How to cut a square grid ?

How to cut a square grid ?

Season 3

Episode 19 Time frame 2 periods

Prerequisites :

^{Coneptofproofbyindution.}

Objectives :

•

^{Conjeturea propertyand useindutionto prove it.}

Materials :

•

^{Retangulargrids.}

•

^Sissors.

1 – Problem 1 : The minimal cut length 10 mins

Students have to nd out ifthereis way tominimize thelength ut.

2 – Problem 2 : The minimal number of cuts 45 mins

Students haveto searh for away to minimizethe number ofuts, autbeingunderstoodasa

omplete lengthwiseand widthwiseutting movement.

Theyshould :

1

^{. notie thatthe number ofuts isalways thesame;}

2

^{. nd theformulafor the number ofuts;}

3

^{. prove thatthe formulaisorret, whatevertheutting methodused.}

3 – Problem 3 : The minimal cutting time 55 mins

In this part, we take into aount that the grid may have to be turned

90 ^◦

^{sometimes. If we}

onsiderthatoneuttakesthesame timeasoneturn,we anomputethetimeneededforeah

utting plan. Students have to nd awayto minimize thatutting time, and prove thevalidity

of their answer.

Document Lesson

Whenpreparingmylessons,Ioftenhavetout

out a squaregridto make individualards.

As itis not a partiularly interesting ativity,

I often wonder what is the best way to do so.

The bestwayould bedened asthe method

using :

•

^{the minimalutlength;}

•

^{the minimalnumberof uts;}

•

^{the minimalutting time.}

Consider a squaregridwithlength

m + 1

^and

width

n + 1

^{, where}

m

^and

n

^{are two whole}

numbers.

n + 1

m + 1

The minimal cut length

This rst problem is easy to solve. No matter how you do it, you'll have to ut

m

^{times along}

thewidth

n + 1

^and

n

^{timesalong the length}

m + 1

^{.Sothe totalutlength isalways equalto}

m(n + 1) + n(m + 1) = mn + m + mn + n = 2mn + m + n.

Therefore, thereisno wayto minimize the totalutlength.

The minimal number of cuts

Now,thereisaproblemthatmaybeworthit.If

m

^and

n

^{arebigenough,therearemanydierent}

waystoutoutthegridompletely.Surelyoneofthemshouldinvolveaminimalnumberofuts.

Of ourse,one ould have theidea of putting two previously utparts one on top of theother,

adjustingthelines, anduttwopieessimultaneously.Experieneshowsthatthisisnot soeasy,

and theutsresulting arealmost neverperfet.So we willforbid this.

It's lear thatfor a

1 × k

^grid,

k

^{uts areneeded. Now, let'sstudy some possibleuting plans}

for a

(m + 1)(n + 1)

^grid.

•

^{If we rst ut along the}

n

^{vertial lines, and thenut the}

n + 1

^{vertial strips along the}

m

horizontal lines, thenthetotal numberofuts is

n + (n + 1) × m = mn + m + n.

•

^{Ifwe rstutalongthe}

m

^{horizontallinesand thenutthe}

m + 1

^{horizontalstrips alongthe}

n

^{horizontallines, thenthe total numberof utsis}

m + (m + 1) × n = mn + m + n.

So for these two extreme situations, the num-

ber of uts is the same. Now, let's try some-

thing abit moreompliated. First,utalong

therst vertial linefrom theleft:

1

^ut.

Then ut along the

m

^{horizontal lines of the}

two vertial strips:

2

^times

m

^uts.

The left-handsidestripisnowompletelyut

out, andthe right-handside gridhasbeenut

into

m + 1

^{strips. Eah of strips must be ut}

along its

n − 1

^{vertial lines.}

Finally,thetotal numberofuts is

1 + m + m + (m + 1)(n − 1)

= 1 + m + m + mn − m + n − 1

= mn + m + n.

So the number of uts seems to be onstant, just like the utlength. It is indeed, and we will

prove this resultbyindution.

Proposition. Any method to ut-out ompletely a

(m + 1) × (n + 1)

^{square grid involves}

mn + m + n

^uts.

Proof.First, onsidera

1 × (k + 1)

^{squaregrid, so}

m = 0

^and

n = k

^{.Any method to utitwill}

involve

k

^uts,and

mn + m + n = 0 × k + 0 + k = k

^{sothepropertyis true.}

Now onsider any

(m + 1) × (n + 1)

^{square grid and assume that the property is true for any}

square grid smaller than that. The rst ut must be a vertial or horizontal one. Let's assume

that it's vertial, the proof being exatlythe same if it is horizontal. This rst utdivides the

gridintwo smallergrids,whose widthare

k + 1

^,where

k

^{isa numberbetween}

0

^and

n − 1

^,and

n + 1 − (k + 1) = n − k

^.

Aording to our indutive hypothesis, any method to ut out the

(m + 1) × (k + 1)

^{grid will}

involve

mk + m + k

^{uts. Inthesame way,anymethodto utout the}

(m + 1) × (n − k)

^gridwill

involve

m(n − k − 1) + m + (n − k − 1)

^{uts. Therefore,thetotal numberofuts isequal to}

1 + mk + m + k + m(n − k − 1) + m + (n − k − 1)

= 1 + mk + m + k + mn − mk − m + m + n − k − 1

= 1 − 1 + mk − mk + m − m + k − k + mn + m + n

= mn + m + n.

Thisproesswill ultimatelyendupwith

1 × (k + 1)

^{squaregrids.So,byomplete indution,the}

propertyisproven.

The minimal cutting time

Now,let'saddatwisttotheproblem.Tomakeaperfetut,it'sbettertoutalongavertialline

than alonga horizontal(wherevertial isunderstoodasthediretion ofthegazeoftheutter).

So we an imagine that before eah ut, it's better to turn thepiee inthe right diretion. To

taje this into aount, let's add 1 for eah turning of a piee of the grid, onsidering that this

takesthesame time asaut.

Inthis setion,all theresults willthereforebegiveninutting timeunits,a unitbeingthetime

needed to do one omplete ut or turn the piee of grid around. We onsider that the utting

timeisnot relatedto thelengthof theutwhih isindeedtheasewhen usinga paper utter

and not apair ofsissors.

Consider oneagain a

(m + 1) × (n + 1)

^squaregrid.

If we start byutting all the vertial lines in the urrent position, then we will have to rotate

eah ofthe

n + 1

^{stripsof squares andutthem one byone. Sotheutting timewill be}

C h = n + (n + 1) × (1 + m) = nm + 2n + m + 1.

Ifwerstrotatethegrid(thisturnisnotounted,astheinitialpositionofthegridisnotdened

a priori),than we getthe symetrialformula:

C v = nm + 2m + n + 1.

Thedierene between thesetwo numbersis

C h − C ^v = n − m

^{.It'sgreaterthan}

0

^if

n

^isstritly

greater than

m

^{(the situation shown on the piture). In that ase, itwouldbe quiker to start}

inthe positionwhere thelongest sideis vertial.

Now, suppose that

m < n

^{and let's study the other utting plan introdued in the previous}

setion, withthe new rule.Itgoeslike this :

•

^{Cutalong the rstvertial line fromtheleft :}

1

^timeunit.

•

^{Rotatetheleft handstripand utitinto}

m + 1

^{unit squares:}

1 + m

^timeunits.

•

^{Rotatetheright handgrid andutit into}

m + 1

^strips:

1 + m

^timeunits.

•

^{For eahofthe}

m + 1

^{strips,rotate itandutitinto}

n

^unitsquares:

(m + 1)(1 + n − 1)

^time

units.

The utting time istherefore

1 + 2 × (1 + m) + n(m + 1) = nm + 2m + n + 3.

Thisis obviously greaterthan

C v

^.

In fat, it may be notied that the number of uts is the same as in the previous setion, the

dierene being in the number of turns. Indeed, if we look at the situation where the longest

length is vertial and we start by doing all the vertial uts, there are

mn + m + n

^{uts and}

m + 1

^{turns(one for eah strip)and}

C v = mn + m + n + m + 1

^{.So our problemis justto nd}

theminimal numberofturns.

Proposition. Any method to ut-out ompletely a

(m + 1) × (n + 1)

^{square grid involves at}

leastMin

(m, n) + 1

^turns.

Proof.First,onsidera

1 × k

^stripof

k

^{squares,inavertialposition,for anynatural number}

k

greaterthan

1

^{.Obviously,to utit out,we need to turnit onerst.So thepropertyis truein}

thatase.

Now, assumethat the property is truefor any

(i + 1) × (j + 1)

^{square grid smaller than}

(m + 1) × (n + 1)

^{,and also that}

m < n

^{.Then we have to prove that thenumber of turns is at least}

m + 1

^.

Weandeideto putthe gridinavertial positionandstart uttingthatway.Therstutwill

be along one of the vertial lines, so thatthere will be a

(k + 1) × (n + 1)

^{grid on the leftand}

a

(m − k) × (n + 1)

^{on the right, with}

0 < k < m

^{.Eah strip is still in a vertial position, as}

k < m < n

^and

m − k < m < n

^{,so,aordingto our indution}hypothesis,we will needat least Min

(k, n) + 1 = k + 1

^{turns for theleft-hand part and Min}

(m − k − 1, n) + 1 = m − k

^{for the}

other. So thetotal numberof turnsfor thewholegrid isat least

k + 1 + m − k = m + 1.

But we an also deide to start with the grid plaed horizontally. Then, the rst ut will be

along one of the shortest lines, so that there will be a

(m + 1) × (k + 1)

^{grid on the left and}

a

(m + 1) × (n − k)

^{on the right, with}

0 < k < n

^{. Let's lookat the}

(m + 1) × (k + 1)

^{grid. If}

k 6 m

^{,than aording to our indution} hypothesis, the minimal number of turns to ut it is Min

(k, m) + 1 = k + 1 = k

^.If

m < k

^{,thanthe minimalnumberofutsisMin}

(k, m) + 1 = m + 1

^.

Now, let'slookat the

(m + 1) × (n − k)

^{grid. If}

n − k − 1 6 m

^{,than aordingto our indution}

hypothesis,theminimalnumberofturnstoutitisMin

(n − k − 1, m)+1 = n − k

^.If

m < n − k − 1

^,

than the minimalnumberof uts isMin

(n − k − 1, m) + 1 = m + 1

^{.Thetotal number of turns}

is given inthetable below:

If

k 6 m m < k

If

n − k − 1 6 m k + 1 + n − k = n + 1 m + 1 + n − k

If

m < n − k − 1 k + 1 + m + 1 m + 1 + m + 1 = 2m + 2

Weassumed that

m < n

^{,and fromits deniton}

0 < k < n

^{.Therefore,we an say that}

n + 1 >

m + 1

^,

k + 1 + m + 1 > m + 1

^and

m + 1 + n − k > m + 1

^{.Also,it'sobviousthat}

2m + 2 > geq

^.

So, in eah ase, the number of turns is more than

m + 1

^{, whih ompletes the proof of the}

proposition, byindutionoverthesize ofthe grid.

The proposition givesalowerboundfor thenumberofturns,and thereforetheutting timefor

any

(m + 1) × (n + 1)

^{grid.Butwealreadyfound autting planthatgivesthatexat value. We}

an thendedue thefollowing theorem.

Theorem. Theminimal utting time for a

(m + 1) × (n + 1)

^{grid, where}

m < n

^,is

mn + 2m + n + 1.

To ut the grid in the minimal time, start by putting the grid vertially, ut along along the

vertial lines, thenturneah stripofsquares and utthemoone by one.

Proof. From the previous theorem, we know that the number of turns is at least

m + 1

^{. Also,}

from theprevious setion, thenumberof utsis

mn + m + 1

^{,sotheutting timeannotbeless}

than thesum

mn + m + 1 + m + 1 = mn + 2m + n + 1.

As we've seen at the beginning of this setion, theutting plan desribed in thetheorem gives

thatexatutting time.Sotheminimaluttintimeis

mn + m + 1 + m + 1 = mn + 2m + n + 1.