the
compartments differently coloured, so that figures with any portion
of common boundary line are differently coloured, then four colours
may be wanted but no more.”
Problems:
”common boundary line”= Hausdorff-dimension non-zero?
4 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
De Morgan / Guthrie: ”. . . if a figure be anyhow divided, and the
compartments differently coloured, so that figures with any portion
of common boundary line are differently coloured, then four colours
may be wanted but no more.”
Problems:
”common boundary line”= Hausdorff-dimension non-zero?
Kaliningrad-like behaviour.
4 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
De Morgan / Guthrie: ”. . . if a figure be anyhow divided, and the
compartments differently coloured, so that figures with any portion
of common boundary line are differently coloured, then four colours
may be wanted but no more.”
Problems:
”common boundary line”= Hausdorff-dimension non-zero?
Kaliningrad-like behaviour.
Lakes of Wada: n ≥ 3 disjoint domains of plane that all have
same boundary line and the closure of their union is the plane.
...
4 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The problem is made a rigorous mathematical question by
translating it to graph theory language.
5 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The problem is made a rigorous mathematical question by
translating it to graph theory language.
5 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Let V be a finite set and E a symmetric relation for which a E a
does not hold for any a ∈ V . Then we say (V , E ) is a simple
graph.
6 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Let V be a finite set and E a symmetric relation for which a E a
does not hold for any a ∈ V . Then we say (V , E ) is a simple
graph.
A simple graph V is planar if it is bijective to a subset of R2 such
that when we draw straight line intervalls (edges) between all the
points a, b for which a E b, then two distinct edges do not intersect
outside the vertices.
6 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Let V be a finite set and E a symmetric relation for which a E a
does not hold for any a ∈ V . Then we say (V , E ) is a simple
graph.
A simple graph V is planar if it is bijective to a subset of R2 such
that when we draw straight line intervalls (edges) between all the
points a, b for which a E b, then two distinct edges do not intersect
outside the vertices.
From now on, the term simple planar graph is replaced by just
graph.
6 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Let V be a finite set and E a symmetric relation for which a E a
does not hold for any a ∈ V . Then we say (V , E ) is a simple
graph.
A simple graph V is planar if it is bijective to a subset of R2 such
that when we draw straight line intervalls (edges) between all the
points a, b for which a E b, then two distinct edges do not intersect
outside the vertices.
From now on, the term simple planar graph is replaced by just
graph.
If V is a graph, then M ⊂ V is also a graph with the induced
relation. We say (M, E|M×M ) is a subgraph of M and write
M ≤ V.
6 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
A graph V is colourable with k colours if V can be decomposed to
k disjoint sets
k
G
V =
Ck
i=1
so that
aEb
⇒
/i such that a, b ∈ Ci
∃
7 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
A graph V is colourable with k colours if V can be decomposed to
k disjoint sets
k
G
V =
Ck
i=1
so that
aEb
⇒
/i such that a, b ∈ Ci
∃
Theorem (Four Colour Theorem)
Any (simple planar) graph is colourable with 4 colours
7 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
Percy Heawood found the mistake on the proof of Kempe
but showed that it proves the Five colour theorem with
slight modifications.
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
Percy Heawood found the mistake on the proof of Kempe
but showed that it proves the Five colour theorem with
slight modifications.
Four colours were shown to be enough for some small maps:
1922 Franklin announced a proof for ≤ 25 regions.
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
Percy Heawood found the mistake on the proof of Kempe
but showed that it proves the Five colour theorem with
slight modifications.
Four colours were shown to be enough for some small maps:
1926 Reynolds announced a proof for ≤ 27 regions.
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
Percy Heawood found the mistake on the proof of Kempe
but showed that it proves the Five colour theorem with
slight modifications.
Four colours were shown to be enough for some small maps:
1940 Winn announced a proof for ≤ 35 regions.
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
Percy Heawood found the mistake on the proof of Kempe
but showed that it proves the Five colour theorem with
slight modifications.
Four colours were shown to be enough for some small maps:
1970 Ore and Stemple announced a proof for ≤ 39 regions.
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
History of the proof
Alfred Bray Kempe had an accepted proof between 1879
and 1890
Percy Heawood found the mistake on the proof of Kempe
but showed that it proves the Five colour theorem with
slight modifications.
Four colours were shown to be enough for some small maps:
1970 Ore and Stemple announced a proof for ≤ 39 regions.
Kenneth Appel and Wolfgang Haken established a
computer assisted proof in 1976 (there were some flaws, finally
corrected by them 1989).
8 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Proof of Six colour theorem.
9 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Proof of Six colour theorem.
Arguing by contradiction, let G be a minimal counterexample to
the theorem.
9 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Proof of Six colour theorem.
Arguing by contradiction, let G be a minimal counterexample to
the theorem. Euler’s formula for graphs
(#of vertices) − (#of edges) + (#of faces) = 2
together with the Handshaking lemma, implies that ∃ vertex v with
at most five edges.
9 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Proof of Six colour theorem.
Arguing by contradiction, let G be a minimal counterexample to
the theorem. Euler’s formula for graphs
(#of vertices) − (#of edges) + (#of faces) = 2
together with the Handshaking lemma, implies that ∃ vertex v with
at most five edges. Recall that G \{v } is colourable with 6 colours.
9 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Proof of Six colour theorem.
Arguing by contradiction, let G be a minimal counterexample to
the theorem. Euler’s formula for graphs
(#of vertices) − (#of edges) + (#of faces) = 2
together with the Handshaking lemma, implies that ∃ vertex v with
at most five edges. Recall that G \{v } is colourable with 6 colours.
There is now one colour left for v .
9 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left. So, we may assume v has 5 neighbours all coloured
differently, let’s say, in red, blue, green, yellow, orange in clockwise
order.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left. So, we may assume v has 5 neighbours all coloured
differently, let’s say, in red, blue, green, yellow, orange in clockwise
order. Change red neighbour of v to green, and make the necessary
consecutive changes to the rest of the graph.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left. So, we may assume v has 5 neighbours all coloured
differently, let’s say, in red, blue, green, yellow, orange in clockwise
order. Change red neighbour of v to green, and make the necessary
consecutive changes to the rest of the graph. This process stops.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left. So, we may assume v has 5 neighbours all coloured
differently, let’s say, in red, blue, green, yellow, orange in clockwise
order. Change red neighbour of v to green, and make the necessary
consecutive changes to the rest of the graph. This process stops. If
the sequence constructed like this did not reach the green
neighbour of v , we just made red a free colour.
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left. So, we may assume v has 5 neighbours all coloured
differently, let’s say, in red, blue, green, yellow, orange in clockwise
order. Change red neighbour of v to green, and make the necessary
consecutive changes to the rest of the graph. This process stops. If
the sequence constructed like this did not reach the green
neighbour of v , we just made red a free colour. If instead it did, do
the same for orange–blue-pair:
10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Sketch of the proof of 5 colour theorem (Kempe–Heawood)
Arguing by contradiction, let G be a minimal counterexample to
the theorem. As before, ∃ vertex v with at most five edges. Recall
that G \{v } is colourable with 5 colours. If strictly less than 5
edges attach to v , we are done: We may colour v by the colour
that is left. So, we may assume v has 5 neighbours all coloured
differently, let’s say, in red, blue, green, yellow, orange in clockwise
order. Change red neighbour of v to green, and make the necessary
consecutive changes to the rest of the graph. This process stops. If
the sequence constructed like this did not reach the green
neighbour of v , we just made red a free colour. If instead it did, do
the same for orange–blue-pair: Starting from orange, it cannot
reach blue (it should cross somehow the red–green path). 10 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
11 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was:
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was: Take a minimal counterexample and find a particular
list of graphs.
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was: Take a minimal counterexample and find a particular
list of graphs. In five colour theorem, this list was the set
(M1 , . . . , M5 ) that are the graphs with one ”primary” vertex and i
edges attached to it.
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was: Take a minimal counterexample and find a particular
list of graphs. In five colour theorem, this list was the set
(M1 , . . . , M5 ) that are the graphs with one ”primary” vertex and i
edges attached to it.
In general, the list should have following properties:
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was: Take a minimal counterexample and find a particular
list of graphs. In five colour theorem, this list was the set
(M1 , . . . , M5 ) that are the graphs with one ”primary” vertex and i
edges attached to it.
In general, the list should have following properties:
The graphs should be unavoidable, in the sense that some of
them necessarily appear in the minimal counterexample as
subgraphs.
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was: Take a minimal counterexample and find a particular
list of graphs. In five colour theorem, this list was the set
(M1 , . . . , M5 ) that are the graphs with one ”primary” vertex and i
edges attached to it.
In general, the list should have following properties:
The graphs should be unavoidable, in the sense that some of
them necessarily appear in the minimal counterexample as
subgraphs.
They should have the property, that a colouring of their
complement is always possible to first adjust and then extend
to the whole graph.
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Strategy was: Take a minimal counterexample and find a particular
list of graphs. In five colour theorem, this list was the set
(M1 , . . . , M5 ) that are the graphs with one ”primary” vertex and i
edges attached to it.
In general, the list should have following properties:
The graphs should be unavoidable, in the sense that some of
them necessarily appear in the minimal counterexample as
subgraphs.
They should have the property, that a colouring of their
complement is always possible to first adjust and then extend
to the whole graph. (We always have for free a colouring of
the complement using a minimal counterexample.)
12 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
a) Ii ≤ Mi .
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
a) Ii ≤ Mi .
b) If X is any graph, it has some Mi as its subgraph.
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
a) Ii ≤ Mi .
b) If X is any graph, it has some Mi as its subgraph.
c) Always when for a graph X it holds Mi ≤ X and X \Ii is
colourable with k colours, X is colourable with k colours.
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
a) Ii ≤ Mi .
b) If X is any graph, it has some Mi as its subgraph.
c) Always when for a graph X it holds Mi ≤ X and X \Ii is
colourable with k colours, X is colourable with k colours.
Appel and Haken found a list of 4-configurations (about 2000
entries long).
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
a) Ii ≤ Mi .
b) If X is any graph, it has some Mi as its subgraph.
c) Always when for a graph X it holds Mi ≤ X and X \Ii is
colourable with k colours, X is colourable with k colours.
Appel and Haken found a list of 4-configurations (about 2000
entries long). A computer checked that their list fulfills the
definition of a list of 4-configurations.
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
Definition
A list of k-configurations is a finite set L = {(M1 , I1 ), . . . , (MN , IN )}
of graph pairs with the following properties:
a) Ii ≤ Mi .
b) If X is any graph, it has some Mi as its subgraph.
c) Always when for a graph X it holds Mi ≤ X and X \Ii is
colourable with k colours, X is colourable with k colours.
Appel and Haken found a list of 4-configurations (about 2000
entries long). A computer checked that their list fulfills the
definition of a list of 4-configurations. The existence of such a list
contradicts the assumption that there would exist a minimal
counter-example to the theorem, thus finishing the proof.
13 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
One cannot verify the proof of Appel and Haken by reading it
through, it (the computer generated part) is (far!) too long.
14 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
One cannot verify the proof of Appel and Haken by reading it
through, it (the computer generated part) is (far!) too long.
The original calculation took 1200 hours (50 days) of
computing time (of supercomputers on 70s).
14 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
One cannot verify the proof of Appel and Haken by reading it
through, it (the computer generated part) is (far!) too long.
The original calculation took 1200 hours (50 days) of
computing time (of supercomputers on 70s).
The computer assisting the proof was not ”doing brute force
calculation” in the strong sense.
14 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
One cannot verify the proof of Appel and Haken by reading it
through, it (the computer generated part) is (far!) too long.
The original calculation took 1200 hours (50 days) of
computing time (of supercomputers on 70s).
The computer assisting the proof was not ”doing brute force
calculation” in the strong sense. Instead, it was running an
algorithm to essentially check the property c of the definition,
i.e., adjusting the colouring of a general map.
14 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
One cannot verify the proof of Appel and Haken by reading it
through, it (the computer generated part) is (far!) too long.
The original calculation took 1200 hours (50 days) of
computing time (of supercomputers on 70s).
The computer assisting the proof was not ”doing brute force
calculation” in the strong sense. Instead, it was running an
algorithm to essentially check the property c of the definition,
i.e., adjusting the colouring of a general map. It was not
possible to determine beforehand, if this algorithm would ever
halt.
14 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The Four colour theorem is true.
15 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The Four colour theorem is true.
”It is much less likely to be incorrect than a typical human-checked
proof of even moderate length.” —Bojan Mohar
15 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The Four colour theorem is true.
”It is much less likely to be incorrect than a typical human-checked
proof of even moderate length.” —Bojan Mohar
But why is this theorem true?
15 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The Four colour theorem is true.
”It is much less likely to be incorrect than a typical human-checked
proof of even moderate length.” —Bojan Mohar
But why is this theorem true?
Should we keep searching for something simpler? Something
answering to the ”why”-question more clearly?
15 / 15
Original problem
Rigorous statement
Proof for six colours
Five colour problem
Treating four colors
Philosophical discussions
The Four colour theorem is true.
”It is much less likely to be incorrect than a typical human-checked
proof of even moderate length.” —Bojan Mohar
But why is this theorem true?
Should we keep searching for something simpler? Something
answering to the ”why”-question more clearly?
Why should we expect there is something simpler out there?
15 / 15